{"keys":[{"path":["title"],"id":"title","weight":1,"src":"title","getFn":null},{"path":["content"],"id":"content","weight":1,"src":"content","getFn":null},{"path":["keywords"],"id":"keywords","weight":1,"src":"keywords","getFn":null}],"records":[{"i":0,"$":[{"v":"Momentum and impulse","n":0.577},{"v":"In everyday life you might notice that it is harder to stop a heavy truck than a light car at the same speed, or a fast tennis ball than a slow one. This is because of momentum \u2013 a measure of how hard it is to stop a moving object. When a force acts for a short time to change an object\u2019s momentum, we describe this using impulse.\r\n\r\nMomentum \\(p\\)\r\nMomentum \\(p\\) describes the motion of an object by combining its mass and velocity. Objects have more momentum if they have more mass or a higher velocity. This makes it harder to stop or change the object\u2019s motion.\r\nMomentum is given by the equation:\r\n\\[p=mv\\]\r\nwhere:\r\n\r\n\t\\(p\\) is the momentum in kilograms metre per second \\(\\text{kg m s}^{-1}\\)\r\n\t\\(m\\) is the mass in kilograms \\(\\text{kg}\\)\r\n\t\\(v\\) is the velocity in metres per second \\(\\text{m s}^{-1}\\).\r\n\r\nMomentum is a vector quantity, so both magnitude and direction must be considered. When solving problems, always choose a sign convention (for example, 'to the right' or 'East' is positive) and stick to it.\r\n\r\nChange in momentum \\(\\Delta p\\)\r\nSuppose an object of mass \\(m\\) changes its velocity from \\(v_{i}\\) (initial velocity) to \\(v_{f}\\) (final velocity) in a time interval \\(\\Delta t\\) under a resultant force \\(\\sum F\\). From Newton\u2019s second law, \\(\\sum F=ma\\), and using \\(a=\\dfrac{v_{f}-v_{i}}{\\Delta t}\\), we get:\r\n\\[\\begin{align*} \\sum F & = m\\left(\\frac{v_{f}-v_{i}}{\\Delta t}\\right) \\\\[6pt]\r\n& = \\frac{m v_{f}-m v_{i}}{\\Delta t} \\end{align*}\\]\r\nMultiplying both sides by \\(\\Delta t\\) gives:\r\n\\[\\sum F\\Delta t=m v_{f}-m v_{i}\\]\r\nThe right-hand side is the change in momentum \\(\\Delta p\\):\r\n\r\n\\[\\begin{align*} \\Delta p & = p_{f}-p_{i} \\\\[6pt]\r\n& = m v_{f}-m v_{i} \\end{align*}\\]\r\nwhere:\r\n\r\n\t\\(p_{f}\\) and \\(p_{i}\\) are the final and initial momentum, respectively, in kilogram metres per second \\(\\text{kg m s}^{-1}\\)\r\n\t\\(v_{f}\\) and \\(v_{i}\\) are the final and initial velocity, respectively, in metres per second \\(\\text{m s}^{-1}\\).\r\n\r\nThese calculations require you to be comfortable with vector subtraction. If you need a refresher, check out this resource.\r\n\r\nImpulse \\(I\\)\r\nThe left-hand side of the equation is the impulse of the force \\(I\\).\r\n\r\n\\[\\begin{align*} I & = \\sum F\\Delta t \\\\[6pt]\r\n& = \\Delta p \\end{align*}\\]\r\nwhere:\r\n\r\n\t\\(I\\) is the impulse in Newton-seconds \\(\\text{N s}\\) (or kilogram metres per second \\(\\text{kg m s}^{-1}\\))\r\n\t\\(\\sum F\\) is the resultant force in Newtons \\(\\text{N}\\)\r\n\t\\(\\Delta t\\) is the time interval in seconds \\(\\text{s}\\).\r\n\r\nImpulse and change in momentum are not just numerically equal. They are two ways of describing the same physical event. Impulse describes it from the perspective of the force that caused the change (how hard and for how long the force acted), while change in momentum describes it from the perspective of the object (how much its motion changed).\r\nLike momentum, change in momentum and impulse are vectors.\r\n\r\n  A negative \\(\\Delta p\\) indicates a decrease in momentum in the chosen positive direction (or a gain in the opposite direction).\r\n  A positive \\(\\Delta p\\) indicates an increase in momentum in the chosen positive direction (or a loss in the opposite direction).\r\n\r\n\r\nUsing force\u2013time graphs to calculate impulse\r\nThe area under a force\u2013time graph gives the impulse of the force.\r\n\r\nThis is especially useful when the force is not constant. For example, in car safety design, we often want to increase the time of impact (such as by using crumple zones or airbags) so that the force on the occupants is reduced.\r\nFor a given change in momentum \\(\\Delta p\\):\r\n\\[\\begin{align*} \\sum F\\Delta t & = \\Delta p\\quad\\text{(a constant)} \\\\[6pt]\r\n& \\Rightarrow F \\propto \\frac{1}{\\Delta t} \\end{align*}\\]\r\n\r\n\tIf the stopping time \\(\\Delta t\\) is small (for example, hitting a rigid wall), the average force is large.\r\n\tIf \\(\\Delta t\\) is larger (for example, hitting a barrier that deforms), the average force is smaller.\r\n\r\n\r\nConservation of momentum\r\nConsider this scenario. According to Newton's third law of motion, when two balls A and B collide, the action of A on B (\\(F_{AB}\\)) is equal and opposite to that of B on A (\\(F_{BA}\\)).\r\n\r\nThis means that the rate of change of momentum of A (\\(\\Delta p_{A}\\)) is equal and opposite to the rate of change of momentum of B (\\(\\Delta p_{B}\\)).\r\nSince the time of contact is the same for both, \\(\\Delta p_{A}\\) is equal and opposite to \\(\\Delta p_{B}\\). That is, the total momentum before impact equals the total momentum after impact. Therefore, \\(\\Delta p_{A}\\) and \\(\\Delta p_{B}\\) cancel out when added together. For example, if two cars collide and stay together after the collision, then the momentum of the two cars before the collision is equal to the momentum of the locked-together cars after the collision.\r\n\\[\\Delta p_{A}+\\Delta p_{B} = 0\\]\r\nThis means that the total change in momentum of the system is zero. In other words, whatever momentum one ball gains, the other loses by exactly the same amount. The total momentum of the system therefore remains constant.\r\n\\[\\sum p_{i}=\\sum p_{f}\\]\r\nThis is known as the law of conservation of momentum. The total momentum is the same before, during and after impact:\r\n\\[\\begin{align*} m_{A}u_{A}+m_{B}u_{B} & =m_{A}v_{A}+m_{B}v_{B} \\end{align*}\\]\r\nwhere:\r\n\r\n\t\\(m_{A}\\) and \\(m_{B}\\) are the masses of A and B, respectively, in kilograms \\(\\text{kg}\\)\r\n\t\\(u_{A}\\) and \\(u_{B}\\) are the velocities of A and B before impact, respectively, in metres per second \\(\\text{m s}^{-1}\\)\r\n\t\\(v_{A}\\) and \\(v_{B}\\) are the velocities of A and B after impact, respectively, in metres per second \\(\\text{m s}^{-1}\\).\r\n\r\n\r\nExample \u2013 calculating momentum and impulse\r\nA cyclist and their bicycle have a combined mass of \\(\\mathbf{90}\\,\\textbf{kg}\\). They are travelling at \\(\\mathbf{8.0}\\,\\textbf{m s}\\mathbf{^{-1}}\\) East along a flat road, when they stop pedalling. Friction brings them to rest in \\(\\mathbf{4.0}\\,\\textbf{s}\\). Taking East as the positive direction, calculate:\r\n\r\n\tthe initial momentum of the cyclist\r\n\r\nMomentum is the product of mass and velocity. Since the cyclist is moving East (our positive direction), the momentum is positive.\r\n\\[\\begin{align*} p_{i} & = mv_{i} \\\\[6pt]\r\n& = 90\\times8.0 \\\\[6pt]\r\n& = 720\\,\\text{kg m s}^{-1}\\text{ East} \\end{align*}\\]\r\n\r\n\r\n\tthe change in momentum of the cyclist\r\n\r\nThe cyclist comes to rest, so the final velocity is \\(v_{f}=0\\). Since the cyclist has lost all their momentum, we expect \\(\\Delta p\\) to be negative, meaning that momentum has decreased in the Eastward direction.\r\n\\[\\begin{align*} \\Delta p & = mv_{f}-mv_{i} \\\\[6pt]\r\n& = \\left(90\\times0\\right)-\\left(90\\times8.0\\right) \\\\[6pt]\r\n& = -720\\,\\text{kg m s}^{-1} \\end{align*}\\]\r\nThe negative sign confirms that the change in momentum is directed West, opposing the original motion. In other words, the cyclist has lost \\(720\\,\\text{kg m s}^{-1}\\) of Eastward momentum.\r\n\r\n\r\n\tthe impulse exerted on the cyclist by friction\r\n\r\nSince impulse equals change in momentum, we do not need to do any additional calculation here. The impulse is whatever force was needed to produce the change in momentum we calculated in part b.\r\n\\[I = \\Delta p = -720\\,\\text{N s}\\]\r\nThe impulse is \\(720\\,\\text{N s}\\) directed West, meaning friction has acted Westward on the cyclist throughout the stop. This makes physical sense since friction always opposes motion.\r\n\r\n\r\n\tthe average friction force acting on the cyclist.\r\n\r\nWe know the impulse and the time over which it acted, so we can find the average force. We rearrange \\(I=\\sum F\\Delta t\\) to:\r\n\\[\\begin{align*} \\sum F & = \\frac{I}{\\Delta t} \\\\[6pt]\r\n& = \\frac{-720}{4.0} \\\\[6pt]\r\n& = -180\\,\\text{N} \\end{align*}\\]\r\nThe average friction force is \\(180\\,\\text{N}\\) directed West. This force acted on the cyclist for \\(4.0\\,\\text{s}\\), producing just enough impulse to bring them to rest. This momentum was not destroyed; it was transferred to the Earth via friction.\r\n\r\n\r\n\r\nExercise \u2013 calculating momentum and impulse\r\n\r\n \tA tennis ball of mass \\(100\\,\\text{g}\\) hits the wall horizontally at \\(8.0\\,\\text{m s}^{-1}\\) East and rebounds at \\(6.0\\,\\text{m s}^{-1}\\) West. The contact time with the wall is \\(0.07\\,\\text{s}\\). Calculate, stating the magnitude and direction:\r\n\t\r\n\t\tthe impulse on the ball by the wall\r\n \t\tthe change in momentum of the ball\r\n\t\tthe force exerted on the ball by the wall\r\n\t\tthe force exerted on the wall by the ball\r\n\t\r\n\r\n\r\n\r\n\r\n \tThe following graph shows how force varies with time for a miniature crash test dummy of mass \\(2\\,\\text{kg}\\) moving to the right. It is involved in a collision with a large concrete block set into the ground, which brings it to rest. Calculate the dummy's initial speed.\r\n\r\n\r\n\r\n\r\n\r\n\r\n \tA truck of mass \\(2500\\,\\text{kg}\\) travelling at \\(20\\,\\text{m s}^{-1}\\) West collides head on with a car of mass \\(800\\,\\text{kg}\\) travelling in the opposite direction at \\(15\\,\\text{m s}^{-1}\\). The two vehicles become locked together.\r\n\t\r\n\t\tWhat is the total momentum of the two vehicles before the collision? Assume the trucks motion (West) is positive.\r\n\t\tWhat is the speed and direction of the car and truck immediately after the collision?\r\n\t\r\n\r\n\r\n\r\n\r\n \tA racing car negotiating a tight bend at \\(30\\,\\text{km h}^{-1}\\) collides with a crash barrier. The air bag in their car inflates and the time taken for it to inflate is \\(0.16\\,\\text{s}\\). The driver's head has a mass of \\(7.0\\,\\text{kg}\\). Explain why the driver is less likely to suffer head injury in a collision with the air bag than if their head collided with the car dashboard, or other hard surface.\r\n\r\n\r\n\r\n \r\nMiniature crash test dummy graph transcript\r\nA graph showing force on the vertical axis and time in milliseconds on the horizontal axis. The graph starts from 0 Newtons at 10 milliseconds and goes up in straight line to 50 Newtons at 20 milliseconds. From  20 to 50 milliseconds, the graph is stable at 50 Newtons, then goes down in a straight line to 0 Newtons at 60 milliseconds.\r\n","n":0.026},[{"v":"Maths","i":2,"n":1},{"v":"Physics","i":1,"n":1},{"v":"Force","i":0,"n":1}]]},{"i":1,"$":[{"v":"Forces on slopes","n":0.577},{"v":"Place a book on a tilted surface and it either slides down or stays put depending on the angle and how rough the surface is. What determines this? This resource helps you to understand the components of gravitational forces that affect this and use them to analyse forces on inclined surfaces. Also learn about how an understanding of forces on slopes can be applied, like in the Budj Bim aquaculture system.\r\n\r\nWhen an object sits on a slope, gravity doesn't act along or against the surface directly; it acts straight down. To analyse the motion, we break that gravitational force into two components:\r\n\r\n\tthe component perpendicular to the slope, which pushes the object into the surface and produces the normal force and friction\r\n\tthe component parallel to the slope, which drives the object down the incline.\r\n\r\n\r\nBreaking a force into components like this is called resolving the force.\r\nJust as we can analyse horizontal and vertical motion separately on a flat surface, here we analyse forces perpendicular and parallel to the sloping surface. The same principle applies: the normal force always acts perpendicular to the surface, while friction and the driving force act parallel to it.\r\n\r\nResolving the components of gravitational force\r\nConsider this object on a plane, where weight (\\(W=mg\\)) acts through the centre of mass.\r\n\r\n\r\nForces parallel to the slope\r\nA frictional force \\(F_{F}\\) acts on the object, stopping it from sliding down the slope. The force driving the object down the slope is given by \\(mg\\sin\\theta\\). The difference between these two forces is the resultant force down the slope:\r\n\\[\\sum F = mg\\sin\\theta -F_{F}\\]\r\nFriction acts to oppose sliding motion, preventing the object from sliding down the plane. For example, if the mass were being dragged uphill, friction would act downhill.\r\n\r\nForces perpendicular to the slope\r\nThe normal force \\(F_{N}\\) is at right angles to the surface. The force pushing the object into the surface is given by \\(mg\\cos\\theta\\). These forces are equal and opposite one another, otherwise the object would fall through the plane. The resultant force perpendicular to the slope is zero because it sits on the slope, hence:\r\n\\[mg\\cos\\theta=F_{N}\\]\r\nTo work out the angles, remember that sum of angles in any triangle is \\(180^{\\circ}\\) and a right angle is \\(90^{\\circ}\\).\r\n\r\n\r\n\r\n\r\nExample 1 \u2013 calculating forces on slopes\r\nA toy car of mass \\(\\mathbf{50}\\,\\textbf{g}\\) travels down a smooth incline at \\(\\mathbf{30^{\\circ}}\\) to the horizontal. Calculate:\r\n\r\n \tthe net force acting on the car as it rolls down the slope\r\n\r\nAlways start with a force diagram. Friction may be ignored as the incline is described as 'smooth'. Gravity, \\(g=9.8\\,\\text{m s}^{-2}\\). As the slope is \\(30^{\\circ}\\), then \\(90-30=60^{\\circ}\\) in top corner and \\(90-60=30^{\\circ}\\) from the normal to vertical force. This allows us to use \\(mg\\sin30^{\\circ}\\) as the component parallel to the slope.\r\n\r\nAs \\(mg\\sin30^{\\circ}\\) is the component of the force parallel to  the slope:\r\n\\[\\begin{align*} \\sum F & = ma \\\\[6pt]\r\n& = mg\\sin30^{\\circ}-F_{f}\\\\[6pt]\r\n& = mg\\sin30^{\\circ}-0\\\\[6pt]\r\n& = mg\\sin30^{\\circ} \\end{align*}\\]\r\nThe surface is frictionless (\\(F_{F}=0\\)). Therefore, the only force allowing the car to roll down the incline is the component of the gravitational force \\(mg\\sin\\theta\\). Remember to convert mass in \\(g\\) to \\(kg\\).\r\n\\[\\begin{align*} \\sum F & = mg\\sin\\theta \\\\[6pt]\r\n& = 50\\times10^{-3}\\times9.8\\times\\sin30^{\\circ}\\\\[6pt] \r\n& = 0.25\\,\\text{N} \\end{align*}\\]\r\nThe net force acting on the car as it rolls down the slope is \\(0.25\\,\\text{N}\\).\r\n\r\n\r\n \tthe force of the incline on the car as it travels down the slope.\r\n\r\nAgain, as the slope is \\(30^{\\circ}\\), the angle in the top corner is \\(90-30=60^{\\circ}\\) and the angle from the normal to vertical force is \\(90-60=30^{\\circ}\\). This allows us to use \\(mg\\cos30^{\\circ}\\) as the component perpendicular.\r\n\r\nThe force of the incline on the car \\(mg\\cos\\theta\\) is equal to the normal force \\(F_{N}\\).\r\n\\[\\begin{align*} F_{N} & = mg\\cos\\theta \\\\[6pt]\r\n& = 50\\times10^{-3}\\times9.8\\times\\cos30^{\\circ}\\\\[6pt]\r\n& = 0.43\\,\\text{N} \\end{align*}\\]\r\nThe force of the incline on the car as it travels down the slope is \\(0.43\\,\\text{N}\\).\r\n\r\n\r\n\r\nExercise \u2013 calculating forces on slopes\r\n\r\n \tA skateboarder riding a skateboard of total mass \\(60\\,\\text{kg}\\) coasts down a frictionless ramp at an angle of \\(30^{\\circ}\\) to the horizontal. Remember, \\(g=9.8\\,\\text{m s}^{-2}\\).\r\n\t\r\n\t\tCalculate the normal force acting on the rider and skateboard.\r\n\t\tCalculate the force acting on the rider and skateboard parallel to the ramp.\r\n\t\r\n\t\r\n\r\n\r\n\r\n\r\n \tThe skateboarder now coasts down another ramp with the same angle, but this time the ramp has a rough surface.\r\n\t\t\r\n\t\t\tCalculate the normal force acting on the rider and skateboard.\r\n\t\t\tCalculate the force acting on the rider and skateboard parallel to the ramp due to gravity.\r\n\t\t\tIf the ramp has a frictional force of \\(54\\,\\text{N}\\), what is the net force acting on the rider and the skateboard?\r\n\t\t\tCalculate the acceleration of the skateboarder.\r\n\t\t\tIf the skateboarder started from rest and the ramp is \\(4\\,\\text{m}\\) long, what was the speed of the skateboarder at the bottom of the ramp?\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n","n":0.036},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Maths","i":2,"n":1},{"v":"Physics","i":1,"n":1},{"v":"Force","i":0,"n":1}]]},{"i":2,"$":[{"v":"Newton's laws of motion","n":0.5},{"v":"In everyday life, forces are all around us, from the push of your feet on the ground when you start walking, to the pull of gravity that keeps you on Earth. Newton\u2019s three laws of motion describe how these forces affect the way objects start moving, keep moving, or come to rest. This resource introduces these laws and shows how they apply in common situations, including objects inside building lifts, sleds being pulled, and using Indigenous tools and watercraft.\r\n\r\nA force \\(F\\), measured in Newtons \\(\\text{N}\\), is any action that makes an object stay still, move or distort. The relationships between the motion of objects and the forces acting on them is described by Newton\u2019s laws.\r\n\r\nNewton's first law of motion\r\nAn object moving at a certain velocity tends to stay moving at that velocity and an object at rest tends to stay at rest unless acted upon by an unbalanced force.\r\nConsider the following force diagram. There is a force of \\(100\\,\\text{N}\\) applied to the left of the object and a force of \\(150\\,\\text{N}\\) to the right.\r\n\r\nForce is a vector quantity, so direction must be taken into consideration. To find the sum of the forces \\(\\sum F\\) or the net force, we need to set one direction as positive and one direction as negative. Here, we will let right be the positive direction and left be the negative direction. This means we have a force of \\(150\\,\\text{N}\\) in the positive direction, and \\(-100\\,\\text{N}\\) in the positive direction.\r\n\\[\\begin{align*} \\sum F & = 150-100 \\\\[6pt]\r\n& = 50\\,\\text{N to the right}\\end{align*}\\]\r\nThere is an unbalanced force causing the object to accelerate to the right.\r\nClick to reveal more examples.\r\n\r\n\r\n\r\n\r\n\r\n\r\nNewton's second law of motion\r\nForce equals mass times acceleration:\r\n\\[\\sum F=ma\\]\r\nwhere:\r\n\r\n\t\\(m\\) is the mass of the object in \\(\\text{kg}\\)\r\n\t\\(a\\) is the acceleration in \\(\\text{m s}^{-2}\\).\r\n\r\n\r\nExample 1 \u2013 using Newton's second law\r\nA toy car of mass \\(\\mathbf{2}\\,\\textbf{kg}\\) has a driving force \\(\\mathbf{F_{D}}\\) of \\(\\mathbf{20}\\,\\textbf{N}\\). The frictional force \\(\\mathbf{F_{F}}\\) acting on the car is \\(\\mathbf{10}\\,\\textbf{N}\\). Find the car's acceleration.\r\nWhen solving problems using Newton's second law, it helps to draw all the forces acting on the object. In this example, the driving and frictional forces are opposing.\r\n\r\nWe take the direction the car is driving in as the positive direction. Therefore, the net force is:\r\n\\[\\begin{align*} \\sum F & = F_{D}-F_{F} \\\\[6pt]\r\n& = 20-10\\\\[6pt]\r\n& = 10\\,\\text{N}\\end{align*}\\]\r\nUsing Newton\u2019s second law:\r\n\\[\\begin{align*} ma & = 10\\\\[6pt]\r\n2a & = 10\\\\[6pt]\r\na & = 5\\,\\text{m s}^{-2}\\end{align*}\\]\r\nTherefore, the magnitude of the car\u2019s acceleration is \\(5\\,\\text{m s}^{-2}\\).\r\n\r\n\r\n\r\nNewton\u2019s third law of motion\r\nFor every action there is an equal and opposite reaction.\r\nWhen two objects interact, the forces they exert on each other are equal in size but opposite in direction. These pairs of forces are called action\u2013reaction forces. Each force in an action\u2013reaction pair acts on a different object \u2013 this is what distinguishes the third law from the second law, where you consider all forces acting on the same object.\r\nConsider a person sitting on a wheelchair. The person pushes down on the wheelchair (action, \\(F_{PW}\\)), and the wheelchair pushes back up on the person with an equal and opposite force (reaction, \\(F_{WP}\\)). These two forces act on different objects\u2014one on the person, one on the wheelchair\u2014which is the defining feature of a Newton's third law pair.\r\n\r\n\r\n\r\nTypes of forces\r\nNewton's laws describe how forces affect motion, but so far we've treated force as a single abstract quantity. In practice, there are many different types of forces, each arising from a specific physical interaction.\r\n\r\nGravitational force and weight \\(W\\)\r\nGravitational force pulls all objects with mass toward each other. For example, all objects near the Earth are pulled towards its centre by the force of gravity. The gravitational field strength near the Earth\u2019s surface is about \\(g=9.8\\,\\text{m s}^{-2}\\).\r\nMass tells you how much matter an object has; weight tells you how strongly gravity pulls on that mass. The weight of an object is the gravitational force acting on it. It is a vector as it acts downwards towards the centre of the Earth.\r\n\\[W=mg\\]\r\nwhere:\r\n\r\n\t\\(W\\) is the weight in Newtons \\(\\text{N}\\)\r\n\t\\(m\\) is the mass in kilograms \\(\\text{kg}\\)\r\n\t\\(g\\) is the gravitational field strength in metres per second squared \\(\\text{m s}^{-2}\\).\r\n\r\n\r\nNormal force \\(F_{N}\\)\r\nThe normal force arises whenever two surfaces are in contact. It acts at right angles to the surface with which it is in contact.\r\nConsider a book resting on a table. Two forces act on the book: its weight \\(W=mg\\) pulling it downward, and the normal force \\(F_{N}\\) from the table pushing it upward. Since the book is not accelerating, Newton's second law tells us that the net force must be zero:\r\n\\[\\sum F = F_{N}-mg = 0\\quad\\therefore F_{N}=mg\\]\r\n\r\nApplying Newton's laws\r\nWe can put these ideas together to analyse real situations. In each of the scenarios below, the approach is the same: identify all the forces acting on the object, assign a positive direction, and apply Newton's second law to find the unknown quantity. Drawing a force diagram before writing any equations is a good habit \u2013 it makes it much easier to see which forces are acting and in which direction.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nNewton's third law of motion tells us that for every action there is an equal and opposite reaction. Two examples from across Australia and the Torres Strait bring this principle to life.\r\n\r\n\r\n\r\nExercise \u2013 applying Newton's laws\r\n\r\n \tDetermine the resultant force needed to give a mass of \\(6.4\\,\\text{kg}\\) an acceleration of \\(2.4\\,\\text{m s}^{-2}\\) West.\r\n\r\n\r\n\r\n\r\n \tIf a resultant force of \\(48\\,\\text{N}\\) produces an acceleration of \\(1.2\\,\\text{m s}^{-2}\\) on an object, calculate the mass of the object.\r\n\r\n\r\n\r\n\r\n \tA resultant force of \\(5.0\\,\\text{N}\\) acts on an object and causes it to reach a velocity of \\(4.0\\,\\text{m s}^{-1}\\) in \\(2.5\\,\\text{s}\\). Determine the mass of the object.\r\n\r\n\r\n\r\n\r\n \tAn object of mass \\(6.0\\,\\text{kg}\\) is at rest on a rough horizontal table. A horizontal force of \\(2.4\\,\\text{N}\\) acts on the mass to the East. The mass reaches a speed of \\(1.2\\,\\text{m s}^{-1}\\) in a distance of \\(2.0\\,\\text{m}\\).\r\n \t\r\n\t\tCalculate the acceleration of the mass.\r\n\t\tCalculate the net horizontal force acting on the object.\r\n\t\tDetermine the frictional force acting on the object.\r\n\t\r\n\r\n\r\n\r\n","n":0.032},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Maths","i":2,"n":1},{"v":"Physics","i":1,"n":1},{"v":"Force","i":0,"n":1}]]},{"i":3,"$":[{"v":"Blocks and pulleys","n":0.577},{"v":"Blocks and pulleys are used throughout engineering: in cranes, elevators, cable-driven machines and stage rigging. Use this resource to learn about pulley systems and how to analyse two-pulley systems.\r\n\r\nPulley systems\r\nA pulley system uses a rope that passes around one or more pulleys to lift or move a load. A single frame containing the wheels through which rope is threaded is called a block. Using several pulleys can change the direction of a force and reduce the force needed to move the load.\r\n\r\nWhen multiple pulleys are connected by a single rope, the motion of one part directly constrains the motion of every other part: if one mass moves down, another must move up or sideways. Engineers often have to analyse how the motion of one part affects the motion of another.\r\n\r\n\r\n\r\nIn any pulley system, the rope itself doesn't stretch, so the total length of each rope stays the same throughout the motion:\r\n\\[L = \\text{constant}\\]\r\nThis is the starting point for every problem. Once you've written \\(L\\) in terms of the positions of the masses and pulleys, you know that its rate of change must be zero:\r\n\\[\\dfrac{dL}{dt} = 0\\]\r\nThis lets you relate the motion of one part of the system to another.\r\n\r\nVelocity\r\nIf \\(x\\) is the position of a mass or pulley (measured from a fixed datum), then differentiating \\(x\\) with respect to time gives its velocity.\r\n\\[\\dfrac{dx}{dt}=v\\]\r\nThis lets us relate length to velocity.\r\n\r\nAcceleration\r\nDifferentiating velocity with respect to time (or differentiating position twice) gives acceleration.\r\n\\[\\dfrac{dv}{dt} = \\dfrac{d^{2}x}{dt^{2}} = a\\]\r\n\r\nMovable pulley factor\r\nWhen a pulley is fixed to a wall or the ceiling doesn't move, the pulley's position is a constant and drops out when you differentiate. A movable pulley, by contrast, 'hangs' in the rope and moves as the rope moves. If the rope wraps around a movable pulley twice (once on each side), that pulley's position is counted twice in the length equation, so it appears with a coefficient of \\(2\\). This factor of \\(2\\) is why the mass hanging from a movable pulley moves at half the speed of the rope pulling it.\r\n\r\nAnalysing pulley systems\r\nThe key principle that enables us to analyse the motion of pulley systems is that rope length is constant. By writing an equation for the total length of each rope in terms of the positions of the masses and pulleys, and then differentiating with respect to time, we can relate the velocities and accelerations of different parts of the system to one another, even when the geometry is complex.\r\n\r\nExample \u2013 analysing a pulley system\r\nConsider the following two-mass pulley system. Mass B is pulled downwards with a velocity of \\(\\mathbf{1.2}\\,\\textbf{m s}\\mathbf{^{-1}}\\). This velocity is decreasing at a rate of \\(\\mathbf{0.6}\\,\\textbf{m s}\\mathbf{^{-2}}\\). Determine the velocity and acceleration of pulley A.\r\n\r\n\r\nFirst, measure the position of all masses from a common datum, and add dimensions to the drawing. These dimensions are position vectors and direction is important. Let the upwards direction (\\(\\uparrow\\)) be positive (\\(+\\text{ve}\\)).\r\n\r\n\r\nFrom the problem, we know that mass B is moving downwards at \\(1.2\\,\\text{m s}^{-1}\\). Since we have defined upwards as positive, \\(v_{B}\\) is negative:\r\n\\[v_{B} = -1.2\\,\\text{m s}^{-1}\\]\r\nThe velocity of B is decreasing \u2013 that is, B is slowing down. Since B moves downwards, slowing down means the acceleration acts in the opposite direction (upwards) which is positive.\r\n\\[a_{B} = 0.6\\,\\text{m s}^{-2}\\]\r\nThe constants \\(L_{1}\\), \\(L_{2}\\), \\(x_{D}\\) and \\(k\\) do not change with time. Pulley D is fixed to the ceiling, so \\(x_{D}\\) is constant. The rope segment \\(k\\) (from the ceiling to pulley D) is also a fixed length. When we differentiate, both terms become zero.\r\nWe write the length of each rope by tracing its path. Rope \\(L_{1}\\) runs from the fixed base, up to the underside of pulley A, back down and under pulley A again (it loops around the pulley), then continues up to pulley C. Tracing this path, the total rope length is the sum of each vertical segment. Using the upward-positive convention with the datum at the floor:\r\n\\[L_{1} = 2x_{C} - x_{B}\\quad(\\text{Equation }1)\\]\r\nThe factor of \\(2\\) appears because the rope passes pulley C twice \u2013 once on each side.\r\nRope \\(L_{2}\\) is anchored at the ceiling and runs down to pulley D (a fixed distance), down again to pulley A (wrapping underneath), back up to pulley D, then down to pulley C. Tracing the variable segments:\r\n\\[\\begin{align*} L_{2} & = 2(x_{D} - x_{A}) + (x_{D} - x_{C}) + k \\\\[6pt]\r\n& = -2x_{A} - x_{C} + 3x_{D} + k\\quad(\\text{Equation }2) \\end{align*}\\]\r\nSince \\(x_{D}\\) and \\(k\\) are constants, the \\(x_{D}+k\\) term disappears when we differentiate.\r\n\r\nSince velocity is the rate of change of position, differentiating a position variable \\(x\\) with respect to time gives velocity \\(v\\). That is, \\(\\dfrac{d}{dt}x=v\\). Differentiating again gives acceleration \\(a\\). Because the rope lengths \\(L_{1}\\) and \\(L_{2}\\) are constant, their derivatives are zero. This is what links the motions of the different parts of the system.\r\n\r\nFor equation 1:\r\n\\[\\begin{align*} \\frac{d}{dt}(L_{1}) & = \\frac{d}{dt}(2x_{C} - x_{B}) \\\\[6pt]\r\n0 & = 2v_{C} - v_{B} \\\\[6pt]\r\nv_{C} & = \\frac{v_{B}}{2} \\\\[6pt]\r\n& = \\frac{-1.2}{2} \\\\[6pt]\r\n& = -0.6\\,\\text{m s}^{-1} \\end{align*}\\]\r\n\r\nFor equation 2:\r\n\\[\\begin{align*} \\frac{d}{dt}(L_{2}) & = \\frac{d}{dt}(-2x_{A} - x_{C} + 3x_{D} + k) \\\\[6pt]\r\n0 & = -2v_{A} - v_{C} + 3(0) \\\\[6pt]\r\nv_{A} & = \\frac{-v_{C}}{2} \\\\[6pt]\r\n& = \\frac{-(-0.6)}{2} \\\\[6pt]\r\n& = 0.3\\,\\text{m s}^{-1} \\end{align*}\\]\r\n\r\nTo find acceleration, we differentiate again. For equation 1:\r\n\\[\\begin{align*} 0 & = 2a_{C} - a_{B} \\\\[6pt]\r\na_{C} & = \\frac{a_{B}}{2} \\\\[6pt]\r\n& = \\frac{0.6}{2} \\\\[6pt] \r\n& = 0.3\\,\\text{m s}^{-2} \\end{align*} \\]\r\nFor equation 2:\r\n\\[\\begin{align*} 0 & = -2a_{A} - a_{C} + 3(0) \\\\[6pt]\r\na_{A} & = \\frac{-a_{C}}{2} \\\\[6pt]\r\n& = \\frac{-(0.3)}{2} \\\\[6pt]\r\n& = -0.15\\,\\text{m s}^{-2} \\end{align*}\\]\r\n\r\nTherefore, pulley A has a velocity of \\(v_{A}=0.3\\,\\text{m s}^{-1}\\) (upwards, since the result is positive) and an acceleration of \\(a_{A}=-0.15\\,\\text{m s}^{-2}\\). The negative sign tells us that pulley A is accelerating downwards. That is, it is slowing down as it moves upwards, consistent with mass B slowing down as it moves downwards.\r\nOnce the velocities and accelerations are known, Newton's second law can be applied to find the tension forces in the ropes, which is typically the next step in a full dynamics analysis of this kind of system.\r\n\r\nExercise \u2013 analysing a pulley system\r\n\r\n\tA single rope passes over one fixed pulley, connecting mass P on one side to mass Q on the other. Mass P is lowered at \\(0.8\\,\\text{m s}^{-1}\\). Find the velocity of mass Q.\r\n\r\n\r\n\r\n\r\n\tUsing the same pulley arrangement as the worked example (pulleys A, C and D), suppose pulley C now moves downwards at \\(0.9\\,\\text{m s}^{-1}\\) with an acceleration of \\(0.4\\,\\text{m s}^{-2}\\) upwards. Determine the velocity and acceleration of pulley A.\r\n\r\n\r\n\r\n\r\n\tMass B moves upwards at \\(2\\,\\text{m s}^{-1}\\), decelerating at \\(1\\,\\text{m s}^{-2}\\). Using Equation 1 from the worked example, find the velocity and acceleration of pulley C.\r\n\r\n\r\n\r\n\r\n\tExplain why a factor of \\(2\\) appears in Equation 1 (\\(L_{1} = 2x_{C} - x_{B}\\)) but pulley D does not appear at all in that equation.\r\n\r\n\r\n\r\n\r\n\tIf pulley A is raised at a constant velocity of \\(0.5\\,\\text{m s}^{-1}\\) (zero acceleration), what does this imply about the velocity and acceleration of mass B?\r\n\r\n\r\n\r\n \r\nBlocks and pulleys problem image transcript\r\nThis image shows a two-mass pulley system with a rope threaded through pulleys A, D and C, with a rectangular mass hanging from pulley A and a triangular mass B hanging from pulley C.\r\n \r\nLabelled blocks and pulleys problem image transcript\r\nThis image presents the same drawing labelled with dimensions.\r\nThe ground is labelled as the datum. The rope coming from the ceiling is \\(L_{2}\\) and the rope coming from the ground is \\(L_{1}\\).\r\nArrows are shown extending upwards from the datum to:\r\n\r\n\tthe centre of pulley A, labelled \\(x_{A}\\)\r\n\tthe top of mass B, labelled \\(x_{B}\\)\r\n\tthe centre of pulley C, labelled \\(x_{C}\\)\r\n\tthe centre of pulley D, labelled \\(x_{D}\\).\r\n\r\nThe distance between the ceiling and pulley D is \\(k\\).\r\n","n":0.028},[{"v":"Physics","i":0,"n":1}]]},{"i":4,"$":[{"v":"Non-constant acceleration","n":0.707},{"v":"In many cases, acceleration changes over time. You see this when a car that has been stopped at a traffic light speeds up to \\(\\mathbf{60}\\,\\textbf{km/h}\\) after the light changes to green. Acceleration also varies throughout a rollercoaster ride. Both of these scenarios demonstrate non-constant (or non-uniform) acceleration. Use this resource to learn about non-constant acceleration, including how it applies to Aboriginal spear-throwing.\r\n\r\nSo far, you've used the constant acceleration equations because acceleration stayed fixed throughout the motion. But when acceleration itself changes, those equations no longer apply, and we need a more general approach.\r\nRecall that velocity is the rate of change of displacement, and acceleration is the rate of change of velocity. Written using calculus notation, these are:\r\n\\[v=\\dfrac{dx}{dt}\\qquad a=\\dfrac{dv}{dt}\\]\r\nThis means we can find velocity and acceleration at any instant by differentiating displacement with respect to time \u2013 once, to get velocity, and again, to get acceleration.\r\nIntegration reverses differentiation, so we can also work backwards: starting from acceleration, we can integrate to find velocity, and integrate again to find displacement.\r\nIf you need a refresher on differentiation or integration before diving in, see the further resources at the end of this page.\r\n\r\n\r\n\r\n\r\nExample \u2013 using non-constant acceleration\r\nA toy car moves along a straight track. Its position is given by \\(\\mathbf{x=t^{3}-6t^{2}}\\), where \\(\\mathbf{t}\\) is in seconds and \\(\\mathbf{s}\\) is in  metres.\r\n\r\n\tCalculate the time(s) when:\r\n\t\t\r\n\t\t\tthe velocity is zero\r\n\t\t\tthe acceleration is zero\r\n\t\t\r\n\t\r\n\r\n\t\r\n\t\t\tWe have the equation for displacement. If we differentiate once with respect to time, we get velocity. Therefore:\r\n\t\t\t\\[\\begin{align*} x & = t^{3}-6t^{2} \\\\[6pt]\r\n\t\t\tv & = \\dfrac{dx}{dt}\\\\[6pt]\r\n\t\t\t& = 3t^{2}-12t\\end{align*}\\]\r\n\t\t\tFor \\(v=0\\), we simply substitute the value in and solve for \\(t\\).\r\n\t\t\t\\[\\begin{align*} 0 & = 3t^{2}-12t \\\\[6pt]\r\n\t\t\t0 & = 3t(t-4)\\\\[6pt]\r\n\t\t\t0=3t\\quad & \\text{ or }\\quad 0=t-4\\\\[6pt]\r\n\t\t\tt=0\\,\\text{s}\\quad & \\text{ or }\\quad t=4\\,\\text{s}\\end{align*}\\] Therefore, the toy car starts from rest and stops at \\(4\\,\\text{s}\\).\r\n\t\t\tThe equation for acceleration is obtained from differentiating the equation for velocity with respect to time.\r\n\t\t\t\\[\\begin{align*} a & = \\dfrac{dv}{dt} \\\\[6pt]\r\n\t\t\t& = 6t-12\\end{align*}\\]\r\n\t\t\tFor \\(a=0\\), we substitute the value in and solve for \\(t\\).\r\n\t\t\t\\[\\begin{align*} 0 & = 6t-12 \\\\[6pt]\r\n\t\t\t12 & = 6t\\\\[6pt]\r\n\t\t\tt & = \\frac{12}{6}\\\\[6pt]\r\n\t\t\tt & = 2\\,\\text{s}\\end{align*}\\]\r\n\t\t\r\n\r\n\tWhat is the total distance travelled by the car after \\(\\mathbf{5}\\,\\textbf{s}\\)?\r\n\r\n\tWe can find the total distance covered using displacement. We can start by finding the starting position, i.e. when \\(t=0\\,\\text{s}\\).\r\n\t\\[\\begin{align*} x & = 0^{3}-6(0)^{2} \\\\[6pt]\r\n\t& = 0\\end{align*}\\]\r\n\tThe car changes direction when \\(v=0\\,\\text{m/s}\\), i.e. at \\(t=0\\,\\text{s}\\) and \\(t=4\\,\\text{s}\\). We can then find the distance it travels before it changes direction at \\(t=4\\,\\text{s}\\).\r\n\t\\[\\begin{align*} x & = (4)^{3}-6(4)^{2} \\\\[6pt]\r\n\t& = 64-96\\\\[6pt]\r\n\t& = -32\\,\\text{m}\\end{align*}\\]\r\n\tFrom \\(t=0\\) to \\(t=4\\), the car travels \\(32\\,\\text{m}\\) in the negative direction.\r\n\tWe can now find the position of the car at \\(t=5\\,\\text{s}\\).\r\n\t\\[\\begin{align*} x & = (5)^{3}-6(5)^{2} \\\\[6pt]\r\n\t& = 125-150\\\\[6pt]\r\n\t& = -25\\,\\text{m}\\end{align*}\\]\r\n\tFrom \\(t=4\\) to \\(t=5\\), the car changes position from \\(-32\\,\\text{m}\\) to \\(-25\\,\\text{m}\\), which is \\(7\\,\\text{m}\\) in the positive direction.\r\n\tThe total distance travelled is the sum of these: \\(32+7=39\\,\\text{m}\\). This is different from the car's displacement, which would be the difference between the displacement at \\(t=5\\) and \\(t=0\\): \\(-25\\,\\text{m}\\).\r\n\r\n\r\n\r\nExercise \u2013 using non-constant acceleration\r\n\r\n\tThe position of a lift (elevator) moving up a shaft is given by \\(x=2t^{3}-3t^{2}+4t\\), where \\(x\\) is in metres and \\(t\\) is in seconds. Calculate:\r\n\t\t\r\n\t\t\tthe lift's velocity when \\(t=3\\,\\text{s}\\)\r\n\t\t\tthe lift's acceleration when \\(t=3\\,\\text{s}\\).\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tA skateboarder's velocity along a straight path is given by \\(v=3t^{2}+2t\\,\\text{m}\\,\\text{s}^{-1}\\), where \\(t\\) is in seconds. Calculate the skateboarder's displacement between \\(t=0\\,\\text{s}\\) and \\(t=2\\,\\text{s}\\).\r\n\r\n\r\n\r\n\r\n\tA rocket sled's acceleration is given by \\(a=6t-4\\,\\text{m}\\,\\text{s}^{-2}\\), where \\(t\\) is in seconds. If the sled's initial velocity (at \\(t=0\\)) is \\(5\\,\\text{m}\\,\\text{s}^{-1}\\), find the sled's velocity when \\(t=3\\,\\text{s}\\).\r\n\r\n\r\n\r\n\r\n\tA model train's position along a straight track is given by \\(x=t^{3}-9t^{2}\\), where \\(x\\) is in metres and \\(t\\) is in seconds. Calculate the time(s) when:\r\n\t\t\r\n\t\t\tthe train's velocity is zero\r\n\t\t\tthe train's acceleration is zero.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tA ball rolls along a straight, grooved track such that its position is given by \\(x=t^{3}-12t\\), where \\(x\\) is in metres and \\(t\\) is in seconds. Calculate the total distance travelled by the ball between \\(t=0\\,\\text{s}\\) and \\(t=5\\,\\text{s}\\).\r\n\r\n\r\n\r\n\r\n\tA sky diver's acceleration after opening their parachute is related to their (downward) velocity by \\(a=-0.25v\\,\\text{m}\\,\\text{s}^{-2}\\). At the moment the parachute opens (\\(t=0\\)), their velocity is \\(40\\,\\text{m}\\,\\text{s}^{-1}\\). Calculate:\r\n\t\t\r\n\t\t\tthe sky diver's velocity when \\(t=4\\,\\text{s}\\)\r\n\t\t\tthe distance fallen by the sky diver between \\(t=0\\,\\text{s}\\) and \\(t=4\\,\\text{s}\\).\r\n\t\t\r\n\t\r\n\r\n","n":0.038},[{"v":"Maths","i":3,"n":1},{"v":"Kinematics","i":2,"n":1},{"v":"Indigenous knowledges","i":1,"n":0.707},{"v":"Physics","i":0,"n":1}]]},{"i":5,"$":[{"v":"Constant acceleration","n":0.707},{"v":"Sometimes an object speeds up or slows down at a steady rate. For example, a car joining a freeway may increase its speed by the same amount every second, or a dropped object (ignoring air resistance) gains about \\(\\mathbf{9.8}\\,\\textbf{m/s}\\) of speed each second as it falls. In these situations, the acceleration stays the same with time \u2013 it is constant, or uniform.\r\n\r\nWhen acceleration is constant, the velocity\u2013time graph is a straight line. This lets us use a set of simple equations to relate displacement, velocity, acceleration and time. The same ideas apply to horizontal motion (like cars or cyclists) and vertical motion under gravity, provided the acceleration is uniform and we clearly choose which direction is positive.\r\n\r\nAcceleration in horizontal motion\r\nWe have already used the equation \\(v = \\dfrac{s}{t}\\) to describe motion where the velocity is constant. This equation cannot be used in situations where the velocity increases over time \u2013 that is, when acceleration occurs. Let's consider objects that accelerate at a constant rate.\r\n\r\nYou might remember that the uniform acceleration \\(a\\) of an object is given by the gradient of a velocity\u2013time graph:\r\n\\[\\begin{align*} a & = \\frac{\\text{rise}}{\\text{run}} \\\\[6pt]\r\n& = \\frac{\\text{change in velocity}}{\\text{time taken}}\\\\[6pt]\r\n& = \\frac{\\Delta v}{t}\\\\[6pt]\r\n& = \\frac{v_{f}-v_{i}}{t}\\\\[6pt]\r\n\\therefore at & = v_{f}-v_{i}\\end{align*}\\]\r\nwhere \\(v_{i}\\) is the initial velocity and \\(v_{f}\\) is the final velocity. This can be rearranged into the form:\r\n\\[v_{f} = v_{i} + at\\]\r\nWe can then use other relationships to obtain the following equations:\r\n\r\n\r\nwhere \\(x\\) is displacement, \\(v_{i}\\) is initial velocity, \\(v_{f}\\) is final velocity and \\(t\\) is time taken.\r\n\r\nOther symbols may be used to denote \\(x\\) (such as \\(s\\)), \\(v_{i}\\) (such as \\(u\\) or \\(v_{0}\\)) and \\(v_{f}\\) (such as \\(v\\)).\r\nWhere relevant, you must specify right or left as the positive or negative direction when doing these problems since they deal with vector quantities.\r\n\r\nExample \u2013 using horizontal acceleration \r\nA truck accelerates at \\(\\mathbf{0.8}\\,\\textbf{m}\\,\\textbf{s}\\mathbf{^{-2}}\\) from a starting velocity of \\(\\mathbf{3}\\,\\textbf{m}\\,\\textbf{s}\\mathbf{^{-1}}\\), until it reaches a velocity of \\(\\mathbf{20}\\,\\textbf{m}\\,\\textbf{s}\\mathbf{^{-1}}\\). Find:\r\n\r\n\tthe time taken\r\n\r\n\tIt is always a good idea to list the data first. We know that \\(v_{i}=3\\,\\text{m}\\,\\text{s}^{-1}\\), \\(v_{f}=20\\,\\text{m}\\,\\text{s}^{-1}\\) and \\(a=0.8\\,\\text{m}\\,\\text{s}^{-2}\\).\r\n\tNow, we need to find the appropriate equation that includes this data, and our unknown \\(t\\).\r\n\t\\[v_{f} = v_{i} + at\\]\r\n\tWe can either transpose the equation for \\(t\\) or substitute, then solve for \\(t\\). In this example, we will transpose first.\r\n\t\\[\\begin{align*} t & = \\frac{v_{f}-v_{i}}{a} \\\\[6pt]\r\n\t& = \\frac{20-3}{0.8}\\\\[6pt]\r\n\t& = 21.25\\,\\text{s}\\end{align*}\\]\r\n\tThe smallest number of significant figures is \\(1\\), so we should write our answer to \\(1\\) significant figure. In other words, the time taken is \\(20\\,\\text{s}\\).\r\n\r\n\tthe distance travelled.\r\n\r\n\tTo find the displacement, we need to find a equation containing our unknown \\(x\\).\r\n\t\\[v_{f}^{2} = v_{i}^{2} + 2ax\\]\r\n\tNow, we can substitute the values into the rearranged equation.\r\n\t\\[\\begin{align*} x & = \\frac{v_{f}^{2}-v_{i}^{2}}{2a} \\\\[6pt]\r\n\t& = \\frac{20^{2}-3^{2}}{2\\times0.8}\\\\[6pt]\r\n\t& = \\frac{400-9}{1.6}\\\\[6pt]\r\n\t& = 244.4\\,\\text{m}\\end{align*}\\]\r\n\r\nAcceleration due to gravity\r\nThe acceleration of a falling object near the Earth\u2019s surface is \\(9.8\\,\\text{m}\\,\\text{s}^{-2}\\). For instance, a coin that is dropped from rest will have a velocity of \\(9.8\\,\\text{m}\\,\\text{s}^{-1}\\) after \\(1\\,\\text{s}\\), \\(19.6\\,\\text{m}\\,\\text{s}^{-1}\\) after \\(2\\,\\text{s}\\), and so on. Acceleration due to gravity can also be expressed in units of \\(\\text{g}\\), where \\(1\\,\\text{g}=9.8\\,\\text{m}\\,\\text{s}^{-2}\\).\r\nSince the acceleration of a freely falling object is uniform (constant), we can use the same equations as we did with horizontal motion. Similarly, we may need to specify up or down as the positive or negative direction as acceleration is a vector quantity.\r\n\r\nExample 1 \u2013 using vertical acceleration\r\nA construction worker accidentally knocks a brick from a building so that it falls vertically a distance of \\(\\mathbf{50}\\,\\textbf{m}\\) to the ground. Calculate:\r\n\r\n\tthe time taken for the brick to reach the ground\r\n\r\n\tList the data first: \\(v_{i}=0\\), \\(x=50\\,\\text{m}\\) and \\(a=9.8\\,\\text{m}\\,\\text{s}^{-2}\\).We need to find \\(t\\), so the equation we need is:\r\n\t\\[x = v_{i}t+\\tfrac{1}{2}at^{2}\\]\r\n\tHere, we will substitute first, then solve for \\(t\\).\r\n\t\\[\\begin{align*} 50 & = \\left(0\\times t\\right)+\\left(\\frac{1}{2}\\times9.8\\times t^{2}\\right) \\\\[6pt]\r\n\t50 & = 0+4.9t^{2}\\\\[6pt]\r\n\t\\frac{50}{4.9} & = t^{2}\\\\[6pt]\r\n\t10.2 & = t^{2}\\\\[6pt]\r\n\t\\sqrt{10.2} & = t\\\\[6pt]\r\n\tt & = 3.2\\,\\text{s}\\end{align*}\\]\r\n\r\n\tthe speed of the brick as it hits the ground.\r\n\r\n\tThe speed of the brick when it hits the ground is the final velocity or \\(v_{f}\\). The relevant equation is:\r\n\t\\[v_{f}=v_{i}+at\\]\r\n\tNext, we take our data and substitute to find \\(v_{f}\\).\t\r\n\t\\[\\begin{align*} v_{f} & = 0+\\left(9.8\\times3.2\\right) \\\\[6pt]\r\n\t& = 31.3\\,\\text{m}\\,\\text{s}^{-1}\\end{align*}\\]\r\n\r\n\r\n\r\nExercise \u2013 using acceleration\r\nFor the following questions, use \\(g=9.8\\,\\text{m}\\,\\text{s}^{-2}\\).\r\n\r\n\tA car, starting from rest, moves with an acceleration of \\(2\\,\\text{m}\\,\\text{s}^{-2}\\). Find:\r\n\t\t\r\n\t\t\tthe velocity at the end of \\(20\\,\\text{s}\\)\r\n\t\t\tthe distance covered in that time.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tDetermine the uniform acceleration required for a spacecraft, starting from rest, to cover \\(1000\\,\\text{m}\\) in \\(10\\,\\text{s}\\).\r\n\r\n\r\n\r\n\r\n\tA cyclist, starting from rest, moves with an acceleration of \\(3\\text{m}\\,\\text{s}^{-2}\\). Find:\r\n\t\t\r\n\t\t\tthe time taken to reach a velocity of \\(30\\,\\text{m/s}\\)\r\n\t\t\tthe distance the cyclist covers in this time.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tAn object starts with a velocity of \\(100\\,\\text{m/s}\\) and decelerates (slows down) at \\(2\\text{m}\\,\\text{s}^{-2}\\). Find:\r\n\t\t\r\n\t\t\tat what time its velocity will be zero\r\n\t\t\thow far it will have gone.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tA book is knocked off a ledge and falls vertically to the floor. If the book takes \\(1.0\\,\\text{s}\\) to fall to the floor, calculate:\r\n\t\t\r\n\t\t\tits speed as it lands\r\n\t\t\tthe height from which it fell\r\n\t\t\tthe distance it falls during the first \\(0.5\\,\\text{s}\\)\r\n\t\t\tthe distance it falls during the final \\(0.5\\,\\text{s}\\).\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tA champagne cork travels vertically into the air. It takes \\(4.0\\,\\text{s}\\) to return to its starting position. Determine:\r\n\t\t\r\n\t\t\thow long the cork takes to reach its maximum height\r\n\t\t\tthe maximum height reached by the cork\r\n\t\t\thow fast the cork was travelling initially\r\n\t\t\tthe speed of the cork as it returned to its starting point\r\n\t\t\tthe acceleration of the cork at each of these times after its launch:\r\n\t\t\t\t\r\n\t\t\t\t\t\\(1.0\\,\\text{s}\\)\r\n\t\t\t\t\t\\(2.0\\,\\text{s}\\)\r\n\t\t\t\t\t\\(3.0\\,\\text{s}\\).\r\n\t\t\t\t\r\n\t\t\t\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n","n":0.034},[{"v":"Maths","i":2,"n":1},{"v":"Kinematics","i":1,"n":1},{"v":"Physics","i":0,"n":1}]]},{"i":6,"$":[{"v":"Work, energy and power","n":0.5},{"v":"Push a box across the floor and you've done work. Drop a ball and it trades height for speed. Run a motor and you measure how fast it delivers energy. Work, energy and power are three ways of describing the same underlying physics. This resource shows you how they connect, how to calculate each one, and presents one real world example of how Australia's first Indigenous female aerospace engineer is applying these principles to make aviation more sustainable.\r\n\r\nWork \\(W\\)\r\nWork \\(W\\) is done when a force causes a displacement in the direction the force is applied. It is a scalar quantity with no direction. For a constant force acting along a straight line, work is given by:\r\n\\[W=Fd\\]\r\nwhere:\r\n\r\n\t\\(W\\) is the work done in joules \\(\\text{J}\\)\r\n\t\\(F\\) is the force in Newtons \\(\\text{N}\\)\r\n\t\\(d\\) is the displacement in metres \\(\\text{m}\\).\r\n\r\n\r\n\r\nConsider a force that acts at an angle. For example, a block is pulled with tension \\(T\\) by a string at an angle \\(\\theta\\) to the horizontal.\r\n\r\nIf the block moves a horizontal distance \\(d\\), only the component of the force in the direction of motion, i.e. horizontal, does work. This is given by the equation:\r\n\\[W=Fd=(T\\cos\\theta)d\\]\r\nwhere:\r\n\r\n\t\\(T\\) is the tension in Newtons \\(\\text{N}\\)\r\n\t\\(\\theta\\) is the angle between the force and the direction of motion in degrees.\r\n\r\nHere, \\(F=T\\cos\\theta\\) is the horizontal component of the tension. We resolve the force along the direction of motion before calculating work. If the force is perpendicular to the displacement, no work is done, since:\r\n\\[W=Fd\\cos\\theta=Fd\\cos90^{\\circ}=0\\]\r\n\r\nUsing force\u2013distance graphs to calculate work\r\nIf the force is not constant, the formula \\(W=Fd\\) does not directly apply. Instead, we can draw a force\u2013distance graph and find the work done from the area under the curve. This is similar to using the area under a force\u2013time graph to find the impulse of a force, but here, the horizontal axis is distance rather than time.\r\n\r\nExample \u2013 calculating work using force\u2013distance graphs\r\nSuppose a force increases linearly from \\(\\mathbf{0}\\,\\textbf{N}\\) to \\(\\mathbf{200}\\,\\textbf{N}\\) over a distance of \\(\\mathbf{5}\\,\\textbf{m}\\). Calculate the work done.\r\nSince the force increases linearly from \\(0\\,\\text{N}\\) to \\(200\\,\\text{N}\\), the force\u2013distance graph is a straight line from the origin, forming a triangle. The work done is the area of this triangle, with base \\(5\\,\\text{m}\\) and height \\(200\\,\\text{N}\\).\r\n\r\n\\[\\begin{align*} W & = \\frac{1}{2}\\times\\text{base}\\times\\text{height} \\\\[6pt]\r\n& = \\frac{1}{2}\\times5\\times200 \\\\[6pt]\r\n& = 500\\,\\text{J} \\end{align*}\\]\r\nThe work done is \\(500\\,\\text{J}\\).\r\n\r\nEnergy and conservation of energy\r\nEnergy is the capacity to do work. When work is done on or by an object, its energy changes.\r\nEnergy exists in many different forms:\r\n\r\n\tkinetic energy, \\(E_{k}\\) or KE\r\n\tgravitational potential energy, \\(U_{g}\\) or GPE\r\n\telectrical energy, E (see the Measuring electricity page)\r\n\telastic potential energy \\(U_{s}\\) or EPE.\r\n\r\nA fundamental principle of nature is:\r\nEnergy cannot be created or destroyed; it can only be transformed or transferred from one form to another.\r\n\r\nWhen a force does work on an object, the work is equal to the change in the object\u2019s energy:\r\n\\[W=\\Delta E\\]\r\nwhere:\r\n\r\n\t\\(W\\) is the work done in joules, \\(\\text{J}\\)\r\n\t\\(\\Delta E\\) is the change in energy, measured in joules \\(\\text{J}\\).\r\n\r\n\r\n\r\nFor example, if an object is acted on by a net force, the work done on it equals the change in its kinetic energy:\r\n\\[\\begin{align*} W & = \\Delta \\text{KE} \\\\[6pt]\r\n& = \\frac{1}{2}mv_f^{2}-\\frac{1}{2}mv_{i}^{2} \\end{align*}\\]\r\nwhere:\r\n\r\n\t\\(m\\) is the mass of the object in kilograms \\(\\text{kg}\\)\r\n\t\\(v_i\\) and \\(v_f\\) are the initial and final speeds, respectively, in metres per second \\(\\text{m s}^{-1}\\).\r\n\r\n\r\nKinetic energy\r\nKinetic energy (KE) is the energy of motion. For a body of mass \\(m\\) moving at speed \\(v\\), its KE is given by:\r\n\\[\\text{KE}=\\frac{1}{2}mv^{2}\\]\r\nIf an object\u2019s speed doubles, its KE increases by a factor of four, since KE is proportional to \\(v^{2}\\). Tripling the speed increases KE by a factor of nine.\r\n\r\nGravitational potential energy\r\nGravitational potential energy (GPE) is energy 'stored' due to an object\u2019s position in a gravitational field. For an object of mass \\(m\\) at height \\(h\\) above a reference level, its GPE is given by:\r\n\\[\\text{GPE}=mgh\\]\r\nwhere \\(g\\) is the gravitational acceleration in metres per second squared \\(\\text{m s}^{-2}\\).\r\n\r\nIn many problems, we use conservation of mechanical energy, where energy transfers between kinetic and potential forms but the total remains (approximately) constant. For example, a ball of mass \\(m\\) dropped from a height \\(h\\) loses gravitational potential energy and gains kinetic energy as it falls.\r\n\\[\\Delta\\text{GPE}+\\Delta\\text{KE}=0\\quad\\Rightarrow\\quad\\Delta(mgh)=\\Delta\\left(\\dfrac{1}{2}mv^{2}\\right)\\]\r\n\r\nElastic potential energy\r\nElastic potential energy (EPE, or strain energy) is the energy stored in an elastic material when it is stretched or compressed. Elastic materials behave differently from rigid materials because they can extend and compress, and then return (approximately) to their original shape.\r\n\r\nHooke\u2019s law\r\nHooke\u2019s law describes how the force needed to stretch or compress a spring is related to its extension. It states that the force applied to a spring is proportional to its extension (or compression):\r\n\r\n\\[F = -k x,\\]\r\nwhere:\r\n\r\n\t\\(F\\) is the force in Newtons \\(\\text{N}\\)\r\n\t\\(k\\) is the spring constant in Newtons per metre \\(\\text{N m}^{-1}\\)\r\n\t\\(x\\) is the extension or compression in metres \\(\\text{m}\\).\r\n\r\n\r\nThe spring constant is equivalent to the slope of the straight-line 'Hookean' region of the force\u2013extension graph. It measures the stiffness of the spring, with a steeper slope meaning a stiffer spring. The negative sign before the \\(k\\) indicates that the force required to restore the spring to its (approximately) original shape acts in the opposite direction to the displacement \u2013 it pulls back towards the natural length.\r\n\r\nMany materials obey Hooke\u2019s Law only up to a certain extension, called the elastic limit. Beyond this point, the behaviour becomes non-linear; the material may stiffen, weaken, or eventually break. Stretching a material beyond its elastic limit can cause permanent deformation so that it does not return to its original length.\r\n\r\nFrom this, the elastic potential energy stored in a spring that obeys Hooke\u2019s law is given by:\r\n\\[\\text{EPE}=\\frac{1}{2}kx^{2}\\]\r\n\r\nUsing force\u2013extension and force\u2013compression graphs to calculate work\r\nJust as the area under a force\u2013distance graph gives the work done by a force, the area under a force\u2013extension (or compression) graph gives the work done in stretching or compressing a spring.\r\nFor a spring that obeys Hooke\u2019s law, the force\u2013extension graph is a straight line from \\(0\\) to \\(F=kx\\). The work done (and energy stored) when the spring is extended from \\(0\\) to \\(x\\) is the area of the triangle under the graph.\r\n\\[\\begin{align*} EPE & = W \\\\[6pt]\r\n& = \\frac{1}{2}\\times x\\times kx \\\\[6pt]\r\n& = \\frac{1}{2}kx^{2} \\end{align*}\\]\r\n\r\nPower \\(P\\)\r\nPower is the rate at which work is done (or energy is transferred):\r\n\\[\\text{Power}=\\frac{W}{t}\\]\r\nwhere:\r\n\r\n\t\\(W\\) is the work done in joules \\(\\text{J}\\)\r\n\t\\(t\\) is the time taken in seconds \\(\\text{s}\\).\r\n\r\nIf a constant force moves an object at constant velocity in the direction of the force, then:\r\n\\[P=\\frac{Fd}{t}=Fv\\]\r\nwhere:\r\n\r\n\t\\(P\\) is the power in watts \\(\\text{W}\\)\r\n\t\\(F\\) is the force in the direction of the motion in Newtons \\(\\text{N}\\)\r\n\t\\(d\\) is the distance in metres \\(\\text{m}\\)\r\n\t\\(t\\) is the time taken in seconds \\(\\text{s}\\)\r\n\t\\(v\\) is the velocity in metres per second \\(\\text{m s}^{-1}\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\nExample \u2013 calculating work, energy and power\r\nThe following graph shows how the total resistance forces acting on a cyclist and her bicycle vary with distance at the start of a race. The cyclist applies a constant force over the first \\(\\mathbf{50}\\,\\textbf{m}\\) of the race and her velocity becomes constant after travelling \\(\\mathbf{40}\\,\\textbf{m}\\). The cyclist and bicycle have a combined mass of \\(\\mathbf{80}\\,\\textbf{kg}\\).\r\n\r\n\r\n\tCalculate the amount of work the cyclist does against the resistance force over the first \\(\\mathbf{40}\\,\\textbf{m}\\) of the race.\r\n\r\nThe area under the graph has two regions: a triangle from \\(0\\) to \\(10\\,\\text{m}\\), and a trapezoid from \\(10\\) to \\(40\\,\\text{m}\\).\r\n\\[\\begin{align*} W & = \\left(\\frac{1}{2}\\times10\\times200\\right) + \\left(\\frac{1}{2}\\times\\left(200+600\\right)\\times30\\right) \\\\[6pt]\r\n& = 1000+12000 \\\\[6pt]\r\n& = 13000\\,\\text{J} \\end{align*}\\]\r\nThe cyclist does \\(13\\,\\text{kJ}\\) of work over the first \\(40\\,\\text{m}\\).\r\n\r\n\r\n\tCalculate the power developed by the cyclist during the first \\(\\mathbf{40}\\,\\textbf{m}\\) of the race if she took \\(\\mathbf{10}\\,\\textbf{s}\\) to cover this distance.\r\n\r\nPower is work done per unit time. Using the work calculated in part a:\r\n\\[\\begin{align*} P & = \\frac{W}{t} \\\\[6pt]\r\n& = \\frac{13000}{10} \\\\[6pt]\r\n& = 1300\\,\\text{W} \\end{align*}\\]\r\nShe develops \\(1300\\,\\text{W}\\) of power during the first \\(40\\,\\text{m}\\).\r\n\r\n\r\n\tCalculate the magnitude of the constant force applied by the cyclist over the first \\(\\mathbf{50}\\,\\textbf{m}\\) of the race.\r\n\r\nWhen velocity becomes constant at \\(40\\,\\text{m s}^{-1}\\), the net force is zero, meaning that the driving force exactly equals the resistance force. From the graph, the resistance force at \\(40\\,\\text{m}\\) is \\(600\\,\\text{N}\\), so the cyclist's constant driving force must also be \\(600\\,\\text{N}\\) throughout.\r\n\r\nExercise \u2013 calculating work, energy and power\r\n\r\n\tA car is at rest at the top of an extremely steep hill (point A) of height \\(100\\,\\text{m}\\). The mass of the car and driver is \\(1200\\,\\text{kg}\\).\r\n\t\r\n\t\t\r\n\t\t\tAssuming frictional forces are ignored, calculate the speed that the car has at point B if it is allowed to roll down the hill without the driver applying the brakes.\r\n\t\t\tAssuming the driver applies brakes and does \\(1.0\\times10^{6}\\,\\text{J}\\) of work against friction as it rolls down the hill, determine the speed of the car at B.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tThe following graph shows how the force applied by a pinball spring plunger changes as it is compressed during a pinball game.\r\n\t\r\n\tThe plunger is compressed \\(1\\,\\text{cm}\\) and then released. If the pinball has a mass of \\(50\\,\\text{g}\\), calculate its speed at the instant it leaves the plunger.\r\n\t\r\n\r\n\r\n\r\n\r\n\tA person pulls a \\(20\\,\\text{kg}\\) suitcase \\(50\\,\\text{m}\\) along a flat airport terminal floor. They apply a force of \\(80\\,\\text{N}\\) at an angle of \\(30^{\\circ}\\) above the horizontal. Calculate the work done.\r\n\r\n\r\n\r\n\r\n\tA spring is stretched \\(0.15\\,\\text{m}\\) by a force of \\(45\\,\\text{N}\\).\r\n\t\t\r\n\t\t\tCalculate the spring constant.\r\n\t\t\tCalculate the elastic potential energy stored in the spring.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tA cyclist rides at a constant speed of \\(0.8\\,\\text{m s}^{-1}\\) against a total resistance force of \\(0.5\\,\\text{m s}^{-1}\\). Calculate the power output of the cyclist.\r\n\r\n\r\n\r\n\r\n\tAn electric motor lifts a \\(200\\,\\text{kg}\\) load at a constant speed of \\(0.5\\,\\text{m s}^{-1}\\). Calculate the power output of the motor, using \\(g=9.81\\,\\text{m s}^{-2}\\).\r\n\r\n\r\n\r\n\r\n\tA \\(15\\,\\text{kg}\\) box is pushed \\(8.0\\,\\text{m}\\) up a slope inclined at \\(25^{\\circ}\\) to the horizontal at a constant speed. The frictional force acting on the box is \\(30\\,\\text{N}\\). Calculate the total work done, using \\(g=9.81\\,\\text{m s}^{-2}\\).\r\n\r\n\r\n\r\n","n":0.025},[{"v":"Physics","i":0,"n":1}]]},{"i":7,"$":[{"v":"Core dynamics","n":0.707},{"v":"Core dynamics explores the forces that shape motion in the world around us \u2013 from the pull of gravity on a hillside to the physics behind a spinning wheel. You'll work with ideas like Newton's laws, friction, momentum and energy, building the skills to tackle the kinds of problems that show up throughout physics and engineering.\r\n","n":0.134},[{"v":"Physics","i":0,"n":1}]]},{"i":8,"$":[{"v":"Ohm's law","n":0.707},{"v":"Why does a cheap phone charger cable get warm and charge your phone slowly, or a longer extension cord sometimes struggle to power a kettle or hairdryer? The answer lies in the relationship between voltage, current and resistance. This resource introduces Ohm's law\u2014one of the most used relationships in electronics\u2014and shows you how to apply it to real circuits.\r\n\r\nResistance \\(R\\)\r\nWhen an electric current flows through a circuit, the components in the circuit can make it harder or easier for the charges to move. This opposition to the flow of current is called resistance.\r\nUnder constant physical conditions (such as temperature), the potential difference (voltage) across a component is proportional to the current through it (\\(V \\propto I\\)).\r\nThis relationship is known as Ohm\u2019s law and is written as:\r\n\\[V = I \\times R\\]\r\nwhere \\(V\\) is the voltage across the component, \\(I\\) is the current through it, and \\(R\\) is the resistance.\r\nThis is exactly what happens with a cheap charger cable. A thinner wire has higher resistance, so for the same voltage, less current reaches your device and the energy lost to resistance is released as heat. A quality cable uses lower-resistance wire, so more current flows through with less wasted energy.\r\n\r\nOhmic conductors\r\nSome components obey Ohm\u2019s law: when the voltage across them increases, the current through them increases in direct proportion. For these components, if you draw a graph of voltage \\(V\\) against current \\(I\\), you get a straight line through the origin.\r\n\r\n\r\nIf the circuit component obeys Ohm's law, it is said to be an ohmic conductor. The slope of its voltage\u2013current graph is a constant and equal to its resistance \\(R\\), measured in ohms \\(\\Omega\\). This resistance is therefore constant.\r\nFrom the definition of power (\\(P = VI\\)) and Ohm\u2019s law (\\(V = IR\\)), we can write:\r\n\\[P = VI = I^{2}R = \\frac{V^{2}}{R}\\]\r\n\r\nA component designed specifically to provide a fixed resistance is called a resistor. Resistors are ohmic conductors\u2014their resistance stays constant regardless of the voltage applied\u2014and are one of the most common components in electronic circuits.\r\n\r\n\r\nExample \u2013 applying Ohm's law to ohmic conductors\r\nAn ohmic conductor of \\(\\mathbf{5\\Omega}\\) is supplied with a voltage which can vary from \\(\\mathbf{1}\\,\\textbf{V}\\) to \\(\\mathbf{100}\\,\\textbf{V}\\).\r\n\r\n\tDetermine the range of current that flows through the conductor.\r\n\r\n\tThe range of current will depend on the range of voltage and the resistance. We can calculate the current for each voltage level.\r\n\tFor \\(1\\,\\text{V}\\):\r\n\t\\[\\begin{align*} V & = I\\times R \\\\[6pt]\r\n\tI & = \\frac{V}{R}\\\\[6pt]\r\n\t& = \\frac{1}{5}\\\\[6pt]\r\n\t& = 0.2\\,\\text{A}\\end{align*}\\]\r\n\tFor \\(100\\,\\text{V}\\):\r\n\t\\[\\begin{align*} V & = I\\times R \\\\[6pt]\r\n\tI & = \\frac{V}{R}\\\\[6pt]\r\n\t& = \\frac{100}{5}\\\\[6pt]\r\n\t& = 20\\,\\text{A}\\end{align*}\\]\r\n\tTherefore, the range of current flowing through the conductor will be \\(0.2\u201320\\,\\text{A}\\).\r\n\r\n\tCalculate the amount of energy that will be dissipated (used up) in the resistor each second.\r\n\r\n\tEnergy dissipated per second is power \\(P\\), since power is the rate of energy transfer/use per unit of time. The question is asking for the range of power values correponding to the range from part a. We use the equation \\(P=V\\times I\\).\r\n\tFor \\(V=1\\,\\text{V}\\) and \\(I=0.2\\,\\text{A}\\):\r\n\t\\[\\begin{align*} P & = V\\times I \\\\[6pt]\r\n\t& = 1\\times0.2\\\\[6pt]\r\n\t& = 0.2\\,\\text{W}\\end{align*}\\]\r\n\tFor \\(V=100\\,\\text{V}\\) and \\(I=20\\,\\text{A}\\):\r\n\t\\[\\begin{align*} P & = V\\times I \\\\[6pt]\r\n\t& = 100\\times20\\\\[6pt]\r\n\t& = 2000\\,\\text{W}\\end{align*}\\]\r\n\tTherefore, the energy dissipated will vary from \\(0.2\u20132000\\,\\text{W}\\) per second or \\(0.2\u20132000\\,\\text{J}\\).\r\n\r\nNon-ohmic conductors\r\nMany components do not behave like ohmic conductors. For these components, the current does not increase in direct proportion to the voltage, so the voltage\u2013current graph is curved. These components are called non-ohmic conductors.\r\nLight bulbs such as car headlamps are common examples of non-ohmic conductors. As the voltage increases, the current will not increase in proportion. Other non-ohmic conductors include devices with resistance that changes with light or temperature. These are particularly useful as detectors in sensors which need to respond to changes in light or temperature levels.\r\n\r\n\r\nExample \u2013 calculating resistance in non-ohmic devices\r\nConsider the graph showing the current\u2013voltage characteristics of a non-ohmic device.\r\n\r\nCalculate the resistance at:\r\n\r\n\t\\(\\mathbf{50}\\,\\textbf{V}\\)\r\n\r\n\tUnlike an ohmic device, the relationship between current and voltage is not linear; the curve shows that resistance changes at different voltages. We can still calculate the resistance at any specific point using \\(\\mathbf{F=\\dfrac{V}{I}}\\), where voltage and current are simply read off the graph at that point. This gives the resistance at that operating point.\r\n\tIn the graph, the current is given in \\(\\text{mA}\\), so we must convert to \\(\\text{A}\\).\r\n\tAt \\(V=50\\,\\text{V}\\), \\(I=150\\,\\text{mA}\\) or \\(0.150\\,\\text{A}\\). Therefore:\r\n\t\\[\\begin{align*} R & = \\frac{V}{I} \\\\[6pt]\r\n\t& = \\frac{50}{0.150}\\\\[6pt]\r\n\t& = 333.33\\,\\Omega\\end{align*}\\]\r\n\r\n\t\\(\\mathbf{150}\\,\\textbf{V}\\)\r\n\r\n\tAt \\(V=150\\,\\text{V}\\), \\(I=210\\,\\text{mA}\\) or \\(0.210\\,\\text{A}\\). Therefore:\r\n\t\\[\\begin{align*} R & = \\frac{V}{I} \\\\[6pt]\r\n\t& = \\frac{150}{0.210}\\\\[6pt]\r\n\t& = 714.2\\,\\Omega\\end{align*}\\]\r\n\r\n\t\\(\\mathbf{200}\\,\\textbf{V}\\).\r\n\r\n\tAt \\(V=200\\,\\text{V}\\), \\(I=240\\,\\text{mA}\\) or \\(0.240\\,\\text{A}\\). Therefore:\r\n\t\\[\\begin{align*} R & = \\frac{V}{I} \\\\[6pt]\r\n\t& = \\frac{200}{0.240}\\\\[6pt]\r\n\t& = 833.3\\,\\Omega\\end{align*}\\]\r\n\r\n\r\nResistivity \\(\\rho\\)\r\nThe resistance of a wire is a measure of how much the wire impedes the flow of electrons along its length. For a wire of a given material, the resistance depends on its length and cross-sectional area.\r\n\r\nThe resistance increases if the wire has a longer length (\\(L\\)): \\(R\\propto L\\)\r\nThe resistance decreases if the wire is thicker or has a larger cross-sectional area (\\(A\\)): \\(R\\propto \\dfrac{1}{A}\\)\r\nThe resistance also depends on the material the wire is made from. This property of the material is called the resistivity \\(\\rho\\) (pronounced rho).\r\n\r\nThese ideas can be summarised using the equation:\r\n\\[R = \\frac{\\rho L}{A}\\]\r\nwhere \\(R\\) is in \\(\\Omega\\), \\(\\rho\\) is in \\(\\Omega\\text{m}\\) and depends on the material used, \\(L\\) is in metres (\\(\\text{m}\\)), and \\(A\\) is in square metres (\\(\\text{m}^{2}\\)).\r\nThis explains the extension cord effect: a longer cord means a larger value of \\(L\\), which means a higher resistance. For a kettle or hairdryer drawing large currents, even a small increase in resistance produces a noticable voltage drop across the cord, leaving less voltage for the appliance itself.\r\nResistivity varies enormously between materials (e.g. copper vs rubber). It also depends on temperature, which\u2014in the scope of this resources\u2014we have not taken into consideration here.\r\n\r\nExample \u2013 using resistivity\r\nNormal household wiring uses \\(\\mathbf{1.8}\\,\\textbf{mm}\\) diameter copper wire. The resistivity of copper is \\(\\mathbf{\\rho=1.7\\times10^{-8}\\Omega}\\textbf{m}\\).\r\n\r\n\tDetermine the resistance of a \\(\\mathbf{10}\\,\\textbf{m}\\) long piece of this copper wire.\r\n\r\n\tThe cross-sectional area of the wire, a circle, is given by \\(A=\\pi\\times r^{2}\\), so:\r\n\t\\[\\begin{align*} A & = \\pi\\times r^{2} \\\\[6pt]\r\n\t& = 3.14\\times\\left(0.9\\times10^{-3}\\right)^{2} \\\\[6pt]\r\n\t& = 2.5\\times10^{-6}\\,\\text{m}^{2}\\end{align*}\\]\r\n\tNote that \\(0.9\\times10^{-3}\\) converts the \\(\\text{mm}\\) units to \\(\\text{m}\\). We are now ready to substitute these values into the equation \\(R=\\dfrac{\\rho L}{A}\\).\r\n\t\\[\\begin{align*} R & = \\frac{\\rho L}{A} \\\\[6pt]\r\n\t& = 1.7\\times10^{-8}\\times\\frac{10}{2.5\\times10^{-6}}\\\\[6pt]\r\n\t& = 0.068\\,\\Omega\\end{align*}\\]\r\n\r\n\tCalculate the drop in voltage drop along the wire if a current of \\(\\mathbf{10}\\,\\textbf{A}\\) is flowing through it.\r\n\r\n\tWe use \\(V=I\\times R\\) to find the voltage drop.\r\n\t\\[\\begin{align*} V & = I\\times R \\\\[6pt]\r\n\t& = 10\\times0.068\\\\[6pt]\r\n\t& = 0.68\\,\\text{V}\\end{align*}\\]\r\n\tTherefore, if the wire is carrying a \\(10\\,\\text{A}\\) current, there will be a voltage drop of \\(0.68\\,\\text{V}\\) along the length of the wire.\r\n\r\n\r\nExercise \u2013 applying Ohm's law\r\n\r\n\tA student finds that the current through a resistor is \\(3.5\\,\\text{A}\\) while a voltage of \\(2.5\\,\\text{V}\\) is applied to it.\r\n\t\t\r\n\t\t\tCalculate the resistance.\r\n\t\t\tThe voltage is then doubled and the current is found to increase to \\(7.0\\,\\text{A}\\). Determine whether the resistor is ohmic across the tested range.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tNahia and Rachel are trying to find the resistance of an electrical device. They find that at \\(5\\,\\text{V}\\), it draws a current of \\(200\\,\\text{mA}\\) and at \\(10\\,\\text{V}\\), it draws a current of \\(500\\,\\text{mA}\\). Nahia says that the resistance is \\(25\\,\\Omega\\), but Rachel maintains that it is \\(20\\,\\Omega\\). Determine who is right and justify your decision.\r\n\r\n\r\n\r\n\r\n\tAnil has an ohmic resistor to which he has applied \\(5\\,\\text{V}\\). He measures the current at \\(45\\,\\text{mA}\\). He then increases the voltage to \\(8\\,\\text{V}\\). Determine the current Anil will find now.\r\n\r\n\r\n\r\n\r\n\tSydney finds that when they increase the voltage across an ohmic resistor from \\(6\\,\\text{V}\\) to \\(10\\,\\text{V}\\), the current increases by \\(2\\,\\text{A}\\).\r\n\t\t\r\n\t\t\tCalculate the resistance of the resistor.\r\n\t\t\tDetermine the current the resistor draws at \\(10\\,\\text{V}\\).\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tThe resistance of a certain piece of wire is found to be \\(0.8\\,\\Omega\\). Calculate the resistance of:\r\n\t\t\r\n\t\t\ta piece of the same wire twice as long\r\n\t\t\ta piece of the same wire double the diameter.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tIf the resistance of a copper wire \\(20\\,\\text{m}\\) long and \\(1\\,\\text{mm}\\) in diameter is \\(0.44\\,\\Omega\\), calculate the resistance of the same length of wire \\(2\\,\\text{mm}\\) in diameter.\r\n\r\n\r\n\r\n","n":0.028},[{"v":"Electricity","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Physics","i":0,"n":1}]]},{"i":9,"$":[{"v":"Measuring electricity","n":0.707},{"v":"Every time you charge your phone, flick a light switch or plug in a kettle, you are interacting with an electrical circuit. But what is actually happening inside it? Use this resource to learn about the key quantities used to describe and measure electricity, the calculations that connect them, and an example of how electrical phenomena are represented in Wandjina rock paintings.\r\n\r\nElectric circuits\r\nAn electric circuit is a closed path that allows electric charge to flow. A simple circuit usually has a power source (such as a battery), conducting wires and one or more components (such as a light globe or motor). A circuit must be closed (with no breaks in the path) for current to flow.\r\nElectricity also occurs naturally in the environment. Lightning is a large electrical discharge caused by the build-up of charge in clouds, while smaller static charges can build up on objects and discharge as a spark.\r\nThe direction of current we use in circuit diagrams is called conventional current flow. It flows from the positive terminal of the battery, around the circuit, to the negative terminal.\r\nIn reality, the moving charges in a metal wire are electrons (\\(\\text{e}^{\u2013}\\)). Electrons flow from the negative terminal to the positive terminal, in the opposite direction to conventional current. If you want to read more about electrons, check out this page.\r\n\r\nElectrons all carry the same tiny amount of negative charge, so the total charge that flows in a circuit depends on how many electrons move past a point. Since the charge on a single electron is so small, electrical charge is usually measured in a large amount called a coulomb \\(\\text{C}\\), rather than counting individual electrons.\r\n\\[1\\,\\text{coulomb of negative charge} = 1\\,\\text{C} = 6.25 \\times 10^{18}\\,\\text{electrons}\\]\r\nor\r\n\\[1\\,\\text{electron} = 1\\,\\text{e}^{\u2013} = 1.6 \\times 10^{-19}\\,\\text{C}\\]\r\n\r\nCurrent \\(I\\) and charge \\(q\\)\r\nCurrent \\(I\\) is the rate of flow of electric charge \\(q\\) past a point in the circuit.\r\n\\[\\text{current} = \\frac{\\text{charge}}{\\text{time}} \\quad \\text{or} \\quad I = \\frac{q}{t}\\]\r\nwhere \\(I\\) is measured in amperes (amps) \\(\\text{A}\\), \\(q\\) is measured in coulombs \\(\\text{C}\\) and \\(t\\) is measured in seconds \\(\\text{s}\\).\r\n\r\nExample \u2013 calculating charge\r\nConsider a torch battery that produces a current of \\(\\mathbf{300}\\,\\textbf{mA}\\). Determine the total amount of charge that has flowed through the circuit if the torch has been left on for \\(\\mathbf{20}\\,\\textbf{min}\\).\r\nThe appropriate equation is \\(q=I\\times t\\).\r\nRemember, in this equation, \\(I\\) needs to be in \\(\\text{A}\\) and \\(t\\) needs to be in \\(\\text{s}\\). Let's complete the conversions separately, first.\r\n\\[\\begin{align*} I & = 300\\,\\text{mA} \\\\[6pt]\r\n& = 300\\times10^{-3}\\,\\text{A}\\\\[6pt]\r\n& = 0.300\\,\\text{A}\\end{align*}\\]\\[\\begin{align*} t & = 20\\,\\text{min} \\\\[6pt]\r\n& = 20\\times60\\,\\text{s}\\\\[6pt]\r\n& = 1200\\,\\text{s}\\end{align*}\\]\r\nSo:\r\n\\[\\begin{align*} q & = I\\times t \\\\[6pt]\r\n& = 0.300\\times1200\\\\[6pt]\r\n& = 360\\,\\text{C}\\end{align*}\\]\r\n\r\nVoltage \\(V\\)\r\nIf electric charges are moving in a circuit, they must be carrying energy. A battery or generator gives electrical potential energy to these charges so they can move around the circuit and do useful work (for example, lighting a globe or turning a motor). As the charges pass through components with resistance, they give up (transfer) some of this energy.\r\nThe voltage of a battery tells us how much energy is given to each coulomb of charge inside the battery. In other words, voltage is the energy per unit of charge (where the unit of charge is a coulomb \\(\\text{C}\\)), measured in volts \\(\\text{V}\\). One volt means that \\(1\\) joule (\\(\\text{J}\\)) of energy is transferred to each coulomb of charge. For example, a \\(6\\,\\text{V}\\) battery transfers \\(6\\,\\text{J}\\) of energy to each coulomb of charge. Voltage is also called electromotive force (EMF).\r\nWhen charges move through any component in a circuit and transfer energy to it, the energy transferred \\(E\\) per coulomb of charge \\(q\\) is called the potential difference (PD) across that component. Potential difference is also measured in volts, and is given by:\r\n\\[\\text{voltage} = \\frac{\\text{energy}}{\\text{charge}} \\quad \\text{or} \\quad V = \\frac{E}{q}\\]\r\nwhere \\(V\\) is in volts \\(\\text{V}\\), \\(E\\) is in joules \\(\\text{J}\\) and \\(q\\) is in coulombs \\(\\text{C}\\).\r\nIn these introductory calculations, you may see all three terms used to refer to voltage \u2013 the distinctions between them matter more in advanced circuit analysis.\r\nSince \\(E = V \\times q\\) and \\(q = I \\times t\\), we can combine them to obtain the equation:\r\n\\[E = V \\times I \\times t\\]\r\n\r\nExample \u2013 calculating energy\r\nThe potential difference across a torch globe is found to be \\(\\mathbf{2.7}\\,\\textbf{V}\\). The current flowing through it is \\(\\mathbf{0.2}\\,\\textbf{A}\\).\r\n\r\n\tDetermine how much charge flows through the torch globe in \\(\\mathbf{1}\\) minute.\r\n\r\n\tThe relevant equation is \\(q=I\\times t\\).\r\n\t\\[\\begin{align*} q & = I \\times t \\\\[6pt]\r\n\t& = 0.2 \\times 60\\text{s}\\\\[6pt]\r\n\t& = 12\\text{C}\\end{align*}\\]\r\n\tIn \\(1\\) minute, \\(12\\,\\text{C}\\) of charge flows through the globe.\r\n\r\n\tCalculate the amount of energy lost by this charge.\r\n\r\n\tThe potential difference of \\(2.7\\,\\text{V}\\) tells us that each coulomb of charge loses \\(2.7\\,\\text{J}\\) of electrical potential energy as it passes through the globe. This energy appears mainly as heat and light. We can calculate the energy loss using \\(V=\\dfrac{E}{q}\\) by first rearranging it to make \\(E\\) the subject.\r\n\t\\[\\begin{align*} E & = q\\times V \\\\[6pt]\r\n\t& = 12\\times2.7\\\\[6pt]\r\n\t& = 32.4\\,\\text{J}\\end{align*}\\]\r\n\tWhen \\(12\\,\\text{C}\\) is flowing through the globe, \\(32.4\\,\\text{J}\\) of energy is lost.\r\n\r\nPower \\(P\\)\r\nPower \\(P\\) is the rate of energy transfer.\r\n\\[\\text{power} = \\frac{\\text{energy}}{\\text{time}} \\quad \\text{or} \\quad P = \\frac{E}{t} = \\frac{V \\times I \\times t}{t} = V \\times I\\]\r\nwhere \\(P\\) is measured in joules per second \\(\\text{J/s}\\) or watts \\(\\text{W}\\), where \\(1\\,\\text{W}=1\\,\\text{J/s}\\).\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 measuring electricity\r\n\r\n\tCalculate the amount of charge that would flow through a:\r\n\t\t\r\n\t\t\t\\(5\\,\\text{mA}\\) pocket calculator in \\(10\\,\\text{min}\\).\r\n\t\t\t\\(200\\,\\text{A}\\) car starter motor in \\(5\\,\\text{s}\\).\r\n\t\t\t\\(400\\,\\text{mA}\\) light bulb in \\(1\\,\\text{h}\\).\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\t\\(1.7\\times10^{26}\\) electrons pass a point in a wire each second. Calculate the electric current that this represents.\r\n\r\n\r\n\r\n\r\n\tA charge of \\(5\\,\\text{C}\\) flows from a battery through an electric water heater and delivers \\(100\\,\\text{J}\\) of heat to the water. Calculate the voltage of the battery.\r\n\r\n\r\n\r\n\r\n\tA portable speaker is powered by a \\(9\\,\\text{V}\\) battery. Calculate the amount of energy transferred from each coulomb of charge as it passes through the circuit that drives the speaker.\r\n\r\n\r\n\r\n\r\n\tA car motor is powered by a \\(12\\,\\text{V}\\) battery. Determine the amount of charge that must have flowed through the circuit for \\(2\\,\\text{kJ}\\) of energy to be delivered to the starter motor.\r\n\r\n\r\n\r\n\r\n\tDetermine the power, in watts, used by a:\r\n\t\t\r\n\t\t\t\\(3\\,\\text{V}\\) torch bulb drawing \\(0.2\\,\\text{A}\\)\r\n\t\t\tstarter motor which takes \\(200\\,\\text{A}\\) from a \\(12\\,\\text{V}\\) battery\r\n\t\t\tmains-powered (\\(240\\,\\text{V}\\)) toaster rated at \\(3\\,\\text{A}\\).\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tCalculate the amount of current used by a:\r\n\t\t\r\n\t\t\t\\(60\\,\\text{W}\\), \\(240\\,\\text{V}\\) light bulb, in milliamps (\\(\\text{mA}\\))\r\n\t\t\t\\(1200\\,\\text{W}\\) mains-powered (\\(240\\,\\text{V}\\)) heater\r\n\t\t\t\\(90\\,\\text{W}\\) car (\\(12\\,\\text{V}\\)) windscreen wiper motor.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tCalculate the voltage of a:\r\n\t\t\r\n\t\t\t\\(100\\,\\text{W}\\) spotlight which draws \\(4\\,\\text{A}\\)\r\n\t\t\t\\(200\\,\\text{mW}\\) radio operating with a current of \\(23\\,\\text{mA}\\)\r\n\t\t\t\\(7500\\,\\text{W}\\) industrial motor using \\(18\\,\\text{A}\\).\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n","n":0.031},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Electricity","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Physics","i":0,"n":1}]]},{"i":10,"$":[{"v":"Centroids","n":1},{"v":"Whenever a force is distributed across a surface, it acts through a single point called the centroid. Finding the centroid tells us where to apply the equivalent resultant force, which we need to analyse structural behaviour. Use this resource to learn how to locate the centroid for simple and composite areas, and explore how Wurundjeri Woi-wurrung people applied an understanding of the underlying physics to make a non-returning boomerang.\r\n\r\nThe centroid of an area is its geometric centre \u2013 the point at which the area would balance perfectly if supported there.\r\n\r\nIt is found using the first moment of area, the sum (or integral) of each small area element multiplied by its distance from a reference axis. If you have a larger area element far from the axis, it contributes more to this than a smaller element close to the axis. This is why the centroid shifts toward whichever part of the shape has more area.\r\n\r\nSimple areas\r\nFor a simple area \\(A\\) bounded by a curve, we use integration to find the centroid.\r\n\r\n\\[\\bar{x} = \\frac{\\int_{A} \\tilde{x}\\,dA}{\\int_{A}\\,dA}, \\qquad \\bar{y} = \\frac{\\int_{A} \\tilde{y}\\,dA}{\\int_{A}\\,dA},\\]\r\nwhere:\r\n\r\n\t\\(\\bar{x}\\) and \\(\\bar{y}\\) are the centroidal \\(x\\) and \\(y\\) coordinates\r\n\t\\(\\int_{A}\\tilde{x}\\,dA\\) is the first moment of area about the \\(x\\) axis\r\n\t\\(\\int_{A}\\tilde{y}\\,dA\\) is the first moment of area about the \\(y\\) axis\r\n\t\\(\\int_{A}\\,dA\\) is the area.\r\n\r\n\r\nTo find the centroid using integration, we divide the area into infinitely thin strips and find the centroid of each strip. The centroidal coordinates of each infinitesimal strip of area \\(dA\\) are denoted by \\(\\tilde{x}\\) and \\(\\tilde{y}\\) (with tildes).\r\nThe centroid of a horizontal strip (width \\(x\\) and thickness \\(dy\\)) is at \\(\\tilde{x}=\\dfrac{x}{2}\\) (halfway across) and \\(\\tilde{y}=y\\) (its own height). We integrate these strip centroids, weighted by their areas, to find the overall centroid \\(\\bar{x}\\) and \\(\\bar{y}\\).\r\n\r\nThe following table helps you understand how to find the area of different shapes, and where the centroid is located.\r\n\r\n\t\r\n\t\t\r\n\t\t\tShape\r\n\t\t\tArea \\(A\\)\r\n\t\t\tPosition of centroid\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tCircle\r\n\t\t\t\\(A=\\dfrac{\\pi d^{2}}{4}\\)\r\n\t\t\twhere \\(d\\) is the diameter of the circle\r\n\t\t\tAt the centre\r\n\t\t\r\n\t\t\r\n\t\t\tSquare\r\n\t\t\t\\(A=l^{2}\\)\r\n\t\t\twhere \\(l\\) is the length of one side\r\n\t\t\tAt the intersection of the diagonals\r\n\t\t\r\n\t\t\r\n\t\t\tRectangle\r\n\t\t\t\\(A=bh\\)\r\n\t\t\twhere \\(h\\) is the length and \\(b\\) is the width\r\n\t\t\tAt the intersection of the diagonals\r\n\t\t\r\n\t\t\r\n\t\t\tTriangle\r\n\t\t\t\\(A=\\dfrac{bh}{2}\\)\r\n\t\t\twhere \\(b\\) is the length of the base and \\(h\\) is the height\r\n\t\t\tAt the intersection of the medians (\\(\\tfrac{1}{3}\\) of height)\r\n\t\t\r\n\t\r\n\r\nIf you need a refresher on integration, visit this page.\r\n\r\nExample \u2013 finding the centroid of a simple area\r\nCalculate the centroid \\((\\bar{x}, \\bar{y})\\) of the shaded area.\r\n\r\nTo find the coordinates of the centroid, we need the area. We will choose a horizontal strip of area \\(dA = x\\,dy\\) because the integration is more straightforward. A vertical strip would use \\(dA = (1 - y^{\\frac{1}{3}})\\,dx\\).\r\n\r\nThe curve is \\(y = x^{3}\\), so \\(x = y^{\\frac{1}{3}}\\). We integrate from \\(y=0\\) to \\(y=1\\). Therefore:\r\n\\[\\begin{align*} \\int_{A}\\,dA & = \\int_{0}^{1}x\\,dy \\\\[6pt]\r\n& = \\int_{0}^{1}y^{\\frac{1}{3}}\\,dy \\\\[6pt]\r\n& = \\left[\\frac{3}{4} y^{\\frac{4}{3}}\\right]_{0}^{1} \\\\[6pt]\r\n& = \\frac{3}{4} \\end{align*}\\]\r\nTo find the \\(\\bar{x}\\)-coordinate, we let \\(\\bar{x}\\) equal to the centroid of the strip of area.\r\n\\[\\begin{align*} \\bar{x} & = \\frac{x}{2} \\\\[6pt]\r\n& = \\frac{y^{\\frac{1}{3}}}{2} \\end{align*}\\]\r\nWe then substitute to find \\(\\bar{x}\\):\r\n\\[\\begin{align*} \\int_{A} \\tilde{x}\\,dA = & \\int_{0}^{1}\\left(\\frac{y^{\\frac{1}{3}}}{2}\\right) x\\,dy \\\\[6pt]\r\n& = \\frac{1}{2}\\int_{0}^{1}y^{\\frac{2}{3}}\\,dy \\\\[6pt]\r\n& = \\frac{1}{2}\\left[\\frac{3}{5}y^{\\frac{5}{3}}\\right]_{0}^{1} \\\\[6pt]\r\n& = \\frac{3}{10} \\end{align*}\\]\r\nDividing this by the area:\r\n\\[\\begin{align*} \\bar{x} & = \\frac{\\frac{3}{10}}{\\frac{3}{4}} \\\\[6pt]\r\n& = \\frac{3}{10}\\times\\frac{4}{3} \\\\[6pt]\r\n& = \\frac{4}{10} \\\\[6pt]\r\n& = 0.4\\,\\text{m} \\end{align*}\\]\r\nWe repeat this for the \\(\\bar{y}\\)-coordinate, where \\(\\tilde{y}=y\\).\r\n\\[\\begin{align*} \\int_{A} \\tilde{y}\\,dA & = \\int_{0}^{1}y\\cdot x\\,dy \\\\[6pt]\r\n& = \\int_{0}^{1}y\\cdot y^{\\frac{1}{3}}\\,dy \\\\[6pt]\r\n& = \\int_{0}^{1}y^{\\frac{4}{3}}\\,dy \\\\[6pt]\r\n& = \\left[\\frac{3}{7} y^{\\frac{7}{3}}\\right]_{0}^{1} \\\\[6pt]\r\n& = \\frac{3}{7} \\end{align*}\\]\r\nDividing this by the area:\r\n\\[\\begin{align*} \\bar{y} & = \\frac{\\frac{3}{7}}{\\frac{3}{4}} \\\\[6pt]\r\n& = \\frac{3}{7}\\times\\frac{4}{3} \\\\[6pt]\r\n& = \\frac{4}{7} \\\\[6pt]\r\n& \\approx 0.57\\,\\text{m} \\end{align*}\\]\r\nTherefore, the coordinates of the centroid are approximately \\((\\bar{x},\\bar{y}) = (0.4,0.57)\\,\\text{m}\\).\r\n\r\nComposite areas\r\nFor composite areas made up of standard shapes (rectangles, triangles, circles) or elements, we can use a simpler weighted-sum method instead, since the centroid of each standard shape is already known.\r\n\\[\\bar{x} = \\frac{\\sum (A x)}{\\sum A}, \\qquad \\bar{y} = \\frac{\\sum (A y)}{\\sum A},\\]\r\nwhere:\r\n\r\n\t\\(A\\) is the area of each element\r\n\t\\(y\\) is the vertical distance of each element's centroid from the \\(x\\)-axis\r\n\t\\(x\\) is the horizontal distance of each element's centroid from the \\(y\\)-axis.\r\n\r\n\r\n\r\nExample \u2013 finding the centroid of a composite area\r\nFind the centroid of the composite area shown with respect to the \\(x\\) and \\(y\\) axes.\r\n\r\nThe L-shaped area can be divided into three rectangles: \\(A_{1}\\) (top-left), \\(A_{2}\\) (middle-left vertical strip) and \\(A_{3}\\) (bottom horizontal strip). For each element, we identify its area and locate its centroid. The centroid of a rectangle always lies at the intersection of its diagonals.\r\n\r\n\r\nReading centroid positions and dimensions from the geometry:\r\n\r\nTo find \\(A\\), \\(Ax\\) and \\(Ay\\), we sum each relevant column.\r\n\\[\\begin{align*} \\sum A & = 600+300+600 \\\\[6pt]\r\n& = 1500\\,\\text{mm}^{2} \\end{align*}\\]\\[\\begin{align*} \\sum (Ax) & = 9000+1500+18000 \\\\[6pt]\r\n& = 28500\\,\\text{mm}^{3} \\end{align*}\\]\\[\\begin{align*} \\sum (Ay) & = 30000+7500+3000 \\\\[6pt]\r\n& = 40500\\,\\text{mm}^{3} \\end{align*}\\]\r\nWe can now apply the composite centroid formula. For \\(\\bar{x}\\):\r\n\\[\\begin{align*} \\bar{x} & = \\frac{\\sum (Ax)}{\\sum A} \\\\[6pt]\r\n& = \\frac{28500}{1500} \\\\[6pt]\r\n& = 19\\,\\text{mm} \\end{align*}\\]\r\nAnd for \\(\\bar{y}\\):\r\n\\[\\begin{align*} \\bar{y} & = \\frac{\\sum (Ay)}{\\sum A} \\\\[6pt]\r\n& = \\frac{40500}{1500} \\\\[6pt]\r\n& = 27\\,\\text{mm} \\end{align*}\\]\r\nTherefore, the centroid of the composite area is at \\((\\bar{x},\\bar{y})=(19,27)\\,\\text{mm}\\).\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding centroids \r\n\r\n\tFind the centroid \\((\\bar{x},\\bar{y})\\) of a rectangle with base \\(b=80\\,\\text{mm}\\) and height \\(h=50\\,\\text{mm}\\), with the origin at the bottom-left corner.\r\n\r\n\r\n\r\n\tFind the centroid \\((\\bar{x},\\bar{y})\\) of an isosceles triangle with base \\(b=60\\,\\text{mm}\\) and height \\(h=90\\,\\text{mm}\\), with the origin at the bottom-left corner.\r\n\r\n\r\n\r\n\tFind the centroid of a T-section with respect to the \\(x\\)-axis. Take the origin at the bottom of the web. The top flange is \\(100\\,\\text{mm}\\) wide and \\(20\\,\\text{mm}\\) thick. The web is \\(20\\,\\text{mm}\\) wide and \\(80\\,\\text{mm}\\) tall.\r\n\t\r\n\r\n\r\n\r\n\tFind the centroid of an L-section with the origin at the bottom-left. The vertical leg is \\(20\\,\\text{mm}\\) wide and \\(120\\,\\text{mm}\\) tall. The horizontal leg is \\(80\\,\\text{mm}\\) wide and \\(20\\,\\text{mm}\\) thick (at the base).\r\n\r\n\r\n\r\n\tA \\(200\\,\\text{mm}\\times150\\,\\text{mm}\\) rectangle has a circular hole of radius \\(30\\,\\text{mm}\\) centred at \\((100,75)\\,\\text{mm}\\). Find the centroid of the remaining area.\r\n\r\n\r\n\r\n\tFind the centroid of the shaded area bounded by \\(y=x^{2}\\) between \\(x=0\\) and \\(x=2\\,\\text{m}\\) and the \\(x\\)-axis.\r\n\r\n\r\n\r\n","n":0.032},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Maths","i":2,"n":1},{"v":"Force","i":1,"n":1},{"v":"Physics","i":0,"n":1}]]},{"i":11,"$":[{"v":"Static friction","n":0.707},{"v":"Without static friction, nothing would stay put. It is the force that keeps objects at rest when a force is applied \u2013 stopping a parked car from rolling down a hill, keeping your feet from sliding when you walk, and holding a stack of books in place on a tilted surface. This resource explores how static friction works, what determines its maximum value, and how it applies to objects on slopes.\r\n\r\nRemember that friction is the force that resists motion when two surfaces come into contact. Static friction is specifically the frictional force acting between two contacting surfaces when they are at rest. It arises from the tiny interactions between the surfaces and is crucial for preventing motion until a certain threshold force is applied.\r\n\r\nStatic frictional force\r\nWhen a block sitting on a table is at rest, static friction is \\(0\\). When a force is applied to the block, static friction imparts an equal and opposite force to keep the block from moving. If more force is applied and this exceeds a value called the maximum static friction force, the block will move \u2013 that is, it is the force required to overcome static friction.\r\nThe static friction force can be calculated using:\r\n\\[F_{\\text{S,max}}=\\mu_{\\text{S}}N\\]\r\nwhere:\r\n\r\n\t\\(F_{\\text{S,max}}\\) is the maximum static friction force in Newtons \\(\\text{N}\\)\r\n\t\\(\\mu_{\\text{S}}\\) is the coefficient of static friction, which is dimensionless (it has no units)\r\n\t\\(N\\) is the normal force in Newtons \\(\\text{N}\\).\r\n\r\nStatic friction is different from kinetic friction, which is the frictional force acting between two contacting surfaces when one or more of them are in motion.\r\n\r\nCoefficient of static friction\r\nThe coefficient of static friction \\(\\mu_{S}\\) describes how 'grippy' two surfaces are with each other. It is determined experimentally and depends on the materials in contact \u2013 for example, rubber on concrete has a high coefficient (between \\(0.6\\) and \\(0.8\\)), meaning a large force is needed to overcome friction, while ice on steel has a very low coefficient (around \\(0.03\\)), meaning very little force is needed.\r\n\r\nA higher \\(\\mu_{S}\\) means the surfaces resist sliding more strongly.\r\nA lower \\(\\mu_{S}\\) means the surfaces slide past each other more easily.\r\n\r\nThe coefficient has no units because it is the ratio of the friction force to the normal force. Both are measured in Newtons, so the units cancel.\r\n\r\nExample \u2013 calculating forces on slopes with static friction\r\nThe crate shown has a mass of \\(\\mathbf{50}\\,\\textbf{kg}\\) and the coefficient of static friction between the crate and the plane is \\(\\mathbf{\\mu_{s}= 0.25}\\). Take \\(\\mathbf{g=9.81}\\,\\textbf{m s}\\mathbf{^{-2}}\\).\r\n\r\n\r\n  Calculate the minimum force \\(\\mathbf{P}\\) required to stop the crate from sliding down the plane.\r\n \r\nFor these problems, we always start by drawing or annotating a free-body diagram which shows only the external forces acting on the one object: the crate.\r\nFor the impending motion of sliding down the plane, friction \\(F_{F}\\) acts up the plane. The other forces on the crate are: weight \\(mg\\), the normal force \\(F_{N}\\), and the applied force \\(P\\).\r\n\r\nThe maximum static friction is \\(F_{S}=\\mu_{S}N\\). To find \\(N\\), we resolve the forces parallel and perpendicular to the plane, using the \\(30^{\\circ}\\) angle between the plane and the horizontal. This gives us simultaneous equations to solve for \\(N\\).\r\nParallel to the plane:\r\n\\[P\\cos30^{\\circ}+F_{S}-mg\\sin30^{\\circ}=0\\]\r\nPerpendicular to the plane:\r\n\\[N=P\\sin30^{\\circ}-mg\\cos30^{\\circ}\\]\r\nTo solve for \\(N\\), we can substitute \\(F_{S}=\\mu_{S}N\\) with \\(\\mu_{S}=0.25\\), \\(m=50\\,\\text{kg}\\) and \\(g=9.81\\,\\text{m s}^{-2}\\). The perpendicular equation becomes:\r\n\\[\\begin{align*} N & = P\\sin30^{\\circ}+mg\\cos30^{\\circ} \\\\[6pt]\r\n& = P\\sin30^{\\circ}+\\left(50\\times9.81\\times\\cos30^{\\circ}\\right)\\\\[6pt]\r\n& = P\\sin30^{\\circ}+490.5\\cos30^{\\circ} \\end{align*}\\]\r\nThe parallel equation becomes:\r\n\\[\\begin{align*} 0 & = P\\cos30^{\\circ}+\\mu_{S}N-mg\\sin30^{\\circ} \\\\[6pt]\r\n& = P\\cos30^{\\circ}+0.25N-\\left(50\\times9.81\\times\\sin30^{\\circ}\\right) \\\\[6pt]\r\n& = P\\cos30^{\\circ}+0.25N-490.5\\sin30^{\\circ} \\end{align*}\\]\r\nSubstituting for \\(N\\):\r\n\\[\\begin{align*}\r\n0 &= P\\cos30^{\\circ}+0.25\\left(0.5P+490.5\\cos30^{\\circ}\\right)-490.5\\sin30^{\\circ} \\\\[6pt]\r\n& = P\\cos30^{\\circ}+0.125P+122.625\\cos30^{\\circ}-245.25 \\\\[6pt]\r\n& = P\\left(\\cos30^{\\circ}+0.125\\right)+122.625\\cos30^{\\circ}-245.25 \\\\[6pt]\r\nP\\left(\\cos30^{\\circ}+0.125\\right) & = 245.25-122.625\\cos30^{\\circ} \\\\[6pt]\r\nP &= \\frac{245.25-122.625\\cos30^{\\circ}}{\\cos30^{\\circ}+0.125} \\\\[6pt]\r\nP &\\approx 140\\,\\text{N}\r\n\\end{align*}\\]\r\nThe minimum horizontal force required to prevent the crate from sliding down the plane is approximately \\(140\\,\\text{N}\\).\r\n\r\n\r\n  Calculate the minimum force \\(\\mathbf{P}\\) required to push the crate up the plane.\r\n\r\nAgain, starting with the free-body diagram. For the impending motion of sliding up the plane, friction \\(F_{F}\\) acts down the plane. The forces are \\(mg\\), \\(F_{N}\\), \\(P\\) and static friction \\(F_{S}\\), with \\(F_{S}=\\mu_{S} N\\).\r\n\r\nParallel to the plane:\r\n\\[P\\cos30^{\\circ}-F_{S}-mg\\sin30^{\\circ}=0\\]\r\nPerpendicular to the plane:\r\n\\[N=P\\sin30^{\\circ}-mg\\cos30^{\\circ}\\]\r\nSolving for \\(N\\), we substitute \\(F_{S}=\\mu_{S}N\\) with \\(\\mu_{S}=0.25\\), \\(m=50\\,\\text{kg}\\) and \\(g=9.81\\,\\text{m s}^{-2}\\). The perpendicular equation becomes:\r\n\\[N=P\\sin30^{\\circ}+mg\\cos30^{\\circ}=0.5P+424.18\\]\r\nThe parallel equation becomes:\r\n\\[\\begin{align*} 0 & = P\\cos30^{\\circ}-\\mu_{S}N-mg\\sin30^{\\circ} \\\\[6pt]\r\n& = 0.87P-0.25N-245.25 \\end{align*}\\]\r\nSubstituting for \\(N\\):\r\n\\[\\begin{align*}\r\n0 & = P\\cos30^{\\circ}-0.25\\left(0.5P+490.5\\cos30^{\\circ}\\right)-490.5\\sin30^{\\circ} \\\\[6pt]\r\n& = P\\cos30^{\\circ}-0.125P-122.625\\cos30^{\\circ}-245.25 \\\\[6pt]\r\n& = P\\left(\\cos30^{\\circ}-0.125\\right)-122.625\\cos30^{\\circ}-245.25 \\\\[6pt]\r\nP\\left(\\cos30^{\\circ}-0.125\\right) & = 122.625\\cos30^{\\circ}+245.25 \\\\[6pt]\r\nP & = \\frac{122.625\\cos30^{\\circ}+245.25}{\\cos30^{\\circ}-0.125} \\\\[6pt]\r\nP & \\approx 474\\,\\text{N} \\end{align*}\\]\r\nThe minimum horizontal force required to push the crate up the plane is approximately \\(474\\,\\text{N}\\).\r\n\r\nExercise \u2013 calculating forces on slopes with static friction\r\n\r\n\tA \\(30\\,\\text{kg}\\) cardboard box is placed on a warehouse loading ramp inclined at \\(20^{\\circ}\\) to the horizontal. The coefficient of static friction between the box and the ramp is \\(0.45\\). Will the box slide or remain stationary?\r\n\r\n\r\n\r\n\r\n\tA \\(1200\\,\\text{kg}\\) car is parked on a hill inclined at \\(15^{\\circ}\\). The coefficient of static friction between the tyres and dry asphalt is \\(0.70\\).\r\n\t\r\n\t\tCalculate the maximum static friction force acting on the car.\r\n\t\tDetermine whether the car will roll.\r\n\t\r\n\r\n\r\n\r\n\r\n\tA \\(75\\,\\text{kg}\\) snowboarder stands stationary on a \\(25^{\\circ}\\) slope. The coefficient of static friction between the board and snow is \\(0.10\\).\r\n\t\r\n\t\tCalculate the normal force acting on the snowboarder.\r\n\t\tCalculate the maximum static friction.\r\n\t\tDetermine whether the snowboarder will begin to slide.\r\n\t\r\n\r\n\r\n\r\n\r\n\tA worker pushes a \\(40\\,\\text{kg}\\) crate up a \\(20^{\\circ}\\) ramp with a force applied parallel to the surface. The coefficient of static friction between the crate and the ramp is \\(0.35\\). Calculate the minimum force required to just start the crate moving up the ramp.\r\n\r\n\r\n\r\n\r\n\tA \\(5\\,\\text{k}\\)g book is placed on a wooden board. The board is slowly tilted until the book just begins to slide at an angle of \\(30^{\\circ}\\) to the horizontal. What is the coefficient of static friction between the book and the board?\r\n\r\n\r\n\r\n","n":0.033},[{"v":"Physics","i":0,"n":1}]]},{"i":12,"$":[{"v":"Forces in three dimensions","n":0.5},{"v":"When a crane lifts a load, a broadcasting tower is guyed by cables, or a lighting rig is suspended above a stage, the support forces act in three dimensions \u2013 not just up, down, left or right. Analysing these forces requires vector methods, which let us resolve, combine and project forces in any direction. This resource introduces you to forces in 3D, and steps you through a worked example involving a 3D beam and cable system.\r\n\r\nIn two dimensions, a force can be resolved into horizontal and vertical components. In three dimensions, we resolve forces into three components along the \\(x\\), \\(y\\) and \\(z\\) axes, written as a vector:\r\n\r\n\\[\\vec{F} = F_x\\hat{i} + F_y\\hat{j} + F_z\\hat{k}\\]\r\nwhere \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\) are the unit vectors in the \\(x\\), \\(y\\) and \\(z\\) directions, respectively.\r\n\r\nIf you need a refresher on vectors and vector operations, visit this page.\r\n\r\nPosition vectors and cable vectors\r\nA position vector describes the location of a point relative to the origin. If point A is at coordinates \\((x_{A}, y_{A}, z_{A})\\), then its position vector is \\(\\vec{r}_{A} = (x_{A}, y_{A}, z_{A})\\).\r\nTo find the vector along a cable from point A to point B, we subtract the position vector of the start point from the end point:\r\n\r\n\\[\\vec{AB} = \\vec{r}_{B} - \\vec{r}_{A}\\]\r\n\r\n\r\nUnit vectors and force vectors\r\nA unit vector has magnitude \\(1\\) and points in the same direction as a given vector. It is found by dividing the vector by its magnitude:\r\n\r\n\\[\\hat{u} = \\frac{\\vec{v}}{|\\vec{v}|}\\quad\\text{where}\\quad|\\vec{v}| = \\sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}}\\]\r\n\r\nIf a cable has a known tension \\(T\\) and direction vector \\(\\vec{AB}\\), the force vector in the cable is:\r\n\r\n\\[\\vec{F} = T\\,\\hat{u}_{AB} = \\frac{T}{|\\vec{AB}|}\\vec{AB}\\]\r\n\r\n\r\nScalar projection\r\nThe scalar projection of a force \\(\\vec{F}\\) along a direction \\(\\hat{u}\\) gives the component of the force in that direction. It is found using the dot product:\r\n\r\n\\[F_{u} = \\vec{F}\\cdot\\hat{u}\\]\r\n\r\nA positive result means the force component acts in the same direction as \\(\\hat{u}\\). A negative result means it acts in the opposite direction.\r\n\r\nPerpendicular component\r\nOnce the parallel (scalar) projection is known, the perpendicular vector component of \\(\\vec{F}\\) to a direction \\(\\hat{u}\\) is:\r\n\r\n\\[\\vec{F}_\\perp = \\vec{F} - F_{u}\\,\\hat{u}\\]\r\n\r\n\r\nAngle between two vectors\r\nThe angle \\(\\theta\\) between two vectors \\(\\vec{a}\\) and \\(\\vec{b}\\) is found using the dot product:\r\n\r\n\\[\\theta = \\cos^{-1}\\left(\\frac{\\vec{a}\\cdot\\vec{b}}{|\\vec{a}|\\,|\\vec{b}|}\\right)\\]\r\n\r\n\r\nExercise \u2013 analysing forces in three dimensions\r\nA beam is held in place by two cables attached at different points.\r\n\r\nDetermine:\r\n\r\n\tthe vector forces in the cables \\(\\mathbf{\\vec{F}_{AB}}\\) and \\(\\mathbf{\\vec{F}_{AC}}\\)\r\n\r\nTo find the forces in each cable, we must first find the position vectors. From the geometry and coordinates shown in the diagram, we can write them in component form:\r\n\\[\\vec{r}_{A} = (0, 3, 0)\\]\\[\\vec{r}_{B} = (-2, 0, 1)\\]\\[\\vec{r}_{C} = (2, 0, 3)\\]\r\nThe position vectors of the cables \\(\\vec{AB}\\) and \\(\\vec{AC}\\) are found by subtracting the position vectors:\r\n\\[\\begin{align*} \\vec{AB} & = \\vec{r}_B - \\vec{r}_A \\\\[6pt]\r\n& = (-2, 0, 1) - (0, 3, 0) \\\\[6pt]\r\n& = (-2, -3, 1) \\end{align*}\\]\\[\\begin{align*} \\vec{AC} & = \\vec{r}_C - \\vec{r}_A \\\\[6pt]\r\n& = (2, 0, 3) - (0, 3, 0) \\\\[6pt]\r\n& = (2, -3, 3) \\end{align*}\\]\r\nUnit vectors \\(\\hat{u}_{AB}\\) and \\(\\hat{u}_{AC}\\) are found by dividing each vector by its magnitude:\r\n\\[\\begin{align*} \\hat{u}_{AB} & = \\frac{1}{\\sqrt{(-2)^2 + (-3)^2 + 1^2}}(-2,-3,1) \\\\[6pt]\r\n& = \\frac{1}{\\sqrt{14}}(-2,-3,1) \\end{align*}\\]\\[\\begin{align*} \\hat{u}_{AC} & = \\frac{1}{\\sqrt{2^2 + (-3)^2 + 3^2}}(2,-3,3) \\\\[6pt]\r\n& = \\frac{1}{\\sqrt{22}}(2,-3,3) \\end{align*}\\]\r\nGiven the cable tensions are \\(500\\,\\text{N}\\), the force vectors are found by multiplying the unit vectors by \\(500\\,\\text{N}\\):\r\n\\[\\begin{align*} \\vec{F}_{AB} & = |\\vec{F}_{AB}|\\,\\hat{u}_{AB} \\\\[6pt]\r\n& = 500 \\frac{1}{\\sqrt{14}}(-2,-3,1) \\\\[6pt]\r\n& \\approx (-267,-401,134)\\,\\text{N} \\end{align*}\\]\\[\\begin{align*} \\vec{F}_{AC} & = |\\vec{F}_{AC}|\\,\\hat{u}_{AC} \\\\[6pt]\r\n& = 500 \\frac{1}{\\sqrt{22}}(2,-3,3) \\\\[6pt]\r\n& \\approx (213,-320,320)\\,\\text{N} \\end{align*}\\]\r\nTherefore, the vector forces in the cables are \\(\\vec{F}_{AB}=(-267,-401,134)\\,\\text{N}\\) and \\(\\vec{F}_{AC}=(213,-320,320)\\,\\text{N}\\).\r\n\r\n\r\n\tthe resultant vector force \\(\\mathbf{\\vec{F}}\\) of \\(\\mathbf{\\vec{F}_{AB}}\\) and \\(\\mathbf{\\vec{F}_{AC}}\\)\r\n\r\nTo find the resultant vector force, we sum \\(\\vec{F}_{AB}\\) and \\(\\vec{F}_{AC}\\) to get the overall (resultant) force:\r\n\\[\\begin{align*} \\vec{F} & = \\vec{F}_{AB} + \\vec{F}_{AC} \\\\[6pt]\r\n& = (-267,-401,134) + (213,-320,320) \\\\[6pt]\r\n& = (-54,-721,454)\\,\\text{N} \\end{align*}\\]\r\nTherefore, the resultant vector force \\(\\vec{F}=(-54,-721,454)\\,\\text{N}\\).\r\n\r\n\r\n\tthe scalar projection of \\(\\mathbf{\\vec{F}}\\) along the beam AO\r\n\r\nThe unit vector \\(\\hat{u}_{AO}\\) along the beam from A to O is found using the position vector of O relative to A:\r\n\\[\\begin{align*} \\hat{u}_{AO} & = \\frac{\\vec{AO}}{|\\vec{AO}|} \\\\[6pt]\r\n& = \\frac{(0,-3,0)}{\\sqrt{0^2 + (-3)^2 + 0^2}} \\\\[6pt]\r\n& = (0,-1,0) \\end{align*}\\]\r\nThe scalar projection of \\(\\vec{F}\\) along AO is:\r\n\\[\\begin{align*} F_{AO} & = \\vec{F} \\cdot \\hat{u}_{AO} \\\\[6pt]\r\n& = (-54,-721,454)\\cdot(0,-1,0) \\\\[6pt]\r\n& = 721\\,\\text{N} \\end{align*}\\]\r\nTherefore, the scalar projection of \\(\\vec{F}\\) along the beam AO is \\(F_{AO}=721\\,\\text{N}\\). Since the result is positive, the force component acts along AO in the direction of \\(\\hat{u}_{AO}\\) (towards the origin).\r\n\r\n\r\n\tthe perpendicular vector component of \\(\\mathbf{\\vec{F}}\\) to the beam AO\r\n\r\nTo find the perpendicular vector component, we first express the parallel component in vector form:\r\n\\[\\begin{align*} \\vec{F}_{AO} & = F_{AO}\\,\\hat{u}_{AO} \\\\[6pt]\r\n& = 721(0,-1,0) \\\\[6pt]\r\n& = (0,-721,0)\\,\\text{N} \\end{align*}\\]\r\nThe perpendicular vector component is found by subtracting the parallel vector component from the resultant force vector:\r\n\\[\\begin{align*} \\vec{F}_{\\perp} & = \\vec{F} - \\vec{F}_{AO} \\\\[6pt]\r\n& = (-54,-721,454) - (0,-721,0) \\\\[6pt]\r\n& = (-54,0,454)\\,\\text{N} \\end{align*}\\]\r\nTherefore, the perpendicular vector component of \\(\\vec{F}\\) to the beam AO is \\(\\vec{F}_{\\perp}=(-54,0,454)\\,\\text{N}\\).\r\n\r\n\r\n\tthe angle between the cables AB and AC.\r\n\r\nThe angle \\(\\theta\\) between \\(\\vec{AB}\\) and \\(\\vec{AC}\\) is found using the dot product formula:\r\n\\[\\begin{align*} \\vec{AB} \\cdot \\vec{AC} & = |\\vec{AB}|\\,|\\vec{AC}| \\cos\\theta \\\\[6pt]\r\n\\theta & = \\cos^{-1}\\left(\\frac{\\vec{AB} \\cdot \\vec{AC}}{|\\vec{AB}|\\,|\\vec{AC}|}\\right) \\\\[6pt]\r\n& = \\cos^{-1}\\left(\\frac{(-2,-3,1)\\cdot(2,-3,3)}{\\sqrt{14}\\,\\sqrt{22}}\\right) \\\\[6pt]\r\n& = \\cos^{-1}\\left(\\frac{-4 + 9 + 3}{\\sqrt{308}}\\right) \\\\[6pt]\r\n& = \\cos^{-1}\\left(\\frac{8}{\\sqrt{308}}\\right) \\\\[6pt]\r\n& \\approx 63^{\\circ} \\end{align*}\\]\r\nTherefore, the angle between cables AB and AC is \\(\\theta=63^{\\circ}\\).\r\n\r\nExercise \u2013 analysing forces in three dimensions\r\n\r\n  A cable runs from point A at \\((0,2,0)\\,\\text{m}\\) to point B at \\((-1,0,3)\\,\\text{m}\\). Find the unit vector \\(\\hat{u}_{AB}\\) along the cable.\r\n\r\n\r\n\r\n\r\n  A cable runs from point A at \\((0,4,0)\\,\\text{m}\\) to point C at \\((-2,0,2)\\,\\text{m}\\) and carries a tension of \\(400\\,\\text{N}\\). Determine the vector force \\(\\vec{F}_{AC}\\).\r\n\r\n\r\n\r\n\r\n  Two cables exert forces \\(\\vec{F}_1=(200,-300,150)\\,\\text{N}\\) and \\(\\vec{F}_2=(-150,-200,100)\\,\\text{N}\\) on the same point. Find the resultant force vector \\(\\vec{F}\\).\r\n\r\n\r\n\r\n\r\n  A resultant force \\(\\vec{F}=(-54,-721,454)\\,\\text{N}\\) acts at point A, which is \\(4\\,\\text{m}\\) along the \\(y\\)-axis from the origin O. Find the scalar projection of \\(\\vec{F}\\) along the beam AO.\r\n\r\n\r\n\r\n\r\n  Two cables leave the same attachment point. Cable AB has direction vector \\((-2,-3,1)\\) and cable AD has direction vector \\((1,-3,2)\\). Find the angle between the cables.\r\n\r\n \r\nForces in three dimensions problem image transcript\r\nA three-dimensional statics diagram showing a beam and two support cables. The coordinate axes are labelled \\(x\\) (pointing to the left), \\(y\\) (pointing to the right foreground) and \\(z\\) (pointing vertically upwards).\r\n\r\n\tA beam extends diagonally from O (the origin) to point A, which is \\(3\\,\\text{m}\\) along the positive \\(y\\)-axis.\r\n\tTwo cables extend from point A:\r\n\t\t\r\n\t\t\tCable AB connects A to point B, which is located \\(2\\,\\text{m}\\) in the negative \\(x\\)-direction and \\(1\\,\\text{m}\\) in the positive \\(z\\)-direction from the origin.\r\n\t\t\tCable AC connects A to point C, which is located \\(2\\,\\text{m}\\) in the positive \\(x\\)-direction and \\(3\\,\\text{m}\\) in the positive \\(z\\)-direction from the origin.\r\n\t\t\r\n\t\r\n\tBoth cables carry a tension of \\(500\\,\\text{N}\\), shown as red arrows pointing from A toward B and from A toward C respectively, and labelled \\(F_{AB}=500\\,\\text{N}\\) and \\(F_{AC}=500\\,\\text{N}\\).\r\n\r\n","n":0.03},[{"v":"Physics","i":0,"n":1}]]},{"i":13,"$":[{"v":"Moment of inertia","n":0.577},{"v":"The resistance of an object to rotation depends not just on how heavy it is, but on how its mass is distributed. This property is called moment of inertia. Use this resource to understand two forms of moment of inertia: mass moment of inertia (used in dynamics) and area moment of inertia (used in structural analysis). Learn how to calculate each one and how they are applied to real world tools, such as Aboriginal weapons, the kodj and leangle.\r\n\r\nMoment of inertia is a measure of how much an object resists rotational motion when a torque is applied. There are three main types:\r\n\r\n\tmass moment of inertia\r\n\tarea moment of inertia (also called second moment of area)\r\n\tpolar area moment of inertia.\r\n\r\n\r\nMass moment of inertia \\(I\\)\r\nMass moment of inertia measures how difficult it is to change an object's rotation. Just as mass resists change to linear motion (Newton's second law: \\(F=ma\\)), mass moment of inertia resists change in rotational motion. The rotational equivalent of Newton's second law is:\r\n\\[\\sum\\tau=I\\times\\alpha\\]\r\nwhere:\r\n\r\n\t\\(\\sum \\tau\\) is the net torque in Newton metres \\(\\text{N m}\\)\r\n\t\\(I\\) is the mass moment of inertia in kilograms metres squared \\(\\text{kg m}^{2}\\) \r\n\t\\(\\alpha\\) is the angular acceleration in radians per second squared \\(\\text{rad s}^{-2}\\).\r\n\r\n\r\nNotice how this parallels Newton's second law. Torque is in place of force, moment of inertia is in place of mass and angular acceleration is in place of linear acceleration.\r\n\r\nCalculating mass moment of inertia\r\nTo calculate mass moment of inertia for a given object, we need to know how its mass is distributed. For a single point mass at a distance from the axis:\r\n\r\n\\[I = m r^{2}\\]\r\nwhere:\r\n\r\n \t\\(m\\) is the mass of the body in kilograms \\(\\text{kg}\\)\r\n\t\\(r\\) is the distance from the mass to the axis of rotation in metres \\(\\text{m}\\).\r\n\r\n\r\n\r\nDifferent objects have different moments of inertia, assuming uniform composition.\r\n\r\n\tFor a disc: \\(I=\\dfrac{1}{2}mr^{2}\\)\r\n\tFor a cylinder: \\(I=\\dfrac{1}{2}mr^{2}\\)\r\n\tFor a thin hoop: \\(I=mr^{2}\\)\r\n\tFor a rod with its axis of rotation on one end: \\(I=\\dfrac{1}{3}mr^{2}\\).\r\n\r\n\r\n\r\n\r\nRadius of gyration\r\nThe radius of gyration is a convenient way to express how far from the axis a body's mass would need to be concentrated (as a thin ring) to produce the same moment of inertia. It is defined by:\r\n\r\n\\[I = m k^{2}\\]\r\nwhere:\r\n\r\n\t\\(m\\) is the mass in kilograms \\(\\text{kg}\\)\r\n\t\\(k\\) is the radius of gyration in metres \\(\\text{m}\\).\r\n\r\n\r\nFor a solid disc, \\(k=\\tfrac{r}{\\sqrt{2}}\\). It is smaller than the disc's actual radius because most of the disc's mass lies closer to the centre than the rim.\r\n\r\n\r\n\r\n\r\n\r\n\r\nArea moment of inertia\r\nThe area moment of inertia, also called the second moment of area, measures how a cross-section resists bending. It is used in beam and structural analysis. Unlike mass moment of inertia, it depends on the shape and size of a cross-section, not on its mass.\r\nThe reference axis may pass through the shape's centroid or may be offset from it \u2013 the centroidal and non-centroidal axes, respectively.\r\n\r\nCentroidal area moment of inertia\r\nWhen the reference axis passes through the centroid of a shape, the result is called the centroidal area moment of inertia, \\(I_c\\).\r\nStandard formulas for common shapes are given in the following table.\r\n\r\n\r\n  \r\n    \r\n      Shape\r\n      Centroidal area moment of inertia \\(I_{c}\\)\r\n    \r\n  \r\n  \r\n    \r\n      Circle\r\n      \\(\\dfrac{\\pi d^{4}}{64}\\)\r\n\t  where \\(d\\) is the diameter of the circle\r\n    \r\n    \r\n      Square\r\n      \\(\\dfrac{l^{4}}{12}\\)\r\n\t  where \\(l\\) is the length of one side\r\n    \r\n    \r\n      Rectangle\r\n      \\(\\dfrac{bh^{3}}{12}\\)\r\n\t  where \\(h\\) is the length and \\(b\\) is the width\r\n    \r\n    \r\n      Triangle\r\n      \\(\\dfrac{b h^{3}}{36}\\)\r\n\t  where \\(b\\) is the length of the base and \\(h\\) is the height\r\n    \r\n  \r\n\r\nThese formulas apply when the reference axis passes through the centroid.\r\nDuring bending, the neutral axis is the line in a beam's cross-section where the bending stress and strain are zero. For a symmetrical cross-section under simple bending, the neutral axis usually passes through the centroid.\r\n\r\nCalculating centroidal area moment of inertia\r\nFor simple shapes, we can use a standard formula. For an irregular shape, we can divide the area into many small elements and add their contributions. This process is called integration.\r\nEach small element \\(dA\\) is multiplied by the square of its perpendicular distance from the reference axis. The contributions are then added across the whole area:\r\n\r\n  \\[I_{x} = \\int_{A} y^{2}\\,dA\\quad\\text{and}\\quad I_{y} = \\int_{A} x^{2}\\,dA\\]\r\nwhere:\r\n\r\n\t\\(I_{x}\\) and \\(I_{y}\\) are the area moment of inertia about the \\(x\\) and \\(y\\) axes, respectively, in \\(\\text{m}^{4}\\) or \\(\\text{mm}^{4}\\)\r\n\t\\(y\\) is the perpendicular distance from the area element \\(dA\\) to the \\(x\\)-axis in metres \\(\\text{m}\\) or millimetres \\(\\text{mm}\\)\r\n\t\\(x\\) is the perpendicular distance from the area element \\(dA\\) to the \\(y\\)-axis in metres \\(\\text{m}\\) or millimetres \\(\\text{mm}\\).\r\n\r\n\r\nThe \\(y{^2}\\) term means that area elements further from the axis contribute much more, which is why deep beams (with more material far from the neutral axis) are so much stiffer in bending than shallow ones.\r\n\r\n\r\n\r\nNon-centroidal area moment of inertia\r\nIn practice, we often need the moment of inertia about an axis that is not through the centroid \u2013 for example, the base of a beam rather than its neutral axis. The parallel axis theorem lets us calculate \\(I\\) about any axis parallel to the centroidal axis, using the centroidal value \\(I_{c}\\) from the previous table.\r\n\r\nCalculating non-centroidal area moment of inertia\r\nMany cross-sections are made from two or more simple shapes. These are called composite shapes. To find the moment of inertia of a composite shape, divide it into simple elements, such as rectangles.\r\nFor an element whose centroidal axis is a distance \\(d\\) from the required parallel axis, the parallel axis theorem is:\r\n\r\n\\[I = I_{c} + A d^{2}\\]\r\nwhere:\r\n\r\n\t\\(I\\) is the moment of inertia about the new offset axis in \\(\\text{m}^{4}\\) or \\(\\text{mm}^{4}\\)\r\n\t\\(I_{c}\\) is the moment of inertia about the centroidal axis in \\(\\text{m}^{4}\\) or \\(\\text{mm}^{4}\\)\r\n\t\\(A\\) is the cross-sectional area in metres squared \\(\\text{m}^{2}\\) or millimetres squared \\(\\text{mm}^{2}\\)\r\n\t\\(d\\) is the distance between the axes in metres \\(\\text{m}\\) or millimetres \\(\\text{mm}\\).\r\n\r\n\r\nThe term \\(Ad^2\\) is called the transfer term. It accounts for the distance between the element's centroidal axis and the required axis. For a composite shape, calculate \\(I_{c}+Ad^{2}\\) for each element and then add the results.\r\n\r\n\r\n\r\nPolar area moment of inertia\r\nThe polar area moment of inertia \\(J\\) measures how a cross-section resists twisting about an axis. It is used when analysing torsion in components such as shafts, axles and drive rods.\r\nA cross-section with more area located far from the axis has a larger polar area moment of inertia. It will therefore resist twisting more strongly. The polar area moment of inertia is a property of the cross-section, not of the material.\r\n\r\nCalculating polar area moment of inertia\r\nFor a circular cross-section, the polar area moment of inertia can be calculated using:\r\n\r\n\\[J=\\frac{\\pi d^{4}}{32}\\quad\\text{or}\\quad\\frac{\\pi r^{4}}{2}\\]\r\nwhere:\r\n\r\n\t\\(J\\) is the polar area moment of inertia in \\(\\text{m}^{4}\\) or \\(\\text{mm}^{4}\\)\r\n\t\\(d\\) is the diameter of the circular cross-section in metres \\(\\text{m}\\) or millimetres \\(\\text{mm}\\)\r\n\t\\(r\\) is the radius of the solid circular cross-section in metres \\(\\text{m}\\) or millimetres \\(\\text{mm}\\).\r\n\r\n\r\nFor more complex cross-sections, the polar area moment of inertia is found by adding the contribution of each small area element:\r\n\r\n\\[J=\\int_A r^{2}\\,dA\\]\r\nwhere \\(r\\) is the distance from the axis of rotation to the small area element \\(dA\\) in metres \\(\\text{m}\\) or millimetres \\(\\text{mm}\\).\r\n\r\n\r\n\r\n\r\nExercise \u2013 calculating moment of inertia\r\n\r\n\tA flywheel is modelled as a solid disc with mass \\(8\\,\\text{kg}\\) and radius \\(0.4\\,\\text{m}\\). Calculate its mass moment of inertia.\r\n\r\n\r\n\r\n\tA motor applies a net torque of \\(45\\,\\text{N m}\\) to a shaft with mass moment of inertia \\(3.6\\,\\text{kg m}^{2}\\). Calculate the angular acceleration.\r\n\r\n\r\n\r\n\tA uniform rod of mass \\(2\\,\\text{kg}\\) and length \\(1.2\\,\\text{m}\\) rotates about one end. Calculate its:\r\n\t\r\n\t\tmass moment of inertia\r\n\t\tradius of gyration.\r\n\t\r\n\r\n\r\n\r\n\tCalculate the centroidal area moment of inertia of a rectangle with length \\(h=120\\,\\text{mm}\\) and width \\(b=80\\,\\text{mm}\\) about its centroidal axis, parallel to its length.\r\n\r\n\r\n\r\n\tUsing the parallel axis theorem, find the moment of inertia about the base of a rectangle with \\(b=60\\,\\text{mm}\\) and \\(h=40\\,\\text{mm}\\), whose centroid is \\(d=20\\,\\text{mm}\\) from the base.\r\n\r\n\r\n\r\n\tA thin hoop has a mass of \\(3\\,\\text{kg}\\) and a radius of \\(0.25\\,\\text{m}\\). Calculate its mass moment of inertia about its central axis.\r\n\r\n\r\n\r\n\r\n\tA solid disc has a mass of \\(6\\,\\text{kg}\\) and a radius of \\(0.2\\,\\text{m}\\). Calculate its radius of gyration about its central axis.\r\n\r\n\r\n\r\n\r\n\tCalculate the centroidal area moment of inertia of a circle with diameter \\(d=100\\,\\text{mm}\\).\r\n\r\n\r\n\r\n\r\n\tA rectangle has a width of \\(50\\,\\text{mm}\\), a height of \\(80\\,\\text{mm}\\), and its centroid is \\(40\\,\\text{mm}\\) from the base. Calculate its area moment of inertia about the base.\r\n\r\n\r\n\r\n\r\n\tA solid circular shaft has a diameter of \\(60\\,\\text{mm}\\). Calculate its polar area moment of inertia.\r\n\r\n\r\n\r\n\r\n\tA solid circular shaft has a radius of \\(25\\,\\text{mm}\\). Calculate its polar area moment of inertia.\r\n\r\n\r\n\r\n \r\nAboriginal Australian weapons image transcript\r\nPhotograph of three Aboriginal Australian weapons\u2014the leangle, parrying shield, and kodj\u2014shown front and side, with annotated dimensions.\r\n\r\n\tThe leangle is a curved hardwood club, 656 mm long and 335 mm wide at its broadest point, with a handle 30 mm wide.\r\n\tThe parrying shield is a narrow, convex hardwood shield, 875 mm long, 104 mm at its widest and 59 mm at its narrowest visible point.\r\n\tThe kodj is a composite hand axe, 365 mm long, with a stone head 114 mm wide and 46 mm deep, hafted onto a wooden handle.\r\n\r\nA map of Australia indicates the geographic origin of each weapon: the kodj (shown in red) originates from south-western Australia around King George Sound, and the leangle and parrying shield (shown in green) originate from south-eastern Australia in the Murray-Darling Basin region. Carpenter's Gap 1 in Western Australia and Narrabeen in New South Wales are also marked on the map.\r\n","n":0.026},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Dynamics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Physics","i":0,"n":1}]]},{"i":14,"$":[{"v":"Moment (torque)","n":0.707},{"v":"Every time you open a door, tighten a bolt with a spanner, or adjust a bike's handlebars, you're creating a turning effect. The further from the pivot you apply a force, and the more directly you push, the stronger that effect. This turning effect is called a moment or torque, and this resource shows you how to calculate and apply it. You will also explore how the physics of moments apples to Aboriginal digging sticks.\r\n\r\nMoment \\(M\\) in 2D\r\nWhen a net force is applied through a point that is not the centre of mass, a moment (or torque) \\(M\\) is applied and rotation occurs. Just as a net force causes linear motion (an object moving in a straight line), a moment causes rotational motion \u2013 it is the rotational equivalent of a pushing or pulling force, also called the rotational analogue of force. \r\n\r\nThe size of a moment depends on:\r\n\r\n\tthe size of the force\r\n\tthe distance from the pivot to where the force is applied\r\n\tthe angle of the force relative to the lever arm.\r\n\r\nThe turning force is most effective when it is acting at right angles to the rotating object. For example, the turning force applied to the beam on the left (approximately \\(45^{\\circ}\\) does not produce as great a moment compared to the beam on the right, where the force is at right angles to the beam.\r\n\r\n\r\n\r\nMoment is found using the equation:\r\n\\[M=F\\sin\\theta\\times r\\quad\\text{or}\\quad M=F_{\\perp}\\times r\\]\r\nwhere:\r\n\r\n\t\\(M\\) is the moment in Newton metres \\(\\text{N m}\\)\r\n\t\\(F\\) is the force applied in Newtons \\(\\text{N}\\)\r\n\t\\(\\theta\\) is the angle between the force and object in degrees\r\n\t\\(r\\) is the distance from the axis of rotation (or moment arm) in metres \\(\\text{m}\\).\r\n\r\n\r\nHere, \\(F\\sin\\theta\\) is the component of \\(F\\) acting perpendicular to the moment arm \u2013\u00a0the part of the force that actually produces rotation.\r\nMoment is a vector quantity. In 2D problems, we describe it as either clockwise or anticlockwise. In 3D problems, it has a direction in space.\r\n\r\n\r\n\r\n\r\nExample \u2013 calculating moment using force components\r\nA force \\(\\mathbf{F}\\) is applied to the handle of a hammer being used to extract a nail. The force must produce a clockwise moment of \\(\\mathbf{75}\\,\\textbf{N m}\\) about point A (the nail head). Determine the magnitude of \\(\\mathbf{F}\\).\r\n\r\nSince moment is a vector, we take clockwise moments as negative. The force \\(F\\) acts \\(30^{\\circ}\\) to the handle, so we resolve it into two components:\r\n\r\n\t\\(F\\cos30^{\\circ}\\) acts along the handle, with a perpendicular moment arm of \\(150\\,\\text{mm}=0.15\\,\\text{m}\\) from point A\r\n\t\\(F\\sin30^{\\circ}\\) acts perpendicular to the handle, with a moment arm of \\(450\\,\\text{mm}=0.45\\,\\text{m}\\) from point A.\r\n\r\nBoth components produce a clockwise (negative) moment about A. Setting up the moment equation:\r\n\\[\\begin{align*} \\sum M_{A} & = \\sum \\left( F_{\\perp}\\times r\\right)_{A} \\\\[6pt]\r\n-75 & = -\\left(F\\cos30^{\\circ}\\times0.15\\right) - \\left(F\\sin30^{\\circ}\\times0.45\\right) \\end{align*}\\]\r\nFactoring out \\(F\\):\r\n\\[\\begin{align*} -75 & = -F\\left(0.15\\cos30^{\\circ}+0.45\\sin30^{\\circ}\\right) \\\\[6pt]\r\n& = -F\\left( (0.15\\times0.866)_+(0.45\\times0.500) \\right) \\\\[6pt]\r\n& = -F\\left(0.1299+0.225\\right) \\\\[6pt]\r\n& = -0.3549F \\\\[6pt]\r\nF & = \\frac{75}{0.3549} \\\\[6pt]\r\n& = 211.3\\,\\text{N} \\end{align*}\\]\r\nThe magnitude of the force applied to the handle of the hammer is \\(211.3\\,\\text{N}\\).\r\n\r\nMoment \\(M\\) in 3D\r\nThe moment of a force can also be written as a vector cross product in the 3D space.\r\n\r\n\\[\\vec{M}=\\vec{r}\\times\\vec{F}\\]\r\nwhere:\r\n\r\n\t\\(\\vec{M}\\) is the moment vector in Newton metres \\(\\text{N m}\\)\r\n\t\\(\\vec{r}\\) is the moment arm vector in metres \\(\\text{m}\\)\r\n\t\\(\\vec{F}\\) is the force vector in Newtons \\(\\text{N}\\).\r\n\r\n\r\n\r\nVector \\(\\vec{M}\\) is perpendicular to the plane containing \\(\\vec{r}\\) and \\(\\vec{F}\\).\r\nTo find the direction of \\(\\vec{M}\\) using the right-hand rule: point your right-hand fingers in the direction of \\(\\vec{r}\\) (the moment arm, from the point O towards where the force is applied), then curl them towards \\(\\vec{F}\\). Your thumb points in the direction of \\(\\vec{M}\\) \u2013 perpendicular to the plane containing \\(\\vec{r}\\) and \\(\\vec{F}\\). The order of the cross product matters: \\(\\vec{r}\\times\\vec{F}\\) gives a different direction to \\(\\vec{F}\\times\\vec{r}\\), so always write the moment arm first.\r\n\r\n\r\nSince \\(\\vec{r}\\) and \\(\\vec{F}\\) are written as \\(x, y, z\\) coordinates, then \\(\\vec{M}\\) can be found by finding the determinant.\r\n\r\n\\[\\mathbf{M} = \\vec{r} \\times \\vec{F} = \\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\\r\nr_{x} & r_{y} & r_{z} \\\\\r\nF_{x} & F_{y} & F_{z} \\end{vmatrix} \\]\r\nwhere:\r\n\r\n\t\\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\) are the unit vectors in the \\(x\\), \\(y\\) and \\(z\\) directions, respectively\r\n\t\\(r_{x}\\), \\(r_{y}\\) and \\(r_{z}\\) are \\(x\\), \\(y\\) and \\(z\\) components of the moment arm vector \\(\\vec{r}\\), respectively\r\n\t\\(F_{x}\\), \\(F_{y}\\) and \\(F_{z}\\) are the \\(x\\), \\(y\\) and \\(z\\) components of the force vector \\(\\vec{F}\\), respectively.\r\n\r\n\r\nIf you need a refresher on vector or cross product, visit this page.\r\n\r\nExample \u2013 calculating moment using vector or cross product\r\nTwo forces act at point A on a 3D frame. If \\(\\mathbf{\\vec{F}_{1}=80\\hat{i}-100\\hat{j}+60\\hat{k}}\\,\\textbf{N}\\) and \\(\\mathbf{\\vec{F}_{2}=-150\\hat{i}+200\\hat{j}+90\\hat{k}}\\,\\mathbf{N}\\), determine the resultant moment produced by these forces about point O. Express your answer as a Cartesian vector.\r\n\r\nFirst, we need to find the resultant force \\(\\vec{F}_{r}\\). Since both forces act at the same point A, we can combine them into a single resultant force before computing the moment. This simplifies the cross product to a single calculation.\r\n\\[\\begin{align*} \\vec{F}_{R} & = \\vec{F}_{1}+\\vec{F}_{2} \\\\[6pt]\r\n& = \\left( 80\\hat{i}-100\\hat{j}+60\\hat{k} \\right) + \\left( -150\\hat{i}+200\\hat{j}+90\\hat{k} \\right) \\\\[6pt]\r\n& = \\left( -70\\hat{i}+100\\hat{j}+150\\hat{k}\\right) \\,\\text{N} \\end{align*}\\]\r\nNow, we identify the position vector \\(\\vec{r}_{A}\\). From the diagram, point A is located at coordinates \\((1,0.6,0.8)\\,\\text{m}\\) from the origin O. The position vector is therefore:\r\n\\[\\vec{r}_{A} = \\{ 1.0\\hat{i}+0.6\\hat{j}+0.8\\hat{k} \\} \\,\\text{m}\\]\r\nWe are ready to calculate the moment using the cross product.\r\n\\[\\begin{align*} \\vec{M} & = \\vec{r}_{A} \\times \\vec{F}_{R} \\\\[6pt]\r\n& = \\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\\r\n1.0 & 0.6 & 0.8 \\\\\r\n-70 & 100 & 150 \\end{vmatrix} \\end{align*}\\]\r\nExpanding the determinant:\r\n\\[\\begin{align*} M_{i} & = \\left( r_{y}\\times F_{z} \\right) - \\left( r_{z}\\times F_{y} \\right) \\\\[6pt]\r\n& = (0.6\\times150)-(0.8\\times100)\\\\[6pt]\r\n& = 90-80\\\\[6pt]\r\n& = 10 \\end{align*}\\]\r\n\\[\\begin{align*} M_{j} & = -[ \\left( r_{x}\\times F_{z} \\right) - \\left(r_{z}\\times F_{x} \\right) ] \\\\[6pt]\r\n& = -[ (1.0\\times150) - (0.8\\times-70) ] \\\\[6pt]\r\n& = -(150+56) \\\\[6pt]\r\n& = -206 \\end{align*}\\]\r\n\\[\\begin{align*} M_{k} & = \\left(r_{x}\\times F_{y}\\right) - \\left(r_{y}\\times F_{x}\\right) \\\\[6pt]\r\n& = (1.0\\times100)-(0.6\\times-70) \\\\[6pt]\r\n& = 100+42 \\\\[6pt]\r\n& = 142 \\end{align*}\\]\r\nThe resultant moment about O is \\(\\vec{M} = \\{ 10\\hat{i}-206\\hat{j}+142\\hat{k} \\}\\,\\text{N m}\\).\r\n\r\nExercise \u2013 calculating moment\r\n\r\n\tTwo workers are using ropes to prevent a vertical pole from rotating about its pin at point A. The worker at point B exerts a force of \\(P=180\\,\\text{N}\\) on their rope at \\(45^{\\circ}\\) to the horizontal, attached \\(5.4\\,\\text{m}\\) above point A. The worker at point C is holding a rope at \\(3.6\\,\\text{m}\\) above point A and pulls in a direction described by a 3\u20134\u20135 ratio (3 vertical, 4 horizontal). Determine the magnitude of the force \\(F\\) the worker at point C must exert to keep the resultant moment about A equal to zero.\r\n\t\r\n\r\n\r\n\r\n\r\n\tA force of \\(F=480\\,\\text{N}\\) acts from point B towards point A along the pipe. Point B is located at \\(\\{0.3\\hat{i}+1.2\\hat{j}+0.6\\hat{k}\\}\\,\\text{m}\\) from point O, and point A is at \\(\\{0\\hat{i}+1.2\\hat{j}+0\\hat{k}\\}\\,\\text{m}\\). Determine the moment of force \\(F\\) about point O. Express the result as a Cartesian vector.\r\n\r\n\r\n\r\n","n":0.03},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Maths","i":2,"n":1},{"v":"Force","i":1,"n":1},{"v":"Physics","i":0,"n":1}]]},{"i":15,"$":[{"v":"Angular motion","n":0.707},{"v":"Now that you are familiar with how an object's velocity changes over time in linear motion, we now shift our focus to angular motion, where similar principles apply to rotational systems. Use this resource to gain insight into the rotational behaviour of objects, and better understand the dynamics of systems in motion, both linearly and rotationally, including how Torres Strait Islander seafarers apply an understanding of angular motion at sea.\r\n\r\nAngular motion is the movement of an object around a fixed point or axis \u2013 the object follows a circular path. This type of motion is characterised by arc length, angular displacement, angular velocity, and angular acceleration. A spinning bicycle wheel, a Ferris wheel carriage, and the hands of a clock are everyday examples of angular motion.\r\n\r\n\r\n\r\n\r\nArc length and angular displacement\r\nArc length \\(s\\) is the distance measured along the curved path of a circle. It is like the length of the \u201cpiece\u201d of the circle\u2019s circumference that an object travels.\r\nAngular displacement \\(\\theta\\) is the angle an object turns through as it moves around a circle, measured from its starting position to its final position (usually in radians).\r\n\r\nConverting between degrees and radians\r\nGenerally, any angle at the centre of a circle is defined in radians by the equation:\r\n\r\n\r\nIn a complete circle, \\(\\theta=360^\\circ\\), so:\r\n\\[\\theta = \\frac{s}{r} = \\frac{2 \\pi r}{r}, \\quad \\text{or} \\quad 360^\\circ=2\\pi\\,\\text{radians}\\]\r\nwhere \\(\\pi=180^\\circ\\) and \\(1\\,\\text{rad}=57.3^\\circ\\).\r\n\r\nTo convert between radians and degrees, use these conversions:\r\n\r\n\\[\\text{From radians to degrees: }\\times\\frac{180}{\\pi}\\]\\[\\text{From degrees to radians: }\\times\\frac{\\pi}{180}\\]\r\n\r\n\r\nAngular velocity \\(\\omega\\)\r\nLinear velocity \\(v\\) is the rate of change of displacement. It is calculated using the equation:\r\n\\[v = \\dfrac{s}{t}=\\dfrac{r \\theta}{t}\\]\r\nAngular velocity \\(\\omega\\) is the rate of change of angular displacement. It is calculated using the equation:\r\n\\[\\omega=\\frac{\\theta}{t}\\]\r\nwhere \\(\\omega\\) is measured in \\(\\text{rad}\\,\\text{s}^{-1}\\).\r\n\r\nFrom these equations, we can derive an equation that links linear and angular velocity.\r\n\\[v=r\\omega\\]\r\nThis equation only applies when angular velocity is constant \u2013 that is, when there is no angular acceleration.\r\n\r\n\r\n\r\n\r\n\r\nConverting between rpm and radians per second\r\nRevolutions per minute \\(\\text{rpm}\\) is a common unit for describing how fast something rotates. It tells you how many complete revolutions an object makes in one minute.\r\nSince our angular equations use radians per second rather than \\(\\text{rpm}\\), we need a way to convert between the two.\r\n\\[\\begin{align*} 1\\,\\text{rpm} & = \\frac{1\\,\\text{rev}}{1\\, \\text{min}} \\\\[6pt]\r\n& = \\frac{2\\pi}{60\\,\\text{s}}\\\\[6pt]\r\n& = \\frac{\\pi}{30}\\,\\text{rad}\\,\\text{s}^{-1}\\end{align*}\\]\r\nMore simply, we can use these conversions:\r\n\\[\\text{From rpm to radians: }\\times\\frac{\\pi}{30}\\]\\[\\text{From radians to rpm: }\\times\\frac{30}{\\pi}\\]\r\n\r\nExample \u2013 converting between rpm and radians per second\r\nA weight on the end of a string describes a circular path of radius \\(\\mathbf{0.5}\\,\\textbf{m}\\). If its linear velocity is \\(\\mathbf{2}\\,\\textbf{m}\\,\\textbf{s}\\mathbf{^{-1}}\\), determine the angular velocity in:\r\n\r\n\t\\(\\textbf{rad}\\,\\textbf{s}\\mathbf{^{-1}}\\)\r\n\r\n\tWe have \\(r=0.5\\,\\text{m}\\) and \\(v=2\\,\\text{m}\\,\\text{s}^{-1}\\).\r\n\tTo find angular velocity, we use the equation \\(v=r\\omega\\). We can rearrange this to make \\(\\omega\\) the subject.\r\n\t\\[\\begin{align*} \\omega & = \\frac{v}{r} \\\\[6pt]\r\n\t& = \\frac{2}{0.5}\\\\[6pt]\r\n\t& = 4\\,\\text{rad}\\,\\text{s}^{-1}\\end{align*}\\]\r\n\r\n\t\\(\\textbf{rpm}\\).\r\n\r\n\tTo convert this to \\(\\text{rpm}\\), we use the conversion factor \\(\\times\\dfrac{30}{\\pi}\\).\r\n\t\\[\\begin{align*} \\omega & = 4\\times\\frac{30}{\\pi} \\\\[6pt]\r\n\t& = 38.2\\,\\text{rpm}\\end{align*}\\]\r\n\r\nAngular acceleration \\(\\alpha\\)\r\nAngular acceleration \\(\\alpha\\) is the rate of change of angular velocity. It is calculated using the equation:\r\n\\[\\alpha=\\frac{\\omega}{t}\\]\r\nSince linear velocity and angular velocity are related by \\(v=r\\omega\\), we can substitute this into the equation for linear acceleration, \\(a=\\dfrac{v}{t}\\) to find a relationship between linear and angular acceleration.\r\n\\[\\begin{align*} a & = \\frac{v}{t} \\\\[6pt]\r\n& = \\frac{r\\omega}{t}\\\\[6pt]\r\n& = r\\alpha\\end{align*}\\]\r\n\r\nLinear and angular equations\r\nWhen angular velocity is not constant\u2014for example, when a motor is speeding up or slowing down\u2014we need a more complete set of equations. Just as the kinematic equations for linear motion account for acceleration, the following angular equations do the same for rotational motion.\r\nThe equations derived for linear motion apply equally to angular motion, provided that we use the angular symbols instead.\r\n\r\n\r\nExample \u2013 using angular equations\r\nA flywheel increases at a constant angular acceleration from rest to \\(\\mathbf{400}\\,\\textbf{rpm}\\) in \\(\\mathbf{15}\\,\\textbf{s}\\). Calculate:\r\n\r\n\tthe angular acceleration\r\n\r\n\tWe know that \\(\\omega_{0}=0\\,\\text{rpm}\\), \\(\\omega=400\\,\\text{rpm}\\) and \\(t=15\\,\\text{s}\\).\r\n\tAngular velocity is measured in \\(\\text{rad}\\,\\text{s}^{-1}\\), so we need to convert \\(400\\,\\text{rpm}\\).\r\n\t\\[400\\times\\dfrac{\\pi}{30}=41.9\\,\\text{rad}\\,\\text{s}^{-1}\\]\r\n\tThe equation for angular velocity is \\(\\omega=\\omega_{0}+\\alpha t\\). We can rearrange to make \\(\\alpha\\) the subject.\r\n\t\\[\\begin{align*} \\alpha & = \\frac{\\omega-\\omega_{0}}{t} \\\\[6pt]\r\n\t& = \\frac{41.9-0}{15}\\\\[6pt]\r\n\t& = 2.79\\,\\text{rad}\\,\\text{s}^{-2}\\end{align*}\\]\r\n\r\n\tthe number of revolutions it makes.\r\n\r\n\tTo find the number of revolutions, we need to find \\(\\theta\\) first.\r\n\t\\[\\begin{align*} \\theta & = \\omega_{0}t+\\frac{1}{2}\\alpha t^{2} \\\\[6pt]\r\n\t& = (0)(15)+\\frac{1}{2}(2.79)(15)^{2}\\\\[6pt]\r\n\t& = 314\\,\\text{rad}\\end{align*}\\]\r\n\tA single revolution is equal to \\(2\\pi\\,\\text{rad}\\), or \\(1\\,\\text{rad}=\\dfrac{1}{2\\pi}\\,\\text{revolutions}\\), so we can convert using:\r\n\t\\[314\\times\\frac{1}{2\\pi}=50\\,\\text{revolutions}\\]\t\r\n\t\r\nExercise \u2013 calculating angular motion\r\n\r\n  The bob of a pendulum of length \\(1.5\\,\\text{m}\\) swings through a \\(20\\,\\text{cm}\\) arc. Calculate the angular displacement.\r\n\r\n\r\n\r\n\r\n\tA motor revolving at \\(1500\\,\\text{rpm}\\) slows down uniformly to \\(1200\\,\\text{rpm}\\) in \\(15\\,\\text{s}\\). Find:\r\n\t\t\r\n\t\t\tthe angular acceleration\r\n\t\t\tthe linear acceleration of a point \\(1.5\\,\\text{m}\\) from the centre.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tA flywheel operates at \\(300\\,\\text{rpm}\\). Calculate:\r\n\t\t\r\n\t\t\tthe angular velocity\r\n\t\t\tthe linear velocity \\(1.25\\,\\text{m}\\) from the centre.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n  Calculate the angular velocity of a car which rounds a curve of radius \\(10\\,\\text{m}\\) at \\(60\\,\\text{km/h}\\).\r\n\r\n\r\n\r\n","n":0.036},[{"v":"Maths","i":3,"n":1},{"v":"Kinematics","i":2,"n":1},{"v":"Indigenous knowledges","i":1,"n":0.707},{"v":"Physics","i":0,"n":1}]]},{"i":16,"$":[{"v":"Circular motion","n":0.707},{"v":"Even at constant speed, circular motion involves constant acceleration. Your velocity is always changing direction, even if its magnitude stays the same. This is what you feel when a car rounds a sharp bend or a rollercoaster carves through a loop. This resource introduces centripetal acceleration and the centripetal force that produces it, shows you how to apply these ideas to real problems, and presents you with examples of how circular motion is applied, such as in Indigenous inventions.\r\n\r\nCentripetal acceleration \\(a\\)\r\nDuring horizontal circular motion, an object moves in a circle on a flat plane. Consider a mass \\(m\\) travelling at a constant speed in a horizontal circle. In the following position, the instantaneous velocity is shown as a vector \\(u\\) in a direction North East.\r\n\r\nA short time later, this mass will be in the following position. The instantaneous velocity is now \\(v\\) in a direction North.\r\n\r\nThe change in velocity \\(\\Delta v\\) is therefore given by:\r\n\\[\\Delta v=v-u\\]\r\nWe can represent this relationship with vectors in 2D space.\r\n\r\n\\(\\Delta v\\) always act radially inwards\u2014or towards the centre of the circle\u2014and the mass is accelerating even though the speed is not changing. This acceleration is called centripetal acceleration. It is the change in velocity with time for an object moving in a circular path at constant speed.\r\nCentripetal acceleration is given by:\r\n\\[a = \\frac{v^{2}}{r} = \\frac{4\\pi^{2} r}{T^{2}} = 4\\pi^{2} r f^{2}\\]\r\nwhere:\r\n\r\n\t\\(a\\) is the centripetal acceleration in metres per second squared \\(\\text{m s}^{-2}\\)\r\n\t\\(v\\) is the velocity in metres per second \\(\\text{m s}^{-1}\\)\r\n\t\\(r\\) is the radius of circle in metres \\(\\text{m}\\)\r\n\t\\(T\\) is the period of motion (how long it takes for one complete cycle of motion) in seconds \\(\\text{s}\\)\r\n\t\\(f\\) is the frequency of motion (the number of cycles occuring in one second) in hertz \\(\\text{Hz}\\).\r\n\r\nConsider a seat on a theme park swing ride. The seat and rider have a combined mass of \\(80\\,\\text{kg}\\), the chain has a length of \\(6.0\\,\\text{m}\\) and the ride completes one revolution every \\(4.0\\,\\text{s}\\). Even though the rider moves at a constant speed, they are continuously accelerating because their direction changes. The centripetal acceleration is:\r\n\\[\\begin{align*} a & = 4\\pi^{2}rf^{2} \\\\[6pt]\r\n& = 4\\pi^{2}\\times6.0\\times0.25^{2} \\\\[6pt]\r\n& = 14.8\\,\\text{m s}^{-2} \\end{align*}\\]\r\nThis acceleration is directed inwards along the chain, towards the centre of the ride.\r\n\r\nCentripetal force \\(F_{C}\\)\r\nAccording to Newton\u2019s second law of motion, the net or unbalanced force on an object is given by \\(\\sum F=ma\\), where the net force is always in the same direction as the acceleration. This means that the net force\u2014as with the centripetal acceleration\u2014acts towards the centre of the circle. This net force is called the centripetal force \\(F_{C}\\).\r\nCentripetal force is given by:\r\n\\[F_{C}=ma=m\\frac{v^{2}}{r}=m\\frac{4\\pi^{2}r}{T^{2}}=4m\\pi^{2}r f^{2}\\]\r\nSince many rotating systems\u2014such as motors, wheels and satellites\u2014are described in terms of angular velocity \\(\\omega\\) rather than linear velocity \\(v\\), it is useful to express centripetal force in angular terms. Using the relationship \\(v=r\\omega\\):\r\n\\[F_{C}=m\\times\\frac{v^{2}}{r}=m\\times\\frac{r^{2} \\omega^{2}}{r}=mr\\omega^{2}\\]\r\n\r\nThe centripetal force always acts inward\u2014towards the centre of the circle\u2014while the object's velocity points tangentially, at right angles to that force. Think of the swing ride again: the chain pulls the seat inwards, and the seat curves around the circle rather than flying off in a straight line. If the chain broke, the seat would immediately travel in a straight line tangent to the circle.\r\n\r\n\r\nForces that cause circular motion\r\nAlways try to identify the force causing the object to turn in a circle. There are three main forces that cause circular motion by acting towards the centre of the circular path.\r\n\r\n\tTensional force: A theme park swing ride seat is pulled inwards by tension in the chain connecting it to the rotating arm above.\r\n\tFrictional force: A car moving in a circle does so because the centripetal force is due to the frictional force between the tyres and the road.\r\n\tGravitational force: The centripetal force causing a satellite (such as the Earth\u2019s natural satellite, the Moon) to orbit the Earth is caused by the gravitational force of the Earth acting on the satellite.\r\n\r\n\r\n\r\nObjects and tools may be designed to generate circular motion, but the true ingenuity lies in using that motion to achieve complex effects. For example, when a boomerang spins, it also generates lift perpendicular to its direction of motion. The centripetal acceleration is directed not just along the curve of flight, but also vertically.\r\n\r\nThe boomerang is just one expression of a deep tradition of physics-based innovation among Aboriginal peoples, which has produced sophisticated tools, technologies and ideas across tens of thousands of years.\r\nHere are two more examples of Aboriginal and Torres Strait Islander inventions that use circular motion.\r\n\r\n\r\n\r\n\r\n\r\nExample \u2013 calculating centripetal acceleration and force\r\nA car of mass \\(\\mathbf{1200}\\,\\textbf{kg}\\) travels around a roundabout at a constant speed of \\(\\mathbf{30}\\,\\textbf{km h}^{-1}\\). The radius of the circular path is \\(\\mathbf{15}\\,\\textbf{m}\\).\r\n\r\n  Determine the period of rotation.\r\n\r\nFirst, we convert the speed to \\(\\text{m s}^{-1}\\).\r\n\\[\\begin{align*} v & = \\frac{30}{3.6} \\\\[6pt]\r\n& = 8.33\\,\\text{m s}^{-1} \\end{align*}\\]\r\nThe period \\(T\\) is the time for one complete revolution \u2013 in other words, the circumference of the circle divided by the speed. We know the circumference is \\(2\\pi r\\), so:\r\n\\[\\begin{align*} T & = \\frac{2\\pi r}{v} \\\\[6pt]\r\n& = \\frac{2\\pi\\times15}{8.33} \\\\[6pt]\r\n& = 11.3\\,\\text{s} \\end{align*}\\]\r\n\r\n\r\n  Calculate the centripetal acceleration of the car.\r\n\r\nEven though the car moves at constant speed, it is continuously changing direction. We use the centripetal acceleration formula, where \\(v\\) is the linear speed and \\(r\\) is the radius of the roundabout.\r\n\\[\\begin{align*} a & = \\frac{v^{2}}{r} \\\\[6pt]\r\n& = \\frac{8.33^{2}}{15} \\\\[6pt]\r\n& = 4.63\\,\\text{m s}^{-2}\\,\\text{towards the centre of the roundabout} \\end{align*}\\]\r\n\r\n\r\n  Determine the magnitude of the centripetal force acting on the car.\r\n\r\nFrom Newton's second law, the net force equals mass times acceleration. Here, the net force is the centripetal force \u2013 the force needed to keep the car turning in a circle rather than travelling in a straight line.\r\n\\[\\begin{align*} F_{C} & = ma \\\\[6pt]\r\n& = 1200\\times4.63 \\\\[6pt]\r\n& = 5556\\,\\text{N} \\end{align*}\\]\r\n\r\n\r\n  Name the force that is responsible for the centripetal acceleration.\r\n\r\nThe road surface exerts a horizontal frictional force on the tyres, directed inward towards the centre of the roundabout.\r\n\r\n\r\n  Describe what would happen if the road was icy and friction was greatly reduced.\r\n\r\nWithout sufficient friction, there is no centripetal force to change the car's direction. According to Newton's first law, the car would continue in a straight line (the direction it was travelling at the moment friction was lost) rather than following the curve of the roundabout. In other words, the car would skid off the roundabout.\r\n\r\nExercise \u2013 calculating centripetal acceleration and force\r\n\r\n\tAn ice skater of mass \\(50\\,\\text{kg}\\) is skating in a horizontal circle of radius \\(1.5\\,\\text{m}\\) at a constant speed of \\(2.0\\,\\text{m s}^{-1}\\).\r\n\t\t\r\n\t\t  Calculate the skater's acceleration.\r\n\t\t  Find the horizontal component of the centripetal force acting on the skater.\r\n\t\t  Name the force that is providing the horizontal component of the net force that enables the skater to move in a circular path.\r\n\t\t\r\n\r\n\r\n\r\n\r\n\tA ball of mass \\(0.3\\,\\text{kg}\\) is attached to a string of length \\(0.5\\,\\text{m}\\) and swung in a horizontal circle at \\(3.0\\) revolutions per second.\r\n\t\t\r\n\t\t\tDetermine the period of rotation.\r\n\t\t\tCalculate the speed of the ball.\r\n\t\t\tCalculate the centripetal acceleration.\r\n\t\t\tFind the tension in the string.\r\n\t\t\tName the type of force providing the centripetal force.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\tA cyclist of mass \\(80\\,\\text{kg}\\) (including bicycle) rides around a circular velodrome track of radius \\(20\\,\\text{m}\\) at \\(8.0\\,\\text{m s}^{-1}\\).\r\n\t\t\r\n\t\t\tCalculate the centripetal acceleration.\r\n\t\t\tFind the centripetal force.\r\n\t\t\tName the type of force providing the centripetal force.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\tThe Moon orbits the Earth at a radius of \\(3.84\\times10^{8}\\,\\text{m}\\) with a period of \\(27.3\\) days.\r\n\t\t\r\n\t\t\tCalculate the orbital speed of the Moon.\r\n\t\t\tCalculate the centripetal acceleration.\r\n\t\t\tName the type of force providing the centripetal force.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\tA satellite of mass \\(420\\,\\text{kg}\\) orbits Earth at a radius of \\(6.6\\times10^{6}\\,\\text{m}\\) with a period of \\(92\\,\\text{min}\\).\r\n\t\t\r\n\t\t\tCalculate the orbital speed of the satellite.\r\n\t\t\tCalculate the centripetal acceleration.\r\n\t\t\tCalculate the centripetal force acting on the satellite.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n","n":0.028},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Maths","i":2,"n":1},{"v":"Force","i":1,"n":1},{"v":"Physics","i":0,"n":1}]]},{"i":17,"$":[{"v":"Homeostasis","n":1},{"v":"Homeostasis keeps your body's internal environment stable and balanced. Understanding homeostasis helps us see how various body systems work together to maintain health and adapt to changes. Use this resources to explore how your body achieves stability and responds to challenges.\r\n\r\n\r\n\r\nHomeostasis (\"homeo\" means \"similar to\" and \"stasis\" means \"standing still\") is the body's way of maintaining a stable internal environment despite external changes. It does this by correcting any imbalances in conditions.\r\nThe steps involved in regulating homeostasis are the same regardless of what conditions are being maintained: \r\n\r\n\tThe stimulus (change in conditions) is detected by a sensor, such as a temperature sensor on the skin or a chemical receptor that detect the pH of blood.\r\n\tThe sensors send signals to the hypothalamus (the \"control centre\" for homeostasis) or other regulatory organs which coordinate a response.\r\n\tEffector organs receive signals from the hypothalamus and carry out the response.\r\n\r\nDepending on whether the response is to decrease the effect of the change or to increase it, a negative or feedback loop may be involved.\r\n\r\n\r\nNegative feedback loops\r\nOne way that homeostasis is maintained is by negative feedback loops. Negative feedback loops reverse changes in the body\u2019s internal environment. When a change occurs, these loops act to bring conditions back to their set points.\r\nSome examples of internal conditions regulated by negative feedback loops are:\r\n\r\n\tblood pressure\r\n\tbody temperature\r\n\tblood sugar levels\r\n\tfluid balance.\r\n\r\nWe will look at body temperature and blood glucose levels in more detail.\r\n\r\n\r\nNegative feedback loops help to keep your internal body temperature close to \\(37^{\\circ}\\textrm{C}\\) whether you're enjoying a topical holiday in \\(40^{\\circ}\\textrm{C}\\) heat or braving the frosty \\(6^{\\circ}\\textrm{C}\\) on your commute to uni one winter's morning. This process is called thermoregulation.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nGlucose is metabolised by the body and converted into ATP, a source of energy, but the levels of glucose in the blood must be carefully maintained because fluctuations can lead to serious health issues.\r\nNormal blood glucose levels should be between \\(60\\textrm{ mg/dL}\\) and \\(140\\textrm{ mg/dL}\\). To coordinate this balance, the pancreas acts as the control centre.\r\n\r\n\r\n\r\n\r\n\r\n\r\nPositive feedback loops\r\n\r\nPositive feedback loops are less common than negative feedback loops. They involve further amplifying the change and bringing the body system further away from homeostasis.\r\nExamples are:\r\n\r\n\tchildbirth\r\n\tlactation\r\n\tblood clotting\r\n\tthe menstrual cycle.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise\r\nSee how well you understand homeostasis with a quick quiz.\r\n\r\n\r\n\r\n\r\n\r\n \r\nFeedback loops\r\nTwo flowcharts showing the negative and positive feedback loops.\r\nOn the left is the negative feedback loop. It shows the stimulus, then the sensor, the control centre and finally, the effector. An inhibitory arrow is shown from the effector to the stimulus. This indicates that the effector decreases the effect of the change.\r\nOn the right is the positive feedback loop. It also shows the same components, but with a regular arrow pointing from the effector to the stimulus. This indicates that the effector increases the effect of the change.\r\n","n":0.047},[{"v":"Biology","i":0,"n":1}]]},{"i":18,"$":{"0":{"v":"Digital skills","n":0.707},"2":[]}},{"i":19,"$":[{"v":"Excretory system","n":0.707},{"v":"The excretory system is the waste removal system of the body. It helps to keep our bodies clean and healthy. Use this resource to learn how the organs of the excretory system work together to support your body's functions.\r\n\r\nThe excretory system (or urinary system) plays a critical role in removing waste from the body and helping to maintain its internal balance.\r\n\r\nThe urinary tract\r\nThe digestive system removes solid waste from digestion, turning food materials that we can't use into faeces. The respiratory system gets rid of waste gases like carbon dioxide when we breathe out. The excretory system is different because it filters the blood. This lets us remove any extra water, salts, and toxins from the body to keep it balanced and hydrated.\r\nThe path through which this happens is called the urinary tract.\r\n\r\nLet's explore the parts of the urinary tract.\r\n\r\nKidneys\r\nThe kidneys are bean-shaped organs that filter waste and extra substances that we don't need from our blood. We have two kidneys. They are found on either side of our spine and each is about the size of a fist. Here is a 3D model of a kidney.\r\n  \r\n\r\nThe kidneys contain tiny filtration units called nephrons, which consist of a glomerulus (a tiny network of capillaries) and tubule. These structures work together to filter blood, reabsorb necessary substances and excrete waste.\r\n\r\nThrough the filtration process, our bodies make urine, a yellowish liquid waste. Urine can then be removed from the body by urination to get rid of toxins, extra minerals and fluid that the body doesn't need. \r\nWatch this video to learn about how your kidneys work.\r\n\r\n\r\nUreters\r\nUreters are two thin tubes that connect the kidneys to the bladder. They transport urine from the kidneys, where it's produced, down to the bladder for storage. \r\nThe walls of the ureters are lined with smooth muscles that contract rhythmically, helping push urine smoothly toward the bladder. This process ensures that urine flows in one direction and reduces the risk of infections.\r\n\r\nBladder\r\nThe bladder is a muscular sac found in the lower abdomen that stores urine until it's ready to be expelled from the body. Its flexible walls stretch to hold urine and contract when it's time to pee.\r\n\r\nA healthy bladder can typically hold about two cups (or \\(473\\) millilitres) of urine. The bladder has sensors that tell the brain when it's full, giving you the urge to urinate.\r\n\r\nUrethra\r\nThe urethra is a tube that carries urine from the bladder out of the body during urination. \r\n\r\n\tIn males, the urethra is longer and also passes through the penis, carrying urine and sperm.\r\n\tIn females, it is shorter and opens just above the vaginal opening.\r\n\r\nMuscles called sphincters help control the release of urine from the body.\r\n\r\nExcretion\r\nExcretion is the process of removing waste from the body to make sure the internal environment remains balanced and healthy. \r\n\r\nThis diagram gives you an overview of the process, but let's step through it in more detail. You can also watch the video under Further resources at the end of this page to see kidney filtration in action.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nSummary\r\nThe key components of the urinary tract and their functions are outlined in the table.\r\n\r\nExercise\r\nSee how well you understand the components of the excretory system and their roles with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nBinh is a PhD student studying the effect of diet on urine composition. For his experiment, he recruited healthy student volunteers and had them follow specific diets (high protein, low sodium, or a standard diet) for one week each. \r\nUrine samples were collected daily and concentrations of urea, creatinine and sodium were measured using spectrophotometry. The average results are shown in the table.\r\n\r\n \r\nThe urinary tract\r\nA diagram of the urinary tract, including differences between the female and male bladders.\r\nOn the left is a diagram of the urinary system within the body. The bones in the lower abdomen, pelvis and upper legs are shown.\r\n\r\n\tTwo kidneys are located in the abdomen.\r\n\tEach kidney is connected to the bladder through ureters.\r\n\tThe bladder is located at the base of the pelvis.\r\n\tThe urethra is below the bladder and leads to the outside of the body.\r\n\r\nIn the top right is a diagram of the female bladder. The two ureters are shown, along with the pelvic floor muscles that surround the urethra.\r\nIn the bottom right is a diagram of the male bladder. It is very similar to the female bladder, but has a prostate gland that wraps around the upper part of the urethra.\r\n \r\nAn interactive three-dimensional model of the kidney.\r\nKidney model:\r\nA cross-section of a kidney. The major blood vessels like the renal artery and vein, and their smaller arterioles, venules and capillaries can be seen. Nephrons are arranged into clusters called renal pyramids. In this particular kidney, there are 7 of them.\r\n \r\nKidney\r\nA labelled cross-section of a kidney.\r\n\r\n\tThe kidney is enclosed in a renal capsule.\r\n\tThe blood vessels enter and leaving the kidney are labelled \"Renal artery\" and \"Renal vein\".\r\n\tThe blood vessels extend into the kidney through the renal columns and into the cortex (outer) region.\r\n\tIn the medulla (inner) region are structures called renal pyramids, where the nephrons are found.\r\n\tThese connect to structures called \"Calyx\", which join together to form the renal pelvis.\r\n\tThe renal pelvis exits the kidney and joins with the ureter.\r\n\r\n \r\nIt's a hot day, and you've just downed several glasses of water, one after the other. Behind the sudden urge that follows are two bean-shaped organs that work as fine-tuned internal sensors. They balance the amount of fluid in your body, detect waste in your blood, and know when to release the vitamins, minerals, and hormones you need to stay alive. Say hello to your kidneys.\r\nThe main role of these organs is to dispose of waste products and to turn them into urine. The body's eight liters of blood pass through the kidneys between 20 and 25 times each day, meaning that, together, these organs filter about 180 liters every 24 hours.\r\nThe ingredients in your blood are constantly changing as you ingest food and drink, which explains why the kidneys need to be on permanent duty. Blood enters each kidney through arteries that branch and branch, until they form tiny vessels that entwine with special internal modules, called nephrons. In each kidney, 1 million of these nephrons form a powerful array of filters and sensors that carefully sift through the blood. This is where we see just how refined and accurate this internal sensing system is.\r\nTo filter the blood, each nephron uses two powerful pieces of equipment: a blob-like structure called a glomerulus, and a long, stringy, straw-like tubule. The glomerulus works like a sieve, allowing only certain ingredients, such as vitamins and minerals, to pass into the tubule. Then, this vessel's job is to detect whether any of those ingredients are needed in the body. If so, they're reabsorbed in amounts that the body needs, so they can circulate in the blood again.\r\nBut the blood doesn't only carry useful ingredients. It contains waste products, too. And the nephrons have to figure out what to do with them. The tubules sense compounds the body doesn't need, like urea, left over from the breakdown of proteins, and redirects them as urine out of the kidneys and through two long sewers called ureters. The tubes empty their contents into the bladder to be discharged, ridding your body of that waste once and for all.\r\nThere's water in that urine, too. If the kidney detects too much of it in your blood, for instance, when you've chugged several glasses at once, it sends the extra liquid to the bladder to be removed. On the other hand, low water levels in the blood prompt the kidney to release some back into the blood stream, meaning that less water makes it into the urine. This is why urine appears yellower when you're less hydrated.\r\nBy controlling water, your kidneys stabilize the body's fluid levels. But this fine balancing act isn't the kidney's only skill. These organs have the power to activate vitamin D to secrete a hormone called renin that raises blood pressure, and another hormone called erythropoietin, which increases red blood cell production.\r\nWithout the kidneys, our bodily fluids would spiral out of control. Every time we ate, our blood would receive another load of unsifted ingredients. Soon, the buildup of waste would overload our systems and we'd expire. So each kidney not only keeps things running smoothly. It also keeps us alive. Lucky then that we have two of these magical beans.\r\n \r\nExcretion\r\nA labelled diagram of the nephron, showing the excretion process. The thin arrows show the direction of fluid movement.\r\n\r\n\tThe glomerulus is where filtration occurs.\r\n\tThe glomerulus is surrounded by a structure called the Bowman's capsule.\r\n\tFiltration occurs in the Bowman's capsule.\r\n\tThe proximal tubule is the next section of the nephron. It is the part of the tubule before the loop of Henle.\r\n\tSecretion and reabsorption happen in the proximal tubule.\r\n\tAt the loop of Henle, reabsorption occurs.\r\n\tAfter the loop is the distal tubule, where reabsorption and secretion occur.\r\n\tTubules from different nephrons connect to a collecting duct, which directs urine to the ureters and bladder.\r\n\r\n","n":0.026},[{"v":"Excretory system","i":3,"n":0.707},{"v":"Anatomy","i":2,"n":1},{"v":"Physiology","i":1,"n":1},{"v":"Human body systems","i":0,"n":0.577}]]},{"i":20,"$":[{"v":"Peripheral nervous system","n":0.577},{"v":"The peripheral nervous system helps the central nervous system communicate with the rest of the body. Learning how it works helps us understand how nerves carry signals to muscles and organs, enabling movement and sensation. Use this resource to learn how the peripheral nervous system keeps us aware and responsive to our environment.\r\n\r\nThe peripheral nervous system (PNS) connects the central nervous system to the rest of the body, allowing for communication and response. It is the network of sensory organs and neurones that extend from the CNS, reaching every part of the body. \r\nBy transmitting signals between the CNS and other regions, the PNS allows our body to respond to stimuli, maintain balance, and coordinate actions, ensuring effective interaction with our environment.\r\n\r\nSensory organs\r\nThe sensory organs are vital components that allow the nervous system to gather information about the environment. These include eyes, ears, nose, tongue, and skin. Each organ is specialised for detecting specific stimuli.\r\n\r\nThe eyes detect light and enable us to see.\r\nThe ears are responsible for hearing and balance.\r\nThe nose senses smells.\r\nThe tongue detects taste.\r\nThe skin senses touch, temperature and pain.\r\n\r\nThese organs send signals to the brain, where the information is processed to form perceptions and guide responses. The sensory organs are essential for interacting with and responding to the world.\r\n\r\n\r\nNeurones\r\n\r\nNeurones (sometimes \"neurons\" or \"nerve cells\") are the building blocks of the nervous system. They send and receive electrical signals or nerve impulses throughout the body. \r\nThere are a few types of neurones, but these cells have the same general structure. Their components are summarised in the table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tComponent\r\n\t\t\tDescription\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tCell body (or soma)\r\n\t\t\tContains the nucleus and other organelles; processes incoming signals and generates outgoing signals\r\n\t\t\r\n\t\t\r\n\t\t\tDendrites\r\n\t\t\tBranch out from the cell body; receive signals from other neurones and sensory receptors and pass information along to the cell body\r\n\t\t\r\n\t\t\r\n\t\t\tAxon\r\n\t\t\tA long, thin projection that transmits signals away from the cell body to other neurones, muscles or glands\r\n\t\t\r\n\t\t\r\n\t\t\tAxon hillock\r\n\t\t\tConnects the cell body to the axon\r\n\t\t\r\n\t\t\r\n\t\t\tMyelin sheath\r\n\t\t\tA fatty layer of insulation covering some neurones; helps increase speed of transmission\r\n\t\t\r\n\t\t\r\n\t\t\tAxon terminals\r\n\t\t\tThe end of an axon from which chemicals called neurotransmitters are released\r\n\t\t\r\n\t\r\n\r\n \r\n\r\n\r\nThe PNS also consists of ganglia (singular: ganglion), which are clusters of neurone cell bodies. Neurones, on the other hand, are individual signalling cells.\r\n\r\nTypes of neurones\r\nThe three types of neurones are:\r\n\r\n\tsensory neurones which carry information from sensory organs to the CNS; also called afferent neurones\r\n\tmotor neurones which send messages from the CNS to the muscles and glands, controlling our movement and reactions; also called efferent neurones\r\n\tinterneurones which link sensory and motor neurones within the CNS and have a role in processing sensory information and making decisions about how our body should respond.\r\n\r\nThe sensory and motor neurones are found in the PNS, but the interneurons are found in the CNS, so the brain and spinal cord.\r\n\r\nThe position of the cell body in these neurones differs, giving them different appearances.\r\n\r\n\r\nNeurotransmitters\r\nNeurotransmitters are chemical substances that neurones use to communicate with sensory organs, muscles, glands, and other neurones. The location of this communication is called the synapse.\r\n\r\nThe synapse includes the neurone releasing the neurotransmitters, called the pre-synaptic neurone; the cell receiving the signals, called the post-synaptic cell; and the space between them, known as the synaptic cleft. Neurotransmitters are packaged into little sacs in the pre-synaptic neurone called vesicles.\r\n\r\nWhen an electrical impulse reaches the axon terminal, the vesicles release the neurotransmitters into the synaptic cleft. The neurotransmitters then travel across the gap and bind to specific targets on the post-synaptic cell, called receptors, to pass along the signal.\r\n\r\n\r\n\r\n\r\n\r\nNerve impulses\r\nNerve impulses are electrical signals that travels along neurones. They allow the nervous system to communicate quickly. \r\nThe process of generating a nerve impulse can be complicated, so let's break it down.\r\n\r\nWatch this video to see how a nerve impulse is propagated through a neurone.\r\n\r\n\r\nNerve impulses begin at a sensory organ, which activates a sensory neurone. This stimulates an interneurone, which in turn stimulates a motor neurone. Finally, the motor neurone connects to a target tissue or effector, coordinating it to respond in the required way.\r\n\r\n\r\nDivision of the PNS\r\nThe PNS is divided into the somatic and autonomic nervous systems. This division allows the body to efficiently control different types of functions.\r\n\r\n\r\n\r\n\r\n\r\n\r\nSummary\r\nThe organisation of the nervous system is summarised in the following flowchart.\r\n\r\n\r\n\r\nExercise\r\nSee how well you understand the peripheral nervous system with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nNatural medicines are commonly used to treat constipation. To investigate the potential for processed Aloe vera powder to be used as a treatment, a group of pharmacology researchers from India and Saudi Arabia conducted a study where they treated isolated sections of rat colon and pig ileum with the dissolved powder and measured smooth muscle contraction. They compared the responses to a standard neurotransmitter, acetylcholine, which produces strong contractions and acts as a positive control.\r\nSome of their results are shown in the graphs.\r\n\r\n\r\n\r\n\r\n\r\n \r\nAn interactive three-dimensional model of a neurone with features labelled.\r\nNeuron model\r\n\r\n\tCell body (soma): The compact area of a nerve cell that constitutes the nucleus and surrounding cytoplasm, excluding the axons and dendrites; the cell body\r\n\tAxon: A tube-like structure that propagates signals to the axon terminals\r\n\tDendrite: A short, branched extension of a nerve cell, along which impulses received from other cells at synapses are transmitted to the cell body\r\n\tMyelin sheath: An insulator that covers an axon in order to minimise dissipation of the electrical signal as it travels down the axon\r\n\tNode of Ranvier: A gap in the myelin sheath of a nerve; sites where the signal is \"recharged\" as it travels along the axon\r\n\tAxon terminal: Branch at the end of the axon that allows signals to be communicated to other cells\r\n\tAxon hillock: Portion of the neuron that connects the cell body, soma, to the axon. The impulses the neuron receives are summed up at the axon hillock to determine whether an action potential will be initiated\r\n\r\n Diagram of a sensory neurone, motor neurone and interneurone\r\n\r\n\tA sensory neurone with its cell body branching off the axon.\r\n\tA motor neurone with an axon extending from the cell body.\r\n\tAn interneurone which has its cell body positioned in the middle of the axon.\r\n \r\nDiagram of the synapse between two neurones\r\n\r\n\tNeurotransmitters are packaged into vesicles at the axon terminal of the pre-synaptic neurone.\r\n\tThe neurotransmiters are released into the synaptic cleft.\r\n\tNeurotransmitters bind to receptors at the dendrites of the post-synaptic neurone.\r\n\r\n \r\nA typical neurone consists of a cell body, plasma membrane extensions called dendrites, an elongated fibre known as an axon, and an axon hillock, the trigger zone that releases a nerve impulse.\r\nThe axon hillock maintains an excitation limit, or threshold, which determines whether or not a neuron will generate a nerve impulse. A nerve impulse is an electrical signal conducted by a neurone, causing a response in another neurone or target cell.\r\nWhen a neurone is at rest, its membrane is polarised because there are more positive ions outside the cell and more negative ions inside the cell, which creates a charge difference across the membrane.\r\nActive transport mechanisms called sodium-potassium pumps carry more sodium and less potassium ions across the membrane to maintain this charged difference. Even in a resting neurone, there is the potential for the charged difference to create an electrical current. This is called a resting membrane potential.\r\nWhen an electrical current flows through a dendrite, this is called a local membrane potential.\r\nWhen a dendrite detects a stimulus, a sodium channel in its plasma membrane opens and lets sodium into the neurone. This influx of positive ions reverses the charge across a particular section of the membrane in a process called depolarisation.\r\nTo repolarise the membrane, potassium channels open and release potassium out of the neurone. Nearby, a sodium-potassium pump transports excess sodium amount and brings potassium in, which restores the resting membrane potential.\r\nThe flow of reversing charges along the dendrite's membrane produces a wavelike electrical current toward the neurone's trigger zone. If the strength of the current meets or exceeds the threshold at the trigger zone, an electrical signal called an action potential or nerve impulse will occur.\r\nIn a nerve impulse, the trigger zone sends an electrical signal down the axon toward the space between neurones called a synapse or to a target cell membrane. \r\n \r\nNeurone pathway\r\nA diagram showing the pathway through which nerve impulses travel.\r\n\r\n\tThe sensory organ, where nerve impulses begin. Pictured in a section of skin.\r\n\tA sensory neurone with dendrites in the sensory organ. The axon terminals of the sensory neurone sends the signal to the dendrites of an interneurone.\r\n\tThe interneurone receives the signal from the sensory neurone and sends the signal on to a motor neurone.\r\n\tThe motor neurone passes this signal to an effector organ. Pictured is a collection of muscle fibres.\r\n\r\n \r\nDiagram showing a reflex arc\r\n\r\n\tThe skin on the person's finger is the sensory organ. It detects the heat of the candle.\r\n\tA sensory (afferent) neurone is activated.\r\n\tThe sensory neurone activates an interneurone in the spinal cord through the release of neurotransmitters.\r\n\tThe interneurone then processes the information and stimulates a motor (efferent) neurone.\r\n\tThe motor neurone forms a synapse with muscle tissue in the arm.\r\n\tThe neurotransmitters released from the motor neurone cause the muscle tissue to contract.\r\n\tThe person withdraws their hand from the flame.\r\n\r\n \r\nThe differences between your sympathetic and parasympathetic nervous systems are so dramatic it can feel a bit like flipping a switch.\r\nSay you're at the beach on a beautiful day, the Sun is shining, there's a gentle breeze. You dive into the water, but struggle to stay afloat. As you fight the current, your sympathetic nervous system or your fight or flight response kicks in. You are in survival mode and your body prepares for conflict.\r\nYour body signals danger by releasing adrenaline. Your heart starts racing. Your pupils expand and your mouth becomes dry. Your airways open up, sending as much oxygen as possible to your brain to keep you alert, and other bodily processes switch gears to help in your time of need. Your stomach and intestines stop digesting, your bladder relaxes, your reproductive system limits blood flow, and your liver releases glucose to give you a burst of energy.\r\nAll these changes happen so quickly, you may not even be aware of them.\r\nWhen you make it back to shore, you start to calm down and relax. This is your parasympathetic nervous system or your rest and digest process at work.\r\nThose bodily processes that were at full attention before change course. Your heart rate slows down, your breathing returns to normal, your pupils constrict and you start to salivate. As your body realises it's no longer under threat, your stomach and intestines begin to digest. Your bladder constricts, and your reproductive system increases its blood flow. In other words, your body returns to its natural state. \r\nThis is an example of your autonomic nervous system at work. The sympathetic nervous system and parasympathetic nervous system perform a vital balancing act that helps you survive and recover. It happens many times throughout the day, whether you're at the beach, hanging out at home, working from an office or anywhere in between.\r\nAnd while you may tend to think of your fight or flight response kicking in during major stressful situations like a car accident, it can also be triggered by minor things like spilling a glass of milk or getting a paper cut. \r\nYou can also thank your rest and digest response for all those loving feelings, big and small, you may have from day to day. From cuddling your puppy, to professing your love to your partner. \r\nSo the next time you are stressed and sense your body tensing up, or you find yourself relaxing on the couch after a busy day of work.\r\n \r\nNervous system flowchart\r\n\r\n\tNervous system\r\n\t\t\r\n\t\t\tCentral nervous system\r\n\t\t\t\t\r\n\t\t\t\t\tBrain\r\n\t\t\t\t\tSpinal cord\r\n\t\t\t\t\r\n\t\t\t\r\n\t\t\tPeripheral nervous system\r\n\t\t\t\t\r\n\t\t\t\t\tSomatic nervous system\r\n\t\t\t\t\tAutonomic nervous system\r\n\t\t\t\t\t\t\r\n\t\t\t\t\t\t\tSympathetic nervous system\r\n\t\t\t\t\t\t\tParasympathetic nervous system\r\n\t\t\t\t\t\t\r\n\t\t\t\t\t\r\n\t\t\t\t\r\n\t\t\t\r\n\t\t\r\n\t\r\n\r\n \r\nLine graph showing rat colon smooth muscle contractility\r\nA line graph titled \"Rat colon smooth muscle contractility\". There are two lines on the graph, a line connecting dots labelled \"Acetylcholine\" and a line connecting crosses labelled \"Aloe vera\".\r\nThe x-axis is labelled \"Concentration (mg/mL)\", starting at 0.0001 mg/mL and going all the way to 1000 mg/mL. It is represented on a logarithmic scale. The y-axis is labelled \"Response (%)\", starting at 0 and going to 100, with 20% increments.\r\nIn the graph, the \"Acetylcholine\" line starts at 0% reponse for a concentration of 0.0001 mg/mL until a concentration of around 0.0003 mg/mL. Then, the response gradually increases until a concentration of 0.3 mg/mL, where a response of 100% is achieved. From 0.3 mg/mL to 1 mg/mL, the response is 100%.\r\nIn the graph, the \"Aloe vera\" line starts at 0% response for a concentration of 0.2 mg/mL until around 0.6 mg/mL. Then, the response gradually increases until a concentration of 60 mg/mL, where a response of around 37% is achieved. From 60 mg/mL to 600 mg/mL, the response seems to mostly plateau, with a maximum of 40% at 600 mg/mL.\r\nThe data is shown in the table.\r\n\r\n\t\r\n    \r\n      Concentration (mg/mL)\r\n      Acetylcholine (% response)\r\n      Aloe vera (% response)\r\n    \r\n  \r\n  \r\n    \r\n      \\(0.0001\\)\r\n      \\(0\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.0003\\)\r\n      \\(0\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.001\\)\r\n      \\(16\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.003\\)\r\n      \\(38\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.01\\)\r\n      \\(61\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.03\\)\r\n      \\(78\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.1\\)\r\n      \\(91\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.3\\)\r\n      \\(100\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(1\\)\r\n      \\(100\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.2\\)\r\n      Not tested\r\n      \\(0\\)\r\n    \r\n    \r\n      \\(0.6\\)\r\n      Not tested\r\n      \\(0\\)\r\n    \r\n    \r\n      \\(2\\)\r\n      Not tested\r\n      \\(6\\)\r\n    \r\n    \r\n      \\(6\\)\r\n      Not tested\r\n      \\(13\\)\r\n    \r\n    \r\n      \\(20\\)\r\n      Not tested\r\n      \\(22\\)\r\n    \r\n    \r\n      \\(60\\)\r\n      Not tested\r\n      \\(37\\)\r\n    \r\n    \r\n      \\(200\\)\r\n      Not tested\r\n      \\(39\\)\r\n    \r\n    \r\n      \\(600\\)\r\n      Not tested\r\n      \\(40\\)\r\n    \r\n  \r\n\r\n\r\nLine graph showing pig ileum smooth muscle contractility\r\nA similar line graph titled \"Pig ileum smooth muscle contractility\". The format is the same as the previous graph, but with different data.\r\nIn the graph, the \"Acetylcholine\" line starts at 0% reponse for a concentration of 0.0001 mg/mL until a concentration of around 0.0003 mg/mL. Then, the response gradually increases until a concentration of 0.3 mg/mL, where a response of 100% is achieved. From 0.3 mg/mL to 1 mg/mL, the response is 100%.\r\nIn the graph, the \"Aloe vera\" line starts at 0% response for a concentration of 0.2 mg/mL until around 0.6 mg/mL. Then, the response gradually increases until a concentration of 200 mg/mL, where a response of around 61% is achieved.\r\nThe data is shown in the table.\r\n\r\n\t\r\n    \r\n      Concentration (mg/mL)\r\n      Acetylcholine (% response)\r\n      Aloe vera (% response)\r\n    \r\n  \r\n  \r\n    \r\n      \\(0.0001\\)\r\n      \\(0\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.0003\\)\r\n      \\(0\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.001\\)\r\n      \\(5\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.003\\)\r\n      \\(20\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.01\\)\r\n      \\(40\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.03\\)\r\n      \\(78\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.1\\)\r\n      \\(85\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.3\\)\r\n      \\(100\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(1\\)\r\n      \\(100\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(0.2\\)\r\n      Not tested\r\n      \\(0\\)\r\n    \r\n    \r\n      \\(0.6\\)\r\n      Not tested\r\n      \\(0\\)\r\n    \r\n    \r\n      \\(2\\)\r\n      Not tested\r\n      \\(2\\)\r\n    \r\n    \r\n      \\(6\\)\r\n      Not tested\r\n      \\(7\\)\r\n    \r\n    \r\n      \\(20\\)\r\n      Not tested\r\n      \\(42\\)\r\n    \r\n    \r\n      \\(60\\)\r\n      Not tested\r\n      \\(51\\)\r\n    \r\n    \r\n      \\(200\\)\r\n      Not tested\r\n      \\(61\\)\r\n    \r\n    \r\n      \\(600\\)\r\n      Not tested\r\n      Not tested\r\n    \r\n  \r\n\r\n","n":0.02},[{"v":"Nervous system","i":3,"n":0.707},{"v":"Anatomy","i":2,"n":1},{"v":"Physiology","i":1,"n":1},{"v":"Human body systems","i":0,"n":0.577}]]},{"i":21,"$":[{"v":"Central nervous system","n":0.577},{"v":"The central nervous system controls the body by processing information and sending signals. Understanding how it works helps us see how our bodies respond to changes and interact with the world. Use this resource to explore how the nervous system keeps everything connected and running smoothly. Also learn about bush medicine that has effects on the central nervous system.\r\n\r\nThe nervous system\r\nThe nervous system is a complex network that controls all the activities in your body. It helps you think, feel, and move by sending messages between your brain and the rest of your body.\r\n\r\nThe nervous system is made up of:\r\n\r\nthe central nervous system (CNS), which consists of the brain and spinal cord\r\nthe peripheral nervous system (PNS), which contains the nerves.\r\n\r\nIt plays a critical role in helping us interpret information from the environment and coordinate a response.\r\n\r\n\r\n\r\nThe central nervous system\r\nThe central nervous system (CNS) is the main control centre of the body. It is made up of the brain and spinal cord, which work together to the interpret information we receive from the environment and coordinate a response.\r\nThe CNS plays a crucial role in everything from simple reflexes to complex cognitive processes.\r\n\r\nThe brain\r\nThe brain is the most complex organ in the body and serves as the command centre. It manages thoughts, emotions, memory, and decision-making. It also regulates vital functions like breathing and heartbeat. \r\n\r\nThe brain is divided into different regions, each with specialised functions. \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nThe spinal cord\r\nThe spinal cord acts as a highway for messages between the brain and the rest of the body. It is responsible for transmitting signals to and from the brain, allowing for fast reflex actions. \r\nThe spinal cord is protected by the bones of the vertebral column and consists of nerve fibres that carry impulses to control movements and respond to sensory inputs. Together, the brain and spinal cord form the central nervous system, ensuring that the body operates in a coordinated and efficient manner.\r\n\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\nExercise\r\nSee how well you understand the components of the central nervous system and their roles with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nIn 2014, researchers at the University of Sousse and the University of Sfax collaborated on a project looking at the effect of caffeine ingestion on reaction time after 36 hours of sleep deprivation. Thirteen healthy male physical education students completed four test sessions at 6 pm:\r\n\r\n\tplacebo during a baseline night (bedtime from 10:30 pm to 7 am)\r\n\tcaffeine ingestion during a baseline night (bedtime from 10:30 pm to 7 am)\r\n\tplacebo during a night of 36 h of sleep deprivation\r\n\tcaffeine ingestion during a night of 36 h of sleep deprivation.\r\n\r\nParticipants were tested on their performance of serveral physical activites, including a reaction time test. For the reaction time test, they were required to press a key on a microcomputer when a visual stimulus appeared. The results for the test are shown in the figure.\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nBrain\r\nA labelled diagram of the brain.\r\n\r\n\tCerebrum (pale blue, pale green, yellow and purple): the largest part of the brain\r\n\tCerebellum (green): the section found at the back of the brain, under the cerebrum\r\n\tBrainstem (blue): the section that connects the brain to the spinal cord\r\n\r\n \r\nAn interactive three-dimensional model of a brain with the lobes of the cerebrum shown in different colours.\r\nBrain model:\r\n\r\n\tFrontal lobe (blue): thinking/imagination\r\n\tTemporal lobe (green): sound and speech processing\r\n\tOccipital lobe (red): visual processing\r\n\tParietal lobe (yellow)\r\n\tPre-frontal lobe (blue): executive and cognitive functions, personality\r\n\r\n \r\nBrainstem\r\nA diagram of the brainstem.\r\n\r\n\tMidbrain (red): the section of the brainstem just below the thalamus (yellow)\r\n\tPons (green) the section of the brainstem that connects the thalamus and midbrain with the cerebellum\r\n\tMedulla oblongata (blue): the base of the brainstem, where the brain connects to the spinal cord\r\n\r\nThe thalamus, sitting above the midbrain, and the spinal cord, positioned directly below the medulla oblongata, are also labeled.\r\n \r\nSpinal cord\r\nA diagram showing the regions of the spinal cord.\r\n\r\n\tThe first 8 segments of the spinal cord below the brain make up the cervical region.\r\n\tThe next 12 segments are the thoracic region.\r\n\tThe following 5 segments are the lumbar region.\r\n\tThen, there are 5 segments in the sacral region and just 1 coccygeal segment at the bottom of the spinal cord, which are not labeled in the figure.\r\n\r\n \r\nBar chart showing results of reaction time test\r\nA vertical bar chart measuring reaction time on with a baseline night of sleep or sleep deprivation, with placebo or caffeine ingestion. \r\nAll groups are shown on the x-axis.\r\n\r\n\tThe first bar is baseline night with placebo.\r\n\tThe second bar is baseline night with caffeine.\r\n\tThe third bar is sleep deprivation with placebo.\r\n\tThe fourth bar is sleep deprivation with caffeine.\r\n\r\nThe y-axis is labelled \"Reaction time (s)\", starting at 0.20 seconds and extending to 0.42 seconds, with 0.02 second increments.\r\nThe data is shown in the table.\r\n\r\n\t\r\n    \r\n      Condition\r\n      Reaction time with placebo (mean \u00b1 SD)\r\n      Reaction time with caffeine (mean \u00b1 SD)\r\n    \r\n  \r\n  \r\n    \r\n      Baseline night\r\n      \\(0.275 \\pm 0.040\\)\r\n      \\(0.278 \\pm 0.030\\)\r\n    \r\n    \r\n      Sleep deprivation\r\n      \\(0.362 \\pm 0.050\\)\r\n      \\(0.268 \\pm 0.030\\)\r\n    \r\n  \r\n\r\n","n":0.035},[{"v":"Indigenous knowledges","i":4,"n":0.707},{"v":"Nervous system","i":3,"n":0.707},{"v":"Anatomy","i":2,"n":1},{"v":"Physiology","i":1,"n":1},{"v":"Human body systems","i":0,"n":0.577}]]},{"i":22,"$":[{"v":"Endocrine system","n":0.707},{"v":"The endocrine system is a network of glands that produce hormones to regulate various bodily functions. Understanding how it works helps us see how these glands and hormones work together to maintain balance and respond to changes in our bodies. Use this resource to learn about the endocrine system, including a culturally responsive model for managing endocrine-related complications during pregancy.\r\n\r\nEndocrine glands\r\nThe endocrine system (or hormonal system) consists of a network of endocrine glands that produce hormones to regulate crucial processes in the body like growth, metabolism, and mood. Together, these glands help to maintain balance in the body and make sure that we can respond effectively to internal and external changes.\r\n\r\n\r\nHormones\r\nHormones are chemical messengers produced by glands in the endocrine system. They allow your glands to communicate with different parts of your body to make sure your organs and systems are working together harmoniously. \r\nHormones are released into your bloodstream, where they can then travel to organs and tissues all around the body.\r\nThe endocrine glands, their location, functions and examples of hormones involved are shown in the table.\r\n\r\nWatch this video for an overview of how hormones work.\r\n\r\n\r\n\r\nWe will look at the hypothalamus, pituitary gland and pancreas in more detail, but you can explore the Further resources at the end of this page to learn more about the other glands and how they function.\r\n\r\n\r\nThe hypothalamus and pituitary gland\r\nThe hypothalamus is a small but crucial part of the brain located at its base. It acts as a bridge between the nervous and endocrine systems, helping to regulate body temperature, hunger, thirst, and emotions.\r\nThe hypothalamus monitors the body's internal environment and responds by releasing hormones that communicate with the pituitary gland. These hormones ensure the body maintains balance and reacts properly to changes, like stress or external temperature shifts. You can learn more about homeostasis and the body's balancing act on the Homeostasis page.\r\n\r\nThe pituitary gland, known as the \"master gland,\" sits just below the hypothalamus. Despite its small size, it plays a significant role in controlling other endocrine glands.\r\n\r\nThe pituitary gland releases hormones that influence growth, metabolism, and reproduction. For example, human growth hormone (HGH) helps control how the body grows, while adrenocorticotropic hormone (ACTH) stimulates the adrenal glands to produce cortisol, a vital stress hormone.\r\nTogether, the hypothalamus and pituitary gland coordinate many essential bodily functions. They ensure that hormones are released in the right amounts and at the right times, keeping the body's systems running smoothly. This teamwork between the brain and the endocrine system is key to overall health and wellbeing.\r\n\r\nThe pancreas\r\nThe pancreas is a vital organ located in the upper abdomen, behind the stomach.\r\n\r\nIt has two main functions:\r\n\r\nproducing enzymes that help digest food in the small intestine\r\nreleasing hormones that regulate blood sugar levels.\r\n\r\nTo perform its function in the endocrine system, the pancreas makes the hormone insulin, which lowers blood sugar, and glucagon, which raises it, helping maintain balance.\r\nThe pancreatic role in digestion and blood sugar control is essential for keeping the body healthy. You can learn more about blood sugar regulation in Homeostasis.\r\n\r\nYou can use this interactive model to see how the pancreas interacts with the duodenum, a part of the digestive system.\r\n  \r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise\r\nSee how well you understand the endocrine system and its function with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nResearchers in Switzerland conducted a study to determine reference values for the key hormones involved in the menstrual cycle. This graph shows the change in progesterone levels (in ng/mL) in blood serum over a 29-day cycle. The solid line represents the median concentration and the dashed lines represent the 95% confidence interval.\r\n\r\n\r\n\r\n\r\n\r\n \r\nThe endocrine system\r\nA diagram of the endocrine system in the female and male bodies, with key organs labelled.\r\n\r\n\tThe pineal gland and pituitary gland are found in the brain.\r\n\tThe thyroid gland is found in the throat, with 4 parathyroid glands located behind it.\r\n\tThe adrenal glands are found on top of each kidney.\r\n\tThe pancreas is found in the abdomen.\r\n\tIn the female figure on the left, the ovaries are labelled.\r\n\tIn the male figure on the right, the testes are labelled.\r\n\r\n \r\nOver the course of our lifetimes, our bodies undergo a series of extraordinary metamorphoses: we grow, experience puberty, and many of us reproduce. Behind the scenes, the endocrine system works constantly to orchestrate these changes.\r\nAlongside growth and sexual maturity, this system regulates everything from your sleep to the rhythm of your beating heart, exerting its influence over each and every one of your cells. \r\nThe endocrine system relies on interactions between three features to do its job: glands, hormones and trillions of cell receptors.\r\nFirstly, there are several hormone-producing glands: three in your brain, and seven in the rest of your body. Each is surrounded by a network of blood vessels, from which they extract ingredients to manufacture dozens of hormones. Those hormones are then pumped out in tiny amounts, usually into the bloodstream.\r\nFrom there, each hormones needs to locate a set of target cells in order to bring about a specific change. To find its targets, it's helped along by receptors, which are special proteins inside or on the cell's surface. Those receptors recognise specific hormones as they waft by and bind to them.\r\nWhen this happens, that hormone-receptor combination triggers a range of effects that either increase or decrease specific processes inside the cell to change the way that cell behaves.\r\nBy exposing millions of cells at a time to hormones in carefully-regulated quantities, the endocrine system drives large-scale changes across the body.\r\nTake, for instance, the thyroid and the two hormones it produces, triiodothyronine and thyroxine. These hormones travel to most of the body's cells, where they influence how quickly those cells use energy and how rapidly they work. In turn, that regulates everything from breathing rate to heartbeat, body temperature, and digestion.\r\nHormones also have some of their most visible\u2014and familiar\u2014effects during puberty. In men, puberty begins when the testes begin secreting testosterone. That triggers the gradual development of the sexual organs, makes facial hair sprout, and causes the voice to deepen and height to increase.\r\nIn women, estrogen secreted from the ovaries signals the start of adulthood. It helps the body develop, makes the hips widen, and thickens the womb's lining, preparing the body for menstruation or pregnancy.\r\nAn enduring misconception around the endocrine system is that there are exclusively male and female hormones. In fact, men and women have estrogen and testosterone, just in different amounts.\r\nBoth hormones play a role in pregnancy, as well, alongside more than 10 other hormones that ensure the growth of the fetus, enable birth, and help the mother feed her child.\r\nSuch periods of hormonal change are also associated with fluctuations in mood. That's because hormones can influence the production of certain chemicals in the brain, like serotonin. \r\nWhen chemical levels shift, they may cause changes in mood, as well. But that's not to say that hormones have unlimited power over us.\r\nThey're frequently viewed as the main drivers of our behaviour, making us slaves to their effects, especially during puberty. But research shows that our behaviour is collectively shaped by a variety of influences, including the brain and its neurotransmitters, our hormones, and various social factors.\r\nThe primary function of the endocrine system is to regulate bodily processes, not control us. Sometimes disease, stress, and even diet can disrupt that regulatory function, however, altering the quantity of hormones that glands secrete or changing the way that cells respond.\r\nDiabetes is one of the most common hormonal disorders, occurring when the pancreas secrets too little insulin, a hormone that managed blood sugar levels. And hypo- and hyperthyroidism occur when the thyroid gland makes too little or too much thyroid hormone.\r\nWhen there's too little thyroid hormone, that results in a slowed heart rate, fatigue and depression, and when there's too much thyroid hormone, weight loss, sleeplessness, and irritability.\r\nBut most of the time, the endocrine system manages to keep our bodies in a state of balance. And through its constant regulation, it drives the changes that ultimately help us become who we are.\r\n \r\nThe hypothalamus and pituitary gland\r\nA diagram of a person's brain, with a zoomed-in look near the hypothalamus and pituitary gland. These structures are located at the base of the brain, at the top of the brainstem. \r\nThe thalamus sits under the cerebrum. The hypothalamus is just underneath it.\r\nThere is a funnel-shaped empty space called the infundibulum, and under that is the pituitary gland. The pituitary gland consists of two regions: the anterior and posterior pituitary.\r\n \r\nThe pancreas\r\nA diagram of the pancreas and surrouding organs. The stomach is shown, leading into the duodenum. Branching off the duodenum is the pancreas with the head and tail of this organ labelled. The liver is also shown with the gallbladder. The tubes leading out of the gallbladder and pancreas combine to form the common bile duct.\r\n \r\nAn interactive three-dimensional model of the pancreas and duodenum, a part of the small intestine. Parts of the pancreas, duodenum and surrounding tissue are annotated with descriptions.\r\n\r\n\tPancreas: The pancreas is an organ located in the abdomen. It plays an essential role in converting the food we eat into fuel for the body's cells. The pancreas has two main functions: an exocrine function that helps in digestion and an endocrine function that regulates blood sugar.\r\n\tDuodenum: The duodenum is the first part of the small intestine. It is located between the stomach and the middle part of the small intestine, or jejunum. After foods mix with stomach acid, they move to the duodenum, where they mix with bile from the gallbladder and digestive juices from the pancreas.\r\n\tPancreatic duct: The pancreatic duct, or duct of Wirsung (also, the major pancreatic duct due to the existence of an accessory pancreatic duct), is a duct joining the pancreas to the common bile duct to supply pancreatic juice provided from the exocrine pancreas, which aids in digestion.\r\n\tCommon bile duct: The common bile duct is a small, tube-like structure formed where the common hepatic duct and the cystic duct join. Its physiological role is to carry bile from the gallbladder and empty it into the upper part of the small intestine (the duodenum). The common bile duct is part of the biliary system.\r\n\tPortal vein: The portal vein or hepatic portal vein is a blood vessel that carries blood from the gastrointestinal tract, gallbladder, pancreas and spleen to the liver. This blood contains nutrients and toxins extracted from digested contents.\r\n\tAbdominal aorta: The abdominal aorta is a continuation of the descending thoracic aorta. It supplies all of the abdominal organs, and its terminal branches go on to supply the pelvis and lower limbs. It also supplies the undersurface of the diaphragm and parts of the abdominal wall.\r\n\tMerging of pancreatic duct and common bile duct\r\n\r\n \r\nLine graph showing serum progesterone levels\r\nLine graph with three lines on the graph, a solid line labelled \"Median\" and two dashed lines labelled \"95% confidence interval\".\r\nThe x-axis is labelled \"Time of cycle\", starting from 0 and ending at 30, with intervals of 2. The y-axis is labelled \"Serum progesterone levels\", starting from 0 and ending at 20, with intervals of 2.\r\nIn the graph, the \"Median\" line begins at day 0 with a progesterone level of 0.5. It remains low until about day 15, where it starts to increase. The progesterone level peaks at 13.5 on day 22. Then, the level drops again to about 1.0 at day 29.\r\nThe \"95% confidence interval\" lines show that the spread of data is very small from day 0 to around day 15. Then, as the median increases, the spread of data also increases. The data is furthest from the median at the peak progesterone concentration. The range of data values is also very high at the 29 day mark.\r\nThe data is summarised in the table.\r\n\r\n","n":0.023},[{"v":"Indigenous knowledges","i":4,"n":0.707},{"v":"Endocrine system","i":3,"n":0.707},{"v":"Anatomy","i":2,"n":1},{"v":"Physiology","i":1,"n":1},{"v":"Human body systems","i":0,"n":0.577}]]},{"i":23,"$":[{"v":"Muscular system","n":0.707},{"v":"The muscular system provides support and facilitates movement of the body. Understanding how it works helps us see how the body performs physical activities and stays stable and strong. Use this resource to explore the components and roles of the muscular system.\r\n\r\nThe muscular system consists of all of the muscules in your body and the tissues that support them. It works with the skeletal system to help you move, maintain your posture and produce heat.\r\nIt also protects the internal organs and assist in critical functions like breathing and digestion.\r\n\r\n\r\nSkeletal muscle\r\nSkeletal muscle is a type of muscle that is attached to bones. It allows us to move voluntarily, like walk and lift things. It also helps us maintain our posture and generates heat to keep us warm.\r\nSkeletal muscle is composed of fibres bundled together by connective tissues. It is striated, meaning that it is made of parallel light and dark bands. Rod-like microfibrils also run parallel to the muscle fibre. These are responsible for muscle contraction.\r\n\r\n\r\nThe contraction of muscles is controlled by the nervous system. The brain coordinates the motor neurones, which send signals to the skeletal muscle to contract. Muscles work in pairs, where one muscle contracts while the other relaxes to coordinate movement. \r\nExamples of skeletal muscle include the triceps, which are at the back of your arm and the gluteus maximus, also known as the glutes or butt, which is the largest muscle in the body. You can go to this link to see a list of the skeletal muscles in the human body.\r\nTo make our skeletal muscle stronger, we need to exercise regularly. This improves flexibility and endurance.\r\n\r\n\r\n\r\nMetabolism\r\nOur skeletal muscles use adenosine triphospate (ATP) for energy during contraction. Learn more about ATP and cellular respiration on this page.\r\nThe body can generate this energy through two types of metabolism:\r\n\r\n\tAerobic respiration occurs when there is a good supply of oxygen. Our muscles efficiently produce ATP by completely breaking down glucose and fat in the mitochondria to form carbon dioxide and water. It is helpful for slow, sustained energy production for activities where high endurance is needed.\r\n\tAnaerobic respiration occurs when oxygen levels are low. Our body quickly produces ATP through the breakdown of glucose alone in the cytoplasm, leading to lactic acid buildup. This can cause muscle fatigue, which explains why when you do intense activites like sprinting or lifting weights, you feel tired and sore more quickly. Anaerobic respiration is helpful for when you quickly need to obtain energy.\r\n\r\nThe process of ATP generation also produces heat, which the body can use to regulate its temperature.\r\n\r\nTendons\r\nTendons connect bones to skeletal muscle. They are mostly made up of collagen fibres, which make them tough and flexible. \r\nTendons work like cords, allowing muscles to pull bones effectively. They help transfer force and stabilise joints during movement.\r\n\r\nSmooth muscle\r\nSmooth muscle is found in the walls of organs like the stomach, intestines and blood vessels. Unlike skeletal muscle, it moves on its own, meaning that we don't control it consciously. \r\nSmooth muscle contracts slowly to move substances through the digestive tract and regulate blood flow. It is particularly important for digestion and blood pressure regulation because it can stay contracted over longer periods of time. Smooth muscle can do this because it uses less ATP.\r\n\r\nWhile skeletal muscles are made up of bundles of fibres that work together, smooth muscle is made up of individual cells that contract as one tissue in organs. Because of this, smooth muscle looks more uniform.\r\n\r\nThe smooth muscle is wrapped in neurons, meaning that it partly relies on the nervous system to coordinate its action. As well as this, it responds to hormones. For example, the hormone gastrin stimulates the contraction of smooth muscle in the stomach to help churn food.\r\n\r\nCardiac muscle\r\nCardiac muscle is found in the walls of the heart and is responsible for its rhythmic contractions. Like smooth muscle, it works involuntarily. It beats tirelessly, pumping blood throughout the body and is crucial for keeping our organs supplied with oxygen and nutrients.\r\nCardiac muscle has a unique structure. It is striated like skeletal muscle but forms interconnected networks called intercalated disks. This allows rapid transmission of electrical signals for the heart to contract efficiently and uniformly. They also have a high density of mitochondria to support constant contraction. Cardiac muscle is regulared by the sinoatrial node.\r\n\r\n\r\nSummary\r\nThe components of the muscular system and their functions are outlined in the table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tComponent\r\n\t\t\tFunction\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tSkeletal muscle\r\n\t\t\tAttached to bones; enables voluntary movements like walking and lifting\r\n\t\t\r\n\t\t\r\n\t\t\tTendons\r\n\t\t\tConnects muscles to bones; transfers force to move bones, stabilising joints\r\n\t\t\r\n\t\t\r\n\t\t\tSmooth muscle\r\n\t\t\tFound in organ walls; controls involuntary movements like digestion and blood flow\r\n\t\t\r\n\t\t\r\n\t\t\tCardiac muscle\r\n\t\t\tLocated in the heart; pumps blood continuously without conscious control\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\nExercise\r\nSee how well you understand the components and functions of the muscular system with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\n\r\nA group of researchers in Japan conducted a meta-analysis of randomised controlled trials to study the relationship between total protein intake and muscle strength, with and without resistance training.\r\nTheir results are shown in the graphs. The solid lines represent the averages and the dashed lines represent the 95% confidence interval.\r\n\r\n\r\n\r\n\r\n\r\n \r\nSkeletal muscle\r\nA diagram of a cross-section of skeletal muscle showing its structure. Each muscle fibre is bundled together to form a fascicle. Each fascicle is then bundled up to form the muscle.\r\n \r\nSmooth muscle\r\nA diagram showing the organisation of smooth muscle, with a zoomed in section of neurones.\r\nOn the right is a section of smooth muscle, as shown as spindle-shaped cells. They are not arranges in tubes like skeletal muscle. The cells are labelled \"Smooth muscle cells\". Wrapped around the cells are thin structures labelled \"Autonomic neurone\". These neurones have sections where they bulge and bulbed areas are labelled \"Varicosity\".\r\nA section of the autonomic neurone with varicosity is highlighted with a rectangle. There is an arrow pointing from this rectangle to the image on the left.\r\nThe image on the left is a zoomed in section of the varicosity. Within the bulbed regions are small balls with the label \"Vesicles with neurotransmitters\".\r\n \r\nCardiac muscle\r\nA diagram showing the organisation of cardiac muscle. There are three images. The first image is of a whole heart. There is a small rectangle drawn on the surface of the heart, corresponding to the second image.\r\nThe second image is a zoomed up look at the heart. It shows the \"Cardiac muscle cells\" labelled. They are connected in a network, with thick black lines separating some strands of muscle. One strand of cardiac muscle is cut to reveal its cross-section. A rectangle is drawn at the cross-section, corresponding to the third image.\r\nThe third image is a zoomed up look at the cardiac muscle fibre. It shows four fibres lined up, each labelled \"Myofibrils\". A thick black line runs across the fibres, labelled \"Intercalated disc\". Between the myofibrils are green blob-like structures, labelled \"Mitochondria\".\r\n \r\nSkeletal, smooth and cardiac muscle\r\nThree microscopy images showing skeletal muscle, smooth muscle and cardiac muscle. Each image has a horizontal bar to indicate the scale. The width of the bar is 25 micron.\r\n\r\n\tThe first image is of skeletal muscle. It consists of long muscle fibres running parallel to each other.\r\n\tThe second image is of smooth muscle. It consists of individual cells that do not run parallel to each other.\r\n\tThe third image is of cardiac muscle. It consists of a network of fibres.\r\n\r\n \r\nLine graph showing change in muscle strength with resistance training\r\nOn the left is a line graph titled \"With resistance training\". There are three data lines on the graph, a solid line labelled \"Average\" and two dashed lines labelled \"95% confidence interval\". There is another line at x equals 0 to show where baseline is.\r\nThe x-axis is labelled \"Total protein intake (grams per kg of bodyweight per day)\", starting at 0.5 g and going all the way to 4, with increments of 0.5. The y-axis is labelled \"Change in muscle strength (%)\", starting at negative 20 and going all the way to 30, with increments of 5.\r\nIn the graph, the \"Average\" line starts at around 0.7 grams per kg of bodyweight per day which gives a 19.5% change in muscle strength. As total protein intake increases, the change in muscle strength increases until a change of 24.5% with 1.5 grams per kg of bodyweight per day. From there, the change in muscle strength gradually decreases until it reaches 15.0% with 3.8 grams per kg of bodyweight per day.\r\nThe 95% confidence interval lines show the spread of data around the mean. As we approach the peak muscle strength, the data becomes less spread. Then, as the total protein intake increases further, the data becomes more spread. At the maximum protein intake of 3.8 grams per kg of bodyweight per day, the 95% confidence interval is between 10% and 20%.\r\n\r\nLine graph showing change in muscle strength with no resistance training\r\nOn the right is a line graph titled \"Without resistance training\". The format is the same as the first graph, but with different data.\r\nIn the graph, the \"Average\" line starts at around 0.7 grams per kg of bodyweight per day which gives a 4% change in muscle strength. As total protein intake increases, the change in muscle strength increases until a change of 7% with 1.3 grams per kg of bodyweight per day. Then, the change in muscle strength gradually decreases until it reaches 3% at 2 grams per kg of bodyweight per day.\r\nThe 95% confidence interval decreases until the peak change in muscle strength is reached, then increases again. The spread of data at 0.7 and 2.0 grams per kg of bodyweight per day is similar.\r\n\r\nThe data is shown in the table.\r\n\r\n\r\n    \r\n      Total protein intake (g/kg bodyweight/day)\r\n      With resistance training (Change in muscle strength %)\r\n      Without resistance training (Change in muscle strength %)\r\n    \r\n  \r\n  \r\n    \r\n      \\(0.7\\)\r\n      \\(19.5\\)\r\n      \\(4.0\\)\r\n    \r\n    \r\n      \\(1.0\\)\r\n      \\(22.0\\)\r\n      \\(6.0\\)\r\n    \r\n    \r\n      \\(1.5\\)\r\n      \\(24.5\\)\r\n      \\(7.5\\)\r\n    \r\n    \r\n      \\(2.0\\)\r\n      \\(23.0\\)\r\n      \\(3.0\\)\r\n    \r\n    \r\n      \\(2.5\\)\r\n      \\(20.0\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(3.0\\)\r\n      \\(18.0\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(3.5\\)\r\n      \\(16.0\\)\r\n      Not tested\r\n    \r\n    \r\n      \\(3.8\\)\r\n      \\(15.0\\)\r\n      Not tested\r\n    \r\n  \r\n\r\n\r\n","n":0.024},[{"v":"Musculoskeletal system","i":3,"n":0.707},{"v":"Anatomy","i":2,"n":1},{"v":"Physiology","i":1,"n":1},{"v":"Human body systems","i":0,"n":0.577}]]},{"i":24,"$":[{"v":"Skeletal system","n":0.707},{"v":"The skeletal system gives our bodies shape and support. Learning how it works helps us understand how our bodies move and stay strong. Use this resource to explore parts of the skeletal system and what they do. Also learn about a culturally responsive movement-based program to improve health outcomes, including skeletal health, for Aboriginal and Torres Strait Islander children.\r\n\r\nMusculoskeletal system\r\nThe musculoskeletal system or locomotor system includes the bones, muscles, tendons and ligaments in the body. They work together to:\r\n\r\n\tkeep us upright\r\n\thelp us move\r\n\tprotect our organs\r\n\tmake new red and white blood cells\r\n\tstore important minerals.\r\n\r\nIt consists of the skeletal and muscular systems. Here, we will look at the skeletal system. The next page will discuss the muscular system.\r\n\r\nThe skeleton\r\nThe skeletal system consists of all the bones in the body and its supporting tissues, including bones, cartilage, ligaments and joints. Together, these form the skeleton. The skeleton gives the body support and shape, and protects the internal organs.\r\n\r\nBones\r\nBones are strong structures that act as the framework for the skeleton. They play a crucial role in protecting the internal organs.\r\nBones have a dense outer layer called compact bone (also known as cortical bone) and a lighter, porous inner layer called spongy bone (also known as cancellous or trebacular bone). Both compact and spongy bone are mostly made of a mineral called hypoxyapatite, which is a form of calcium phosphate. This makes up around \\(70\\%\\) of bone; the remaining \\(30\\%\\) is a mix of cells and water.\r\n\r\nThe spaces in the spongy bone are where bone marrow is found. Red marrow makes blood cells. Yellow marrow stores fat.\r\nBlood vessels run through the compact and spongy bone, playing an important role in bone development, repair and maintenance.\r\n\r\nThe outer surface of the bone is covered in a thin membrane called the periosteum (\u201cperi\u201d meaning \u201caround\u201d and \u201costeon\u201d meaning \u201cbone\u201d). It has blood vessels and nerves that help the bone grow and heal.\r\nThe overall structure of bone gives it strength, but also flexibility and the ability to absorb shock.\r\nThe image shows some of the key bones in the skeletal system.\r\n\r\n\r\nCartilage\r\nCartilage is a flexible type of tissue that cushions the joints and reduces friction between bones so you can move easily. You typically find cartilage covering the surfaces of bones at your joints. It also covers structures like the nose, ears and parts of the rib cage.\r\nCartilage is made of cells called chondrocytes (\u201cchrondro\u201d meaning \u201ccartilage\u201d and \u201ccyte\u201d meaning \u201ccells\u201d). These produce a gel-like substance that is both strong and flexible.\r\nUnlike bone, the cartilage does not contain blood vessels. It gets nutrients from surrounding tissues. This means that when cartilage is damaged, it takes a lot longer to heal.\r\nThere are three main types of cartilage:\r\n\r\n\thyaline cartilage which provides smooth surfaces for joint movement, like between the ribs\r\n\telastic cartilage which allows for flexibility, like in the ear\r\n\tfibrocartilage which provides strong support and absorbs shock, like in the spine.\r\n\r\n\r\nLigaments\r\nBones are connected at joints by strong bands of tissue called ligaments. They provide stability by holding the bones together and allowing you to move in a controlled way, while also stopping you from making overextended or abnormal movements that could injure you.\r\nLigaments are made up of dense, fibrous connective tissue. This gives them strength and flexibility. They support movement and keep your joints aligned.\r\n\r\nJoints\r\nThe joints connect bones to let us move. They are complex structures containing various tissue types. Joints include cartilage and ligaments.\r\nWe can classify joints based on their structure and the material that holds the bones together. There are three primary categories:\r\n\r\nfibrous joints connect the bones with dense fibrous connective tissue. Most fibrous joints are immovable.\r\ncartilaginous joints connect bones with cartilage. Most cartilaginous joints allow for slight movement.\r\nsynovial joints are characterised by a joint cavity containing synovial fluid, a slippery substance that reduces friction and helps the joint to move smoothly. All synovial joints can move freely.\r\n\r\nSynovial joints are the most common and familiar joints in the body. There are different types of synovial joints, which we have summarised in the following table.\r\n\r\n\r\n\r\nNot all joints in your body can move. There are fibrous joints in your skull called 'sutures', which are not mobile... thank goodness!\r\n[ll-image url=\"https://rmitlibrary.github.io/cdn/learninglab/illustration/subject-support/life-science/human-body-systems/skull-sutures.png\" alt=\"Skull showing coronal, lambdoid and squamous sutures.\" caption=\"Skull by  \r\nInside the bone\r\nA diagram of the end of a bone. Where the bone has been cut, there is red bone marrow. Extending through the core of the bone are blood vessels. The outer layer of bone is labelled \"Compact bone\" which is dense. The inner area of bone is labelled \"Spongy bone\", which is porous, as shown by the tiny holes throughout the structure.\r\n \r\nBones in the skeletal system\r\nA diagram of a labelled skeleton.\r\n\r\n\tThe head of the skeleton, or \"Skull\", is labelled. It consists of the top part labelled \"Cranium\" and the jaw, labelled \"Mandible\".\r\n\tThe \"Spinal column\" extends from the head down to the pelvis. It consists of the \"Cervical vertebrae\", which is the neck portion, the \"Thoracic vertebrae\", which starts at the shoulders and extends down to the bottom of the ribs, the \"Lumbar region\", which is the lower back, and the \"Sacrum\" and \"Coccyx\" which are located in the pelvis.\r\n\tThe collarbone is labelled \"Clavicle\". It connects the body to the arms.\r\n\tThe \"Scapula\" is the shoulder blade.\r\n\tThe \"Sternum\" is the breastplate from which the ribs extend.\r\n\tThe \"Ribs\" are curved bones that surround the core of the body.\r\n\tThe \"Humerus\" is the bone in the arm between the shoulder and the forearm.\r\n\tThe forearm is made up of the \"Ulna\" and \"Radius\", which are the bones on the outside and inside of the arm, respectively.\r\n\tThe \"Carpals\" are the bones in the wrist. The \"Metacarpals\" are the bones in the hand. The \"Phalanges\" are the bones in the fingers.\r\n\tThe \"Pelvic girdle\" is the main pelvic bone.\r\n\tThe \"Femur\" is the thigh bone.\r\n\tThe \"Patella\" is the kneecap.\r\n\tThe \"Tibia\" is the larger bone in the lower leg and the \"Fibula\" is the smaller bone.\r\n\tThe \"Tarsals\" are the bones in the ankles. The \"Metatarsals\" are the bones in the foot. The \"Phalanges\" are the bones in the toes.\r\n\r\n \r\nJoints in the skull\r\nA diagram of the skull showing the joints holding together the plates.\r\n\r\n\tCoronal suture: the joint between the front skull bone and the pair of bones in the front roof of the skull\r\n\tSquamous suture: the joint between side of the skull and the pair of bones in the back roof of the skull\r\n\tLambdoid suture: the joint between the back skull bone and the pair of bones in the back roof of the skull\r\n\r\n","n":0.031},[{"v":"Indigenous knowledges","i":4,"n":0.707},{"v":"Musculoskeletal system","i":3,"n":0.707},{"v":"Anatomy","i":2,"n":1},{"v":"Physiology","i":1,"n":1},{"v":"Human body systems","i":0,"n":0.577}]]},{"i":25,"$":[{"v":"Circulatory system","n":0.707},{"v":"\r\nClick to reveal more detail about each component.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nBlood vessels\r\nThe blood vessels are the network of pathways that carry blood throughout the body. There are three main types.\r\n\r\n\tArteries generally transport oxygen-rich blood away from the heart to the body's tissues.\r\n\tVeins usually carry oxygen-depleted blood back to the heart for reoxygenation.\r\n\tCapillaries are the tiny vessels where gas exchange occurs between the blood and tissues.\r\n\r\nIn the figure shown, arteries branch into smaller arterioles, which then branch into capillaries. The arteries and veins join at the capillary bed, then transition into venules, which then join up to form veins.\r\n\r\nTogether, these vessels ensure that blood flows efficiently, delivering vital substances and removing waste from the body.\r\n\r\nThe heart\r\nThe heart is a muscular organ that pumps blood throughout the body to distribute oxygen, nutrients and hormones.\r\n  \r\n\r\n\r\nIt has four chambers:\r\n\r\n\tthe atria \u2013 two upper chambers\r\n\tthe ventricles \u2013 two lower chambers.\r\n\r\n\r\nThe heart works in a coordinated way, with the right side of the heart involved in the pulmonary circulation and the left side involved in the systemic circulation.\r\n\r\nGas exchange occurs at the capillary beds, which are networks of tiny blood vessels where oxygen is delivered to cells and carbon dioxide is picked up for removal. These capillary beds are most dense around major organs, like the heart, brain, liver and kidneys.  \r\n\r\n\r\n\r\n\r\n\r\n\r\n The atria receives blood and the ventricles pump it out.\r\n\r\n\r\nBlood in the circulatory system is kept moving in one direction by valves. The heart's natural pacemaker, known as the sinoatrial node regulates the heartbeat, coordinating contractions of the heart muscles to keep a steady rhythm.\r\n\r\nWatch this video to learn more about the heart and see it in action.\r\n\r\n\r\n\r\nSummary\r\nThe components of the circulatory system and their functions are outlined in the table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tComponent\r\n\t\t\tFunction\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tBlood\r\n\t\t\tTransports oxygen, nutrients and hormones around the body\r\n\t\t\r\n\t\t\r\n\t\t\tBlood vessels\r\n\t\t\tCarry blood throughout the body\r\n\t\t\r\n\t\t\r\n\t\t\tHeart\r\n\t\t\tPumps blood around the body\r\n\t\t\r\n\t\r\n\r\n\r\nExercise\r\nSee how well you understand the circulatory system with a quick quiz.\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nBlood composition\r\nA diagram of a blood sample tube with the composition of blood shown.\r\n\r\n\tPlasma, 55%\r\n\t\r\n\t\tProteins, 4%\r\n\t\tWater, 50%\r\n\t\tOther solutes, 1%\r\n\t\r\n\tWhite blood cells and platelets, less than 1%\r\n\t\r\n\t\tPlatelets, less than 1%\r\n\t\tLeukocytes, less than 1%\r\n\t\r\n\tRed blood cells, 45%\r\n\t\r\n\t\tErythrocytes, 45%\r\n\t\r\n\r\n \r\nBlood vessels\r\nA diagram of the blood vessels showing how they are structured and connected.\r\n\r\n\tAn \"Artery\" is shown on the left. It branches into \"Arterioles\" and then \"Capillaries\".\r\n\tAn arrow towards the artery shows the flow of blood from the heart.\r\n\tA \"Vein\" is shown on the right. It branches into \"Venules\", and then also \"Capillaries\". \r\n\tAn arrow away from the vein shows the flow of blood to the heart.\r\n\tThe capillaries meet in the middle to form a continuous network of vessels.\r\n\r\n \r\nAn interactive three-dimensional model of a beating human heart. A section of the front of the heart has been removed to show the atria, ventricles and valves inside. The walls of the heart are shown expanding and relaxing as the muscles pump and valves open and close to push blood around the body.\r\n \r\nChambers of the heart\r\nA diagram of the heart with the chambers visible.\r\n\r\n\tRight atrium\r\n\tRight ventricle\r\n\tLeft atrium\r\n\tLeft ventricle\r\n\r\n \r\nBlood circulation\r\nA diagram showing the pathway of blood circulation. Blue pathways leading to the heart show deoxygenated blood and red pathways leading from the heart show oxygenated blood. The capillary beds are shown in purple to indicate gas exchange.\r\nThe \"Pulmonary circulation\" involves the lungs, heart and pulmonary artery and vein. There is an arrow from the heart through the pulmonary artery to the lungs, then back to the heart via the pulmonary vein.\r\nThe \"Systemic circulation\" involves the other major organs in the body. Arrows show movement of blood from the heart through the aorta to the capillary beds in the upper body, then back into the heart through the vena cava.\r\nBlood from the aorta also goes to the liver through the hepatic artery, then returns to the heart through the hepatic vein. Blood goes to the stomach and intestines, which gets filtered into the liver through the hepatic portal vein.\r\nThe kidneys receives blood from the renal artery, which is returned to the heart via the renal vein.\r\nThe lower body also has arteries and veins that transport blood to and from the lower body.\r\n \r\nVeins are blood vessels containing blood flowing to the heart, while arteries have blood flowing from the heart. The blue is representative of blood vessels containing deoxygenated blood, while the red blood vessels have oxygenated blood fresh from the lungs.\r\nNow, let's follow the path of the blood through the heart. The superior vena cava receives blood from the head, neck, upper limbs and chest. Meanwhile, the inferior vena cava receives blood from the trunk, viscera and lower limbs. Both the superior and inferior vena cava end up in the right atrium, one of the four chambers of the heart.\r\nThe heart not only has four chambers, it also has four valves. The purpose of the valves is to keep blood moving in the right direction and not flow backwards.\r\nBlood exits the right atrium through the tricuspid valve, so called because it has three flaps, and enters the right ventricle. The blood exits the right ventricle through the pulmonary valve and enters the right pulmonary artery.\r\nAgain, it is an artery because blood is flowing away from the heart, but it is blue because it lacks oxygen.\r\nThe pulmonary artery then splits into the left and right pulmonary arteries, which go to each respective lung. In the lungs, gas exchange occurs. The blood discards carbon dioxide and picks up oxygen. \r\nNow, blood comes back from the lungs through the pulmonary veins, entering the left atrium. Next, the blood is pumped into the left ventricle through the mitral, or bicuspid valve.\r\nFinally, the oxygenated blood leaves the left ventricle through the aortic semi-lunar valuve, entering the aortic arch. \r\nThe aorta, which is the largest of all the arteries, distributes the oxygenated blood to the rest of the body. The aortic arch has three major branches, which supply the head and arms with blood. \r\nThen the aorta curls downward behind the heart, forming the descending aorta, which descends through the chest and continues down through the abdomen. In the abdomen, the descending aorta splits to the supply the pelvis and legs with blood.\r\n","n":0.031},[{"v":"Circulatory system","i":3,"n":0.707},{"v":"Anatomy","i":2,"n":1},{"v":"Physiology","i":1,"n":1},{"v":"Human body systems","i":0,"n":0.577}]]},{"i":26,"$":[{"v":"Respiratory system","n":0.707},{"v":"The respiratory system brings oxygen into the body and removes carbon dioxide. Understanding how it works helps us see how the body breathes and keeps cells healthy. Use this resource to learn about the organs and functions of the respiratory system.\r\n\r\nCellular respiration\r\nWe need oxygen to survive. It is an important element used by the body during a process called cellular respiration. Cellular respiration is a chemical reaction where glucose is converted to energy in the form of adenosine triphosphate (ATP). This occurs in the mitochondria, as you have learned when looking at organelles.\r\n\r\nATP is used by cells to carry out their functions. Everything from muscle contraction and nerve impulse transmission, to keeping body temperature steady and building new tissues relies on ATP.\r\nCellular respiration is given by the following word and chemical equations:\r\n\\[\\textrm{glucose}+\\textrm{oxygen}\\rightarrow\\textrm{carbon dioxide}+\\textrm{water}+\\textrm{energy}\\]\\[\\ce{C6H12O6}+\\ce{6O2}\\rightarrow\\ce{6CO2}+\\ce{6H2O}+\\textrm{ATP}\\]\r\nBut how do we get oxygen to the cells? First, we have to get oxygen into the bloodstream. This is where the respiratory system comes in. \r\n\r\nThe respiratory tract\r\nWe need a respiratory system to breathe and maintain healthy oxygen levels in the body. This system of organs and tissues is contained inside the ribcage.\r\nThe respiratory tract is the pathway that inhaled (and exhaled) air follows to and from the lungs.\r\n\r\nOur respiratory tract starts with our nostrils, nasal and oral cavities, and ends with the alveoli deep in our lungs. It works with other organs of the respiratory system, like the diaphragm, to exchange oxygen and carbon dioxide.\r\nLet's look at each part of the respiratory system in more detail.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nGas exchange\r\nThe air we inhale (breathe in) contains a mixture of different gases: \r\n\t\\(78\\%\\) nitrogen, \\(\\ce{N2}\\)\r\n\t\\(21\\%\\) oxygen, \\(\\ce{O2}\\)\r\n\t\\(0.04\\%\\) carbon dioxide, \\(\\ce{CO2}\\)\r\n\t\\(1\\%\\) other gases.\r\n\r\n...but we really just need oxygen. So how do we extract just the oxygen we need and get rid of any waste gases? The answer is through a process called gas exchange. This occurs deep within the alveoli of the lungs.\r\n\r\nThis efficient exchange makes sure that our cells receive the oxygen they need for energy production, while also expelling carbon dioxide to maintain healthy function.\r\nAlthough nitrogen makes up majority of the air we breathe, it does not play a big role in gas exchange. In fact, nitrogen is an inert gas, meaning that it's not reactive. It is, however, important for us to breathe it in and out anyway, because it helps to keep our alveoli inflated and maintain the pressure balance in our lungs.\r\nThe air we exhale is made up of:\r\n\r\n\t\\(78\\%\\) nitrogen, \\(\\ce{N2}\\)\r\n\t\\(17\\%\\) oxygen, \\(\\ce{O2}\\)\r\n\t\\(4\\%\\) carbon dioxide, \\(\\ce{CO2}\\)\r\n\t\\(1\\%\\) other gases.\r\n\r\n\r\nRespiration\r\nWe need the help of our muscles to draw breaths in and push them out of our lungs. Most important are the diaphragm and intercostal muscles.\r\nThe diaphragm is a large, dome-shaped muscle found at the base of the ribcage. The intercostal muscles are small muscles found between the ribs. The following table summarises how they behave when you inhale and exhale.\r\n\r\n\t\r\n\t\t\r\n\t\t\t\r\n\t\t\tInhalation\r\n\t\t\tExhalation\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tDiaphragm\r\n\t\t\tContracts and moves downward to make more space in your chest cavity so your lungs can expand and fill with air\r\n\t\t\tRelaxes and returns to a dome shape, pushing air out of your lungs\r\n\t\t\r\n\t\t\r\n\t\t\tIntercostal muscles\r\n\t\t\tContract to raise the ribcage and expand the chest cavity\r\n\t\t\tRelax and let the ribcage move inward, helping push air out\r\n\t\t\r\n\t\r\n\r\nOne way to remember when the muscles contract and when they relax is to think of inhalation as an active process. Active movement requires muscle contraction. So, these muscles contract when we inhale and relax when we exhale.\r\n\r\nWatch this video to see how the organs and tissues of the respiratory system work together to take in oxygen and expel carbon dioxide.\r\n\r\n\r\nSummary\r\nThe components of the respiratory system and their functions are outlined in the table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tComponent\r\n\t\t\tFunction\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tNasal and oral cavities\r\n\t\t\tFilter, warm and humidify inhaled air\r\n\t\t\r\n\t\t\r\n\t\t\tPharynx\r\n\t\t\tActs as a pathway for air from the nose and mouth to the larynx\r\n\t\t\r\n\t\t\r\n\t\t\tLarynx\r\n\t\t\tDirects air into the trachea and helps produce sound for speech\r\n\t\t\r\n\t\t\r\n\t\t\tTrachea\r\n\t\t\tProvides a clear pathway for air to enter and exit the lungs\r\n\t\t\r\n\t\t\r\n\t\t\tBronchi\r\n\t\t\tCarries air from the trachea into each lung\r\n\t\t\r\n\t\t\r\n\t\t\tBronchioles\r\n\t\t\tDistribute air throughout the lungs to the alveoli\r\n\t\t\r\n\t\t\r\n\t\t\tAlveoli\r\n\t\t\tFacilitate gas exchange by allowing oxygen and carbon dioxide to move between the air and bloodstream\r\n\t\t\r\n\t\t\r\n\t\t\tDiaphragm\r\n\t\t\tContracts and relaxes to change the volume of the chest cavity to facilite inhalation and exhalation\r\n\t\t\r\n\t\t\r\n\t\t\tIntercostal muscles\r\n\t\t\tHelp expand and contract the ribcage to support breathing\r\n\t\t\r\n\t\r\n\r\n\r\nExercise\r\nSee how well you understand respiration and the components of the respiratory system with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nResearchers from the Royal Darwin Hospital in the Northern Territory of Australia and the University of Sousse in Tunisia investigated sex differences in pulmonary function among Indigenous Australian with and without chronic airway disease.\r\nThe forced vital capacity (FVC) was measured for each individual. This is the maximum amount of air that a person can force out of their lungs after taking in a deep breath. \r\n\r\nThe FVC data for males and females, without chronic airway disease only, is summarised in the table. The p-values are from two-tailed Students t-tests, with p<0.05 considered as statistically significant. The values are averages. The publication also shows the 95% confidence interval for measurements, but they are not shown on this page.\r\n\r\n\t\r\n\t\t\r\n\t\t\t\r\n\t\t\tFemale (n=179)\r\n\t\t\tMale (n=111)\r\n\t\t\tp-value\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tFVC\r\n\t\t\t2.59\r\n\t\t\t3.63\r\n\t\t\t<0.001\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n \r\nThe respiratory tract\r\nA diagram of the respiratory tract, with each component labelled.\r\n\r\n\tThe \"Nostril\" and \"Oral cavity\" are the opening points and the start of the respiratory tract.\r\n\tThe \"Nasal cavity\" sits beyond the nostrils.\r\n\tThe nasal and oral cavities connect to form the \"Pharynx\".\r\n\tThe \"Larynx\" is located in the throat.\r\n\tThe \"Trachea\" is a tube with cartilage rings that extends down into the chest.\r\n\tThe trachea branches into two lungs, labelled \"Right lung\" and \"Left lung\". It connects to the lungs through the \"Right main bronchus\" and \"Left main bronchus\".\r\n\tThe \"Diaphragm\" is a sheet of muscle that sits under the lungs.\r\n\r\n \r\nThe trachea and bronchi\r\nA diagram of the trachea and bronchi, with key components labelled.\r\n\r\n\tThe \"Larynx\" is shown a the top.\r\n\tThe \"Trachea\" extends down from the larynx. It is surrounded by cartilage.\r\n\tThe trachea branches into two \"Primary bronchi\".\r\n\tThese then branch further into \"Secondary bronchi\" and \"Tertiary bronchi\".\r\n\tThe smallest branch is labelled \"Bronchiole\".\r\n\r\n \r\nAlveoli\r\nA diagram of an alveoli at the end of a bronchiole, with key components labelled.\r\n\r\n\tThe wall of the alveoli is labelled \"Mucosal lining\", with small structures attached to the surface labelled \"Mucous gland\".\r\n\tThe \"Alveolar sacs\" split from the bronchiole through \"Alveolar ducts\".\r\n\tThe central space inside the alveoli is called the \"Atrium\", with \"Alveoli\" being the individual sacs.\r\n\tThe \"Pulmonary artery\" and \"Pulmonary vein\" extend around the outside of the alveolar sacs, forming a network labelled \"Capillary beds\".\r\n\tThe alveoli are surrounded by \"Connective tissue\".\r\n\r\n \r\nMany of us have hundreds of things on our minds at any moment, often struggling to keep track of everything we need to do. But fortunately, there's one important thing we don't have to worry about remembering: breathing.\r\nWhen you breathe, you transport oxygen to the body's cells to keep them working and clear your system of the carbon dioxide that this work generates.\r\nBreathing, in other words, keeps the body alive. So, how do we accomplish this crucial and complex task without even thinking about it? The answer lies in our body's respiratory system.\r\nLike any machinery, it consists of specialised components, and requires a trigger to start functioning. Here, the components are the structures and tissues making up the lungs, as well as the various other respiratory organs connected to them. And to get this machine moving, we need the autonomic nervous system, our brain's unconscious control centre, for the vital functions.\r\nAs the body prepares to take in oxygen-rich air, this system sends a signal to the muscles around your lungs, flattening the diaphragm and contracting the muscles between your ribs to create more space for the lungs to expand. Air then whooshes into your nose and mouth, through your trachea, and into the bronchi that split at the trachea's base, with one entering each lung.\r\nLike tree branches, these small tubes divide into thousands of tinier passages called bronchioles. It's tempting to think of the lungs as huge balloons, but instead of being hollow, they're actually spongy inside, with the bronchioles running throughout the parenchyma tissue.\r\nAt the end of each bronchiole is a little air sac called an alveolus, wrapped in capillaries full of red blood cells containing special proteins called haemoglobin. \r\nThe air you've breathed in fills these sacs, causing the lungs to inflate. Here is where the vital exchange occurs.\r\nAt this point, the capillaries are packed with carbon dioxide, and the air sacs are full of oxygen. But due to the basic process of diffusion, the molecules of each gas want to move to a place where there's a lower concentration of their kind.\r\nSo as oxygen crosses over to the capillaries, the haemoglobin grabs it up, while the carbon dioxide is unloaded into the lungs. The oxygen-rich haemoglobin is then transported throughout the body via the bloodstream.\r\nBut what do our lungs do with all that carbon dioxide? Exhale it, of course.\r\nThe autonomic nervous system kicks in again, causing the diaphragm to ball up, and the intercostal muscles to relax, making the chest cavity smaller and forcing the lungs to compress. The carbon dioxide-rich air is expelled, and the cycle begins again.\r\nSo that's how these spongy organs keep our bodies efficiently supplied with air. Lungs inhale and exhale between 15 and 25 times a minute, which amounts to an incredible 10,000 litres of air each day. That's a lot of work, but don't sweat it. Your lungs and your autonomic nervous system have got it covered.\r\n","n":0.025},[{"v":"Respiratory system","i":3,"n":0.707},{"v":"Anatomy","i":2,"n":1},{"v":"Physiology","i":1,"n":1},{"v":"Human body systems","i":0,"n":0.577}]]},{"i":27,"$":[{"v":"Common assessment mistakes","n":0.577},{"v":"On this page, we'll look at a few common mistakes that students make on assessments and give you some tips to avoid them.\r\n\r\nLack of proofreading\r\nWe often make spelling or grammatical mistakes or do not express ourselves clearly the first time we write something. This could lead to answers being marked as unsatisfactory.\r\n\r\nWhat to do\r\n\r\n\tReread your answers: \r\n\tBefore you hit submit, give yourself some time to reread your answers and make sure what you have written is clear and free from spelling and grammatical mistakes.\r\n\tKnow your common errors:\r\n\tDo you often leave the \"s\" off plurals? Sometimes forget to put \"ed\" on past tense verbs? Do you occasionally forget to capitalise? Look out for these mistakes in your answers.\r\n\tRead your answers out loud:\r\n\tThis is a technique that is commonly used which helps you catch mistakes that you may miss when you read the text in your head. If you are doing the assessment under exam conditions, you could try mouthing the words without making any noise.\r\n\tRead your writing backwards:\r\n\tThis is a great way to catch spelling errors because you are not concentrating on meaning or grammar. It allows you to look at each word in isolation. So, if you sometimes have issues with spelling, this is a great technique.\r\n\r\nTry this quiz to test your proofreading skills.\r\n\r\n\t\t\r\nMisinterpreting the task\r\nIf you don't read assessment questions and instructions carefully before answering, you could accidentally answer the question in the wrong way.\r\n\r\nLet's explore an example:\r\nDiego needs to answer the following:\r\n\"Explain the process of installing a sink trap and its function in plumbing.\"\r\n\r\nHe answers:\r\n\"A sink trap is a U-shaped section of pipe located beneath the sink. Its primary function is to prevent sewer gases from entering the building by trapping a small amount of water in the bend of the pipe. This water acts as a barrier to block gases from coming back through the drain.\"\r\n\r\nIn this answer, Diego has misinterpreted the question. He has answered what a sink trap is, not how it can be installed. Therefore, his answer is not satisfactory.\r\n\r\nWhat to do\r\nHere are some tips for understanding questions and prompts:\r\n\t\r\n\twork through the other pages in this section to understand assessment instructions, beginning with VE assessments\r\n\tread the questions or prompts carefully, and underline the key words that tell you what you need to do\r\n\tafter you have written your answer, reread the question or prompt to check if you have answered it correctly.\r\n\r\n\r\nNot allocating enough time\r\nIn timed assessments, using your time wisely is very important. If you spend too long on some questions, you might rush others or not finish.\r\n\r\nWhat to do\r\nHere are some steps you can take to use your time effectively:\r\n\r\n\tread your assessment carefully before you begin so that you understand each question\r\n\tidentify questions that need longer answers and those that need short answers\r\n\teven when a longer answer is needed, choose your words carefully. Writing unnecessarily long answers may take up a lot of time.\r\n\tFor more information, check the Understanding instruction words page which explains when to provide simple answers and when to give more details.\r\n\r\n\r\nTest your knowledge\r\nJoanna is a construction student. She is taking an assessment that she has one hour to complete. Decide if her answers are  appropriate for the following assessment questions.\r\n\r\n","n":0.043},[{"v":"Assignment planning","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":28,"$":[{"v":"Digestive system","n":0.707},{"v":"The digestive system is responsible for converting food into a form that our bodies can use for energy and growth. Knowing how the digestive system works helps us understand how the body processes food and absorbs vital nutrients. Use this resource to explore the organs and functions that make up the digestive system, and how Indigenous peoples made cycad seeds safe to eat.\r\n\r\nA lot of the nutrients we need to survive cannot be produced by the body, so we need to obtain them from food. But how does, say, the egg in our breakfast bagel get converted into proteins that we can use? This is where the digestive system comes in.\r\nThe digestive system:\r\n\r\n\ttakes in food\r\n\tbreaks down food into nutrients\r\n\tbrings nutrients from food into the bloodstream\r\n\tremoves undigested food from the body.\r\n\r\n The digestive system consists of the digestive tract and accessory organs.\r\n\r\n\r\n\r\nThe digestive tract\r\n\r\nThe digestive tract (also known as the alimentary or gastrointestinal (GI) tract) is a long tube that starts at the mouth and ends at the anus, and also includes the mouth, oesophagus, stomach, small intestine and large intestine. The teeth, tongue, salivary glands, liver, gallbladder and pancreas are known as accessory organs.\r\nLet's explore what each of these organs does during digestion.\r\n\r\n\r\n\r\nDigestion\r\nDigestion is the process by which the body breaks down food into smaller components that can be absorbed and used for energy, growth and repair.\r\nThere are two ways we can do this: mechanically or chemically.\r\n\r\n\tMechanical digestion is the physical breakdown of food into smaller pieces. It occurs purely because of the forces created by motion.\r\n\tChemical digestion is when chemicals like enzymes or acids participate in chemical reactions that break down molecules into simpler substances.\r\n\r\nMechanical and chemical digestion happen at different stages of food digestion. Let's look at the steps involved.\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nSummary\r\nThe key components of the digestive tract and their functions are outlined in the table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tComponent\r\n\t\t\tFunction\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tMouth\r\n\t\t\tChews food and mixes it with saliva to begin digestion\r\n\t\t\r\n\t\t\r\n\t\t\tOesophagus\r\n\t\t\tTransports food from the mouth to the stomach\r\n\t\t\r\n\t\t\r\n\t\t\tStomach\r\n\t\t\tMixes food with gastric juices to break it down further\r\n\t\t\r\n\t\t\r\n\t\t\tSmall intestine\r\n\t\t\tUses enzymes and bile to further digest food and then absorb most nutrients\r\n\t\t\r\n\t\t\r\n\t\t\tLarge intestine\r\n\t\t\tAbsorbs water and forms waste for excretion\r\n\t\t\r\n\t\t\r\n\t\t\tAnus\r\n\t\t\tExcretes waste from the body\r\n\t\t\r\n\t\r\n\r\n\r\nThe components of the biliary tract and their functions are outlined in the following table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tComponent\r\n\t\t\tFunction\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tLiver\r\n\t\t\tProduces bile and processes nutrients absorbed from the intestine\r\n\t\t\r\n\t\t\r\n\t\t\tGallbladder\r\n\t\t\tStores bile produced by the liver\r\n\t\t\r\n\t\t\r\n\t\t\tPancreas\r\n\t\t\tProduces digestive enzymes and bicarbonate to assist digestion\r\n\t\t\r\n\t\t\r\n\t\t\tBile ducts\r\n\t\t\tTransports bile from the gallbladder to the small intestine to assist digestion\r\n\t\t\r\n\t\r\n\r\nExercise\r\nSee how well you understand digestion and the components of the digestive system with a quick quiz.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nThe digestive system\r\nA diagram of the digestive system.\r\n\r\n\tThe digestive system starts with the mouth, where salivary glands and the tongue are found.\r\n\tThe epiglottis is the flap at the entry of the digestive tract.\r\n\tThe oesophagus is the channel from the mouth to the stomach.\r\n\tThe liver, pancreas and gallbladder sit around the stomach.\r\n\tThe first part of the small intestine is the duodenum.\r\n\tThe next part of the small intestine is the ileum.\r\n\tAt the start of the large intestine is the appendix and caecum.\r\n\tThe large intestine then leads to the rectum.\r\n\tFinally, the anus is the exit point of the digestive system.\r\n\r\n \r\nPeristalsis\r\nA diagram showing peristalsis acting on a section of the oesophagus. On the left, the food is near the top of the oesophagus. The muscles directly above the food contract and the muscles directly under the food relax. A downward arrow shows the direction of food movement.\r\nOn the right, the food has moved further down the oesophagus. The muscles directly above the food contract and the muscles directly under the food relax. A downward arrow shows the food direction.\r\n \r\nWe don't have chunks of cake or bits of burger floating around our bloodstream. So how do we get the energy from our food to our cells? Two things have to happen before we can make use of our lunch: it needs to be broken down into smaller pieces or digested and then it needs to be absorbed into the bloodstream.\r\nDigestion takes place in lots of places along the alimentary canal, and absorption happens in the small intestine, which is found just after the stomach along the length of the digestive system. To learn more about the digestive system, watch this video [Digestion - What Is It?], and to learn more about absorption, watch this video [Small intestine and food absorption].\r\nIn this video, we're going to look at the enzymes involved in digestion in more detail. The large molecules which make up our food, like lipids, proteins and carbohydrates are too big to be moved into our blood. So, they need to be digested into smaller molecules by physical processes like chewing, and chemically by special proteins called enzymes. \r\nDifferent types of enzymes digest the different types of food. So let's look at the food groups in turn. First, carbohydrates.\r\nFoods like rice and pasta are made of carbohydrates. The simplest carbohydrates are sugars which can be joined into big chains to make complex carbohydrates like starch. Carbohydrates are digested by carbohydrase enzymes such as amylase. Amylase is a special type of carbohydrase which breaks down starch, a big carbohydrate, into smaller molecules. These can then be broken down further into glucose which is small enough to be moved into the blood. Amylase is found in your saliva which is where carbohydrates will first start to be broken down chemically, and another amylase is released into the small intestine from the pancreas, so much further down your digestive system. \r\nNow, let's look at proteins. Proteins [are] found in meat, fish, beans and pulses. They are made up of amino acids and are digested by protease enzymes. The protease enzymes break down proteins into amino acids. Protease called pepsin breaks down proteins in the stomach, but this doesn't work once the food moves into the small intestine. In the small intestine, the conditions are different, so a different protease called trpysin is released into the small intestine from the pancreas for continued protein digestion.\r\nWhat conditions do you think might be different in the small intestine compared to the stomach? Whilst the stomach is very acidic and has a pH of about 2, the small intestine has a higher pH of about 8. Protease enzymes which work well at pH 2 don't work at pH 8, which is why different protease enzymes are needed in the small intestine.\r\nAnd the last group of food, lipids. Lipids digested by enzymes called lipases into glycerol and fatty acids, but it's not an easy job. It requires a process called emulsification to take place first. Emulsification breaks the lipids into smaller droplets. The smaller droplets have a larger surface area for the lipase enzymes to work on. To learn more about bile and emulsification, watch this video [Bile and Emulsification].\r\nSo, to recap. The main enzymes involved in digestion are:\r\n\r\ncarbohydrase called amylase is released in the saliva; this digests starch into smaller sugars\r\na protease called pepsin digest proteins in the stomach\r\nthen a protease called trypsin further digests proteins into amino acids in the small intestine\r\na carbohydrase called amylase and lipase are also released into the small intestine from the pancreas. They digest carbohydrates into sugars and lipids into fatty acids and glycerol.\r\nThese molecules are small enough to be absorbed into the blood.\r\n \r\nCan you believe your digestive system is as long as a bus? Yep, it's about nine metres long, all wrapped up inside your body. A lot of things go on in your digestive system, not just digestion. What else do you think happens?\r\nDigestion goes hand in hand with absorption which happens in your small intestine, which is what we're going to look at in this video. \r\nDigestion is the breakdown of food, from large insoluble molecules into small molecules which can be used by your body. Digestion involves many organs before it reaches the small intestine. For more information on the roles of these organs in the digestive system, watch this video [Digestion - What Is It?]. Enzymes and other chemical such as bile break down food; for more on the role of enzymes and bile in digestion, watch these videos [Digestive enzymes; Bile and Emulsification].\r\nAs they move through the digestive system, carbohydrates are broken down into sugars, proteins are broken down into amino acids and fats are broken down into fatty acids and glycerol. But the job of the digestive system doesn't stop at digestion. For our body to make use of these small, soluble molecules, we need to absorb them into our bloodstream. This happens in the small intestine.\r\nThe small intestine is made up of lots of tiny, finger-like projections called villi. The villi increase the surface area of the small intestine. The larger the surface area, the more absorption that can take place.\r\nEach villus has a tiny blood vessel called a capillary and another vessel called a lacteal. The food molecules pass from the villi of the small intestine into either the blood capillary or the lacteal. \r\nOnce absorbed into the blood vessel in the small intestine, the digestive food molecules can then be transported around the body to where they are needed.\r\nSo why did those large molecules need to be broken down in the first place? The small intestine is kind of like a sieve; it has small holes in it. Small molecules are able to pass through those holes, whilst large molecules cannot. So carbohydrates, proteins and fats are all too large to pass through the holes in the small intestine, but sugars, amino acids, fatty acids and glycerol are all small enough to pass through.\r\nSugars and amino acids pass through the holes into the blood capillary, while fatty acids and glycerol pass into the lacteal of the villi. These small molecules move by diffusion, moving from a high to a low concentration. For example, the small intestine has a high concentration of sugar molecules, whereas the blood supplying the small intestine has a low concentration of sugar. So the sugar moves down the concentration gradient, from a high concentration in the small intestine, through the holes in the villi, into the blood capillary where it's in low concentration. Sugar, amino acids, fatty acids and glycerol all use diffusion to move into the blood within the small intestine. \r\nSo after digestion breaks down large molecules into small molecules, absorption of the small molecules happens inside our small intestine by diffusion into the blood vessels. What an amazing system!\r\n","n":0.024},[{"v":"Indigenous knowledges","i":4,"n":0.707},{"v":"Digestive system","i":3,"n":0.707},{"v":"Anatomy","i":2,"n":1},{"v":"Physiology","i":1,"n":1},{"v":"Human body systems","i":0,"n":0.577}]]},{"i":29,"$":[{"v":"Human body systems","n":0.577},{"v":"The study of human body systems looks at how different organs and tissues work together keep us healthy. Understanding these systems helps us see how the body functions and how different parts support life. Use this resource to learn how systems like the circulatory, respiratory, and digestive systems work together to keep the body running smoothly.\r\n\r\nOverview\r\n\r\n\t\tDigestive system \u2013 how does the digestive system help extract nutrients from the food we eat?\r\n\t\tCentral nervous system \u2013 how do the brain and spinal cord coordinate responses?\r\n\t\tPeripheral nervous system \u2013 how do the neurones in the body work with the central nervous system to mount responses?\r\n\t\tSkeletal system \u2013 what structures provide the body with shape and protection?\r\n\t\tMuscular system \u2013 what structures help the body move?\r\n\t\tRespiratory system \u2013 how do we obtain the oxygen we need to survive?\r\n\t\tCirculatory system \u2013 how does the body deliver oxygen and nutrients to the cells?\r\n\t\tEndocrine system \u2013 how do hormones work?\r\n\t\tExcretory system \u2013 how do we get rid of waste from the body?\r\n\t\tHomeostasis \u2013 how does the body maintain a stable internal environment?\r\n\r\n\r\n","n":0.076},[{"v":"Disease","i":4,"n":1},{"v":"Digestive system","i":3,"n":0.707},{"v":"Anatomy","i":2,"n":1},{"v":"Physiology","i":1,"n":1},{"v":"Human body systems","i":0,"n":0.577}]]},{"i":30,"$":[{"v":"Glossary of life science terms","n":0.447},{"v":"Life science uses various terms and concepts to explain the functions and processes of living organisms. Use this glossary to understand any unfamiliar terminology you come across.\r\n\r\nTo help you navigate, the glossary terms have been separated based on the topic they fall under. If you are unsure, just hit CTRL+F to find the term you are looking for!\r\n\r\nStructure and function of cells\r\nThe table outlines the definitions for key terms in Structure and function of cells.\r\n\r\n\t\r\n\t\t\r\n\t\t\tTerm\r\n\t\t\tDefinition\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tactive transport\r\n\t\t\ta transport process that requires energy in which molecules move from a region of lower concentration to an area of higher concentration\r\n\t\t\r\n\t\t\r\n\t\t\tanaphase\r\n\t\t\tthe stage of mitosis in which sister chromatids are pulled apart by spindle fibres to opposide sides of the cell\r\n\t\t\r\n\t\t\r\n\t\t\tantiporter\r\n\t\t\ta type of carrier protein that transports two molecules across the cell membrane at a time, in opposite directions\r\n\t\t\r\n\t\t\r\n\t\t\tbinary fission\r\n\t\t\tthe process by which a single parent reproduces to form offspring\r\n\t\t\r\n\t\t\r\n\t\t\tcell cycle\r\n\t\t\ta series of events that a cell goes through as it grows and divides\r\n\t\t\r\n\t\t\r\n\t\t\tcell membrane\r\n\t\t\ta flexible phospholipid bilayer with embedded proteins and other molecules; also known as the plasma membrane\r\n\t\t\r\n\t\t\r\n\t\t\tcell theory\r\n\t\t\tthe cell is the most basic unit of life, all living organisms are composed of one or more cells and all cells arise only from pre-existing cells\r\n\t\t\r\n\t\t\r\n\t\t\tcell wall\r\n\t\t\tthe rigid outer layer that provides structural support and protection for plant cells\r\n\t\t\r\n\t\t\r\n\t\t\tcellular respiration\r\n\t\t\tthe chemical process that converts glucose and oxygen into energy in the form of ATP for the cell to function\r\n\t\t\r\n\t\t\r\n\t\t\tcentrosome\r\n\t\t\tstructures found near the cell nucleus that are important for cell division\r\n\t\t\r\n\t\t\r\n\t\t\tchlorophyll\r\n\t\t\tthe green pigment that captures light energy\r\n\t\t\r\n\t\t\r\n\t\t\tchloroplast\r\n\t\t\tthe organelle in which photosynthesis occurs\r\n\t\t\r\n\t\t\r\n\t\t\tchromatin\r\n\t\t\ta complex made of DNA and proteins\r\n\t\t\r\n\t\t\r\n\t\t\tcisternae\r\n\t\t\tstacked membrane structures that make up the Golgi apparatus\r\n\t\t\r\n\t\t\r\n\t\t\tcristae\r\n\t\t\tthe structures that the inner membrane of the mitochondria are folded into\r\n\t\t\r\n\t\t\r\n\t\t\tcytokinesis\r\n\t\t\tthe stage after mitosis in which the cell's cytoplasm divides to form two distinct daughter cells\r\n\t\t\r\n\t\t\r\n\t\t\tcytoplasm\r\n\t\t\tthe gel-like filling inside a cell\r\n\t\t\r\n\t\t\r\n\t\t\tcytoskeleton\r\n\t\t\ta network of protein fibres that help organiste the contents of a cell\r\n\t\t\r\n\t\t\r\n\t\t\tendoplasmic reticulum\r\n\t\t\ta network of membranes involved in protein and lipid synthesis; abbreviated to ER\r\n\t\t\r\n\t\t\r\n\t\t\teukaryotic cell\r\n\t\t\ta cell that has a well-defined nucleus and membrane-bound organelles\r\n\t\t\r\n\t\t\r\n\t\t\tfacilitated diffusion\r\n\t\t\ta passive transport process in which molecules move from an area of higher concentration to an area of lower concentration with the help of special proteins\r\n\t\t\r\n\t\t\r\n\t\t\tfluid mosaic model\r\n\t\t\tstates that the cell membrane is a dynamic, flexible, semi-permeable phospholipid bilayer with embedded proteins\r\n\t\t\r\n\t\t\r\n\t\t\tG1 stage\r\n\t\t\tthe stage of interphase in which the cell grows and carries out regular metabolic functions\r\n\t\t\r\n\t\t\r\n\t\t\tG2 stage\r\n\t\t\tthe stage of interphase in which the cell continues to grow and produce what it needs for mitosis\r\n\t\t\r\n\t\t\r\n\t\t\tGolgi apparatus\r\n\t\t\tthe organelle that modifies, sorts and packages proteins and lipids for transport\r\n\t\t\r\n\t\t\r\n\t\t\tgrana\r\n\t\t\tstacks of thylakoids\r\n\t\t\r\n\t\t\r\n\t\t\tinterphase\r\n\t\t\tthe phase in which the cell grows and its genetic material is replicated\r\n\t\t\r\n\t\t\r\n\t\t\tlumen\r\n\t\t\tchannel\r\n\t\t\r\n\t\t\r\n\t\t\tlysosome\r\n\t\t\tthe organelle that contains digestive enzymes to break down excess or old cell parts, food particles and invading viruses or bacteria\r\n\t\t\r\n\t\t\r\n\t\t\tmatrix\r\n\t\t\t(in mitochondria) the space inside the cristae\r\n\t\t\r\n\t\t\r\n\t\t\tmetaphase\r\n\t\t\tthe stage of mitosis in which the chromosomes align at the equator of the cell\r\n\t\t\r\n\t\t\r\n\t\t\tmitochondria\r\n\t\t\tthe powerhouse of the cell that produces energy in the form of ATP for the cell to function\r\n\t\t\r\n\t\t\r\n\t\t\tmitosis\r\n\t\t\tthe process by which a cell divides into two identical daughter cells\r\n\t\t\r\n\t\t\tmulticellular\r\n\t\t\tan organism consisting of multiple cells\r\n\t\t\r\n\t\t\r\n\t\t\tnuclear envelope\r\n\t\t\ta double membrane that surrounds the nucleus\r\n\t\t\r\n\t\t\r\n\t\t\tnucleolus\r\n\t\t\ta dense area within the nucleus where the ribosomal RNA is found\r\n\t\t\r\n\t\t\r\n\t\t\tnucleoplasm\r\n\t\t\tthe gel-like filling inside a nucleus\r\n\t\t\r\n\t\t\r\n\t\t\tnucleus\r\n\t\t\tthe control centre of the cell that houses the cell's genetic material\r\n\t\t\r\n\t\t\r\n\t\t\torganelle\r\n\t\t\ta specialised structure in eukaryotic cells that performs a distinct function\r\n\t\t\r\n\t\t\r\n\t\t\tphotosynthesis\r\n\t\t\tthe chemical process that converts carbon dioxide, water and energy from sunlight into oxygen and glucose\r\n\t\t\r\n\t\t\r\n\t\t\tprokaryotic cell\r\n\t\t\ta cell that lacks a cell nucleus and membrane-bound organelles\r\n\t\t\r\n\t\t\r\n\t\t\tprophase\r\n\t\t\tthe stage of mitosis in which the chromosomes condense and the nuclear envelope begins to dissolve\r\n\t\t\r\n\t\t\r\n\t\t\trate of diffusion\r\n\t\t\tthe amount of material that is transported over a specific amount of time\r\n\t\t\r\n\t\t\r\n\t\t\tribosome\r\n\t\t\tthe organelle that reads instructions from the messenger RNA to assemble proteins \r\n\t\t\r\n\t\t\r\n\t\t\trough endoplasmic reticulum\r\n\t\t\tendoplasmic reticulum with ribosomes attached; involved in protein folding and modification\r\n\t\t\r\n\t\t\r\n\t\t\tS stage\r\n\t\t\tthe stage of interphase in which DNA replication occurs\r\n\t\t\r\n\t\t\r\n\t\t\tsimple diffusion\r\n\t\t\ta passive transport process in which molecules move from an area of higher concentration to an area of lower concentration\r\n\t\t\r\n\t\t\r\n\t\t\tsmooth endoplasmic reticulum\r\n\t\t\tendoplasmic reticulum with no ribosomes attached; involved in lipid synthesis and detoxification\r\n\t\t\r\n\t\t\r\n\t\t\tstoma\r\n\t\t\tthe gel-like fluid surrounding the grana\r\n\t\t\r\n\t\t\r\n\t\t\tsurface area\r\n\t\t\tthe total area of the cell's outer membrane\r\n\t\t\r\n\t\t\r\n\t\t\tsurface area to volume ratio\r\n\t\t\ta measure that compares the size of a cell's surface to its volume\r\n\t\t\r\n\t\t\r\n\t\t\tsymporter\r\n\t\t\ta type of carrier protein that transports two molecules across the cell membrane at a time, in one direction\r\n\t\t\r\n\t\t\r\n\t\t\ttelophase\r\n\t\t\tthe stage of mitosis in which the nuclear envelope reforms around the separated chromosomes and they recondense\r\n\t\t\r\n\t\t\r\n\t\t\tthylakoid\r\n\t\t\ta membrane in the chloroplast\r\n\t\t\r\n\t\t\r\n\t\t\tunicellular\r\n\t\t\tan organism consisting of a single cell\r\n\t\t\r\n\t\t\r\n\t\t\tuniporter\r\n\t\t\ta type of carrier protein that transports one molecule across the cell membrane in one direction\r\n\t\t\r\n\t\t\r\n\t\t\tvacuole\r\n\t\t\ta storage organelle that stores water, nutrients and waste products in animal cells, and nutrients and waste products in animal cells\r\n\t\t\r\n\t\t\r\n\t\t\tvesicle\r\n\t\t\ta small compartment into which molecules can be packaged into for transport\r\n\t\t\r\n\t\t\r\n\t\t\tvolume\r\n\t\t\tthe amount of space inside the cell\r\n\t\t\r\n\t\r\n\r\n\r\nMolecular biology\r\nThe table outlines the definitions for key terms in Molecular biology.\r\n\r\n\t\r\n\t\t\r\n\t\t\tTerm\r\n\t\t\tDefinition\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\t5' capping\r\n\t\t\ta post-transcriptional modification that involves adding a cap to the 5' end of the mRNA\r\n\t\t\r\n\t\t\r\n\t\t\tacetylation\r\n\t\t\tthe process in which an acetyl group is added\r\n\t\t\r\n\t\t\r\n\t\t\tactive site\r\n\t\t\tthe site on an enzyme in which the substrates bind\r\n\t\t\r\n\t\t\r\n\t\t\taminoacyl site\r\n\t\t\tthe A site of the tRNA where the tRNA with the correct anticodon binds\r\n\t\t\r\n\t\t\r\n\t\t\tanticodon\r\n\t\t\ta complementary codon to the sense codon\r\n\t\t\r\n\t\t\r\n\t\t\tcarbohydrate\r\n\t\t\ta macromolecule made up of simple sugar monomers\r\n\t\t\r\n\t\t\r\n\t\t\tcatalyst\r\n\t\t\tsomething that speeds up a chemical reaction\r\n\t\t\r\n\t\t\r\n\t\t\tcentral dogma\r\n\t\t\ta fundamental theory in molecular biology that describes the flow of genetic material\r\n\t\t\r\n\t\t\r\n\t\t\tcentromere\r\n\t\t\tthe region of a chromatid in which the short and long arm meet\r\n\t\t\r\n\t\t\r\n\t\t\tchromatid\r\n\t\t\tthe structure of a chromosome which has a short arm and a long arm\r\n\t\t\r\n\t\t\r\n\t\t\tchromatin\r\n\t\t\ta fibre of coiled solenoids\r\n\t\t\r\n\t\t\r\n\t\t\tchromosome\r\n\t\t\ta condensed structure formed from the packaging and coiling of DNA\r\n\t\t\r\n\t\t\r\n\t\t\tcodon\r\n\t\t\ta set of three nucleotides that corresponds to an amino acid\r\n\t\t\r\n\t\t\r\n\t\t\tcompetitive inhibitor\r\n\t\t\ta molecule that binds to the active site of an enzyme and blocks its natural substrate from binding\r\n\t\t\r\n\t\t\r\n\t\t\tcomplementary base pairing\r\n\t\t\tthe tendency for adenine to pair with thymine, and cytosine to pair with guanine\r\n\t\t\r\n\t\t\r\n\t\t\tcondensation\r\n\t\t\ta reaction in which two molecules come together and react to form one new molecule and release one water molecule\r\n\t\t\r\n\t\t\r\n\t\t\tconformational change\r\n\t\t\ta change in the 3D structure of a molecule\r\n\t\t\r\n\t\t\r\n\t\t\tdenaturation\r\n\t\t\ta permanent change in the enzyme's structure that causes it to lose its ability to catalyse reactions\r\n\t\t\r\n\t\t\r\n\t\t\tdeoxyribonucleic acid\r\n\t\t\ta nucleic acid that carries the genetic instructions for the cell\r\n\t\t\r\n\t\t\r\n\t\t\tenzyme\r\n\t\t\ta protein that speeds up chemical reactions\r\n\t\t\r\n\t\t\r\n\t\t\texit site\r\n\t\t\tthe E site of the tRNA where the tRNA breaks away from the amino acid\r\n\t\t\r\n\t\t\r\n\t\t\tglycosylation\r\n\t\t\tthe process in which a sugar molecule is added\r\n\t\t\r\n\t\t\r\n\t\t\thistone\r\n\t\t\tpackaging proteins that bind to DNA\r\n\t\t\r\n\t\t\r\n\t\t\tinactive\r\n\t\t\thaving low activity\r\n\t\t\r\n\t\t\r\n\t\t\tinduced fit model\r\n\t\t\tstates that an enzyme's active site changes shape slightly to bind substrates\r\n\t\t\r\n\t\t\r\n\t\t\tinitiation factor\r\n\t\t\ta protein that helps the mRNA template bind to the ribosome\r\n\t\t\r\n\t\t\r\n\t\t\tinitiator tRNA\r\n\t\t\tthe tRNA that recognises the start codon\r\n\t\t\r\n\t\t\r\n\t\t\tlipid\r\n\t\t\ta hydrophobic macromolecule involved in energy storage\r\n\t\t\r\n\t\t\r\n\t\t\tlock-and-key model\r\n\t\t\tstates that an enzyme's active site is rigid and specifically matches the shape of its substrates\r\n\t\t\r\n\t\t\r\n\t\t\tmacromolecule\r\n\t\t\ta large, complex molecule that is essential for life\r\n\t\t\r\n\t\t\r\n\t\t\tmethylation\r\n\t\t\tthe process in which a methyl group is added\r\n\t\t\r\n\t\t\r\n\t\t\tmonosaccharide\r\n\t\t\ta simple monomer of sugar\r\n\t\t\r\n\t\t\r\n\t\t\tmRNA\r\n\t\t\tmessenger RNA that acts as a template for protein production\r\n\t\t\r\n\t\t\r\n\t\t\tnon-competitive inhibitor\r\n\t\t\ta molecule that binds to a different part of an enzyme and causes a conformational change that reduces it activity\r\n\t\t\r\n\t\t\r\n\t\t\tnonsense codon\r\n\t\t\ta codon that does not correspond to an amino acid\r\n\t\t\r\n\t\t\r\n\t\t\tnucleic acid\r\n\t\t\ta macromolecule that stores genetic information\r\n\t\t\r\n\t\t\r\n\t\t\tnucleosome\r\n\t\t\ta section of DNA that is wrapped around a cluster of eight histones\r\n\t\t\r\n\t\t\r\n\t\t\tnucleotide\r\n\t\t\ta simple monomer of nucleic acids; composed of a nitrogenous base, sugar and phosphate group\r\n\t\t\r\n\t\t\r\n\t\t\tpeptide bond\r\n\t\t\tthe bond that links amino acids\r\n\t\t\r\n\t\t\r\n\t\t\tpeptidyl site\r\n\t\t\tthe P site fo the tRNA where the amino acid forms a peptide bond with the previous amino acid in the peptide chain\r\n\t\t\r\n\t\t\r\n\t\t\tpeptidyl transferase\r\n\t\t\tan enzyme that helps join amino acids together\r\n\t\t\r\n\t\t\r\n\t\t\tphosphodiester bond\r\n\t\t\ta bond that links nucleotides\r\n\t\t\r\n\t\t\r\n\t\t\tphosphorylation\r\n\t\t\tthe process in which a phosphate group is added\r\n\t\t\r\n\t\t\r\n\t\t\tpolyadenylation\r\n\t\t\ta post-transcriptional modification that involves adding a poly-A tail to the mRNA\r\n\t\t\r\n\t\t\r\n\t\t\tpre-mRNA\r\n\t\t\tthe product of transcription\r\n\t\t\r\n\t\t\r\n\t\t\tprimary structure\r\n\t\t\tthe linear sequence of amino acids linked by peptide bonds\r\n\t\t\r\n\t\t\r\n\t\t\tpromoter region\r\n\t\t\tthe region of the DNA that initiates transcription\r\n\t\t\r\n\t\t\r\n\t\t\tprotein\r\n\t\t\ta chain of amino acids\r\n\t\t\r\n\t\t\r\n\t\t\tquaternary structure\r\n\t\t\tthe arrangement of multiple protein subunits into a more complex structure\r\n\t\t\r\n\t\t\r\n\t\t\tribonucleic acid\r\n\t\t\ta nucleic acid that plays a role in translating DNA into proteins\r\n\t\t\r\n\t\t\r\n\t\t\tRNA polymerase\r\n\t\t\tan enzyme that binds to a region of the DNA to start transcription\r\n\t\t\r\n\t\t\r\n\t\t\tsecondary structure\r\n\t\t\tthe alpha helices or beta pleated sheets formed by hydrogen bonding between amino acid side chains\r\n\t\t\r\n\t\t\r\n\t\t\tsense codon\r\n\t\t\ta codon that corresponds to an amino acid\r\n\t\t\r\n\t\t\r\n\t\t\tsolenoid\r\n\t\t\ta coiled spiral of nucleosomes\r\n\t\t\r\n\t\t\r\n\t\t\tsplicing\r\n\t\t\ta post-transcriptional modification that involves cutting introns out of the mRNA and rejoining exons together\r\n\t\t\r\n\t\t\r\n\t\t\tstart codon\r\n\t\t\tthe codon AUG that corresponds to the amino acid methionine and also signals that start of translation\r\n\t\t\r\n\t\t\r\n\t\t\tstop signal\r\n\t\t\tthe region of the DNA that terminates transcription\r\n\t\t\r\n\t\t\r\n\t\t\tsubstrate\r\n\t\t\ta molecule that binds to an enzyme\r\n\t\t\r\n\t\t\r\n\t\t\ttertiary structure\r\n\t\t\tthe overall 3D shape of a single protein molecule created by bonds between amino acid side chains\r\n\t\t\r\n\t\t\r\n\t\t\ttranscription\r\n\t\t\tthe process in which a specific segment of DNA is copied into mRNA\r\n\t\t\r\n\t\t\r\n\t\t\ttranslation\r\n\t\t\tthe process in which an mRNA template is used to form a protein\r\n\t\t\r\n\t\t\r\n\t\t\ttRNA\r\n\t\t\ttransfer RNA that binds the mRNA template and reads it to form a protein\r\n\t\t\r\n\t\t\r\n\t\t\tubiquitination\r\n\t\t\tthe process in which a ubiquitin molecule is added\r\n\t\t\r\n\t\t\r\n\t\t\tuncompetitive inhibitor\r\n\t\t\ta molecule that binds to the enzyme\u2013substrate complex and prevents the product from being released\r\n\t\t\t\t\r\n\t\r\n\r\n\r\nGenetics\r\nThe table outlines the definitions for key terms in Genetics.\r\n\r\n\t\r\n\t\t\r\n\t\t\tTerm\r\n\t\t\tDefinition\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tactivator\r\n\t\t\ta regulatory protein that activates transcription\r\n\t\t\r\n\t\t\r\n\t\t\tallele\r\n\t\t\tdifferent version of a gene\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tautosomal dominant trait\r\n\t\t\tan autosomal trait that requires only one dominant allele for the trait to be expressed\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tautosomal recessive trait\r\n\t\t\tan autosomal trait that requires two copies of the recessive allele for the trait to be expressed\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tautosomal trait\r\n\t\t\ta genetic characteristic determined by genes located on the 22 pairs of autosomes\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tautosome\r\n\t\t\ta non-sex chromosome\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tcarrier\r\n\t\t\tan individual that has the allele for the gene but does not express it\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tcell reproduction\r\n\t\t\tthe division of cells to create new cells\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tconstitutive gene\r\n\t\t\ta gene that is expressed at all times because it is continually being transcribed\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tcross\r\n\t\t\tbreeding two organisms together to produce offspring\r\n\t\t\r\n\t\t\r\n\t\t\tcrossing over\r\n\t\t\tthe process by which genetic material is exchanged between chromosomes; also called recombination\r\n\t\t\r\n\t\t\r\n\t\t\tdiploid\r\n\t\t\ta cell with two sets of chromosomes\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tDNA ligase\r\n\t\t\tan enzyme that joins together Okazaki fragments\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tDNA replication\r\n\t\t\tthe process by which a cell makes an exact copy of its DNA before cell divsiion\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tdominant trait\r\n\t\t\ta trait that will show if one copy of the dominant allele is present\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tenhancer\r\n\t\t\ta region of the DNA where activators bind to increase transcription levels of the gene\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tfacultative gene\r\n\t\t\ta gene that is expressed only when it is needed\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tgamete\r\n\t\t\tsex cells like sperm and eggs\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tgene\r\n\t\t\ta sequence of DNA that code for a protein\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tgene expression\r\n\t\t\tthe process by which genes are used to produce proteins\r\n\t\t\r\n\t\t\r\n\t\t\tgene regulation\r\n\t\t\tthe process of controlling when and how much of a gene is expressed\r\n\t\t\r\n\t\t\r\n\t\t\tgenome\r\n\t\t\tan organism's entire set of DNA\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\tgenotype\r\n\t\t\tthe allele specific to an individual\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\thaploid\r\n\t\t\ta cell with a single set of chromosomes\r\n\t\t\r\n\t\t\r\n\t\t\thelicase\r\n\t\t\tan enzyme that unwinds the DNA double helix\r\n\t\t\r\n\t\t\r\n\t\t\theterozygous\r\n\t\t\ta genotype consisting of two different alleles\r\n\t\t\r\n\t\t\r\n\t\t\thomozygous\r\n\t\t\ta genotype consisting of two identical alleles\r\n\t\t\t\t\r\n\t\t\r\n\t\t\thousekeeping gene\r\n\t\t\ta type of constitutive gene that is expressed at stable levels\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tkaryotype\r\n\t\t\ta visual representation of an organism's complete set of chromosomes\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tlagging strand\r\n\t\t\tthe strand of DNA that points away from the replication fork during DNA replication\r\n\t\t\r\n\t\t\r\n\t\t\tlaw of dominance\r\n\t\t\tstates that when there are two alleles of a gene, one can be dominant and hide the other\r\n\t\t\r\n\t\t\r\n\t\t\tlaw of independent assortment\r\n\t\t\tstates that alleles for different traits are distributed to gametes independently of one another\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tlaw of segregation\r\n\t\t\tstates that during the formation of gametes, the two alleles of each trait separate so that each gamete receives only one allele\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tleading strand\r\n\t\t\tthe strand of DNA that points towards the replication fork during DNA replication\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tmeiosis\r\n\t\t\ta type of cell division that produces sex cells\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tmeiosis I\r\n\t\t\tthe first round of division during meiosis to produce two diploid daughter cells\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tmeiosis II\r\n\t\t\tthe second round of division during meiosis to produce four haploid daughter cells\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tMendel's laws\r\n\t\t\tthe principles of inheritance developed by Gregor Mendel\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tOkazaki fragment\r\n\t\t\ta short segment of DNA that is replicated from the lagging strand/td>\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tpedigree\r\n\t\t\ta chart that shows how a trait is passed down generations\r\n\t\t\r\n\t\t\r\n\t\t\tphenotype\r\n\t\t\ta physical trait\r\n\t\t\r\n\t\t\r\n\t\t\tpromotor\r\n\t\t\ta region of the DNA where RNA polymerase binds to start transcription\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tPunnett square\r\n\t\t\ta diagram used to predict the genetic outcomes of a cross\r\n\t\t\t\t\r\n\t\t\r\n\t\t\trecessive trait\r\n\t\t\ta trait that will show only if two copies of the recessive allele are present\r\n\t\t\t\t\r\n\t\t\r\n\t\t\trecombination\r\n\t\t\tthe process by which genetic material is exchanged between chromosomes; also called crossing over\r\n\t\t\t\t\r\n\t\t\r\n\t\t\treplication fork\r\n\t\t\twhere two single strands of DNA are exposed during DNA replication\r\n\t\t\t\t\r\n\t\t\r\n\t\t\trepressor\r\n\t\t\ta regulatory protein that blocks transcription\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tsex chromosome\r\n\t\t\tan X or Y chromosome\r\n\t\t\r\n\t\t\r\n\t\t\tsex-linked trait\r\n\t\t\ta genetic characteristic determined by genes located on the sex chromosomes\r\n\t\t\t\t\r\n\t\t\r\n\t\t\tsilencer\r\n\t\t\ta region of the DNA where repressors bind to reduce transcription levels of the gene\r\n\t\t\r\n\t\t\r\n\t\t\tX-linked dominant trait\r\n\t\t\ta sex-linked trait that is caused by dominant alleles on the X chromosome\r\n\t\t\r\n\t\t\r\n\t\t\tX-linked recessive trait\r\n\t\t\ta sex-linked trait that is caused by recessive alleles on the X chromosome\r\n\t\t\r\n\t\t\r\n\t\t\tY-linked trait\r\n\t\t\ta sex-linked trait that is associated with genes on the Y chromosome\r\n\t\t\t\t\r\n\t\r\n\r\n\r\nHuman body systems\r\nThe table outlines the definitions for key terms in Human body systems.\r\n\r\n  \r\n\t\r\n\t\tTerm\r\n\t\tDefinition\r\n\t\r\n  \r\n  \r\n  \t\r\n\t\tabsorption\r\n      the process through which nutrients are transported from the digestive system into the bloodstream\r\n    \r\n    \r\n      action potential\r\n      a series of voltage changes across a cell membrane\r\n    \r\n    \r\n      adenosine triphosphate\r\n      the body's energy currency\r\n    \r\n    \r\n      aerobic respiration\r\n      a process in which energy (in the form of ATP) is produced in abundant oxygen\r\n    \r\n    \r\n      afferent neurone\r\n      neurones that carry information from sensory organs to the central nervous system; also called sensory neurones\r\n    \r\n    \r\n      alveolus\r\n      a balloon-like structure covered in blood vessels where gas exchange occurs\r\n    \r\n    \r\n      anaerobic respiration\r\n      a process in which energy (in the form of ATP) is produed in the low oxygen conditions\r\n    \r\n    \r\n      anus\r\n      the end of the digestive tract; an opening through which faeces is removed\r\n    \r\n    \r\n      aorta\r\n      the largest artery in the body\r\n    \r\n    \r\n      artery\r\n      a blood vessel that transports blood away from the heart to the body's tissues\r\n    \r\n    \r\n      atrium\r\n      an upper chamber in the heart\r\n    \r\n    \r\n      autonomic nervous system\r\n      the division of the peripheral nervous system that controls involuntary movements\r\n    \r\n    \r\n      biconcave\r\n      curved inwards on both sides\r\n    \r\n    \r\n      bile\r\n      a digestive fluid produced by the liver that helps break down fats\r\n    \r\n    \r\n      bile duct\r\n      a tube that connects the liver and gallbladder to the small intestine\r\n    \r\n    \r\n      biliary tract\r\n      the network of tubes and associated organs that make and transport bile from the liver to the small intestine\r\n    \r\n    \r\n      bladder\r\n      a muscular sac that stores urine\r\n    \r\n    \r\n      blood\r\n      the fluid that delivers nutrients and oxygen to the body's cells\r\n    \r\n    \r\n      blood vessel\r\n      a tube that carries blood throughout the body\r\n    \r\n    \r\n      bolus\r\n      a mass of food and saliva\r\n    \r\n    \r\n      bone\r\n      a strong structure that acts as the framework for the skeleton\r\n    \r\n    \r\n      bone marrow\r\n      soft tissue found in spongy bone\r\n    \r\n    \r\n      brain\r\n      the command centre for the body\r\n    \r\n    \r\n      brainstem\r\n      the structure that connects the brain to the spinal cord\r\n    \r\n    \r\n      bronchi\r\n      the branches from the trachea that lead to the lungs\r\n    \r\n    \r\n      bronchiole\r\n      the next branch of the bronchi\r\n    \r\n    \r\n      bronchus\r\n      a single bronchi\r\n    \r\n    \r\n      capillary\r\n      a tiny blood vessel where gas exchange occurs between the blood and body's tissues\r\n    \r\n    \r\n      capillary bed\r\n      a network of capillaries where gas exchange occurs\r\n    \r\n    \r\n      cardiac muscle\r\n      a type of muscle found in the walls of the heart\r\n    \r\n    \r\n      cardiovascular system\r\n      the body system that transports blood, oxygen and nutrients around the body; also called the circulatory system\r\n    \r\n    \r\n      cartilage\r\n      a flexible tissue that covers the ends of bones to cushion joints\r\n    \r\n    \r\n      cartilage\r\n      a strong connective tissue\r\n    \r\n    \r\n      cellular respiration\r\n      the process by which glucose is converted to energy in the form of adenosine triphosphate (ATP)\r\n    \r\n    \r\n      central nervous system\r\n      the main control centre of the body; includes the brain and spinal cord\r\n    \r\n    \r\n      cerebellum\r\n      the back part of the brain; responsible for coordination and balance\r\n    \r\n    \r\n      cerebrum\r\n      the largest part of the brain; responsible for higher thinking processes and voluntary movement\r\n    \r\n    \r\n      chemical digestion\r\n      the chemical breakdown of food caused by enzymes and acids\r\n    \r\n    \r\n      chondrocyte\r\n      a cartilage cell\r\n    \r\n    \r\n      chyme\r\n      the semi-liquid mixture that forms when food is broken down in the stomach\r\n    \r\n    \r\n      circulatory system\r\n      the body system that transports blood, oxygen and nutrients around the body; also called the cardiovascular system\r\n    \r\n    \r\n      compact bone\r\n      the dense outer layer of bone\r\n    \r\n    \r\n      defecation\r\n      the process of removing waste from the body through the anus\r\n    \r\n    \r\n      depolarisation\r\n      the process by which the membrane potential gets closer to 0\r\n    \r\n    \r\n      diaphragm\r\n      a large dome-shaped muscle found at the base of the ribcage\r\n    \r\n    \r\n      digestion\r\n      the process by which the body breaks down food into smaller components\r\n    \r\n    \r\n      digestive system\r\n      the body system responsible for breaking down food into a form that the body can use for energy\r\n    \r\n    \r\n      digestive tract\r\n      the long tube and associated organs that start at the mouth and end at the anus; also called the alimetary tract or gastrointestinal tract\r\n    \r\n    \r\n      duodenum\r\n      the first part of the small intestine, where most of the chemical digestion occurs\r\n    \r\n    \r\n      ear\r\n      a sensory organ that detects sound and lets us hear\r\n    \r\n    \r\n      effector\r\n      the target tissue\r\n    \r\n    \r\n      efferent neurone\r\n      neurones that send information from the central nervous system to the muscles and glands; also called motor neurones\r\n    \r\n    \r\n      elastic cartilage\r\n      a type of cartilage that provides strength and elasticity/flexibility\r\n    \r\n    \r\n      endocrine gland\r\n      an organ that produces hormones\r\n    \r\n    \r\n      endocrine system\r\n      the body system that produces hormones; also called the hormonal system\r\n    \r\n    \r\n      erythrocyte\r\n      a type of blood cell that carries oxygen; also called a red blood cell\r\n    \r\n    \r\n      excretion\r\n      the process of removing waste from the body\r\n    \r\n    \r\n      excretory system\r\n      the body system that removes waste from the body in the form of urine; also known as the urinary or renal system\r\n    \r\n    \r\n      eye\r\n      a sensory organ that detects light and lets us see\r\n    \r\n    \r\n      faeces\r\n      solid waste produced by the digestive system\r\n    \r\n    \r\n      fibre\r\n      a single thread\r\n    \r\n    \r\n      fibrocartilage\r\n      a type of cartilage that provides support and absorbs shock\r\n    \r\n    \r\n      filtrate\r\n      a liquid that has passed through a filter\r\n    \r\n    \r\n      filtration\r\n      (in excretion) the process by which waste products, excess water and unwanted small molecules are passively filtered out of the blood and into the kidney tubule\r\n    \r\n    \r\n      ganglia\r\n      clusters of neurone cell bodies\r\n    \r\n    \r\n      gas exchange\r\n      the process through which oxygen moves from the lungs into the bloodstream, and waste gases like carbon dioxide move from the bloodstream to the lungs\r\n    \r\n    \r\n      gastric juice\r\n      a complex acidic mixture of water, hydrochloric acid and digestive enzymes\r\n    \r\n    \r\n      glomerulus\r\n      a tiny network of capillaries inside the nephron\r\n    \r\n    \r\n      glucagon\r\n      a hormone produced by the pancreas that raises blood glucose levels\r\n    \r\n    \r\n      gluconeogenesis\r\n      the process by which new glucose is formed in the liver\r\n    \r\n    \r\n      glycogenesis\r\n      the process by which glucose is converted to glycogen for storage\r\n    \r\n    \r\n      glyconeogenesis\r\n      the process by which glucagon is broken down into glucose for use\r\n    \r\n    \r\n      heart\r\n      a muscular organ that pumps blood throughout the body\r\n    \r\n    \r\n      homeostasis\r\n      the maintenance of a stable internal environment despite external changes\r\n    \r\n    \r\n      hormone\r\n      a chemical substance produced by an endocrine gland that is used to communicate to different parts of the body\r\n    \r\n    \r\n      hyaline cartilage\r\n      a type of cartilage that is found on many joint surfaces to reduce friction\r\n    \r\n    \r\n      hyperpolarised\r\n      when the membrane potential is more negative that its resting potential\r\n    \r\n    \r\n      hypothalamus\r\n      a small structure in the brain that bridges the nervous and endocrine system; it helps regulate body temperature, hunger, thirst and emotions\r\n    \r\n    \r\n      hypoxyapatite\r\n      a mineral composed of calcium phosphate that makes up a large amount of bone\r\n    \r\n    \r\n      ileum\r\n      the third part of the small intestine, where vitamin B12, bile salts and any remaining nutrients are absorbed\r\n    \r\n    \r\n      inert\r\n      non-reactive\r\n    \r\n    \r\n      ingestion\r\n      the process of taking food into the mouth\r\n    \r\n    \r\n      insulin\r\n      a hormone produced by the pancreas that lowers blood glucose levels\r\n    \r\n    \r\n      intercalated disc\r\n      a specialised structure found between cardiac muscle cells\r\n    \r\n    \r\n      intercostal muscle\r\n      a small muscle found between the ribs\r\n    \r\n    \r\n      interneurone\r\n      neurones that link sensory and motor neurones with the centra nervous system\r\n    \r\n    \r\n      jejunum\r\n      the second part of the small intestine, where most of the nutrient absorption occurs\r\n    \r\n    \r\n      joint\r\n      a region where two bones meet\r\n    \r\n    \r\n      kidney\r\n      a bean-shaped organ that filters waste from the blood\r\n    \r\n    \r\n      large intestine\r\n      the part of the digestive system between the small intestine and the anus; also called the colon\r\n    \r\n    \r\n      layrnx\r\n      the voice box\r\n    \r\n    \r\n      leukocyte\r\n      a type of blood cell in the immune system that helps the body fight infection and disease; also called a white blood cell\r\n    \r\n    \r\n      ligament\r\n      a strong band of tissue that holds bones together\r\n    \r\n    \r\n      liver\r\n      a metabolic organ that has a range of important functions in the body, like detoxification through filtration of blood and assisting in digestion\r\n    \r\n    \r\n      lobe\r\n      a division of an organ\r\n    \r\n    \r\n      locomotor system\r\n      the body system that includes the bones, muscles, tendons and ligaments in the body; also called the musculoskeletal system\r\n    \r\n    \r\n      lung\r\n      a spongy, air-filled organ where gas exchange occurs\r\n    \r\n    \r\n      mechanical digestion\r\n      the physical breakdown of food caused by forces of motion\r\n    \r\n    \r\n      medulla oblongata\r\n      the lower part of the brainstem that regulates unconscious processes like heartbeat and breathing\r\n    \r\n    \r\n      membrane potential\r\n      the difference in charge between the inside and outside of the cell\r\n    \r\n    \r\n      microfibril\r\n      a fine fibre\r\n    \r\n    \r\n      midbrain\r\n      the top part of the brainstem that controls eye movement and visual and auditory processing\r\n    \r\n    \r\n      motor neurone\r\n      neurones that send information from the central nervous system to the muscles and glands; also called efferent neurones\r\n    \r\n    \r\n      muscular system\r\n      the body system that provides support and facilitates movement\r\n    \r\n    \r\n      musculoskeletal system\r\n      the body system that includes the bones, muscles, tendons and ligaments in the body; also called the locomotor system\r\n    \r\n    \r\n      nasal cavity\r\n      the space inside the nose through which most of the air filters through on the way into and out of the body\r\n    \r\n    \r\n      negative feedback loop\r\n      a mechanism for regulating internal body conditions where the biological reaction reverses the change caused by a stimulus\r\n    \r\n    \r\n      nephron\r\n      a filtration unit inside the kidney\r\n    \r\n    \r\n      nerve fibre\r\n      a bundle of neurones\r\n    \r\n    \r\n      nerve impulse\r\n      an electrical signal that travels along a neurone\r\n    \r\n    \r\n      nervous system\r\n      the body system that controls the body by processing information and sending signals\r\n    \r\n    \r\n      neurone\r\n      a specialised cell that sends and receives electrical signals; also called a neuron or nerve cell\r\n    \r\n    \r\n      neurotransmitter\r\n      a chemical substance that neurones use to communicate with sensory organs, muscles, glands and other neurones\r\n    \r\n    \r\n      nose\r\n      a sensory organ that detects smells and lets us smell\r\n    \r\n    \r\n      nostrils\r\n      the entry points to the nasal cavity\r\n    \r\n    \r\n      oesophagus\r\n      the tube in the throat that connects the mouth to the stomach\r\n    \r\n    \r\n      oral cavity\r\n      the mouth\r\n    \r\n    \r\n      pancreas\r\n      the organ that produces digestive enzymes, and hormones like insulin and glucagon\r\n    \r\n    \r\n      parasympathetic nervous system\r\n      the divison of the autonomic nervous system that plays a role in the \"rest-and-digest\" response\r\n    \r\n    \r\n      periosteum\r\n      a thin membrane that covers the surface of bone\r\n    \r\n    \r\n      peripheral nervous system\r\n      the system that helps the central nervous system communicate with the rest of the body\r\n    \r\n    \r\n      peristalsis\r\n      rhythmic smooth muscle contractions that move food through the digestive system\r\n    \r\n    \r\n      pharynx\r\n      the section of the respiratory tract that connects the nasal and oral cavities to the larynx and oesophagus; also called the throat\r\n    \r\n    \r\n      pituitary gland\r\n      a small structure that sits below the hypothalamus and controls many endocrine glands; also called the master gland\r\n    \r\n    \r\n      plasma\r\n      (in blood) the biggest component of blood consisting of water, proteins and other solutes\r\n    \r\n    \r\n      platelet\r\n      a small cell fragment involved in blood clotting; also called a thrombocyte\r\n    \r\n    \r\n      pons\r\n      the middle part of the brainstem that regulates sleep\u2013wake cycles, breathing and facial expressions\r\n    \r\n    \r\n      positive feedback loop\r\n      a mechanism for regulating internal body conditions where the biological reaction amplifies the change caused by a stimulus\r\n    \r\n    \r\n      post-synaptic neurone\r\n      the neurone receiving neurotransmitters\r\n    \r\n    \r\n      pre-synaptic neurone\r\n      the neurone that releases neurotransmitters into a gap\r\n    \r\n    \r\n      propulsion\r\n      the process of moving food through the digestive tract\r\n    \r\n    \r\n      pulmonary circulation\r\n      the part of the circulatory system where blood is transported between the heart and lungs\r\n    \r\n    \r\n      reabsorption\r\n      (in excretion) the process by which essential nutrients and water are returned to the blood from the kidney tubule\r\n    \r\n    \r\n      rectum\r\n      where faeces is stored until it is ready for removal\r\n    \r\n    \r\n      red blood cell\r\n      a type of blood cell that carries oxygen; also called an erythrocyte\r\n    \r\n    \r\n      reflex arc\r\n      a nerve pathway involved in reflex actions\r\n    \r\n    \r\n      repolarisation\r\n      the process by which the membrane potential returns to its resting potential\r\n    \r\n    \r\n      respiratory system\r\n      the body system that brings oxygen into the body and removes waste gases\r\n    \r\n    \r\n      respiratory tract\r\n      the pathway through which air travels into and out of the body\r\n    \r\n    \r\n      saliva\r\n      a fluid produced in the mouth that moistens the food\r\n    \r\n    \r\n      salivary gland\r\n      an organ that produces saliva\r\n    \r\n    \r\n      secretion\r\n      (in excretion) the process by which additional waste products and toxins are actively transported from the blood into the kidney tubule\r\n    \r\n    \r\n      sensor\r\n      something that detects stimuli\r\n    \r\n    \r\n      sensory neurone\r\n      neurones that carry information from sensory organs to the central nervous system; also called afferent neurones\r\n    \r\n    \r\n      sensory organ\r\n      an organ that detects stimuli from the environment\r\n    \r\n    \r\n      sinoatrial node\r\n      the heart's natural pacemaker that is responsible for regulating heartbeat\r\n    \r\n    \r\n      skeletal muscle\r\n      a type of muscle that is attached to bones\r\n    \r\n    \r\n      skeletal system\r\n      the body system that gives the body support and shape, protecting the internal organs\r\n    \r\n    \r\n      skeleton\r\n      the structural frame of the body\r\n    \r\n    \r\n      skin\r\n      a sensory organ that detects touch, temperature and pain\r\n    \r\n    \r\n      small intestine\r\n      the part of the digestive tract between the stomach and the large intestine, composed of the duodenum, jejunum and ileum; where most of the nutrient absorption occurs\r\n    \r\n    \r\n      smooth muscle\r\n      a type of muscle that lines some organs, like the stomach, intestine and blood vessels\r\n    \r\n    \r\n      somatic nervous system\r\n      the division of the peripheral nervous system that controls voluntary movements\r\n    \r\n    \r\n      sphincter\r\n      a ring of muscle at the end of a tube that controls the passage of substances into or out of the tube\r\n    \r\n    \r\n      spinal cord\r\n      the tube of nerve fibres that extends from the brain to the lower back\r\n    \r\n    \r\n      spongy bone\r\n      the lighter, porous outer layer of bone\r\n    \r\n    \r\n      stomach\r\n      the muscular, hollow organ that receives food from the oesophagus\r\n    \r\n    \r\n      striated\r\n      organised in parallel streaks\r\n    \r\n    \r\n      sympathetic nervous system\r\n      the division of the autonomic nervous system that plays a role in the \"fight-or-flight\" response\r\n    \r\n    \r\n      synapse\r\n      the gap between a neurone and what it is communicating with\r\n    \r\n    \r\n      synovial fluid\r\n      the liquid substance inside a joint that helps it move smoothly\r\n    \r\n    \r\n      systemic circulation\r\n      the part of the circulatory system where blood is transported between the heart and the rest of the body\r\n    \r\n    \r\n      tendon\r\n      a collagen fibre that connects bones to skeletal muscle\r\n    \r\n    \r\n      thermoregulation\r\n      the regulation of internal body temperature\r\n    \r\n    \r\n      thrombocyte\r\n      a small cell fragment involved in blood clotting; also called a platelet\r\n    \r\n    \r\n      tongue\r\n      a sensory organ that detects taste\r\n    \r\n    \r\n      trachea\r\n      a sturdy, cartilage-lined tube that connects the layrnx to the bronchi; also called the windpipe\r\n    \r\n    \r\n      tubule\r\n      a small tube\r\n    \r\n    \r\n      ureter\r\n      a thin tube that connects the kidneys to the bladder\r\n    \r\n    \r\n      urethra\r\n      a tube that carries urine from the bladder out of the body during urination\r\n    \r\n    \r\n      urinary system\r\n      the body system that removes waste from the body in the form of urine; also known as the excretory or renal system\r\n    \r\n    \r\n      urinary tract\r\n      the pathway through which urine is formed and removed from the body\r\n    \r\n    \r\n      urination\r\n      the process of removing urine from the body\r\n    \r\n    \r\n      urine\r\n      a yellowish liquid waste formed through kidney filtration\r\n    \r\n    \r\n      valve\r\n      a structure that controls the flow of a fluid\r\n    \r\n    \r\n      vasoconstriction\r\n      the narrowing of blood vessels\r\n    \r\n    \r\n      vasodilation\r\n      the widening of blood vessels\r\n    \r\n    \r\n      vein\r\n      a blood vessel that transports blood back to the heart from the body's tissues\r\n    \r\n    \r\n      vena cava\r\n      the largest vein in the body\r\n    \r\n    \r\n      ventricle\r\n      a lower chamber in the heart\r\n    \r\n    \r\n      vertebral column\r\n      the chain of bones that protect the spinal cord\r\n    \r\n    \r\n      vesicle\r\n      a capsule containing neurotransmitters\r\n    \r\n    \r\n      villi\r\n      tiny finger-like projections that help increase surface area\r\n    \r\n    \r\n      voice box\r\n      the structure that houses the vocal cords and connects the pharynx to the trachea\r\n    \r\n    \r\n      white blood cell\r\n      a type of blood cell in the immune system that helps the body fight infection and disease; also called a leukocyte\r\n    \r\n  \r\n","n":0.014},[{"v":"Glossary","i":0,"n":1}]]},{"i":31,"$":[{"v":"Citing and referencing images","n":0.5},{"v":"If you're unsure how to cite or reference an image, this page is for you. It explains the different types of images you might include in your assignment and offers tips on citing and referencing them. It also introduces Easy Cite, a great referencing tool.\r\nStudents often include images in their assignments, such as:\r\n\r\n \tartwork\r\n \tgraphs, charts and tables\r\n \tphotographs\r\n \tdiagrams\r\n \tdecorative images.\r\n\r\nAnd the images they use can come from various sources. For example, they could be:\r\n\r\n \tbooks, journals, websites, etc.\r\n \ta student's image made from data taken from another source\r\n \ta student's own original artwork or image\r\n \tAI generated\r\n \tsocial media\r\n \tdatabases.\r\n\r\nRegardless of the type of image you are using, or where you have found it, you will need to cite those images in your academic work. This forms part of acting with academic integrity and giving credit to the creator of any images you use.\r\n\r\nLet's explore some of the factors we need to take into consideration.\r\nReferencing style\r\nAt RMIT, different courses require specific referencing styles such as AGLC4, APA 7th Edition, Chicago A, Chicago B, IEEE, RMIT Harvard, and Vancouver. Each style has unique formatting rules that should be followed to enhance your work and avoid losing marks and this applies to the images that you use. Check your assignment instructions for the required style, and contact academic staff if you\u2019re unsure.\r\n\r\nUse Easy Cite for detailed information on citing and referencing images.\r\nAcknowledgement\r\nImages need to be cited in three places in assignments:\r\n\r\n \tin-text\r\n \tas captions\r\n \tin your reference list.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nReference lists\r\nAny images from books, websites, databases, etc. need to be included in the reference list also and referenced accordingly. Include the details of the source where the image was found, for example the details of the book or website you where located the image.\r\n\r\nFor non-creative images like tables, graphs, and charts sourced from other works, reference them according to the style of the source material (e.g., a book, journal, or website) in the required referencing style.\r\n\r\nFormatting references for artistic works differs between referencing styles. Please check Easy Cite for more detailed information.\r\nReferring to images not included in your work\r\nYou might encounter situations where you discuss an image without including it in your assignment. This often occurs with images from books, webpages, or articles. In such cases, you should include:\r\n\r\n \tan in-text citation: Follow the usual citation format for the source where you viewed the image, which depends on your referencing style.\r\n \ta reference list entry: Add an entry for the source where viewed the image. For example, if the image was in a book, provide a reference entry for the book; if it was on a webpage, use the webpage referencing format.\r\n\r\nHere's an example using the APA 7th edition referencing style to cite and reference an image from an ebook:\r\nExample citation:\r\nIn Drowsy Dedication, Bagdonas depicts a father fighting back fatigue to spend quality time with his children (Muller, 2023).\r\n\r\nExample reference list entry: \r\nMuller, J. (2023). Artistic depictions of modern parenting. Artscape. https://doi.org/10.2224/93h33415306315\r\nNote: The referencing and citation examples on this page are in the APA 7th edition style. For instructions on how to reference and cite images in other styles, please visit Easy Cite.\r\n\r\nImages on this page by RMIT, licensed under CC BY-NC 4.0.","n":0.043},[{"v":"Citation","i":1,"n":1},{"v":"Referencing","i":0,"n":1}]]},{"i":32,"$":[{"v":"Second order non-homogeneous","n":0.577},{"v":"Second order non-homogeneous differential equations are used to model systems like climate dynamics with variable weather patterns, economic models with fluctuating market forces, and chemical reactions with external catalysts. Understanding how to solve them helps predict system responses to external stimuli.\r\n\r\nSecond order non-homogeneous differential equations\r\nA second order non-homogeneous differential equation is an equation that involves a function and its derivatives up to the second order. They are called \"non-homogeneous\" because they have a non-zero function on the output side\u2014that is, \\(f(x)\\neq0\\).\r\nSecond order non-homogeneous differential equations have the general form:\r\n\\[f(x)=a_{1}\\dfrac{d^{2}y}{dx^{2}}+a_{2}\\dfrac{dy}{dx}+a_{3}y\\]\r\nwhere \\(a_{1}\\), \\(a_{2}\\) and \\(a_{3}\\) are constants, and \\(f(x)\\neq0\\).\r\n\r\nSolving second order non-homogeneous differential equations\r\nThe method for solving second order non-homogeneous differential equations builds on the method for homogeneous equations. We find the auxiliary and complementary functions in the same way.\r\nOne key difference is that we use a particular integral, \\(y_{p}(x)\\). The particular integral is a type of function based on the general form of the function \\(f(x)\\); it is the function on the other side of the equals sign. The table shows some examples.\r\n\r\n\t\r\n\t\t\r\n\t\t\tFunction \\(f(x)\\)\r\n\t\t\tParticular integral \\(y_{p}(x)\\)\r\n\t\t\r\n\t\t\r\n\t\r\n\t\t\r\n\t\t\t\\(f(x)=4\\)\r\n\t\t\t\\(y_{p}(x)=a\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(f(x)=2x+3\\)\r\n\t\t\t\\(y_{p}(x)=ax+b\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(f(x)=x^{2}-2x+7\\)\r\n\t\t\t\\(y_{p}(x)=ax^{2}+bx+c\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(f(x)=2e^{\\alpha x}\\)\r\n\t\t\t\\(y_{p}(x)=ae^{\\alpha x}\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(f(x)=3\\sin(px)\\), \\(f(x)=7\\cos(px)\\) or \\(f(x)=4\\sin(px)+2\\cos(px)\\)\r\n\t\t\t\\(y_{p}(x)=a\\sin(px)+b\\cos(px)\\)\r\n\t\t\r\n\t\r\n\r\nFor sine and cosine functions, the particular integral is ALWAYS \\(a\\sin(px)+b\\cos(px)\\), regardless or whether the original function only consists of sine or cosine functions, or a sum of both.\r\nThe general solution for second order non-homogeneous differential equations is:\r\n\\[y(x)=y_{c}(x)+y_{p}(x)\\]\r\n\r\nExample 1 \u2013 solving second order non-homogeneous differential equations\r\nDetermine the general solution \\(y(x)\\) for the equation \\(2\\dfrac{d^{2}y}{dx^{2}}-\\dfrac{dy}{dx}-6y=3x+2\\).\r\nFind the auxiliary equation and solve for \\(m\\).\r\n\\[\\begin{align*} 2m^{2}-m-6 & = 0\\\\\r\n(2m+3)(m-2) & = 0\\\\\r\n2m+3=0 & \\textrm{ or } m-2=0\\\\\r\nm_{1}=-\\dfrac{3}{2} & \\textrm{ or } m_{2}=2\r\n\\end{align*}\\]\r\nThere are two real and different solutions, so the complementary function is:\r\n\\[y_{c}(x)=Ae^{-\\frac{3}{2}}+Be^{2x}\\]\r\nFor this example, \\(f(x)=3x+2\\) which is a linear function. Therefore, the particular integral is:\r\n\\[y_{p}(x)=ax+b\\]\r\nTo find \\(a\\) and \\(b\\), we need to use the original equation:\r\n\\[2\\frac{d^{2}y}{dx^{2}}\\frac{dy}{dx}-6y=3x+2\\]\r\nWe can find \\(\\dfrac{d^{2}y}{dx^{2}}\\) and \\(\\dfrac{dy}{dx}\\).\r\n\\[\\frac{dy}{dx}=a\\] \\[\\frac{d^{2}y}{dx^{2}}=0\\] \r\nLet's substitute these values into the equation, group the \"like\" powers and solve for \\(a\\) and \\(b\\).\r\n\\[\\begin{align*} 2(0)-(a)-6(ax+b) & = 3x+2\\\\\r\n(-6a)x+(-a-6b) & = 3x+2\\\\\r\n-6a=3 & \\textrm{ and } -a-6b=2\\\\\r\na=-\\frac{1}{2} & \\textrm{ and } -\\frac{1}{2}-6b=2\\\\\r\n& \\quad\\quad\\quad\\quad b=-\\frac{1}{4}\r\n\\end{align*}\\]\r\nThe particular integral is therefore:\r\n\\[y_{p}(x)=-\\frac{1}{2}x-\\frac{1}{4}\\]\r\nFinally, we can find the general solution.\r\n\\[\\begin{align*} y(x) & = y_{c}(x)+y_{p}(x)\\\\\r\n& = Ae^{-\\frac{3}{2}x}+Be^{2x}-\\frac{1}{2}x-\\frac{1}{4}\r\n\\end{align*}\\]\r\n\r\nRepeated solutions\r\nSometimes, a term in the complementary function will also appear in the form you chose for the particular integral. This overlap causes the particular integral to not work; you will not be able to solve the question effectively. \r\nTo fix this, we just need to multiply the initial guess for the particular integral by increasing powers of \\(x\\) until it is no longer matches any part of the complementary solution.\r\nBut what does this all mean? Let's look at Example 2.\r\n\r\nExample 2 \u2013 solving second order non-homogeneous differential equations\r\nSolve the differential equation \\(y''-y'-2y=6e^{-x}\\).\r\nThe auxiliary equation is \\(m^{2}-m-2=0\\). The solutions for \\(m\\) are \\(m=-1\\) and \\(m=2\\). These are two real and different solutions.\r\nThe complementary function is therefore:\r\n\\[y_{c}(x)=Ae^{-x}+Be^{2x}\\]\r\nThe particular integral is \\(y_{p}(x)=ke^{-x}\\), but this term is already contained within the complementary function.\r\nWe start by multiplying by \\(x\\) to get:\r\n\\[y_{p}(x)=kxe^{-x}\\]\r\nNow, we can find the value of \\(k\\) by substituting \\(y_{p}(x)\\) and its derivatives into the original equation. Here, we need to use The product rule.\r\n\\[\\frac{dy}{dx}=ke^{-x}-kxe^{-x}\\] \\[\\frac{dy^{2}}{dx^{2}}=-2ke^{-x}+kxe^{-x}\\]\r\nSubstituting these into the original equation and solving for \\(k\\), we get:\r\n\\[\\begin{align*} -2ke^{-x}+kxe^{-x}-(ke^{-x}-kxe^{-x})-2(kxe^{-x}) & = 6e^{-x}\\\\\r\n-3ke^{-x} & = 6e^{-x}\\\\\r\nk & =-2\r\n\\end{align*}\\]\r\nThe particular integral is therefore:\r\n\\[y_{p}(x)=-2xe^{-x}\\]\r\nThe general solution is:\r\n\\[\\begin{align*} y(x) & = y_{c}(x)+y_{p}(x)\\\\\r\n& = Ae^{-x}+Be^{2x}-2xe^{-x}\r\n\\end{align*}\\]\r\n\r\nExercise \u2013 solving second order non-homogeneous differential equations\r\n\r\n\tFind the general solutions for the following.\r\n\t\r\n\t\t\\(\\dfrac{d^{2}y}{dx^{2}}-2\\dfrac{dy}{dx}+y=5e^{x}\\)\r\n\t\t\\(2\\dfrac{d^{2}y}{dx^{2}}+5\\dfrac{dy}{dx}-3y=4\\sin(2x)\\)\r\n\t\t\\(\\dfrac{d^{2}y}{dx^{2}}-4\\dfrac{dy}{dx}+4y=4x+3\\cos(2x)\\)\r\n\t\t\\(\\dfrac{d^{2}y}{dx^{2}}+2\\dfrac{dy}{dx}+2y=6e^{x}\\sin(2x)\\)\r\n\t\r\n\tFind the particular solutions for the following.\r\n\t\r\n\t\t\\(\\dfrac{d^{2}y}{dx^{2}}+4\\dfrac{dy}{dx}=6\\) given that \\(y(0)=0\\) and \\(\\dfrac{dy}{dx}(0)=0\\)\r\n\t\t\\(\\dfrac{d^{2}y}{dx^{2}}+9y=12\\cos(3x)\\) given that \\(y(0)=2\\) and \\(\\dfrac{dy}{dx}(0)=3\\)\r\n\t\r\n\tWhen a mass of \\(2\\textrm{ kg}\\) is attached to a spring whose constant is \\(32\\textrm{ N/m}\\), it comes to rest in the equilibrium position. Starting at \\(t=0\\), a force equal to \\(f(t)=68e^{-2t}\\cos(4t)\\) is applied to the system. Find the equation of motion in the absence of damping (dissipation of energy).\r\n\tA mass of \\(40\\textrm{ kg}\\) stretches a spring \\(10\\textrm{ cm}\\). A damping device imparts a resistance to motion equal to \\(560\\) times the instantaneous velocity. Find the equation of motion if the mass is released from the equilibrium position with a downward velocity of \\(2\\textrm{ cm/sec}\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.039},[{"v":"Differential equations","i":1,"n":0.707},{"v":"Maths","i":0,"n":1}]]},{"i":33,"$":[{"v":"Second order homogeneous","n":0.577},{"v":"Second order homogeneous differential equations are crucial for modelling natural phenomena, like mechanical vibrations, electrical circuits and wave propagation. By learning how to solve these equations, you can understand and predict how systems behave under various conditions.\r\n\r\nSecond order homogeneous differential equations\r\nA second order homogeneous differential equation is an equation that involves a function and its derivatives up to the second order. They are called \"homogeneous\" because they do not have any standalone constants; this means they are equal to \\(0\\) on the opposite side of the function and its derivatives.\r\nThe general form for a second order homogeneous equation is:\r\n\\[f(x)=a_{1}\\dfrac{d^{2}y}{dx^{2}}+a_{2}\\dfrac{dy}{dx}+a_{3}y\\]\r\nwhere \\(a_{1}\\), \\(a_{2}\\) and \\(a_{3}\\) are constants, and \\(f(x)=0\\). \r\n\r\nSolving second order homogeneous differential equations\r\nTo solve second order homogeneous differential equations, we need an auxiliary equation. To generate an auxiliary equation, we let:\r\n\\[y''=\\dfrac{d^{2}y}{dx^{2}}=m^{2}\\] \\[y'=\\dfrac{dy}{dx}=m\\] \\[y=1\\]\r\nWe substitute these values into our differential equation, then solve for \\(m\\).\r\nDepending on our solutions for \\(m\\), we get a different complementary function, \\(y_{c}x\\), as shown in the table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tSolution to auxillary equation\r\n\t\t\tComplementary function\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tTwo real and different solutions\r\n\t\t\t\\[m_{1}\\textrm{ and }m_{2}\\]\r\n\t\t\t\\(y_{c}(x)=Ae^{m_{1}x}+Be^{m_{2}x}\\)\r\n\t\t\r\n\t\t\r\n\t\t\tOne real, repeated solution\r\n\t\t\t\\[m_{1}=m_{2}\\]\r\n\t\t\t\\(y_{c}(x)=(At+B)e^{mx}\\)\r\n\t\t\r\n\t\t\r\n\t\t\tComplex solution\r\n\t\t\t\\[m=\\alpha\\pm\\beta i\\]\r\n\t\t\t\\(y_{c}(x)=e^{\\alpha x}(A\\cos(\\beta x)+B\\sin(\\beta x)\\)\r\n\t\t\r\n\t\r\n\r\n\\(A\\) and \\(B\\) are constants.\r\nYou can see that some solutions to the auxilliary equation may involve complex numbers, so make sure you are confident working with them!\r\nJust like first order differential equations, we can solve for a general solution with a constant. To solve for a particular solution, we need to know initial or boundary values:\r\n\r\n\t\\(y(0)\\)\r\n\t\\(\\dfrac{dy}{dx}(0)\\)\r\n\r\nYou can see how to find a particular solution in Example 2.\r\n\r\nExample 1 \u2013 solving second order homogeneous differential equations\r\nSolve for \\(y(x)\\) given \\(2\\dfrac{d^{2}y}{dx^{2}}-5\\dfrac{dy}{dx}-3y=0\\).\r\nThe auxiliary equation is:\r\n\\[2m^{2}-5m-3=0\\]\r\nWe can solve for \\(m\\) by factorising.\r\n\\[\\begin{align*} 2m^{2}-5m-3 & = 0\\\\\r\n(2m+1)(m-3) & = 0\\\\\r\n2m+1=0 & \\textrm{ or } m-3=0\\\\\r\nm_{1}=-\\dfrac{1}{2} & \\textrm{ or } m_{2}=3\r\n\\end{align*}\\]\r\nNext, we generate the complementary function. Here, we have two real and different solutions, so the complementary function is:\r\n\\[\\begin{align*} y_{c}(x) & = Ae^{m_{1}x}+Be^{m_{2}x}\\\\\r\n& = Ae^{-\\frac{1}{2}x}+Be^{3x}\r\n\\end{align*}\\]\r\nwhere \\(A\\) and \\(B\\) are constants.\r\n\r\n\r\n\r\nExercise \u2013 solving second order homogeneous differential equations\r\n\r\n\tFind the general solutions for the following.\r\n\t\r\n\t\t\\(\\dfrac{d^{2}y}{dx^{2}}-6\\dfrac{dy}{dx}+13y=0\\)\r\n\t\t\\(\\dfrac{d^{2}y}{dx^{2}}-6\\dfrac{dy}{dx}+9y=0\\)\r\n\t\t\\(2\\dfrac{d^{2}y}{dx^{2}}-\\dfrac{dy}{dx}-6y=0\\)\r\n\t\t\\(y''-10y'+25y=0\\)\r\n\t\t\\(y''+y'+y=0\\)\r\n\t\r\n\tFind the particular solutions for the following.\r\n\t\r\n\t\t\\(6\\dfrac{d^{2}y}{dx^{2}}+5\\dfrac{dy}{dx}-6y=0\\) given that \\(y=5\\) and \\(\\dfrac{dy}{dx}=-1\\) when \\(x=0\\)\r\n\t\t\\(4\\dfrac{d^{2}y}{dx^{2}}-5\\dfrac{dy}{dx}+y=0\\) given that \\(y=1\\) and \\(\\dfrac{dy}{dx}=-2\\) when \\(x=0\\)\r\n\t\t\\(x''-6x'+9x=0\\) given that \\(x(0)=2\\) and \\(x'(0)=0\\)\r\n\t\t\\(y''+6y'+13y=0\\) given that \\(y(0)=4\\) and \\(y'(0)=0\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.053},[{"v":"Differential equations","i":1,"n":0.707},{"v":"Maths","i":0,"n":1}]]},{"i":34,"$":[{"v":"First order linear","n":0.577},{"v":"First order linear differential equations are are widely used in various fields, such as physics, engineering, and biology, to model processes like cooling, population growth, and electrical circuits. Understanding how to solve first order linear differential equations helps you analyse and predict the behavior of systems in the real world.\r\n\r\nFirst order linear differential equations\r\nAn equation is considered to be a first order linear differential equation if it can be written in the standard form:\r\n\\[\\frac{dy}{dx}+p(x)y=q(x)\\]\r\nThe variables in this type of equation cannot be separated into opposite sides of the equation.\r\n\r\nIntegrating factor\r\nTo solve first order linear differential equations, we use the integrating factor method. The integrating factor is often denoted by \\(I\\).\r\n\r\n\tRearrange the equation into the standard form \\(\\dfrac{dy}{dx}+p(x)y=q(x)\\).\r\n\tLet the integrating factor \\(I=e^{\\int p(x)dx}\\).\r\n\tMultiply both sides by \\(I\\) to get \\(yI=\\int(I\\times q(x))dx\\).\r\n\tIntegrate both sides and solve for \\(y\\).\r\n\tIf a coordinate is provided, find the particular solution.\r\n\r\n\r\nExample \u2013 solving first order linear differential equations\r\nSolve for \\(y(x)\\) give \\(\\dfrac{dy}{dx}+5y=e^{2x}\\) and \\(y(0)=0\\).\r\nThe equation is already in the standard form, so we let \\(p(x)=5\\) and \\(q(x)=e^{2x}\\).\r\nSince \\(I=e^{\\int p(x)dx}\\) and \\(\\int p(x)dx=5x\\), then \\(I=e^{5x}\\).\r\nWe can substitute this into \\(yI=\\int(I\\times q(x)dx\\) and solve for \\(y\\).\r\n\\[\\begin{align*} ye^{5x} & = \\int e^{5x}\\times e^{2x}dx\\\\\r\n& = \\int e^{7x}dx\\\\\r\n& = \\frac{1}{7}e^{7x}+c,\\,c\\in\\mathbb{R}\\\\\r\ny & = \\frac{1}{7}e^{2x}+ce^{-5x}\r\n\\end{align*}\\]\r\nWe know that \\(y(0)=0\\), so we can find the particular solution.\r\n\\[\\begin{align*} 0 & = \\frac{1}{7}e^{0}+ce^{0}\\\\\r\n& = \\frac{1}{7}+c\\\\\r\nc & = -\\frac{1}{7}\r\n\\end{align*}\\]\r\nThe solution is \\(y=\\dfrac{1}{7}e^{2x}-\\dfrac{1}{7}e^{-5x}\\).\r\n\r\nExercise \u2013 solving first order linear differential equations\r\n\r\n\tFind the general solution for the following.\r\n\t\r\n\t\t\\(3\\dfrac{dy}{dx}+12y=4\\)\r\n\t\t\\(x\\dfrac{dy}{dx}+2y=3\\)\r\n\t\t\\(\\dfrac{dy}{dx}+y=e^{3x}\\)\r\n\t\t\\(y'+3x^{2}y=x^{2}\\)\r\n\t\t\\(x^{2}y'+xy=1\\)\r\n\t\t\\(xdy=(x\\sin(x)-y)dx\\)\r\n\t\t\\(\\cos(x)\\dfrac{dy}{dx}+y\\sin(x)=1\\)\r\n\t\t\\(x\\dfrac{dy}{dx}+4y=x^{3}-x\\)\r\n\t\r\n\tFind the particular solution for the following.\r\n\t\r\n\t\t\\(\\dfrac{dy}{dx}+2xy=x\\) where \\(y(0)=-3\\)\r\n\t\t\\(\\cos^{2}x\\dfrac{dy}{dx}+y=1\\) where \\(y(0)=-3\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.064},[{"v":"Differential equations","i":1,"n":0.707},{"v":"Maths","i":0,"n":1}]]},{"i":35,"$":[{"v":"First order separable","n":0.577},{"v":"The \\(x\\) and \\(y\\) terms in first order separable differential equations can be separated to make them easier to solve. These equations are seen when modelling population growth or chemical reactions. Use this resource to learn how to solve them.\r\n\r\nFirst order separable differential equations\r\nA first order separable differential equation is a special type of differential equation in which we can isolate the variables on opposite sides of the equation.\r\nThe general form for a first order separable equation is:\r\n\\[\\frac{dy}{dx}=g(x)h(y)\\]\r\nThis is considered separable because we can rearrange it to:\r\n\\[\\frac{1}{h(y)}dy=g(x)dx\\]\r\nOnce we separate it out, we can integrate each side independently to find the solution. This gives us a simpler understanding of how the variables relate to each other.\r\n\r\nExercise \u2013 identifying first order separable differential equations\r\nWhich of the following are first order separable differential equations?\r\n\r\n\t\\(\\dfrac{dy}{dx}=xy^{2}\\)\r\n\t\\(y'+xy=\\sin(x)\\)\r\n\t\\(\\dfrac{dy}{dx}=\\dfrac{x}{y}\\)\r\n\t\\(x^{2}\\dfrac{dy}{dx}=y+x\\)\r\n\r\n\r\n\r\nSolving first order separable differential equations\r\nThe method for solving first order separable differential equations is straightforward. \r\n\r\n\tRearrange the differential equation so that the \\(y\\) terms and \\(dy\\) are on one side, and the \\(x\\) terms and \\(dx\\) are on the other side.\r\n\tIntegrate each side separately.\r\n\tSolve for \\(y\\) and don't forget to include a constant.\r\n\r\nIn some cases, we don't have enough information to find an exact solution, so we give a general solution that includes a constant.\r\nBut if we are given a set of coordinates that satisfy the equation, we can find a particular solution. All we need to do is substitute the \\(x\\) and \\(y\\) values into our general solution to find the constant.\r\n\r\nExample 1 \u2013 solving first order separable differential equations\r\nSolve for \\(y(x)\\) given \\(\\dfrac{dy}{dx}-x^{2}y=0\\).\r\nRearrange the differential equation so that the \\(y\\) terms and \\(dy\\) are on one side, and the \\(x\\) terms and \\(dx\\) are on the other side.\r\n\\[\\begin{align*} \\frac{dy}{dx}-x^{2}y & = 0\\\\\r\n\\frac{1}{y}\\frac{dy}{dx} & = x^{2}\\\\\r\n\\frac{1}{y}dy & = x^{2}dx\r\n\\end{align*}\\]\r\nNow that we have separated the variables, we can integrate each side separately.\r\n\\[\\begin{align*} \\int\\frac{1}{y}dy & = \\int x^{2}dx\\\\\r\n\\ln(y) & = \\frac{x^{3}}{3}+c\\\\\r\ne^{\\ln(y)} & = e^{\\frac{x^{3}}{3}+c}\\\\\r\ny & = e^{\\frac{x^{3}}{3}}\\times e^{c}\\\\\r\n& = Ae^{\\frac{x^{3}}{3}}\r\n\\end{align*}\\]\r\nwhere \\(A=e^{c}\\). We can do this because \\(e^{c}\\) will just be a constant.\r\n\r\n\r\n\r\nExercise \u2013 solving first order separable differential equations\r\n\r\n\tSolve the following.\r\n\t\r\n\t\t\\(\\dfrac{dy}{dx}=\\sin(5x)\\)\r\n\t\t\\(\\dfrac{dy}{dx}=\\dfrac{y^{3}}{x^{2}}\\)\r\n\t\t\\(\\dfrac{dy}{dx}=\\dfrac{x^{2}y^{2}}{1+x}\\)\r\n\t\t\\(\\dfrac{dy}{dx}=e^{3x+2y}\\)\r\n\t\r\n\tFind the general solutions for the following.\r\n\t\r\n\t\t\\(dx+e^{3x}dy=0\\)\r\n\t\t\\((x+1)\\dfrac{dy}{dx}=x+6\\)\r\n\t\t\\(xy'=4y\\)\r\n\t\r\n\tFind the particular solutions for the following.\r\n\t\r\n\t\t\\(\\dfrac{dy}{dx}=yx\\sin(5x)\\) where \\(y(0)=1\\)\r\n\t\t\\(\\dfrac{dy}{dx}+ty=y\\) where \\(y(1)=3\\)\r\n\t\t\\(\\dfrac{dy}{dx}=y^{2}-1\\) where \\(y(0)=2\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.053},[{"v":"Differential equations","i":1,"n":0.707},{"v":"Maths","i":0,"n":1}]]},{"i":36,"$":[{"v":"Introduction to differential equations","n":0.5},{"v":"Differential equations help us understand and predict behaviours in STEM, where they are used to model phenomena like motion, heat transfer, population dynamics, and financial markets. Use this resource to learn about first order separable differential equations.\r\n\r\nDifferential equations\r\nA differential equation is an equation that includes a function and one or more of its derivatives. Differential equations shows how a quantity changes over time or space. They are used to model real-world situations where things are changing, like how fast a car is moving, how heat spreads through a metal, or how populations grow.\r\nConsider the function \\(y=f(x)\\), where \\(y'\\) is the derivative of \\(y\\). Some examples of differential equations that involve this function are:\r\n\r\n\t\\(y'=3x\\)\r\n\t\\(y'+2y=x+7\\)\r\n\t\\(y''-3y+8y=21e^{-8x}\\)\r\n\r\n\r\nOrder of differential equations\r\nThe order of a differential equation is the highest order of any derivative of the unknown function \\(y\\). \r\nIf \\(y'\\) has an order of \\(1\\):\r\n\r\n\t\\(y''\\) has an order of \\(2\\)\r\n\t\\(y'''\\) has an order of \\(3\\)\r\n\t\\(y^{4}\\) has an order of \\(4\\)\r\n\r\n...and so on.\r\nThe order of a differential equation helps us understand how complex it is to find a solution to the equation. If we can figure out the order, we can choose the right method to solve the differential equation.\r\n\r\nExercise \u2013 determining the order of differential equations\r\nFind the order of the following differential equations.\r\n\r\n\t\\(y'=3x\\)\r\n\t\\(y'+2y=x+7\\)\r\n\t\\(y''-3y+8y=21e^{-8x}\\)\r\n\r\n\r\n\r\nSolving differential equations\r\nTo solve a differential equation, we look for a function or set of functions that satisfy the equation. This involves determining the function that, when subsituted into the differential equation, makes the equation true.\r\nThe method we use to solve the equation depends on the type of differential equation it is. We will cover:\r\n","n":0.062},[{"v":"Differential equations","i":1,"n":0.707},{"v":"Maths","i":0,"n":1}]]},{"i":37,"$":{"0":{"v":"Vectors and matrices","n":0.577},"2":[]}},{"i":38,"$":[{"v":"Complex numbers","n":0.707},{"v":"Expand your understanding of complex numbers by learning how to express them in terms of magnitude and angle, and simplify the process of finding the powers and roots of complex numbers.\r\n\r\n\tPolar form of complex numbers\r\n\t\tDe Moivre's theorem\r\n\r\n","n":0.164},[]]},{"i":39,"$":[{"v":"Equation solving","n":0.707},{"v":"Solving algebraic equations involves finding the values of variables that satisfy mathematical statements. Build on your foundational understanding of algebra by tackling how to solve various types of equations, from linear to quadratic and beyond.\r\n\r\n\r\n\tSolving quadratic equations\r\n\tQuadratic formula\r\n\tSolving cubic equations\r\n \tSolving exponential equations\r\n\r\n","n":0.152},[]]},{"i":40,"$":[{"v":"Algebra","n":1},{"v":"Delve into more complex factorisation techniques to further enhance your algebraic abilities. Discover new methods to break down expressions and tackle challenging equations with confidence.\r\n\r\n\r\n\tPartial fractions\r\n\tCubic factorisation\r\n\r\n","n":0.192},[{"v":"Algebra","i":0,"n":1}]]},{"i":41,"$":[{"v":"Algebra","n":1},{"v":"Algebra focuses on understanding and manipulating mathematical symbols and expressions. Here, you'll explore key techniques such as simplification, factorisation, and rearranging formulas, which are essential for solving equations and understanding mathematical relationships.\r\n\r\n\r\n\tQuadratic factorisation\r\n\tPerfect squares\r\n\tDifference of two squares\r\n\tCompleting the square\r\n\tQuadratics in algebraic fractions\r\n\r\n\r\n","n":0.154},[{"v":"Algebra","i":0,"n":1}]]},{"i":42,"$":{"0":{"v":"Algebra","n":1},"2":[]}},{"i":43,"$":[{"v":"Matrices","n":1},{"v":"Enhance your understanding of matrices by exploring augmented matrices, elementary row operations, and finding the inverse of \\(3\\times3\\) matrices. Delve into the powerful concepts of eigenvalues and eigenvectors, which are key to solving complex systems and understanding linear transformations.\r\n\r\n\r\n\tAugmented matrices and elementary row operations\r\n \tInverse of a 3×3 matrix\r\n \tEigenvalues and eigenvectors\r\n\r\n\r\n","n":0.139},[]]},{"i":44,"$":[{"v":"Matrices","n":1},{"v":"Now that you have the basics down, explore more complex topics in matrices, like special matrices and finding the inverse of a matrix.\r\n\r\n\r\n \tSpecial matrices\r\n\tInverse of a 2×2 matrix\r\n\r\n\r\n","n":0.186},[]]},{"i":45,"$":[{"v":"Vectors","n":1},{"v":"Advance your vector skills by exploring planes in 3D and directional derivatives.\r\n\r\n\r\n\t\tEquation of a plane in 3D\r\n\t\tIntersecting planes in 3D\r\n\t\tDistance from a point to a plane\r\n\t\tDirectional derivatives\r\n\r\n\r\n","n":0.192},[{"v":"Vectors","i":0,"n":1}]]},{"i":46,"$":[{"v":"Vectors","n":1},{"v":"Consolidate your understanding of vectors with more complex calculations involving vectors, and start to look at vectors in 3D.\r\n\r\n\r\n\t\tVector or cross product\r\n\t\tEquation of a line in 3D\r\n\t\tIntersecting lines in 3D\r\n\t\r\n\r\n","n":0.183},[{"v":"Vectors","i":0,"n":1}]]},{"i":47,"$":{"0":{"v":"Trigonometry","n":1},"2":[]}},{"i":48,"$":[{"v":"Calculus","n":1},{"v":"\r\n\r\n\r\n\r\n\r\nGetting started\r\nDifferentiation\r\n\r\n \tLimit of a function\r\n \tGradients, tangents and derivatives\r\n \tDifferentiation from first principles\r\n \tRules for differentiation\r\n \tThe chain rule\r\n \tThe product rule\r\n \tThe quotient rule\r\n\r\nIntegration\r\n\r\n \tAntidifferentiation\r\n\r\n\r\n\r\n\r\n\r\nNext steps\r\nDifferentiation\r\n\r\n \tMaxima and minima\r\n \tCurve sketching\r\n \tHigher order derivatives\r\n\r\nIntegration\r\n\r\n \tIntegration of polynomials\r\n \tIntegration of reciprocal functions\r\n \tIntegration of exponential functions\r\n \tIntegration of trigonometric functions\r\n \tArea under a curve\r\n\r\n\r\n\r\n\r\n\r\nDive deeper\r\nDifferentiation\r\n\r\n \tImplicit differentiation\r\n \tLogarithmic differentiation\r\n \tRates of change\r\n \tLinear approximation and differentials\r\n \tPartial differentiation\r\n\r\nIntegration\r\n\r\n \tIntegration by substitution\r\n \tIntegration using partial fractions\r\n \tIntegration by parts\r\n \tDouble integrals\r\n\r\nDifferential equations\r\n\r\n \tIntroduction to differential equations\r\n \tFirst order separable\r\n \tFirst order linear\r\n \tSecond order homogeneous\r\n \tSecond order non-homogeneous\r\n\r\n\r\n\r\n\r\n\r\n\r\n@media (min-width: 992px) {\r\n.my-column-2 {\r\n\tborder-right: 0;\r\n\tmargin-right: 0;\r\n}\r\n.my-column-3 {\r\n\tborder-left: 1px solid #bfbfbf;\r\n\tpadding-left: 24px;\r\n}\r\n}\r\n\r\n\r\n\r\n\r\n\r\nFurther resources\r\n\r\n\r\n","n":0.095},[]]},{"i":49,"$":[{"v":"Reporting measurements","n":0.707},{"v":"Building on your foundational knowledge, this resource delves into the quality of measurements and how to report values. This will enhance your precision and accuracy in STEM applications. Explore these concepts to refine your skills and make more reliable measurements.\r\n\r\n\tSignificant figures\r\n\tSignificant figures in calculations\r\n\tQuality of measurements\r\n\tPropagating uncertainty\r\n\r\n\r\n","n":0.146},[]]},{"i":50,"$":{"0":{"v":"Numbers and measurement","n":0.577},"2":[]}},{"i":51,"$":[{"v":"Interrogatives","n":1},{"v":"Interrogatives are frequently used in assessment questions. They are also known as question words and 'wh' words. Understanding what each type of question requires will help you answer assessment questions more effectively.\r\nWho, what, when, where, why, which and how\r\n\r\n\r\n\r\n\r\n\t\t\r\n\r\n\r\n\r\n\r\n\t\r\n\t\t\r\n\r\n\t\r\nWhich questions\r\n\r\n\t \r\nHow often\r\n\t\t\r\n\t\t\tSimple: describe the frequency (e.g. daily, weekly, once a month)\r\n\t\t\tAdd more detail: include any patterns, exceptions, or why this particular frequency is correct\r\n\t\t\r\n\tHow much / many\r\n\t\t\r\n\t\t\tSimple: give the quantity or number (e.g. 5 units, 20 people)\r\n\t\t\tAdd more detail: explain why the quantity or number is the correct answer\r\n\t\t\r\n\tHow do you / can you (do something)\r\n\t\t\r\n\t\t\tSimple: describe the basic steps or method\r\n\t\t\tAdd more detail: provide detailed steps or explanation about why these steps or this method should be used\r\n\t\t\t\r\n\tHow is / are\r\n\t\t\r\n\t\t\tSimple: describe a current state, method or difference\r\n\t\t\tAdd more detail: explain why the state is the way it is, why a method should be used or detail the differences\r\n\t\t\r\n\r\n\r\n","n":0.082},[{"v":"Assessment","i":1,"n":1},{"v":"Understanding assignments","i":0,"n":0.707}]]},{"i":52,"$":[{"v":"Scenario-based question","n":0.707},{"v":"For some assessment questions, you will be given a situation, scenario or problem that you need to answer. This is called a 'hypothetical' situation where you need to imagine how you would act, what advice you would give, or decide what is causing a problem. On this page, you will learn some strategies for understanding and answering these questions.\r\nScenarios \r\n\r\nScenario-based questions often have two parts. For example:\r\n\r\n\tyou are presented with the scenario AND what would you do?\r\n\tyou are presented with the scenario AND what is the cause of the problem?\r\n\r\n\r\nIn these types of questions, you will often see language like:\r\n\r\n\tImagine that you are... \t\r\n\tWhat decisions would you make?\r\n\tIf you could change one thing about... what would it be?\r\n\tWhat would happen if...?\r\n\tHow would you solve this problem?\r\n\tWhat actions/steps would you take if...?\r\n\tIf you had to..., what would you do?\r\n\r\n\r\nTo successfully answer these questions, show how you'd use what you've learned from your course in the situation. It's helpful to explain your thinking; don't just say what you'd do or what's wrong\u2014make sure to say why you think that, too.\r\n\r\nHere is an example of a scenario question: \r\n\r\n\r\n\r\n\r\nHere are some steps that you can use to answer this question effectively:\r\n\r\n\tRead the question carefully. Read it more than once.\r\n\tIdentify what you are being asked to do by highlighting the keywords. Pay attention to details. For example, in this question, you need to mention the 'actions' you would take. This means that if you only mention one action, you are not answering the question properly.\r\n\tThink about what you have learned in your course. Have you done any exercises or reading that could be applied to this situation?\r\n\tThink about how much you should write. Sometimes you are given a word limit. When there is no word limit, write your answer with enough explanation that others will be able to understand it clearly.\r\n\tWrite your answer simply and clearly. Avoid using complex words where possible and avoid long, complicated sentences.\r\n\tGo back to the question and check that your answer covers everything you have been asked to do.\r\n\r\n\r\nAnother example\r\n\r\nJuan is studying youth work and he receives this question in an assessment. It says:\r\n\r\n\"Sylvia, a teenager in your youth group, has recently become more withdrawn during activities. She often arrives late and seems distracted when participating. You've noticed she hasn't been engaging with her usual group of friends. What steps would you take to support Sylvia in re-engaging with the group?\"\r\n\r\nSample answer\r\nTo answer this question successfully, outline the steps you would take and explain why they would be effective. Here's an example of an effective answer:\r\n\r\n\"To re-engage Sylvia with her youth support group, I would talk to her privately to understand her situation and offer support. This would help to build trust. I would encourage peer support by suggesting activities and groups she might try, that could help her make supportive friendships. If these steps don\u2019t work, I would recommend she seeks guidance from a counsellor to address her issues and learn management strategies.\"\r\n\r\n\r\n\r\n","n":0.045},[{"v":"Assignment planning","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":53,"$":[{"v":"Instructions, explanations and reflection","n":0.5},{"v":".gap {   width: 100px; display: inline-block;border-bottom: 2px dotted #333333; /* Creates an underscored line */ width: 8rem; /* Sets the length of the line */ height: 1em; /* Keeps the line inline with text */ }\r\nOn this page, we'll explore three types of instruction words that you may come across in your assessments. Learning about these instruction words will help you complete your assessments with confidence.\r\nWe'll focus on instruction words that:\r\n\r\n \texplain activities\r\n \tindicate that you need to provide information or ideas\r\n \ttell you that you need to review or reflect.\r\n\r\nActivity instructions\r\nIn some assessments, you will be asked to complete activities. These activities are often signalled by words such as:\r\n\r\n \tlabel\r\n \tmatch\r\n \tfill in\r\n \tcomplete\r\n\r\nFor example:\r\nLabel the diagram with the correct chair parts.\r\n\r\nA satisfactory response to this prompt would be to simply place the correct names of chair parts into the correct points of the diagram. No explanation or other information is needed.\r\n\r\nProvide information and ideas\r\nWatch this video to learn about instruction words that indicate you need to provide information, ideas and examples.\r\n\r\n\r\n\r\nReview and reflect\r\nThis group of instruction words indicates that you need to consult other sources of information for your answer.\r\n\r\n \trefer (to)\r\n \texamine\r\n \treflect\r\n \tresearch\r\n \tinvestigate\r\n \tconsider\r\n \tread\r\n\r\nThey will usually be accompanied by a second instruction word that tells you what you need to do once you have referred to the other piece of information.\r\nHere are some examples.\r\n\r\n \r\nInstruction words like \u2018come up with,\u2019 \u2018provide,\u2019 and \u2018suggest\u2019 ask you to deliver more than a simple answer. Your teacher will expect further information, examples and ideas. \r\n\r\nHere's an example assignment prompt: \u201cFandangle Fashion has faced a year of falling sales. The director asks your group to find a solution to improve customer engagement and boost sales. Come up with one solution to this issue.\" \r\n \r\nHow can you respond effectively? \r\nThis prompt has no single correct answer, however a simple response like \"Increase advertising\" isn't enough. \r\n \r\nIf you suggest a solution and explain how and why it will solve the problem, you are more likely to get a satisfactory mark. \r\n\r\nFor example, you might answer:  \r\n\u201cA solution for this issue is to invest in market research to help the company to identify their target audience and create a campaign that speaks directly to that audience. Targeted advertising can then increase product visibility, enhance brand recognition and drive customer interest, ultimately contributing to increased sales.\" \r\n\r\nThis answer is more suitable because it clearly states the solution, explains how it would be done, and why it would be effective.  \r\n \r\nThis way, you\u2019re more likely to convince the Director of Fandangle Fashion and impress your teacher. \r\n","n":0.048},[{"v":"Assignment planning","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":54,"$":[{"v":"Understanding instruction words","n":0.577},{"v":"Have you ever been unsure if you're answering an assessment question or prompt in the right way? On this page, we will explain instruction words — the words that tell you what you need to do. Understanding these words will help you complete your assessments successfully.\r\n\r\nPlay this video to find out more.\r\n\r\n\r\n\r\nQuestion and prompt types\r\nNow, we'll explore question and prompt instruction words in more detail.\r\n\r\nSimple answer questions and prompts\r\nLet's start with words that require you to respond with specific information:\r\n\r\nidentifylistnameoutline\r\n\r\nFor questions or prompts with these words, you need to provide the correct information. You do not need to provide an explanation or analysis.\r\n\r\nFor example:\r\n\"List\u00a0the steps required to create a care plan.\"\r\n\r\nTo answer this question correctly, all you need to do is write the correct steps. You do not need to describe why the steps are important or analyse if the steps are effective.\r\n\r\n\r\n\r\nQuestions and prompts that require explanation\r\n\r\n\r\nThe following words require you to provide explanation, description and/or analysis in your answer.\r\n\r\nanalysecomparedescribediscussexplain\r\n\r\nHere's an example prompt:\r\n\r\n\"Explain why sharps should be disposed of correctly.\"\r\nNow, let's explore some sample answers.\r\n\r\n  \r\n    \"It is very important that sharps are disposed of correctly.\"\r\n    The student has not answered this question properly. They have said that it is important to dispose of sharps correctly, but not explained why it is important.\r\n  \r\n\r\n\r\n\r\n  \r\n    \"Sharps should be disposed of correctly for safety.\"\r\n    This answer has explained why it is important to dispose of sharps correctly, but they have not given enough detail. A more effective answer would provide detail about how sharps disposal and safety are related.\r\n  \r\n\r\n\r\n\r\n  \r\n    \"Sharps should be disposed of correctly to prevent needle stick injuries, which can transmit infections. Proper disposal ensures a safe environment for healthcare workers and the public by reducing these risks.\"\r\n    This answer has provided an explanation of why it is important to dispose of sharps correctly and the risks of incorrect disposal, so it meets the requirements of the prompt.\r\n  \r\n\r\n\r\n\r\n\r\n \r\nAssignment instructions can be confusing.  \r\n\r\nLet's say you\u2019ve been given a series of prompts, but you're not sure exactly what you need to provide in an answer. \r\n\r\nHere are two examples.  \r\n\r\nPrompt 1: Name the law that protects personal data in social care.  \r\n\r\nFor this prompt, the key words are \"Name the law\". It's asking for a straightforward answer.  \r\n\r\nSo, responding with \"the Privacy Act 1988\" is all that's required. You don\u2019t need to explain or analyse anything.  \r\n\r\nPrompt 2: Describe one way that communication skills can impact client outcomes in social care.  \r\n\r\nThe key word here is \"describe.\" So, you need to provide an explanation. \r\n\r\nFor example, if your answer is that trust can impact client outcomes, simply answering \"trust\" isn\u2019t enough; you need to explain how it impacts outcomes.  \r\n\r\nA better answer might be:  \r\n\r\n\"Communication skills can impact the level of trust a client has towards their carer. Clear, respectful, and polite communication can build trust, while rude or disrespectful communication can decrease it.\"  \r\n\r\nBy describing clearly how trust impacts outcomes, you will have fully answered the prompt. \r\n","n":0.045},[{"v":"Understanding assignments","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":55,"$":[{"v":"Life science","n":0.707},{"v":"\r\n\r\n\r\n\r\n\tStructure and function of cells\r\n\tMolecular biology\r\n\tGenetics\r\n\tHuman body systems\r\n\r\n\r\n\r\n\r\nFurther resources\r\n\r\n\r\n\r\n","n":0.333},[{"v":"Biology","i":1,"n":1},{"v":"Life science","i":0,"n":0.707}]]},{"i":56,"$":[{"v":"Predicting traits","n":0.707},{"v":"Punnett squares are vital tools for understanding how genetic traits are inherited. They provide a simple way to predict the probabilities of offspring inheriting certain traits based on the genetic makeup of the parents. This knowledge is crucial in fields like medicine, where it helps assess the risk of genetic disorders, and in agriculture, where it aids in breeding for desirable traits. Use this resource to learn how Punnett squares can simplify complex genetic predictions. \r\n\r\nWhen two organisms are crossed, the produce offspring that inherit genetic information from both parents. Punnett squares are diagrams used to predict the genetic outcomes of a cross. \r\n\r\nThey illustrate how alleles from each parent combine, showing the possible genotypes of their offspring. Punnett squares are helpful for visualising how traits are inherited, and for understanding the probability of offspring expressing certain traits.\r\n\r\nUsing Punnett squares\r\nTo set up and use a Punnett square:\r\n\r\n\tIdentify the parent genotypes.\r\n\tSet up the grid by drawing a square and dividing it into four smaller squares.\r\n\tAssign letters for the dominant and recessive alleles, if they are not already given. Write one parent's alleles across the top and the other's on the side. Each allele gets its own row or column.\r\n\tFill in the square. For each box, combine the allele from the top with the allele on the side. This will show the possible genotypes of the offspring.\r\n\tLook at the completed Punnett square. Determine the possible genotypes and phenotypes of the offspring, including their ratios or probabilities.\r\n\r\n\r\nExample\r\n\r\nIn pea plants, purple flowers \\((P)\\) are dominant over white flowers \\((p)\\). If a plant with heterozygous purple flowers \\((Pp)\\) is crossed with a plant with white flowers \\((pp)\\), what are the possible genotypes and phenotypes of the offspring?\r\n\r\n\r\n\tIdentify the parent genotypes.\r\n\r\nOne parent is heterozygous \\((Pp)\\) and the other is homozygous recessive \\((pp)\\).\r\n\r\n\tSet up the grid by drawing a square and dividing it into four smaller squares.\r\n\tAssign letters for the dominant and recessive alleles, if they are not already given. Write one parent's alleles across the top and the other's on the side. Each allele gets its own row or column.\r\n\r\nThe dominant allele for purple is \\(P\\) and the recessive allele for white is \\(p\\). Let's write the alleles for the heterozygous parent (\\(P\\) and \\(p\\)) across the top. We can then write the alleles for the homozygous recessive parent (\\(p\\) and \\(p\\)) down the side.\r\n\r\n  \r\n    \r\n    P\r\n    p\r\n  \r\n  \r\n    p\r\n    Pp\r\n    pp\r\n  \r\n  \r\n    p\r\n    Pp\r\n    pp\r\n  \r\n\r\n\r\n\tFill in the square. For each box, combine the allele from the top with the allele on the side. This will show the possible genotypes of the offspring.\r\n\r\n\r\n  \r\n    \r\n    P\r\n    p\r\n  \r\n  \r\n    p\r\n    \r\n    \r\n  \r\n  \r\n    p\r\n    \r\n    \r\n  \r\n\r\n\r\n\tLook at the completed Punnett square. Determine the possible genotypes and phenotypes of the offspring, including their ratios or probabilities.\r\n\r\nThe genotypes are \\(Pp\\) and \\(pp\\). The phenotypes are purple \\((Pp)\\) and white \\((pp)\\). The ratios for the genotypes are \\(2\\,Pp:2\\,pp\\) or \\(1:1\\). The ratios for the phenotypes are \\(2\\) purple : \\(2\\) white or \\(1:1\\).\r\nThis means that \\(50\\%\\) of the offspring will have purple flowers and \\(50\\%\\) will have white flowers.\r\n\r\nExercise\r\n\r\n\tUse a Punnett square to determine the genotypes and phenotypes, along with their ratios, for the following situations.\r\n\t\r\n\t\tIn mice, black fur \\((B)\\) is dominant over white fur \\((b)\\). If a heterozygous black-furred mouse \\((Bb)\\) is crossed with a white-furred mouse \\((bb)\\), what are the possible fur colours of the offspring?\r\n\t\tIn flowers, tall stems \\((T)\\) are dominant over short stems \\((t)\\). Cross a homozygous tall plant \\((TT)\\) with a heterozygous tall plant \\((Tt)\\). What are the possible stem heights of the offspring?\r\n\t\tIn guinea pigs, rough coat \\((R)\\) is dominant over smooth coat \\((r)\\). If two heterozygous rough-coated guinea pigs \\((Rr)\\) mate, what are the possible coat textures of their offspring?\r\n\t\tIn humans, having a cleft chin \\((C)\\) is dominant over having a smooth chin \\((c)\\). If one parent is homozygous for a cleft chin \\((CC)\\) and the other parent has a smooth chin \\((cc)\\), what are the possible chin types of their children?\r\n\t\r\n\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nLisa, a genetics researcher, is investigating the distribution of a particular trait in a local bird population. She focuses on the trait controlled by a single gene with two alleles: A (dominant) and a (recessive). The possible genotypes are AA, Aa, and aa.\r\nLisa collected data on the genotype frequencies within the population to better understand how genetic diversity influences trait prevalence. This is shown in the pie chart.\r\n\r\n\r\n\r\n\r\n \r\nPunnett square\r\nAn example of a Punnett square. The genotype across the top, capital P and lower case p, represents a pea plant with purple flowers and the genotype down the side, lower case p and lower case p, represents a pea plant with white flowers.\r\n\r\n\t\r\n\t\t\r\n\t\t\t\r\n\t\t\tP\r\n\t\t\tp\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tp\r\n\t\t\tPp\r\n\t\t\tpp\r\n\t\t\r\n\t\t\r\n\t\t\tp\r\n\t\t\tPp\r\n\t\t\tpp\r\n\t\t\r\n\t\r\n\r\n \r\nA pie chart showing genotype frequencies\r\nA pie chart the percentage of different genotypes within a population. The genotypes are AA (homozygous dominant), Aa (heterozygous) and aa (homozygous recessive).\r\n\r\n\t\r\n\t\t\r\n\t\t\tGenotype\r\n\t\t\tFrequency\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tAA (homozygous dominant)\r\n\t\t\t\\(45\\%\\)\r\n\t\t\r\n\t\t\r\n\t\t\tAa (heterozygous)\r\n\t\t\t\\(40\\%\\)\r\n\t\t\r\n\t\t\r\n\t\t\taa (homozygous recessive)\r\n\t\t\t\\(10\\%\\)\r\n\t\t\r\n\t\r\n\r\n","n":0.035},[{"v":"Inheritance","i":1,"n":1},{"v":"Genetics","i":0,"n":1}]]},{"i":57,"$":[{"v":"Patterns of inheritance","n":0.577},{"v":" Patterns of inheritance describe how genetic traits are passed from one generation to the next. Understanding these patterns is crucial for predicting how traits will appear in offspring and for studying genetic disorders. Use this resource to explore how genes combine and interact to create the diversity of life.\r\n\r\nBefore we explore how genetic traits are passed from one generation to the next, we must understand the genetic structures that are important for inheritance, and how inheritance through generations is represented. Both are shown using visual representations: karyotypes and pedigrees. \r\n\r\nKaryotypes\r\nA karyotype is a visual representation of an organism's complete set of chromosomes, arranged in pairs. It provides a snapshot of the genome, showing all the chromosomes found in the nucleus of a cell. \r\nA human karyotype shows a complete set of \\(46\\) chromosomes in a typical body cell, arranged in \\(23\\) pairs. Of these, \\(22\\) pairs are autosomes, labelled from \\(1\\) to \\(22\\) by size. Chromosome \\(1\\) is the largest. \r\nThe \\(23\\)rd pair is the sex chromosomes, \\(X\\) and \\(Y\\), which determine a person's sex. Females usually have two \\(X\\) chromosomes \\((XX)\\), while males have one \\(X\\) and one \\(Y\\) chromosome \\((XY)\\).\r\n\r\nA karyotype gives important insights into genetic health and variations, such as by showing changes like extra or missing chromosomes.\r\n\r\n\r\nPedigrees\r\nA pedigree is a chart that shows how a trait is passed down generations. It is a family tree where:\r\n\r\n\tMales are shown as squares.\r\n\tFemales are shown as circles.\r\n\tA horizontal marriage line is drawn between parents.\r\n\tA vertical line of descent is drawn from the marriage line to the next generation. \r\n\tChildren are shown in the next generation, with a sibling line linking them.\r\n\r\nDifferent colours are used to show whether an individual is affected or unaffected by a given trait\u2014in other words, whether they have or do not have the trait. Here, we will show affected individuals with red and unaffected individuals with white. \r\n\r\n\r\n\r\nAutosomal traits\r\nAutosomal traits are genetic characteristics determined by genes located on the \\(22\\) pairs of non-sex chromosomes, known as autosomes. These traits can be inherited as either dominant or recessive, influencing whether a single copy or two copies of an allele are needed to express a trait.\r\nAutosomal traits affect individuals regardless of their sex.\r\n\r\nAutosomal dominant traits\r\n\r\nAn autosomal dominant trait requires only one copy of the dominant allele for the trait to be expressed. This means if a child inherits a dominant allele from at least one parent, they will display the trait. \r\nCommon examples include traits like widow's peak, freckles or certain genetic conditions such as Huntington's disease.\r\nEven if only one parent carries the dominant allele, there's a \\(50\\%\\) chance for the child to inherit the trait.\r\n\r\nTo spot an autosomal dominant trait, check the parents of the affected individuals. If one or more parent is affected, then it is an autosomal dominant trait.\r\n\r\nAutosomal recessive traits\r\nAn autosomal recessive trait requires two copies of the recessive allele for the trait to be expressed. This means if a child displays a trait that neither parent shows, both parents must be carriers of the recessive allele. A carrier is a person who has the allele for the gene but does not express it.\r\nExamples include traits like straight hairline or conditions such as cystic fibrosis. Albinoism, where the body makes little or no melanin, is also an autosomal recessive trait.\r\n\r\nIf both parents are carriers, there's a \\(25\\%\\) chance that their child will inherit and express the trait. \r\n\r\nTo spot an autosomal recessive trait, check the parents of the affected individuals. If neither of them are affected, then it is an autosomal recessive trait. Autosomal recessive traits also skips generations.\r\n\r\nSex-linked traits\r\nSex-linked traits are associated with genes located on the sex chromosomes, particularly the \\(X\\) chromosome. These traits often exhibit different inheritance patterns in males and females due to the presence of one \\(X\\) chromosome in males and two in females.\r\n\r\n\\(X\\)-linked dominant traits\r\n\\(X\\)-linked dominant traits are caused by dominant alleles on the \\(X\\) chromosome. A child inherits the trait when they have a single copy of a dominant allele, shown as \\(X^{A}\\). \r\n\r\n\tIf a female has \\(X^{A}X\\) or \\(X^{A}X^{A}\\), they will show the trait because of the dominant allele. An affected father (\\(X^{A}Y\\)) will pass the trait on to all daughters (\\(X^{A}X\\)) but no sons.\r\n\tIf a male has \\(X^{A}Y\\), they will show the trait because they have one \\(X\\) chromosome with the dominant allele. An affected mother (\\(X^{A}X\\) or \\(X^{A}X^{A}\\)) can pass the trait to both sons (\\(X^{A}Y\\)) and daughters (\\(X^{A}X\\)).\r\n\r\nAicardi syndrome, where an individual has a partial or absent structure in their brain called the corpus callosum, is \\(X\\)-linked dominant. There are not many well-documented \\(X\\)-linked traits that are not disorders. \r\n\r\nTo spot an \\(X\\)-linked dominant trait, check the affected individuals. If a father is affected, all his daughters will be affected. If a mother is affected, her sons and daughters can be affected.\r\n\r\n\\(X\\)-linked recessive traits\r\n\\(X\\)-linked recessive traits result from recessive alleles on the \\(X\\) chromosome. Males are more affected, as they have only one \\(X\\) chromosome and only need one recessive allele (\\(X^{a}\\)) to express the trait. \r\n\r\n\tMales with \\(X^{a}Y\\) express the trait because they have one \\(X\\) chromosome with the recessive allele. They usually inherit the trait from a carrier (\\(X^{A}X^{a}\\)) or affected mother (\\(X^{a}X^{a}\\))\r\n\tAffected females must inherit two recessive alleles to express the trait; they must have the genotype \\(X^{a}X^{a}\\). If they have \\(X^{A}X^{a}\\), they are carriers but do not show the trait.\r\n\r\nSince males only have one \\(X\\) chromosome, they are more likely to express \\(X\\)-linked recessive traits.\r\nAn example of an \\(X\\)-linked recessive trait is red\u2013green colour blindness.\r\n\r\n\r\nTo spot an \\(X\\)-linked recessive trait, check the affected individuals. If mostly males are affected and the trait skips generations, it is likely \\(X\\)-linked recessive. An affected female usually has an affected father.\r\n\r\n\\(Y\\)-linked traits\r\nY-linked traits are associated with genes located on the \\(Y\\) chromosome. They are only present in males as females do not have a \\(Y\\) chromosome. These traits are passed directly from father to son, as the \\(Y\\) chromosome is only inherited through the paternal line.\r\nExamples of \\(Y\\)-linked traits are webbed toes and hairy ears.\r\n\r\nTo spot a \\(Y\\)-linked trait, check whether the trait only affects males and is passed from father to son. No females are affected by \\(Y\\)-linked traits.\r\n\r\nSummary\r\nTo help you interpret pedigrees and identify whether a trait is autosomal dominant, recessive, or sex-linked, you can use this flowchart.\r\n\r\nExercise\r\nSee how well you understand patterns of inheritance with a quick quiz.\r\n\r\n\r\nReady for some more? Practise your skills by determining the pattern of inheritance of traits based on pedigrees.\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nAlesander is a second year masters student studying the inheritance patterns of Duchenne Muscular Dystrophy (DMD), a severe type of muscular dystrophy that affects mostly boys. DMD is caused by a mutation in the DMD gene on the X chromosome. \r\nAlesander has collected data from several families to understand how this trait is passed down through generations.\r\n\r\n\r\n\r\n \r\nPedigree\r\nA pedigree with three generations is shown. Females are shown as circles and males as squares. If they are affected by a trait, the shape is shaded in red. A marriage line is used between mothers and fathers. If they have children, a line of descent is shown. If there are multiple children, a sibling line is used to join them.\r\n \r\nPedigree\r\n\r\n\tFirst generation:\r\n\t\r\n\t\tFemale is married to male. Neither are affected by the trait.\r\n\t\tThey have three children.\r\n\t\r\n\tSecond generation:\r\n\t\r\n\t\tTwo sons and one daughter are all affected by the trait.\r\n\t\tThe first son has three children with an unaffected female.\r\n\t\tThe second son does not have children.\r\n\t\tThe daughter has one child with an unaffected male.\r\n\t\r\n\tThird generation:\r\n\t\r\n\t\tThe first family has one affected son and two unaffected sons.\r\n\t\tThe affected son has two children with an unaffected female.\r\n\t\tNeither of the unaffected sons have children.\r\n\t\tThe second family has one unaffected daughter.\r\n\t\tThe unaffected daughter has one child with an unaffected male.\r\n\t\r\n\tFourth generation:\r\n\t\r\n\t\tThe first family has one affected daughter and one affected son.\r\n\t\tThe second family has one affected daughter.\r\n\t\r\n\r\n \r\nPedigree for autosomal dominant trait\r\n\r\n\tFirst generation:\r\n\t\r\n\t\tAn affected female is married to an unaffected male.\r\n\t\tThey have three children.\r\n\t\r\n\tSecond generation:\r\n\t\r\n\t\tThe two sons are affected and the daughter is unaffected.\r\n\t\tThe first son has three children with an unaffected female.\r\n\t\tThe second son does not have children.\r\n\t\tThe daughter has one child with an unaffected male.\r\n\t\r\n\tThird generation:\r\n\t\r\n\t\tThe first family has two affected sons and one unaffected son.\r\n\t\tThe first affected son has two children with an unaffected female.\r\n\t\tThe second affected son and unaffected son do not have children.\r\n\t\tThe second family has one unaffected daughter.\r\n\t\tThe unaffected daughter has one child with an unaffected male.\r\n\t\r\n\tFourth generation:\r\n\t\r\n\t\tThe first family has one affected daughter and one unaffected son.\r\n\t\tThe second family has one unaffected daughter.\r\n\t\r\n\r\n \r\nPedigree for autosomal recessive trait\r\n\r\n\tFirst generation:\r\n\t\r\n\t\tAn affected female is married to an unaffected male.\r\n\t\tThey have three children.\r\n\t\r\n\tSecond generation:\r\n\t\r\n\t\tThe two sons and one daughter are unaffected.\r\n\t\tThe first son has three children with an unaffected female.\r\n\t\tThe second son does not have children.\r\n\t\tThe daughter has one child with an unaffected male.\r\n\t\r\n\tThird generation:\r\n\t\r\n\t\tThe first family has one affected son and two unaffected sons.\r\n\t\tThe first unaffected son has two children with an affected female.\r\n\t\tThe second unaffected son and affected son do not have children.\r\n\t\tThe second family has one affected daughter.\r\n\t\tThe affected daughter has one child with an unaffected male.\r\n\t\r\n\tFourth generation:\r\n\t\r\n\t\tThe first family has one unaffected daughter and one affected son.\r\n\t\tThe second family has one unaffected daughter.\r\n\t\r\n\r\n \r\nPedigree for X-linked dominant trait\r\n\r\n\tFirst generation:\r\n\t\r\n\t\tAn affected female is married to an unaffected male.\r\n\t\tThey have three children.\r\n\t\r\n\tSecond generation:\r\n\t\r\n\t\tThe two sons are affected and the daughter is unaffected.\r\n\t\tThe first son has three children with an unaffected female.\r\n\t\tThe second son does not have children.\r\n\t\tThe daughter has one child with an unaffected male.\r\n\t\r\n\tThird generation:\r\n\t\r\n\t\tThe first family has two affected sons and one unaffected son.\r\n\t\tThe first unaffected son has two children with an unaffected female.\r\n\t\tThe second unaffected son and affected son do not have children.\r\n\t\tThe second family has one unaffected daughter.\r\n\t\tThe unaffected daughter has one child with an unaffected male.\r\n\t\r\n\tFourth generation:\r\n\t\r\n\t\tThe first family has one unaffected daughter and one unaffected son.\r\n\t\tThe second family has one unaffected daughter.\r\n\t\r\n\r\n \r\nPedigree for X-linked recessive trait\r\n\r\n\tFirst generation:\r\n\t\r\n\t\tAn unaffected female is married to an unaffected male.\r\n\t\tThey have three children.\r\n\t\r\n\tSecond generation:\r\n\t\r\n\t\tOne son is affected and the other son and the daughter are unaffected.\r\n\t\tThe affected son has three children with an unaffected female.\r\n\t\tThe unaffected son does not have children.\r\n\t\tThe daughter has one child with an affected male.\r\n\t\r\n\tThird generation:\r\n\t\r\n\t\tThe first family has three unaffected sons.\r\n\t\tThe first unaffected son has two children with an affected female.\r\n\t\tThe other unaffected sons do not have children.\r\n\t\tThe second family has one unaffected daughter.\r\n\t\tThe unaffected daughter has one child with an unaffected male.\r\n\t\r\n\tFourth generation:\r\n\t\r\n\t\tThe first family has one unaffected daughter and one affected son.\r\n\t\tThe second family has one unaffected daughter.\r\n\t\r\n\r\n \r\nPedigree for Y-linked trait\r\n\r\n\tFirst generation:\r\n\t\r\n\t\tAn unaffected female is married to an unaffected male.\r\n\t\tThey have three children.\r\n\t\r\n\tSecond generation:\r\n\t\r\n\t\tOne son is affected and the other son and the daughter are unaffected.\r\n\t\tThe unaffected son has three children with an unaffected female.\r\n\t\tThe affected son does not have children.\r\n\t\tThe daughter has one child with an affected male.\r\n\t\r\n\tThird generation:\r\n\t\r\n\t\tThe first family has three unaffected sons.\r\n\t\tThe first unaffected son has two children with an unaffected female.\r\n\t\tThe other unaffected sons do not have children.\r\n\t\tThe second family has one unaffected daughter.\r\n\t\tThe unaffected daughter has one child with an affected male.\r\n\t\r\n\tFourth generation:\r\n\t\r\n\t\tThe first family has one unaffected daughter and one affected son.\r\n\t\tThe second family has one unaffected daughter.\r\n\t\r\n\r\n \r\nWhat is the pattern of inheritance of a trait?\r\n\r\nDo all affected individuals have an affected parent?\r\n\r\n  Yes: Dominant\r\n    \r\n      Do all affected males have an affected mother?\r\n        \r\n          Yes: Are the daughters of an affected father all affected?\r\n            \r\n              Yes: Are the sons of an affected father all affected?\r\n                \r\n                  Yes: Autosomal dominant\r\n                  No: X-linked dominant\r\n                \r\n              \r\n              No: Autosomal dominant\r\n            \r\n          \r\n          No: Autosomal dominant\r\n        \r\n      \r\n    \r\n  \r\n  No: Recessive\r\n    \r\n      Does the trait mostly affect males?\r\n        \r\n          Yes: Do all affected fathers have affected sons?\r\n            \r\n              Yes: Y-linked\r\n              No: X-linked recessive\r\n            \r\n          \r\n          No: Autosomal recessive\r\n        \r\n      \r\n    \r\n  \r\n\r\n\r\n\t\r\n \r\n\tPedigree for part a\r\n\t\r\n\t\tFirst generation\r\n\t\t\r\n\t\t\tAn unaffected female is married to an affected male.\r\n\t\t\tThey have four children.\r\n\t\t\r\n\t\t\r\n\t\tSecond generation\r\n\t\t\r\n\t\t\tTwo daughters are affected and two sons are unaffected.\r\n\t\t\tOne affected daughter is married to an unaffected male.\r\n\t\t\tThe other children do not have any children of their own.\r\n\t\t\r\n\t\t\r\n\t\tThird generation\r\n\t\t\r\n\t\t\tThe family has one affected son, one unaffected son, one affected daughter and one unaffected daughter.\r\n\t\t\r\n\t\t\r\n\t\r\n\t \r\n\tPedigree for part b\r\n\t\r\n\t\tFirst generation\r\n\t\t\r\n\t\t\tAn unaffected female is married to an affected male.\r\n\t\t\tThey have six children.\r\n\t\t\r\n\t\t\r\n\t\tSecond generation\r\n\t\t\r\n\t\t\tThree sons are affected and none of the three daughters are unaffected.\r\n\t\t\tOne unaffected daughter is married to an unaffected male.\r\n\t\t\tThe other children do not have any children of their own.\r\n\t\t\r\n\t\t\r\n\t\tThird generation\r\n\t\t\r\n\t\t\tThe family has two sons and two daughters. None of them are affected.\r\n\t\t\r\n\t\t\r\n\t\r\n\t \r\n\tPedigree for part c\r\n\t\r\n\t\tFirst generation\r\n\t\t\r\n\t\t\tAn affected female is married to an unaffected male.\r\n\t\t\tThey have six children.\r\n\t\t\r\n\t\t\r\n\t\tSecond generation\r\n\t\t\r\n\t\t\tTwo daughters are affected and one son is affected. The other two sons and one daughter are unaffected.\r\n\t\t\tOne affected daughter is married to an unaffected male.\r\n\t\t\tThe other children do not have any children of their own.\r\n\t\t\r\n\t\t\r\n\t\tThird generation\r\n\t\t\r\n\t\t\tThe family has two affected sons, one affected daughter and one unaffected daughter.\r\n\t\t\r\n\t\t\r\n\t\r\n\t \r\n\tPedigree for part d\r\n\t\r\n\t\tFirst generation\r\n\t\t\r\n\t\t\tAn unaffected female is married to an affected male.\r\n\t\t\tThey have four children.\r\n\t\t\r\n\t\t\r\n\t\tSecond generation\r\n\t\t\r\n\t\t\tBoth of their sons are affected.\r\n\t\t\tOne affected son is married to an affected female.\r\n\t\t\tThe other affected son is married to an unaffected female.\r\n\t\t\r\n\t\t\r\n\t\tThird generation\r\n\t\t\r\n\t\t\tThe first family has one unaffected son.\r\n\t\t\tThe second family has one affected daughter.\r\n\t\t\r\n\t\t\r\n\t\tFourth generation\r\n\t\t\r\n\t\t\tThe unaffected son in the first family marries an unaffected female, with whom he has three unaffected sons.\r\n\t\t\tThe affected daughter in the second family marries an unaffected male, with whom she has one affected son.\r\n\t\t\r\n\t\t\r\n\t\r\n\t \r\n\tPedigree for part e\r\n\t\r\n\t\tFirst generation\r\n\t\t\r\n\t\t\tAn unaffected female is married to an affected male.\r\n\t\t\tThey have four children.\r\n\t\t\r\n\t\t\r\n\t\tSecond generation\r\n\t\t\r\n\t\t\tTwo sons and two daughters are unaffected.\r\n\t\t\tOne unaffected daughter is married to an unaffected male.\r\n\t\t\tThe other children do not have any children of their own.\r\n\t\t\r\n\t\t\r\n\t\tThird generation\r\n\t\t\r\n\t\t\tThe family has one affected son, one unaffected son, and two unaffected daughters.\r\n\t\t\r\n\t\t\r\n\t\r\n\t \r\n\tPedigree part f\r\n\t\r\n\t\tFirst generation\r\n\t\t\r\n\t\t\tAn affected female is married to an unaffected male.\r\n\t\t\tThey have three children.\r\n\t\t\r\n\t\t\r\n\t\tSecond generation\r\n\t\t\r\n\t\t\tTwo sons are unaffected and one son is affected.\r\n\t\t\tOne unaffected son marries an affected female. They have three children together.\r\n\t\t\tThe affected son marries an affected female. They have one child together.\r\n\t\t\tThe other son does not have any children of their own.\r\n\t\t\r\n\t\t\r\n\t\tThird generation\r\n\t\t\r\n\t\t\tThe first family has three unaffected sons.\r\n\t\t\tThe second family has one affected daughter.\r\n\t\t\r\n\t\t\r\n\t\r\n\t \r\n\tPedigree part g\r\n\t\r\n\t\tFirst generation\r\n\t\t\r\n\t\t\tAn affected female is married to an unaffected male.\r\n\t\t\tThey have four children.\r\n\t\t\r\n\t\t\r\n\t\tSecond generation\r\n\t\t\r\n\t\t\tTwo sons are affected and two daughters are unaffected.\r\n\t\t\tOne affected son is married to an affected female.\r\n\t\t\tOne unaffected daughter is married to an unaffected male.\r\n\t\t\tThe other two children do not have any children of their own.\r\n\t\t\r\n\t\t\r\n\t\tThird generation\r\n\t\t\r\n\t\t\tThe first family has two affected sons and one affected daughter.\r\n\t\t\tThe second family has one affected son, one unaffected son, and two unaffected daughters.\r\n\t\t\r\n\t\t\r\n\t\r\n\t","n":0.02},[{"v":"Inheritance","i":1,"n":1},{"v":"Genetics","i":0,"n":1}]]},{"i":58,"$":[{"v":"Mendelian genetics","n":0.707},{"v":"Mendelian genetics help to explain how traits are inherited. This knowledge is vital in fields like medicine, where it helps identify patterns of genetic disorders, and in agriculture, where it guides the breeding of plants and animals. Use this resource to learn about the foundational concepts that explain genetic inheritance.\r\n\r\nMendel's laws\r\nMendelian genetics explores how traits are inherited based on the work of Austrian biologist Gregor Mendel. His principles of inheritance\u2014segregation, independent assortment and dominance\u2014are key to understanding how traits pass from parents to offspring. These are called Mendel\u2019s laws.\r\n\r\nLaw of segregation\r\nThe law of segregation is one of Mendel\u2019s main ideas in genetics. It states that during the formation of reproductive cells (gametes), the two alleles for each trait separate, or segregate, so that each gamete receives only one allele.\r\n\r\n\r\nThis means that offspring inherit one allele for each trait from each parent. The law of segregation explains how genetic variation occurs, as the mixing of alleles can lead to different trait combinations in offspring.\r\nAn example of the law of segregation is seen in human blood type inheritance. Consider a parent with the genotype AO for blood type. During gamete formation, the A and O alleles segregate so that one gamete receives the A allele and another receives the O allele. \r\nIf two AO parents have children, their offspring can inherit different combinations of these alleles, resulting in blood types A, O, or even AA, depending on the alleles contributed by each parent.\r\n\r\nLaw of independent assortment\r\nThe law of independent assortment states that alleles for different traits are distributed to gametes independently of one another. This means the inheritance of one trait generally does not affect the inheritance of another. \r\nFor example, the genes for human eye colour and hair colour are inherited separately. They are located on different chromosomes. As a result, the allele a person inherits for eye colour (such as blue or brown) is inherited independently of the allele for hair colour (such as blonde or black). This is why you can have various combinations, like people with blue eyes and black hair or brown eyes and blonde hair, showing the independent assortment of these traits.\r\nIndependent assortment occurs because genes are located on different chromosomes that align and segregate independently during meiosis. This process is enhanced by crossing over, where segments of chromosomes exchange genetic material, further increasing genetic diversity by mixing alleles for genes on the same chromosome.\r\n\r\nLaw of dominance\r\n\r\nThe law of dominance is one of Mendel\u2019s main ideas in genetics. It states that when there are two different versions of a gene (or two alleles), one can be dominant and hide the other, which is recessive.\r\nThis means that if an organism has one dominant and one recessive allele for a trait, the dominant one will show up in its appearance. The recessive one will only show if both alleles are recessive. This law helps to explain why some traits appear in offspring even if a parent doesn\u2019t show it.\r\nA classic example of the law of dominance is seen in pea plant flower colour. In a famous experiment, Mendel crossed a pure-breeding purple-flowered plant \\((PP)\\) with a pure-breeding white-flowered plant \\((pp)\\). All of the offspring \\((Pp)\\) had purple flowers.\r\nHere, the purple allele \\((P)\\) is dominant over the white allele \\((p)\\), so even though the offspring have each of one allele, they have the purple flower phenotype.\r\n\r\nExercise\r\nSee how well you understand Mendelian genetics with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nDr. Pomelo conducts a field study to determine whether there is a relationship between pea plant flower colour and petal size. He surveys the phenotypes of six pea plants and records them in a table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tPlant ID\r\n\t\t\tFlower colour\r\n\t\t\tPetal size\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\t1\r\n\t\t\tPurple\r\n\t\t\tLarge\r\n\t\t\r\n\t\t\r\n\t\t\t2\r\n\t\t\tPurple\r\n\t\t\tSmall\r\n\t\t\r\n\t\t\r\n\t\t\t3\r\n\t\t\tWhite\r\n\t\t\tLarge\r\n\t\t\r\n\t\t\r\n\t\t\t4\r\n\t\t\tWhite\r\n\t\t\tSmall\r\n\t\t\r\n\t\t\r\n\t\t\t5\r\n\t\t\tPurple\r\n\t\t\tLarge\r\n\t\t\r\n\t\t\r\n\t\t\t6\r\n\t\t\tWhite\r\n\t\t\tSmall\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n \r\nLaw of segregation\r\nA flowchart showing using blood types as an example of the law of segregation.\r\n\r\n\tTop row shows two heterozygous parents. Both have the genotype AO for blood type.\r\n\tEach parent's genotype split into individual alleles: A and O. These correspond to the alleles carried by their gametes.\r\n\tGametes carrying the O allele from the first parent and the A allele from the second parent combine to create an offspring with the genotype AO. They are heterozygous.\r\n\r\n \r\nIndependent assortment\r\n\r\n\tParent cell has two sets of chromosomes.\r\n\tParent cell goes through interphase and the chromosomes duplicate.\r\n\tAfter meiosis I, two daughter cells are formed with a single set of chromosomes that are duplicated. The chromosomes have swapped DNA in a process called crossing over.\r\n\tAfter meiosis II, four daughter cells are formed with a single set of chromosomes. Each is genetically different from one another.\r\n\r\n","n":0.036},[{"v":"Inheritance","i":1,"n":1},{"v":"Genetics","i":0,"n":1}]]},{"i":59,"$":[{"v":"VE assessments","n":0.707},{"v":"\r\nSo, you've got a new assessment and you're not sure where to start. On the next few pages, we'll look at the conditions for your assessment (due date, submission, etc.) and how to understand the instructions (what you need to do and how you need to do it). We will also learn about what educators want to see in your submission.\r\n\r\nFirst steps\r\nWhat should I do first for my assessment? Watch this short video to find out.\r\n\r\n\r\nPossible assessment conditions\r\nThe instructions should also tell you essential information such as:\r\n\r\n \thow to apply for extensions or special consideration\r\n \thow your assessment will be graded\r\n(e.g. SAT / satisfactory v NYS / not satisfactory)\r\n \twhat to do if you have an Equitable Learning Plan (ELP).\r\n\r\nYou can also find out about:\r\n\r\nQuestions\r\n\r\n \thow many need to be answered\r\n \tthe types of questions (multiple choice, short answer, etc.)\r\n \thow many questions must be answered correctly\r\n \tthe word count range (if applicable) for answering questions (e.g. 50-100 words).\r\n\r\nSubmission\r\n\r\n \thow to submit the assessment (e.g. via Canvas)\r\n \tif there is a specific file or file type you need to submit (e.g. a completed template, Word doc, PDF, etc.)\r\n \tthe conditions for resubmission.\r\n\r\n\r\nMaryam has an assessment that she needs to complete. Read the assessment instructions and then answer the questions.\r\n\r\nAssessment instructions\r\nWhat\r\nAn online quiz with multiple-choice and short-answer questions.\r\n\r\nWhere\r\nThis assessment is to be completed in the student's own time.\r\n\r\nHow\r\nYou must complete all of the 22 questions in this assessment to be assessed as satisfactory. You must complete this task within the allocated time (90 minutes).\r\n\r\nThe assessment is accessed through the Canvas course shell, and you must complete and submit your responses within the allocated assessment timeframe.\r\n\r\nConditions\r\n\r\n \tThis assessment should be completed individually.\r\n \tClass notes may be used to help answer the questions.\r\n \tYou must answer all questions correctly to be deemed satisfactory in this assessment.\r\n \tYou will be assessed as satisfactory or not yet satisfactory.\r\n \tThe task must be completed in 90 minutes.\r\n\r\n\r\nQuiz\r\n\r\n \r\nNo matter what type of assessment you're given, your first step should always be to read the assessment instructions carefully.\r\n\r\nThey tell you exactly what you need to do so that the work you submit meets your teacher's expectations.\r\n\r\nThe assessment instructions will include conditions like:\r\n\r\nWhat you need to submit, that is, the type of assessment.\r\nFor example, it might be a knowledge task, a project task, or a practical task such as a role play.\r\n\r\nWhere the assessment will take place.\r\nIt might be online, in class, or in your own time and space.\r\nPractical tasks might require a specific location.\r\n\r\nHow you should complete the task.\r\nFor example, you might be required to complete every question. If you miss a condition like this and only complete some questions, you will not receive a 'satisfactory' mark.\r\n\r\nThe instructions will also include important details like: whether it's a group or individual assessment, if there is a time limit, and whether you're allowed to use things like notes, class materials, or the internet.\r\n\r\nIf you are unsure about anything in the assessment instructions, check your course materials and notes for supporting information.\r\n\r\nIf you are still unsure, contact your teacher.\r\n\r\n","n":0.044},[{"v":"Understanding assignments","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":60,"$":[{"v":"Cell reproduction","n":0.707},{"v":"For an organism to develop, repair tissues and stay healthy, its cells must reproduce. You need a good understanding of cell reproduction in medicine for studying cancer and developing treatments, and in agriculture for improving crops. Use this resource to explore the replication of DNA and the process of meiosis.\r\n\r\nCell reproduction is how cells divide to create new cells, allowing for growth, repair, and continuation of life. It includes processes like:\r\n\r\n\tDNA replication, where genetic material is copied\r\n\tmitosis, where two identical cells are produced\r\n\tmeiosis, where reproductive cells are produced.\r\n\r\nTogether, these processes ensure healthy development and reproduction. You have already learned about mitosis, so let\u2019s discuss DNA replication and meiosis.\r\n\r\nDNA replication\r\nDNA replication is the process by which a cell makes an exact copy of its DNA before cell division. This occurs during interphase.\r\nJust like mRNA transcription and protein translation, DNA replication follows the three main steps: initiation, elongation and termination.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nWatch this video to see DNA replication in action.\r\n\r\nMeiosis\r\nMeiosis is a special type of cell division that produces sex cells called gametes, such as sperm and eggs. Gametes are haploid cells, meaning that they have a single set of chromosomes. Compare this to mitosis, which produces diploid cells with two sets of chromosomes.\r\nMeiosis is crucial for sexual reproduction\u2014that is, reproduction involving two parents\u2014ensuring that offspring receive a consistent number of chromosomes.\r\nThis process involves two divisions and results in four genetically unique daughter cells. \r\n\r\n\r\nMeiosis I\r\nThe first division is called meiosis I. This begins after interphase, when the DNA has been replicated. The stages of meiosis I are:\r\n\r\n\tprophase I \u2013 The chromosomes condense and pair up. They exchange genetic material through a process called crossing over and recombination. The nuclear envelope breaks down and the spindle fibres begin to form.\r\n\tmetaphase I \u2013 Chromosome pairs line up along the equator of the cell.\r\n\tanaphase I \u2013 Each chromosome in the pair is pulled to opposite poles of the cell by the spindle fibres.\r\n\ttelophase I and cytokinesis \u2013 The chromosomes gather at each pole and the cell divides into two diploid daughter cells.\r\n\r\n\r\nAt the end of meiosis I, there are two haploid cells with duplicated chromosomes.\r\n\r\nMeiosis II\r\nThe second division is called meiosis II. The stages of meiosis II are:\r\n\r\n\tprophase II \u2013 The spindle fibres form again and the nuclear envelop breaks down in each daughter cell.\r\n\tmetaphase II \u2013 Chromosomes line up along the equator of the cell.\r\n\tanaphase II \u2013 Sister chromatids are separated and move to opposite poles of the cell.\r\n\ttelophase II and cytokinesis \u2013 The nuclear envelopes reform around the separated chromatids and the cell membranes pinch to form four haploid cells with unique genetic combinations.\r\n\r\n\r\nAt the end of meiosis II, there are four haploid cells, each with single chromosomes.\r\n\r\nMeiosis makes sure there is genetic diversity. Watch this video to see meiosis in action.\r\n\r\nExercise\r\nSee how well you understand cell reproduction with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nA group of students are writing a literature review on the cell biology of cancer. Anton has put his hand up to find some information on DNA replication errors.\r\nAnton spends a few hours reading research papers and comes up with the following data by bringing together values from five different publications.\r\n\r\n\t\r\n\t\t\r\n\t\t\tCell type\r\n\t\t\tError rate (mutations per 10 million bases)\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tHuman somatic cells\r\n\t\t\t\\(1\\)\r\n\t\t\r\n\t\t\r\n\t\t\tMouse somatic cells\r\n\t\t\t\\(0.8\\)\r\n\t\t\r\n\t\t\r\n\t\t\tHuman germ cells\r\n\t\t\t\\(0.5\\)\r\n\t\t\r\n\t\t\r\n\t\t\tMouse germ cells\r\n\t\t\t\\(0.3\\)\r\n\t\t\r\n\t\t\r\n\t\t\tHuman stem cells\r\n\t\t\t\\(0.8\\)\r\n\t\t\r\n\t\t\r\n\t\t\tMouse stem cells\r\n\t\t\t\\(0.7\\)\r\n\t\t\r\n\t\t\r\n\t\t\tHuman cancer cells\r\n\t\t\t\\(5\\)\r\n\t\t\r\n\t\t\r\n\t\t\tMouse cancer cells\r\n\t\t\t\\(4.5\\)\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n \r\nThe initiation step of DNA replication\r\nA diagram showing the initiation of DNA replication. A segment of double-stranded DNA is shown. On the right, a turqoise blob labelled \"Helicase\" is attached to the bottom DNA strand at a region called the \"Replication fork\". The segment of DNA to the left of the replication fork has been unwound.\r\n \r\nThe elongation step of DNA replication\r\nA diagram showing the elongation step of DNA replication. One purple blobs, \"DNA polymerase\", had bound to each unwound single strand of DNA.\r\nThe top strand of DNA has a DNA polymerase moving towards the replication fork. This enzyme makes a new strand of DNA to complement the top unwound strand, leaving behind double-stranded DNA called a \"Leading strand\".\r\nThe bottom strand of DNA has a DNA polymerase moving away from the replication fork. This enzyme makes a new strand of DNA to complement the bottom unwound strand, leaving behind sections of \"Lagging strand\". Each individual section of DNA created by the DNA polymerase at the bottom is called an \"Okazaki fragment\".\r\n \r\nDuring DNA replication, both strands of the double helix act as templates for the formation of new DNA molecules. Copying occurs at a localised region called the replication fork, which is a Y shaped structure where new DNA strands are synthesised by a multi-enzyme complex.\r\nHere, the DNA to be copied enters the complex from the left. One new strand is leaving at the top of frame and the other new strand is leaving at bottom. \r\nThe first step in DNA replication is the separation of the two strands by an enzyme called helicase. This spins the incoming DNA to unravel it: at ten thousand RPM in the case of bacterial systems. The separated strands are called three prime and five prime, distinguished by the direction in which their component nucleotides join up.\r\nThe 3 prime DNA strand, also known as the leading strand, is diverted to a DNA polymerase and is used as a continuous template for the synthesis of the first daughter DNA helix.\r\nThe other half of the DNA double helix, known as the lagging strand, has the opposite 3 prime to 5 prime orientation and consequently requires a more complicated copying mechanism. As it emerges from the helicase, the lagging strand is organised into sections called Okazaki fragments. These are then presented to a second DNA polymerase enzyme in the preferred 5 prime to 3 prime orientation. These sections are then effectively synthesised backwards.\r\nWhen the copying is complete, the finished section is released and the next loop is drawn back for replication. Intricate as this mechanism appears, numerous components have been deliberately left out to avoid complete confusion.\r\nThe exposed strands of single DNA are covered by protective binding proteins. And in some systems, multiple Okazaki fragments may be present. The molecular reality is very different from the iconic image of the double helix neatly separating into two DNA copies as so often depicted.\r\n \r\nMeiosis summary\r\n\r\n\tParent cell has two sets of chromosomes.\r\n\tParent cell goes through interphase and the chromosomes duplicate.\r\n\tAfter meiosis I, two daughter cells are formed with a single set of chromosomes that are duplicated.\r\n\tAfter meiosis II, four daughter cells are formed with a single set of chromosomes.\r\n\r\n \r\nMeiosis I\r\n\r\n\tProphase I (top left): The nuclear envelope dissolves and chromatin condenses into chromosomes.\r\n\tMetaphase I (top right): The chromosomes align along the equator of the cell and spindle fibres extend from the centrosomes.\r\n\tAnaphase I (bottom left): Spindle fibres pull sister chromatids to opposite ends of the cell.\r\n\tTelophase I and cytokinesis (bottom right): Two haploid cells with duplicated chromosomes are formed.\r\n\r\n \r\nMeiosis II\r\n\r\n\tProphase II (top left): The nuclear envelope in each daughter cell breaks down.\r\n\tMetaphase II (top right): Chromosomes align along the equator of each cell.\r\n\tAnaphase II (bottom left): Sister chromatids are pulled to opposite poles of the cell.\r\n\tTelophase II and cytokinesis (bottom right): Four haploid cells, each with one set of chromosomes, are formed.\r\n\r\n \r\nSperm meets egg, which then undergoes multiple cell divisions to make a living organism with trillions of cells, all doing their own specialised function. Impressive, but where do the egg and sperm come from. The answer is as you may have guessed from the title, meiosis.\r\nFirst, it's good to know that there are two types of cell division processes.\r\nThe simpler one is mitosis which produces two identical cells with exactly the same genetic information. You can think of them as clones of each other. And to find out more, why not look at our videos about mitosis?\r\nThe other process is meiosis, is a much more complicated process creating not 2 but 4 cells with only half the number of chromosomes, and crucially, all genetically different from each other.\r\nBoth mitosis and meiosis include the same phases: prophase, metaphase, anaphase and telophase. Except, in meiosis, they happen twice, so they usually referred to as one and two. \r\nThe easiest way to remember these phases' names is to remember IPMAT. So let's look at meiosis in more detail.\r\nAs always, cellular division starts with a process called DNA replication. This involves making two identical copies of the original DNA molecule. The cell ends up temporarily with double the normal number of chromosomes. You can learn more about DNA replication in this video.\r\nIn prophase 1, the duplicated chromosomes join up with the pair from the other parnent. So the mother's pair bind with the parther's pair, forming a group of two chromosomes called homologous chromosomes. \r\nWhat happens next is vital for successful meiosis. As each chromosome is lined up next to its partner's pair, one chromatid from each side gets entangled with a corresponding chromatid from the other side. This is called crossing over. \r\nDuring this brief period, the two chromatids swap certain sections of DNA. This is called recombination. The sections that they trade correspond to the same location so that each chromatid retains the correct number of genes. \r\nRecombination is really important because it creates variety. The new cells aren't identical to the other parents and they also are different to one another as well. They are new genetic combinations. \r\nIn fact, that's the whole point of sexual reproduction. It increases genetic variability.\r\nEach chromatid is now different and as each one will end up in a separate gamete, it means each sex cell is genetically different from all others. This explains why brothers and sisters are different despite having the same appearance. Only identical twins have the same genetic makeup as they both originate from the exact same egg and sperm.\r\nNow, back to meiosis. Next comes metaphase 1 as the chromosomes align themselves up in the middle of the cell. And in anaphase 1, the spindle fibres pull the chromosomes apart to opposite ends. And during telophase 1 and cytokinesis, the cell pinches apart in the middle and the nuclear membrane reforms around the two new daughter cells. That's the end of meiosis 1.\r\nNow, for part 2. We start with our recombined daughter cells, each still with 46 chromosomes,. But sperm and egg cells only have 23 chromosomes, so we need to cut these cells in half.\r\nHere goes a second round of division. The process is exactly the same as before, except that there is no DNA replication. We start straight with prophase 2 with chromatin clumping again to form chromosomes, then align in the middle of the cell during metaphase 2 and chromatids are pulled apart during anaphase 2 by the spindle fibres. Telophase and cytokinesis pinch the cells together with four new granddaughter cells being formed.\r\nThe end of meiosis gives us four different sex cells each, each with only 23 chromosomes ready for future fertilisation. But that's a story for another video.\r\nIf you like this video, give it a thumbs up and don't forget to subscribe. Comment below if you have any questions. Why not check out our Fuse School apps as well. Until next time!\r\n","n":0.023},[{"v":"DNA replication","i":2,"n":0.707},{"v":"Meiosis","i":1,"n":1},{"v":"Genetics","i":0,"n":1}]]},{"i":61,"$":[{"v":"Genes and alleles","n":0.577},{"v":" Genes and genomes are the blueprints of life. Understanding how genes function and how the entire genome is organised helps us learn how traits are inherited and how organisms develop. This knowledge is essential for advancements in medicine, agriculture, and biotechnology. Use this resource to explore the roles genes and genomes play in biology, and learn about how genetic variations in Aboriginal and Torres Strait Islander peoples could inform precision cancer medicine.\r\n\r\nGenes\r\nEvery living thing relies on genetic instructions to function and grow. To do this, we need proteins which are coded by sequences of DNA called genes. Genes determine our physical traits like eye colour, height, skin colour and facial features.\r\n\r\n\r\n\r\n  \r\nAn organism\u2019s entire set of DNA, including all their genes and also non-coding sequences, is called its genome.\r\n\r\nGene expression\r\nGene expression is the process by which the instructions in a gene are used to produce proteins*, which perform essential functions in the cell. (*Find out more about Protein synthesis).\r\nNot all genes are active at all times. Genes that are \u201con\u201d at all times are called constitutive genes; they are continually transcribed. Important examples are ribosomal protein genes which are constantly needed for protein production.\r\nSome constitutive genes, known as housekeeping genes, are expressed at stable levels. They are needed for basic cellular functions. \r\nGenes that are expressed only when needed are called facultative genes. These genes allow cells to respond to environment changes or developmental signals. \r\n\r\nGene regulation\r\nGene regulation involves controlling when and how much a gene is expressed. This ensures that proteins are produced at the right time, in the right amount, and in the right cells. Gene regulation is crucial for normal growth, development, and adaptation.\r\nThere are regulatory proteins that control this process.\r\n\r\n\tRepressors block transcription.\r\n\tActivators induce (or activate) transcription.\r\n\r\nThese proteins bind to specific regions of DNA to control transcription.\r\n\r\n\tPromotors are found near the start of a gene. This is where RNA polymerase binds to start transcription.\r\n\tEnhancers can be located very far away from the actual gene. When activators bind to enhancer regions, they increase transcription levels of the gene.\r\n\tSilencers can also be found very far away from the gene. When repressors bind to silencer regions, they reduce transcription levels of the gene.\r\n\r\n\r\nEnhancers and silencers can be found between genes and in the introns of other genes, which are non-coding segments of DNA that do not produce proteins. Because DNA is folded and coiled, these regions can actually be close to the transcription start site for the gene.\r\n\r\n\r\n\r\n\r\n\r\nAlleles and traits\r\nEach gene can have different versions, called alleles. A single gene usually has two alleles, with one given to you from each parent. As the human has more than \\(20,000\\) protein-coding genes, think about all the possible combinations of alleles. This is what gives humans such diversity.\r\nThe alleles specific to you as an individual result is your genotype. These are indicated using a two-letter code, with each letter representing an allele.\r\nFor example, a gene for the colour of corn has two alleles, \\(P\\) and \\(p\\). The possible genotypes are \\(PP\\), \\(pp\\) and \\(Pp\\).\r\n\r\n\tA homozygous genotype has two identical alleles, such as \\(PP\\) or \\(pp\\).\r\n\tA heterozygous genotype has two different alleles, like \\(Pp\\).\r\n\r\nEach genotype may give a different physical trait or phenotype. In this case, \\(P\\) causes purple corn phenotype and \\(p\\) causes yellow corn phenotype.\r\n\r\n\r\nExercise\r\nSee how well you understand genes and alleles with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nDr Mango conducts an experiment to investigate the levels of mRNA and protein expression for a specific gene, Gene X, across different tissues (liver, brain, and heart) under normal and stress conditions. Dr Mango records her results in the following table.\r\n\r\n \r\nLocations on the DNA for gene regulation\r\nA diagram showing the different regions where proteins can bind on DNA to control transcription. Two segments of double-stranded DNA are shown. They belong to the same DNA molecule, as indicated by the label \"Single DNA molecule\", but are shown folded together.\r\nRNA polymerase, shown as a purple blob, is attached to a section of the DNA called the \"Promoter region\". It is also bound to a green blob labelled \"Activator\". The activator has bound to a section of the DNA called the \"Enhancer region\".\r\n","n":0.038},[{"v":"Indigenous knowledges","i":2,"n":0.707},{"v":"Genes","i":1,"n":1},{"v":"Genetics","i":0,"n":1}]]},{"i":62,"$":[{"v":"Genetics","n":1},{"v":"Genetics is the study of how traits are passed from parents to their children. Understanding genetics helps us learn how traits are inherited and how genes affect living things. Use this resource to explore how DNA and genes work in growth and development.\r\n\r\nOverview\r\n\r\n\tGenes and alleles \u2013 what are genes and what do they code for?\r\n\tCell reproduction \u2013 how do cells reproduce?\r\n\tMendelian genetics \u2013 what are Mendel's laws and how do they explain changes in our genes?\r\n\tPredicting traits \u2013 what are Punnett squares and how do we use them to predict inherited traits?\r\n\tPatterns of inheritance \u2013 how are genetic differences passed through families?\r\n\r\n\r\n","n":0.099},[{"v":"Genetics","i":0,"n":1}]]},{"i":63,"$":[{"v":"De Moivre's theorem","n":0.577},{"v":"De Moivre's theorem is a powerful tool in mathematics for working with complex numbers. In particular, it helps when dealing with powers and roots of complex numbers. You might apply De Moivre's theorem in a range of disciplines, including AC circuit analysis and signal processing in engineering, computer graphics and navigation, and even in finance to analyse patterns in market data. Use this resource to learn about De Moivre's theorem.\r\n\r\nPowers of complex numbers\r\nDe Moivre's theorem provides a simple way to raise complex numbers to a power. It states that:\\(\\DeclareMathOperator{\\cis}{cis}\\)\r\nIf \\(z=r\\cis(\\theta)\\), then \\(z^{n}=r^{n}\\cis(n\\theta)\\). \r\n\r\nExample 1 \u2013 using De Moivre's theorem to find the powers of complex numbers\r\nExpress \\((1-i)^{6}\\) in the form \\(x+yi\\).\r\nFirst, convert the complex number to its polar form. Go back to Polar form of complex numbers to review how to do this.\r\n\\[1-i = \\sqrt{2}\\cis\\left(-\\frac{\\pi}{4}\\right)\\]\r\nNow, we can apply De Moivre's theoreom.\r\n\\[\\begin{align*} \\left( \\sqrt{2}\\cis\\left(-\\frac{\\pi}{4}\\right)\\right)^{6} & = (\\sqrt{2})^{6}  \\cis \\left(6\\times-\\frac{\\pi}{4}\\right)\\\\\r\n& = 8\\cis\\left(-\\frac{3\\pi}{2}\\right)\\\\\r\n& = 8i\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExercise \u2013 using De Moivre's theorem to find the powers of complex numbers\r\n\r\n\tEvaluate the following, giving your answers in polar form, where \\(-\\pi\\leq\\theta\\leq\\pi\\).\r\n\t\r\n\t\t\\((\\sqrt{3}+i)^{3}\\)\r\n\t\t\\((1-i)^{5}\\)\r\n\t\t\\((-2\\sqrt{3}+2i)^{2}\\)\r\n\t\r\n\tSimplify the following, giving your answers in polar form.\r\n\t\r\n\t\t\\((1+i)^{4}(2-2i)^{3}\\)\r\n\t\t\\(\\dfrac{(2-2\\sqrt{3}i)^{4}}{(-1+i)^{6}}\\)\r\n\t\r\n\r\n\r\n\r\n\r\nRoots of complex numbers\r\nDe Moivre's theorem is also helpful for finding all of the possible roots of a complex number. For \\(z^{n}=r\\cis(\\theta)\\), there will be \\(n\\) solutions in the form:\r\n\\(z^{\\frac{1}{n}} = r^{\\frac{1}{n}}\\cis \\left( \\dfrac{\\theta+2\\pi k}{n} \\right)\\) where \\(k=0,1,\\ldots n-1\\).\r\n\r\nExample \u2013 using De Moivre's theorem to find the roots of complex numbers\r\nSolve for the roots of \\(z^{n}=1-\\sqrt{3}i\\).\r\nFirst, convert the complex number to its polar form.\r\n\\[1-\\sqrt{3}i = 2\\cis\\left(-\\frac{\\pi}{3}\\right)\\]\r\nSince \\(n=4\\), this means there are four solutions. Since \\(k=0,1,\\ldots n-1\\), we let \\(k=0\\), \\(k=1\\), \\(k=2\\) and \\(k=3\\).\r\nFor \\(k=0\\):\r\n\\[\\begin{align*} z & = 2^{\\frac{1}{4}}\\cis \\left( \\frac{-\\dfrac{\\pi}{3}+2\\pi(0)}{4} \\right)\\\\\r\n& = 2^{\\frac{1}{4}}\\cis\\left( -\\frac{\\pi}{12} \\right) \r\n\\end{align*}\\]\r\nFor \\(k=1\\):\r\n\\[\\begin{align*} z & = 2^{\\frac{1}{4}}\\cis \\left( \\frac{-\\dfrac{\\pi}{3}+2\\pi(1)}{4} \\right)\\\\\r\n& = 2^{\\frac{1}{4}}\\cis\\left( \\frac{5\\pi}{12} \\right) \r\n\\end{align*}\\]\r\nFor \\(k=2\\):\r\n\\[\\begin{align*} z & = 2^{\\frac{1}{4}}\\cis \\left( \\frac{-\\dfrac{\\pi}{3}+2\\pi(2)}{4} \\right)\\\\\r\n& = 2^{\\frac{1}{4}}\\cis\\left( \\frac{11\\pi}{12} \\right) \r\n\\end{align*}\\]\r\nFor \\(k=3\\):\r\n\\[\\begin{align*} z & = 2^{\\frac{1}{4}}\\cis \\left( \\frac{-\\dfrac{\\pi}{3}+2\\pi(3)}{4} \\right)\\\\\r\n& = 2^{\\frac{1}{4}}\\cis\\left( \\frac{17\\pi}{12} \\right) \\\\\r\n& = 2^{\\frac{1}{4}}\\cis\\left( \\frac{-7\\pi}{12} \\right)\r\n\\end{align*}\\]\r\nThe four solutions for \\(z\\) are \\(2^{\\frac{1}{4}}\\cis\\left( -\\dfrac{\\pi}{12} \\right)\\), \\(2^{\\frac{1}{4}}\\cis\\left( \\dfrac{5\\pi}{12} \\right)\\), \\(2^{\\frac{1}{4}}\\cis\\left( \\dfrac{11\\pi}{12} \\right) \\) and \\(2^{\\frac{1}{4}}\\cis\\left( \\dfrac{17\\pi}{12} \\right)\\).\r\nThis can be represented graphically:\r\n\r\n\r\n\r\nExercise \u2013 using De Moivre's theorem to find the roots of complex numbers\r\n\r\n\tSolve the following, giving your answers in polar form, where \\(-\\pi\\leq\\theta\\leq\\pi\\).\r\n\t\r\n\t\t\\(z^{3}=-1\\)\r\n\t\t\\(z^{4}=16i\\)\r\n\t\t\\(z^{3}=\\sqrt{6}-\\sqrt{2}i\\)\r\n\t\r\n\tIf \\(\\sqrt{3}+i\\) is one solution of \\(z^{3}=8i\\), use a diagram to find the other solutions in rectangular form.\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.051},[{"v":"Complex numbers","i":0,"n":0.707}]]},{"i":64,"$":[{"v":"Polar form of complex numbers","n":0.447},{"v":"The polar form of complex numbers is a different way of writing them that makes some math problems easier to solve. Understanding how to convert between forms and learning a new way to handle operations with complex numbers is helpful in many disciplines, like engineering and physics where you often deal with cycles or rotations. Use this resource to learn how to express complex numbers in their polar form and complete calculations using this form.\r\n\r\nRectangular and polar forms\r\nIn Introduction to complex numbers, you learned about complex numbers expressed in the form \\(z=x+yi\\). This is called the rectangular form. \\(\\DeclareMathOperator{\\cis}{cis} \\DeclareMathOperator{\\Arg}{Arg}\\)\r\nYou also learned about Argand diagrams, which graphically represent complex numbers. Cartesian coordinates in the form \\((x,y)\\) can be used to indicate a point \\(P\\) on a plane, but so can polar coordinates in the form \\((r,\\theta)\\), where \\(r\\) is the distance of \\(P\\) from the origin and \\(\\theta\\) is the angle that \\(\\overrightarrow{OP}\\) makes with the positive \\(x\\)-axis.\r\nThe polar form gives us direct information about the magnitude (distance) and angle of the complex number.\r\nWe can convert the polar coordinate of a complex number to the polar form of the number using:\r\n\\[\\begin{align*} z & = x+yi\\\\\r\n& = r\\cos(\\theta) + r\\sin(\\theta)\\\\\r\n& = r(\\cos(\\theta) + \\sin(\\theta))\\quad\\textrm{which we abbreviate to}\\\\ \\DeclareMathOperator{\\cis}{cis}\r\n& = r\\cis(\\theta)\r\n\\end{align*}\\]\r\nTherefore, the polar form of the complex number \\(z\\) is:\r\n\\[z=r\\cis(\\theta)\\]\r\nwhere \\(r=\\sqrt{x^{2}+y^{2}}\\) and \\(\\theta=\\tan^{-1}\\left(\\dfrac{y}{x}\\right)\\).\r\n\r\nModulus, argument and Argument\r\nTo find \\(r\\), we need to find the modulus of \\(z\\) or \\(\\left| z \\right|\\). \r\n\\[\\begin{align*} \\mod(z) & = \\left| z \\right|\\\\\r\n& = \\left| x+yi \\right|\\\\\r\n& = \\sqrt{x^{2}+y^{2}}\\\\\r\n& = r\r\n\\end{align*}\\]\r\nTo find \\(\\theta\\), we need to find the argument (lower case \"a\") of \\(z\\), or \\(\\arg(z)\\). If \\(\\arg(z)=\\theta\\), then:\r\n\r\n\t\\(\\sin(\\theta)=\\dfrac{y}{\\left| z \\right|}\\)\r\n\t\\(\\cos(\\theta)=\\dfrac{x}{\\left| z \\right|}\\)\r\n\t\\(\\tan(\\theta)=\\dfrac{y}{x}\\).\r\n\r\nAn infinite number of arguments of \\(z\\) exist. For example, if \\(z=i\\), then \\(\\arg(z)=\\dfrac{\\pi}{2}+2\\pi n\\), where \\(n\\in Z\\) (i.e. \\(n\\) can be any integer).\r\nWe can also use a value called the Argument (capital \"A\"), where \\(\\Arg(z)=\\theta\\) for \\(-\\pi\\leq\\theta\\leq\\pi\\). Because the Argument is restricted to \\(-\\pi\\leq\\theta\\leq\\pi\\), \\(\\Arg(z)\\) has only one value, compared to \\(\\arg(z)\\) which has many values.\r\n\r\nExample 1 \u2013 expressing complex numbers in polar form\r\nExpress \\(z=1-i\\) in polar form.\r\nWe start by finding the coordinates on the Cartesian plane. For \\(z=1-i\\), \\(x=1\\) and \\(y=-1\\). The point is in the fourth quadrant.\r\nRemember that the polar form of \\(z=r\\cis(\\theta)\\). We can start by finding \\(r\\).\r\n\\[\\begin{align*} r & = \\sqrt{x^{2}+y^{2}}\\\\\r\n& = \\sqrt{1+1}\\\\\r\n& = \\sqrt{2}\r\n\\end{align*}\\]\r\nNext, let's find \\(\\theta\\).\r\n\\[\\begin{align*} \\theta & = \\tan^{-1}\\left(\\frac{y}{x}\\right)\\\\\r\n& = \\tan^{-1}\\left(\\frac{-1}{1}\\right)\\\\\r\n& = \\tan^{-1}(-1)\\\\\r\n& = -\\frac{\\pi}{4}\r\n\\end{align*}\\]\r\nPutting it altogether:\r\n\\[\\begin{align*} z & = r\\cis(\\theta)\\\\\r\n& = \\sqrt{2}\\cis\\left(-\\frac{\\pi}{4}\\right)\r\n\\end{align*}\\] \r\n\r\nExercise \u2013 expressing complex numbers in polar form\r\n\r\n\tFind the polar form (in radians) of the following complex numbers.\r\n\t\r\n\t\t\\(z=-1+i\\)\r\n\t\t\\(z=-\\sqrt{3}+i\\)\r\n\t\t\\(z=-3i\\)\r\n\t\t\\(z=-2-4i\\)\r\n\t\r\n\tExpress the following complex numbers in rectangular form.\r\n\t\r\n\t\t\\(3\\cis\\left(\\dfrac{\\pi}{4}\\right)\\)\r\n\t\t\\(\\sqrt{7}\\cis(\\pi)\\)\r\n\t\t\\(8\\cis\\left(\\dfrac{\\pi}{2}\\right)\\)\r\n\t\t\\(10\\cis(0.41)\\)\r\n\t\r\n\tIf \\(z=2+i\\) and \\(w=1-4i\\), find the polar form (in radians) of the following complex numbers.\r\n\t\r\n\t\t\\(\\left| z \\right|\\)\r\n\t\t\\(\\left| w \\right|\\)\r\n\t\t\\(\\Arg(z)\\)\r\n\t\t\\(\\left| \\overline{w} \\right|\\)\r\n\t\t\\(\\Arg(zw)\\)\r\n\t\t\\(zw\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\nOperations on complex numbers in polar form\r\nTo add and subtract complex numbers, they must be converted from polar form to rectangular form.\r\nFor multiplication and division, we use a trigonometric identity. \r\nTo multiply:\r\n\\[z_{1}z_{2} = r_{1}r_{2}\\cis(\\theta_{1}+\\theta_{2})\\]\r\nTo divide:\r\n\\[\\dfrac{z_{1}}{z_{2}} = \\dfrac{r_{1}}{r_{2}}\\cis(\\theta_{1}-\\theta_{2})\\]\r\n\r\nExample 1 \u2013 performing operations on complex numbers in polar form\r\nIf \\(z_{1}=2\\cis\\left(\\dfrac{\\pi}{4}\\right)\\) and \\(z_{2}=3\\cis\\left(\\dfrac{11\\pi}{6}\\right)\\), find \\(z_{1}z_{2}\\) in polar form, where \\(-\\pi\\leq\\theta\\leq\\pi\\).\r\nNote that \\(z_{2}=3\\cis\\left(\\dfrac{11\\pi}{6}\\right)=3\\cis\\left(-\\dfrac{\\pi}{6}\\right)\\).\r\n\\[\\begin{align*} z_{1}z_{2} & = 2\\cis\\left(\\frac{\\pi}{4}\\right) \\times 3\\cis\\left(-\\frac{\\pi}{6}\\right)\\\\\r\n& = 6\\cis\\left(\\frac{3\\pi}{12}-\\frac{2\\pi}{12}\\right)\\\\\r\n& = 6\\cis\\left(\\frac{\\pi}{12}\\right)\\quad\\textrm{since }-\\pi\\leq\\theta\\leq\\pi\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExercise \u2013 performing operations on complex numbers in polar form\r\n\r\n\tSimplify the following.\r\n\t\r\n\t\t\\(4\\cis\\left(\\dfrac{\\pi}{3}\\right) \\times 3\\cis\\left(\\dfrac{\\pi}{4}\\right)\\)\r\n\t\t\\(\\dfrac{3\\cis \\left( \\dfrac{5\\pi}{6} \\right)}{12 \\cis \\left( \\dfrac{\\pi}{6} \\right)}\\)\r\n\t\r\n\tIf \\(u=6\\cis\\left(\\dfrac{3\\pi}{4}\\right)\\) and \\(v=4\\cis\\left(-\\dfrac{\\pi}{4}\\right)\\), express \\(\\dfrac{u}{v}\\) in polar form.\r\n\tIf \\(z=1-\\sqrt{3}i\\), find \\(\\overline{z}\\) and express both \\(z\\) and \\(\\overline{z}\\) in polar form (in radians).\r\n\r\n\t\r\n","n":0.042},[{"v":"Complex numbers","i":0,"n":0.707}]]},{"i":65,"$":[{"v":"Introduction to complex numbers","n":0.5},{"v":"Complex numbers might seem daunting at first, but they open a whole new dimension in maths. Understanding complex numbers allows you to explore their applications in engineering (like signalling in cellular and wireless technologies) and science (like studying brain waves in biology). Use this resource for an introduction to the basics.\r\n\r\nImaginary numbers\r\nIf you have tried to solve algebraic equations, you might notice there are sometimes, there are no \"real\" solutions. For example, no real number satisfies \\(x^{2}+1=0\\).\r\nTo find solutions to such equations, we use complex numbers. This concept centres around the imaginary number \\(i\\), where:\r\n\\[i=\\sqrt{-1}\\]\r\nThe powers of \\(i\\) follow a pattern, as shown in the tabke, \\(i^{2}=-1\\), \\(i^{3}=-i\\), \\(i^{4}=1\\), \\(i^{5}=i\\), and so on.\r\n\r\n\t\r\n\t\t\r\n\t\t\t\\(n\\)\r\n\t\t\t\\(i^{n}\\)\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\t\\(1\\), \\(5\\), \\(9\\), \\(13\\), etc.\r\n\t\t\t\\(-1\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(2\\), \\(6\\), \\(10\\), \\(14\\), etc.\r\n\t\t\t\\(-i\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(3\\), \\(7\\), \\(11\\), \\(15\\), etc.\r\n\t\t\t\\(1\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(4\\), \\(8\\), \\(12\\), \\(16\\), etc.\r\n\t\t\t\\(i\\)\r\n\t\t\r\n\t\r\n\r\n\r\nComplex numbers\r\nA complex number \\(z\\) is written as:\r\n\\(z=x+yi\\) where \\(x\\) and \\(y\\) are real numbers.\r\n\r\n\t\\(x\\) is called the real part of \\(z\\), denoted by \\(\\Re(z)\\) or \\(\\textrm{Re}(z)\\).\r\n\t\\(y\\) is called the imaginary part of \\(z\\), denoted by \\(\\Im(z)\\) or \\(\\textrm{Im}(z)\\).\r\n\r\nTwo complex numbers are equal only if their real parts are equal and their imaginary parts are equal.\r\n\\(a+bi=c+di\\) only if \\(a=c\\) and \\(b=d\\).\r\n\r\nExample 1 \u2013 finding solutions for complex numbers\r\nFind the real and imaginary parts of \\(z=5-3i\\).\r\nComplex numbers are written in the form \\(x+yi\\). \\(x\\) is the real part of \\(z\\), so \\(\\Re(z)=5\\). \\(y\\) is the imaginary part of \\(z\\), so \\(\\Im(z)=-3\\).\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding solutions for complex numbers\r\n\r\n\tExpress the following in terms of \\(i\\) in simplest surd form.\r\n\t\r\n\t\t\\(\\sqrt{-9}\\)\r\n\t\t\\(\\sqrt{-2}\\)\r\n\t\t\\(\\sqrt{-5}\\times\\sqrt{3}\\)\r\n\t\t\\(\\sqrt{-5}\\times\\sqrt{10}\\)\r\n\t\t\\(\\sqrt{-6}\\times\\sqrt{12}\\)\r\n\t\r\n\tEvaluate:\r\n\t\r\n\t\t\\(i^{4}\\)\r\n\t\t\\(i^{9}\\)\r\n\t\t\\(i^{7}-i^{11}\\)\r\n\t\t\\(i^{5}+i^{6}-i^{7}\\)\r\n\t\t\\(2i-i^{6}+2i^{7}\\)\r\n\t\r\n\tState the value of \\(\\Re(z)\\) and \\(\\Im(z)\\) for the following complex numbers.\r\n\t\r\n\t\t\\(2+7i\\)\r\n\t\t\\(10-i\\)\r\n\t\t\\(\\pi+3i\\)\r\n\t\t\\(\\dfrac{i}{6}\\)\r\n\t\t\\(-8\\)\r\n\t\r\n\tFind the values of \\(x\\) and \\(y\\).\r\n\t\r\n\t\t\\(x+yi=4+9i\\)\r\n\t\t\\(x+yi=3-i\\)\r\n\t\t\\(x+yi=23\\)\r\n\t\t\\(x+yi=-\\sqrt{2}i\\)\r\n\t\t\\(x+i=-5+yi\\)\r\n\t\r\n\r\n\r\n\r\n\r\nAdding and subtracting complex numbers\r\nTo add or subtract complex numbers, we add or subtract the real and imaginary parts separately.\r\n\\[(a+bi)+(c+di)=(a+c)+(b+d)i\\] \\[(a+bi)-(c+di)=(a-c)+(b-d)i\\]\r\nThis is similar to the way we group like terms in algebraic expressions.\r\n\r\nExample 1 \u2013 adding and subtracting complex numbers\r\n\\[\\begin{align*} (2+3i)+(4-i) & = (2+4)+(3-1)i\\\\\r\n& = 6+2i\r\n\\end{align*}\\] \r\n\r\nExercise \u2013 adding and subtracting complex numbers\r\n\r\n\t\r\n\t\t\\((3+4i)+(5+2i)\\)\r\n\t\t\\((-1+6i)+(2-3i)\\)\r\n\t\tIf \\(z_{1}=7+i\\) and \\(z_{2}=2+3i\\), find \\(z_{1}+z_{2}\\).\r\n\t\t\\((4-5i)+(-3+7i)\\)\r\n\t\tIf \\(z_{1}=2i\\) and \\(z_{2}=-4+i\\), find \\(z_{1}+z_{2}\\).\r\n\t\r\n\t\r\n\t\t\\((5+9i)-(8+3i)\\)\r\n\t\t\\((-1+7i)-(2-4i)\\)\r\n\t\t\\((6+2i)-(3+4i)\\)\r\n\t\tIf \\(z_{1}=4+8i\\) and \\(z_{2}=10-3i\\), find \\(z_{2}-z_{1}\\).\r\n\t\tIf \\(z_{1}=9-2i\\) and \\(z_{2}=7+i\\), find \\(z_{2}-z_{1}\\).\r\n\t\r\n\r\n\r\n\r\n\r\nMultiplying complex numbers\r\nWe also multiply complex numbers like how we multiply algebraic expressions. To multiply by a constant \\(k\\), we just expand the expression.\r\n\\[k(a+bi)=ka+kbi\\]\r\nWhen we multiply one complex number by another, we also expand.\r\n\\[\\begin{align*} (a+bi)(c+di) & = ac+adi+bci+bdi^{2}\\\\\r\n& = ac+adi+bci-bd\\quad\\textrm{since }i^{2}=-1\\\\\r\n& = (ac-bd)+(ad+bc)i\r\n\\end{align*}\\]\r\n\r\nExample 1 \u2013 multiplying complex numbers\r\nExpand and simplify \\(i(3+4i)\\).\r\n\\[\\begin{align*} i(3+4i) & = 3i+4i^{2}\\\\\r\n& = 3i-4\\\\\r\n& = -4+3i\r\n\\end{align*}\\] \r\n\r\nExercise \u2013 multipying complex numbers\r\n\r\n\tExpand and simplify the following.\r\n\t\r\n\t\t\\(i(3-2i)\\)\r\n\t\t\\(2i^{3}(1-5i)\\)\r\n\t\t\\(8-3i)(2-5i)\\)\r\n\t\t\\((4-3i)^{2}\\)\r\n\t\t\\((3+2i)(3-2i)\\)\r\n\t\r\n\tIf \\(z_{1}=-1+3i\\) and \\(z_{2}=2-i\\), find the following.\r\n\t\r\n\t\t\\(z_{1}z_{2}\\)\r\n\t\t\\(2z_{1}-z_{2}\\)\r\n\t\t\\((z_{1}-z_{2})^{2}\\)\r\n\t\r\n\tFind the values of \\(x\\) and \\(y\\) if \\((x+yi)(2-3i)=-13i\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\nComplex conjugates\r\nA pair of complex numbers of the form \\(a+bi\\) and \\(a-bi\\) are called complex conjugates. This is similar to conjugate surds. \r\nThe complex conjugate of \\(z=x+yi\\) is denoted by \\(\\overline{z}\\), where \\(\\overline{z}=x-yi\\).\r\nWhen you multiply a pair of complex conjugates, you get a real number. \r\n\\[z\\overline{z}=(x+yi)(x-yi)=x^{2}+y^{2}\\]\r\nThis looks like the difference of two squares rule, but the sign is a plus rather than a minus. This is because \\(i^{2}=-1\\), so we invert the sign.\r\nThere are two properties of complex conjugates worth remembering:\r\n\r\n\t\\(\\overline{z_{1}+z_{2}}=\\overline{z_{1}}+\\overline{z_{2}}\\)\r\n\t\\(\\overline{z_{1}z_{2}}=\\overline{z_{1}}\\times\\overline{z_{2}}\\)\r\n\r\n\r\nExample \u2013 multiplying complex conjugates\r\nIf \\(z=2-i\\) and \\(w=-3+4i\\), find:\r\n\r\n\t\\(\\overline{z}\\)\r\n\t\\(\\overline{z}-\\overline{w}\\)\r\n\t\\(\\overline{z+w}\\).\r\n\r\n\\[\\overline{z} = 2+i\\] \\[\\begin{align*} \\overline{z}-\\overline{w} & = (2+i)-(-3-4i)\\\\\r\n& = 2+i+3+4i\\\\\r\n& = 5+5i\r\n\\end{align*}\\] \\[\\begin{align*} \\overline{z+w} & = \\overline{(2-i)+(-3+4i)}\\\\\r\n& = \\overline{-1+3i}\\\\\r\n& = -1-3i\r\n\\end{align*}\\]\r\n\r\nExercise \u2013 multiplying complex conjugates\r\n\r\n\tFind the conjugate of the following complex numbers.\r\n\t\r\n\t\t\\(4+9i\\)\r\n\t\t\\(-3-15i\\)\r\n\t\t\\(\\sqrt{3}-4i\\)\r\n\t\r\n\tFind the conjugate of \\((2-i)(4+7i)\\).\r\n\tIf \\(z=2-i\\) and \\(w=1+2i\\), express the following in the form \\(x+yi\\).\r\n\t\r\n\t\t\\(\\overline{z}\\)\r\n\t\t\\(\\overline{z+w}\\)\r\n\t\t\\(\\overline{z}+\\overline{w}\\)\r\n\t\t\\(\\overline{zw}\\)\r\n\t\t\\(\\overline{\\overline{z}-\\overline{w}}\\)\r\n\t\r\n\r\n\r\n\r\n\r\nDividing complex numbers\r\nTo divide complex numbers, we rewrite them as fraction and rationalise the same way we do with surds. We multiply both the numerator and denominator by the complex conjugate of the denominator, then simply to the form \\(x+yi\\).\r\n\\[\\frac{a+bi}{c+di}\\times\\frac{c-di}{c-di}\\]\r\n\r\nExample 1 \u2013 dividing complex numbers\r\nExpress \\(\\dfrac{2-i}{1+3i}\\) in the form \\(x+yi\\).\r\n\\[\\begin{align*} \\frac{2-i}{1+3i}\\times\\frac{1-3i}{1-3i}\\\\\r\n& = \\frac{(2-i)(1-3i)}{(1+3i)(1-3i)}\\\\\r\n& = \\frac{2-6i-i+3i^{2}}{1+9}\\\\\r\n& = \\frac{2-7i-3}{10}\\\\\r\n& = \\frac{-1-7i}{10}\\\\\r\n& = -\\frac{1}{10}-\\frac{7}{10}i\r\n\\end{align*}\\] \r\n\r\nExercise \u2013 dividing complex numbers\r\n\r\n\tExpress the following in the form \\(x+yi\\).\r\n\t\r\n\t\t\\(\\dfrac{4-9i}{3}\\)\r\n\t\t\\(\\dfrac{1}{3-i}\\)\r\n\t\t\\(\\dfrac{5+i}{2-7i}\\)\r\n\t\r\n\tSimplify \\(\\dfrac{2}{1-i}+\\dfrac{3+i}{i}\\).\r\n\tIf \\(w=-1+6i\\), express \\(\\dfrac{w+1}{w-i}\\) in the form \\(x+yi\\).\r\n\r\n\r\n\r\n\r\nArgand diagrams\r\nYou can present complex numbers graphically using Argand diagrams. For the complex number \\(z=ax+byi\\), the \\(x\\) coordinate will be \\(a\\) and the \\(y\\) coordinate will be \\(bi\\). In other words, the point is \\((a,bi)\\).\r\n\r\n\r\nExercise \u2013 drawing Argand diagrams\r\nIf \\(z=2-3i\\) and \\(w=1+4i\\), illustrate the following on an Argand diagram.\r\n\r\n\t\\(z\\)\r\n\t\\(w\\)\r\n\t\\(z+w\\)\r\n\t\\(\\overline{z+w}\\)\r\n\t\\(2z-w\\)\r\n\r\n\r\n\r\n\r\n","n":0.039},[{"v":"Complex numbers","i":0,"n":0.707}]]},{"i":66,"$":[{"v":"Structure and function of enzymes","n":0.447},{"v":"\r\nEnzymes are special proteins that are critical in many chemical reactions that occur in living organisms. Knowing how enzymes work is necessary for understanding how metabolic processes are controlled and maintained within cells. This has broader applications in agriculture, medicine and even in textiles. Use this resource to learn about the structure and function of enzymes.\r\n\r\nBiological catalysts\r\nEnzymes are proteins that act as catalysts; they speed up chemical reactions by providing another pathway for the reaction to occur that requires less energy. This makes processes occur more efficiently.\r\n\r\nEnzymes play critical roles in a huge range of cellular functions, like digestion, energy production, protein synthesis and DNA replication. Examples of enzymes include RNA polymerase which catalyses the transcription of DNA into RNA, amylase which breaks down complex sugars from your diet into simple sugars that your body can use, and lactase which catalyses the digestion of lactose found in dairy foods.\r\nUsually, enzymes work by bringing together two or more molecules, called substrates and joining them, or breaking one substrate into multiple products. Their ability to do this depends on their structure. The site where the enzyme binds the substrates, called the active site needs to the correct shape to bind the substrates.\r\nOnce the substrates bind to the enzyme, they form an enzyme\u2013substrate complex. The reaction occurs to form an enzyme\u2013products complex, then the products are released and the enzyme is free to catalyse the next reaction. Enzymes are not consumed in catalysis.\r\n  \r\n\r\n\r\n\r\nModels of enzyme function\r\nThe function of enzymes is described by two models: the lock-and-key model and the induced fit model.\r\nLock-and-key model\r\nThe lock-and-key model of enzyme action was proposed by German chemist Emil Fisher in 1984. He theorised that the shape of an enzyme\u2019s active site was rigid and specifically matched the shape of its substrates, just like a lock has a unique key.\r\n\r\nInduced fit model\r\nThe induced fit model of enzyme action was proposed by American biochemist Daniel E. Koshland in 1958. It is an extension of Fisher\u2019s lock-and-key model, where the enzyme changes shape slightly (or undergoes a conformation change) to bind substrates. This model helped to explain how enzymes can bind more than one type of substrate.\r\n\r\nFactors affecting enzyme action\r\nSeveral factors can affect enzyme action or activity, which then affects the efficiency of chemical reactions. These include:\r\n\r\n\ttemperature\r\n\tpH\r\n\tsubstrate concentration\r\n\tenzyme concentration\r\n\tthe presence or absence of inhibitors and activators. \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise\r\nSee how well you understand the structure and function of enzymes with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nDr Catherine de List is studying the effect of pH on the activity of three enzymes: enzyme A, B and C. She measures their activity in enzyme units per millilitre over the whole pH spectrum. This is presented in the following graph.\r\n\r\n\r\n\r\n\r\n \r\nLine graph showing energy changes during a reaction\r\nA line graph showing how energy changes as a reaction occurs. There are two lines on the graph: a solid line labelled \"Without enzyme\" and a dashed line labelled \"With enzyme\".\r\nThe x-axis is labelled \"Reaction progress\" and the y-axis is labelled \"Energy\". Neither have values, as the graph shows a rough profile of how energy changes.\r\nThe without enzyme line begins at about two-fifths of the way up the y-axis and continues at this energy level as the reaction progresses, until it increases and reaches a peak close to the maximum value of the y-axis. The energy then decreases gradually until it plateaus at a lower energy than when the reaction started. \r\nThe with enzyme line looks very similar to the without enzyme one. The only difference is that the peak occurs at the lower energy value.\r\n \r\nThe process of enzyme action\r\n\r\n\tAn enzyme is shown with a single substrate (green) matching the shape of its active site nearby.\r\n\tThe substrate binds to the active site of the enzyme to form an enzyme/substrate complex.\r\n\tA reaction occurs to form an enzyme/product complex. This is clear because the original green substrate is now two products coloured in shades of pink.\r\n\tThe two products are released from the active site of the enzyme.\r\n\r\n \r\nLock-and-key model of enzyme action\r\n\r\n\tAn enzyme is shown with two substrates (yellow and blue) matching the shape of its active site nearby.\r\n\tThe substrates bind to the active site of the enzyme to form an enzyme/substrate complex.\r\n\tA reaction occurs to form an enzyme/product complex. This is clear because the two original substrates are now a single product coloured in green.\r\n\tThe product is released from the active site of the enzyme.\r\n\r\n \r\nInduced fit model of enzyme action\r\n\r\n\tAn enzyme is shown with two substrates (yellow and blue) nearby.\r\n\tThe substrates bind to the active site of the enzyme to form an enzyme/substrate complex. A conformational change in the enzyme active site is required to perfectly fit the substrates.\r\n\tA reaction occurs to form an enzyme/product complex. This is clear because the two original substrates are now a single product coloured in green.\r\n\tThe product is released from the active site of the enzyme. The active site returns to its original shape.\r\n\r\n \r\nLine graph showing effect of temperature on enzyme action\r\nA line graph showing how enzyme action is affected by temperature.\r\nThe x-axis is labelled \"Temperature\" and the y-axis is labelled \"Enzyme activity\". Neither have values, as the graph shows a rough profile of how enzyme activity is affected.\r\nThe line starts at 0, gradually increases to a maximum, then more quickly decreases back down to 0. At the far left end is the label \"Enzyme inactivated (reversible)\" and at the far right end is the label \"Enzyme denatured (permanent)\".\r\nThe peak of the curve is labelled \"Optimum temperature\", with a dashed line extending from the line down to the temperature axis.\r\n \r\nLine graph showing effect of pH on enzyme action\r\nA line graph showing how enzyme action is affected by pH.\r\nThe x-axis is labelled \"pH\" and the y-axis is labelled \"Enzyme activity\". Neither have values, as the graph shows a rough profile of how enzyme activity is affected.\r\nThere are two curves. The red curve extends over the lower pH range and is labelled with \"High H+\". It increases from no enzyme activity to a maximum, before decreasing back down to 0. The peak of the curve is labelled \"Optimum pH for enzyme 1\", with a dashed line extending from the line down to the pH axis.\r\nThe blue curve extends over the higher pH range, and is labelled \"Low H+\". It has a similar bell shape as the red curve, but is a bit wider. The peak of the curve is labelled \"Optimum pH for enzyme 2\", with a dashed line extending from the line down to the pH axis.\r\n \r\nLine graph showing effect of enzyme or substrate concentration on enzyme action\r\nA line graph showing how enzyme action is affected by concentrations of enzyme or substrate.\r\nThe x-axis is labelled \"Enzyme or substrate concentration\" and the y-axis is labelled \"Enzyme activity\". Neither have values, as the graph shows a rough profile of how enzyme activity is affected.\r\nThere is one curve, beginning at 0 and increasing at a steady rate before it plateaus.\r\n \r\nEnzyme inhibitors\r\n\r\n\tCompetitive inhibitor (left): A competitive inhibitor (green) is bound to the active site of the enzyme, stopping the blue and yellow substrates from binding.\r\n\tNon-competitive inhibitor (middle): A non-competitive inhibitor (green) is bound to another site on the enzyme, causing a change in the shape of the active site. This stops the substrates from binding.\r\n\tUncompetitive inhibitor (right): An uncompetitive inhibitor (green) is bound to the enzyme/substrate complex, stopping the products from being released.\r\n\r\n \r\nLine graph showing effect of pH on enzyme action\r\nA line graph showing pH affects the action of three enzymes: Enzyme A, Enzyme B and Enzyme C.\r\nThe x-axis is labelled \"pH\", starting at 0 and ending at 14, with increments of 14. The y-axis is labelled \"Enzyme activity (U/mL)\", starting at 0 and ending at 60, with increments of 10.\r\nThere are three curves, each representing one enzyme. \r\n\r\n\tThe curve for Enzyme A starts with no enzyme activity at a pH of 1. Activity gradually increases until it reaches a peak of around 55 U/mL at a pH of 2.5. After the peak, the enzyme activity gradually decreases until it returns to no enzyme activity at a pH of 4.\r\n\tThe curve for Enzyme B starts with no enzyme activity at around pH 3.7. Activity gradually increases until it reaches a peak of around 15 U/mL at a pH of 7. After the peak, the enzyme activity gradually decreases until it returns to no enzyme activity at a pH of 10.5.\r\n\tThe curve for Enzyme C starts with no enzyme activity at a pH of 7. Activity gradually increases until it reaches a peak at around 40 U/mL at a pH of 9. After the peak, the enzyme activity gradually decreases until it returns to no enzyme actvity at a pH of 11.\r\n\r\n","n":0.026},[{"v":"Enzymes","i":2,"n":1},{"v":"Proteins","i":1,"n":1},{"v":"Molecular biology","i":0,"n":0.707}]]},{"i":67,"$":[{"v":"DNA","n":1},{"v":"\r\n\r\nDNA carries the genetic instructions needed for the growth, development and functioning of all living organisms. You must understand the structure of DNA to grasp how genetic information is passed down and used in biological processes. This is crucial in fields like medicine, where it aids in developing treatments and in biotechnology, where it drives innovation. Use this resource to learn about the structure of DNA.\r\n\r\nDNA is composed of two long strands of nucleotides twisted around each other. Go back to Macromolecules if you need to review nucleic acids and nucleotides.\r\nThe nucleotides in a strand of DNA are connected by phosphodiester bonds between the third carbon (3\u2019 or three prime) of a sugar of one nucleotide and the phosphate attached to the fifth carbon (5\u2019 or five prime) of the sugar on the next nucleotide. Nucleotides are only added to the 3\u2019 end, so the direction of DNA is 5\u2019 to 3\u2019. DNA replication and transcription also occur in this direction.\r\n\r\nSpecific sequences of nucleotides are called genes, which code for specific proteins. You will learn about how DNA is used to make proteins in Protein synthesis.\r\n\r\nThe double helix\r\nDNA has a double helix structure. This is formed through complementary base pairing. Adenine (A) pairs specifically with thymine (T), and cytosine (C) pairs specifically with guanine (G). DNA is often referred to as a twisted ladder as these bases form hydrogen bonds with each other to create \"rungs\". The two strands run in opposite directions causing the sugar-phosphate backbone to alternate. \r\nGuanine and cytosine pair via three hydrogen bonds, whereas there are only two hydrogen bonds between adenine and thymine. The stronger association between guanine and cytosine gives DNA its twisted structure.\r\n\r\nYou can use this interactive to see a 3D model of the DNA structure.\r\n \r\n\r\n\r\n\r\nChromosomes\r\nIn the nucleus, the DNA is coiled up into a condensed structure which protects the DNA during cell division.\r\nPackaging proteins called histones bind to the DNA molecule. When a section of DNA is wrapped around a cluster of eight histones, it is called a nucleosome. Chains of nucleosomes are further coiled into a spiral called a solenoid. The solenoids coil to form the fibres of chromatin.\r\nWhen the cell is not dividing, DNA is present in the nucleus as chromatin. When it is ready to divide, it further condenses to form chromosomes. \r\n \r\n\r\n\r\nA chromosome consists of a chromatid, which has a short arm and a long arm. Between the arms is a constricted region called a centromere. One chromosome is one DNA molecule.\r\nWhen DNA is being replicated, the chromosome duplicates and consists of two sister chromatids joined at the centromere.\r\n\r\n\r\nExercise\r\nSee how well you understand the structure of DNA with a quick quiz.\r\n\r\n\r\n\r\nReady for some more? Practise your skills by determining the sequence of the DNA strand that is complementary to the following strands:\r\n\r\n\tTAC\u2013CGA\u2013CCA\u2013ATT\u2013TAG\u2013TTA\u2013CCC\r\n\tCCG\u2013AAG\u2013CTG\u2013ATG\u2013GGA\u2013CAG\u2013ACG\r\n\tGTT\u2013AGA\u2013CTA\u2013TAA\u2013GCG\u2013ACT\u2013TTT\r\n\tAAT\u2013ACC\u2013CAG\u2013TTG\u2013CGG\u2013AAA\u2013TCA\r\n\tAGA\u2013CTC\u2013TGA\u2013CGG\u2013TAC\u2013GAA\u2013CTG\r\n\tTCT\u2013GAG\u2013ACT\u2013GCC\u2013ATG\u2013CTT\u2013GAC\r\n\tTGC\u2013AAG\u2013TTC\u2013GGA\u2013CCA\u2013ATG\u2013TAC\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\n\r\nA class of biology students have recently completed a practical where they extracted DNA from strawberries. Each group of students used half a strawberry each, completed the extraction and measured the concentration of DNA using a spectrophotometer.\r\nThe class data is collated in the following table.\r\n\r\n\r\n\t\r\n\t\t\r\n\t\t\tGroup\r\n\t\t\tDNA concentration (ng/\u00b5L)\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tA\r\n\t\t\t\\(28.5\\)\r\n\t\t\r\n\t\t\r\n\t\t\tB\r\n\t\t\t\\(32.0\\)\r\n\t\t\r\n\t\t\r\n\t\t\tC\r\n\t\t\t\\(29.5\\)\r\n\t\t\r\n\t\t\r\n\t\t\tD\r\n\t\t\t\\(33.0\\)\r\n\t\t\r\n\t\t\r\n\t\t\tE\r\n\t\t\t\\(31.5\\)\r\n\t\t\r\n\t\t\r\n\t\t\tF\r\n\t\t\t\\(27.5\\)\r\n\t\t\r\n\t\t\r\n\t\t\tG\r\n\t\t\t\\(13.8\\)\r\n\t\t\r\n\t\t\r\n\t\t\tH\r\n\t\t\t\\(34.5\\)\r\n\t\t\r\n\t\t\r\n\t\t\tI\r\n\t\t\t\\(35.0\\)\r\n\t\t\r\n\t\t\r\n\t\t\tJ\r\n\t\t\t\\(30.5\\)\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n \r\nPhosphodiester bond\r\nA section of a nucleic acid strand is shown, consisting of two nucleotides. The phosphodiester bond highlighted. This is a phosphate group that connects the two nucleotides.\r\nThe phosphate group forms a bridge between the 3 prime carbon of the top sugar and the 5 prime sugar of the bottom sugar. This group is made up of a central phosphorus atom bonded to four oxygen atoms: one double-bonded oxygen (O=P), two single-bonded oxygens linking to the sugars (O-P-O) and one negatively charged oxygen (O-). \r\nComplementary base pairing\r\nA segment of double-stranded DNA, with each strand containing two nucleotides. The left strand has a cytosine nucleotide and an adenine nucleotide. The right strand has a guanine nucleotide and a thymine nucleotide.\r\nThe cytosine nucleotide (highlighted in green) pairs with a guanine nucleotide (highlighted in yellow) through three hydrogen bonds, shown in dashed lines. The adenine nucleotide (highlighting in red) pairs with a thymine nucleotide (highlighted in blue) through two hydrogen bonds, also shown as dashed lines.\r\n \r\nA three-dimensional model of a segment of DNA with atoms shown as coloured sticks.\r\nDNA double helix model\r\nPhosphate is shown is gold, oxygen is red, nitrogen is blue, carbon is teal and hydrogen is white. The atoms are arranged to form two strands of nucleotides. The strands form a twisted ladder structure called the double helix.\r\n \r\nA three-dimensional model of DNA structure, showing the different levels of organisation.\r\nDNA structure model\r\n\r\n\tStrands of DNA are organised into a double helix structure.\r\n\tThe double helix is coiled around clusters of four histone proteins, shown in pink and yellow.\r\n\tThe histones are further coiled to form solenoids.\r\n\tSolenoids are coiled even further to eventually form chromatin.\r\n\tChromatin condenses into chromosomes.\r\n\r\n \r\nChromatids\r\nA simplified drawing of chromatids. A single chromatid is shown on the left. It has a pinched region in the middle, labelled the \"Centromere\". The centromere is slightly higher on the chromatid, so the top section is shorter than the bottom. These correspond to the \"Short arm\" and \"Long arm\", respectively.\r\nOn the right, sister chromatids are shown. It consists of two chromatids joined at the constricted region. The centromere, short arms and long arms are also labelled, matching the drawing on the left.\r\n","n":0.034},[{"v":"DNA","i":1,"n":1},{"v":"Molecular biology","i":0,"n":0.707}]]},{"i":68,"$":[{"v":"Protein synthesis","n":0.707},{"v":"The central dogma of molecular biology describes the flow of genetic material within a biological system. This process is fundamental to understanding how proteins are made using genetic instructions in DNA. Understanding protein synthesis is key to developing therapies in medicine, enhancing crop yields in agriculture and creating new biotechnologies.\r\n\r\nThe central dogma\r\nThe synthesis of proteins is explained using the central dogma (\u201ccentral dogma\u201d just means it is a fundamental principle or concept). It tells us that DNA is copied into messenger RNA (mRNA) through a process called transcription and then mRNA is used to make proteins in a process called translation.\r\n\r\n\r\n\r\nYou will soon learn that this process is very complex, with a number of modifications that occur to the mRNA and protein before it is ready for use by the body. Any errors with these processes could lead to a malfunctioning protein.\r\n\r\nTranscription\r\nTranscription (\u201ctrans\u201d means \u201cacross\u201d and \u201cscribe\u201d means \u201cwrite\u201d) is the process in which a specific segment of DNA is copied into mRNA. This mRNA strand contains the genetic blueprints for making a protein. \r\nThere are three steps to transcribing DNA, which occur in the nucleus of eukaryotic cells: initiation, elongation and termination.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nWatch this video to see transcription in action.\r\n\r\nPost-transcriptional modifications\r\nIn eukaryotes, transcription forms pre-mRNA which needs to undergo additional modifications like 5\u2019 capping, polyadenylation and splicing to prepare it for translation.\r\n\r\n\t5\u2019 capping \u2013 to improve the stability of the mRNA transcript and protect it from degradation, a cap can be added to the 5\u2019 end via a phosphate linkage\r\n\tpolyadenylation \u2013 just like 5\u2019 capping, polyadenylation helps to stabilise the mRNA transcript; it involves adding a poly-A tail consisting of about 200 adenine nucleotides\r\n\tsplicing \u2013 segments of genetic code called exons code for functional proteins but mRNA transcribed from DNA also contains segments called introns which no not code for protein; introns are cut out from the mRNA and the exons are joined together again.\r\n\r\n\r\nTranslation\r\nTranslation is the conversion of the genetic code from the mRNA into a specific sequence of amino acids, forming a polypeptide or protein. It occurs within ribosomes, which consist of two subunits. The small subunit binds the mRNA transcript and the large subunit binds transfer RNA (tRNA).\r\n \r\n\r\n\r\n\r\n\r\ntRNA molecules read the mRNA template in sets of three nucleotides, called codons, with each codon corresponding to a specific amino acid. They bring the appropriate amino acids to the ribosome so they can be added to the polypeptide chain.\r\nThe codon chart is shown.\r\n\r\nLike transcription, translation involves an initiation, elongation and termination step. \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nPost-translational modifications\r\nJust like after transcription, there are modifications that are often made after translation. This depends on the specific role of the protein, its destination within or outside the cell, and the needs of the cell.\r\nThere are many modifications that can be made to alter the protein\u2019s function, activity, stability or location. Common examples are:\r\n\r\n\tphosphorylation \u2013 addition of a phosphate group to activate or deactivate enzymes and regulate cell signalling\r\n\tglycosylation \u2013 addition of sugar molecules to help protein folding, stability and regular cell signalling\r\n\tacetylation \u2013 addition of an acetyl group to affect gene expression\r\n\tubiquitination \u2013 addition of ubiquitin molecules to mark proteins for degradation\r\n\tmethylation \u2013 addition of a methyl group to affect gene expression and protein function.\r\n\r\nThese modifications occur in the Golgi apparatus and endoplasmic reticulum. Then, the protein is folded into its secondary, tertiary and sometimes, quarternary structure.\r\nHumans have over 100,000 different proteins. Together with the unique sequence of amino acids, post-translational modifications give proteins their specificity. This accounts for the diversity of proteins in the body.\r\n\r\nExercise\r\nSee how well you understand how proteins are synthesised with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nDr Jean Nome wishes to investigate how differences in transcription affects the final protein production in cells.\r\nShe looks at five different genes and measures mRNA abundance (the number of copies of messenger RNA in each cell) and protein yield (the mass of protein in milligrams produced by each cell). Dr Nome measures the protein yield using colorimetry. \r\n\r\nThe results are presented in the following table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tGene ID\r\n\t\t\tmRNA abundance (copies per cell)\r\n\t\t\tProtein yield (mg per cell)\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tGene A\r\n\t\t\t\\(500\\)\r\n\t\t\t\\(50\\)\r\n\t\t\r\n\t\t\r\n\t\t\tGene B\r\n\t\t\t\\(300\\)\r\n\t\t\t\\(15\\)\r\n\t\t\r\n\t\t\r\n\t\t\tGene C\r\n\t\t\t\\(700\\)\r\n\t\t\t\\(25\\)\r\n\t\t\r\n\t\t\r\n\t\t\tGene D\r\n\t\t\t\\(200\\)\r\n\t\t\t\\(40\\)\r\n\t\t\r\n\t\t\r\n\t\t\tGene E\r\n\t\t\t\\(600\\)\r\n\t\t\t\\(30\\)\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n \r\nFlowchart of the central dogma\r\nA three-step flowchart, starting with \"DNA\" on the left, \"RNA\" in the middle and \"Protein\" on the right. DNA is represented by a double-helix. RNA is represented with a single strand of DNA. Protein is represented by a thick coiled line.\r\n\r\n\tA solid arrow is shown pointing from \"DNA\" to \"RNA\", labelled \"Transcription\".\r\n\tA dashed arrow is shown pointing from \"RNA\" back to \"DNA\", labelled \"Reverse transcription\".\r\n\tA solid curved arrow is shown pointing from \"DNA\" back to itself, labelled \"Replication\".\r\n\tA solid arrow is shown pointing from \"RNA\" to \"Protein\", labelled \"Translation\".\r\n\tA dashed curved arrow is shown pointing from \"RNA\" back it itself, labelled \"Replication\".\r\n\r\n \r\nInitiation step of transcription\r\nA diagram showing the initiation of transcription. A segment of double-stranded DNA is shown, with part of it on the far left labelled \"Promoter region\". At the promoter region, a purple blob is attached. It is labelled \"RNA polymerase\".\r\nThis diagram shows the binding of the RNA polymerase enzyme to the promoter region of the DNA.\r\n \r\nElongation step of transcription\r\nA diagram showing the elongation stage of transcription. A segment of double-stranded DNA is shown. The RNA polymerase has attached to the bottom strand and moved along it, as shown by an arrow.\r\nFrom the promoter region on the far left and extending to the right up until where the RNA polymerase is, the strands of DNA have separated from each other, or unwound. \r\nComing out from the RNA polymerase is a new strand of nucleotides labelled \"mRNA\". The nucleotide squence matches the top strand and is complementary to the strand that the polymerase has moved along and read.\r\n \r\nTermination step of transcription\r\nA diagram showing the termination of transcription. A segment of double-stranded DNA is shown, with a \"Stop signal\" labelled on the far right. Between the promoter region and stop signal, the DNA double helix has unwound.\r\nThe purple RNA polymerase enzyme has detached from the DNA, shown by an arrow.\r\nAbove the double-stranded DNA is a singe strand labelled \"pre-mRNA\". Its sequence matches the top strand in the DNA.\r\n \r\nWhat you are about to see is DNA's most extraordinary secret \u2014 how a simple code is turned into flesh and blood. It begins with a bundle of factors assembling at the start of a gene. A gene is simply a length of DNA instructions stretching away to the left. The assembled factors trigger the first phase of the process, reading off the information that will be needed to make the protein. Everything is ready to roll: three, two, one, GO! The blue molecule racing along the DNA is reading the gene. It's unzipping the double helix, and copying one of the two strands. The yellow chain snaking out of the top is a copy of the genetic message and it's made of a close chemical cousin of DNA called RNA. The building blocks to make the RNA enter through an intake hole. They are matched to the DNA - letter by letter - to copy the As, Cs, Ts and Gs of the gene. The only difference is that in the RNA copy, the letter T is replaced with a closely related building block known as \"U\". You are watching this process - called transcription - in real time. It's happening right now in almost every cell in your body.\r\n \r\nPost-transcriptional modifications\r\n\r\n\t5 prime capping (top): The pre-mRNA strand has an extra yellow nucleotide added to the left end.\r\n\tPolyadenylation (middle): The pre-mRNA strand has an additional string of red nucleotides added to the right end.\r\n\tSplicing (bottom): Two sections of the pre-mRNA template are labelled introns. In the final mRNA, they are removed.\r\n\r\n \r\nA three-dimensional model of a single ribosome with the two subunits shown in different colours.\r\nRibosome model\r\n\r\n\tSmall subunit (pink): decodes the genetic messages sent by the mRNA.\r\n\tLarge subunit (purple): the ribosome reads the messenger RNA (mRNA) and uses the information to string together amino acids into a protein.\r\n\r\n \r\nThe image is a codon chart, used to identify amino acids coded by mRNA sequences during translation.\r\nCodon chart\r\nIt organises the codons by the three bases that make them up:\r\n\r\n\tFirst base: displayed on the left, includes U (uracil), C (cytosine), A (adenine), and G (guanine)\r\n\tSecond base: Displayed along the top, with the same bases\r\n\r\nEach cell within the chart represents a codon made up of a combination of bases, specifying a particular amino acid or a start/stop signal. Here's a breakdown:\r\n\r\n\tPhenylalanine: UUU and UUC\r\n\tLeucine: UUA, UUG, CUU, CUC, CUA and CUG\r\n\tIsoleucine: AUU, AUC and AUA\r\n\tMethionine: AUG, the start codon (indicated in green as START)\r\n\tValine: GUU, GUC, GUA, GUG\r\n\tSerine: UCU, UCC, UCA and UCG\r\n\tProline: CCU, CCC, CCA and CCG\r\n\tThreonine: ACU, ACC, ACA and ACG\r\n\tAlanine: GCU, GCC, GCA and GCG\r\n\tTyrosine: UAU and UAC\r\n\tStop codons: UAA, UAG, UGA (indicated in red as STOP)\r\n\tHistidine: CAU and CAC\r\n\tGlutamine: CAA and CAG\r\n\tAsparagine: AAU and AAC\r\n\tLysine: AAA and AAG\r\n\tAspartic acid: GAU and GAC\r\n\tGlutamic acid: GAA and GAG\r\n\tCytosine: UGU and UGC\r\n\tTryptophan: UGG\r\n\tArginine: CGU, CGC, CGA and CGG\r\n\tSerine: AGU and AGC\r\n\tArginine: AGA and AGG\r\n\tGlycine: GGU, GGC, GGA and GGG\r\n\r\n \r\nInitiation step of protein translation\r\nA diagram showing a segment of mRNA with a ribosome and transfer RNA attached. The small subunit of the ribosome is attached to the mRNA at the start codon. The large subunit is attached to an initiator tRNA molecule, which has methionine associated with it. The large subunit has three sites for tRNA to bind. The middle site is occupied by the initiator tRNA.\r\n \r\n\t\tElongation step of protein translation\r\n\t\tA diagram showing a segment of mRNA with a ribosome and two tRNA molecules occupying the middle and right-hand sites. The methionine is shown attaching to the amino acid on the tRNA in the right-hand site, which is labelled \"tRNA with anticodon\". The anticodon matches with the codon, labelled \"Sense codon\", on the mRNA strand. The right-hand site is labelled the \"A site\". Freely floating to the right is another tRNA molecule with a different amino acid, labelled the \"Charged tRNA\".\r\n\t \r\n\t\tElongation step of protein translation\r\n\t\tA diagram showing a segment of mRNA with a ribosome and three tRNA molecules occupying the left-hand, middle and right-hand sites. The amino acid is shown attaching to the growing polypeptide chain on the tRNA in the right-hand site. Each tRNA has moved one site to the left from the previous diagram. The middle site is labelled the \"P site\".\r\n\t \r\n\t\tElongation step of protein translation\r\n\t\tA diagram showing a segment of mRNA with a ribosome and three tRNA molecules occupying the left-hand, middle and right-hand sites. The right-hand site contains a tRNA molecule with a blue amino acid. The middle site contains a tRNA molecule with the rest of the polypeptide chain. The left-hand site is labelled the E site. Each tRNA has moved one site to the left from the previous diagram. Freely floating to the left is the tRNA molecule that was in the E site in the previous diagram. It is labelled the \"Uncharged tRNA\".\r\n\t \r\nTermination step of translation\r\nA diagram showing the termination of translation. The right-hand site contains a tRNA molecule with no amino acid. It binds to a segment of the mRNA called the \"Stop codon\". The middle site contains a tRNA molecule with the final polypeptide chain. Each tRNA has moved one site to the left from the previous diagram. \r\n \r\nWhen the RNA copy is complete, it snakes out into the outer part of the cell. Then in a dazzling display of choreography, all the components of a molecular machine lock together around the RNA to form a miniature factory called a ribosome. It translates the genetic information in the RNA into a string of amino acids that will become a protein. Special transfer molecules, the green triangles, bring each amino acid to the ribosome. The amino acids are the small red tips attached to the transfer molecules. There are different transfer molecules for each of the twenty amino acids. Each transfer molecule carries a three letter code that is matched with the RNA in the machine. Now we come to the heart of the process. Inside the ribosome, the RNA is pulled through like a tape. The code for each amino acid is read off, three letters at a time, and matched to three corresponding letters on the transfer molecules. When the right transfer molecule plugs in, the amino acid it carries is added to the growing protein chain. Again, you are watching this in real time. And after a few seconds the assembled protein starts to emerge from the ribosome. Ribosomes can make any kind of protein. It just depends what genetic message you feed in on the RNA. In this case, the end product is hemoglobin. The cells in our bone marrow churn out a hundred trillion molecules of it per second! And as a result, our muscles, brain and all the vital organs in our body receive the oxygen they need.\r\n","n":0.022},[{"v":"DNA","i":2,"n":1},{"v":"Proteins","i":1,"n":1},{"v":"Molecular biology","i":0,"n":0.707}]]},{"i":69,"$":[{"v":"Macromolecules","n":1},{"v":"Macromolecules play key roles in the structure and function of cells. Understanding macromolecules helps us learn how organisms grow, develop, and maintain their systems. Beyond the body, they are used in medical treatments, nutrition and biotechnology. Use this resource to learn about the types of macromolecules.\r\n\r\nMacromolecules are large, complex molecules essential for life. They include carbohydrates, proteins, lipids, and nucleic acids. Each type has a different role within a cell. For example, proteins help build and repair tissues, while nucleic acids like DNA store genetic information.\r\n\r\nCarbohydrates\r\nCarbohydrates are macromolecules made up of simple sugar monomers (units) called monosaccharides (\u201cmono\u201d meaning \u201cone\u201d and \u201csaccharide\u201d meaning \u201csugar\u201d). These can be linked together to form repeating polymers called disaccharides (\u201cdi\u201d meaning \u201ctwo\u201d, i.e. two sugar monomers) or polysaccharides (\u201cpoly\u201d meaning \u201cmany\u201d, i.e. many sugar monomers).\r\n\r\nThere are three different types of carbohydrates:\r\n\r\n\tsugars \u2013 such as glucose and fructose in fruits, vegetables and honey\r\n\tstarches \u2013 such as amylose and amylopectin found in potatoes, rice and corn\r\n\tfibre \u2013 such as cellulose, oligosaccharides and chitins found in leafy vegetables, grains and crustaceans.\r\n\r\nCarbohydrates are composed of carbon (\\(\\ce{C}\\)) atoms, hydrogen (\\(\\ce{H}\\)) atoms and oxygen (\\(\\ce{O}\\)) atoms. The \u201ccarbo\u201d part of its name refers to the carbon atoms and the \u201chydrate\u201d part refers to the presence of water (\\(\\ce{H2}\\)). For example, glucose has the formula \\(\\ce{C6H12O6}\\) and amylose \\(\\ce{(C6H10O5)_{n}}\\), where \\(n\\) represents the number of repeating glucose units.\r\nIn living organisms, carbohydrates are the primary source of energy. They are crucial for cell function and metabolism.\r\n\r\n\r\nProteins\r\nProteins are polymer chains of amino acids, which are linked by peptide bonds.  Amino acids have an amino (\\(\\ce{NH2}\\)) group, a carboxyl (\\(\\ce{COOH}\\)) group and a unique side chain (R group) which gives the amino acid its properties. There are \\(20\\) standard amino acids that combine in different sequences to form proteins.\r\n\r\nProteins are essential for bodily functions like building and repairing tissues, facilitating chemical reactions and supporting the immune response. They are composed of carbon (\\(\\ce{C}\\)), hydrogen (\\(\\ce{H}\\)), oxygen (\\(\\ce{O}\\)) and nitrogen (\\(\\ce{N}\\)) atoms.\r\nThe 3D structure of a protein is very important for its function, so it is important to understand how we arrive at its final shape. Proteins are organised into four levels of structure:\r\n\r\n\tPrimary structure: the linear sequence of amino acids linked by peptide bonds\r\n\tSecondary structure: the alpha helices or beta pleated sheets formed by hydrogen bonds between amino acid side chains\r\n\tTertiary structure: the overall 3D shape of a single protein molecule created by disulfide bonds, hydrogen bonds, ionic bonds and hydrophobic interactions between amino acid side chains\r\n\tQuaternary structure: the arrangement of multiple protein subunits into a more complex structure. Not all proteins have a quaternary structure.\r\n\r\n\r\n\r\nAn example of a protein is haemoglobin, which transports oxygen in the blood. It is able to bind oxygen-carrying heme molecules because of its complex quaternary structure. Use the 3D model shown to look at the structure of haemoglobin.\r\n\r\n\r\n\r\n\r\n\r\n\r\nLipids\r\nLipids are hydrophobic (\"hydro\" meaning \"water\" and \"phobic\" meaning \"hating\") macromolecules, including fats, oils and steroids. They are composed of primarily of carbon (\\(\\ce{C}\\)), hydrogen (\\(\\ce{H}\\)) and oxygen (\\(\\ce{O}\\)) atoms, but can contain other atoms.\r\nLipids are essential for storing energy, forming cell membranes, insulation, producing hormones and serving as signalling molecules. Unlike proteins and carbohydrates, lipids don't have a standard monomer but are often made up of fatty acids and glycerol.\r\nCommon examples include triglycerides which store energy in fat cells, phospholipids which form the protective outer layer of cells, and cholesterols which help your body create bile.\r\n\r\n\r\n\r\nNucleic acids\r\nNucleic acids are macromolecules that store genetic information. The two main types of nucleic acids are: \r\n\r\n\tdeoxyribonucleic acid (DNA) which carries genetic instructions for the cell\r\n\tribonucleic acid (RNA) which plays a role in translating DNA into proteins.\r\n\r\n\r\nNucleic acids are composed of nucleotide monomers. These consist of a sugar, phosphate group and nitrogenous base. \r\n\r\nThe bases are adenine, thymine (or uracil in RNA), cytosine and guanine. These four bases form pairs to encode genetic instructions for the cell. The atoms in these macromolecules are carbon (\\(\\ce{C}\\)), hydrogen (\\(\\ce{H}\\)), oxygen (\\(\\ce{O}\\)), nitrogen (\\(\\ce{N}\\)) and phosphate (\\(\\ce{P}\\)).\r\n\r\nYou will learn more about how DNA is transcribed into RNA, then translated into proteins in protein synthesis. \r\n\r\nSummary\r\nThe features of the four types of macromolecules are summarised in the table.\r\n\r\n\r\nExercise\r\nSee how well you understand macromolecules and their characteristics with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nSochanda is a first year student enrolled in the Bachelor of Food Technology and Nutrition at RMIT. As part of her Chemistry course, she conducts a practical comparing the energy released from proteins, lipids and carbohydrates using calorimetry. This involves combusting each macronutrient, measuring the heat released, and using this to calculate the energy content.\r\nSochanda's results are shown in the table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tMacronutrient\r\n\t\t\tEnergy content (kJ g\\(^{-1}\\))\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tLipid\r\n\t\t\t\\(32\\)\r\n\t\t\r\n\t\t\r\n\t\t\tProtein\r\n\t\t\t\\(13\\)\r\n\t\t\r\n\t\t\r\n\t\t\tCarbohydrate\r\n\t\t\t\\(10\\)\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n \r\nStructure of proteins\r\n\r\n\tPrimary structure (top left): coloured balls connected in a chain.\r\n\tSecondary structure (top right): two corrugated sheets which are beta pleated sheets or two helical shapes which called alpha helices.\r\n\tTertiary structure (bottom left): beta pleated sheets and alpha helices connected together in a tangled chain.\r\n\tQuaternary structure (bottom right): four units of the tertiary structure.\r\n\r\n \r\nA three-dimensional model of oxyhaemoglobin.\r\nOxy-haemoglobin model\r\nFour protein subunits (coloured green, light blue, pink and yellow) are represented using coils of ribbon. Each subunit contains a heme molecule with its full structure shown using ball-and-stick models. Each heme molecule interacts with iron, represented by an orange sphere. Each iron interacts with an oxygen molecule, shown using its ball-and-stick model. All interactions are shown using dashed white lines.\r\n","n":0.033},[{"v":"DNA","i":4,"n":1},{"v":"Lipids","i":3,"n":1},{"v":"Carbohydrates","i":2,"n":1},{"v":"Proteins","i":1,"n":1},{"v":"Molecular biology","i":0,"n":0.707}]]},{"i":70,"$":[{"v":"Molecular biology","n":0.707},{"v":"Molecular biology is the study of biological processes and interactions at the molecular level within living organisms. It focuses on how molecules like proteins, lipids, carbohydrates, and nucleic acids contribute to the functions of cells and tissues. Molecular biology is important for medicine, agriculture, and biotechnology as it helps us understand how life operates at a molecular level and how genetic and cellular processes are controlled.\r\n\r\nOverview\r\n\r\n\tMacromolecules \u2013 what are four major classes of macromolecules and how do they differ in structure?\r\n\tDNA \u2013 what is DNA and how is it organised?\r\n\tProtein synthesis \u2013 how does a cell use instructions from DNA to make proteins?\r\n\tStructure and function of enzymes \u2013 what are enzymes and how do they work?\r\n\r\n\r\n","n":0.093},[{"v":"Molecular biology","i":0,"n":0.707}]]},{"i":71,"$":[{"v":"Propagating uncertainty","n":0.707},{"v":"When multiple measurements are used to calculate a quantity, each of these measurements may carry its own uncertainty. By being careful and taking into account these uncertainties, scientists and engineers can better assess the reliability and precision of their conclusions. Use this resource to learn how to consider, or propagate, uncertainty.\r\n\r\nPropagation of uncertainty involves estimating the uncertainty in the calculated value. The way you propagate uncertainty depends on whether you have added, subtracted, multiplied or divided the measured quantities.\r\n\r\nUncertainty in sums and differences\r\nFor addition and subtraction of measurements, the uncertainties are treated the same way.\r\nWhen measurements are added or subtracted, the absolute uncertainties are added.\r\n\r\nExample 1 \u2013 propagating uncertainty for sums and differences\r\nConsider the measurements:\r\n\\[A=4.0\\pm0.2\\textrm{ cm}\\quad B=2.0\\pm0.1\\textrm{ cm}\\]\r\nFind the sum of \\(A\\) and \\(B\\), including the absolute uncertainty.\r\nWe simply add the measurements together, and then their uncertainties.\r\n\\[\\begin{align*} A+B & = (4.0+2.0)\\pm(0.2+0.1)\\textrm{ cm}\\\\\r\n& = 6.0\\pm0.3\\textrm{ cm}\r\n\\end{align*}\\]\r\nThe absolute uncertainty is \\(0.3\\textrm{ cm}\\).\r\n\r\n\r\n\r\nUncertainty in products and quotients\r\nFor multiplication and division of measurements, the way we find the uncertainty is different. In fact, we don't look at absolute uncertainty, but instead, percentage uncertainty.\r\nWhen measurements are multiplied or divided, the percentage uncertainties are added.\r\n\r\nExample \u2013 propagating uncertainty for products and quotients\r\nCalculate the area of a rectangle and its associated absolute uncertainty from the following data:\r\n\\[\\textrm{Length}=14.26\\pm0.02\\textrm{ cm}\\quad \\textrm{Width} = 5.94\\pm0.02\\textrm{ cm}\\]\r\nThe area of the rectangle is calculated by length times width.\r\n\\[\\begin{align*} A & = =14.26\\times5.94\\\\\r\n& = 84.7\\textrm{ cm}^{2}\r\n\\end{align*}\\]\r\nThe percentage uncertainty is the sum of the uncertainty in the length and the uncertainty in the width.\r\n\\[\\begin{align*} \\% \\textrm{Uncertainty in length} & = \\frac{\\textrm{Absolute error in length}}{\\textrm{Length}} \\times100\\\\\r\n& = \\frac{0.02}{14.26}\\\\\r\n& = 0.14\\%\r\n\\end{align*}\\] \\[\\begin{align*} \\% \\textrm{Uncertainty in width} & = \\frac{\\textrm{Absolute error in width}}{\\textrm{Width}} \\times100\\\\\r\n& = \\frac{0.02}{5.94}\\\\\r\n& = 0.34\\%\r\n\\end{align*}\\] \\[\\begin{align*} \\% \\textrm{Uncertainty in area} & = 0.14\\% + 0.34\\%\\\\\r\n& = 0.48\\%\r\n\\end{align*}\\]\r\nPutting these together, the area of the rectangle is \\(84.7\\textrm{ cm}^{2}\\pm0.48\\%\\). But, the question is asking for absolute uncertainty, so we should find this using the percentage.\r\n\\[0.48\\%\\times84.7=0.4\\textrm{ cm}^{2}\\]\r\nSo, the area of the rectangle is \\(84.7\\pm0.4\\textrm{ cm}^{2}\\).\r\n\r\nUncertainty in powers\r\nFor measurements raised to a power, the percentage uncertainty is added as many times as the power. For example, a value raised to the power of \\(3\\) will have its percentage uncertainty trebled. This is the same as multiplying the value by itself three times, so the three percentage uncertainty values are added.\r\nWhen measurements are raised to a power \\(n\\), the percentage uncertainties are multiplied by \\(n\\).\r\n\r\nExample \u2013 propagating uncertainty for powers\r\nFor a sphere of radius \\(5.0\\pm0.1\\textrm{ cm}\\), calculate its volume and associated uncertainty.\r\nThe equation for the volume of a sphere is \\(\\textrm{Volume} = \\dfrac{4}{3}\\pi r^{3}\\). Let's calculate the volume first.\r\n\\[\\begin{align*} \\textrm{Volume} & = \\frac{4}{3}\\pi (5.0)^{3}\\\\\r\n& = 524\\textrm{ cm}^{3}\\\\\r\n\\approx 520\\textrm{ cm}^{3}\r\n\\end{align*}\\]\r\nThe percentage uncertainty for the radius is:\r\n\\[\\begin{align*} \\% \\textrm{Uncertainty in radius} & = \\frac{\\textrm{Absolute uncertainty}}{\\textrm{Radius}}\\times 100\\\\\r\n& = \\frac{0.1}{5}\\times100\\\\\r\n& = 2\\%\r\n\\end{align*}\\]\r\nTo find the percentage uncertainty for the volume, we multiply by \\(n\\) which is the power. Here, the power is \\(3\\), so:\r\n\\[\\begin{align*} \\% \\textrm{Uncertainty in volume} & = 3\\times \\% \\textrm{Uncertainty in radius}\\\\\r\n& = 3\\times2\\%\\\\\r\n& = 6\\%\r\n\\end{align*}\\]\r\nThe final volume is \\(520\\textrm{ cm}^{3}\\pm 6\\%\\) or \\(520\\pm30\\textrm{ cm}^{3}\\).\r\n\r\nExercise \u2013 propagating uncertainty\r\n\r\n\tIf \\(A=6.0\\pm0.1\\textrm{ cm}\\) and \\(B=3.4\\pm0.2\\textrm{ cm}\\), calculate:\r\n\t\r\n\t\t\\(A+B\\)\r\n\t\t\\(A-B\\)\r\n\t\t\\(A\\times B\\)\r\n\t\t\\(A\\div B\\)\r\n\t\r\n\tThe mass of a square cube was found to be \\(60.7\\pm0.05\\textrm{ g}\\). A side length of this cube is \\(1.5\\pm0.05\\textrm{ cm}\\). Calculate the density of the material and its associated uncertainty. \\(\\left(\\textrm{Density}=\\dfrac{\\textrm{Mass}}{\\textrm{Volume}}\\right)\\)\r\n\tA cylinder has a height of \\(10.3\\pm0.05\\textrm{ cm}\\). Its radius is \\(4.5\\pm0.05\\textrm{ cm}\\). Calculate the volume of the cylinder and its associated error. \\(\\left( \\textrm{Volume}=\\pi r^{2}h \\right)\\).\r\n\r\n\r\n\r\n\r\n","n":0.042},[{"v":"Measurement","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":72,"$":[{"v":"Copyright","n":1},{"v":"While studying, you will use works written or created by others in your assignments, presentations and assessments. Many of those works will be protected by copyright law.\r\nCopyright is the legal right of the creator to determine if and how their works can be used by others, and this includes the right to reproduce, distribute, display, and make adaptations. In Australia, certain forms of original expression in material form are automatically protected by copyright. These include:\r\n\r\n\r\n \tfilm and sound recording\r\n \timages\r\n \ttext\r\n \tmusic.\r\n\r\nUnderstanding how you can use copyright protected material will save you time, and help you avoid potentially serious consequences during your studies at RMIT and beyond.\r\n\r\nKnowing what material you can use and how to use it correctly, is an important aspect of academic integrity.\r\nThe Fair dealing provision for Research or study\r\nThe Fair dealing provision for Research or Study in the Australian Copyright Act enables students and researchers to rely on using a reasonable portion of a copyrighted work, such as 10% or one chapter of a book, for research or study.\r\n\r\nFor artistic works, videos, films, and sound recordings, there is no simple defined rule on the amount you can use. Instead, \"reasonable\" is key the key factor \u2014 reasonable in terms of the amount of works used and in terms of the nature of use.\r\n\r\nThe fair dealing provision is closed, which means:\r\n\r\n \tIt applies only during a course of study or research.\r\n \tCopyrighted materials must be deleted from RMIT platforms like Canvas once studies conclude.\r\n \tThe Research or Study provision does not permit sharing someone else's copyrighted work on public sites such as LinkedIn or Instagram.\r\n \tSharing copright protected works within a public digital portfolio requires permission from the copyright holder.\r\n\r\nOther key points to understand are:\r\n\r\n \t\r\n\r\n \tGenerally, students own the copyright to original works created in a course at RMIT.\r\n \tIf someone else, such as an instructor, wishes to reuse your assignment or artwork, as an example or promotional activity, they must have your consent.\r\n \tFor works created as part of a group, copyright is shared. You need all group members' permission for public sharing.\r\n\r\n\r\n\r\nMoral rights and the right of attribution\r\nAll uses of copyright works, including under the fair dealing provisions, must contain a credit statement or attribution referencing the creator/author of the work.\r\n\r\nAnd remember, you should not falsely claim to be an author of a work you did not create.\r\n\r\nFor more information, visit the Referencing page.\r\nTest your knowledge\r\n","n":0.05},[{"v":"Referencing","i":2,"n":1},{"v":"Assessment","i":1,"n":1},{"v":"Academic integrity","i":0,"n":0.707}]]},{"i":73,"$":[{"v":"Quality of measurements","n":0.577},{"v":"The quality of measurements is very important in science and engineering because it affects how reliable the results are. There are a number of factors that affect the quality of measurements, like accuracy, precision, uncertainty and error. Use this resource to learn about these concepts.\r\n\r\nAccuracy and precision\r\nAccuracy shows how close a measurement is to the true value, while precision shows how consistent the measurements are.\r\nConsider a target board used for archery.\r\n\r\n\tAccuracy refers to how close the arrows are to the bullseye. If the arrows are consistently hitting near the bullseye, they are accurate.\r\n\tPrecision refers to how close the arrows are to each other. If all arrows land in a small cluster, even if they're not near the bullseye, they are precise.\r\n\r\nIdeally, you want both accuracy and precision, meaning the arrows are tightly grouped around the bullseye.\r\n\r\n\r\nUncertainty\r\nWe measure physical quantities using devices called instruments. A ruler measures length, a thermometer measures temperature, a scale measures weight, a spring balance measures force and a stopwatch measures time.\r\nEach of these instruments has uncertainty, meaning they can't give exact measurements. This uncertainty comes from things like the instrument's limits or changes in the environment. For example:\r\n\r\n\tA ruler marked with millimetres cannot measure smaller details like fractions of a millimetre.\r\n\tHigh humidity and moisture in the environment can affect a scale's sensors, leading to inaccurate weight measurements.\r\n\tTemperature affects the pressure reading on a barometer.\r\n\tAn analogue stopwatch can only be read to the nearest tenth of a second. It is not able to measure finer time intervals accurately.\r\n\r\nKnowing about these uncertainties helps to make sure that your measurements are accurate and reliable.\r\n\r\nReporting measurement uncertainty\r\nMeasurement uncertainty can be expressed as the absolute uncertainty of the mean. This is the uncertainty in the average value obtained from multiple measurements and gives an indication of how much the average is expected to vary.\r\nAbsolute uncertainty is denoted by \\(\\Delta\\bar{x}\\) and is calculated using the maximum and minimum measurement values:\r\n\\[\\textrm{Absolute uncertainty}\\,(\\Delta\\bar{x}) = \\pm \\frac{x_{\\textrm{max}}-x_{\\textrm{min}}}{2}\\]\r\nFrom this, the percentage uncertainty can be calculated. This represents the uncertainty as a proportion of the measurement average.\r\n\\[\\textrm{Percentage uncertainty} = \\frac{\\textrm{Absolute uncertainty}}{\\textrm{Measurement average}} \\times 100\\]\r\n\r\nExample \u2013 calculating uncertainty\r\nSuppose you measure the length of a table five times and get the values \\(200.1\\textrm{ cm}\\), \\(200.3\\textrm{ cm}\\), \\(199.9\\textrm{ cm}\\), \\(200.2\\textrm{ cm}\\) and \\(200.0\\textrm{ cm}\\). The average measurement is \\(200.1\\textrm{ cm}\\). Find the absolute and percentage uncertainty of the measurement.\r\nUsing the formula, you would calculate the absolute uncertainty to be:\r\n\\[\\begin{align*} \\Delta\\bar{x} & = \\frac{x_{\\textrm{max}}-x_{\\textrm{min}}}{2}\\\\\r\n& = \\frac{200.3-199.9}{2}\\\\\r\n& = 0.2\\textrm{ cm}\r\n\\end{align*}\\]\r\nWe would therefore state the measurement as \\(200.1\\pm0.2\\textrm{ cm}\\).\r\nIn terms of percentage uncertainty:\r\n\\[\\begin{align*} \\textrm{Percentage uncertainty} & = \\frac{\\textrm{Absolute uncertainty}}{\\textrm{Measurement average}} \\times 100\\\\\r\n& = \\frac{0.2\\textrm{ cm}}{200.1\\textrm{ cm}}\\times100\\\\\r\n& \\approx 0.1\\%\r\n\\end{align*}\\]\r\nSo, the measurement would be \\(200.1\\textrm{ cm}\\pm0.1\\%\\). This means that the measurement deviates from the average of \\(200.1\\textrm{ cm}\\) by \\(0.2\\textrm{ cm}\\) or \\(0.1\\%\\). That is, there is a \\(0.2\\textrm{ cm}\\) or \\(0.1\\%\\) uncertainty in the result.\r\n\r\n\r\nError\r\nError refers to the difference between the measured value and the true value. Errors are classified as systematic or random. \r\n\r\nSystematic error\r\nA systematic error is caused by consistent problems with the measurement instruments or the methods used. Common systematic errors are: \r\n\r\n\tParallax error, which occurs when reading a measurement from an angle, rather than directly in line with the scale\r\n\t\r\n\tCalibration error, which happens when an instrument hasn't been adjusted to the correct standard, leading to consistently high or low readings\r\n\tZero error, which occurs when an instrument, like a balance, does not start from a true zero, affecting all measurements taken with that instrument.\r\n\r\nTo reduce systematic errors, you should make sure your instruments are calibrated, zeroed (or \"tared\") before use, and that you are using the correct techniques.\r\n\r\nRandom error\r\nRandom errors happen due to unpredictable changes during the experiment. Examples include:\r\n\r\n\tEnvironmental fluctuations, like random changes in temperature, humidity, voltage, wind, etc, that can affect readings\r\n\tHuman reaction time, which can introduce small variations when you have to manually start and stop a timing device.\r\n\r\nWe can limit the effect of random errors by taking repeat measurements and averaging them, improving the way we measure things (like using an automated tool rather than relying on your own reaction speed), and controlling environmental factors where we can (e.g. maintaining constant temperature).\r\n\r\nReporting measurement error\r\nThe error in a measurement can be expressed as absolute error. This is calculated using:\r\n\\[\\textrm{Absolute error} = \\left| \\textrm{Measured value}-\\textrm{True value} \\right| \\]\r\nThis gives an indication of the accuracy of the result; the lower the absolute error, the more accurate the result, i.e. the closer it is to the true value.\r\nError can also be represented by percentage error, which represents the accuracy as a proportion of the true value.\r\n\\[\\textrm{Percentage error} = \\left| \\frac{\\textrm{Measured value}-\\textrm{True value}}{\\textrm{True value}} \\right| \\times100 \\]\r\n\r\nExample \u2013 calculating error\r\nIf the true value of a length measurement is \\(15.0\\textrm{ cm}\\) and the measured value is \\(14.6\\textrm{ cm}\\), calculate the absolute and percentage error.\r\n\\[\\begin{align*} \\textrm{Absolute error} & = \\left| \\textrm{Measured value}-\\textrm{True value} \\right| \\\\\r\n& = \\left| 14.6-15.0 \\right| \\\\\r\n& = 0.4\\textrm{ cm}\r\n\\end{align*}\\]\\[\\begin{align*} \\textrm{Percentage error} & = \\left| \\frac{\\textrm{Measured value}-\\textrm{True value}}{\\textrm{True value}} \\right| \\times100 \\\\\r\n& = \\frac{0.02}{1.00}\\times 100\\\\\r\n& = 2\\%\r\n\\end{align*}\\]\r\n\r\nExercise \u2013 calculating error\r\n\r\n\tA student measures the mass of an object as \\(48\\textrm{ g}\\), but the actual mass is \\(50\\textrm{ g}\\). Calculate the percentage error of the measurement.\r\n\tIf a volume is measured as \\(250\\textrm{ mL}\\) but the true volume is \\(245\\textrm{ mL}\\), what is the absolute error?\r\n\tA student measures the density of a liquid as \\(1.02\\textrm{ g cm}^{-3}\\), while the true density is \\(1.00\\textrm{ g cm}^{-3}\\). Calculate the percentage error.\r\n\r\n\t\r\n\r\n\r\n\r\nMistakes\r\nUnlike uncertainty and error, which are natural in the measurement process, mistakes are usually accidental slip-ups that can be fixed with careful checking and practice.\r\nMistakes include:\r\n\r\n\twriting down the wrong value, e.g. the scale reads \\(72\\) but you write down \\(27\\) by mistake\r\n\tusing the wrong instrument, e.g. picking the wrong pipette and adding \\(1\\textrm{ mL}\\) instead of \\(10\\textrm{ mL}\\)\r\n\tmisusing the calculator, e.g. pressing the \\(\\div\\) button instead of \\(\\times\\).\r\n\r\n\r\n\r\n\r\n","n":0.032},[{"v":"Error","i":2,"n":1},{"v":"Measurement","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":74,"$":[{"v":"Significant figures in calculations","n":0.5},{"v":"Significant figures are important in calculations because they help keep the precision of measurements accurate in the results. When you do math operations, the number of significant figures in your values decides how precise the final answer should be. Use this resource to learn the rules for using significant figures correctly in calculations.\r\nThe rules for significant figures depend on whether you are adding and subtracting, or multipling and dividing.\r\n\r\n\r\nAddition and subtraction\r\nWhen you add and subtract numbers with a different number of significant figures:\r\n\r\n\r\n \tAlign the numbers by their decimal points.\r\n \tIdentify the number with the fewest decimal places.\r\n \tRound off the other numbers so that they all have the same number of decimal places as this number.\r\n \tComplete the calculation.\r\n \tThe answer will automatically have the correct number of significant figures.\r\n\r\n\r\nRounding numbers\r\nRemember the rules for rounding numbers.\r\n\r\n \tIf the last digit is greater than \\(5\\), round up the second last digit by \\(1\\).For example, \\(5.78\\) is rounded up to \\(5.8\\).\r\n \tIf the last digit is less than \\(5\\), round down the second last digit by \\(1\\).For example, \\(1.304\\) is rounded down to \\(1.30\\).\r\n \tIf the last digit is \\(5\\), round off the second last digit to the nearest even number.For example, \\(4.85\\) becomes \\(4.8\\) (not \\(4.9\\)) and \\(4.55\\) becomes \\(4.6\\) (not \\(4.5\\)).\r\n\r\nExercise \u2013 rounding numbers\r\nRound off each of the following numbers to one less significant figure, then to two less significant figures.\r\n\r\n \t\\(7.668\\)\r\n \t\\(0.085\\,4\\)\r\n \t\\(21.092\\)\r\n \t\\(255.6\\)\r\n \t\\(35.3\\)\r\n \t\\(6.75\\)\r\n \t\\(2.85\\)\r\n\r\n\r\nExample \u2013 using significant figures in addition and subtraction\r\nAdd \\(6.703\\), \\(2.49\\) and \\(11.736\\,8\\) and give your answer to the correct number of significant figures.\r\n\r\nFirst, let's align the numbers.\r\n\\[\\begin{align*} 6&.703\\\\\r\n2&.49\\\\\r\n11&.736\\,8\r\n\\end{align*}\\]\r\n\r\n\\(2.49\\) has the fewest number of decimal places: \\(2\\). We need to round the other numbers to \\(2\\) decimal places. \\(6.703\\) would become \\(6.70\\) and \\(11.7368\\) would become \\(11.74\\). We can then add the numbers.\r\n\\[\\begin{array}{r}\r\n6.70\\\\\r\n2.49\\\\\r\n\\underline{+11.74}\\\\\r\n20.93\r\n\\end{array}\\]\r\n\r\nThe answer, given to the correct number of significant figures, is \\(20.93\\).\r\nExercise \u2013 using significant figures in addition and subtraction\r\nCalculate the following and give your answers to the correct number of significant figures.\r\n\r\n \t\\(6.9+0.35+12.625\\)\r\n \t\\(27.1+13+64.51\\)\r\n \t\\(8.72-3.001-0.2\\)\r\n \t\\(67.4-18.37+0.55\\)\r\n \t\\(0.003+0.0125-0.003\\,78\\)\r\n\r\n\r\nMultiplication and division\r\nWhen you multiply and divide numbers with a different number of significant figures:\r\n\r\n\r\n \tIdentify the number with the fewest significant figures.\r\n \tComplete the calculation.\r\n \tRound the final answer to the same number of significant figures as the number identified.\r\n\r\n\r\nExample 1 \u2013 using significant figures in multiplication and division\r\nMultiply \\(5.2\\) by \\(6.3\\) and give your answer to the correct number of significant figures.\r\n\r\nBoth numbers have \\(2\\) significant figures. We multiply as is:\r\n\\[5.2\\times6.3=33\\]\r\n\r\nThe answer is \\(33\\), to \\(2\\) significant figures.\r\n\r\n\r\n\r\n\r\nExercise \u2013 using significant figures in multiplication and division\r\nCalculate the following and give your answers to the correct number of significant figures.\r\n\r\n \t\\(8.3\\times0.25\\)\r\n \t\\(0.2\\times1.3\\)\r\n \t\\(1.13\\times3.5\\times0.964\\)\r\n \t\\(6.71\\times3.4\\)\r\n \t\\(3.000\\times91\\div72.60\\)\r\n \t\\(450\\times3\\div0.671\\)\r\n\r\n\r\n\r\n","n":0.047},[{"v":"Measurement","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":75,"$":[{"v":"Important considerations for portfolios","n":0.5},{"v":"On this page, we will examine some important considerations for your portfolio, such as file size and quality, the privacy and protection of your work, and accessibility and referencing.\r\n\r\nUploading content to your portfolio\r\nWhen uploading files to your portfolio, there are some factors to keep in mind. \r\n\r\nCompatibility It is recommended that you upload files in commonly used formats like PDF, JPEG, PNG, and MP4 to ensure that they can be opened on different devices and platforms without issues.\r\n\r\nQuality \r\n\r\nSome file formats can lead to a loss of quality. If you would like to display images in high resolution, choose formats that preserve high quality, such as PNG for images with transparency, and JPEGs for photographs.\r\n\t\r\nImages may need to be edited to ensure that they are the right resolution and size for the platform you are uploading them to. For example, mobile phones these days often take such high resolution images that they are too big for use on standard websites. This can lead to files taking a long time to load.\r\n\t\r\nIn addition, if you have concerns that others may take your high resolution images and use them without permission, you may like to consider other options that will make it difficult for your images to be reused by others such as watermarks.\r\n\r\nFile size\r\nKeep the file size of individual high-resolution images below a couple of megabytes (MB) to ensure that your content loads quickly and easily.\r\n\r\nAccessibility\r\n\r\nIt is good practice to ensure that your work will be accessible to people with a range of needs. Here are some tips:\r\n\r\n\tImages with alt text: ensure all images in your portfolio have descriptive alt text to aid users who rely on screen readers.\r\n\tReadable fonts: use clear, easily readable fonts and avoid cursive or overly small text to enhance readability for all users.\r\n\tAvoid colour-only indicators: do not use colour alone to convey meaning, as this can be challenging for those with colour vision deficiencies.\r\n\tColour contrast: maintain strong colour contrast between text and the background to improve visibility and readability.\r\n\tHeading hierarchy: use a logical heading hierarchy to help users navigate your content more easily and understand the structure of your portfolio.\r\n\r\nFor more information, visit the Accessibility in assignments page where you can learn about colour contrast, captions and transcripts, alternative text, and more.\r\n\r\n\r\nEthics, referencing and attribution\r\nJust as you would for any other assignment, if you use ideas, theories, artwork, images, or computer code  from another author, you need to provide a citation and a reference. Check your assignment instructions to see which style you need to use and use Easy Cite to guide you on formatting your citations and references.\r\n\r\nYou should also be aware of your responsibilities relating to the use of copyrighted materials. In Australia, under the fair dealing provision, students and researchers can rely on using a reasonable portion of a copyright work (e.g. 10% or 1 chapter) as part of their research or study. For more information, visit the Copyright page.\r\n\r\nAside from avoiding academic integrity and legal issues, referencing and attribution are just the right things to do. Crediting shows respect for other people's creative integrity and intellectual property. Additionally, make sure that their work isn't used to send messages the original artist would not agree with.\r\n\r\nPromoting your portfolio\r\nSharing your portfolio on social networks like LinkedIn and Instagram can be a great way to promote your work and showcase your skills. \r\n\r\nIf you wish to share your work with an audience beyond your instructor and fellow students in the course you are enrolled, the fair dealing provision for research and study does not allow posting someone else's copyright material to public sites such as Portfolium, Pebblepad or other open sharing or publishing platforms. For example, sharing copyright works as part of a public facing digital portfolio is not covered by fair dealing for research and study and will require the permission of the copyright holder. \r\n\r\nIs your creative piece 'inspired' by someone else's work?\r\nIdeas, styles, slogans, names and facts are not protected by copyright.  \r\n\r\nAn infringement of copyright is when you do something which only the author/owner has the right to do without their consent (i.e. copy, reproduce, adapt, transfer, etc.). \r\n\r\nFor it to be considered an infringement, the part reproduced needs to be a \"substantial part in essence\" of the original work. A \"substantial part\" is not defined in the Copyright Act and is measured on the quality rather than the quantity copied. \r\n\r\nIf you can recognise the source of the adapted work, the creator probably will too, and they may assert their rights if they have been infringed. \r\n\r\nThis can be a tricky topic to understand, so be aware of your responsibilities relating to the use of copyrighted materials.  \r\n\r\nPrivacy of your work\r\nIt is important to understand the Terms of Use of any portfolio sharing platforms you sign up to use. For many portfolio sharing sites, you retain copyright of any original works you create and post to the site, but you need to be aware of what licence you are agreeing to, that is, what rights you are giving the platform in relation to your work.\r\n\r\nYou also need to be aware of what activities the platforms allow or prohibit. Most portfolio sharing platforms will prohibit you uploading others' copyright protected material. Make sure you read the platform privacy information carefully. \r\n\r\nIf you have concerns about access or use of your work, consult your educators to understand how they want to share your work; some privacy settings might restrict their access. \r\n\r\n","n":0.033},[{"v":"Digital skills","i":2,"n":0.707},{"v":"Assignment planning","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":76,"$":[{"v":"Significant figures","n":0.707},{"v":"In our everyday world, we measure things with precision. This is where significant figures come in. They help us communicate how precise our measurements are. Use this resource to learn how to identify and apply significant figures.\r\n\r\nA significant figure (or \"sig fig\") is a digit in a measurement that gives us an idea of how precise the measurement is. It tells us which digits are reliable and meaningful and give us an indication of how accurate the instrument or method used is.\r\nAll the digits in a measured value are called the significant figures. The last digit is often an estimate.\r\nIf a number has more significant figures, this suggests that a quantity has been measured more precisely, If it has fewer significant figures, this indicates less certainty in the measurement.\r\n\r\nRules for significant figures\r\nThere are \\(5\\) rules to help you identify the number of significant values in a measured value.\r\n\r\n\r\n\tAny non-zero digit is significant. The position of a decimal point makes no difference.\r\n\tFor example, \\(15.7\\) and \\(1.57\\) both have \\(3\\) significant figures.\r\n\tZeroes between numbers are significant.\r\n\tFor example, \\(1.05\\) has \\(3\\) significant figures, and \\(200.708\\) has \\(6\\).\r\n\tZeroes at the right-hand end of whole numbers are not significant, unless otherwise stated.\r\n\tFor example, \\(2860\\) has \\(3\\) significant figures, while \\(70\\) has only \\(1\\).\r\n\tZeroes at the left-hand end of decimal numbers are not signicant.\r\n\tFor example, \\(0.28\\) has \\(2\\) significant figures, and so does \\(0.0039\\).\r\n\tZeroes at the right-hand end of decimal numbers are significant.\r\n\tFor example, \\(12.0\\) has \\(3\\) significant figures and \\(0.48300\\) has \\(5\\).\r\n\r\n\r\n\r\nExample 1 \u2013 identifying the number of significant figures\r\nFind the number of significant figures in \\(0.000\\,3\\).\r\nAccording to rule \\(4\\), zeroes at the left-hand end of decimal numbers are not signicant, so \\(0.000\\,3\\) has only \\(1\\) significant figure.\r\n\\[0.000\\,\\underline{3}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 identifying the number of significant figures\r\nState the number of significant figures in each of the following values.\r\n\r\n\t\\(345\\)\r\n\t\\(17\\,642\\)\r\n\t\\(0.003\\,3\\)\r\n\t\\(0.000\\,306\\)\r\n\t\\(870\\)\r\n\t\\(20\\,000\\)\r\n\t\\(140.600\\)\r\n\t\\(710.0\\)\r\n\t\\(0.040\\,80\\)\r\n\t\\(0.005\\,0\\)\r\n\r\n\r\n\r\n\r\nSignificant figures and scientific notation\r\nUsing scientific notation is a helpful way to manage significant figures, especially with very large or small numbers. It clearly expresses the precision of a measurement by keeping only the significant digits. Scientific notation is also handy when you need to round a value to a specific number of significant figures.\r\nConsider the number \\(34\\,218\\,043\\), which has \\(8\\) significant figures. We want to round it to \\(3\\) significant figures. \r\nOne way we could do this is by writing it as:\r\n\\[34\\,200\\,000\\]\r\nbut this can lead to ambiguity about the precision of the number; someone else reading the number may not know if the zeroes that come after the \\(2\\) are significant.\r\nInstead, we can write it in scientific notation:\r\n\\[3.42\\times10^{7}\\]\r\nwhich clearly shows \\(3\\) significant figures.\r\n\r\nSignificant figures for measurements\r\nWhen recording measurements, the number of significant figures depends on the scale of the instrument.\r\nConsider this ruler.\r\n\r\nLet's say a measurement is taken between \\(6.4\\) and \\(6.5\\) centimetres.\r\n\r\n\tWe can say with certainty that the measurement is greater than \\(6\\) and less than \\(7\\) centimetres.\r\n\tWe can say with certainty that the measurement is greater than \\(6.4\\) and less than \\(6.5\\) centimetres.\r\n\tWe CANNOT say with certainty that the measurement is greater than \\(6.43\\) and less than \\(6.45\\) centimetres.\r\n\r\nThe result would be recorded as \\(6.44\\) centimetres, to \\(3\\) significant figures. The first two digits are certain and the last is a good estimate.\r\nWriting \\(6.44\\) implies \\(6.44\\pm0.005\\), i.e. between \\(6.435\\) and \\(6.445\\), unless otherwise stated.\r\nIt is important to report your measurements to the appropriate number of significant figures. This will help you to avoid:\r\n\r\n\toverstating or understating the precision of the result\r\n\tmaking calculating errors\r\n\tforming misleading conclusions about your results.\r\n\r\nReporting your values to the correct number of significant figures helps to maintain the integrity of the data you generate.\r\n\r\n","n":0.041},[{"v":"Measurement","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":77,"$":[{"v":"Notation","n":1},{"v":"\r\nImagine trying to write out the mass of the Earth in kilograms; that would be \\(5\\,972\\,000\\,000\\,000\\,000\\,000\\,000\\,000\\textrm{ kg}\\). Expressing such big numbers this way is hard and can lead to mistakes. In science and engineering, notation plays a key role in simplifying calculations and conveying information efficiently. Use this resource to learn how to write numbers in scientific and engineering notation.\r\n\r\nScientific notation\r\nScientific notation is a way of writing numbers in a compact form. It uses a number from \\(1\\) to less than \\(10\\), multiplied by a power of \\(10\\).\r\nFor example, the mass of the Earth would be concisely written as \\(5.972\\times10^{24}\\). The charge on one electron, \\(0.000\\,000\\,000\\,000\\,000\\,000\\,160\\,2\\textrm{ C}\\) is \\(1.602\\times10^{-19}\\textrm{ C}\\).\r\nBy using scientific notation:\r\n\r\n\tthe size of the number is more easily seen\r\n\terrors are less likely to occur\r\n\tcalculations can be simplified using index laws.\r\n\r\n\r\nExpressing numbers in scientific notation\r\nTo write a number in scientific notation:\r\n\r\n\r\n\tMove the decimal point so that there is one non-zero digit before the decimal point.\r\n\tMultiply by a power of \\(10\\) equal to the number of places that the decimal point has been moved. If:\r\n\t\r\n\t\tthe decimal point is moved to the left, the power of \\(10\\) is positive\r\n\t\tthe decimal point is moved to the right, the power of \\(10\\) is negative.\r\n\t\r\n\r\n\r\nIf the number is less than \\(1\\), the power of \\(10\\) will be negative. If it is more than \\(1\\), the power of \\(10\\) will be positive.\r\n\r\nExample 1 \u2013 expressing numbers in scientific notation\r\nWrite \\(5630\\) in scientific notation.\r\nLocate the decimal point in the number. For whole numbers, there is an imaginary decimal point at the end.\r\n\\[5630.\\]\r\nWe need to move the decimal point so that we have one non-zero digital before the decimal point. In this case, it needs to be between the \\(5\\) and \\(6\\).\r\n\\[5\\underleftarrow{630} = 5.630\\]\r\nWe have moved the decimal point \\(3\\) places to the left, so the power of \\(10\\) is \\(+3\\).\r\n\\[5630=5.6\\times10^{3}\\]\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 expressing numbers in scientific notation\r\n\r\n\tWrite the following numbers in scientific notation.\r\n\t\r\n\t\t\\(58000\\)\r\n\t\t\\(0.0026\\)\r\n\t\t\\(70.6\\)\r\n\t\t\\(0.3\\)\r\n\t\t\\(2\\,400\\,000\\)\r\n\t\t\\(0.000\\,000\\,684\\)\r\n\t\t\\(0.0704\\)\r\n\t\t\\(0.260\\)\r\n\t\t\\(17\\,600\\)\r\n\t\t\\(0.04080\\)\r\n\t\t\\(0.000\\,500\\)\r\n\t\t\\(357\\,000\\,000\\,000\\)\r\n\t\r\n\tWrite the following quantities in scientific notation.\r\n\t\r\n\t\t\\(6\\,370\\,000\\textrm{ m}\\) \u2013 the mean radius of the Earth\r\n\t\t\\(86\\,400\\textrm{ s}\\) \u2013 the time for the Earth to orbit the Sun\r\n\t\t\\(0.0018\\textrm{ s}\\) \u2013 the half-time of a radioactive isotope\r\n\t\t\\(380\\,000\\,000\\textrm{ m}\\) \u2013 the average distance between the Moon and the Earth\r\n\t\t\\(0.000\\,000\\,000\\,001\\,1\\textrm{ s}\\) \u2013 the time for a ray of light to pass through a window pane\r\n\t\t\\(637\\,000\\,000\\,000\\,000\\,000\\,000\\,000\\textrm{ kg}\\) \u2013 the mass of the planet Mars.\r\n\t\r\n\tWrite the following numers in decimal form or in whole numbers.\r\n\t\r\n\t\t\\(7\\times10^{2}\\)\r\n\t\t\\(4\\times10^{0}\\)\r\n\t\t\\(3.46\\times10^{3}\\)\r\n\t\t\\(5.96\\times10^{-5}\\)\r\n\t\t\\(9\\times10^{7}\\)\r\n\t\t\\(3.98\\times10^{1}\\)\r\n\t\t\\(2.78\\times10^{5}\\)\r\n\t\t\\(6.78\\times10^{-1}\\)\t\t\r\n\t\r\n\r\n\r\n\r\n\r\nEngineering notation\r\nIn engineering, a similar notation is used to handle practical applications. This is called engineering notation and it uses powers of \\(10\\) in multiples of \\(3\\), i.e. \\(10^{3}\\), \\(10^{6}\\), \\(10^{12}\\), \\(10^{-6}\\), etc.\r\nAs an example, the number \\(74\\,000\\) is \\(7.4\\times10^{4}\\) in scientific notation, but in engineering notation, we write it as \\(74\\times10^{3}\\).\r\n\r\nExpressing numbers in engineering notation\r\nTo write a number in engineering notation:\r\n\r\n\r\n\tMove the decimal point in groups of \\(3\\) to give a number between \\(1\\) and \\(1\\,000\\).\r\n\tMultiply by a power of \\(10\\) equal to the number of places that the decimal point has been moved. This will be a multiple of \\(10\\). If:\r\n\t\r\n\t\tthe decimal point is moved to the left, the power of \\(10\\) is positive\r\n\t\tthe decimal point is moved to the right, the power of \\(10\\) is negative.\r\n\t\r\n\r\n\r\n\r\nExample 1 \u2013 expressing numbers in engineering notation\r\nWrite \\(1\\,636\\,000\\,000\\) in engineering notation.\r\nWe need to move the decimal point in groups of \\(3\\).\r\n\\[ 1\\,\\underleftarrow{636}\\,\\underleftarrow{000}\\,\\underleftarrow{000} = 1.636\\]\r\nWe have \\(3\\) groups of \\(3\\) places, so \\(9\\) places in total, to the left. The power is \\(+9\\).\r\n\\[1\\,636\\,000\\,000 = 1.636\\times10^{9}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 expressing numbers in engineering notation\r\nWrite the following numbers in engineering notation.\r\n\r\n\t\\(53\\,800\\)\r\n\t\\(145\\,000\\,000\\)\r\n\t\\(0.761\\)\r\n\t\\(0.000\\,534\\)\r\n\t\\(0.002\\,8\\)\r\n\t\\(9\\,620\\)\r\n\t\\(0.348\\,20\\)\r\n\t\\(0.6\\)\r\n\t\\(5.6\\times10^{-2}\\)\r\n\t\\(54.8\\times10^{8}\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\nUsing prefixes and SI units\r\nEngineering notation is more compatible with metrix prefixes like kilo (\\(10^{3}\\)), mega (\\(10^{6}\\)) and milli (\\(10^{-3}\\)), which are common in engineering contexts. The preferred prefixes are shown in the table. They all have a multiple of \\(3\\).\r\n\r\nThis helps engineers easily relate numerical values to practical units, making calculations and communication clearer.\r\n\r\nExample 1 \u2013 using prefixes and SI units\r\nWrite the following quantities in SI units, using a preferred prefix.\r\n\r\n\t\\(6\\,000\\textrm{ m}\\)\r\n\t\\(0.005\\textrm{ V}\\)\r\n\t\\(0.000\\,3\\textrm{ s}\\).\r\n\r\nFirst, write each quantity in engineering notation.\r\n\\[6,\\underleftarrow{000}\\textrm{ m} = 6\\times10^{3}\\textrm{ m}\\] \\[0\\underrightarrow{.005}\\textrm{ V} = 5\\times10^{-3}\\textrm{ V}\\] \\[0\\underrightarrow{.000}\\,\\underrightarrow{300}\\textrm{ s} = 300\\times10^{-6}\\textrm{ s}\\]\r\nMatch the power to the correct prefix and use it in place of the power.\r\n\\[6\\times10^{3}\\textrm{ m} = 6\\textrm{ km}\\] \\[5\\times10^{-3}\\textrm{ V}=5\\textrm{ mV}\\] \\[300\\times10^{-6}\\textrm{ s}=300\\textrm{ \u00b5s}\\]\r\n\r\n\r\n\r\nExercise \u2013 using prefixes and SI units\r\n\r\n\tWrite the following in scientific notation, engineering notation, and using the correct prefix and SI unit.\r\n\t\r\n\t\t\\(450\\textrm{ m}\\)\r\n\t\t\\(63\\,200\\textrm{ W}\\)\r\n\t\t\\(0.000\\,007\\textrm{ F}\\)\r\n\t\t\\(37\\,808\\,000\\textrm{ m}\\)\r\n\t\t\\(0.000\\,000\\,083\\textrm{ m}\\)\r\n\t\t\\(0.800\\textrm{ s}\\).\r\n\t\r\n\tWrite the following quantities in engineering notation, using the correct preferred prefix.\r\n\t\r\n\t\t\\(0.002\\textrm{ s}\\)\r\n\t\t\\(0.075\\textrm{ km}\\)\r\n\t\t\\(0.05\\textrm{ mA}\\)\r\n\t\t\\(0.003\\,3\\textrm{ \u00b5F}\\)\r\n\t\t\\(4\\,500\\textrm{ mm}\\)\r\n\t\t\\(0.25\\textrm{ kV}\\).\r\n\t\r\n\r\n\r\n","n":0.037},[{"v":"Measurement","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":78,"$":[{"v":"Dimensions","n":1},{"v":"Our everyday world is filled with things we can measure; each is defined by dimensions in maths and physics. By understanding dimensions, we can accurately describe and compare different physical quantities. Use this resource to learn about dimensions. \r\n\r\nDimensions describe the fundamental units that make up physical properties. When we write the dimensions of the physical quantities mass, length, time and current, we use the symbols \\(\\left[ M \\right]\\), \\(\\left[ L \\right]\\), \\(\\left[ T \\right]\\) and \\(\\left[ I \\right]\\), respectively. \r\nAs an example, speed combines two of these dimensions. It is expressed as the distance (length) traveled per unit of time. We can write this out to form a dimensional formula:\r\n\\[\\left[\\textrm{speed}\\right] = \\frac{\\left[\\textrm{length}\\right]}{\\left[\\textrm{time}\\right]} = \\frac{\\left[ L \\right]}{\\left[ T \\right]} = \\left[ L \\right] \\left[ T \\right]^{-1}\\]\r\nSpeed, therefore, has dimensions of length divided by time.\r\n\r\nUsing dimensions to determine the units\r\nThe unit of a derived quantity can be found by substituting the fundamental unit for each dimension into the dimensional formula.\r\nLet's consider a calculation for density using mass in kilograms and volume in cubic metres. Metres is a measure of length. The dimension equation for density is:\r\n\\[\\left[ \\textrm{density} \\right] = \\frac{ \\left[ \\textrm{mass} \\right] }{ \\left[ \\textrm{volume} \\right] } = \r\n\\frac{ \\left[ M \\right] }{ \\left[ L \\right]^{3} } = \\left[ M \\right] \\left[ L \\right]^{-3}\\]\r\nWe can then subsitute the fundamental units \\(\\textrm{kg}\\) and \\(\\textrm{m}^{3}\\) into \\(\\left[ M \\right]\\) and \\(\\left[ V \\right]\\), respectively.\r\n\\[\\textrm{density} = \\frac{\\textrm{kg}}{\\textrm{m}^{3}} = \\textrm{kg}\\,\\textrm{m}^{-3} \\]\r\n\r\nExample \u2013 using dimensions to determine the units\r\nDr. Pepper wants her student to measure the rate at which carbon dioxide escapes from a canned soda drink. The student sets up an experiment where they record the time in minutes and the volume of carbon dioxide gas in millilitres. \r\nDetermine the units that the student's final result will be in.\r\nFirst, we should determine the dimensional formula. For the rate of change of carbon dioxide, a gas, it will be:\r\n\\[\\textrm{rate of change} = \\frac{\\left[ \\textrm{volume} \\right]}{\\left[ \\textrm{time} \\right]} = \\frac{\\left[ V \\right]}{\\left[ T \\right]} = \\left[ V \\right] \\left[ T \\right]^{-1} \\]\r\nVolume is measured in millilitres and time is measured in minutes. Therefore:\r\n\\[\\textrm{rate of change} = \\frac{\\textrm{mL}}{\\textrm{min}} = \\textrm{mL}\\,\\textrm{min}^{-1} \\]\r\n\r\nChecking dimensions\r\nIn your studies, you will derive physical quantities from formulas or equations. You can use dimensions to check whether these expressions make sense. This involves making sure that the dimensions of the left-hand side are balanced with the dimensions on the right-hand side.\r\nThis will make more sense with an example.\r\n\r\nExample \u2013 checking dimensions\r\nA student claims that the equation for the circumference of a circle is \\(C=2\\pi r\\). Determine whether this is dimensionally correct.\r\nThe circumference is a measure of length, so the dimension is \\(\\left[ L \\right]\\). This is the left-hand side of the equation.\r\nOn the right-hand side, we have \\(2\\pi\\), which does not have units, and \\(r\\), which is a measure of length. \\(r\\) therefore has a dimension of \\(\\left[ L \\right]\\).\r\nPutting these together, we have:\r\n\\[\\begin{align*} C & = 2\\pi r\\\\\r\n\\left[ L \\right] & = \\left[ L \\right] \r\n\\end{align*}\\]\r\nThe left-hand side and right-hand side are dimensionally correct.\r\n\r\nExercise \u2013 checking dimensions\r\nCheck that the following equations are dimensionally correct.\r\n\r\n\t\\(V=\\pi r^{2} h\\)\r\n\t\\(A=4\\pi r^{2}\\)\r\n\t\\(A=2\\pi r^{2} + 2\\pi rh\\)\r\n\r\n\r\n\r\n\r\n","n":0.044},[{"v":"Measurement","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":79,"$":[{"v":"Revising, organising and accessibility","n":0.5},{"v":"You've finished your assignment and now it's time to put on the finishing touches on. In this section, you'll find tips for editing your work, using tables and figures, and for including appendices in your assignments. You'll also learn about another important consideration: accessibility.\r\n\r\nRevising your work\r\nAfter you finish writing your assignment, it's time to submit, right? Wrong!  Here are some steps you can follow to make sure you submit quality work.\r\n\r\n\r\n\r\nWait a bit of time to review, if possible\r\nWhen you are working on an assignment for a long time, it can be difficult to see the problems or errors that it may contain. By waiting a day or two before you review your work, you will be able to look at it with fresh eyes. This often leads to catching more mistakes.\r\n\r\nHave you met your objectives?\r\nIt's important to check that what you have written or produced is relevant to your assignment topic. An effective way to do this is to:\r\n\r\n\tRevisit the assignment prompt to ensure your content addresses all the required topics.\r\n\tVerify that the points you outlined in your introduction are thoroughly covered in your body paragraphs and align with the prompt.\r\n\r\nHave met assignment criteria?\r\nTo check this, it can be useful to turn the assignment's instructions and rubric into a checklist. This can be done by summarising the main themes from the assignment instructions and rubric into brief bullet points.\r\n\r\nFor example: (give example assignment instructions and how they can be put into a checklist)\r\n\r\nCheck the flow\r\nTo ensure your assignment flows well:\r\n\r\n\tReread your introduction and list the points you said would be to covered in your assignment. Then, check the first sentence of each body paragraph to ensure these topics have been addressed.\r\n\tExamine your body paragraphs to ensure they are logically grouped by theme. Avoid jumping back and forth between topics.\r\n\tConfirm that each paragraph focuses on a single theme.\r\n\r\nAre your ideas expressed clearly?\r\nHave you avoided using long, overly complicated sentences? Have you avoided jargon? Have you provided necessary explanation?\r\n\r\nProofread\r\nWhen you proofread, you check your writing for mistakes such as grammatical, punctuation, and spelling errors. If your writing contains too many of these errors, it could make your writing difficult to understand and appear less professional.\r\n\r\nSome effective techniques for proofreading are:\r\n\r\nRead your work aloud. This can help identify places where your writing does not flow well.\r\nUnderstand the errors that you often make. For example, are there any words that you frequently misspell? Are there any grammatical errors you commonly make?\r\nRead slowly and carefully.\r\nTake breaks to ensure your concentration is sharp.\r\nSome people like to read their work backwards. This helps them focus on spelling errors, as they only concentrate on spelling rather than the meaning of the text.\r\n\r\n\r\nRead this paragraph and see how many grammatical and spelling mistakes you can find. Use some of the techniques you have just learnt.\r\n\r\n\r\n\r\nCheck referencing and citation\r\nReferencing and citation are often included in the marking criteria for assignments. That's why it is important to make sure that you have included both a citation and a reference for  all ideas, theories, studies, computer code, information and images that have come from other sources. \r\n\r\nEnsure you use the referencing style specified in your assignment instructions and that citations and references are correctly formatted. This includes using the correct order of information and proper punctuation.\r\nStill not sure if you're on the right track?\r\nIf you still have any doubts about your assignment, contact RMIT's Study Support. This team has members who specialise in academic writing, maths, science, referencing and finding reliable sources of information.\r\n","n":0.041},[{"v":"Researching","i":2,"n":1},{"v":"Assessment","i":1,"n":1},{"v":"Assignments","i":0,"n":1}]]},{"i":80,"$":[{"v":"The cell cycle","n":0.577},{"v":"The cell cycle is a series of events that a cell goes through as it grows and divides. You need to understand the cell cycle to grasp how organisms grow, repair tissues, and stay healthy. Use this resource to learn about the cell cycle.\r\n\r\nThe cell cycle consists of several phases: interphase, mitosis and cytokinesis. We can compare these phases to preparing for and executing a theatrical play.\r\n\r\nInterphase\r\nInterphase is like the rehearsal period before the big night. During this time, all the necessary planning, growth and organisation occur to make sure everything is ready for the show. It is when the cell grows and its genetic material (DNA) is replicated.\r\nInterphase is the longest phase in the cell cycle. It consists of three stages:\r\n\r\n\tG1 \u2013 The cell grows and carries out regular metabolic functions.\r\n\tS \u2013 DNA replication occurs to make sure that each new cell has an identical set of genetic material. The centrosome also duplicates.\r\n\tG2 \u2013 The cell continues to grow and produce what it needs to mitosis.\r\n\r\nInterphase is crucial for make sure the cell has all the parts it needs to divide successfully and accurately.\r\nThe stages of interphase as shown in the figure.\r\n\r\n\r\n\r\nMitosis\r\nMitosis is like the performance night. All the preparations made during interphase come together\u2014the actors take the stage, deliver their lines and the story unfolds in front of an audience. Just as the play transitions through various scenes and acts, mitosis involves a series of stages where the cell divides its replicated DNA and splits into two identical daughter cells.\r\nMitosis consists of several stages:\r\n\r\n\tprophase \u2013 The chromosomes condense and the nuclear envelope begins to dissolve. \"Pro\" means \"before\", so you can think of it as the stage before the chromosomes start to move.\r\n\tmetaphase \u2013 The chromosomes align along the equator of the cell. \"Meta\" means \"in between\", so it is the middle stage. You might also like to think of it as when the chromosomes are lined up between the centrosomes.\r\n\tanaphase \u2013 The sister chromatids are pulled apart by spindle fibres to opposite sides (or poles) of the cell. \"Ana\" means \"backwards\", referring to the pulling of chromatids back to the poles.\r\n\ttelophase \u2013 The nuclear envelope reforms around the separated chromosomes and the chromosomes recondense. \"Telo\" means \"end\", so it is the phase at the end of cell division.\r\n\r\nMitosis makes sure that each daughter cell has an exact copy of the parent cell's DNA, maintaining genetic consistency across cell generations.\r\nThe stages of mitosis are shown in the figure.\r\n\r\n\r\nUse this mnemonic to remember the order of the phases in mitosis: Post Memes And Tiktoks.\r\n\r\nCytokinesis\r\nIn the theatrical play analogy, cytokinesis would be like the stage crew closing the curtains and preparing for the next show. After the actors have finished their performance (mitosis), the stage is carefully reset and divided, ensuring everything is neatly organised. This process ensures that the two separate \"productions\" (or cells) are ready for future events.\r\nCytokinesis is the final step of the cell cycle, where the cell's cytoplasm divides to form two distinct daughter cells.\r\n\r\n\tIn animal cells, the cell membrane pinches inwards, creating a groove that eventually splits the cell in two.\r\n\tIn plant cells, a new structure called a cell plate forms in the middle, which becomes the new cell wall separating the two daughter cells.\r\n\r\nCytokinesis ensures that each daughter cell receives not only an equal and complete set of genetic material but also the necessary organelles and resources to function independently.\r\n\r\n\r\nSummary\r\nThe cell cycle is summarised in the figure.\r\n\r\n\r\nExercise\r\nSee how well you understand the cell cycle with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nTelomeres are protective caps on the ends of chromosomes that protect the DNA from damage. They also play important roles in DNA replication. \r\nResearchers at the Institute of Dee and Aye published a study looking at telomere length and human aging. Telomere length in white blood cells was plotted against age, in years, and the p- and r-values are shown.\r\n\r\n\r\n\r\n\r\n \r\nThe four stages of mitosis\r\n\r\n\tProphase (top left): The nuclear envelope dissolves and chromatin condenses into chromosomes.\r\n\tMetaphase (top right): The chromosomes align along the equator of the cell and spindle fibres extend from the centrosomes.\r\n\tAnaphase (bottom right): Spindle fibres pull sister chromatids to opposite ends of the cell.\r\n\tTelophase (bottom left): The chromosomes recondense into chromatin and the nuclear envelopes reform.\r\n\r\n \r\nCytokinesis\r\n\r\n\tTop: The membrane pinches inwards to form a grove in the centre.\r\n\tBottom: The cells have split into two identical daughter cells.\r\n\r\n \r\nThe cell cycle\r\n\r\n\tInterphase:\r\n\t\t\r\n\t\t\tG1\r\n\t\t\tS\r\n\t\t\tG2\r\n\t\t\r\n\t\r\n\tMitosis:\r\n\t\t\r\n\t\t\tProphase\r\n\t\t\tMetaphase\r\n\t\t\tAnaphase\r\n\t\t\tTelophase\r\n\t\t\r\n\t\r\n\tCytokinesis\r\n\r\n \r\nScatter plot showing telomere length with age\r\nScatter plot showing the relationship between telomere length and age. The x-axis is labelled \"Age (years)\", starting at 0 years and ending at 80 years, with increments of 10 years. The y-axis is labelled \"Telomere length (arbitrary units)\", beginning at 0 and ending at 1.4, with increments of 0.1. \r\nThere are 60 to 70 data points plotted on the graph. All of these points fall between ages 20 and 80. The line of best fit is shown, starting at about 1.14 arbitrary units at age 0, and gradually declines until it reaches about 1.01 arbitrary units at age 80.\r\nThe results of statistical analysis are shown, with a p-value less than 0.0001 and an r-value of negative 0.4047.\r\nThe data based on the line of best fit is summarised in the following table.\r\n\r\n  \r\n    \r\n      Age (years)\r\n      Telomere length (arbitrary units)\r\n    \r\n  \r\n  \r\n    \r\n      \\(10\\)\r\n      \\(1.124\\)\r\n    \r\n    \r\n      \\(20\\)\r\n      \\(1.108\\)\r\n    \r\n    \r\n      \\(30\\)\r\n      \\(1.091\\)\r\n    \r\n    \r\n      \\(40\\)\r\n      \\(1.075\\)\r\n    \r\n    \r\n      \\(50\\)\r\n      \\(1.059\\)\r\n    \r\n    \r\n      \\(60\\)\r\n      \\(1.043\\)\r\n    \r\n    \r\n      \\(70\\)\r\n      \\(1.026\\)\r\n    \r\n    \r\n      \\(80\\)\r\n      \\(1.010\\)\r\n    \r\n  \r\n\r\n \r\nThe three stages of interphase\r\n\r\n\tG1 (top left): Normal cell shown with condensed chromatin, nucleus and centrosome labelled\r\n\tS (right): DNA and centrosome duplicate\r\n\tG2 (bottom): Cell continues to grow and produce what it needs for mitosis\r\n\r\n","n":0.032},[{"v":"Cell cycle","i":1,"n":0.707},{"v":"Cells","i":0,"n":1}]]},{"i":81,"$":[{"v":"Surface area to volume ratio","n":0.447},{"v":"Understanding the surface area to volume ratio helps us understand how cells efficiently manage transport and exchange with the environment. This affects how cells take in nutrients and get rid of waste. Use this resource to learn how this ratio impacts cell size and function in biology.\r\nThe surface area to volume ratio is a measure that compares the size of a cell's surface to its volume. Surface area is the total area of the cell's outer membrane, and volume is the amount of space inside the cell. Sometimes, you may see surface area to volume ratio being written as SA:V.\r\nThe difference in the surface area and volume of a cube is shown.\r\n\r\n\r\nEffect of surface area to volume ratio on cell function\r\nA high surface area to volume ratio is important for cell function. The following table highlights specific ways in which a high surface area to volume ratio supports cells.\r\n\r\n\t\r\n\t\t\r\n\t\t\tHigh SA:V ...\r\n\t\t\tby ...\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tenhances nutrient uptake\r\n\t\t\tallowing for more efficient absorption of nutrients from the environment.\r\n\t\t\r\n\t\t\r\n\t\t\timproves waste removal\r\n\t\t\tenabling quicker and more effective transport of waste products out of the cell.\r\n\t\t\r\n\t\t\r\n\t\t\tincreases gas exchange\r\n\t\t\toptimising the exchange of gases like oxygen and carbon dioxide, which are needed for cellular respiration and survival.\r\n\t\t\r\n\t\t\r\n\t\t\tsupports faster communication\r\n\t\t\tallowing signals and materials to be distributed more quickly within the cell and between cel.\r\n\t\t\r\n\t\t\r\n\t\t\tallows the cell to adapt more quickly to change\r\n\t\t\tenabling cells to react more quickly to changes by speeding up internal and external exchanges.\r\n\t\t\r\n\t\r\n\r\nA high surface area to volume ratio makes the exchange of nutrients and waste more efficient between the cell and its environment; it maximises the rate of diffusion.\r\n\r\nEffect of surface area to volume ratio on cell size \r\nWhy are cells so small? The surface area to volume ratio is a key factor in determining the size of a cell.\r\nAs a cell grows, its volume increases faster than its surface area. This reduces the ratio and makes it harder for the cell to efficiently take up nutrients and expel waste. To function well, most cells stay small or have adaptations like microvilli (small projections) to increase surface area. Think of them like the bristles of a brush. Just as the many bristles increase the surface area of the brush for cleaning, microvilli increase the cell's surface area to enhance absorption and exchange of materials.\r\n\r\n\r\nIntestinal epithelial cells are examples of cells that have microvilli to increase their surface area. These cells line the intestine and are involved in absorption of nutrients. The presence of microvilli makes the absorption process more efficient.\r\nThis ratio helps explain why cells are usually so small.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nLeo is having an argument with his classmate, Lydia. Leo claims that that larger cells have a higher surface area to volume ratio, but Lydia disagrees. \r\nTheir teacher Mr Poprocks suggests that they collect some data on the size of different immune cells and do some calculations.\r\nLeo and Lydia compile what they've found in a table.\r\n\r\n \t\r\n\t\t\r\n\t\t\tImmune cell type\r\n\t\t\tDiameter (\u00b5m)\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tNeutrophils\r\n\t\t\t\\(14\\)\r\n\t\t\r\n\t\t\r\n\t\t\tEosinophils\r\n\t\t\t\\(17\\)\r\n\t\t\r\n\t\t\r\n\t\t\tBasophils\r\n\t\t\t\\(16\\)\r\n\t\t\r\n\t\t\r\n\t\t\tLymphocytes\r\n\t\t\t\\(9\\)\r\n\t\t\r\n\t\t\r\n\t\t\tMacrophages\r\n\t\t\t\\(21\\)\r\n\t\t\r\n\t\r\n\r\n\r\n\r\nTo calculate the surface area (SA) and volume (V) of a cell, we assume that all cells are spheres and use the formulas:\r\n\\[\\textrm{SA}=4\\pi r^{2}\\quad\\textrm{and}\\quad\\textrm{V}=\\frac{4}{3}\\pi r^{3}\\]\r\n\r\n \r\nSurface area of a cube\r\nA three-dimensional cube and a two-dimensional net of a cube. The sides of the cube are 2 centimetres long. The volume of the cube is 2 times 2 times 2 equals 8 centimetres cubed. The net of the cube also has sides of 2 centimetres. The surface area of the cube, based on its net, is 2 times 2 centimetres times 6 faces, which is equal to 24 centimetres squared.\r\n","n":0.041},[{"v":"Cells","i":0,"n":1}]]},{"i":82,"$":[{"v":"Organelles and their functions","n":0.5},{"v":"Organelles are specialised structures in eukaryotic cells. Understanding the different types of organelles and their roles helps us comprehend how cells operate and maintain life processes. Use this resource to learn more.\r\n\r\nOrganelles perform distinct functions needed for cell survival and efficiency. They are held within the cytoplasm (\"cyto\" meaning \"cell\" and \"plasm\" meaning \"something molded or created\"), which is a gel-like filling the inside of the cell. The cytoplasm helps move materials within the cell and maintain the cell's shape.\r\nThe cytoplasm also contains the cytoskeleton, a network of protein fibres that helps organise the contents of a cell.\r\nThere are many types of organelles, with some that are unique to each type of cell. Let's explore the organelles of animal and plant cells.\r\n\r\n\r\n\r\nNucleus\r\nThe nucleus, present in both animal and plant cells, is the control centre of the cell. It manages gene expression, cell growth and replication, and houses the cell's genetic material in the form of deoxyribonucleic acid, or DNA. DNA is organised into chromatin, a complex made of DNA and proteins.\r\nWithin the nucleus is the nucleolus, a dense area where ribosomal ribonucleic acid (rRNA) is found. The nucleolus provides the blueprints for making and assembling ribosomes.\r\nA double membrane called the nuclear envelope surrounds the nucleus. It has pores (holes) that regulate the movement of molecules in and out of the nucleus.\r\nJust like the cell has cytoplasm, the nucleus contains a similar gel-like substance called nucleoplasm.\r\n\r\n  \r\n\r\n\r\n\r\n\r\nLocated near the nucleus, and only in animal cells, is a structure called a centrosome. Centrosomes are important for cell division. They duplicate before a cell starts to divide and move to opposite ends of the cell. This will be explained more in The cell cycle.\r\n\r\nMitochondria\r\nThe mitochondria are known as the powerhouse of the cell \u2013 this might sound familiar!  They produce energy through cellular respiration, which involves converting glucose and oxygen into adenosine triphosphate (ATP), the energy the cell needs to function.\r\nMitochondria have a double membrane, with the inner membrane folded into structures called cristae. The high level of folding provides more surface area for respiration, allowing the mitochondria to generate more ATP. The space inside the cristae is called the matrix.\r\nMitochondria are special because they contain their own DNA. This allows them to replicate independently within the cell and adapt to increased energy needs, such as when a muscle cell is working harder during exercise.\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\nRibosomes\r\nRibosomes are the cell's protein factories. These tiny structures read the genetic instructions from messenger RNA (mRNA) and assemble proteins from amino acids. Proteins are crucial for many cell functions, including growth and repair.\r\nRibosomes are the only organelles that are not membrane-bound. They can be found floating freely in the cytoplasm or attached to the endoplasmic reticulum.\r\nYou will delve into the structure of ribosomes and learn more about protein synthesis later.\r\n\r\nEndoplasmic reticulum\r\nThe endoplasmic reticulum (\"endoplasmic\" meaning \"within the cell\" and \"reticulum\" meaning \"small net\") is a network of membranes that play a key role in the synthesis of proteins and lipids. It is often shortened to \"ER\".\r\nIt also helps transport materials within the cell and ensures that cells function smoothly.\r\nThere are two forms of ER:\r\n\r\n\tRough ER has ribosomes attached and helps fold and modify new proteins. It prepares them for transport within or out of the cell via the Golgi apparatus. Because it processes proteins, rough ER is typically located near the nucleus.\r\n\t\r\n\tSmooth ER lacks ribosomes and is involved in lipid synthesis and detoxification. It detoxifies by helping drugs and toxins dissolve better in water so they can be cleared from the body. Often, smooth ER is located near the cell membrane.\r\n\t\r\n\r\n\r\n\r\nGolgi apparatus\r\nLike the endoplasmic reticulum, the Golgi apparaus modifies, sorts and packages proteins and lipids for transport. It consists of flattened, stacked membranes called cisternae.\r\nOnce proteins and lipids have been processed within the lumen (channel) of a cisterna, they pinch off into small vesicles which are transported to their destinations inside or outside the cell.\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\nIn plant cells, the Golgi apparatus helps make components of the cell wall.\r\nYou may see the Golgi apparatus also being referred to as the Golgi body or Golgi complex; they all mean the same thing.\r\n\r\nVacuoles\r\nVacuoles are storage organelles found in both plant and animal cells, but they serve different roles and vary in size.\r\n\r\n\tIn plant cells, there is a central vacuole that is large and takes up most of the cell space. It stores water, nutrients and waste products to help maintain the structure and pressure inside the cell. The pressure control is important because it helps the plant stand upright.\r\n\tIn animal cells, the vacuoles are smaller and there are more of them. They store nutrients and waste products. Vacuoles in animal cells are less important for maintaining cell shape.\r\n\r\n\r\nCell membrane\r\nThe cell membrane (or plasma membrane) acts as a protective barrier for cells. It is made up of a flexible phospholipid bilayer with embedded proteins and other molecules. The cell membrane helps the cell communicate with its surroundings, and controls what moves in and out of the cell.\r\n\r\n\r\n\r\n\r\n\r\nWe will look at the cell membrane in more detail later.\r\n\r\nLysosomes\r\nLysosomes (\"lys\" meaning \"loosening\" and \"some\" meaning \"body\") are the waste disposal system of animal cells. They contain digestive enzymes that break down (or lyse) excess or old cell parts, food particles and invading viruses or bacteria. \r\nThey help cells stay healthy by recycling cellular components and keeping the cell clean.\r\nPlant cells don't need lysosomes because their central vacuole performs this function.\r\n\r\nCell wall\r\nPlant cells contain a cell wall, a rigid outer layer that provides structural support and protection. It also regulates interactions with the environment.\r\nA plant cell wall is mainly made up of cellulose, a strong carbohydrate that helps the cell stay firm and stable so that it can grow. Cell walls are attached to the plasma membrane and help maintain the rectangular shape of plant cells. Since animal cells lack a cell wall, their shape is much more variable due to their flexible plasma membrane.\r\n\r\nChloroplasts\r\nPlant cells also produce their own energy through photosynthesis. This process converts carbon dioxide, water and energy from sunlight into oxygen and glucose.\r\nPhotosynthesis occurs in chloroplasts (\"khloros\" meaning \"green\" and \"plastos\" meaning \"formed\"). Chloroplasts contain a green pigment called chlorophyll which captures light energy.\r\nThey have membranes called thylakoids (\"thylakos\" meaning \"sac\"), which form stacks known as grana. The thylakoids are where the light-dependent reactions of photosynthesis occur. The gel-like fluid surrounding the grana is called stroma.\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n\r\nChloroplasts, a large vacuole and a cell wall are unique to plant cells. Centrosomes and lysosomes are unique to animal cells.\r\n\r\nExercise\r\nSee how well you understand organelles and their functions with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nJia's PhD project involves studying the role of mitochondria in fibroblast energy metabolism. She needs mouse embryonic fibroblasts (or MEF cells), so she grows and characterises them to make sure that they are suitable for her experiments.\r\nThis involves fixing the cells and staining them with fluorescent dye to see the nuclei and mitochondrial networks. She then uses high-resolution fluorescent microscopy to capture detailed images of the cells. One representative image is shown, with cell nuclei in blue and the mitochondrial matrix in red.\r\n\r\n\r\n \r\nAn interactive three-dimensional model of an animal cell and a plant cell, side by side.\r\nAnimal cell model (Left):\r\n\r\n\tShape: Spherical\r\n\t\tKey features:\r\n\t\t\t\r\n\t\t\t\tNucleus: Large, centrally located with a nuclear envelope.\r\n\t\t\t\tMitochondria: Blue oval-shaped structures responsible for energy production.\r\n\t\t\t\tEndoplasmic reticulum (ER): Orange, ribbon-like structures; smooth and rough sections visible.\r\n\t\t\t\tGolgi apparatus: Stack of curved membranes near the nucleus.\r\n\t\t\t\tCell membrane: Outer yellow layer, selectively permeable.\r\n\t\t\t\r\n\t\t\r\n\t\r\n\r\nPlant cell model (Right):\r\n\r\n\tShape: Rectangular with a rigid green cell wall.\r\n\t\tKey features:\r\n\t\t\t\r\n\t\t\t\tNucleus: Located near the cell wall, enclosed in a double membrane.\r\n\t\t\t\tCentral vacuole: Large blue, occupying most of the cell\u2019s interior.\r\n\t\t\t\tChloroplasts: Green ovals with internal stacks, for photosynthesis.\r\n\t\t\t\tGolgi apparatus: Located near ER, involved in packaging.\r\n\t\t\t\tCell wall: Outer green layer, provides structure and support.\r\n\t\t\t\r\n\t\t\r\n\t\r\n\r\n \r\nAn interactive three-dimensional model of the cell nucleus with features labelled.\r\nNucleus model:\r\n\r\n\tNuclear envelope: The outermost reddish layer surrounding the nucleus.\r\nMembrane that encloses the nucleus and protects the DNA inside.\r\n\tNuclear pore: Small openings in the nuclear envelope, depicted as lighter spots. Channels that allow the passage of specific molecules into the nucleus, such as nucleic acid.\r\n\tNucleoplasm: The dense, brownish material within the nucleus. Gel-like substance to provide structure and further protection.\r\n\tChromatin: Fibres present around the nucleolus, depicted as orange lines. These fibres are coiled and condensed to form chromosomes.\r\n\tNucleolus: A distinct, darker central structure. Here, RNA is synthesised and assembled so it can be transported to other regions of the cell.\r\n\r\n \r\nAn interactive three-dimensional model of a mitochondrion with features labelled.\r\nMitochondrion model:\r\n\r\n\tOuter membrane: The outermost smooth blue layer surrounding the entire structure.\r\n\tInner membrane: The folded orange membrane located within the blue outer membrane.\r\n    Cristae: The infoldings of the inner membrane, shown as orange extensions into the green interior.\r\n\tMitochondrial DNA (mtDNA): Shown as small, circular, dark structures within the green matrix.\r\n\tMatrix: The innermost green area of the structure.\r\n    Matrix (function): Also within the green matrix area, related to its function.\r\n    Ribosomes: Small blue dots within the green matrix area.\r\n    Intermembrane space: The space shown between the blue outer membrane and the orange inner membrane.\r\n    Intermembrane space full of H+ Ions: Highlighted with a concentration of ions within the intermembrane space.\r\n    ATP synthase: Located in the orange inner membrane as part of the electron transport chain complex.\r\n\r\n \r\nAn interactive three-dimensional model of a mitochondrion with features labelled.\r\nMitochondrion model:\r\n\r\n\tOuter membrane: The outermost smooth blue layer surrounding the entire structure.\r\n\t\t\r\n\t\t\tDescription: Smooth, surrounds the mitochondrion, composed of a phospholipid bilayer with embedded proteins.\r\n\t\t\tFunction: Acts as a barrier, controlling the entry and exit of substances.\r\n\t\t\r\n\t\r\n\tInner membrane: The folded orange membrane located within the blue outer membrane.\r\n\t\t\r\n\t\t\tDescription: Highly folded into cristae, increasing surface area.\r\n\t\t\tFunction: Contains the electron transport chain and ATP synthase, key for ATP production. It is highly impermeable, managing ion and molecule passage.\r\n\t\t\r\n\t\r\n    Cristae: The infoldings of the inner membrane, shown as orange extensions into the green interior.\r\n\t\t\r\n\t\t\tDescription: Infoldings of the inner membrane.\r\n\t\t\tFunction: Increase surface area for ATP production through oxidative phosphorylation.\r\n\t\t\r\n\t\r\n\tMitochondrial DNA (mtDNA): Shown as small, circular, dark structures within the green matrix.\r\n\t\t\r\n\t\t\tDescription: Circular DNA molecules in the matrix.\r\n\t\t\tFunction: Encodes proteins, ribosomal RNA, and transfer RNA needed by the mitochondrion.\r\n\t\t\r\n\t\r\n\tMatrix: The innermost green area of the structure.\r\n\t\t\r\n\t\t\tDescription: Innermost compartment of the mitochondrion, enclosed by the inner membrane.\r\n\t\t\tComponents: Contains Krebs cycle enzymes, mtDNA, ribosomes, and enzymes.\r\n\t\t\r\n\t\r\n    Matrix (function): Also within the green matrix area, related to its function.\r\n\t\t\r\n\t\t\tFunction: Site of the Krebs cycle, contributing to ATP production through metabolic reactions.\r\n\t\t\r\n\t\r\n    Ribosomes: Small blue dots within the green matrix area.\r\n\t\t\r\n\t\t\tDescription: Small, protein-synthesizing particles in the matrix.\r\n\t\t\tFunction: Translate mitochondrial DNA into proteins required for mitochondrial function.\r\n\t\t\r\n\t\r\n    Intermembrane space: The space shown between the blue outer membrane and the orange inner membrane.\r\n\t\t\r\n\t\t\tDescription: Space between the outer and inner membranes.\r\n\t\t\tImportance: Contains a high concentration of H\u207a ions, crucial for oxidative phosphorylation.\r\n\t\t\r\n\t\r\n    Intermembrane space full of H+ Ions: Highlighted with a concentration of ions within the intermembrane space.\r\n\t\t\r\n\t\t\tDescription: Electron transport chain pumps H\u207a from the matrix to create a gradient.\r\n\t\t\tFunction: This electrochemical gradient, or proton motive force, is essential for ATP production.\r\n\t\t\r\n\t\r\n    ATP synthase: Located in the orange inner membrane as part of the electron transport chain complex.\r\n\t\t\r\n\t\t\tDescription: Enzyme complex in the inner membrane.\r\n\t\t\tFunction: Uses proton flow to synthesize ATP from ADP and inorganic phosphate (Pi) through chemiosmosis.\r\n\t\t\r\n\t\r\n\r\n \r\nAn interactive three-dimensional model of the endoplasmic reticulum with features labelled.\r\nEndoplasmic reticulum model:\r\n\r\n    General structure: The model shows a network of interconnected, flattened sacs and tubules with a blue colour, representing the extensive membranous structure connected to the nuclear envelope.\r\n        \r\n            Description: The endoplasmic reticulum (ER) is a network of membranes in the cell that looks like flattened sacs and tubules. It\u2019s connected to the nuclear envelope. There are two main types of ER: rough ER (RER) and smooth ER (SER).\r\n        \r\n    \r\n    The RER has ribosomes: Shown with small, dark bumps on its surface.\r\n        \r\n            Description: The RER has ribosomes (small particles) on its surface, making it look bumpy.\r\n        \r\n    \r\n    Lumen: The inner space of the RER, highlighted as hollow spaces within the formations.\r\n        \r\n            Protein folding: Newly made proteins enter the lumen where they are folded into their correct shapes.\r\n            Modification: Some proteins are chemically modified in the lumen before they are transported to their next destination.\r\n        \r\n    \r\n    Smooth endoplasmic reticulum (SER): Shown as smooth, tubular structures without ribosomes, located towards the bottom of the image.\r\n        \r\n            Lipid production: The SER synthesizes lipids like phospholipids and steroids, essential for cell membranes and hormone production.\r\n            Detoxification: Enzymes in the SER membrane help detoxify harmful substances, particularly in liver cells.\r\n        \r\n    \r\n    SER: Part of the smooth structures, representing sites where calcium ions are stored.\r\n        \r\n            Calcium storage: The SER stores calcium ions inside its lumen, crucial for muscle contraction and other cellular activities.\r\n        \r\n    \r\n    SER function: Related to the rest of the smooth ER structures, symbolising its metabolic roles.\r\n        \r\n            Carbohydrate metabolism: The SER aids in carbohydrate metabolism by converting glucose-6-phosphate to glucose, helping maintain blood sugar levels.\r\n        \r\n    \r\n\r\n \r\nAn interactive three-dimensional model of the Golgi apparatus with features labelled.\r\nGolgi apparatus model:\r\n\r\n    Trans face: Side of the Golgi apparatus that sends vesicles off to other parts of the cell. It is the side farthest from the ER.\r\n    Secretory vesicles: Vesicles that transport proteins such as hormones.\r\n    Cisternae: Flattened membrane vesicles that can be found in the ER and the Golgi apparatus. They help package and modify certain proteins.\r\n    Transport vesicles: Vesicles that deliver proteins from the endoplasmic reticulum to the Golgi apparatus.\r\n    Cis face: The side nearest to the ER. Receives transport vesicles from it.\r\n\r\n \r\nAn interactive three-dimensional model of the Golgi apparatus with features labelled.\r\nGolgi apparatus model:\r\n\r\n    Lumen: The hollow areas within each of the pink, flattened sac-like structures in the model.\r\n    Incoming transport vesicle: Small spherical structures approaching the Golgi from the top-right side in the model.\r\n    Cisternae: The stacked, curved layers of pink membranes forming the main body of the Golgi.\r\n    Newly forming vesicle: Small budding structures emerging from the edges of the cisternae in the model.\r\n    Secretory vesicle: Round bubbles detaching and moving away from the lower part of the Golgi model.\r\n    Cis face: The top portion of the Golgi model, where vesicles arrive.\r\n    Trans face: The lower portion of the Golgi model, from where vesicles depart.\r\n\r\n \r\nAn interactive three-dimensional model of the cell membrane.\r\nCell membrane model:\r\n\r\n    Phospholipid bilayer: Composed of two layers of blue spheres (hydrophilic heads) with red tails (hydrophobic tails) in between, forming a semi-permeable barrier.\r\n    Integral proteins: Large, purple structures spanning the entire membrane thickness.\r\n    Peripheral proteins: Smaller purple structures located on the membrane surface.\r\n    Glycoproteins and glycolipids: Green chains extending from the membrane surface.\r\n    Cholesterol: Not clearly shown in this model, typically interspersed among phospholipid tails.\r\n\r\n \r\nAn interactive three-dimensional model of a chloroplast with features labelled.\r\nChloroplast model:\r\n\r\n\tStroma lamallae: Flat extensions connecting thylakoid stacks, appearing as connections between the grana.\r\n\tThylakoid: Circular green discs stacked together.\r\n\tLumen: The space inside the thylakoid, highlighted in the cut-away section of a thylakoid.\r\n        \r\n            This cut view of the thylakoid reveals the lumen inside the thylakoid membrane.\r\n        \r\n    \r\n\tIntermembrane space: The space between the inner and outer membranes of the chloroplast.\r\n\tOuter membrane: The smooth, transparent green outer layer of the chloroplast.\r\n\tGranum: A stacked column of thylakoids.\r\n\r\n \r\nAn interactive three-dimensional model of a chloroplast with features labelled.\r\nChloroplast model:\r\n\r\n    Outer membrane: The smooth, transparent green outer layer of the chloroplast.\r\n    Inner membrane: The second, inner green layer just beneath the outer membrane.\r\n    Stroma: The fluid-filled space within the inner membrane, shown as the lighter green background.\r\n    Lumen: The space inside the thylakoid, highlighted in the cut-away section of a thylakoid.\r\n        \r\n            This cut view of the thylakoid reveals the lumen inside the thylakoid membrane.\r\n        \r\n    \r\n    Thylakoid: Circular green discs stacked together.\r\n        \r\n            Multiple stacked thylakoids make up a grana.\r\n        \r\n    \r\n    Lamellae: Flat extensions connecting thylakoid stacks, appearing as connections between the grana.\r\n    Lipid: Small structures scattered within the stroma, depicted as light \r\n    Starch granule: Larger, round structures, shown in light green within the stroma.\r\n    Ribosomes: Tiny blue dots within the stroma, representing ribonucleic structures.\r\n        \r\n            Molecules consisting of RNA.\r\n        \r\n    \r\n    DNA: Circular loop shown within the stroma, indicating the genetic material.\r\n\r\n \r\nAn interactive drawing of a plant cell and an animal cell\r\n\r\nThe cells have information hotspots added to them to indicate different organelles. When each hotspot icon is clicked, a pop-up appears with a brief description on it.\r\n\r\nChloroplast:\r\n\t\r\n\tSolar panels of the cell\r\n\tConverts energy from sunlight into oxygen and glucose\r\n\tPresent in only plant cells\r\n\t\r\n\r\nNucleus:\r\n\t\r\n\tControl centre of the cell\r\n\tHouses cell's genetic material\r\n\tPresent in both animal and plant cells\r\n\t\r\n\r\nMitochondria:\r\n\t\r\n\tPowerhouse of the cell\r\n\tProduces energy in the form of ATP via cellular respiration\r\n\tPresent in both animal and plant cells\r\n\t\r\n\r\nRibosomes:\r\n\t\r\n\tProtein factories of the cell\r\n\tAssembles proteins\r\n\tPresent in both animal and plant cells\r\n\t\r\n\r\nRough endoplasmic reticulum:\r\n\t\r\n\tManufacturing and packaging plant for proteins\r\n\tPresent in both animal and plant cells\r\n\t\r\n\r\nSmooth endoplasmic reticulum:\r\n\t\r\n\tManufacturing and packaging plant for lipids\r\n\tPresent in both animal and plant cells\r\n\t\r\n\r\nGolgi apparatus:\r\n\t\r\n\tPost office of the cell\r\n\tModifies, sorts and packages proteins and lipids for delivery\r\n\tPresent in both animal and plant cells\r\n\t\r\n\r\nVacuoles:\r\n\t\r\n\tWarehouses of the cell\r\n\tPresent in both animal and plant cells\r\n\t\t\r\n\t\tIn plants, there is a single, large, central vacuole.\r\n\t\tIn animals, there are many small vacuoles.\r\n\t\t\r\n\t\r\n\t\r\n\r\nPlasma membrane:\r\n\t\r\n\tBorder control of the cell\r\n\tRegulates entry and exit of materials to and from the cell\r\n\tPresent in both animal and plant cells\r\n\t\r\n\r\nLysosomes:\r\n\t\r\n\tWaste disposal centre of the cell\r\n\tBreaks down and recycles cellular debris\r\n\tPresent only in animal cells\r\n\t\r\n\r\nCell wall:\r\n\t\r\n\tFortress of the cell\r\n\tProvides protection and rigidity to the cell\r\n\tPresent only in plant cells\r\n\t\r\n\r\nChloroplasts:\r\n\t\r\n\tSolar panels of the cells\r\n\tConverts energy from sunlight into oxygen and glucose\r\n\tPresent only in plant cells\r\n\t\r\n\r\nCytoplasm:\r\n\t\r\n\tWorkshop floor of the cell\r\n\tHelps move materials within the cell\r\n\tMaintains the cell's shape\r\n\tPresent in both animal and plant cells\r\n\t\r\n\r\nCentrosomes:\r\n\t\r\n\tConstruction managers of the cell\r\n\tInvolved in cell replication\r\n\tPresent only in animal cells\r\n\t\r\n\r\n\r\n \r\nFluorescence microscopy image of mouse embryonic fibroblast cells\r\nA fluorescence microscopy image showing two mouse embryonic fibroblast (MEF) cells. The cell nuclei are stained blue using DAPI, clearly highlighting their rounded structures. Surrounding the nuclei and extending throughout the cytoplasm, the mitochondrial network is visible in red, shown with a DsRed derivative dye that specifically stains the mitochondrial matrix. The mitochondria appear elongated and interconnected. A scale bar indicating 10 \u00b5m is present in the bottom right corner of the image.\r\n","n":0.018},[{"v":"Organelles","i":1,"n":1},{"v":"Cells","i":0,"n":1}]]},{"i":83,"$":[{"v":"The cell membrane","n":0.577},{"v":"The cell membrane is the border security of a cell. Understanding transport across the cell membrane is crucial for grasping how cells interact with their environment. Use this resource to learn about the fluid mosaic model and the different transport methods.\r\nThe fluid mosaic model\r\nThe structure of the cell membrane can be described using the fluid mosaic model. This model represents the membrane as a dynamic, flexible layer composed of a phospholipid bilayer with embedded proteins.\r\n  \r\n\r\n\r\nPhospholipids are molecules that have a hydrophilic (water-loving) head and a hydrophobic (water-hating) tail. \"Hydro\" means \"water\", \"philos\" means loving and \"phobos\" means \"hating\". \r\nThey are arranged in a way that creates a semi-permeable barrier. This means that the membrane selectively allows substances to move in and out of the cell. Within this bilayer, proteins float like boats on a sea, facilitating communication and transport of molecules.\r\nIt is called the \"fluid\" because the lipid and protein molecules can move sideways within the layer, giving the membrane flexibility and allowing it to self-heal. The \"mosaic\" part refers to the patchwork of proteins that drift in or out of the bilayer, providing a range of functions like transport, structural support and cell recognition.\r\nThe nature of the cell membrane lets the cell interact with the environment, adapt to changes and perform functions that are essential for life.\r\nTransport across the cell membrane\r\nThere are many different ways for materials to be transported into and out of a cell. Let\u2019s look at simple diffusion, facilitated diffusion and active transport.\r\nSimple diffusion\r\nSimple diffusion is a passive transport process, meaning that it doesn\u2019t use any energy from the cell. In this process, molecules move from an area of higher concentration to an area of lower concentration until they are evenly spread out, or when \"equilibrium\" is reached.\r\n  \r\n\r\n\r\nThis type of diffusion relies on a difference in concentration on the inside and outside of a cell. It happens without help from proteins.\r\nExamples of substances that are transported across the cell membrane by simple diffusion are the gases oxygen and carbon dioxide.\r\nFacilitated diffusion\r\nFacilitated diffusion is also a type of passive transport. But unlike simple diffusion, it requires some help from special proteins. These transport proteins act as gateways (or channels) to let molecules move from an area of higher concentration to an area of lower concentration. \r\nAs well as protein channels, diffusion can be facilitated by carrier proteins, which bind to molecules and release them on the other side of the membrane.\r\n \r\n\r\n\r\nThis process allows substances to move in and out of a cell efficiently, even if they cannot pass through the bilayer on their own.\r\nExamples of substances that are transported across the cell membrane by facilitated diffusion are glucose and ions.\r\nActive transport\r\nWhat about when substances need to move against a concentration gradient to cross the cell membrane? How do substances move from a region of lower concentration to a region of higher concentration? This is where active transport comes in.\r\nActive transport requires energy, usually in the form of adenosine triphosphate (ATP), which is the energy currency of a cell. It works against the natural flow of molecules. Specialised carrier proteins in the cell membrane, called pumps, facilitate this movement.\r\n \r\n\r\n\r\nThis animation shows one type of active transport where one molecule is moved across the cell membrane in one direction. The type of carrier protein that facilitates this is called a uniporter (\"uni\" means \"one\" and \"porter\" means \"to carry\"). There are also symporters which transport two molecules across the cell membrane at a time in one direction, and antiporters which transport two molecules across the cell membrane in opposite directions.\r\nExamples of substances transported across the cell membrane by active transport are sodium and potassium ions.\r\n\r\nExercise\r\nSee how well you understand transport across the cell membrane with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nTo compare the rate of transport of different nutrients (glucose, amino acids and fatty acids) across the cell membrane, mouse muscle cells were treated with \\(100\\textrm{ \u00b5L}\\) of equal \\(100\\textrm{ mg/dL}\\) solutions of each nutrient.\r\nEvery \\(15\\) minutes, a sample of the extracellular medium (the liquid outside the cells) was collected and the remaining concentration of each nutrient was measured using a range of test kits. The graph of the data is shown.\r\n\r\n\r\n \r\nAn interactive three-dimensional model of the cell membrane.\r\nCell membrane model:\r\n\r\n\tPhospholipid bilayer: Represented by the two layers of yellow spheres (hydrophilic heads) with yellow tails (hydrophobic tails) in between.\r\n\tIntegral proteins: Shown as large, varied-coloured structures (green, blue) spanning the membrane.\r\n\tPeripheral proteins: Depicted as smaller coloured blobs (orange, grey, purple) on the membrane surface.\r\n\tGlycoproteins: Illustrated by the red hexagon chains attached to a protein.\r\n\tGlycolipids: Shown as blue and red chains on the lipid surface.\r\n\tCholesterol: Depicted as grey ovals integrated into the bilayer.\r\n\r\n \r\nAn interactive three-dimensional animated model of the simple diffusion.\r\nSimple diffusion model\r\n\r\n\tPhospholipid bilayer: Represented by the two layers of yellow spheres (hydrophilic heads) with yellow tails (hydrophobic tails) in between.\r\n\tMolecules moving: Green spheres present on either side of the bilayer representing substances moving across the cell membrane. The region above the bilayer has more green spheres than the region below. Green spheres from the region above are shown passing through the bilayer to the region below.\r\n\r\n \r\nAn interactive three-dimensional animated model of the facilitated diffusion.\r\nFacilitated diffusion model\r\n\r\n\tPhospholipid bilayer: Represented by the two layers of yellow spheres (hydrophilic heads) with yellow tails (hydrophobic tails) in between.\r\n\tProtein channels: Depicted as pink, orange and blue structures spanning the bilayer, forming channels.\r\n\tMolecules moving: Green, purple and red spheres present on either side of the bilayer representing substances moving across the cell membrane.\r\n\t\r\n\t\tThe region below the bilayer has more green spheres than the region above. Green spheres from the region below are shown passing through the blue protein channel to the region above.\r\n\t\tThe region above the bilayer has more red spheres than the region below. Red spheres from the region above are shown passing through the pink protein channel to the region below.\r\n\t\tThe region above the bilayer has more purple spheres than the region below. Purple spheres from the region above are shown passing through the orange protein channel to the region below.\r\n\t\r\n\t\r\n\r\n \r\nAn interactive three-dimensional animated model of the active transport through a uniporter protein.\r\nActive transport model\r\n\r\n\tPhospholipid bilayer: Represented by the two layers of yellow spheres (hydrophilic heads) with yellow tails (hydrophobic tails) in between.\r\n\tCarrier protein: Depicted as a green structure spanning the bilayer, forming a channel.\r\n\tMolecule moving: Shown as a blue sphere moving through the green structure.\r\n\r\n \r\nLine graph showing nutrient uptake by muscle cells\r\nLine graph titled \"Uptake of nutrients by muscle cells\". There are three lines on the graph, a solid line labelled \"Glucose\", a dashed line labelled \"Amino acids\" and a dotted line labelled \"Fatty acids\".\r\nThe vertical axis is labelled \"Extracellular concentration\", beginning with 0 to 120, in increments of 20. The horizontal axis is labelled \"Time\", beginning at 0 and ending with 120, in increments of 10.\r\nIn the graph, the \"Glucose\" line begins at time 0 with an extracellular concentration of close to 100, gradually decreasing to about 24 by the time it reaches a time of 120. The \"Amino acids\" line also begins at time 0 with an extracellular concentration of around 100, decreasing to about 53 by the time it reaches a time of 120. The \"Fatty acids\" line similarly starts at around 100, and decreases to 69 at time 120.\r\nThe data is summarised in the following table.\r\n\r\n\t\r\n    \r\n      Time\r\n      Glucose\r\n      Amino acids\r\n      Fatty acids\r\n    \r\n  \r\n  \r\n    \r\n      \\(0\\)\r\n      \\(98\\)\r\n      \\(98\\)\r\n      \\(98\\)\r\n    \r\n    \r\n      \\(10\\)\r\n      \\(90\\)\r\n      \\(92\\)\r\n      \\(95\\)\r\n    \r\n    \r\n      \\(20\\)\r\n      \\(80\\)\r\n      \\(87\\)\r\n      \\(92\\)\r\n    \r\n    \r\n      \\(30\\)\r\n      \\(70\\)\r\n      \\(83\\)\r\n      \\(89\\)\r\n    \r\n    \r\n      \\(40\\)\r\n      \\(65\\)\r\n      \\(80\\)\r\n      \\(86\\)\r\n    \r\n    \r\n      \\(50\\)\r\n      \\(55\\)\r\n      \\(76\\)\r\n      \\(84\\)\r\n    \r\n    \r\n      \\(60\\)\r\n      \\(45\\)\r\n      \\(68\\)\r\n      \\(81\\)\r\n    \r\n    \r\n      \\(70\\)\r\n      \\(35\\)\r\n      \\(65\\)\r\n      \\(79\\)\r\n    \r\n    \r\n      \\(80\\)\r\n      \\(30\\)\r\n      \\(62\\)\r\n      \\(77\\)\r\n    \r\n    \r\n      \\(90\\)\r\n      \\(28\\)\r\n      \\(59\\)\r\n      \\(75\\)\r\n    \r\n    \r\n      \\(100\\)\r\n      \\(26\\)\r\n      \\(57\\)\r\n      \\(73\\)\r\n    \r\n    \r\n      \\(110\\)\r\n      \\(25\\)\r\n      \\(55\\)\r\n      \\(71\\)\r\n    \r\n    \r\n      \\(120\\)\r\n      \\(24\\)\r\n      \\(53\\)\r\n      \\(69\\)\r\n    \r\n  \r\n\r\n","n":0.028},[{"v":"Cell membrane","i":1,"n":0.707},{"v":"Cells","i":0,"n":1}]]},{"i":84,"$":[{"v":"Cell theory","n":0.707},{"v":"Cells are the building blocks of life. You need to understand cell theory and how cells are classified to grasp the basics of life science. Use this resource to start your journey into the world of biology, then discover how contemporary research has investigated the antibiotic activity of microorganisms associated with Aboriginal bush medicines.\r\nCell theory is a key idea in biology that changed how we view living things. It helps us understand that cells are essential building blocks for all life.\r\nCell theory states that:\r\n\r\n\tThe cell is the most basic unit of life.\r\n\tAll living organisms are composed of one or more cells.\r\n\tAll cells arise only from pre-existing cells.\r\n\r\nGerman physician and physiologist Theodor Schwann (1810\u20131882) and botanist Matthias Jakob Schleiden (1804\u20131881) are credited as the founders of cell theory, but many scientists from the 1800s helped develop our current understanding of cells.\r\n\r\n\r\nClassification of cells\r\nAt the most basic level, cells are classified into two types, based on their structure and function: prokaryotic or eukaryotic. \r\n\r\nProkaryotic cells\r\n\r\nProkaryotic cells lack a cell nucleus and membrane-bound organelles. They contain genetic material that gives them instructions on how to function, grow and develop. However, this genetic material is found in the central area of the cell and is not surrounded by any membrane.\r\nProkaryotic cells are very small and simple, typically existing as single-celled (or unicellular) organisms. This means that one cell makes up the entire organism. Imagine you were made of just one cell!\r\nThese cells reproduce quickly and from a single parent through a process called binary fission.\r\nProkaryotic cells include organisms from the bacteria and archaea domains, such as:\r\n\r\n\tE. coli bacteria\r\n\tcyanobacteria (blue-green algae)\r\n\tmethanogenic (methane-generating) archaea\r\n\tthermophiles (heat-loving archaea).\r\n\r\n\r\nThe biological domains are archaea, bacteria and eukarya. This is the first level of classifying life.\r\n\r\nEukaryotic cells\r\n\r\nEukaryotic cells are bigger and more complex than prokaryotic cells. They have a well-defined cell nucleus. This means that their genetic material is inside a nuclear membrane. \r\nEukaryotic cells also have membrane-bound organelles. This allows different cellular processes to happen separately and at the same time. This allows eukaryotic cells to specialise, giving organisms more complex functions.\r\nUnlike prokaryotic cells, eukaryotic cells can form both unicellular and multicellular organisms. These organisms usually need two parents to reproduce, but can also reproduce with one parent.\r\nSome examples of organisms formed from eukaryotic cells are:\r\n\r\n\tcapybaras\r\n\toak trees\r\n\tamoebas\r\n\toyster mushrooms.\r\n\r\n\r\n\r\n\r\n\r\n\r\nSummary\r\nIn life sciences, many terms come from Latin or Greek roots. Knowing these can help you remember complex biological words. \r\nFor example, \"pro\u2013\" and \"eu\u2013\" come from Ancient Greek, meaning \"before\" and \"true\", respectively.  \"Karyon\" means \"nut\" or \"seed\", referring to the cell nucleus. So:\r\n\r\n\tProkaryotic cells are \"before the nucleus\", meaning they lack a nucleus.\r\n\tEukaryotic cells have a \"true nucleus\", meaning they have a defined cell nucleus.\r\n\r\nThe difference between prokaryotic and eukaryotic cells is summarised in the table.\r\n\r\nExercise\r\nSee how well you understand cell theory with a quick quiz.\r\n\r\n\r\n\r\nData drill\r\nRead the scenario and use the information provided to answer the questions in the quiz.\r\nPramod is a biology student researching the differences between prokaryotic and eukaryotic cells. He obtains a dataset of cell sizes from an online database to analyse these differences.\r\nWhile exporting the data, he accidentally omits the classification of each cell type (prokaryotic or eukaryotic) and only has the average cell diameters. The data is shown in the table.\r\n\r\n\t\r\n\t\t\r\n\t\t\tCell\r\n\t\t\tAverage diameter (\u00b5m)\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\t\\(1\\)\r\n\t\t\t\\(2\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(2\\)\r\n\t\t\t\\(10\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(3\\)\r\n\t\t\t\\(7\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(4\\)\r\n\t\t\t\\(100\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(5\\)\r\n\t\t\t\\(7\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\\(6\\)\r\n\t\t\t\\(1\\)\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n","n":0.043},[{"v":"Aboriginal and Torres Strait Islander knowledges","i":2,"n":0.408},{"v":"Indigenous knowledges","i":1,"n":0.707},{"v":"Cells","i":0,"n":1}]]},{"i":85,"$":[{"v":"Structure and function of cells","n":0.447},{"v":"All living things are made of cells. They are the basic building blocks of life. Use this resource to learn about what cells are made of, how materials are transported in and out of cells, and how they grow and divide.\r\nOverview\r\n\r\n\t\tCell theory \u2013 what is cell theory and how are cells classified?\r\n\t\tOrganelles and their functions \u2013 what organelles are found within animal and plant cells and what are their functions?\r\n\t\tThe cell membrane \u2013 how is the cell membrane structured and how are materials transported between the inside and outside of a cell?\r\n\t\tSurface area to volume ratio \u2013 why is surface area to volume ratio important?\r\n\t\tThe cell cycle \u2013 how do cells grow and divide?\r\n\r\n\r\n","n":0.094},[{"v":"Cells","i":0,"n":1}]]},{"i":86,"$":[{"v":"Digital portfolios","n":0.707},{"v":"Digital portfolio assignments are used in many subject areas, including fine art, law, business, IT, and health. They offer an engaging alternative to traditional essays by providing students with the opportunity to develop essential career skills. Let's explore the steps to create an effective portfolio.\r\nDigital portfolio assignments require you to collect and arrange artefacts or items (work that demonstrates your skill) on a digital platform, and reflect on your professional growth. This enables you to review experiences, analyse successes and failures, and identify lessons learned.\r\n\r\nYou can showcase your achievements in your portfolio. However, a digital portfolio is not just about presenting a perfect final product. It's just as important to document the steps taken to create the final outcome, including successes and failures.\r\nWhat will you be asked to do?\r\nThis can vary from assignment to assignment, but generally speaking, you'll be required to:\r\n\r\n \tcreate or collect items for your portfolio\r\n \tupload them to a digital platform and arrange them logically and/or aesthetically\r\n \treflect, explain and/or analyse your items and the process used to create them.\r\n\r\nWhat will you get out of it?\r\nProducing a portfolio provides you with the opportunity to hone your skills and map your progress. Other benefits include:\r\n\r\n \tlearning to reflect on your practice\r\n \texperience using a digital portfolio platform\r\n \tpresenting your work in a logical and visually appealing way\r\n \tfor art and design students, experience presenting digitised forms of your work on the web.\r\n\r\nWhat the person marking your portfolio wants to see\r\nWhen marking your work, evaluators often look for some or all of the following:\r\n\r\n \tprofessional growth\r\n \tyour ability to reflect\r\n \tthe work you are capable of producing\r\n \tthe process you use to create your work.\r\n\r\nProfessional portfolios v assignment portfolios\r\nPortfolios produced for assessment differ from professional portfolios in a number of ways:\r\n","n":0.058},[{"v":"Assessment","i":0,"n":1}]]},{"i":87,"$":[{"v":"Accessibility in assessment","n":0.577},{"v":"Digital accessibility means creating content in a way that users of all abilities can engage with them. Let's discover some tips that will help make your assignments accessible for everyone.\r\n\r\nWhy is accessibility important?\r\nHere are just some of the many reasons:\r\n\r\n\tInclusion: Accessibility recognises that people experience the world differently, with varying abilities affecting how they access content. For example, individuals with vision impairments may rely on screen readers or require specific font sizes and styles. To ensure our content is truly accessible, we must consider the diverse needs of all individuals.\r\n\tSocial responsibility: By creating accessible content, we help build a world where individuals of all abilities feel valued and included.\r\n\tPreparation for future careers: Understanding accessibility is an essential skill in many careers today. People who are knowledgeable about accessibility are often highly valued by employers.\r\n\tAudience consideration: Consider the diverse needs of your audience when creating assignments. Ensuring accessibility allows a wider range of people to engage with your work.\r\n\r\nWhat can I do to make my assignments accessible?\r\nThis is a complex and evolving area, but everything you do will make a difference to many people. Here are some small considerations that can have an instant impact.\r\nColour contrast\r\nConsider the colours of your text and your background. If there is poor colour contrast, it will be difficult for some members of your audience to read.\r\n\r\nIt is best not to assume that your text and images have good colour contrast — using an accessibility checker like WebAim is much more reliable.\r\n\r\nCaptions and transcripts\r\n\r\nFor any videos in your assignment, make sure that you include captions. This allows people who are not able to hear the audio to access its content. Captions are also useful for people who may have difficulty understanding the speakers in the video.\r\n\r\nFor videos, audio and complicated images, transcripts can be provided to allow your audience to read a text-based version of the dialogue.\r\n\r\nDon't use colour alone to convey meaning\r\nFor those with colour vision deficiency (colour blindness), it can be difficult to distinguish between some colours. Therefore, when colour is used to convey meaning, this information can be missed. \r\n\r\nAn example of this is a graph that has a key showing that different colours represent different piece of information. In this case, it is more inclusive to use patterns as well as colour. Also, it is good practice to provide a transcript of graphs to assist screen reader users.\r\n\r\nAlternative text\r\n\r\nMembers of your audience using screenreaders will not be able to percieve the images you have included unless you provide alternative text (also known as alt text). This is text added to your images that will be read out by screen readers to describe the image.\r\n\r\nPlease note that alt text should only be included for images that add meaning to your assignment. If the image is purely decorative, alternative text does not need to be provided. In some programs, you can mark these images as 'decorative', and if you are creating a web page, you should include an empty alt tag (e.g. alt=\"\").\r\n\r\nWhen writing alternative text:\r\n\r\n\tbe concise: avoid wordy descriptions\r\n\tthink about the context: what is it about the image that the reader needs to know to understand how the image fits into your assignment?\r\n\r\n\r\nAvoid text in images\r\nWhere possible, avoid including images that contain text in your assignments. The text in the image will not be picked up by screen readers. It is better to put the text into the page.\r\n\r\nThat said, sometimes text in an image is not practically avoidable \u2013 for example in a screen-shot, a detailed diagram, or a chart.  In this case, you avoid using text in images as the only method of conveying important information. For example, you could provide alt text or a transcript of the text from the image.\r\n\r\nMeaningful link text\r\nIf your assignment contains links to other web pages, it is best practice to use meaningful link text. What does this mean exactly? \r\n\r\nAvoid using text in your links such as 'click here for more' or full URLs to pages — readers of your work won't know where they'll be taken by the link.\r\n\r\nInstead, use text that makes it clear which page they will be taken to. Here are some examples:\r\nLink destination: RMIT Equitable Learning Services homepage. \r\nExample: For more information, visit RMIT's Equitable Learning Services page.\r\n\r\nLink destination: Web Content Accessibility Guidelines 2 Overview\r\nExample: Learn about the Web Content Accessibility Guidelines (WCAG).\r\n\r\nText font and size\r\n\r\nText that is small is difficult for people with vision imapairments to read. Use a minimum of 12pt in documents and 16px for web page for your assignments.\r\n\r\nSome font styles can be difficult to read for people with low vision. Increase the readability of your assignment by avoiding cursive, decorative and serif fonts.\r\n\r\nHeading hierarchy\r\nUsing heading styles such as 'title', 'heading 1', 'heading 2', etc. makes your work easier to skim and scan. Also, for people with low vision and for people who use screen readers, it makes your work more navigable. \r\n\r\nHeadings should be nested semantically. A page should only have one heading 1. Heading 2 should be used for main sections and heading 3 for sub-sections. Choose your heading level for its purpose, not how it looks, and be consistent.\r\n\r\nWhere can I learn more?\r\nAccessibility is an evolving concept. What is considered best practice may change in the future as technology and practices develop. A reliable source of the best current practices are the Web Content Accessibility Guidelines (WCAG).\r\n\r\nUseful accessibility tools\r\n\r\n\tWebAim \r\n\tWCAG Color Contrast Checker\r\n\tWeb Accessibility Evaluation Tool List\r\n\tAccessibility Checker\r\n\tColor Vision Deficiency (CVD) Simulator\r\n","n":0.033},[{"v":"Assignment planning","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":88,"$":[{"v":"Infographics","n":1},{"v":"Dry data is often hard to digest. Infographics turn information into engaging visuals, capturing attention and improving communication. On this page, we'll cover key considerations for presenting your information effectively.\r\nSuccessful infographics:\r\n\r\n \tconvey complex messages clearly and concisely\r\n \tcontain visual elements that communicate a message\r\n \tuse a minimum amount of text, which should be carefully worded\r\n \thave a logical flow of information\r\n \trequire the audience to spend little time and effort to receive the message.\r\n\r\nWhere to start\r\nBefore starting, analyse the assignment instructions. Check for details like:\r\n\r\n \tRequired topic areas\r\n \tFormatting requirements\r\n \tSoftware to use\r\n \tPresentation format (printed or online, specific file formats)\r\n \tReferencing requirements\r\n\r\nRemember that your content and explanations should be tailored to your audience \u2014 this could be teachers, classmates, industry professionals, or the general public. For example, the general public often needs more context than professionals.\r\nWhat is your objective?\r\nYour objective will influence how you communicate your message. For example, if you are trying to catch people's attention, then bold colours might be a good choice. For sensitive topics, subtle colours may be more appropriate.\r\n\r\nSome examples of objectives are:\r\n\r\n \tcommunication of data/statistics/information\r\n \tpresentation of a timeline\r\n \ttelling a story\r\n \toutlining a process\r\n \tvisualisation of a theory or method.\r\n\r\nWhat content do I include?\r\nMake sure your message is focused. You can't include an entire data set or explain complex theories in depth in an infographic. Select a few key points to highlight. Too much information can reduce audience engagement and dilute your main message.\r\n\r\nUse simple language to increase the readability of your work. Avoid long sentences, complicated language and jargon.\r\nDesign\r\nHere are some tips to get maximum impact from your infographic:\r\n\r\n \tThe flow of information should be logical and help tell the story. Use elements such as numbers, colours, lines and arrows to guide the reader from one point to the next.\r\n \tKeep your infographic uncluttered. Space around visual elements and text will help your main message stand out.\r\n \tPut the most important point or information at the top. In western culture, we read top left to top right and then down, losing focus and interest.\r\n \tIn a huge infographic, people might only read some of it. Consider this when creating your infographic.\r\n \tEmphasis can catch your audience\u2019s eye and their attention.\r\n \tIf you do really want to emphasise a number, make sure you emphasise the what just as much, with a bold word or a picture.\r\n\r\nLayout\r\nThere are many ways that you can organise your infographic. Here are just a few of them:\r\nStacked layout\r\nInformation is \"stacked\" on top of each other in a single column, guiding the reader from the top of the page to the bottom.\r\n\r\nUse this layout for:\r\n\r\n \ttimelines and processes\r\n \tpresenting information from most important to least important.\r\n\r\n\r\n\r\nComparison layouts\r\nThese can be used if you want to highlight the differences between two or more items. Items are placed side by side so that similarities and differences can be easily identified.\r\n\r\nUse this layout to:\r\n\r\n \tcompare the benefits and drawbacks of different theories\r\n \tshow the differences between two policies or products\r\n \trepresent results of different test groups.\r\n\r\n\r\n\r\nTimeline layout\r\nTimelines are great for conveying time-related information in a clear way. Points or steps are organised chronologically so that\r\naudiences can move from one step to the next in a logical way. This format is great for storytelling.\r\n\r\nUse this layout to:\r\n\r\n \tDescribe the stages in the development of a theory or a product\r\n \tShow how a company was established.\r\n\r\n\r\nCircular layout\r\nThis layout helps to show how a process move from one stage to another and guides the audience's eyes in a logical manner.\r\n\r\nUse this layout to:\r\n\r\n \tdescribe a process or procedure\r\n \trepresent a lifecycle of an animal or a product.\r\n\r\n\r\nData-heavy layout\r\nDo you have A LOT of numbers and figures to communicate? Use a data-heavy layout to group information into logical chunks.\r\nSeparate information into categories by using colours and accompany text with meaningful icons.\r\n\r\nUse this layout for:\r\n\r\n \thighlight the results of a research report\r\n \tconvey statistical information.\r\n\r\n\r\nImages and visual elements\r\nVisual aids are a must for infographics and they can come in many forms including graphs (e.g. line, bar, pie), tables, maps, images and icons.\r\n\r\nWhen choosing or designing visual elements, opt for clear and simple designs that can be easily understood by the audience. Avoid packing too much information into graphs and consider how you can make key information from graphs stand by using larger text and icons.\r\n\r\nWhen it comes to choosing images for your infographic:\r\n\r\n \tchoose images that are appropriate to your topic that help the audience to understand your message.\r\n \tmake sure that images accurately represent the information. For example, if your content relates to sizes, make sure you have the correct scale.\r\n \tkeep your style consistent. For example, try using images from one artist as using images from multiple artists and mixing illustrations and photographs often result in an inconsistent style.\r\n \tif an image was created by someone else, don't forget to reference or credit the creator.\r\n\r\n\r\nText\r\nText should tell some of your story, but not all of it. Keep text to a minimum and use key words and phrases rather than full sentences where possible.\r\n\r\nChoose clear font styles and sizes that can be easily read.\r\n\r\nIf you plan to print your infographic, ensure that text is readable at arms distance. Use a font size of at least 14pt for body text and larger for headings. Test readability by printing a small section to confirm the text is clear and easy to read.\r\nPlatform\r\nFor some assignments, a specific platform is required; for others, you can choose. Popular options include Miro, Canva, and Adobe Express. Check if the service is free and if there are any limitations to the features you can use when accessing the free version of the tool.\r\nStudent example\r\nHere is an example of an infographic that has been made for an assignment.\r\nCarlos is a health sciences student who has been asked to make an infographic to help combat the recent rise of diabetes in the Australian population. Instead of proving exhaustive background information about what diabetes is and how it is caused, he decided to create an infographic that could be placed in the waiting rooms of medical clinics that gives simple pieces of advice that could help people lower their risk of diabetes.\r\n\r\n\r\n \r\nDoggy facts\r\n\r\n\tPopularity: Australians keep over 6 million dogs as pets.\r\n\tFood: Large dogs can cost $60 a week to feed on average.\r\n\tPlay: Dogs require interaction and play time with their owners.\r\n\tSleep: Dogs need around 11 hours sleep a day, less than cats do.\r\n\r\n \r\nComparison\r\nFeeding Costs\r\n\r\n\tCats cost $20 a week to feed on average.\r\n\tLarge dogs can cost $60 a week to feed on average.\r\n\r\nIndependence\r\n\r\n\tCats are largely independent and amuse themselves.\r\n\tDogs require interaction and play time with you.\r\n\r\nSleep\r\n\r\n\tCats sleep around 18 hours a day. Mostly chill.\r\n\tDogs need around 11 hours sleep a day. Zero chill.\r\n\r\n \r\nA doggy day\r\n\r\n\t7:00am: Wake up\r\n\tMuch bork. If hooman not awake, jump on hooman.\r\n\t7:30am: Breakfast\r\n\tMany noms. If noms insufficient, many mlems. More bork.\r\n\t8:00am: Play time\r\n\tMany zoomies. Do not eat furniture. Get pats from hooman.\r\n\t9:30am: Nap time\r\n\tFeeling safe. Lie on favourite chair upside down. Much lolling.\r\n\r\n Kitty lifecycle\r\n\r\n\tNewborn: Birth to two weeks old\r\n\tWeaning: Two to twelve weeks old\r\n\tJuvenile: Two to twelve weeks old\r\n\tAdolescent: Around 6-9 months old\r\n\tAdult: Around 7-10 years old is 'middle-aged'\r\n\tSenior: 11 years and older\r\n\r\n Australian pets\r\nThere are 28.7 million pets in Australia living in 6.9 million households.\r\nPopularity of pets by types (most popular to least):\r\n\r\n\tDogs\r\n\tCats\r\n\tFish\r\n\tBirds\r\n\tOther (small mammals, reptiles)\r\n\r\nAdopt!\r\nIn 2022, the RSPCA received 18,770 dogs and 33,747 cats.\r\n\r\n\tAdopt\r\n\tTrain\r\n\tCare\r\n\tLove\r\n\r\n75% of pet owners live in a multi-person household, and 86% of pet owners are families with children under 18.\r\n85% of owners report that pets have positively impacted their daily lives.\r\nPopular collar colours:\r\n\r\n\tRed\r\n\tBlue\r\n\tGreen\r\n\r\n \r\nDiabetes prevention\r\nThese five lifestyle changes can help prevent the onset of type 2 diabetes.\r\n\r\n \tLose extra weight\r\n \tBe physically active\r\n \tEat healthy plant foods\r\n \tEat healthy fats\r\n \tAvoid fad diets\r\n\r\nTalk to your doctor about setting some reasonable goals. It\u2019s never too late to start!\r\n\r\n","n":0.028},[{"v":"Presentations","i":1,"n":1},{"v":"Assessment","i":0,"n":1}]]},{"i":89,"$":[{"v":"Selecting artefacts portfolios","n":0.577},{"v":"Carefully selecting artefacts for your portfolio assignment allows you to effectively showcase your skills and growth. This page provides tips to help you choose work that best demonstrates your abilities.\r\n\r\nWhat should I put into my portfolio?\r\nFor professional portfolios, the focus is on showcasing a polished, finished product. However, for portfolio assignments, the process and the lessons learnt are as important as the final product. In other words, it's essential to demonstrate your ability to produce quality work and to show how you've grown professionally from the experience.\r\n\r\nHere are some tips for achieving this balance through the items you select to go into your portfolio.\r\nHighlight experimentation\r\nShowcasing experimentation in your portfolio is vital, as it reflects your readiness to step out of your comfort zone and learn. Failure is a natural part of the learning process, and demonstrating how these experiences have fostered your growth is important.\r\n\r\nEvaluators value seeing how you've turned setbacks into a path towards success. While not every attempt leads to immediate success, highlighting your resilience and the insights gained from experimentation emphasises your determination and innovation.\r\nShow your process\r\nIncluding drafts can be beneficial to demonstrate the process leading to the final product, but ensure you check your assignment instructions to see if this is required.\r\n\r\nIf you've received feedback on your work from peers or teaching staff, it can be a good idea to select pieces that demonstrate how you've incorporated that advice. This shows your willingness to learn and adapt.\r\n\r\nAdditionally, show how you have applied any methodologies and theories that you've encountered during your studies. This displays how you are able to turn theory into practice.\r\n\r\n\r\nDemonstrate your skills\r\nYour portfolio should clearly convey your capabilities and highlight your strengths. It's important for readers to understand what you are capable of producing.\r\n\r\nDisplaying the diversity of your work is advantageous, as it demonstrates your ability to work across various areas and apply your skills to different subjects. By presenting a range of projects, you can illustrate your versatility and adaptability.\r\nReflect your interests\r\nPortfolios offer you the chance to deepen your understanding of areas that interest you. You can tailor your artefacts to align with what you would like to do professionally after your studies. This way, your portfolio will not only demonstrate your current skills but also reflect your passion and future career goals.\r\n\r\n","n":0.051},[{"v":"Assignment planning","i":2,"n":0.707},{"v":"Digital skills","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":90,"$":[{"v":"Writing in portfolios","n":0.577},{"v":"Regardless of the type of course you're studying, it is extremely likely that what you display in your portfolio will need to be accompanied by some form of writing. Let\u2019s explore the types of writing often found in portfolios.\r\nReflection\r\nReflective writing means looking back on an experience, describing the experience, analysing its value and importance, and planning how it will be used to inform future practice. \r\n\r\nThere are a number of frameworks that can be used for reflective writing. One of these is DIEP, where you: \r\n\r\n\tDescribe the experience that you had\r\n \tInterpret the meaning of the experience\r\n\tEvaluate why the experience was significant\r\n\tPlan how you will use what you have used in the future. \r\n\r\nA more detailed explanation can be found in the Academic reflection section. \r\n\r\nHere are some examples of how students from different study areas have used DIEP for their reflective writing: \r\n\r\n\r\n\r\n\r\n\r\n\r\nArtist statements\r\nArtist statements briefly describe a piece of work that you have created, and they can accompany pieces included in a portfolio. \r\n\r\nThey can help your audience to understand your motivations and intentions, and the techniques you have employed. \r\n\r\nFor more information, visit the Artist statement section on the Learning Lab. \r\n\r\n\r\nProcess description / materials and methods\r\nIn some cases, you may be required to include a process description or materials and methods section. This describes how you created a design or artwork, or how you conducted a procedure.\r\n\r\nTo do this, you'll need to communicate to your reader the materials, equipment and measurements used, and clearly describe the steps you took.\r\n\r\nSome considerations for this type of writing are:\r\n\r\n\tuse the past tense (the solution was heated to 200\u00b0C; an acrylic paint was applied)\r\n\tlist the steps in chronological order. In other words, in the order they were completed.\r\n\tuse signaling words like then, next, subsequently, afterwards and finally to show the sequence of the steps\r\n\tprovide enough detail that a reader would be able to replicate your process\r\n\tprocess descriptions can be written in paragraph or bullet-point form. Check with your educator for their preference\r\n\tin some cases, you will need to avoid using pronouns like 'I' and 'me'. Check with your educator about the level of formality they would like you to use.\r\n\r\n\r\n\r\n\r\n\r\n","n":0.053},[{"v":"Understanding assignments","i":2,"n":0.707},{"v":"Assessment","i":1,"n":1},{"v":"Digital skills","i":0,"n":0.707}]]},{"i":91,"$":[{"v":"Digital portfolios: the first steps","n":0.447},{"v":"If you have a portfolio assignment to work on and you are not sure where to begin, this page gives you some practical tips to get you off to a strong start.\r\nCarefully read your assignment instructions and rubric\r\nFor any assignment, your first task should be getting to know your assignment instructions. By doing this you can discover information such as:\u00a0\r\n\r\n\tthe scope: understand the specific requirements, such as the number of artworks, draft drawings, or lesson plans you need to produce.\r\n\tthe platform: identify if you are required to use a specific platform like WordPress or PebblePad, or if you have the flexibility to choose your own.\r\n\twritten or reflective components: establish if any writing should accompany your work, what its purpose is and how many words are required. For instance, you may need to reflect on the creation process of an artwork or explain the choices you made to produce a lesson plan.\r\n\trequired artefacts: regardless of your area of study, check if you need to include only completed works or also drafts to show the progression of your work.\r\n\t\r\nAdditionally, familiarise yourself with the assignment rubric, which outlines the marking criteria. Understanding the rubric will help ensure that you have included all required components and met the expectations from the marker\u2019s perspective.\u00a0\r\nDo you have doubts about what is required?\r\nIf you're unclear about any assignment instructions or have unanswered questions, contact your teacher early. This gives them time to respond and for you to implement their suggestions.\r\n\r\nAs they or their colleagues will be marking your assignment, they are the best source for answers. An alternative is to contact Study Support, where staff members can assist you with understanding assignment instructions and planning the steps of your assignment.\u00a0\r\nConsider your audience\r\n\r\nYou will be producing your work for a specific target audience, and this will influence the work you produce and how it is presented. Therefore, it is worthwhile spending some time contemplating your target audience.\u00a0\u00a0\r\nSome examples of audiences for portfolios assessments are:\u00a0\r\n\r\n\tmock companies or clients: you will be asked to imagine that you have received a brief from a client and you need to work to the brief.\r\n\tteachers/lecturers: in these cases, they will assess your work against criteria set out in a rubric and assignment instructions.\r\n\tclassmates: peer reviews are sometimes used so that students can see their work from their classmates\u2019 perspectives and feedback can be given.\r\n\tindustry experts: some courses arrange for experts to provide feedback on student work to help students understand real-world expectations and to give them a taste of working in their industry.\r\n\ta panel of judges or reviewers: if you are applying for a grant, your work may be reviewed by a panel who have been appointed to assess the work of applicants.\r\n\tyourself: you may decide to continue your portfolio after the assignment, so that you can continue to map your professional growth and reflect on your experiences.\r\n\r\nConsider your purpose\r\nThe purpose of your portfolio shapes the artefacts you select as well as the accompanying writing. Here are some common purposes:\r\n\r\n\r\n\tcompleting an assignment task: For a course assignment, your goal is to meet the criteria and aim for a good grade. Follow instructions and the rubric closely. Experiment by applying course theories to practice and reflect on the outcomes to show your engagement. You'll often need to outline your process. For example, an art student might describe how they moved from initial stages to a finished product.\r\n\tapplying for a job: When using a portfolio for a job application, focus on showcasing your abilities and final products, but don't forget to outline your process \u2014 this can display your problem solving skills to potential employers.\r\n\tapplying for a scholarship: Scholarship applications may require a portfolio highlighting career and study achievements, academic background, career objectives, and your CV. Explain why you're an excellent candidate for the scholarship.\r\n\tcontinuous learning If you enjoy the process of creating your portfolio, you might decide to continue beyond your studies. It\u2019s a great way to grow, experiment, and reflect. Sharing it with peers can be valuable to them, too.\r\n\r\nStart producing/collecting artefacts\r\nWhether you need to produce or collect artefacts will depend on the course you are studying. Let\u2019s consider two scenarios:\u00a0\r\n\r\n\tthroughout the course: Portfolios can be used as a continuous assessment tool, meaning that students compile and reflect on their work over time. This provides opportunities for students to make improvements based on feedback at each stage.\u00a0\r\n\r\nIf your assignment runs throughout the course, then it's likely you will be producing artefacts and updating your portfolio with them on a regular basis.\r\n\tend of the course: Portfolios may be assigned as a project at the end of a course. You might be asked to organise existing artefacts into a portfolio to showcase your cumulative learning and skills. In this case, your task will be to organise the artefacts into your portfolio in a logical way that also showcases your learning journey.\r\n\r\nKeep the reflective process in mind while you work\r\n\r\nIf you are required to write a reflective piece to include in your portfolio, it\u2019s a good idea to keep a journal. Record your thoughts and feelings regularly because they can often be forgotten later on.\u00a0\u00a0\r\nTo reflect, ask yourself questions such as:\u00a0\r\n\r\n\tHow does completing this lesson plan / draft drawing / sculpture / etc. make me feel?\r\n\tHave I found the process easy / hard / frustrating / rewarding / enlightening? Why?\r\n\r\n","n":0.034},[{"v":"Understanding assignments","i":2,"n":0.707},{"v":"Assessment","i":1,"n":1},{"v":"Digital skills","i":0,"n":0.707}]]},{"i":92,"$":[{"v":"Literature review checklist","n":0.577},{"v":"Use this checklist to revise your literature review before you submit it.\r\n\r\nTopic\r\nHave you:\r\n\r\nselected an appropriate topic? Not too broad/narrow?\r\n\r\nIntroduction\r\nHave you:\r\n\r\ngiven background to your subject area?\r\nclearly stated the topic?\r\nincluded a thesis statement?\r\n\r\nBody\r\n\r\nDo you paragraphs begin with a clear topic sentence?\r\nHave you compared and contrasted ideas of authors?\r\nAre your paragraphs organised in a logical way?\r\n\r\nConclusion\r\nHave you:\r\n\r\nrestated your aims?\r\nsummarised the main points?\r\nsuggested areas for future study?\r\n\r\nCitation and referencing\r\nHave you:\r\n\r\nused the correct referencing style?\r\nprovided a citation for information from other sources?\r\nincluded a properly formatted reference list?\r\nchecked your reference formatting with Easy Cite?\r\nused sources that are reliable, current and appropriate?\r\n\r\nOther\r\nHave you:\r\n\r\nchecked for errors in spelling, punctuation and sentence structure?\r\nincluded your own voice?\r\nhave you done this in an academic way?\r\n\r\n ","n":0.094},[{"v":"Literature review","i":2,"n":0.707},{"v":"Assessment","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":93,"$":[{"v":"Your voice","n":0.707},{"v":"\r\nA successful literature review does more than just represent the ideas of other authors. It also includes the voice of the person writing the review. Let's find out how we can do this.\r\n\r\nHighlighting gaps, strengths and limitations\r\n\r\nThis shows your evaluation of the studies you have engaged with. This can be done with attitude markers. Here are some tips to get started:\r\n\r\nAttitude towards topics or issues\r\n        \r\n                 Abdul (2020) explores the critical issue of... \r\n\t\t\t\t Lubov (2021) examines the alarming trend of... \r\n\t\t\t\t Jones (2024) describes the pressing need for...\r\n          \r\nAttitude through adverbs\r\n          \r\n                   Interestingly, Hope (2022) discusses that...\r\n                    Unfortunately, the sample size was too small to...\r\n                    The results clearly show that...\r\n                     Importantly, the study took into account the fact that...\r\n             \r\nAttitude through 'that' phrases\r\n            \r\n                    It is evident that...\r\n                    It is worth noting that...\r\n                    It is possible that...\r\n             \r\nAttitude towards studies\r\n            \r\n                    Khalid (2023)  provides a compelling argument that...\r\n                    Chen (2024) thoroughly describes...\r\n                    Matherson (2022) convincingly argues that...\r\n                    Anuman et al. (2023) astutely observe that...\r\n                    Alim (2021) underestimates the importance of...\r\n                    In their comprehensive study from 2025, Drummond states that...\r\n            \r\n\r\nProviding criticism\r\nNaturally, at times you will find yourself disagreeing with what an author has said. However, critiquing other authors should be done professionally, respectfully and constructively. Simply saying that you disagree with someone's ideas is not constructive, as you need to detail why you disagree with them. Similarly, using strong, emotive language to describe their ideas such as saying it is 'wrong' or 'ridiculous' is not respectful.\r\n\r\nHere are some techniques that can be used to soften criticism of the views of other authors.\r\n\r\n\r\nAcknowledging strengths first\r\nMention a positive, before expressing a negative.\r\n\r\nWhile Rodriguez (2022) provides a comprehensive overview of the topic, there are areas that could benefit from further exploration.\r\nUse careful language\r\n Employ hedging words to soften the critique, such as \"might,\" \"may\", \"could,\" \"appears to,\" or \"suggests.\" \r\n\r\nCallao (2022) may have overlooked the possibility of...\r\nSuggest further research\r\nIf you have not been convinced by an author's theory, then you may suggest further research in the area is needed.\r\n\r\nWhile Chen (2021) raises some interesting points, further investigation into the root causes of academic misconduct is required.\r\nAcknowledge complexity\r\nIf you believe an author has not provided a satisfactory answer to a problem, you can state that you disagree but also acknowledge that the issue is complex.\r\n\r\nWhile Hernan (2023) has suggested some helpful steps, given the complexity of the issue, these steps cannot be considered a conclusive solution.\r\nExpress uncertainty\r\nA polite way of showing disagreement can be to express uncertainty. \r\n\r\nThere appear to be some inconsistencies in the results obtained in the 2022 study by Mraz and Webster.\r\nAvoid emotional language\r\nOverly emotional or 'emotive' language can make you seem biased. Instead of using words like 'terrible' or 'ridiculous' to describe someone's ideas, you could instead say that they are 'problematic' or 'inadequate'.\r\n\r\nThis approach is problematic because it fails to take into account the people who are affected by the situation. \r\n\r\nHere are some more examples of emotive language that should be avoided in academic writing and some more neutral words that can be used instead:\r\n\r\n  \r\n    Emotive\r\n    Neutral\r\n  \r\n  \r\n    Awful\r\n    Unfavourable, suboptimal\r\n  \r\n  \r\n    Fantastic\r\n    Noteworthy, effective\r\n  \r\n\r\n  \r\n    Disaster\r\n    Problematic\r\n  \r\n  \r\n    Perfect\r\n    Ideal\r\n  \r\n  \r\n    Dreadful\r\n    Unsatisfactory\r\n  \r\n  \r\n    Unbelievable\r\n    Improbable\r\n  \r\n  \r\n    One-sided\r\n    Biased, influenced\r\n  \r\n","n":0.043},[{"v":"Assessment","i":2,"n":1},{"v":"Academic writing","i":1,"n":0.707},{"v":"Literature review","i":0,"n":0.707}]]},{"i":94,"$":[{"v":"Integrating information","n":0.707},{"v":"A key component of literature reviews is incorporating the information, ideas and theories of other scholars. To do this, we need to use paraphrasing, and to a lesser extent, quoting. On this page, you'll find examples of how information has been integrated into a literature review paragraph.\r\nParaphrasing and quoting\r\nLet's imagine that you have found information from a number of authors that you would like to use in a body paragraph of your literature review. How can you integrate these ideas into your writing?\r\n\r\nThe answer is paraphrasing and quoting:\r\n\r\n\r\nParaphrasing involves rewording an original text while retaining its meaning. It requires citing the original author(s) with details like surnames and publication year.\r\n\r\nQuoting means using an author's exact words, enclosed in quotation marks, with a citation including the author's surname, publication year, and page number.\r\n\r\nCitation formats for both paraphrasing and quoting depend on the required referencing style. For guidance, consult Easy Cite and the Paraphrasing and Quoting pages.\r\n\r\nExample\r\nNow, let's explore an example of how one student integrated information from four articles into a paragraph on the topic of emotional intelligence.\r\n\r\nOriginal sentences:\r\n\r\n\tCerra (2024): \"Emotional intelligence in leadership not only enhances interpersonal relationships but also drives strategic decision-making processes, leading to superior organisational outcomes.\"\r\n\tMichaels (2021):  \"Emotional intelligence enhances leaders' ability to understand team dynamics, leading to improved team cohesion and performance.\"\r\n\tKhan et al. (2024): \"The application of emotional intelligence in leadership roles is linked to higher employee satisfaction and reduced turnover rates.\"\r\n\tKoroka & Jones (2023): \"Leaders with high emotional intelligence tend to foster environments where communication is open, and conflicts are resolved more effectively.\"\r\n\r\n\r\nParagraph with integrated information\r\n\r\n\r\nNotice how for the paraphrases:\r\n\r\n   the sentences use a number of different words from the originals\r\n   the structures of the sentences have changed from the originals\r\n   citations have been included that provide author names and years of publication.\r\n\r\n\r\nNotice how for the direct quote:\r\n\r\n  quotation marks have been used\r\n  a page number has been provided.\r\n\r\n\r\nInformation prominent / author prominent\r\nAnother thing that you may have noticed from the example paragraph is that in some cases that the author or authors' names are included in the sentence, and in others, their names are in brackets at the end of the sentence.\r\n\r\nThese two styles are called information prominent and author prominent.\r\n\r\nInformation prominent sentences are used when we want to emphasise the information rather than the author. The citation is placed at the end of the sentence in brackets.\r\nFor example:\r\nA number of studies have found that leaders who possess high emotional intelligence are better equipped to comprehend team dynamics, which in turn fosters greater team cohesion and performance (Cerra, 2024; Michaels, 2021).\r\n\r\nAuthor prominent sentences are used when we want to emphasise who came up with the theory or conducted the study. The citation is placed at the beginning of the sentence.\r\nFor example:\r\nThis is supported by Khan et al. (2024), who found that emotionally intelligent leaders often see \"increased employee satisfaction and reduced turnover rates\", suggesting that such leaders create more stable and committed teams (p. 49).\r\n\r\nGenerally speaking, most academic writing uses a mixture of these two styles.\r\n\r\n*Please note that the APA 7th edition referencing style has been used for the examples on this page. Check with your teacher or instructor which referencing style they require you to use.","n":0.043},[{"v":"Literature review","i":2,"n":0.707},{"v":"Academic writing","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":95,"$":[{"v":"Structuring a literature review","n":0.5},{"v":"By organising your literature review effectively, you'll enhance comprehension and engagement, making your work stand out. Discover how to structure your review logically, ensuring that your arguments flow smoothly, making it easy for your readers to follow your analysis.\r\n\r\n\r\nThere are different ways that your information can be organised in a literature review. They include: \r\n\r\nthematic: grouping literature according to themes or topics.\r\nchronological: arranging the literature according to the order in which the studies were published.\r\ntheoretical: organising the literature by theoretical frameworks or models that can be applied to the topic. \r\n\r\n\r\nWhichever way you choose to group your information, remember that you need to compare and contrast the ideas of authors, rather than writing one paragraph about one study then another paragraph about another study, and so on. \r\n\r\nIntroduction \r\nIn your introduction, begin by providing your reader with an overview of the general context of your literature review. This is where you introduce the topic in broad terms. Example\r\nThe rise of online gaming has emerged as a significant cultural phenomenon worldwide, with Australian high school students increasingly participating in this digital activity.\r\nNext, provide more specific details about the topic you will explore, and explain its importance and relevance.\r\n\r\nExample\r\nWhile gaming can offer entertainment and foster social connections, there is growing concern about its potential mental health impacts, especially when gaming becomes excessive. Understanding these impacts is crucial, as they can influence students' academic performance, social relationships, and overall well-being.\r\nFollowing this, clearly state the aim of your review, also known as your 'thesis statement'.\r\n\r\nExample\r\nThis literature review aims to critically analyse existing research on the mental health effects of excessive online gaming among Australian high school students, identifying key findings and highlighting gaps in the current literature.\r\nFinally, define the scope of your review. Specify what will be covered and how it will be approached.\r\n\r\nExample\r\nThe review will focus on studies conducted over the past decade, incorporating both quantitative and qualitative research, to provide a comprehensive overview of this pressing issue.\r\n\r\nThe body\r\nThis is where you explore your topic in detail. The information that you found in your research is presented, showing the most relevant and important themes,  ideas and studies. You highlight the areas of agreement and disagreement  between authors and researchers and include your own perspective. Where possible, identify any potential gaps in information.\r\n\r\nThings to note:\r\n\r\nYour body paragraphs should begin with a topic sentence that tells the reader the purpose of the paragraph.\r\nIn your body paragraphs, it is important to use linking words and phrases to show how different ideas connect with each other. \r\nBody paragraphs often end with a concluding sentence that briefly summarises the information from the paragraph.\r\n\r\n\r\nExample\r\nResearch has consistently shown that excessive online gaming can have adverse effects on the mental health of high school students. For instance, Nguyen (2021) found that students who engage in prolonged gaming sessions often experience higher levels of anxiety and depression compared to their peers. Similarly, Hassan et al. (2022) emphasise that excessive gaming can lead to social withdrawal, which further exacerbates feelings of loneliness and isolation among adolescents. In agreement with these findings, O'Connor (2023) highlights that the immersive nature of online games can contribute to addictive behaviours, ultimately impacting students' emotional well-being and academic performance. Collectively, these studies underscore the potential negative mental health implications of excessive online gaming among high school students.\r\n\r\nHowever, not all researchers concur with these conclusions. Kimura (2021) argues that online gaming can serve as a valuable outlet for stress relief and social interaction, offering benefits that may outweigh potential negative impacts. Furthermore, Rodriguez (2022) points out that the context in which gaming occurs is crucial, suggesting that when balanced with other activities, gaming does not necessarily lead to adverse mental health outcomes. Additionally, Ahmad (2023) challenges the notion of gaming addiction, proposing that the majority of players engage in gaming as a healthy, recreational activity without significant negative effects. These contrasting perspectives highlight the complexity of the issue and suggest that further research is needed to fully understand the varied impacts of online gaming on adolescent mental health.\r\n\r\nConclusion\r\nAfter reading your review, there will be some things that your audience will forget. The conclusion is your opportunity to remind them about your key points and concepts so that they will go away with these things in mind. \r\n\r\nIn your conclusion, you should:\r\n\r\nrestate the research aim from your introduction. This should be paraphrased; do not use the same words.\r\nsummarise the key points and concepts covered\r\nsuggest areas for future study based on what has been uncovered by your review\r\n\r\n\r\nExample\r\nIn summary, this literature review critically assessed the mental health impacts of excessive online gaming on Australian high school students, identifying key findings and gaps in the research. The review highlighted a general agreement that excessive gaming is associated with negative outcomes such as anxiety, depression, and social withdrawal. However, it also presented differing views, with some studies suggesting potential benefits of gaming for stress relief and social interaction when balanced with other activities. The research also revealed that themes like addiction and academic performance are significant, and showed how understanding has evolved in response to technological and social changes.\r\n\r\nFuture studies should consider longitudinal research to explore the long-term effects of gaming. Additionally, investigating individual differences and expanding the scope to include diverse cultural contexts within Australia could provide a more comprehensive understanding of the issue.","n":0.034},[{"v":"Academic writing","i":2,"n":0.707},{"v":"Assessment","i":1,"n":1},{"v":"Literature review","i":0,"n":0.707}]]},{"i":96,"$":[{"v":"Organising information","n":0.707},{"v":"Literature reviews require students to collect, synthesise and interpret large amounts of information. This means that keeping organised from the start is very important. On this page, you will find practical tips for arranging the content for your review.\r\nA common problem that students run into when writing a literature review is that they collect a lot of information without organising it logically. Using a system that allows you to do the following will save you a lot of time and frustration later:\r\n\r\n \tfind what you are looking for quickly\r\n \tsort information into topics\r\n \tcompare and contrast views, theories and studies.\r\n\r\nAlso keep in mind that you will need to record the referencing details of the sources you find immediately; there is nothing worse than finding great information and not remembering where it came from.\r\nSearchable tables\r\nOne way to get organised is to create a table in a program such as Microsoft Word or Excel. This way, you can use control + f (Windows) or command + f (Mac) to search the table and locate the information you need quickly and easily.\r\nTo illustrate, the following headings could be used:\r\n\r\n \tAuthor(s): It's essential to record the author or authors' names so that you do not lose track of who the information comes from. Make sure that their names are correctly spelled.\r\n \tTopic: As you decide on the topics you want to cover in your review, create two-to-three word tags for them that can be placed in this column. Then, when you come to write about a particular topic, you can find all of the information from different authors on that topic quickly.\r\n \tKey points: In this column, write the information from the source text that you would like to include in your review. It's a good idea to record the page number of each piece of information so that you can locate it easily in the future, should you need to.\r\n \tAgreement/disagreement: For literature reviews, you need to find areas of agreement and disagreement. Using a searchable table can help you to compare the views of authors side-by-side.\r\n \tReference: Referencing often makes up part of your grade, so it is important to record this information as you go.\r\n\r\n\r\nVisual diagrams\r\nIf you are someone who prefers to see how information and ideas connect visually, then creating a mind map or a flow chart can be a great way to organise the information you have collected. To do this, you can arrange topics around your central theme and use arrows to show the areas of agreement and disagreement between the authors.\r\n\r\nVisit the Mind mapping page to learn more.","n":0.048},[{"v":"Academic writing","i":2,"n":0.707},{"v":"Assessment","i":1,"n":1},{"v":"Literature review","i":0,"n":0.707}]]},{"i":97,"$":[{"v":"Literature reviews","n":0.707},{"v":"Have you been asked to write a literature review for an assignment but you are not sure what you are required to do? The following pages will cover researching for your literature review, organising information, structuring your work and integrating information from other sources into your writing.\r\nOn this page, you will learn what literature reviews are and how they differ from other types of academic writing.\r\n\r\nLiterature reviews are a common type of assignment at universities. They can be given as stand-alone assignments, or they can be a section of a larger research report. They are useful because they help students to:\r\n\r\n \tdevelop research, organisational, and analytical skills\r\n \tinvestigate a topic deeply and compare the different theories and ideas associated with it\r\n \tthink critically about the information they encounter.\r\n\r\nWhat do I need to do?\r\nLiterature reviews involve selecting a topic of interest and conducting research to uncover existing information, ideas, theories, and practices in that area. Your task is to:\r\n\r\n \tsynthesise the information\r\n \tidentify areas of agreement and disagreement among authors\r\n \tspot potential gaps in information\r\n \tidentify any problems or issues that remain unresolved\r\n \toffer your evaluation of the literature.\r\n\r\nWhat to avoid\r\nKeep in mind that a literature review is not just a summary of the articles and texts you find. It is not about simply stating what each author believes in sequence. Instead, you should identify connections between the ideas you encounter, noting whether they support each other or offer contrasting viewpoints.\r\n\r\nConsider the following paragraph from a literature review. Has the author successfully linked the concepts discussed by the source text authors?\r\nOn the topic of artificial intelligence (AI), researchers tend to fall into two camps: those who focus on its potential benefits, and those who worry for the future. M\u00fcller (2020) discusses how AI can improve efficiency in data processing, highlighting advancements in machine learning algorithms. Batbayar (2019) explores the ethical implications of AI, particularly concerning privacy and decision-making.\r\n\r\n\r\nUnlike argumentative essays, literature reviews do not require you to make an argument to prove a particular point and you do not need to come up with 'the' answer to a problem. While your critique of the information is important, the main objective is to review the literature and present a comprehensive picture of the most significant views and theories in the field to your reader.\r\n\r\nRead the following paragraphs. Which paragraph would be more suitable for an argumentative essay and which would be more appropriate for a literature review?\r\n\r\nParagraph 1\r\nRecent studies have explored the impact of project-based learning (PBL) on student engagement and achievement in high school education. For instance, Garc\u00eda and Thompson (2020) found that PBL can significantly increase student motivation by allowing learners to explore real-world problems and collaborate with peers. This approach aligns with the findings of Lee (2021), who observed that students engaged in PBL demonstrated higher levels of critical thinking and problem-solving skills compared to those in traditional lecture-based settings. However, Kimura (2023) points out that while PBL can boost engagement, its effectiveness might change depending on the subject and how well the teacher manages the PBL activities. Kimura suggests that without proper implementation, PBL might not lead to the expected academic improvements. Together, these studies suggest that while project-based learning has potential benefits, its success largely depends on how it is applied in the classroom.\r\nParagraph 2\r\nOne of the most compelling reasons to incorporate digital literacy into high school curricula is its role in preparing students for the modern workforce. According to Williams (2021), the digital economy is expanding rapidly, and employers are increasingly seeking candidates who possess strong digital skills. By integrating digital literacy into education, schools can equip students with the necessary tools to navigate and succeed in a technology-driven job market. Moreover, Ahmed (2022) highlights that students who are proficient in digital skills are better positioned to adapt to new technologies and platforms, which is crucial in a world where technological advancements are constant. Therefore, ensuring that high school students graduate with a strong understanding of digital tools is not only beneficial but essential for their future career prospects.\r\n\r\nWhat do your markers want to see\r\n\r\nAssignments that are similar to literature reviews\r\nSystematic reviews:\r\nLike literature reviews, systematic reviews involve gathering and examining existing research on a specific topic. They differ because systematic reviews use a strict and organised method to choose and evaluate studies to ensure the conclusions are reliable and unbiased.\r\n\r\nFor more information, visit the Research and writing skills for masters level students book.\r\n\r\nScoping reviews:\r\nScoping reviews, similar to literature reviews, aim to explore and summarise what research is available on a topic. The difference is that scoping reviews are usually broader and focus on identifying what research has been done and what gaps exist, often to see if a more detailed review is possible.\r\n\r\nSimply put, systematic reviews tend to be more in depth, and scoping reviews more general, while literature reviews usually are somewhere in the middle.","n":0.035},[{"v":"Academic writing","i":2,"n":0.707},{"v":"Literature review","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":98,"$":[{"v":"Choosing a topic and researching","n":0.447},{"v":"For some literature review assignments, you will be given a topic to explore. For others, you will need to find your own topic. Keep reading to find out how to select a topic to research  that will help you meet the requirements of your assignment and how to gather the information you need.\r\n\r\nToo broad? Too narrow? Just right?\r\n\r\nSelecting the right topic for your literature review is vital for your assignment's success and can be more challenging than it seems.\r\n\r\nA topic that is too broad makes it difficult to cover all important literature within your assignment's scope, while a topic that is too narrow may not provide enough relevant information.\r\n\r\nStriking the right balance is key. You can achieve this by adjusting limiting words and phrases in your topic, such as:\r\n\r\ncategories: solar/wind/coal energy, casual/permanent/contract employment\r\ntime period: in the last five years, in 2024, since January\r\ndemographic: people living in the south-eastern suburbs, women aged from 24-36, children under 5, retirees, first-home buyers\r\ngeographical location: Melbourne, the southern hemisphere, Latin America\r\norganisations or industries: multinational companies, the film industry, Australian universities\r\n\r\n\r\nToo broad\r\nGiani is studying architecture and has been asked to write a 2,000-word literature review. She has decided she would like to explore the topic of \"conservation strategies in sustainable architecture\". \r\n\r\nAfter doing a little research on her topic, Giani shows her topic to her teacher for feedback. Her teacher suggests that it is \"too broad\" and recommends that she \"narrows\" it. \r\n\r\nGiani has now refined her topic and has decided to focus on a few specific aspects of sustainability in event management. This leaves her with:\r\n\r\n\r\n\r\nSimilarly, some literature review topics may be too narrow and it may be difficult to find enough information for your assignment. In these cases, it may be necessary to remove limiting words and phrases to make the topic more broad.\r\n\r\n\r\nToo narrow\r\nEdmond has chosen the following topic for their literature review: \r\nThe mental health impacts of excessive gaming on Discord on male Australian students from 13-15 years of age.\r\n\r\nAfter doing a search for information, Edmond was unable to find many relevant articles or studies. They contacted their teacher, who suggested that they make their topic broader. \r\n\r\nAfter revising their topic, Edmond settled on:\r\nThe mental health impacts of excessive online gaming on Australian high school students.\r\n\r\nThe topic has been made broader by:\r\n\r\nremoving the specific gaming platform\r\nremoving the gender focus\r\nbroadening the demographic from 13-15 to all high school students\r\n\r\n\r\n\r\nBrainstorming\r\nOnce you have a good idea of your topic, it's time to do some brainstorming about what potential subtopics you would like to cover.\r\nThink about what you already know about the topic and information that's been covered in your course.\r\n\r\nIt's a good idea to use a mindmap so that you can make a visual plan that shows how topics relate to each other.\r\n\r\nResearching your topic\r\nNow, start your search for the information that will be included in your literature review. You can start by doing some basic background reading and familiarising yourself with the core concepts or principles in the area. \r\n\r\nAfter that, it's time to do a more in-depth search. Visit the Find information section for some great tips for how to do this. \r\n\r\nHow much time should I spend researching?\r\nJudging how much information and how many sources to include in your literature review can be challenging. A common trap is spending excessive time searching for information, which can reduce the time available for writing and result in gathering more information than can be included in your assignment.\r\n\r\nTo avoid this, consider these practical tips:\r\n\r\n\tSet a research deadline, such as a few days or a week, and stick to it.\r\n\tLet the word count guide you. For instance, you don't need to cite 50 authors for a 1,500-word literature review.\r\n\r\nEvaluating information\r\nThe quality of your sources is more important than the quantity. Your literature review should focus on the most relevant theories and studies in your field. Including unreliable or outdated information won't impress your marker.\r\n\r\nConsider these factors:\r\n\r\n\tCurrency: When was the information published or updated? Is it current?\r\n\tRelevance: Is it significant in your area of study and relevant to your topic?\r\n\tAuthority: Who is the author\u2014are they an expert? Do they have affiliations that suggest bias?\r\n\tAccuracy: Is the information supported by reliable evidence and verifiable by other sources? Has it been peer-reviewed?\r\n\tPurpose: Does the information aim to inform, sell, entertain, or persuade? How does this impact its reliability, and can you identify any biases?\r\n\r\nLearn more on the Choose valid sources page.\r\n\r\nTest your knowledge\r\n\r\n","n":0.037},[{"v":"Literature review","i":2,"n":0.707},{"v":"Academic writing","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":99,"$":[{"v":"Rubrics and learning outcomes","n":0.5},{"v":"Rubrics are tools used to grade students' work by setting clear criteria and performance levels. They show the aspects that will be taken into consideration when the marker evaluates a student's work.\r\n\r\nThe aim of a rubric is to provide explicit and clear communication of assignment expectations. Therefore, reading and understanding a rubric can improve students' chances of success on their assignments.\r\n\r\nFeatures of rubrics\r\nRubrics can come in many different forms. However, here are some of their common features:\r\n\r\nassignment title\r\ncriteria\r\ncriterion weight\r\nscore range\r\nscore descriptor.\r\n\r\nHere's a single criterion from a rubric with the criterion, criterion weight, score range and score descriptor highlighted. The full rubric can be viewed in the Sample rubric accordion folder.\r\n\r\n\r\n\r\nUnderstanding the weighting of a rubric\r\nEach criterion in an assignment has a \"weight\" which indicates its contribution to your overall grade. The total of all criteria in assignments usually adds up to 100%, and you will usually be given a grade out of 100. For example, if the \"Analysis\" section of an assignment is worth 40%, then your performance in that area will contribute 40% to your final grade for the assignment.\r\n\r\nBy understanding your assignment's rubric and the weight of each criterion, you can better understand where to invest your time and energy to maximise your chances of success on the assignment.\r\n\r\nUsing rubrics to your advantage\r\nNow that we understand what rubrics are, let's explore how they can be used to improve our assignments.\r\n\r\n\tReview the criteria: Read through each criterion carefully. Understand the distinctions between the different levels of achievement.\r\n\tAlign your work: Use the rubric as a checklist to guide your work. Aim for the description from the  column aligned to the highest score to guide you.\r\n\tFocus on analysis and application: Your ability to apply concepts to real-world situations  is critical. Don't just summarise; analyse and justify your recommendations.\r\n\tOrganise your work: Make sure your assignment is clear, logical, and well-structured, with easy transitions between ideas.\r\n\tProofread your work: Ensure that your writing is clear, concise, and well-structured. If the rubric asks for an academic register, you'll need to use formal language.\r\n\tReferencing: Ensure that you are using RMIT Harvard referencing style throughout your assignment for citations and the reference list. Incorrect referencing will impact your score.\r\n\tPrioritise: Focus on heavily weighted sections, as they significantly impact your grade. However, low-weight sections also matter, so aim for consistent performance to avoid lowering your overall grade.\r\n\r\n\r\nWhen assignments don't have a rubric\r\nAt RMIT, rubrics must be provided for all assignments that are worth 20 per or more of the total course mark. If you have not been provided with a rubric for your assignment, you can:\r\n\r\n\tReview the Assignment Prompt: Understand the task requirements by identifying key instruction verbs to ensure you are following the instructions as closely as possible.\r\n\tAsk Questions: Seek clarification from your teacher or tutor to find out what their expectations for the assignment are.\r\n\tReview the Course Outline: This will give you an idea of what skills and abilities you are expected to develop from the course.\r\n\r\nCourse learning outcomes\r\nUniversity courses outline skills for students to acquire, known as course learning outcomes (CLOs). Assignment instructions often specify which CLOs are addressed, helping students understand what assessors expect. \r\n\r\nHere's an example of CLOs from a Bachelor of Education course linked to an assignment, with key content and instruction terms highlighted to guide students on the skills to demonstrate.\r\n\r\nTest your knowledge\r\nAnswer the following questions to test your understanding of what you've learnt.\r\n","n":0.042},[{"v":"Understanding assignments","i":2,"n":0.707},{"v":"Assignment planning","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":100,"$":[{"v":"Accessibility","n":1},{"v":"Accessibility has been a key focus in our redevelopment of the Learning Lab.\r\nImproved accessibility features for all users include:\r\n\r\n \tDark mode \u2013 to reduce eye strain (change settings in the hamburger menu)\r\n \tConsistent navigation including breadcrumb navigation at the top of the page, a hamburger global navigation system in the top right corner that can be accessed at any time, and a right-hand sidebar which keeps users oriented and allows them to easily find content within each section\r\n \tAn accessible colour palette for text highlighting and other visual content elements\r\n \tFunctionality that may not be visible to many users, but is essential for assistive technologies such as alt text for images, \u2018skip to text\u2019, Aria labels and more to allow for a smoother experience\r\n \tImproved responsive layout elements that work for desktop and mobile screens - our testing revealed that some sight imparied users prefer to use mobile devices so it's essential that all content works just as well for these small screens\r\n \tConsistent transcripts across all videos as well as interactive activities and visual elements that require text-based explanation\r\n \tReinterpreting selected activities, examples, and visual elements from legacy content that was not previously accessible, to create accessible content elements that don't require separate transcripts\r\n\r\n\r\nOf course, accessibility is not just about technical elements that allow access to this website. It is also about how the content is structured and presented for comprehension by students with different needs and at different levels. We have placed an emphasis on improving these aspects of the content to create a more equitable experience for everyone.\r\n\r\nThese improvements have been informed by the IDEA Framework, RMIT\u2019s Open Scholarship policy, and our user testing with students and accessibility user testers including neurodiverse, sight-impaired, blind, and deaf users.","n":0.059},[{"v":"Accessibility","i":0,"n":1}]]},{"i":101,"$":[{"v":"About the Learning Lab","n":0.5},{"v":"The Learning Lab is a free and open resource where students can learn and apply foundational skills needed to succeed in tertiary education. \r\n\r\nMaterial on the Learning Lab is designed to promote active learning. Each section includes activities so that students may apply their knowledge to tasks.\r\n\r\nWhile the Learning Lab does not replace curriculum, it can complement courses by providing short tutorials and pages where students can gain confidence in areas of assumed knowledge.\r\n\r\nThe Learning Lab is separated into sections accessible by the tiles on the home page. There are also keywords on each page that can be used to navigate content.\r\n\r\nThe Learning Lab is in ongoing development. Be sure to bookmark or check back regularly to see what\u2019s new and updated.\r\n\r\nFor educators: sharing Learning Lab content with your students\r\n\r\nThere are many ways to familiarise your students with the Learning Lab:\r\n\r\nEmbed a resource or page from Learning Lab into your Canvas course (contact the Digital Learning Team for assistance with this)\r\nInclude links to specific Learning Lab resources on your Canvas pages or other learning resources\r\nInclude a Canvas page devoted to helping students navigate the Learning Lab\r\nEvery Canvas course should have an overview of Library services page included. Ensure students are familiar with the page as a part of orientation.\r\n\r\n\r\nDocumentation\r\n\r\nThe Learning Lab is built on a design system developed in 2024 with a focus on accessibility, engagement, readability and creating an equitable experience for a broad range of users.\r\n\r\nIn keeping with this ethos, the website is also self-documenting, with guidelines for the creation and editing of content, and the sustainable management of the site over time.\r\n\r\nDocumentation of Learning Lab functional elements","n":0.061},[]]},{"i":102,"$":[{"v":"Page not found","n":0.577},{"v":"This page shares code with 404.php. This shared code is found in /includes/redirect.php. This setup exists because Sitesucker does not capture a 404 page. By using this structure, we can demonstrate most of the functionality within the WordPress environment and generate a static page. This static page is renamed to 404.html during the publishing process.\r\n\r\nThe page then scans an array. If there's a match, the user us redirected after 5 seconds. Otherwise, a 404 message is displayed. \r\n\r\nThe content for the redirect 404 page is located in the following file:\r\n\r\n/wp-content/themes/picostrap5-child-base/page-redirect-404.php\r\n\r\nTemplate Hierarchy: WordPress uses a template hierarchy to determine which template file to use. For specific pages, you can create custom templates by naming them page-{slug}.php or page-{ID}.php, where {slug} is the page's slug and {ID} is the page's ID.","n":0.088},[]]},{"i":103,"$":[{"v":"Delivery","n":1},{"v":"Part of the mark you receive for your presentation will be given for the way your information is delivered. For this, your marker will consider things like your voice, your body language and gestures, your slides and your engagement with the audience. Here are some tips for maximising your marks for delivery.\r\n\r\nVisit the space before your presentation\r\nIf you're unfamiliar with the presentation venue, try to visit it before the day of your presentation. This allows you to decide where to stand so you don't block the slides and where you can maintain eye contact with the entire audience, not just the teacher.\r\n\r\nIt's also a good opportunity to test the room's technology to prevent any surprises on the day.\r\nFace your audience\r\nIt can be tempting to turn around and read from your slides, but this should be avoided. Every time you turn your back on your audience, you break the engagement. This is why positioning yourself to the side of your slides and making hand gestures towards important points is preferred.\r\n\r\nAlso, having fewer words on your slides will reduce the temptation to read from them.\r\nUse your voice\r\nThe voice we use when giving presentations should be different from our normal speech. When presenting, we should:\r\n\r\n \tSlow down: This will improve the clarity of your voice by allowing you to focus more on pronunciation and it gives the audience more time to process what you are telling them.\r\n \tProject: This means speaking more loudly than you usually would so that everyone in the room can hear you clearly.\r\n \tEmphasise: By using emphasis, you put a stress on key words and concepts, which will make them stand out more for your audience. For example, the following came from a presentation that a student gave on deforestation.\r\n\r\nIn this example, the emphasised words are in bold:\r\n\"Deforestation is a major issue because it leads to the loss of biodiversity and contributes significantly to climate change.\"\r\nBody language\r\nA good posture and positive body language tell your audience that you are confident, interested and excited about what you have to say. On the other hand, if you have closed body language and do not seem interested in your topic, it will be hard for your audience to be engaged.\r\n\r\nWhere possible, try to maintain good posture, give eye contact to each person in the room and use hand gestures that complement your talk. For example, you can signal to key points in your slides or to emphasise certain key words.\r\n\r\nAvoid crossing your arms, looking at the ground, turning your back to the audience and only focusing on one section of the audience.\r\nPractice\r\nIf you want to build up your confidence before a talk, then you should know your topic inside and out. Here are some tips for getting prepared:\r\n\r\n \tRehearse your talk a number of times and time yourself. When you know your talk well, you will deliver it well.\r\n \tPractise with a friend or family member and get feedback from them.\r\n \tPractise delivering your talk using the main points and brief notes only on cue cards. It is important that you know your topic thoroughly so that you can deliver your presentation confidently.\r\n","n":0.044},[{"v":"Presentations","i":1,"n":1},{"v":"Assessment","i":0,"n":1}]]},{"i":104,"$":[{"v":"Visuals","n":1},{"v":"Most presentations are accompanied by visuals in the form of slides. When well-designed, they can make a presentation come to life and complement what you say. When designed poorly, they can confuse the audience. On this page, we unpack the elements that make visuals effective. \r\n\r\nLess is more\r\nBusy slides take audience members longer to read. The more time they spend reading your slides, the more they have to divide their attention between what you have on the slide and what you are saying. \r\n\r\nTherefore:\r\n\r\navoid full sentences in your slides. Instead, use key words and phrases, and give extra detail orally\r\nkeep the design of your slides clear and simple. Cluttered slides take longer to read and understand.\r\n\r\n\r\nKeep accessibility in mind\r\nWhen preparing your slides, it's important to consider that your audience may have diverse abilities. To ensure that individuals with visual impairments can access the information, you can:\r\n\r\nuse high colour contrast. Aim for 4.5:1 for normal text and 3:1 for large text. You can use tools such as Web Aim's contrast checker to help you with this.\r\nuse a minimum of 18pt font size for body text\r\nchoose font styles that are easily readable\r\nconsider describing any visual elements on your slides aloud to make the content accessible to everyone.\r\n\r\n\r\nDesign\r\nYour slides should be designed in a clear and consistent way. This means that the colours and styles of visuals and fonts should be uniform throughout the presentation, giving your presentation a cohesive and professional appearance.\r\n\r\nTransitions and animation can add movement to your presentation. However, do not overuse them on your slides, as they can become distracting for your audience. \r\n\r\nWhere graphs and tables are used to represent data, they should be easy to interpret for your audience. Keep the design simple and visually engaging.","n":0.059},[{"v":"Assessment","i":1,"n":1},{"v":"Presentations","i":0,"n":1}]]},{"i":105,"$":[{"v":"Creating cohesion","n":0.707},{"v":"The use of signalling and transition words (also referred to as signposting), helps you show your audience how the presentation is structured and how ideas relate to each other.\r\n\r\nConnecting the parts\r\nWe can also create cohesion in our talks by linking the different sections together. One way that we can do this is by:\r\n\r\ntelling the audience what the main points of the presentation will be in the introduction\r\nexpanding on those points and providing explanation in the body of the talk\r\nreminding the audience what the main points of the talk were in the conclusion\r\n\r\n\r\nLet's use this presentation outline from a marketing student as an example:\r\n\r\n","n":0.099},[{"v":"Assessment","i":1,"n":1},{"v":"Presentations","i":0,"n":1}]]},{"i":106,"$":[{"v":"Logarithmic differentiation","n":0.707},{"v":"\r\nLogarithmic differentiation makes it easier to differentiate complex functions. This method is particularly useful for functions with variable exponents or intricate products. You might encounter logarithmic differentiation when calculating compound interest, analysing population growth models or optimising processes in engineering and physics.\r\n\r\nSometimes it is easier to differentiate the logarithm of a function than the original function. This is called logarithmic differentiation. By taking the natural logarithm (\\(\\log_{e}\\) or \\(\\ln\\)) of both sides of an equation, we can transform these functions into a form that is easier to differentiate.\r\nSuppose you have to differentiate the following function.\r\n\\[\\begin{align*}\r\ny & =\\frac{x^{2}-1}{x^{8}\\sqrt{x^{4}+1}}\r\n\\end{align*}\\]\r\nAt first sight, you would see that you need to use the quotient rule. You may also realise that you will need to use the product rule to differentiate the denominator. But if you take this approach, you will find that it gets very tricky very quickly. Logarithmic differentiation makes this much easier to handle.\r\nTo master logarithmic differentiation, you need to be confident with using the logarithm laws. Go back to Logarithms if you need a refresher.\r\n\r\nUsing logarithmic differentiation\r\nSuppose you want to differentiate \\(y=f(x)\\) where \\(f(x)>0\\) for all \\(x\\) values. \r\nWe can take the natural logs of both sides. But only if \\(y>0\\).\r\n\\[\\ln(y)=\\ln(f(x))\\]\r\nWhen we differentiate with respect to \\(x\\), we use the chain rule.\r\n\\[\\begin{align*} \\frac{d}{dx}(\\ln(y)) & = \\frac{d}{dx}(\\ln(f(x))\\\\\r\n& = \\frac{d}{dy}(\\ln(y))\\frac{dy}{dx}\\\\\r\n& = \\frac{1}{y}\\frac{dy}{dx}\r\n\\end{align*}\\]\r\nThe basis of logarithmic differentiation is the equation:\r\n\\[\\frac{d}{dx}(\\ln(f(x))=\\frac{1}{y}\\frac{dy}{dx}\\]\r\nThe effectiveness of logarithmic differentiation depends on whether it is easier to differentiate the original \\(f(x)\\) or \\(\\ln(f(x))\\). \r\nLet's go back to the original function at the beginning of this page and see how logarithmic differentiation can be used (Example 1).\r\n\r\nExample 1 \u2013 using logarithmic differentiation\r\nDifferentiate \\(y=\\dfrac{x^{2}-1}{x^{8}\\sqrt{x^{4}+1}}\\).\r\nWe assume that \\(y<0\\), which allows us to take the natural log of both sides. We can then use the log rules to break the function into more manageable parts.\r\n\\[\\begin{align*} \\ln y & = \\ln\\left(\\frac{x^{2}-1}{x^{8}\\sqrt{x^{4}+1}}\\right)\\\\\r\n& = \\ln(x^{2}-1)-\\ln(x^{8}\\sqrt{x^{4}+1})\\\\\r\n& = \\ln(x^{2}-1)-\\ln(x^{8})-\\ln(\\sqrt{x^{4}+1})\r\n\\end{align*}\\]\r\nWe can now differentiate, using the chain rule on the left.\r\n\\[\\begin{align*} \\frac{d}{dx}\\ln(y) & = \\frac{d}{dy}(\\ln(y))\\frac{dy}{dx}\\\\\r\n& = \\frac{1}{y}\\frac{dy}{dx}\\\\\r\n& = \\frac{x^{8}\\sqrt{x^{4}+1}}{x^{2}-1}\\frac{dy}{dx}\r\n\\end{align*}\\]\r\nWe will look at each term on the right separately, too. For \\(\\ln(x^{2}-1)\\), we can use the chain rule. If we let \\(u=x^{2}-1\\), \\(\\dfrac{du}{dx}=2x\\). Using the chain rule:\r\n\\[\\begin{align*} \\frac{d}{dx}\\ln(x^{2}-1) & = \\frac{d}{du}\\ln(u)\\frac{du}{dx}\\\\\r\n& = \\frac{1}{u}\\frac{du}{dx}\\\\\r\n& = \\frac{1}{x^{2}-1}2x\\\\\r\n& = \\frac{2x}{x^{2}-1}\r\n\\end{align*}\\]\r\nThe second term on the right is calculated by setting \\(u=x^{8}\\). \\(\\dfrac{du}{dx}=8x^{7}\\) and using the chain rule:\r\n\\[\\begin{align*} \\frac{d}{dx}\\ln(x^{8}) & = \\frac{d}{du}\\ln(u)\\frac{du}{dx}\\\\\r\n& = \\frac{1}{u}\\frac{du}{dx}\\\\\r\n& = \\frac{1}{x^{8}}8x^{7}\\\\\r\n& = \\frac{8}{x}\r\n\\end{align*}\\]\r\nThe final term on the right is calculated by setting \\(u=\\sqrt{x^{4}+1}\\). Here, we use the chain rule twice. \r\n\\[\\begin{align*} u & = (x^{4}+1)^{\\frac{1}{2}}\\\\\r\n\\frac{du}{dx} & = \\frac{1}{2}(x^{4}+1)^{-\\frac{1}{2}}4x^{3}\\\\\r\n& = \\frac{4}{2}x^{3}(x^{4}+1)^{-\\frac{1}{2}}\\\\\r\n& = 2x^{3}(x^{4}+1)^{-\\frac{1}{2}}\\\\\r\n& = \\frac{2x^{3}}{\\sqrt{x^{4}+1}}\r\n\\end{align*}\\] \\[\\begin{align*} \\frac{d}{dx}\\ln(\\sqrt{x^{4}+1}) & = \\frac{d}{du}\\ln(u)\\frac{du}{dx}\\\\\r\n& = \\frac{1}{u}\\frac{du}{dx}\\\\\r\n& = \\frac{1}{\\sqrt{x^{4}+1}}\\frac{2x^{3}}{\\sqrt{x^{4}+1}}\\\\\r\n& = \\frac{2x^{3}}{x^{4}+1}\r\n\\end{align*}\\]\r\nPutting it all together:\r\n\\[\\begin{align*} \\frac{x^{8}\\sqrt{x^{4}+1}}{x^{2}-1}\\frac{dy}{dx} & = \\frac{2x}{x^{2}-1}-\\frac{8}{x}-\\frac{2x^{3}}{x^{4}+1}\\\\\r\n\\frac{dy}{dx} & = \\frac{x^{2}-1}{x^{8}\\sqrt{x^{4}+1}}\\left(\\frac{2x}{x^{2}-1}-\\frac{8}{x}-\\frac{2x^{3}}{x^{4}+1}\\right)\r\n\\end{align*}\\]\r\nJust remember that depending on how you simplify, you may get a different answer.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using logarithmic differentiation\r\n\r\n \tDifferentiate \\(y=3^{x}\\) with respect to \\(x\\) using logarithmic differentiation.\r\n \tUsing logarithmic differentiation, differentiate \\(y=\\dfrac{(x-1)^{3}}{(x+1)^{4}(2x+3)}\\) with respect to \\(x\\), for \\(x>1\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.046},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":107,"$":[{"v":"Tables and figures","n":0.577},{"v":"Tables and figures are useful in your assignment writing because they can summarise data or complex information in a more readable way. However, you should only use them if they assist the reader in understanding.\r\n\r\nWhat's the difference between a table and a figure?\r\nThere are different writing conventions for tables and figures:\r\n\r\nTables have vertical columns and horizontal rows.\r\n\r\n\t\r\n\t\r\nFigures include all other graphic representations, e.g. graphs, pie charts, diagrams, maps, photographs.\r\n\r\n\r\nPositioning tables and figures\r\n\r\n\r\nHere are some tips for finding the most effective place in your assignment for tables and figures:\r\n\r\n \tPlace tables and figures as close to the relevant text as possible.\r\n \tTry to make each table or figure fit onto one page, changing to landscape orientation if necessary. If it won\u2019t fit onto one page, begin at the top of a page and continue on following pages.\r\n \tDon\u2019t leave a large blank space on a page. If you discuss a table that appears on the following page, just refer to the table and continue with your text.\r\n \tIf you have a very large table or figure, consider simplifying it to illustrate the key data. Put complex, raw data in an appendix.\r\n \tEach table and each figure needs to be numbered sequentially, even if you only have one.\r\n\r\nTitles, labels and legends\r\nThe way that you provide titles and labels depends on the referencing style you have been asked to use for your assignment. We recommend that you use Easy Cite to help you with this. However, some general rules are:\r\n\r\n \tTables and figures all need brief but informative titles (don\u2019t be creative).\r\n \tTitles for tables are generally written above the table; titles for figures are usually written below the figure (check your assignment instructions for which referencing style to use).\r\n \tIf your table covers more than one page, begin each subsequent page with the table number and write \u2018continued\u2019. For example: Table 4 (continued).\r\n \tLabel the axes of graphs, and of columns and rows in tables, in order to clearly show units of measurement or analysis.\r\n \tUse legends where necessary to define symbols, abbreviations or terminology.\r\n\r\nReferring to tables and figures in your text\r\nKeep in mind that your writing must refer to each table or figure. We do this by referring to them in the text by \u2018Table 1\u2019, Figure 5\u2019, etc., not by their titles. Remember that tables and figures are included to complement the information that you have written in your text, not to replace it.\r\nYou can do this by using your writing to:\r\n\r\n \tinterpret the table or figure\r\n \thighlight the main or important points\r\n \thighlight any unusual or unexpected findings\r\n \tshow relationships between data\r\n \tdraw attention to a relationship or trend\r\n \tsummarise information in the table or figure.\r\n\r\nHere are some ways of introducing tables and figures to your text:\r\n\r\n \tIdentify the table number first and make some statement about it.For example: The data in Table 5.4 indicates that...\r\n \tUse \u2018as\u2019, either at the beginning or end of the sentence.As shown in Table 3.3, student satisfaction increased by 20 per cent in 2022.\r\n\r\nThe number of students who withdrew from their course decreased in 2024, as shown in Figure 3.8.\r\n \tUse the passive voice.Further demographic and contextual data of the PhD (thesis) candidates are shown in Table 4.1.\r\n \tRefer to the table or figure in brackets at the end of the sentence.If the market continues on its current trend, spending will need to be decreased significantly (Figure 3.3).\r\n\r\nTest your knowledge\r\nAnswer these three questions to test what you have learnt.\r\n\r\n","n":0.042},[{"v":"Academic writing","i":2,"n":0.707},{"v":"Assignments","i":1,"n":1},{"v":"Assessment","i":0,"n":1}]]},{"i":108,"$":[{"v":"Inverse hyperbolic functions","n":0.577},{"v":"\r\nProviding a function is one to one, it is possible to find an inverse function. This module discusses inverse hyperbolic functions, which are used in advanced calculus for integration and in the solution of differential equations.\r\n\r\nSee Hyperbolic functions for a list of definitions.\r\n\r\nInverse hyperbolic functions may also be defined.\r\n\r\nThe inverse sinh function is \\(\\sinh^{-1}\\) and is sometimes denoted\r\nas \\(\\text{asinh or $\\text{arcsinh.}$ }\\)The latter is pronounced\r\n\u201carc shine\u201d.\r\n\r\nThe inverse cosh function is \\(\\cosh^{-1}\\) and is sometimes denoted\r\nby \\(\\text{acosh}\\left(x\\right)\\) or \\(\\text{arccosh$\\left(x\\right)$ .}\\)\r\nThe latter is pronounced \u201carc cosh\u201d. Note that \\(\\cosh\\left(x\\right)\\)\r\nis not one to one. Restricting its domain to \\([0,\\infty)\\), it is\r\none to one and the inverse function is defined for this restricted\r\nfunction.\r\n\r\nFinally, the inverse \\(\\tanh\\) function is \\(\\tanh^{-1}\\) or \\(\\text{atanh}\\)\r\nor \\(\\text{arctanh.}\\) The latter is pronounced \u201carc than\u201d.\r\n\r\nThe inverse hyperbolic functions may be written in terms of the natural\r\nlog function as\r\n\\[\\begin{align*}\r\n\\sinh^{-1}\\left(x\\right) & =\\ln\\left(x+\\sqrt{x^{2}+1}\\right)\\\\\r\n\\cosh^{-1}\\left(x\\right) & =\\ln\\left(x+\\sqrt{x^{2}-1}\\right),\\ x\\geq1\\\\\r\n\\tanh^{-1}\\left(x\\right) & =\\frac{1}{2}\\ln\\left(\\frac{1+x}{1-x}\\right),\\ \\left|x\\right|<1\\\\\r\n\\text{csch$^{-1}\\left(x\\right)$ } & =\\ln\\left(\\sqrt{\\frac{1}{x^{2}}+1}+\\frac{1}{x}\\right),\\ -\\infty<x<\\infty,\\ x\\neq0\\\\\r\n\\text{sech$^{-1}\\left(x\\right)$ } & =\\ln\\left(\\sqrt{\\frac{1}{x^{2}}-1}+\\frac{1}{x}\\right),\\ 0<x\\leq1\\\\\r\n\\text{coth$^{-1}$ }\\left(x\\right) & =\\frac{1}{2}\\ln\\left(\\frac{x+1}{x-1}\\right),\\ -\\infty<x<-1\\text{ or $1<x<\\infty.$ }\r\n\\end{align*}\\]\r\nDerivative of inverse hyperbolic functions\r\nThe derivatives of the inverse hyperbolic functions are:\r\n\\[\\begin{align*}\r\n\\frac{d}{dx}\\text{sinh$^{-1}\\left(x\\right)$ } & =\\frac{1}{\\sqrt{x^{2}+1}},\\text{ for all real }x\\\\\r\n\\frac{d}{dx}\\text{cosh$^{-1}$ $\\left(x\\right)$ } & =\\frac{1}{\\sqrt{x^{2}-1}},\\text{for all real }x>1\\\\\r\n\\frac{d}{dx}\\tanh^{-1}\\left(x\\right) & =\\frac{1}{1-x^{2}},\\text{ for all real }|x|<1\\\\\r\n\\frac{d}{dx}\\text{csch$^{-1}\\left(x\\right)$ } & =\\frac{-1}{|x|\\sqrt{1+x^{2}}},\\text{ for all real }x,x\\neq0\\\\\r\n\\frac{d}{dx}\\text{sech$^{-1}\\left(x\\right)$ } & =\\frac{-1}{x\\sqrt{1-x^{2}}},\\text{ for all real }x\\in(0,1)\\\\\r\n\\frac{d}{dx}\\text{coth$^{-1}$ }\\left(x\\right) & =\\frac{1}{1-x^{2}},\\text{ for all real }|x|>1\r\n\\end{align*}\\]\r\nExample 1\r\nWhat is the derivative of \\(\\sinh^{-1}\\left(4x\\right)\\)?\r\nSolution\r\nSince \\(\\frac{d}{dx}\\text{sinh$^{-1}\\left(x\\right)$ }=\\frac{1}{\\sqrt{x^{2}+1}}\\),\r\nwe use the chain rule with \\(u=4x.\\) So\r\n\\[\\begin{align*}\r\n\\frac{du}{dx} & =4\r\n\\end{align*}\\]\r\nand\r\n\\[\\begin{align*}\r\n\\frac{d}{dx} & \\sinh^{-1}\\left(4x\\right)=\\\\\r\n\\frac{d}{dx}\\sinh^{-1}\\left(4x\\right) & =\\frac{d}{du}\\sinh^{-1}\\left(u\\right)\\frac{du}{dx}\\\\\r\n& =\\frac{1}{\\sqrt{u^{2}+1}}\\cdot4\\\\\r\n& =\\frac{4}{\\sqrt{\\left(4x\\right)^{2}+1}}\\\\\r\n& =\\frac{4}{\\sqrt{16x^{2}+1}}.\r\n\\end{align*}\\]\r\nAlternative solution\r\nUsing the definition of sinh\\(^{-1}\\left(x\\right)\\) in terms of the\r\nnatural log function,\r\n\\[\\begin{align*}\r\n\\sinh^{-1}\\left(4x\\right) & =\\ln\\left(4x+\\sqrt{\\left(4x\\right)^{2}+1}\\right)\\\\\r\n& =\\ln\\left(4x+\\sqrt{16x^{2}+1}\\right).\r\n\\end{align*}\\]\r\nSetting \\(u=4x+\\sqrt{16x^{2}+1}\\),\r\n\\[\\begin{align*}\r\n\\frac{du}{dx} & =4+\\frac{1}{2}\\left(16x^{2}+1\\right)^{-\\frac{1}{2}}\\left(32x\\right)\\\\\r\n& =4+16x\\left(16x^{2}+1\\right)^{-\\frac{1}{2}}.\r\n\\end{align*}\\]\r\nUsing the chain rule,\r\n\\[\\begin{align*}\r\n\\frac{d}{dx}\\left(\\sinh^{-1}\\left(4x\\right)\\right) & =\\frac{d}{dx}\\left(\\ln\\left(4x+\\sqrt{16x^{2}+1}\\right)\\right)\\\\\r\n& =\\frac{d}{du}\\left(\\ln\\left(u\\right)\\right)\\frac{du}{dx}\\\\\r\n& =\\frac{1}{u}\\left(4+16x\\left(16x^{2}+1\\right)^{-\\frac{1}{2}}\\right)\\\\\r\n& =\\frac{4+16x\\left(16x^{2}+1\\right)^{-\\frac{1}{2}}}{4x+\\sqrt{16x^{2}+1}}.\r\n\\end{align*}\\]\r\nNow, rationalise the denominator by multiplying the top and bottom\r\nby \\(4x-\\sqrt{16x^{2}+1}\\),\r\n\\[\\begin{align*}\r\n\\frac{d}{dx}\\left(\\sinh^{-1}\\left(4x\\right)\\right) & =\\frac{4+16x\\left(16x^{2}+1\\right)^{-\\frac{1}{2}}}{4x+\\sqrt{16x^{2}+1}}\\times\\frac{4x-\\sqrt{16x^{2}+1}}{4x-\\sqrt{16x^{2}+1}}\\\\\r\n& =\\frac{16x-4\\sqrt{16x^{2}+1}+64x^{2}\\left(16x^{2}+1\\right)^{-\\frac{1}{2}}-16x}{16x^{2}-16x^{2}-1}\\\\\r\n& =4\\sqrt{16x^{2}+1}-64x^{2}\\left(16x^{2}+1\\right)^{-\\frac{1}{2}}\\\\\r\n& =4\\left(\\sqrt{16x^{2}+1}-\\frac{16x^{2}}{\\sqrt{16x^{2}+1}}\\right)\\\\\r\n& =4\\left(\\frac{16x^{2}+1}{\\sqrt{16x^{2}+1}}-\\frac{16x^{2}}{\\sqrt{16x^{2}+1}}\\right)\\\\\r\n& =\\frac{4}{\\sqrt{16x^{2}+1}}.\r\n\\end{align*}\\]\r\nExercises\r\n\r\n \tFind the derivative, with respect to \\(x,\\) of\r\n\\(\\quad\\)a) \\(y=6\\cosh^{-1}\\left(x/3\\right)\\)\r\n\\(\\quad\\)b) \\(y=\\frac{1}{2}\\sinh^{-1}\\left(2x+1\\right)\\).\r\n \tExpress \\(\\sinh^{-1}\\left(-\\frac{5}{12}\\right)\\) in terms\r\nof the natural log function.\r\n\r\n\r\n\r\n","n":0.058},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":109,"$":[{"v":"Derivatives and integrals of hyperbolic functions","n":0.408},{"v":"\r\nThe differentiation and integration of hyperbolic functions allow us to understand how these functions change and accumulate values which teaches us the essential tools for analysing complex mathematical and real-world systems. Keep reading to learn about these functions and put your new knowledge to the test with some exercises.\r\n\r\nSee Hyperbolic functions for a list of definitions.\r\n\r\n\r\nThe derivatives of the hyperbolic functions may be found using their\r\ndefinitions.\r\n\r\nFor \\(\\sinh,\\) let \\(a\\) be a constant, then:\r\n\\[\\begin{align*}\r\n\\frac{d}{dx}\\sinh\\left(ax\\right) & =\\frac{d}{dx}\\left(\\frac{e^{ax}-e^{-ax}}{2}\\right)\\\\\r\n& =\\frac{ae^{ax}+ae^{-ax}}{2}\\\\\r\n& =a\\cosh\\left(ax\\right).\r\n\\end{align*}\\]\r\nFor cosh,\r\n\\[\\begin{align*}\r\n\\frac{d}{dx}\\cosh\\left(ax\\right) & =\\frac{d}{dx}\\left(\\frac{e^{ax}+e^{-ax}}{2}\\right)\\\\\r\n& =\\frac{ae^{ax}-ae^{-ax}}{2}\\\\\r\n& =a\\sinh\\left(ax\\right).\r\n\\end{align*}\\]\r\n\r\nThe quotient, product and chain rules can be applied to functions\r\ninvolving hyperbolic functions. For example, using the quotient rule,\r\n\\[\\begin{align*}\r\n\\frac{d}{dx}\\tanh\\left(ax\\right) & =\\frac{d}{dx}\\left(\\frac{\\sinh\\left(ax\\right)}{\\cosh\\left(ax\\right)}\\right)\\\\\r\n& =\\frac{d}{dx}\\left(\\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}\\right)\\\\\r\n& =\\frac{\\left(e^{ax}+e^{-ax}\\right)\\frac{d}{dx}\\left(e^{ax}-e^{-ax}\\right)-\\left(e^{ax}-e^{-ax}\\right)\\frac{d}{dx}\\left(e^{ax}+e^{-ax}\\right)}{\\left(e^{ax}+e^{-ax}\\right)^{2}}\\\\\r\n& =\\frac{\\left(e^{ax}+e^{-ax}\\right)a\\left(e^{ax}+e^{-ax}\\right)-a\\left(e^{ax}-e^{-ax}\\right)\\left(e^{ax}-e^{-ax}\\right)}{\\left(e^{ax}+e^{-ax}\\right)^{2}}\\\\\r\n& =\\frac{a\\left(e^{ax}+e^{-ax}\\right)^{2}-a\\left(e^{ax}-e^{-ax}\\right)^{2}}{\\left(e^{ax}+e^{-ax}\\right)^{2}}\\\\\r\n& =a-a\\frac{\\left(e^{ax}-e^{-ax}\\right)^{2}}{\\left(e^{ax}+e^{-ax}\\right)^{2}}\\\\\r\n& =a\\left(1-\\tanh^{2}\\left(ax\\right)\\right)\\\\\r\n& =a\\text{sech$^{2}\\left(ax\\right).$ }\r\n\\end{align*}\\]\r\n\r\nIn summary,\r\n\\[\\begin{align*}\r\n\\frac{d}{dx}\\sinh\\left(ax\\right) & =a\\cosh\\left(ax\\right)\\\\\r\n\\frac{d}{dx}\\cosh\\left(ax\\right) & =a\\sinh\\left(ax\\right)\\\\\r\n\\frac{d}{dx}\\tanh\\left(ax\\right) & =a\\text{sech$^{2}$ $\\left(ax\\right)$ .}\r\n\\end{align*}\\]\r\nExample 1\r\nFind the derivative, with respect to \\(x\\), of \\(\\cosh\\left(x^{2}+3x\\right).\\)\r\nSolution\r\nLet \\(u=x^{2}+3x\\) then \\(du/dx=2x+3\\) and by the chain rule,\r\n\\[\\begin{align*}\r\n\\frac{d}{dx}\\cosh\\left(x^{2}+3x\\right) & =\\frac{d}{du}\\cosh\\left(u\\right)\\frac{du}{dx}\\\\\r\n& =\\sinh\\left(u\\right)\\left(2x+3\\right)\\\\\r\n& =\\left(2x+3\\right)\\sinh\\left(x^{2}+3x\\right).\r\n\\end{align*}\\]\r\nHence the derivative, with respect to \\(x,\\) of \\(\\cosh\\left(x^{2}+3x\\right)\\)\r\nis \\(\\left(2x+3\\right)\\sinh\\left(x^{2}+3x\\right)\\).\r\nExample 2\r\nFind the approximate slope of the tangent to \\(y=\\sinh\\left(4x\\right)\\)\r\nat \\(x=0.5\\) to three decimal places.\r\nSolution\r\nDifferentiating,\r\n\\[\\begin{align*}\r\n\\frac{dy}{dx} & =4\\cosh\\left(4x\\right).\r\n\\end{align*}\\]\r\nAt \\(x=0.5,\\)\r\n\\[\\begin{align*}\r\n\\left.\\frac{dy}{dx}\\right|_{x=0.5} & =4\\cosh\\left(4\\left(0.5\\right)\\right)\\\\\r\n& =4\\cosh\\left(2\\right)\\\\\r\n& =4\\left(\\frac{e^{2}+e^{-2}}{2}\\right)\\\\\r\n& =2\\left(e^{2}+\\left(\\frac{1}{e^{2}}\\right)\\right)\\\\\r\n& \\approx15.049\\,.\r\n\\end{align*}\\]\r\nThe slope of the tangent is approximately \\(15.049\\) at \\(x=0.5\\).\r\nIntegrals of hyperbolic functions\r\nLet \\(a\\) be a constant. Using the definitions,\r\n\\[\\begin{align*}\r\n\\int\\sinh\\left(ax\\right)dx & =\\int\\left(\\frac{e^{ax}-e^{-ax}}{2}\\right)dx\\\\\r\n& =\\frac{\\frac{1}{a}e^{ax}+\\frac{1}{a}e^{-ax}}{2}+c,\\quad c\\in\\mathbb{R}\\\\\r\n& =\\frac{1}{a}\\cosh\\left(ax\\right)+c.\r\n\\end{align*}\\]\r\nSimilarly,\r\n\\[\\begin{align*}\r\n\\int\\cosh\\left(ax\\right)dx & =\\int\\left(\\frac{e^{ax}+e^{-ax}}{2}\\right)dx\\\\\r\n& =\\frac{\\frac{1}{a}e^{ax}-\\frac{1}{a}e^{-ax}}{2}+c,\\quad c\\in\\mathbb{R}\\\\\r\n& =\\frac{1}{a}\\sinh\\left(ax\\right)+c.\r\n\\end{align*}\\]\r\nFor the integral of \\(\\tanh,\\) we use integration by substitution.\r\nLet \\(u=\\cosh\\left(ax\\right),\\) then \\(du/dx=a\\,\\sinh\\left(x\\right)\\)\r\nand\r\n\\[\\begin{align*}\r\n\\int\\tanh\\left(ax\\right)dx & =\\int\\frac{\\sinh\\left(ax\\right)}{\\cosh\\left(ax\\right)}dx\\\\\r\n& =\\frac{1}{a}\\int\\frac{1}{\\cosh\\left(ax\\right)}\\frac{du}{dx}dx\\\\\r\n& =\\frac{1}{a}\\int\\frac{1}{u}du\\\\\r\n& =\\frac{1}{a}\\ln\\left(\\cosh\\left(ax\\right)\\right)+c,\\quad c\\in\\mathbb{R}.\r\n\\end{align*}\\]\r\n\r\nSummarising\r\n\\[\\begin{align*}\r\n\\int\\sinh\\left(ax\\right)dx & =\\frac{1}{a}\\cosh\\left(ax\\right)+c\\\\\r\n\\int\\cosh\\left(ax\\right)dx & =\\frac{1}{a}\\sinh\\left(ax\\right)+c\\\\\r\n\\int\\tanh\\left(ax\\right)dx & =\\frac{1}{a}\\ln\\left(\\cosh\\left(ax\\right)\\right)+c.\r\n\\end{align*}\\]\r\nExample 3\r\nFind \\(\\int\\cosh\\left(3x\\right)dx.\\)\r\nSolution\r\n\\[\\begin{align*}\r\n\\int\\cosh\\left(3x\\right) & =\\frac{1}{3}\\sinh\\left(3x\\right)+c\r\n\\end{align*}\\]\r\nwhere \\(c\\) is a constant.\r\nExample 4\r\nFind the value of \\(\\int_{0}^{\\ln\\left(2\\right)}\\sinh\\left(x\\right)dx.\\)\r\nSolution\r\n\\[\\begin{align*}\r\n\\int_{0}^{\\ln\\left(2\\right)}\\sinh\\left(x\\right)dx & =\\left[\\cosh\\left(x\\right)\\right]_{x=0}^{x=\\ln\\left(2\\right)}\\\\\r\n& =\\cosh\\left(\\ln\\left(2\\right)\\right)-\\cosh\\left(0\\right)\\\\\r\n& =\\frac{1}{2}\\left(e^{\\ln\\left(2\\right)}+e^{-\\ln\\left(2\\right)}\\right)-\\frac{1}{2}\\left(e^{0}+e^{-0}\\right)\\\\\r\n& =\\frac{1}{2}\\left(e^{\\ln\\left(2\\right)}+e^{\\ln\\left(\\frac{1}{2}\\right)}\\right)-1\\\\\r\n& =\\frac{1}{2}\\left(2+\\frac{1}{2}\\right)-1\\\\\r\n& =\\frac{1}{2}\\left(\\frac{5}{2}\\right)-1\\\\\r\n& =\\frac{1}{4}.\r\n\\end{align*}\\]\r\nExample 5\r\nFind \\(\\int\\left(12x^{3}-2\\right)\\tanh\\left(3x^{4}-2x\\right)dx.\\)\r\nSolution\r\nWe use the substitution method. Let\r\n\\[\\begin{align*}\r\nu & =3x^{4}-2x\r\n\\end{align*}\\]\r\nthen\r\n\\[\\begin{align*}\r\n\\frac{du}{dx} & =12x^{3}-2.\r\n\\end{align*}\\]\r\nNow, the integral may be written\r\n\\[\\begin{align*}\r\n\\int\\left(12x^{3}-2\\right)\\tanh\\left(3x^{4}-2x\\right)dx & =\\int\\frac{du}{dx}\\tanh\\left(u\\right)dx\\\\\r\n& =\\int\\tanh\\left(u\\right)du\\\\\r\n& =\\ln\\left(\\cosh\\left(u\\right)\\right)+c'\\,,\\ \\text{where $c'\\ \\text{is a constant.}$ }\\\\\r\n& =\\ln\\left(\\cosh\\left(3x^{4}-2x\\right)\\right)+c\\,,\\ \\text{where $c\\ \\text{is a constant.}$ }\r\n\\end{align*}\\]\r\nHence\r\n\\[\\begin{align*}\r\n\\int\\left(12x^{3}-2\\right)\\tanh\\left(3x^{4}-2x\\right)dx & =\\ln\\left(\\cosh\\left(3x^{4}-2x\\right)\\right)+c\\,,\\ \\text{where $c\\ \\text{is a constant.}$ }\r\n\\end{align*}\\]\r\nExercises\r\n\r\n \tFind the derivative, with respect to \\(x,\\) of\r\n\\(\\quad\\)a) \\(y=6\\cosh\\left(x/3\\right)\\)\r\n\\(\\quad\\)b) \\(y=\\frac{1}{2}\\sinh\\left(2x+1\\right)\\)\r\n \tEvaluate:\r\n\\(\\quad\\)a) \\(\\int\\cosh\\left(3x\\right)dx\\)\r\n\\(\\quad\\)b) \\(\\int_{1}^{2}\\frac{\\cosh\\left(\\ln\\left(t\\right)\\right)}{t}dt.\\)\r\n\r\n\r\n\r\n","n":0.058},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":110,"$":[{"v":"Hyperbolic functions","n":0.707},{"v":"\r\nThe hyperbolic functions are analogous to the circular (trigonometric) functions and are widely used in engineering, science and mathematics.\r\nThis module introduces hyperbolic functions, their graphs and similarities to the circular functions.\r\n\r\nWhereas circular functions are defined on a unit circle, the hyperbolic functions are defined on a hyperbola.\r\nHyperbolic functions are used to describe a cable or chain that is suspended at its end points.\r\n\r\nFor example, these functions can be used to describe the curve adopted by electrical supply lines.\r\nDefinitions\r\nThe basic hyperbolic functions are sinh (pronounced \"shine\") and cosh.\r\n\r\nThe hyperbolic sine function is\r\n\r\n\\[\\begin{align*}\r\n\\sinh\\left(x\\right) & =\\frac{e^{x}-e^{-x}}{2}.\r\n\\end{align*}\\]\r\n\r\nThe hyperbolic cosine function is defined as\r\n\\[\\begin{align*}\r\n\\cosh\\left(x\\right) & =\\frac{e^{x}+e^{-x}}{2}.\r\n\\end{align*}\\]\r\n\r\nIn addition to these we also define:\r\n\\[\\begin{align*}\r\n\\tanh\\left(x\\right) & =\\frac{\\sinh\\left(x\\right)}{\\cosh\\left(x\\right)}\\\\\r\n& =\\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}.\r\n\\end{align*}\\]\r\n\r\nTanh is pronounced \"than\" like the beginning of \"thank\".\r\n\r\nJust as for the circular functions, there are reciprocal hyperbolic\r\nfunctions. They are:\r\n\r\n\\[\\begin{align*}\r\n\\text{$\\text{coshec$\\left(x\\right)$ }$ } & =\\frac{1}{\\sinh\\left(x\\right)}\\\\\r\n\\text{sech$\\left(x\\right)$ } & =\\frac{1}{\\cosh\\left(x\\right)}\\\\\r\n\\coth\\left(x\\right) & =\\frac{1}{\\tanh\\left(x\\right)}.\r\n\\end{align*}\\]\r\nExample 1\r\nWhat is the exact value of \\(\\sinh\\left(0\\right)?\\)\r\nSolution\r\nBy definition,\r\n\\[\\begin{align*}\r\n\\sinh\\left(0\\right) & =\\frac{e^{0}-e^{-0}}{2}\\\\\r\n& =\\frac{1-1}{2}\\\\\r\n& =0.\r\n\\end{align*}\\]\r\n\r\nHence the value of \\(\\sinh\\left(0\\right)=0.\\)\r\nExample 2\r\nWhat is the exact value of \\(\\cosh\\left(\\ln\\left(2\\right)\\right)\\)?\r\nSolution\r\nBy definition,\r\n\\[\\begin{align*}\r\n\\cosh\\left(\\ln\\left(2\\right)\\right) & =\\frac{1}{2}\\left(e^{\\ln\\left(2\\right)}+e^{-\\ln\\left(2\\right)}\\right)\\\\\r\n& =\\frac{1}{2}\\left(e^{\\ln\\left(2\\right)}+e^{\\ln\\left(2^{-1}\\right)}\\right)\\\\\r\n& =\\frac{1}{2}\\left(e^{\\ln\\left(2\\right)}+e^{\\ln\\left(\\frac{1}{2}\\right)}\\right)\\\\\r\n& =\\frac{1}{2}\\left(2+\\frac{1}{2}\\right)\\\\\r\n& =\\frac{1}{2}\\cdot\\frac{5}{2}\\\\\r\n& =\\frac{5}{4}.\r\n\\end{align*}\\]\r\n\r\nHence \\(\\cosh\\left(\\ln\\left(2\\right)\\right)=5/4.\\)\r\nExample 3\r\nSolve for \\(x,\\) \\(\\sinh\\left(x\\right)=3/4\\).\r\nSolution\r\nUsing the definition of \\(\\sinh,\\)\r\n\\[\\begin{align*}\r\n\\frac{1}{2}\\left(e^{x}-e^{-x}\\right) & =\\frac{3}{4}\\\\\r\ne^{x}-e^{-x} & =\\frac{3}{2}.\r\n\\end{align*}\\]\r\nMultiplying both sides by \\(2e^{x}\\) and rearranging,\r\n\\[\\begin{align*}\r\n2e^{2x}-2 & =3e^{x}\\\\\r\n2e^{2x}-3e^{x}-2 & =0\\\\\r\n2\\left(e^{x}\\right)^{2}-3e^{x}-2 & =0,\r\n\\end{align*}\\]\r\nwhich is a quadratic in \\(e^{x}.\\) Using the quadratic formula1 For a quadratic\r\n\\[\\begin{align*}\r\nax^{2}+bx+c & =0\r\n\\end{align*}\\]\r\nwhere \\(a,\\) \\(b\\) and \\(c\\) are constants,\r\n\\[\\begin{align*}\r\nx & =\\frac{-b\\pm\\sqrt{b^{2}-4ac}}{2a}.\r\n\\end{align*}\\]\r\n\\[\\begin{align*}\r\ne^{x} & =\\frac{3\\pm\\sqrt{\\left(-3\\right)^{2}-4\\left(2\\right)\\left(-2\\right)}}{2\\left(2\\right)}\\\\\r\n& =\\frac{3\\pm\\sqrt{9+16}}{4}\\\\\r\n& =\\frac{3\\pm5}{4}.\r\n\\end{align*}\\]\r\nSince \\(e^{x}>0\\) for all \\(x,\\) we ignore the negative sign and\r\n\\[\\begin{align*}\r\ne^{x} & =\\frac{3+5}{4}\\\\\r\n& =2.\r\n\\end{align*}\\]\r\nTaking logs of both sides gives \\(x=\\ln\\left(2\\right).\\) Hence, the\r\nrequired value of \\(x=\\ln\\left(2\\right).\\)\r\nGraphs of hyperbolic functions\r\nThe graphs of \\(\\sinh\\left(x\\right),\\) \\(\\cosh\\left(x\\right)\\) and \\(\\tanh\\left(x\\right)\\)\r\nare shown below in red, blue and green respectively.\r\n\r\n\r\n\r\nThe domain for each function is \\(\\mathbb{R}\\). The range of \\(\\sinh\\)\r\nis \\(\\mathbb{R}\\) while the range of \\(\\cosh\\) is \\(\\left[\\left.1,\\infty\\right)\\right..\\)\r\nThe graph of \\(\\tanh\\left(x\\right)\\) has asymptotes at \\(y=\\pm1\\) so\r\n\\(\\tanh\\) has range \\(\\left(-1,1\\right).\\)\r\n\r\nNote that\r\n\\[\\begin{align*}\r\n\\sinh\\left(-x\\right) & =-\\sinh\\left(x\\right)\\\\\r\n\\cosh\\left(-x\\right) & =\\cosh\\left(x\\right),\r\n\\end{align*}\\]\r\nthat is, the functions \\(\\sinh\\) and \\(\\cosh\\) are odd and even functions,\r\nrespectively.\r\n\r\nNote also that the function \\(\\cosh\\left(x\\right)\\) is not one to one.\r\nHyperbolic identities\r\nThe hyperbolic functions have identities that are similar, though\r\nnot the same, as circular functions.\r\n\r\nThe most important are:\r\n\\[\\begin{align*}\r\n\\cosh^{2}\\left(x\\right)-\\sinh^{2}\\left(x\\right) & =1\\\\\r\n1-\\tanh^{2}\\left(x\\right) & =\\text{sech$^{2}\\left(x\\right)$ }\\\\\r\n\\sinh\\left(x+y\\right) & =\\sinh\\left(x\\right)\\cosh\\left(y\\right)+\\cosh\\left(x\\right)\\sinh\\left(y\\right)\\\\\r\n\\cosh\\left(x+y\\right) & =\\cosh\\left(x\\right)\\cosh\\left(y\\right)+\\sinh\\left(x\\right)\\sinh\\left(y\\right)\\\\\r\n\\sinh\\left(2x\\right) & =2\\sinh\\left(x\\right)\\cosh\\left(x\\right).\r\n\\end{align*}\\]\r\nNote that these are similar to the trigonometric identities:\r\n\\[\\begin{align*}\r\n\\cos^{2}\\left(x\\right)+\\sin^{2}\\left(x\\right) & =1\\\\\r\n1+\\tan^{2}\\left(x\\right) & =\\text{$\\sec^{2}$ $\\left(x\\right)$ }\\\\\r\n\\sin\\left(x+y\\right) & =\\sin\\left(x\\right)\\cos\\left(y\\right)+\\cos\\left(x\\right)\\sin\\left(y\\right)\\\\\r\n\\cos\\left(x+y\\right) & =\\cos\\left(x\\right)\\cos\\left(y\\right)-\\sin\\left(x\\right)\\sin\\left(y\\right)\\\\\r\n\\sin\\left(2x\\right) & =2\\sin\\left(x\\right)\\cos\\left(x\\right).\r\n\\end{align*}\\]\r\nOther identities may be derived from these. For example,\r\n\\[\\begin{align*}\r\n\\cosh\\left(2x\\right) & =\\cosh\\left(x+x\\right)\\\\\r\n& =\\cosh\\left(x\\right)\\cosh\\left(x\\right)+\\sinh\\left(x\\right)\\sinh\\left(x\\right)\\\\\r\n& =\\cosh^{2}\\left(x\\right)+\\sinh^{2}\\left(x\\right)\\\\\r\n& =1+\\sinh^{2}\\left(x\\right)+\\sinh^{2}\\left(x\\right)\\\\\r\n& =1+2\\sinh^{2}\\left(x\\right).\r\n\\end{align*}\\]\r\nExample 4\r\nIf \\(\\sinh\\left(x\\right)=3/5,\\) what is the value of \\(\\cosh\\left(x\\right)?\\)\r\nSolution\r\nUsing the identity\r\n\\[\\begin{align*}\r\n\\cosh^{2}\\left(x\\right)-\\sinh^{2}\\left(x\\right) & =1\r\n\\end{align*}\\]\r\nwe have\r\n\\[\\begin{align*}\r\n\\cosh^{2}\\left(x\\right) & =1+\\sinh^{2}\\left(x\\right)\\\\\r\n& =1+\\left(\\frac{3}{5}\\right)^{2}\\\\\r\n& =\\frac{34}{25}\\\\\r\n\\cosh\\left(x\\right) & =\\pm\\sqrt{\\frac{34}{25}}\\\\\r\n& =\\frac{\\sqrt{34}}{5}.\r\n\\end{align*}\\]\r\nNote that we reject \\(-\\sqrt{34}/5\\) as the range of \\(\\cosh\\) is \\(\\left[\\left.1,\\infty\\right)\\right..\\)\r\nExample 5\r\nIf \\(\\cosh\\left(x\\right)=5/3,\\) what is the value of \\(\\sinh\\left(x\\right)\\)?\r\nSolution\r\nUsing the identity\r\n\\[\\begin{align*}\r\n\\cosh^{2}\\left(x\\right)-\\sinh^{2}\\left(x\\right) & =1\r\n\\end{align*}\\]\r\nwe have\r\n\\[\\begin{align*}\r\n\\sinh^{2}\\left(x\\right) & =\\cosh^{2}\\left(x\\right)-1\\\\\r\n& =\\left(\\frac{5}{3}\\right)^{2}-1\\\\\r\n& =\\frac{25}{9}-1\\\\\r\n& =\\frac{16}{9}\\\\\r\n\\sinh\\left(x\\right) & =\\pm\\sqrt{\\frac{16}{9}}\\\\\r\n& =\\pm\\frac{4}{3}.\r\n\\end{align*}\\]\r\nNote that there are two solutions in this case because the range of\r\n\\(\\sinh\\) is \\(\\left(-\\infty,\\infty\\right)\\) and \\(\\cosh\\left(x\\right)\\)\r\nis not one to one.\r\nExample 6\r\nIf \\(\\sinh\\left(x\\right)=3/4,\\) what is the exact value of \\(x?\\)\r\nSolution\r\nFrom the definition of \\(\\sinh\\left(x\\right)\\)\r\n\\[\\begin{align*}\r\n\\frac{3}{4} & =\\frac{1}{2}\\left(e^{x}-e^{-x}\\right).\r\n\\end{align*}\\]\r\nMultiplying both sides by 4 and rearranging,\r\n\\[\\begin{align*}\r\n2e^{x}-3-2e^{-x} & =0.\r\n\\end{align*}\\]\r\nMultiplying both sides2 No problems arise from this as \\(e^{x}>0\\) for all \\(x.\\)\r\nby \\(e^{x},\\)\r\n\\[\\begin{align*}\r\n2e^{2x}-3e^{x}-2 & =0.\r\n\\end{align*}\\]\r\nThis is a quadratic in \\(e^{x}\\) and may be factorised to give\r\n\\[\\begin{align*}\r\n\\left(2e^{x}+1\\right)\\left(e^{x}-2\\right) & =0.\r\n\\end{align*}\\]\r\nThe first bracket is never zero as \\(e^{x}>0\\) for all \\(x.\\) Hence\r\n\\[\\begin{align*}\r\ne^{x}-2 & =0\\\\\r\ne^{x} & =2\\\\\r\nx & =\\ln\\left(2\\right).\r\n\\end{align*}\\]\r\n\r\nThe value of \\(x\\) is \\(\\ln\\left(2\\right).\\)\r\nExample 7\r\nShow that \\(\\sinh\\left(2x\\right)=2\\sinh\\left(x\\right)\\cosh\\left(x\\right).\\)\r\nSolution\r\nIn this solution, we use the difference of two squares formula3 Difference of two squares formula is\r\n\\[\\begin{align*}\r\na^{2}-b^{2} & =\\left(a-b\\right)\\left(a+b\\right).\r\n\\end{align*}\\]. Using the definition of \\(\\sinh,\\) the left hand side may be written\r\nas\r\n\\[\\begin{align*}\r\n\\sinh\\left(2x\\right) & =\\frac{1}{2}\\left(e^{2x}-e^{-2x}\\right)\\\\\r\n& =\\frac{1}{2}\\left(\\left(e^{x}\\right)^{2}-\\left(e^{-x}\\right)^{2}\\right)\\\\\r\n& =\\frac{1}{2}\\left(e^{x}-e^{-x}\\right)\\left(e^{x}+e^{-x}\\right)\\\\\r\n& =\\sinh\\left(x\\right)\\left(e^{x}+e^{-x}\\right)\\\\\r\n& =2\\sinh\\left(x\\right)\\frac{\\left(e^{x}+e^{-x}\\right)}{2}\\\\\r\n& =2\\sinh\\left(x\\right)\\cosh\\left(x\\right)\r\n\\end{align*}\\]\r\nas required.\r\nExample 8\r\nGiven that \\(\\tanh\\left(x\\right)=4/5,\\) find the value of all the other\r\nhyperbolic functions: \\(\\sinh,\\) \\(\\cosh,\\) \\(\\ \\text{sech, coshec and coth.}\\)\r\nSolution\r\nUsing the identity \\(1-\\tanh^{2}\\left(x\\right)=\\text{sech$^{2}\\left(x\\right)$ }\\),\r\n\\[\\begin{align*}\r\n\\text{sech$\\left(x\\right)$ } & =\\sqrt{1-\\left(\\frac{4}{5}\\right)^{2}}\\\\\r\n& =\\sqrt{\\frac{9}{25}}\\\\\r\n& =\\frac{3}{5}\r\n\\end{align*}\\]\r\nand\r\n\\[\\begin{align*}\r\n\\cosh\\left(x\\right) & =\\frac{5}{3}.\r\n\\end{align*}\\]\r\n\r\nUsing the identity \\(\\cosh^{2}\\left(x\\right)-\\sinh^{2}\\left(x\\right)=1\\),\r\n\\[\\begin{align*}\r\n\\sinh\\left(x\\right) & =\\sqrt{\\left(\\frac{5}{3}\\right)^{2}-1}\\\\\r\n& =\\sqrt{\\frac{16}{9}}\\\\\r\n& =\\frac{4}{3}\r\n\\end{align*}\\]\r\nand\r\n\\[\\begin{align*}\r\n\\text{cosech$\\left(x\\right)$ } & =\\frac{3}{4}.\r\n\\end{align*}\\]\r\n\r\nNote that we could also use\r\n\\[\\begin{align*}\r\n\\tanh\\left(x\\right) & =\\frac{\\sinh\\left(x\\right)}{\\cosh\\left(x\\right)}\\\\\r\n\\sinh\\left(x\\right) & =\\tanh\\left(x\\right)\\cosh\\left(x\\right)\\\\\r\n& =\\frac{4}{5}\\times\\frac{5}{3}\\\\\r\n& =\\frac{4}{3}\r\n\\end{align*}\\]\r\nto get \\(\\sinh\\left(x\\right)\\) and hence cosech\\(\\left(x\\right).\\)\r\n\r\nFinally, given that \\(\\tanh\\left(x\\right)=4/5\\),\r\n\\[\\begin{align*}\r\n\\text{coth$\\left(x\\right)$ } & =\\frac{5}{4}.\r\n\\end{align*}\\]\r\nExercises\r\n\\(1\\). Simplify\r\n\\(\\quad\\)a) \\(2\\cosh\\left(\\ln\\left(x\\right)\\right)\\)\r\n\\(\\quad\\)b) \\(\\sinh\\left(2\\ln\\left(x\\right)\\right)\\).\r\n\r\n\\(2\\). Using the definitions of \\(\\cosh\\) and \\(\\sinh,\\) in\r\nterms of exponential functions, show that\r\n\\[\\begin{align*}\r\n\\cosh^{2}\\left(x\\right)-\\sinh^{2}\\left(x\\right) & =1.\r\n\\end{align*}\\]\r\n\r\n\\(3\\). If \\(\\sinh\\left(x\\right)=3/4,\\) what is the value of\r\n\\(\\tanh\\left(x\\right)\\)?\r\n\r\n\r\n\r\n","n":0.039},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":111,"$":[{"v":"Hyperbolic functions","n":0.707},{"v":"Information about functions and graphs to improve your maths skills in these areas.\r\n\r\n\r\n \tHF1 Hyperbolic functions\r\nThis module introduces hyperbolic functions, their graphs and similarities to the circular functions.\r\n \tHF2 Derivatives and integrals of hyperbolic functions\r\nThe hyperbolic functions are widely used in engineering, science and mathematics.\r\n \tHF3 Inverse hyperbolic functions\r\nProviding a function is one to one, it is possible to find an inverse function.\r\n","n":0.126},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":112,"$":[{"v":"Artist statement: video","n":0.577},{"v":"A good artist statement communicates ideas with clarity. You don't have to follow a formal structure, and there is no right or wrong way to write your statement.\r\nThe examples from professional artists and designers in this section are all different in length, format, communication style, structure, and use of vocabulary.\r\nThe statements presented here were written as part of a creative or research project, or as part of a presentation, exhibition, assessment, or documentation of creative work.\r\n\r\nView the videos, read these statements, analyse their content and structure, to improve your understanding of how to write an engaging and effective artist statement.\r\nAll videos have auto-generated closed captions and a transcript of the spoken statement.\r\nElizabeth Gower\r\nhe loves me, he loves me not, 2016, RMIT Gallery\r\n\r\n\r\nParul Thaker\r\nUnfolding: New Indian Textiles 2015, RMIT Gallery\r\n\r\n\r\nKhaled Sabsabi\r\nExperimenta Recharge 2015, RMIT Gallery\r\n\r\n\r\nAnasia Franco\r\nExperimenta Recharge 2015, RMIT Gallery\r\n \r\n\u2018he loves me he loves me not\u2019 is written 21,319 times which represents the number of days I have lived and asked this question.\r\n\r\nThe number is calculated to symbolically represent a lifetime of re-evaluation and wavering, resilience and resolve. The phrase is written on behalf of women across cultures who are conditioned to seek approval, permission and sanction from the generic he.\r\n\r\n\u2018he loves me he loves me not\u2019 can also have more negative connotations that signify the trauma of domestic instability and violence. The monumental scale of the installation and the endurance of the task amplifies the futility of continually asking the question.\r\n This work came from me observing my grandmother for years, and one fine morning it was just an idea to photograph her hands because her hands were completely, kind of, they became one with the material and the material became one with her hands.\r\nShe had this continuous movement. It was it was like I was, I will see the in-between energies which were surrounding her hands.\r\nMohammed has been a very important influence; she\u2019s been everything to me in my life and I have grown up with her. It\u2019s been a journey with her, however where she\u2019s been working with textiles and with yarns and with threads all her life, for me it was like part of growing up in that environment.\r\nI work with her fibers \u2013 textiles, which are back in my country, which the craftsmen and weavers have been working for years. The techniques I use kind of, they\u2019re invented by us, so they don\u2019t exist in in like history.\r\nAnd it\u2019s not about the textile, it\u2019s about the language. It\u2019s about an abstract language and trying to communicate through the works, of me wanting to search within the answers which I never could resolve in day-to-day living my life so it was more about me wanting to go within \u2013 wanting to ask questions about living, about life, feelings, the physicality, the tactility of the materials, which we are just not aware, which we\u2019ve lost in our day-to-day livings.\r\nSo yes there is a so the process of me doing these works which made me search, you know, different ways which made me engage with textiles in a very different approach. The work \u201870,000 veils\u2019 has taken seven years to make. Of course, initially I didn\u2019t have an idea what the work will look like or what will eventuate in that time. The work in terms of the process, consists of ten thousand still images taken from my travels of four or five years and then these images are taken and then deconstructed layer for layer to form ten thousand different layers, and then those layers are reconfigured and composited again to form 70,000 veils then they\u2019re animated.\r\nWhat inspires the work: R\u016bm\u012b, Jal\u0101l ad-D\u012bn R\u016bm\u012b \u2013 known as a poet in the West but he was a mystic, a Sufi mystic said, \u201cbetween the individual and the divine there are 70,000 layers and none between the divine and the individual\u2019, and this inspired me and got me thinking about not just from a metaphysical view but also from physical life. I mean you walk into the work; you can choose not to put on the 3d glasses you can choose to put them on.\r\nFor me that that symbolizes life generally \u2013 you can walk through life and have a rewarding life and you would never have to think about if there is anything out there, or you can cross a threshold, a rite of passage and behind the work is all the wires and the sticky tape and the glue and the bolts and the nuts, and the wood frame and so on. Essentially, it\u2019s the workings behind the universe, held by nails and bolts.\r\n I\u2019m very inspired by psychology and cognitive science and it\u2019s a kind of analytic relationship, psychoanalysis relationship between you and the object. So, I\u2019m showing wave of happiness, frustration and paranoia.\r\nSo, psychosomatic series it\u2019s something that you build in your mind and then you react in your body. The idea of the works is to create relationships with people and objects so the objects are in bed with emotions like the happiness or frustration or paranoia, and generally objects doesn\u2019t have emotions only us, and computers doesn\u2019t have emotions.\r\nSo the idea is to create emotions in the objects that are activated by humans so then you could be able to have a relationship with the object.\r\n","n":0.034},[{"v":"Assignments","i":2,"n":1},{"v":"Art","i":1,"n":1},{"v":"Artist statement","i":0,"n":0.707}]]},{"i":113,"$":[{"v":"Appendices","n":1},{"v":"Do you have information such as raw data or graphs that you think are too long or distracting to put into the body of your assignment? Keep reading to find out how appendices could be used to solve this problem.\r\nAn appendix is a section at the end of an assignment that contains supplementary information which may help your reader to better understand your research. However, this information should not be essential to the meaning of your assignment. In other words, your assignment should be complete without your appendices.\r\nWhat is included in an appendix?\r\nAn appendix is where you put information that would allow your reader to better understand your research, but which is too long or distracting to include in the body of the assignment. Some examples are:\r\n\r\n \traw data including tables and figures\r\n \tgraphs, charts, statistics\r\n \ttechnical figures or drawings\r\n \tsupporting evidence\r\n \tquestionnaires and surveys\r\n \tinterview transcripts\r\n \tfield notes\r\n \tcorrespondence such as letters and emails\r\n \tsample calculations\r\n \tdescriptions of research instruments\r\n \tmaps\r\n \tphotographs.\r\n\r\nOrganising appendices\r\nHere are some considerations for how to organise your appendices:\r\n\r\n \tYou should have a separate appendix for each different theme or set of data.\r\n \tTable and figure numbers are required in the appendices, just as they are in the body of the assignment.\r\n \tYour appendices should be listed in the table of contents.\r\n\r\nStyle and format\r\nHow your appendices are formatted will depend on the style guide you have been asked to use in your assignment instructions. Some common styles are APA 7th Edition, Chicago, IEEE and Harvard. However, here are some general formatting instructions for appendices:\r\n\r\n \tYour appendices should not be overcrowded. Try to make it as easy as possible for the reader to access the information.\r\n \tUse graphs, tables and charts that are easy to understand.\r\n \tYour appendices will come either before or after your references, depending on the style you are using.\r\n \tA heading of \u2018Appendix\u2019 should be used for each section, followed by a number or letter (e.g. Appendix 1 / Appendix C), depending on the style you are using.\r\n \tOrder your appendices in the order that the information appears in the body of your work.\r\n \tEach appendix should begin on a new page.\r\n\r\nTest your knowledge\r\n","n":0.053},[{"v":"Writing structures","i":2,"n":0.707},{"v":"Academic writing","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":114,"$":[{"v":"Receiving peer review feedback","n":0.5},{"v":"Receiving feedback is a valuable part of the learning process. It helps you improve your work, develop critical thinking skills and improve your communication skills. Here are some tips for making the most out of the feedback you receive from a peer.\r\nBe open-minded\r\nApproach feedback with an open mind. Remember that the reviewer is trying to help you improve. Even if the feedback is not what you expected, consider it carefully before dismissing it. One of your reviewer\u2019s comments might spark a new idea for your work.\r\nSeek clarification\r\nIf any part of the feedback is unclear, ask for clarification or extra explanation. This will help you get the most out of the suggestions the reviewer has made. Clarifying misunderstandings can prevent errors in your work.\r\nWhat if you disagree?\r\nIt\u2019s okay to disagree with feedback, but it\u2019s important to do so respectfully. First, consider the reviewer\u2019s points and decide if they have merit. If you still disagree, you can choose not to implement the suggestions.\r\nExpress gratitude\r\nWhether or not you believe you have received sound feedback, thank your reviewer for the time they took to consider your work. This helps create a positive, collaborative learning environment in your class.\r\nDon't take it personally\r\nThe feedback you receive is about your work, not about you as a person. Try to separate your personal feelings from the critique. This will help you view the feedback objectively and use it to enhance your work. \r\n\r\nIf you are not happy with some of the suggestions, avoid responding in a rude or defensive manner. Try to respond professionally. It is likely that they were trying their best, even if you did not like their advice.","n":0.061},[{"v":"Group work","i":2,"n":0.707},{"v":"Understanding assignments","i":1,"n":0.707},{"v":"Assignments","i":0,"n":1}]]},{"i":115,"$":[{"v":"Providing peer review feedback","n":0.5},{"v":"Providing peer feedback is an art that requires sensitivity and thoughtful explanations. Mastering this skill will enable you to give respectful and constructive feedback, which will enhance the learning experience for both you and your peers.\r\n\r\nUse the rubric\r\nThe assignment rubric tells us what the person marking our work is looking for in the assignment. By understanding the rubric, you are more likely to provide useful feedback for your classmates.\r\nIf you find many grammatical or spelling errors that make the work hard to understand, you can point these out. However, you do not need to proofread your peer\u2019s assignment and catch all of these mistakes. Comments relating to the rubric, the structure, and the logic of arguments will be more beneficial.\r\nRespect privacy\r\nBe mindful that the student\u2019s work you review is not to be shared with others or posted online. This could lead to other students using the ideas from the assignment you reviewed, which is considered collusion \u2014 a form of academic misconduct.\r\nBe polite\r\nHaving their work reviewed by a classmate could make some students feel uncomfortable. To minimise this, keep your comments polite and professional.\r\n\r\nPersonal comments or direct criticisms are best avoided.\r\nIllustrate your points\r\nWhen providing feedback, it will be more beneficial to your classmate if you clearly explain your comments and provide details of why you have made certain suggestions. For example, instead of saying, \"This sentence doesn\u2019t make sense,\" let them know what makes it unclear. Is there a grammatical error? Has a word been used incorrectly? Is the sentence too long and complicated? This will help your classmate make improvements to their work.\r\nActivity: Appropriate or inappropriate feedback\r\nRead the following sentences. Decide whether each sentence is either appropriate or inappropriate based on whether you think the feedback has been written in a respectful way.\r\n\r\n","n":0.058},[{"v":"Group work","i":2,"n":0.707},{"v":"Assignments","i":1,"n":1},{"v":"Assessment","i":0,"n":1}]]},{"i":116,"$":[{"v":"Peer review in assignments","n":0.5},{"v":"During your time at university, you may be asked to complete an assignment that includes a peer review stage. This involves providing feedback on other students\u2019 assignments to help them improve their work.\r\n\r\nWhy are peer reviews used?\r\nThese assignments help students develop several important skills. By reviewing your colleagues\u2019 work, you can improve your critical thinking and learn to be an effective communicator, as your feedback should be given in a sensitive and constructive manner.\r\n\r\nPeople can often feel offended when they perceive criticism. By participating in peer reviews within the safe setting of your course, you can learn to accept and integrate suggestions professionally. This helps you understand that feedback is a collaborative approach to learning rather than personal criticism.\r\nHow are peer reviews used in assignments?\r\nThere are a number of ways peer reviews can be used in assignments. Here are the two most common:\r\nPre-submission peer review\r\nIn some assignments, a peer review stage is included before the final submission. This means that after completing a draft, students are expected to provide constructive feedback on their peers\u2019 drafts, highlighting strengths and suggesting areas for improvement.\r\n\r\nOnce a student has received feedback, they then revise their drafts to enhance the quality of their work.\r\n\r\nThis process allows students to improve their work before the final submission and helps them learn from each other, incorporating diverse perspectives.\r\nCollaborative assignments\r\nIn group assignments, peer review can be used to evaluate individual contributions and overall performance within the group. This peer review stage typically occurs during the assignment process, rather than at the end.\r\n\r\nIt can involve:\r\n\r\n \tFeedback on contributions: Students assess the work of their group members, providing constructive comments on their contributions.\r\n \tA performance review: This may also include providing feedback on their performance and collaboration within the group assignment.\r\n\r\nWhat is expected?\r\nIn assignments with a peer review stage, you are expected to engage actively and thoughtfully in the process. This means providing constructive feedback on your peers' work and being open to receiving feedback on your own. The goal is to foster a collaborative learning environment where everyone benefits from shared insights and constructive criticism.\r\nProvide feedback\r\nThis involves reading your peers' assignment thoroughly and assessing it based on the criteria provided by your instructor. You will need to comment on various aspects such as content, structure, and clarity. Your goal is to help your peer improve their work by offering insights and suggestions.\r\nComplete a form / submit feedback\r\nWhile you are reviewing your peers' assignment, you will need to complete a feedback form, which you will often be asked to submit to your class instructor. This form will guide you in structuring your feedback and ensuring it covers all necessary areas.\r\nReceive feedback\r\nThis involves reviewing the comments and suggestions provided by your peer. You will need to read through their feedback carefully to understand their perspective and identify areas for improvement.\r\nAction feedback\r\nThis requires making revisions to your assignment based on the feedback you received. This step involves critically evaluating the suggestions and incorporating changes to enhance the quality of your work. The revised assignment is then prepared for final submission.","n":0.044},[{"v":"Group work","i":2,"n":0.707},{"v":"Understanding assignments","i":1,"n":0.707},{"v":"Assignments","i":0,"n":1}]]},{"i":117,"$":[{"v":"How does AI work?","n":0.5},{"v":"Artificial intelligence can do some amazing (and sometimes not so amazing) things. But how does it work? This page explains how text and images are produced by AI.\r\n\r\nHow does an AI tool generate text?\r\nTools that generate text are a type of AI called a \u2018large language model\u2019 (LLM). They have this name because they are large, very complex computer programs that are fed billions of words of text (a process called \u2018training\u2019) until they become good at understanding the patterns of language\u2014not only grammatical patterns, but also which words and phrases are related to each other and how they tend to appear together in a text.\r\n\r\nWhen you ask one of these tools a question, it analyses your question and tries to answer you based on the patterns it knows. These generative-text tools are based on statistical models of language patterns, so as they write your answer, they are continually analysing probabilities to guess which word or phrase is most likely to appear next in a piece of clear and coherent writing.\r\n\r\nWhile LLMs are not themselves search engines, many new LLMs and generative-text AI tools use what is called 'retrieval-augmented generation', meaning that they search the web in real time to help provide a current and accurate response. They can still make mistakes, however, and anything that they produce needs to be carefully evaluated.\r\nHow does an AI tool generate an image?\r\nLike generative-text tools, AI tools that create images based on a text description are trained on very large sets of data until they\u2019ve learned what patterns make an image look realistic\u2014for example, the usage of colour, texture and lighting, and the way objects are arranged\u2014as well as what text descriptions match different types of images.\r\n\r\nImage-generating tools can create art in different styles because of the large number of images they\u2019ve already analysed. Similar to LLMs, these tools mimic the visual patterns of millions of other images. It should be noted that in many cases, artists have not been consulted before their work was used to train these tools, leading to ethical concerns.\r\n\r\n\r\n\r\nAs AI-generated image tools have become more advanced, it has become easier and easier to create realistic-looking fake images and even videos. It is important to apply critical thinking skills when encountering images and videos online, now more than ever. And acknowledging when you create images with AI is one way to ensure you are engaging your audience with honesty and clarity. This can be done by adding a watermark or image caption.","n":0.049},[{"v":"Media literacy","i":1,"n":0.707},{"v":"Artificial intelligence","i":0,"n":0.707}]]},{"i":118,"$":[{"v":"Trigonometric identities","n":0.707},{"v":"Trigonometric identities are essential tools for simplifying expressions and solving equations in mathematics. They are used in a wide range of disciplines, including architecture and engineering, physics, computer graphics, astronomy and medical imaging. Use this resource to learn about some of the fundamental identities, double angle formulas, sums and difference, and other trigonometric functions.\r\n\r\n\r\nTrigonometric identities are equations involving trigonometric functions that are true for all values of the involved variables. For example, an expression such as:\r\n\\[\\tan (\\theta) =\\frac{\\sin (\\theta)}{\\cos (\\theta)}\\] is true for ALL values of \\(\\theta\\). So, we call this an identity.\r\nTo see how this is derived, remember that in any right-angled triangle:\r\n\\[\\sin(\\theta)=\\frac{\\textrm{Opposite}}{\\textrm{Hypotenuse}}\\] \\[\\cos(\\theta)=\\frac{\\textrm{Adjacent}}{\\textrm{Hypotenuse}}\\] \\[\\tan(\\theta)=\\frac{\\textrm{Opposite}}{\\textrm{Adjacent}}\\]\r\nTherefore:\r\n\\[\\frac{\\sin(\\theta)}{\\cos(\\theta)}=\\frac{\\frac{\\textrm{Opposite}}{\\textrm{Hypotenuse}}}{\\frac{\\textrm{Adjacent}}{\\textrm{Hypotenuse}}}=\\frac{\\textrm{Opposite}}{\\textrm{Adjacent}}=\\tan(\\theta)\\]\r\n\r\nFundamental trigonometric identities\r\nThe expression \\(\\tan(\\theta)=\\dfrac{\\sin(\\theta)}{\\cos(\\theta)}\\) is called a fundamental trigonometric identity. There are others, including one that is derived from the unit circle.\r\n\r\n\r\nWhen we apply Pythagoras\u2019 theorem to the right-angled triangle within the\r\nunit circle, we find that \\(\\left(\\sin\\theta\\right)^{2}+\\left(\\cos\\theta\\right)^{2}=1^{2}\\), that is:\r\n\\[\\sin^{2}(\\theta)+\\cos^{2}(\\theta)=1\\]\r\nThis is true for any value of \\(\\theta\\).\r\n\r\nBesides these two fundamental identities, there are many other trigonometric identities\r\nthat can be useful to simplify or rearrange trigonometric expressions.\r\n\r\nDouble angle formulas\r\nAs the name suggests, the double angle formulas are helpful when you deal with single angles \\(\\theta\\) that have been doubled to \\(2\\theta\\).\r\nThe sine double angle formula is:\r\n\\[\\sin(2\\theta) = 2\\sin(\\theta)\\cos(\\theta)\\]\r\nThe cosine double angle formula is:\r\n\\[\\begin{align*} cos(2\\theta) & = \\cos^{2}(\\theta)-\\sin^{2}(\\theta)\\\\\r\n& = 1-2\\sin^{2}(\\theta)\\\\\r\n& = 2\\cos^{2}(\\theta)-1\r\n\\end{align*}\\]\r\nThe tangent double angle formula is:\r\n\\[\\tan(2\\theta) = \\frac{2\\tan(\\theta)}{1-\\tan^{2}(\\theta)}\\]\r\n\r\nSums and differences formulas\r\nThe sums and differences formulas allow you to express trigonometric functions of sums or differences of angles in terms of the functions of the individual angles.\r\nThe sine sums and differences formulas are:\r\n\\[\\begin{align*} \\sin(x+y) & = \\sin (x)\\cos (y)+\\cos (x)\\sin (y)\\\\\r\n\\sin(x-y) & = \\sin (x)\\cos (y)-\\cos (x)\\sin(y)\r\n\\end{align*}\\]\r\nThe cosine sums and differences formulas are:\r\n\\[\\begin{align*} \\cos(x+y) & = \\cos (x)\\cos (y)-\\sin (x)\\sin (y)\\\\\r\n\\cos(x-y) & = \\cos (x)\\cos (y)+\\sin (x)\\sin (y)\r\n\\end{align*}\\]\r\nThe tangent sums and differences formulas are:\r\n\\[\\begin{align*} \\tan(x+y) & = \\frac{\\tan (x)+\\tan (y)}{1-\\tan (x)\\tan (y)}\\\\\r\n\\tan(x-y) & = \\frac{\\tan (x)-\\tan (y)}{1+\\tan (x)\\tan (y)}\r\n\\end{align*}\\]\r\n\r\nReciprocal formulas\r\nThe reciprocal formulas are used for inverse trigonometric functions.\r\nThe reciprocal of cosine is cosecant, denoted by cosec or \\(\\csc\\). It is given by:\r\n\\[\\csc(\\theta) = \\frac{1}{\\sin(\\theta)}\\]\r\nThe reciprocal of sine is secant, denoted by \\(\\sec\\). It is given by:\r\n\\[\\sec(\\theta) = \\frac{1}{\\cos(\\theta)}\\]\r\nThe reciprocal of tangent is cotangent, denoted by \\(\\cot\\). It is given by: \r\n\\[\\cot(\\theta) = \\frac{1}{\\tan(\\theta)}=\\frac{\\cos(\\theta)}{\\sin(\\theta)}\\]\r\n\r\nExample 1 \u2013 using trigonometric identities\r\nSimplify \\(\\tan(x)\\cos(x)\\).\r\nWe can use the fundamental trigonometric identity: \\(\\tan(\\theta)=\\dfrac{\\sin(\\theta)}{\\cos(\\theta)}\\).\r\n\\[\\begin{align*} \\tan (x)\\cos (x) & = \\frac{\\sin (x)}{\\cos (x)}\\times\\cos (x)\\\\\r\n& = \\sin (x)\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using trigonometric identities\r\n\r\n \tShow that \\((\\sec(x))(\\cot(x))=\\csc(x)\\).\r\n \tSimplify:\r\n\t\r\n\t\t\\(\\dfrac{\\sin(2\\theta)}{\\sin(\\theta)}\\)\r\n\t\t\\(\\cos(2\\theta)+2\\sin^{2}(\\theta)\\)\r\n\t\r\n\tSolve \\(\\sin^{2}(x)-\\cos(x)-1=0\\) for \\(0\\leq x\\leq2\\pi\\).\r\n\tGiven that \\(\\sin\\left(\\dfrac{\\pi}{3}\\right)=\\dfrac{\\sqrt{3}}{2}\\), \\(\\sin\\left(\\dfrac{\\pi}{4}\\right)=\\dfrac{1}{\\sqrt{2}}\\), \\(\\cos\\left(\\dfrac{\\pi}{3}\\right)=\\dfrac{1}{2}\\) and \\(\\cos\\left(\\dfrac{\\pi}{4}\\right)=\\dfrac{1}{\\sqrt{2}}\\), find the exact value of \\(\\sin\\left(\\dfrac{\\pi}{12}\\right)\\).\r\n\r\n\r\n\r\n\r\n","n":0.049},[{"v":"Trigonometry","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Functions","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":119,"$":[{"v":"Circular functions","n":0.707},{"v":"\r\nTrigonometric functions such as sin, cos and tan are usually defined as the ratios of sides in a right-angled triangle. These ratios can be extended to angles greater than \\(90^{\\circ}\\), using angles in a unit circle. Circular functions have applications in many STEM disciplines, such modelling oscillating motion in engineering, or analysing wave patterns in physics.\r\n\r\nThe unit circle\r\nA unit circle is a circle that is centred at the origin \\((0,0)\\) and has a radius of \\(1\\).\r\n\r\nAn angle measured from the positive \\(x\\)-axis may be used to define any point on the unit circle. They can be positive or negative.\r\n\r\n\tPositive angles are in an anti-clockwise direction.\r\n\tNegative angles are in a clockwise direction.\r\n\r\n\r\nAlthough there are \\(360^{\\circ}\\) in a circle, it is possible to rotate through more than \\(360^{\\circ}\\). For example a rotation of \\(45^{\\circ}\\) identifies the same point as a rotation of \\(45^{\\circ}+360^{\\circ}=405^{\\circ}\\).\r\n\r\n\r\n\r\nQuadrants\r\nThe unit circle sits on the coordinate plane, which is divided into an upper and lower section by the \\(x\\)-axis. It is further divided into quadrants by the \\(y\\)-axis. These four quadrants are numbered from one to four in an anti-clockwise direction.\r\n\r\nKnowing exact values for some angles in quadrant \\(1\\), allows you to find exact values in other quadrants.\r\n\r\nSine and cosine functions on the unit circle\r\nImagine a point \\(P\\left(x,y\\right)\\) on the unit circle.\r\n\r\nIn the triangle \\(POQ\\):\r\n\r\n\tthe hypotenuse \\(OP\\) has a length of \\(1\\)\r\n\tthe adjacent side to the angle \\(\\theta\\) is \\(OQ\\)\r\n\tthe opposite side to the angle \\(\\theta\\) is \\(QP\\).\r\n\r\nUsing the trigonometric ratios for sine and cosine in a right-angled triangle, we get:\r\n\\[\\begin{align*} \\cos(\\theta) & = \\frac{\\textrm{adjacent}}{\\textrm{hypotenuse}}\\\\ \r\n& = \\frac{OQ}{OP}\\\\ \r\n& = \\frac{x}{1}\\\\ \r\n& = x\r\n\\end{align*}\\]\r\nand\r\n\\[\\begin{align*} \\sin(\\theta) & = \\frac{\\textrm{opposite}}{\\textrm{hypotenuse}}\\\\\r\n& = \\frac{QP}{OP}\\\\\r\n& = \\frac{y}{1}\\\\\r\n& = y\r\n\\end{align*}\\]\r\n\r\nAny point on the unit circle has coordinates: \\[x=\\cos(\\theta)\\quad y=\\sin(\\theta)\\] where \\(\\theta\\) is the angle (positive or negative) measured from the positive \\(x\\)-axis to the point \\(P\\).\r\n\r\nFrom the unit circle, we can see that:\r\n\r\n\t\\(\\sin\\left(0^{\\circ}\\right)=0\\)\r\n\t\\(\\sin\\left(90^{\\circ}\\right)=\\sin\\left(\\dfrac{\\pi}{2}\\right)=1\\)\r\n\t\\(\\cos\\left(0\\right)=1\\)\r\n\t\\(\\cos\\left(\\dfrac{\\pi}{2}\\right)=\\cos\\left(90^{\\circ}\\right)=0\\)\r\n\r\n\r\nTangent function on the unit circle\r\nNow, consider point \\(N\\), which is an extension of line \\(OP\\).\r\n\r\nThe triangles \\(OMN\\) and \\(OQP\\) are similar.\r\n\\[\\begin{align*} \\frac{MN}{OM} & =\\frac{PQ}{OQ}\\\\ \r\nMN & = \\frac{PQ}{OQ}\\quad\\textrm{since OM=1}\r\n\\end{align*}\\]\r\nLength \\(MN\\) is the tangent of the angle \\(\\theta\\), or \\(\\tan\\theta\\). It is the distance of the vertical tangent to the circle that passes through the point \\(M\\left(1,0\\right)\\).\r\nFrom triangle \\(OMN\\), we can also see that \\(\\sin\\left(\\theta\\right)=PQ\\) and \\(\\cos\\left(\\theta\\right)=OQ\\).\r\nWhen we bring all of these together, we get an equation that connects the sin, cos and tan functions.\r\n\\[\\tan(\\theta)=\\frac{\\sin(\\theta)}{\\cos(\\theta)}\\]\r\nThis is consistent with the tangent function for right-angled triangles. We can prove this by substituting SOH and CAH into the equation.\r\n\\[\\begin{align*} \\tan(\\theta) & = \\frac{\\frac{\\textrm{O}}{\\textrm{H}}}{\\frac{\\textrm{A}}{\\textrm{H}}}\\\\\r\n& = \\frac{\\textrm{O}}{\\textrm{H}} \\times \\frac{\\textrm{H}}{\\textrm{O}}\\\\\r\n& = \\frac{\\textrm{O}}{\\textrm{A}}\r\n\\end{align*}\\]\r\n\r\nExact values of sine, cosine and tangent functions\r\nWe would usually use a calculator to evaluate \\(\\sin(\\theta)\\), \\(\\cos(\\theta)\\) and \\(\\tan(\\theta)\\), but there are some special angles you should remember. These are shown in the following table.\r\n\r\n\\(\\tan(\\theta)\\) is undefined for \\(90^{\\circ}\\) and \\(270^{\\circ}\\). This is because \\(\\tan(\\theta)=\\dfrac{\\sin(\\theta)}{\\cos(\\theta)}\\) and dividing by \\(\\cos(\\theta)=0\\) is undefined.\r\nKnowing these exact values will help you to work out the exact values for other angles using the symmetry of the unit circle.\r\n\r\n\r\n\r\nExample 1 \u2013 using exact values in circular functions\r\nFind the exact values of \\(\\sin\\left(\\dfrac{2\\pi}{3}\\right)\\) and \\(\\cos\\left(\\dfrac{2\\pi}{3}\\right)\\).\r\nSketch a unit circle and locate the point where the angle is \\(\\dfrac{2\\pi}{3}\\). Remember that \\(\\dfrac{2\\pi}{3}=120^{\\circ}\\).\r\n\r\n\\(\\dfrac{2\\pi}{3}\\) is shown in blue along with \\(\\sin\\left(\\dfrac{2\\pi}{3}\\right)\\) and \\(\\cos\\left(\\dfrac{2\\pi}{3}\\right)\\).\r\nWe can use the symmetry of the unit circle to draw in an angle that is related to \\(\\dfrac{2\\pi}{3}\\) but for which you know the exact value. In this case, it is \\(\\dfrac{\\pi}{3}\\), shown in red.\r\nYou can see that length \\(NQ\\) is the same as \\(MP\\) which means:\r\n\\[\\begin{align*} NQ & = MP\\\\ \r\n\\sin\\left(\\frac{2\\pi}{3}\\right) & = \\sin\\left(\\frac{\\pi}{3}\\right)\\\\ \r\n& = \\frac{\\sqrt{3}}{2}\\quad\\textrm{ from the table} \\end{align*}\\]\r\nLengths \\(ON\\) and \\(MO\\) are the same, but \\(MO=-ON=-\\cos\\left(\\dfrac{\\pi}{3}\\right)\\), so:\r\n\\[\\begin{align*} \\cos\\left(\\frac{2\\pi}{3}\\right) & = -ON\\\\\r\n& = -\\cos\\left(\\frac{\\pi}{3}\\right)\\\\ \r\n& = -\\frac{1}{2}\\quad\\textrm{ from the table} \\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using exact values in circular functions\r\n\r\n\tDetermine the coordinates for the points on the unit circle that make the following angles with the positive \\(x\\)-axis.\r\n\t\r\n\t\t\\(30^{\\circ}\\)\r\n\t\t\\(125^{\\circ}\\)\r\n\t\t\\(-60^{\\circ}\\)\r\n\t\t\\(270^{\\circ}\\)\r\n\t\t\\(-180^{\\circ}\\)\r\n\t\t\\(720^{\\circ}\\)\r\n\t\r\n\tFind exact values for:\r\n\t\r\n\t\t\\(\\sin(330^{\\circ})\\)\r\n\t\t\\(\\cos(210^{\\circ})\\)\r\n\t\t\\(\\sin(-30^{\\circ})\\)\r\n\t\t\\(\\cos(90^{\\circ})\\)\r\n\t\t\\(\\tan(300^{\\circ})\\)\r\n\t\t\\(\\cos(180^{\\circ})\\)\r\n\t\t\\(\\sin(-120^{\\circ})\\)\r\n\t\t\\(\\cos(315^{\\circ})\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n","n":0.039},[{"v":"Trigonometry","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Functions","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":120,"$":[{"v":"Active reading","n":0.707},{"v":"Have you ever read something without really absorbing it? Do you feel like you spend too much time reading the wrong texts? On this page, we will learn about active reading — a reading technique designed to help you get more out of your reading experience.\r\n\r\nGetting started with active reading\r\nActive reading means making choices about what you read and how you read, and different strategies should be used for different texts. Here are some things to take into consideration:\r\nHave a purpose in mind when you read\r\nSome purposes for reading are learning about new concepts in depth, searching for topics to include in an assignment, and looking for information to support a specific point.\r\n\r\nHow much will you read?\r\nIf the article or book is a core text, then you may decide to read it from cover to cover while taking notes. However, if you are searching for specific information, surveying, skimming, and scanning techniques may be more appropriate.\r\n\r\nHow does it relate to other texts\r\nIf you find that a text has a connection to something else you have read, then as an active reader you would make note of this. As time passes and you read more texts, it is easy to forget this type of detail.\r\n\r\nTake notes and keep organised\r\nActive readers take notes as they read and represent their information in visual ways, such as using tables, diagrams, and mind maps.\r\nEvaluating the text\r\nGet to know the writer\r\nGetting to know the author can tell you a lot about the purpose of the text. Check their affiliations, that is, any companies, groups or parties that they are associated with. For example, an article written about the benefits of cow\u2019s milk by the CEO of a dairy company should not be the only source used if you are discussing the health impact of dairy products.\r\nThink about the title\r\nThe title of a text can give you useful information about the writer\u2019s purpose, tone, and attitude. Compare the two titles below and consider the information they reveal about the writer:\r\n\r\n    Title 1: The effects of social media usage on adolescent mental health\r\n    Title 2: How social media is disrupting the lives of today's teens\r\n\r\nYou may have identified that the first title is objective. In other words, it does not indicate a bias from the author, and it is probable that the author will be reporting on the observations they have uncovered through the course of their study. On the other hand, title 2 appears to be more subjective, and could be more based on the author's personal feelings.\r\nCritical thinking\r\nActive readers evaluate the strengths and weaknesses of the text as they go. To do this, they may ask themselves questions such as:\r\n\r\n    Has the article/report fulfilled its purpose?\r\n    Is the argument clearly stated and supported by evidence?\r\n    Is the research valid? (Sufficient, appropriate, adequate, objective, valid methodology?)\r\n    Has the argument been developed logically?\r\n    Does it present solutions based on evidence?\r\n    Are recommendations appropriate to the purpose?\r\n    Has the article/book been published by a reputable publisher?\r\n\r\n\r\nActivity\r\n\r\nLet's test what you know about active reading.\r\n\r\n","n":0.045},[{"v":"Media literacy","i":2,"n":0.707},{"v":"Information sources","i":1,"n":0.707},{"v":"Reading skills","i":0,"n":0.707}]]},{"i":121,"$":[{"v":"Presentations","n":1},{"v":"This section is designed to help you master the skills needed for effective face-to-face and online presentations. You will also learn how to create poster presentations and infographics. Let's get started!\r\n\r\nOral presentation\r\nMost students will need to give an oral presentation at some stage of their study. For some, presentations can be a scary experience, while others prefer them to written assignments. Whichever group you belong to, preparation is the key to a successful presentation.\r\n\r\nIn fact, the best way to cope with nerves is to know your material really well and practise, practise, practise until you can say it easily.\r\n\r\nTo deliver a strong presentation, you need to combine two elements: delivery and content. On this page, we will suggest some ways that you can plan your content.\r\n\r\nThe first steps\r\n\r\nUnderstand the task\r\nYour first step for any assignment should be to read and understand the assignment instructions. There, you will find useful information such as:\r\n\r\nthe required length of the presentation.\r\nwhether the presentation should be delivered individually or in a group.\r\nif visuals are required.\r\nif there are any special considerations for the assignment. For example, some assignments require presenters to answers a specific number of questions after their presentation.\r\nrequirement around referencing and citation, such as the required number of references and the referencing style you should use.\r\n\r\n\r\nUnderstand your audience\r\nThe people who make up your audience will shape what you include in your presentation and how you present it. Some presentations will be for your teacher and your classmates. Others may be delivered to industry professionals. And sometimes you will be given a specific scenario, such as presenting information to the board members of a company.\r\n\r\nWhatever the situation, think about what your audience already knows and what they may not know. For example, for professional audiences, you can assume that they will already be familiar with certain concepts, but for general audiences, more explanation may be required.\r\n\r\nResearching\r\nCarefully plan and organise your information. Use the following steps to develop a clear plan for your talk:\r\n\r\n\r\nBrainstorm your topic to clarify what you know\r\nOrganise your information logically, e.g. chronologically, spatially (diagrams and maps), sequentially (first, second)\r\nDevelop a plan. For more detailed information go to the Mind mapping page.\r\n\r\n\r\nWrite and edit\r\n Write your talk in full or write detailed notes. Think about:\r\n\r\nContent: Make sure the ideas are logical and clear. Use examples and facts to illustrate your points.\r\nLanguage: Use language appropriate to the audience and purpose, generally a formal but conversational tone is best (no slang or colloquial language).\r\nStructure: Summarise your talk into meaningful chunks (e.g. numbered or bullet points). Consisting of main points and supporting details (similar to an essay plan). Use these brief notes to guide you through your talk.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.048},[{"v":"Presentations","i":0,"n":1}]]},{"i":122,"$":[{"v":"Getting started with assignments","n":0.5},{"v":"Planning and organising an assignment for the first time may seem daunting to new students. This short video walks you through easy steps for analysing assessment tasks, reading, taking notes, and planning assignments.\r\n\r\nSlam your assignment: plan it, research it, write it\r\nThis short tutorial gives you an overview of the process of\u00a0planning,\u00a0researching\u00a0and\u00a0writing\u00a0an assignment with quick links to the resources you can use at each stage.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nPlanning an assignment.\r\n\r\nWhen you get an assignment, it is not effective just to sit down and write. You need to plan, and here are three important steps for planning an assignment.\r\n\r\nYou need to analyse the task, so that you know exactly what you are being asked to do. You need to read and take notes in an organised way. And, you need to plan the structure of your writing.\r\n\r\nFirst, read the task really carefully to work out what the topic is and how they want you to write the assignment. Here is an example. First, find the topic through the content words. Then, look for instructional words that tell you how to write the assignment.\r\n\r\nNow, think carefully about what this means. Having two instructional words in the question probably means two parts to the essay. It\u2019s helpful to rewrite the question in your own words to really understand it. Getting it right at this point can save a lot of pain later on. Then, you need to find all the information you need. You need to read a lot and take notes. This takes lots of time. This is why you cannot leave things to the last minute.\r\n\r\nRead the general stuff first, and then the more specific. Reading generally is really important if you are not familiar with the topic. Start with books. Academic journal articles tend to be quite specific \u2013 these need some general understanding before you read them. Writing notes is important, too, because it helps you remember. But you need to keep your notes organised so you can easily find information and the sources they come from.\r\n\r\nHere is a system. The left-hand margin is for scanning the page, and if you organise it like this, you can keep a record of your references. Once you have all the information you need, plan by using a mind map. Mind maps have three steps: (1) brainstorm everything you know about the topic, (2) group, prioritise, and maybe even delete some of the ideas, (3) organise all of this and show the connections between the ideas.\r\n\r\nBut the last step in planning is turning this mind map into a linear plan. Work out the structure of the body first before any other section. This can take a lot of thought. Work from big structures like sections, to finer structures like paragraphs, finding the most logical order for everything can be quite tricky, too. After all of this is done, then write your assignment.\r\n\r\n","n":0.046},[{"v":"Assignment planning","i":1,"n":0.707},{"v":"Assignments","i":0,"n":1}]]},{"i":123,"$":[{"v":"Study essentials","n":0.707},{"v":"Find out about key study strategies needed for successful tertiary learning.\r\n\r\n\r\n\r\n\r\n","n":0.302},[{"v":"Study skills","i":0,"n":0.707}]]},{"i":124,"$":[{"v":"Assessments","n":1},{"v":"\r\n\r\n\r\n\r\n\r\n\r\n \tGetting started with assignments\r\n \tEssays\r\n \tAcademic reflection\r\n \tPresentations\r\n \tCase studies\r\n\r\n\r\n\r\n\r\n\r\n\r\n \tReports\r\n \tLiterature reviews\r\n \tAnnotated bibliographies\r\n \tVE assessments\r\n\r\n\r\n\r\n\r\n","n":0.236},[]]},{"i":125,"$":{"0":{"v":"Writing fundamentals","n":0.707},"2":[]}},{"i":126,"$":{"0":{"v":"University essentials","n":0.707},"2":[]}},{"i":127,"$":[{"v":"Negligence law: Language focus","n":0.5},{"v":"This language section outlines the use of modal and conditional language. Both are important in legal argument as they allow arguments to be constructed in terms of possibility, probability and conditions. It is the difference between saying 'it is because' versus 'it could be because'.\r\n\r\nModal verbs\r\nThe conditional nature of statements in legal case studies means that you may need to use words that express degrees of probability, possibility and certainty. The following chart on modal verbs may be useful when writing case studies.\r\n\r\n\r\nConditional language\r\nOften the advice in legal answers outlines a possible outcome in a future court case. Conditional language is very useful in case studies, especially when giving advice on possible or probable solutions to legal problems. A conditional is a two part sentence, e.g. if this happens, then that may happen.\r\nCommon conditional terms\r\n\r\n\r\n \tUnless and except\r\n \tIf clauses mean on the condition that. It is a two part sentence; an if usually followed by a then.\r\n\r\n \te.g. 'If the damage was reasonably foreseeable by the defendant, then liability will flow. If proper warning signs had been erected by the council... [then]...Brooke's injury would not have occurred.'\r\n\r\n\r\n \tproviding\r\n \tprovided that (on the understanding that)\r\n \twhether (if)\r\n\r\n \te.g. 'The question to be asked is whether a reasonable person would foresee that damage might result from the defendants action.'\r\n\r\n\r\n \tIt could be argued... (a possible argument is...)\r\n \tIt seems likely that... Introduces sentence/s expressing probability.\r\n\r\n\r\nActivity\r\n\r\nIn the tort of negligence, the plaintiff must prove that the defendant owed them a duty of care, breached that duty and that damages were suffered as a result of a breach of that duty.\r\n\r\n\r\n  \r\n\r\nWhere the text says [select], choose an option from these missing words: must, would, might, must be, could be, whether.\r\n\r\nFor Brooke to make a successful claim against the Yarra Valley City Council she [select] establish that a duty of care existed. Here the test of reasonable foreseeability [select] applied. The question to be asked is [select] a reasonable person [select] foresee that damage [select] result from the defendant's action. It [select] argued in Brooke's case that the signs put up by the Council created a reasonably foreseeable risk of injury of some kind to someone such as herself (See Chapman v Hearse 1961).\r\n\r\nSolution:\r\nFor Brooke to make a successful claim against the Yarra Valley City Council she must establish that a duty of care existed. Here the test of reasonable foreseeability must be applied. The question to be asked is whether a reasonable person would foresee that damage might result from the defendant's action. It could be argued in Brooke's case that the signs put up by the Council created a reasonably foreseeable risk of injury of some kind to someone such as herself (See Chapman v Hearse 1961).\r\n\r\n","n":0.047},[{"v":"Negligence","i":2,"n":1},{"v":"English language skills","i":1,"n":0.577},{"v":"Business Law","i":0,"n":0.707}]]},{"i":128,"$":[{"v":"Poster presentations","n":0.707},{"v":"Depending on your program of study, you might be asked to create a poster presentation as part of an assessment. Learn how to design an effective poster presentation that balances text with graphics and helps your audience understand the main points of your work.\r\n\r\nWhat is a poster presentation?\r\nA poster presentation is a visual display, usually on a large poster board, that communicates the key points of a study, project, or topic. It often includes images, diagrams, flowcharts and other graphics, and bullet points are often used to highlight the main ideas. Poster presentations are common at conferences and other academic and professional events.\r\n\r\nWhile your poster might be presented to your peers with a short talk or discussion, the content and message of your poster need to be expressed clearly and concisely, in a way that is easy for your audience to understand without further explanation.\r\nPosters are a great way for you to get your message across with effective visual displays of data and small blocks of supporting text.\r\nAn effective poster is:\r\n\r\n \ta source of information\r\n \ta conversation starter\r\n \ta summary of your work\r\n \tan advertisement for your work.\r\n\r\nGetting started\r\nRead your assignment brief carefully\r\nIt is always a good idea to have a clear understanding of the assessment criteria of your assignment. This way, you will know what the person who is marking your work will be using to grade your assignment. It is recommended that you read through the assignment instructions carefully and highlight keywords and keep these in mind throughout the planning phase of your assignment. You could use the criteria as a checklist when you have finished your draft, to ensure that you have covered everything that you needed to cover.\r\n\r\nAlso, check if there are any requirements regarding the dimensions, font sizes or layout of your poster because if these requirements are not met, this could lead to a lower grade.\r\nContact the library\r\nDon\u2019t know where to look for information? Get in touch with a Librarian through the Ask the Library chat service.\r\n\r\nThey can help you find the best search terms to use, show you how to filter results, and help you identify which databases to use for your subject area.\r\nKnow your audience\r\nThink about who it is that you will present the information to. Are they professionals in your field? Are they members of the general public?\r\n\r\nUnderstanding your audience will help you to decide what content to include and the language you should use to communicate your work. Technical language and excessive jargon is not always appropriate for every assignment. Ensure that your message is clearly communicated to the target audience.\r\nWhat to include\r\nThis will depend on the purpose of your poster. For example, are you telling an audience about research that you conducted? You might decide that your audience needs to know why you conducted the research, how you did it, who participated in it, what the results were and what the implications of the results are.\r\n\r\nIn this case, you might decide to follow a typical report structure of an introduction, methods, results, conclusion and discussion sections. For another type of project, you may need to include a literature review section or recommendations. And don\u2019t forget to include your references.\r\n\r\nIf you are confused about how to structure your work, you can contact an Academic Skills Advisor through the Ask the Library service to get some feedback and advice.\r\nDesign\r\nCarefully consider the design of your poster, as this will affect how your audience engages with your information. Remember that you should include a mixture of writing and visual aids such as images, graphs, tables and charts. Your poster should be sized roughly at A0 size (841 x 1189 mm).\r\nOverall appearance\r\nYour poster should be neat and uncluttered.\r\n\r\n \tUse a pleasing arrangement of graphics, text, colours.\r\n \tOrganise sections by using white space.\r\n \tBalance the placement of text and figures.\r\n\r\nUse headings to help readers find what they\u2019re looking for: objective, results, conclusions, etc.\r\n\r\n \tMinimise text \u2013 use graphics.\r\n \tKeep text in blocks of no more than 50-75 words \u2013 don\u2019t create large, monolithic paragraphs.\r\n \tUse a columnar format.\r\n\r\nConsider text size.\r\n\r\n \tAll text should be large enough to read from 1\u20132 metres, including the text in figures.\r\n \tTitle should be larger, to attract attention from far away.\r\n \tContent should be written from top to bottom (first) and left to right.\r\n\r\nUse colour cautiously.\r\n\r\n \tDark letters on light background are easiest to read.\r\n \tStick to a theme of 2\u20133 colours.\r\n \tAvoid overly bright colours. They attract attention but wear out readers\u2019 eyes.\r\n\r\nInclude full contact information. You want to be found \u2013 the reader should not have to look up anything to find you.\r\nLayout\r\n\r\n \tConsider whether your poster would be best in landscape or portrait orientation.\r\n \tYour information should flow in a logical way, e.g., left to right, from the centre outwards, or in columns.\r\n\t\r\n\r\n \tMake sure that your poster is not overcrowded. There should be space around each section, as this will increase the readability of your work.\r\n \tInclude a border of 3cm around your poster.\r\n\t\r\n\r\n\r\nImages\r\nPart of your message will be conveyed through images, so they should be chosen carefully.\r\n\r\nAlways add citations, attributions and captions to images. If you are not sure how to properly cite an image or include a caption, visit Easy Cite for guidance on a variety of referencing styles.\r\n\r\n\r\nRemember, you can't include everything in your poster. Choose the most important data, graphs, and images to include. You can expand upon your poster in an accompanying presentation.\r\n\r\nAvoid overly complicated graphs, charts or tables. Rather, aim to include graphic information that is clear and easy to understand. Focus on relationships \u2013 exact values are usually less important.\r\n\r\n \tKeep the focus on data.\r\n \tGrid lines, detailed ticks on axes, data markers, and backgrounds are often not needed.\r\n \tLabel data directly, when possible. Legends force readers to look back and forth to decode the graph.\r\n\r\n\r\n\r\nInclude a title that stands out and catches your audience\u2019s attention. In physical settings, your title should be able to be read from 2-3 metres away.\r\n\r\nFor font sizes, it is recommended that you use:\r\n\r\n \t80-96 for titles\r\n \t30-36 for headings\r\n \t18-24 for body text\r\n\r\nOther factors to consider are:\r\n\r\n \tWhere possible, use bullet points to make your text more concise and help your reader to focus on the main ideas of your poster.\r\n \tChoose your text and background colours carefully to ensure that your work is readable.\r\n\r\n \tDon't use decorative or overly stylised fonts such as Comic Sans, as they are not considered academic.\r\n\r\n\r\n\r\nTechnology\r\nThere are many apps and programs that can help you to design and create your poster. They include:\r\n\r\n \tMicrosoft PowerPoint\r\n \tCanva\r\n \tAdobe Illustrator\r\n \tAdobe InDesign\r\n \tMicrosoft Publisher\r\n\r\nThey each have their pros and cons, so it is recommended that you try a few and see what works best for you. Some free software and apps are available for RMIT students.\r\n\r\nOnce you have made your first successful poster presentation, especially if you receive good feedback for it, you can use it as a template for future presentations.\r\nAcademic writing\r\nIt is highly likely that your tutor or lecturer will request that you follow academic style conventions in your poster, such as avoiding personal pronouns and contractions, and using formal language. For help with academic style, you can make an appointment with an Academic Skills Advisor to review your work and get advice.\r\nPresenting your poster\r\nYou may be asked to \"walk\" people through your poster, that is, to prepare a 3\u20135 minute verbal explanation of your work.\r\n\r\nWhen giving your presentation, don't just read the poster from top to bottom. Instead:\r\n\r\n \tgive the big picture,\r\n \texplain why the problem is important, and\r\n \texplain how the graphics on your poster illustrate and support your findings and recommendations.\r\n\r\nA summary handout is something you can refer to when talking to people about your work. Prepare a handout that provides a written record for readers. You want people to remember your work. You can include:\r\n\r\n \ta miniature version of your poster\r\n \tmore detailed graphics, tables, and text explanations\r\n \tcomplete contact information.\r\n","n":0.028},[{"v":"Presentations","i":1,"n":1},{"v":"Poster presentations","i":0,"n":0.707}]]},{"i":129,"$":[{"v":"Digital literacy skills","n":0.577},{"v":"Your ability to use technology skillfully and confidently, also known as digital literacy, is essential for success in tertiary studies.\r\nStrong digital literacy skills facilitate better study, work and lifelong learning. These skills are valuable for a range of activities, including:\r\n\r\nLearning and development\r\nMost of your course and support materials will be available online, usually accessed through a Learning Management System (LMS) like Canvas. The LMS will offer course materials, readings, video content, discussions, and manage your online assessment. Being comfortable in this online environment is critical to your success in your studies.\r\nYou can use other educational websites such as simulations, learning activities, or the library website. You'll also use digital tools like videoconferencing and discussion boards to interact with your classmates and teachers.\r\n\r\nCommunication, collaboration and participation\r\nYou'll participate in a variety of networks and platforms for work together with your peers. When working in groups, you could use project management tools and shared documents. To share information and participate in discussions, you might use social media platforms like wikis and blogs.\r\n\r\nDigital creation, problem solving and innovation\r\nYou can create digital resources including websites, images and video.\r\n\r\n\r\n\r\n\r\nInformation literacy\r\nThe ability to find and evaluate information for relevance in a timely manner is important. Knowing when information is reliable, credible and suitable for assessments will improve the quality of your work. Navigating online resources such as databases with effective search strategies will help to locate credible information sources. This will enable you to produce better work and achieve higher marks.\r\nMethods to develop your information literacy\r\nWhen searching the web, it's important to think critically about what you're reading. Be careful about misinformation and baseless opinions.\r\n\r\nThere are different methods for determining the quality and reliability of a resource. These include the SIFT and CRAAP methods. Consider one for your practice\u2014they both offer helpful approaches to testing the integrity of information you find online.\r\n\r\nSIFT\r\nThe SIFT (Four Moves) method is great for investigating online content. Its steps include:\r\n\r\n \tStop\r\n \tInvestigate the Source\r\n \tFind Better Coverage, and\r\n \tTrace claims, quotes and media to the original context.\r\n\r\n\r\n\r\nCRAAP\r\nThe CRAAP method is a checklist you can use to evaluate your source. It explains how you can determine its quality and appropriateness. The criteria includes:\r\n\r\n \tCurrency\r\n \tRelevance\r\n \tAccuracy\r\n \tAuthority\r\n \tPurpose\r\n\r\n\r\n\r\nResearching your assignment\r\nInformation literacy is part of the research process. When looking for information for an assignment, applying these skills will mean that you can find more relevant and valuable information, and this can deepen your knowledge of the subject.\r\nDigital identity and wellbeing\r\nWhen you use social media, remember that your profile can be seen by many people, including future employers. The way that you interact, and the information you present online makes an impression. Be careful about how you engage with online networks, ensuring that anything you contribute is not discriminatory, defamatory or embarrassing.\r\n\r\nEnsure your privacy and security online with tactics to protect against threats like phishing1 and hacking2. Some of these include:\r\n\r\n \tStrong passwords\r\n \tTwo-factor authentication\r\n \tNot using public wifi\r\n \tUpdating device and network security\r\n \tUsing a VPN\r\n \tNot sharing personal information on email or other unsecured channels\r\n\r\nIt's also important that you manage your use of technology to prevent it from interfering with other activities or affecting your mental health. Overuse of social media, for instance, can lead to anxiety, depression and low self-esteem. Engaging with technology in a thoughtful and balanced manner can greatly reduce these harmful effects. Find out more about online safety risks at the eSafety Toolkit for Universities.\r\n\r\n1 Phishing:\u00a0the fraudulent practice of sending emails or other messages purporting to be from reputable companies in order to induce individuals to reveal personal information, such as passwords and credit card numbers.\r\n2 Hacking:\u00a0the gaining of unauthorized access to data in a system or computer.\r\nSource: Oxford Languages","n":0.04},[{"v":"Media literacy","i":2,"n":0.707},{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":130,"$":[{"v":"Types of classes and assessments","n":0.447},{"v":"Tertiary study will likely include different class types and styles of assessment that are different from what you experienced during your secondary education. Learn what to expect so that you can prepare for classes and assessments in a way that maximses your learning and utilises your time effectively.\r\n\r\n\r\n\r\nClass types\r\nThere are a variety of class types that might form part of your timetable. These are chosen by your teachers as the most effective way to learn the material. You may have to prepare differently for each type of session.\r\n\r\nSome class types in your tertiary study may include:\r\n\r\n \tTutorials: Small group activity focused on discussion, including non-specialist lab classes, with a up of 40 students\r\n \tLectorials: Lecture model of delivery with up to 120 students, there may be a participation element of this mode.\r\n \tPracticals: Teacher-led practice in laboratory or studio spaces\r\n \tWorkshops: Activities similar to tutorials but with the addition of a practical component.\r\n \tDigital Interactive sites: These might include pre-recorded lecture content, discussion forums, and interactive digital activities such as demonstrations, assessment tasks and submissions.\r\n\r\n\r\n\r\n\r\nAssessment types\r\nDuring tertiary study you will have different forms of assessments that aim to test your knowledge and skills in each course. Assessment marks and feedback are also a way to keep track of your academic progress and achievements. It's possible to speak with your teacher about your progress if you're having difficulties.\r\n\r\n\r\n\r\nKeep in mind that you may have multiple assessments due around the same time, so planning and working on them early are pivotal to positive assessment outcomes. Some of popular assessments in your course might include but are not limited to the following types:\r\n\r\n \tessays\r\n \treports\r\n \tcase studies\r\n \toral presentations\r\n \tgroup projects\r\n \tcreative development (exegesis, artefacts, artwork, etc.)\r\n \tlaboratory work\r\n \tself and peer-to-peer assessments\r\n \tonline quizzes\r\n \texams\r\n\r\nLearn more about how to work effectively on a range of assessments.\r\n","n":0.057},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":131,"$":[{"v":"New ways of learning","n":0.5},{"v":"You're expected to learn independently at tertiary study. Take charge of your learning experience by motivating yourself. This page has tips to keep you on top of your study.\r\n\r\nLearn independently\r\n\r\nTime management skills are key in being a success in your learning. Take an independent approach in managing class timetables, attending classes, and reading course materials before class.\r\n\r\nIf you need an extra hand to improve your study skills, there are supports available for you to use.\r\n\r\n\r\n\r\nLearn actively\r\nActive learning is another core skill in tertiary study. In active learning, you take active control of your study by choosing the suitable learning strategies that make study meaningful for you. These active learning strategies vary but can include:\r\n\r\n \tunderstanding, assessing and questioning information\r\n \tfinding tricks and tips to help you get more organised\r\n \treading additional resources beyond required readings\r\n \tdeveloping a habit of reading, understanding, writing, synthesising materials.\r\n\r\nVisit Goal setting and focus to learn how to take an active learning approach.\r\nThink critically\r\nYou'll use critical thinking in higher education to read, assess, question, challenge and consider the pros and cons of the information and opinions you encounter in your research and study. This skill can be developed once you have considered the different viewpoints of an issue. Critical thinking can help you stand out from the crowd in your class and in the workplace.\r\n\r\nVisit Critical thinking to practise your knowledge.\r\nLearn collaboratively\r\nIt's essential to learn to work collaboratively with others. Expect that you will study within the constraints of different class groups. You may be in a class of 300 or 25, and it's likely that you'll participate in group work. How you work within different class sizes and in groups might impact your approach with your peers.\r\n\r\nGet prepared for collaborative learning with tips at Group work which will help you with completing class discussion activities or assessments when several students are contributing.\r\n\r\nLearn with integrity\r\nIn tertiary study you are expected to work with a high level of academic integrity. Academic integrity is defined as being when the \"wider academic community is built on shared values and norms of behaviour, including honesty, fairness and responsibility\" (La Trobe University 2023:para. 1). This integrity requirement applies to you as an active member of the academic community. You can demonstrate your knowledge and adherence to academic integrity rules by attending to and applying these points to your study:\r\n\r\n \tbe honest, fair and respectful in all academic activities, from completing your assignments to sitting exams\r\n \temploy good academic practice, whether you are working alone or with a group of classmates\r\n \tensure your work, both individual and collaborative, acknowledges any information that has influenced your thinking. This includes everything from books and articles, to websites and images\r\n \ttake personal responsibility for ensuring you act appropriately in your studies\r\n \tunderstand that these practices are applicable to the whole academic community, including students, staff and other professionals.\r\n\r\nLearn more about academic integrity and RMIT policies around academic integrity.\r\nReferencing\r\nIn tertiary study you'll learn how to acknowledge other people's ideas in your work by using referencing. You'll need to reference the source(s) both in your writing (in-text citation) and at the end of your writing (in a bibliography or reference list).\r\n\r\nReferencing styles vary from one discipline to another, so it's good to check with your teachers which style is used in your course. Visit Easy Cite for a list of referencing styles. If you don't have much experience with referencing, it's good to know that it takes time to master all the formats of your style \u2013 you don't need be an expert right away. Library staff are experts in referencing and can help you.\r\nTime management\r\nIt's common to have commitments beyond studying, such as work, family, and social life, that require your time. It is good to keep in mind that if you do a full-time study load, it can equal a full-time job.\r\n\r\nIt's important to manage your time in ways that can often prioritise your studies, amongst other things. There are things you can do to use your time wisely and keep the load moving.\r\n\r\n \tUse time management tools such as Weekly Planner to control, allocate and keep track of your time to different commitments.\r\n \tStay focused on your tasks by avoiding distractions such as screens and noise (if you need time to read and write).\r\n \tAvoid procrastination. You might think that you can work better under last-minute pressure. The reality is, it will get a lot more stressful when multiple study and work deadlines arrive at the same time. It's always better to plan and act early.\r\n\r\nLearn more about time management.\r\n\r\n\r\nReferences\r\nLaTrobe University (2023) What is academic integrity?, Academic Integrity website, accessed 07 May 2023. https://latrobe.libguides.com/academic-integrity\r\n","n":0.036},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":132,"$":[{"v":"Social connections","n":0.707},{"v":"A new place often equals meeting new people. In higher education, those new relationships can have even more meaning.\r\n\r\n\r\n\r\nGet to know your teachers\r\nDevelop an understanding of youre teachers' expectations. This can help you to learn more effectively in classes as you'll know how to study and interact with them.\r\n\r\nEducators can also help you to figure out what career pathway you might want to take and how to get desired internships and jobs. Your teachers have industry experience and networks that may help you connect with your chosen field.\r\nGet to know your peers\r\nIt's beneficial for you to get connected with other students in your course. You can find them at orientation events and in class (online and on-campus) .\r\n\r\nReach out to say hello and introduce yourself to others. You're likely to share similar study interests and expectations that will make an exchange of ideas and a chance of networking easier.\r\n\r\n\r\nJoin groups, activities or cultural clubs\r\nThere are a many ways to become involved in social life at tertiary study. Group activities on campus or in the wider community can help you establish friendships and support networks. Participation can provide you with valuable experience and skills in collaborating, negotiating and reducing conflicts with others. You'll also find that you'll have a lot of fun and interesting experiences.\r\n\r\nThis will set you up with good people skills later in your studies when working with other students. Group participation can help build essential collaborative skills for the workplace and other areas in life.","n":0.064},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":133,"$":[{"v":"Changing habits and developing self-direction","n":0.447},{"v":"One of the biggest challenges to learning at a tertiary level is having to develop successful habits and adopting new ways of learning and working. While secondary school does some of the work to prepare you for the world of work and study, higher education expects more of you.\r\n\r\n\r\n\r\nHabits, good and bad\r\nA habit is a tendency of your behaviour that repeats until it starts to happen automatically. Good habits take some time to form and might be difficult at first, until we do them daily without thinking.\r\n\r\nMany of the study habits you form at university will develop in the same way, but they will support you through the rest of your life so it's never too early to explore ways of working that suit you. You'll have a range of class and assessment types, and teaching styles also vary. It's worth keeping an open mind that one way of studying may not suit all classes and assessment types.\r\nSelf-direction, doing it yourself\r\nA key factor in being a success in your studies will be your ability to be self-directed. You may not have immediate access to teachers to answer questions, or there may be a shortage of support services when you need them most. Assessment periods at your institution often mean that everyone needs help at the same time. Organising your time, developing good habits early and having a 'Plan B' for some of the challenges presented will ensure that you're ready for anything.\r\nPulling it all together with a portfolio\r\nA portfolio is a collection of documents and pieces of work that best represent your achievements. It's a record of your academic accomplishments and progress, as well as professional development and reflections on your work.\r\n\r\nYour studies are intended to change you in ways that help you develop the skills and potential to be a success. This progress can be demonstrated by a compilation of your best assessments and related material and activities throughout your studies, providing evidence of your progression and success.\r\n\r\nDeveloping a portfolio like this can help you maintain confidence and can support your learning and capability post-study. After graduation, you may want to use it for seeking further study or employment. With all the varied types of activities you undertake as part of your study, there's a lot you can capture to tell the story of your skill development and ability.\r\n","n":0.051},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":134,"$":[{"v":"Wellbeing, health and safety","n":0.5},{"v":"Tertiary institutions encourage and support a welcoming and inclusive environment for all students both online and on campus. Wellbeing, health and safety services are available to students who may need additional support.\r\nThese services support students with diverse social and cultural identities as well as physical and mental needs. Some of the support available includes services for:\r\n\r\n \tequitable learning and disability\r\n \tIndigenous students\r\n \tinternational students\r\n \tunder 18 students\r\n \tLGBTIQ+ students.\r\n\r\nServices are also designed to support your wellbeing and ensure that you have a safe experience at university. Many institutions will have emergency and crisis support, feedback and complains processes, campus security and financial and legal support.\r\n\r\nSecurity\r\nUniversities are big places, and they have a duty of care for people who work and study there. Even though making a safe environment for others is everyone's responsibility, it is the campus security team's top priority to ensure you feel safe and protected. They're often available 24/7 365 and have a phone number if you need them.\r\nCultural safety\r\nEveryone deserves to feel safe, respected, valued and treated as an equal in their place of study.\r\nIf students experience prejudice and discrimination or feel uncomfortable in learning situations, it's up to everyone to change this. Institutions should create a safe environment for everyone, an address inequities by recognizing and respecting cultural identities. This can be achieved through knowledge sharing and collaborative learning.\r\nAccessibility\r\nUniversities should be inclusive and accessible environments where students find digital and physical resources easy to understand and navigate. Although some spaces, especially in older buildings, may not be accessible to everyone, there are services available to address this issue. These supports may include physical adjustments, or staff who can help in person or online.\r\n\r\nCounselling\r\nCounselling services are available at most tertiary institutions. You can make an appointment to visit a counsellor online or in person. There may also be a range of personal development and wellbeing workshops available to help you with mental health and wellbeing issues. Visit your institution's website and explore the counselling support services on offer.","n":0.055},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":135,"$":[{"v":"Study support and library services","n":0.447},{"v":"Study support services are available online and face-to-face to help you build on your academic skills.\r\nOnline channels can include library chat or online consultations with an academic skills advisor, and there are often online self-help materials and tutorials across a range of topics and subjects. Face-to-face help might include workshops, drop-in sessions, and talking to your peers. Seek out support services by visiting your insitution's website and searching for 'study support'. In Vietnam, use the keywords 'student support' to learn more about the wide range of services available to you.\r\n\r\nOnline chat\r\nMost libraries have an online chat service that you can use to contact the library. Use the online chat to ask questions about library services and activities, and to find information for your assignments, referencing, and even support for your studies (depending on your campus).\r\nAcademic skills and research support\r\nExpert Academic Skills Advisors and Librarians (Melbourne) are available to help you with your studies: from help researching and writing assignments, to referencing and effective study strategies. Online and in person appointments with librarians and academic skills advisors are available. You can find out how to book an appointment by visiting the website, chat with an online chat service at your institution, or visit the library for more information.\r\n\r\nIn Vietnam, you can find out how to meet a Learning Advisor by visiting the SAS student site.\r\nCourse-specific help\r\nAcademic Skills Advisors and Librarians (Melbourne) are available to help with specific content in courses and will work with your teacher to identify areas where they can provide support. This help can include workshops based on assessment tasks and course content, or targeted instruction that is embedded in classes and can be for individuals or groups of students.\r\n\r\nIn Vietnam, Program Tutors are available to help with specific content in courses and will work with you to review key theories and concepts. They also offer workshops that will help you prepare for assignments. You can find out how to meet a Program Tutor or attend a workshop by visiting the SAS student site.\r\nPeer mentoring\r\nMany universities provide a service where students help their fellow students. These students have achieved high marks and can offer course-specific advice as well as general study tips.\r\nTo find a peer mentor, check your institution's website, or ask your students' group or teacher.\r\n\r\nPeer support can also be obtained through informal networks, by meeting with friends to discuss your coursework, or with group project team members. Remember to use the platforms available, like discussions boards and social media, to ask questions and contribute to conversations.\r\nOnline resources\r\nCheck your institution's student and library pages for online resources that can support your study. You're likely to find common tools such as referencing tools and academic skills tutorials. Many libraries also host subject guides that can direct you to the best resources in your area of study.\r\n\r\nThere are plenty of resources on the RMIT Library website to help you with using the library, academic writing, referencing and assignment planning. You can also visit RMIT Study Support online.\r\nAsk the Library\r\nAsk the Library is a great way to access the services of the Study Support Hub online. Chat with our friendly staff if you need help with your coursework or using the Library.\r\n\r\nExpert Learning Advisors and Librarians are available via Ask the Library online chat to help you with:\r\n\r\n \tassignment planning and structure\r\n \twriting, referencing and online resources\r\n \tgeneral study skills\r\n \tmaths, engineering, and physics coursework\r\n \tand more!\r\n\r\n\r\n\r\n \r\n\r\nStudent 1: I don't understand any of this!\r\n\r\nI thought I did yesterday during the lecture, but now I just don't get it!\r\n\r\nI wish I could get some help!\r\n\r\nStudent 2: You can get help from teachers experienced in maths, statistics, physics and chemistry!\r\n\r\nI found them through Ask the Library on the RMIT library website.\r\n\r\nThey were great! They worked through the questions with me and showed me how to tackle my worksheet.\r\n\r\nStudent 1: That sounds great! Did you have to wait long?\r\n\r\nStudent 2: Not at all! Ask the Library usually finds a teacher who'll contact you within a couple of hours, or early the next day.\r\n\r\nStudent 1: I'll get on to them right now!\r\n\r\n","n":0.038},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":136,"$":[{"v":"University structure","n":0.707},{"v":"Navigating any space is easier when you understand how it's organised. Learn more about the way universities are structured and the types of study programs that are available, as well as the role of research at a university.\r\n\r\nOrganisation of university activities\r\nA university is generally led by the Vice-Chancellor and governed by the council and the academic board. Each university has its own values, policies, and legislation developed according to the national standards, which they are required to obey and respect.\r\n\r\nAt RMIT, the colleges Business and Law (CoBL), Design and Social Context (DSC) and STEM are made up of schools. The college of Vocational Education (CoVE) is made up of industry clusters.\r\n\r\n\r\n\r\nCollege of Business and Law\r\n\r\n\r\n \tThe Business School, RMIT Vietnam\r\n \tSchool of Accounting, Information Systems & Supply Chain\r\n \tSchool of Economics, Finance and Marketing\r\n \tSchool of Law\r\n \tSchool of Management\r\n\r\n\r\nCollege of Design and Social Context\r\n\r\n\r\n\r\n \tSchool of Architecture and Urban Design\r\n \tSchool of Art\r\n \tSchool of Design\r\n \tSchool of Education\r\n \tSchool of Fashion and Textiles\r\n \tSchool of Global, Urban and Social Studies\r\n \tSchool of Media and Communication\r\n \tSchool of Property, Construction and Project Management\r\n \tSchool of Communication and Design, RMIT Vietnam\r\n\r\n\r\nSTEM College\r\n\r\n\r\n \tSchool of Computing Technologies\r\n \tSchool of Engineering\r\n \tSchool of Health and Biomedical Sciences\r\n \tSchool of Science\r\n\r\n\r\nCollege of Vocational Education\r\n\r\n\r\n \tCreative Industries\r\n \tFuture Technologies\r\n \tBusiness and Enterprise\r\n \tBuilt Environment and Sustainability\r\n \tSocial Care and Health\r\n\r\n\r\nStudy programmes\r\nTertiary institutions offer a range of study programs from vocational education and higher education to postgraduate education in many disciplines. Here is a list of the levels of education in Australia:\r\n\r\nPre-university study: Programs for domestic and international students to prepare them for tertiary studies or students who require extra support to get into a university-level program or who are keen to brush up their skills before starting at the university level.\r\n\r\nVocational study: Many Australian universities offer vocational education or TAFE (technical and further education) programs, such as certificate programs, diplomas and advanced diplomas.\r\n\r\nShort courses: Some universities offer fully online short courses. The core study areas include artificial intelligence, design, marketing, business and finance, health, science and engineering, data analytics and information technology.\r\nUndergraduate study\r\nAssociate degrees: Two-year programs that combine both theoretical and practical knowledge, with a significant focus on work-based learning. This is a stand-alone qualification to enter the workforce as well as a pathway to enter a relevant bachelor degree.\r\n\r\nBachelor degrees: Usually three-year study programs with a focus on both theory and practice. However, some professional degrees can take up to four or five years, depending on the study discipline. These degrees are open to both domestic and international students.\r\n\r\nHonours degree: The honours degree is a one-year specialised study that is generally undertaken upon the completion of the bachelor degree. Typically, students are required to have a high GPA from a bachelor to enter an honours degree. With an honours degree, you specialise in a specific area of your discipline. This degree is comprised of a larger portion of research or placement and coursework components. Honours degrees generally prepare you for masters by research or doctoral degree.\r\n\r\nPostgraduate study\r\nPostgraduate studies by coursework: Generally any study done after the completion of a bachelor degree. Occasionally, with extensive industry experience, you might be eligible to undertake a postgraduate program without a bachelor degree. Different levels of postgraduate programs are available, such as: graduate certificates, graduate diplomas and masters.\r\n\r\nResearch degrees: Research degrees are master and doctoral degrees across many disciplines. These programs are comprehensive research conducted independently under the guidance of an expert academic supervisor. You can conduct research as a thesis model or project model, depending on the discipline of study. Masters by research degrees are two years of full-time work, and PhDs are four years of full-time work.\r\n\r\nResearch: Research and innovation are central to university culture, addressing national and global challenges through collaboration with industry, government, and other institutions. Universities employ dedicated research staff, such as research students, assistants, postdoctoral researchers, and senior fellows. Many lecturers also engage in research alongside their teaching duties.\r\nRecognised prior learning\r\nRecognised prior learning (also known as credit) is the recognition of previous study or learning that can be counted towards a current program.\r\n\r\nThis can reduce the number of units (courses) needed to complete your program, which means the length and cost of your studies are reduced and you don't need to spend time learning things you already know.\r\n\r\n\r\n","n":0.037},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":137,"$":[{"v":"Modes of study","n":0.577},{"v":"Unlike secondary education, studying at a university can take many different forms. You might have classes that meet both on campus and online, and you will likely engage with a variety of online resources.\r\n\r\n\r\n\r\nThe digital campus\r\nNot all learning takes place at the physical location where a tertiary institution operates. The tertiary community will have to work across the internet, online resources, digital learning environments and social platforms. The institution where you're enrolled will aim to use these locations as well as the physical ones to best deliver a quality educational experience so you can develop the skills and knowledge to prepare you for work and life.\r\n\r\nBlended learning\r\nThe blended learning experience is a mix of online and face-to-face activities. Blended learning is flexible, accessible, inclusive and connected, supported by specialist spaces, services and technologies.\r\n\r\nDigital learning is a core part of every student's experience. Blended learning programs may include a mixture of face-to-face and/or online learning activities incorporating different media such as text, audio or visual media. This provides opportunities to explore different learning methods and activities.\r\n\r\nA typical example of a blended learning timetable usually means that your lecturers will create in-person classes in the physical classroom and combine them with online classes. \r\nThe flipped classroom\r\nIn a flipped classroom, students engage with the materials before or outside of the class. It's flipped because you have done the homework before rather than after the class. This gives you the opportunity to discuss the learning material you have explored, contextualise what you have learnt, and clarify any concepts with your teacher.\r\n\r\nConcurrent teaching\r\nConcurrent teaching is when a class is both face-to-face and online at the same time. Students who need to attend university remotely can tune into a lecture or tutorial via live video stream on their own device.","n":0.058},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":138,"$":[{"v":"Educators","n":1},{"v":"At RMIT, all educators share a common goal: to help students succeed in their studies.\r\nLet's explore the different roles that educational staff may play in your degree.\r\nAcademics\r\nAcademics at a university can include tutors, lecturers, professors, or deans. Some focus solely on research, while others combine teaching and research. Typically, academics are not involved in Vocational Education (VE).\r\n\r\n\r\n\r\nProgram coordinators / Program managers\r\nProgram coordinators are responsible for the overall quality of a degree program or certificate program.\r\n\r\nTheir main responsibilities include:\r\n\r\n \tbeing the first point of contact for course advice\r\n \tdeveloping short and long-term goals for the program\r\n \tscheduling courses, teaching teams and timetabling of the program\r\n \tworking closely with course coordinators to organise off-campus programs such as internships, placements, study tours, etc.\r\n \tmanaging program resources such as budgets, facilities, and types of equipment.\r\n\r\nCourse coordinators\r\nCourse coordinators are responsible for designing, planning and delivering a particular course within a degree/certificate program. You can contact them if you have queries about learning materials, timetables, assessment dates, etc.\r\n\r\nCourse coordinators also:\r\n\r\n \twork with the program manager and the department head to prepare course objectives\r\n \tlead the course team, recruitment, training and management of the tutors for the particular course\r\n \torganise guest lectures or industry expert seminars.\r\n\r\nLecturers / VE teachers\r\nLecturers facilitate the learning process, and assess and mentor students according to the course curriculum. They are content experts in a specific discipline. Often your course coordinator and/or program coordinator will also be one of the lecturers delivering the course.\r\n\r\nVE educators are usually referred to as teachers. They deliver learning materials in a classroom/lecture theatre (face-to-face) or online (via Teams or Collaborate Ultra). When you have difficulties understanding course materials, you can always contact your teacher and ask for further explanation.\r\n\r\nThe main responsibilities of lecturers and VE teachers include:\r\n\r\n \tdelivery of quality education and training to the students\r\n \tengaging students in the learning process\r\n \ttraining for technical skills associated with the course, such as laboratory skills, computer skills, etc.\r\n \tassessment preparation and marking.\r\n\r\nGuest lecturers and industry professionals\r\nMost university-level courses offer lectures or seminars from guest lecturers or industry professionals, experts in their fields, as work-integrated learning. They share insights and experiences from their career as well as current knowledge in the industry or profession. These seminars provide an opportunity to ask questions about the industry or profession you will enter after your studies, as well as a networking platform.\r\n\r\nTutors\r\nTutors usually deliver tutorial sessions, workshops and lectorials with or without a lecturer. They help you understand tutorial questions, complex concepts from the course material and provide feedback on your learning. Most importantly, they are approachable during the class/session. Most tutors in undergraduate and postgraduate courses are current PhD students.\r\nLibrarians\r\nLibrarians are information professionals who support learning, teaching and research. They are experts in information literacy, which is the ability to identify and evaluate reliable sources of information. They're available to support students via online and on-campus consultations, as well as delivering workshops and classes.\r\n\r\nLibrarians can guide you to the best resources for your assignments. They can also assist you in developing research skills such as evaluating the reliability of information sources, which is extremely valuable for your assignments.\r\nAcademic skills advisors / Learning advisors\r\nAcademic skills advisors (Melbourne) and Learning advisors (Vietnam) provide valuable advice and support across a range of subject areas and academic skills. This includes writing, assignment planning and structure, English language, presentation skills and more. Some are specialists in subject areas including science and maths. They are available for online on-campus consultations with studrnts, as well as delivering in-class workshops.\r\nLaboratory demonstrators / Technicians\r\nLaboratory demonstrators play a vital role in the delivery of practical-based classes. They teach and assist you with laboratory skills. They are available to answer student questions during the class, provide clear explanations that combining theory and the practical aspect and provide feedback on your laboratory skills and laboratory-based written work. In VE, laboratory demonstrators are called technicians.\r\nPlacement supervisors\r\nA placement supervisor is a professional who works in a specific field and is assigned by the organisation to guide, teach, and support students during their placement. They are usually willing to share knowledge, provide close guidance to develop professional practice skills and create a supportive environment to work.\r\nHDR supervisors\r\nHigher Degree by Research (HDR) students have completed an undergraduate degree and are continuing their studies through completing a postgraduate study. This may include a Master's by research or PhD (Doctor of Philosophy).\r\n\r\nThe HDR supervisory team guides, mentors, and supports students throughout the HDR candidature. These teams comprise a primary supervisor, co-supervisors and associate supervisors.\r\nPrimary supervisors:\r\n\r\n \tare mainly responsible for the student's candidature\r\n \tare responsible for the student's ongoing progress\r\n \tarrange milestone panels and prepare the student for the milestone review\r\n \tensure the student has access to the required resources to complete the study.\r\n \tare the main point of contact for students.\r\n\r\nAssociate supervisors provide support with a particular area of the research.","n":0.035},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":139,"$":[{"v":"Glossary of maths symbols","n":0.5},{"v":"Maths uses many symbols, or notations, which represent different functions and operations. You will come across lots of these as you explore the Learning Lab. Use this glossary to learn what any unfamiliar symbols mean.\r\n\r\nFoundational notations\r\nFoundational notations appear across different areas of maths and science.\r\n\r\n\r\n\r\nSymbol\r\nExplanation\r\n\r\n\r\n\r\n\r\n\\(+\\)\r\nplus\r\n\r\n\r\n\\(-\\)\r\nminus\r\n\r\n\r\n\\(\\times\\)\r\nmultiplied by, times\r\n\r\n\r\n\\(\\div\\)\r\ndivided by\r\n\r\n\r\n\\(\\pm\\)\r\nplus-or-minus\r\n\r\n\r\n\\(=\\)\r\nequals\r\n\r\n\r\n\\(\\gt\\)\r\ngreater than\r\n\r\n\r\n\\(\\lt\\)\r\nless than\r\n\r\n\r\n\\(\\ge\\)\r\ngreater than or equal to\r\n\r\n\r\n\\(\\le\\)\r\nless than or equal to\r\n\r\n\r\n\\(\\neq\\)\r\ndoes not equal\r\n\r\n\r\n\\(\\approx\\)\r\nis approximately\r\n\r\n\r\n\\(\\propto\\)\r\nis proportional to\r\n\r\n\r\n\\(\\therefore\\)\r\ntherefore\r\n\r\n\r\n\\(\\sqrt{}\\)\r\nroot\r\n\r\n\r\n\\(\\%\\)\r\npercent\r\n\r\n\r\n\\(\\infty\\)\r\ninfinity\r\n\r\n\r\n\\(\\theta\\)\r\ntheta, represents an angle\r\n\r\n\r\n\\(\\Delta\\)\r\ndelta, represents change\r\n\r\n\r\n\\(\\vert{x}\\vert\\)\r\nabsolute value of x\r\n\r\n\r\n\r\n\r\nSpecialised notations\r\nEach area of maths has specific notations.\r\n\r\nTrigonometry, Indices, surds and logarithms, and Functions and graphs\r\n\r\n\r\n\r\n\r\nSymbol\r\nExplanation\r\n\r\n\r\n\r\n\r\n\\(\\angle\\)\r\nangle\r\n\r\n\r\n\\(A\\)\r\narea\r\n\r\n\r\n\\(V\\)\r\nvolume\r\n\r\n\r\n\\(l\\)\r\nlength\r\n\r\n\r\n\\(r\\)\r\nradius\r\n\r\n\r\n\\(C\\)\r\ncircumference\r\n\r\n\r\n\\(\\pi\\)\r\npi, a constant approximately equal to 3.14\r\n\r\n\r\n\\(e\\)\r\nEuler\u2019s number, a constant approximately equal to 2.718\r\n\r\n\r\n\\(\\log\\)\r\nlogarithm\r\n\r\n\r\n\\(\\ln\\)\r\nnatural log\r\n\r\n\r\n\\(\\in\\)\r\nis an element of\r\n\r\n\r\n\\(\\Bbb{R}\\)\r\nthe set of real numbers\r\n\r\n\r\n\r\n\r\nDifferentiation, Integration, and Laplace transforms\r\n\r\n\r\n\r\nSymbol\r\nExplanation\r\n\r\n\r\n\r\n\r\n\\(f'(x)\\)\r\nderivative of the function of x\r\n\r\n\r\n\\(\\int\\)\r\nintegral\r\n\r\n\r\n\\(\\partial\\)\r\npartial derivative\r\n\r\n\r\n\\(L\\)\r\nLaplace transform\r\n\r\n\r\n\\(\\lambda\\)\r\neigenvalue\r\n\r\n\r\n\r\n\r\nStatistics\r\n\r\n\r\n\r\nSymbol\r\nExplanation\r\n\r\n\r\n\r\n\r\n\\(\\sum\\)\r\nsum\r\n\r\n\r\n\\(\\bar{x}\\)\r\nx-bar, the mean of a set of data\r\n\r\n\r\n\\(\\mu\\)\r\nmu, the mean of the population\r\n\r\n\r\n\\(s\\)\r\nstandard deviation of a sample\r\n\r\n\r\n\\(\\sigma\\)\r\nstandard deviation of the population\r\n\r\n\r\n\\(s^2\\)\r\nvariance of a sample\r\n\r\n\r\n\\(\\sigma^2\\)\r\nvariance of the population\r\n\r\n\r\n\\(Pr\\)\r\nprobability\r\n\r\n\r\n\\(C\\)\r\ncombination\r\n\r\n\r\n\\(!\\)\r\nfactorial\r\n\r\n\r\n\\(\\cap\\)\r\nintersection\r\n\r\n\r\n\\(\\cup\\)\r\nunion\r\n\r\n\r\n\\(z\\)\r\nz-score\r\n\r\n\r\n\\(n\\)\r\nsize of a sample\r\n\r\n\r\n\\(H_0\\)\r\nnull hypothesis\r\n\r\n\r\n\\(\\alpha\\)\r\nsignificance level\r\n\r\n\r\n\\(t\\)\r\nt-test\r\n\r\n\r\n\\(\\hat{p}\\)\r\np-hat, a sample proportion\r\n\r\n\r\n\\(p\\)\r\npopulation proportion\r\n\r\n\r\n\\(q\\)\r\nprobability of failure\r\n\r\n\r\n\\(\\lambda\\)\r\nmean or expected occurrences per unit time\r\n\r\n\r\n\r\n\r\nVectors\r\n\r\n\r\n\r\nSymbol\r\nExplanation\r\n\r\n\r\n\r\n\r\n\\(\\vec{}\\)\r\nvector\r\n\r\n\r\n\\(\\vert{\\vec{a}}\\vert\\)\r\nlength or magnitude of a vector\r\n\r\n\r\n\\(\\hat{a}\\)\r\nunit vector a\r\n\r\n\r\n\\(\\cdot\\)\r\ndot product\r\n\r\n\r\n\\(\\nabla\\)\r\ndirectional derivative\r\n\r\n\r\n","n":0.077},[{"v":"Maths","i":0,"n":1}]]},{"i":140,"$":[{"v":"Systems thinking quiz","n":0.577},{"v":"Take this quiz to test your knowledge of systems thinking and the circular economy.\r\n\r\nTake the quiz to test your knowledge of systems thinking and the circular economy.\r\n","n":0.192},[{"v":"Systems thinking","i":1,"n":0.707},{"v":"Sustainability","i":0,"n":1}]]},{"i":141,"$":[{"v":"Engaging critically with social media","n":0.447},{"v":"The SIFT method can help you to critically evaluate the validity and reliability of information before sharing it with your online community. Let's explore how it can be used to evaluate information from social media.\r\n\r\n\r\nSocial media is one way that people learn what\u2019s happening in the world. But anyone can share just about anything online. Before you repost or retweet, how can you be sure that what you\u2019re reading isn\u2019t fake news?\r\n\r\nWhen we see or hear information that aligns with a belief we already hold, we generally don\u2019t question it. \r\nTo illustrate, if you think a certain political party is plagued by corruption, you\u2019re likely to believe a post saying one of their members was involved in a scandal.\r\n\r\n\r\nOur tendency to trust information that supports what we already believe, and to ignore other facts and evidence, is called confirmation bias. Combine confirmation bias with a provocative social media post, and misinformation can spread rapidly. It\u2019s important to consider whether what you\u2019re reading and seeing is factual, especially if you\u2019re going to share it with your online community.\r\nThe SIFT method\r\nThe SIFT method is a set of four simple steps to help you determine whether something you are reading online, or anywhere, is trustworthy.\r\n\r\nSIFT stands for\r\n\r\n \tStop\r\n \tInvestigate the source\r\n \tFind better coverage\r\n \tTrace claims, quotes and media back to their original context.\r\n\r\nThese steps are all about understanding the source and context of the information, so that you can decide before you even read it whether it is valid and reliable.\r\n\r\n\r\n\r\n\r\nStop\r\nThis first step reminds you to pause and consider whether you know and trust the source of the information you\u2019re reading. If you have any doubts, continue with the next steps.\r\n\r\n\r\n\r\nInvestigate the source\r\n\r\nBefore you dig into an idea you find online, it\u2019s helpful to know where it\u2019s coming from. If you\u2019re not familiar with a website, look it up on Wikipedia for an overview. Knowing the expertise, perspective and agenda of the source can help you to better analyse and interpret the information provided.\r\n\r\n\r\n\r\nFind better coverage\r\nTo determine whether a claim is true, find more information about the topic from trusted sources. You can try to find the best, most authoritative source on a topic, or you can read multiple sources to see how different publications treat a claim you are investigating.\r\n\r\n\r\n\r\nTrace claims, quotes and media back to their original context\r\nSome people or organisations might not provide the whole picture of a topic. Quotes can be taken out of context, claims distorted, and media altered to support a particular perspective. Tracing these back to the original source allows you to see the original context and decide if the ideas have been presented accurately.\r\n\r\n\r\n\r\nApply your understanding\r\nRead the content shared below and consider whether the information is factual and trustworthy. Use the SIFT method to help with your analysis.\r\n\r\n\r\n","n":0.046},[{"v":"Media literacy","i":2,"n":0.707},{"v":"Critical reading","i":1,"n":0.707},{"v":"Critical thinking","i":0,"n":0.707}]]},{"i":142,"$":[{"v":"Artificial intelligence","n":0.707},{"v":"Discussions of artificial intelligence, or AI, are everywhere these days. With each new week there seems to be a new tool that can write text, create images, or solve other problems. But what exactly is AI, and just what can it\u2014and can it not\u2014accomplish?\r\nArtificial intelligence is a branch of computer science. Its goal is to create computer programs or tools that can mimic human intelligence and complete tasks that generally require human skills. AI tools are helping with critical thinking and analysis tasks in many fields, such as medicine and engineering. You might be familiar with tools that can produce writing and create art, which are referred to as 'generative-AI' tools.\r\n\r\n\r\nCan I use generative-AI tools to help with my assignments?\r\nIt\u2019s important check with your instructor whether you\u2019re permitted to use AI tools when completing an assignment. Even if using AI tools is allowed, relying on them too much could mean that you don\u2019t learn the necessary skills of your course. And if you use AI tools when it isn\u2019t permitted, that would be a breach of academic integrity.\r\n\r\nYou should also be aware that the companies who have developed AI tools will often save any information that you enter to continue training their tools. You should not upload any documents that you don't own, such as course materials or readings, to external AI tools. To keep your data private and secure, RMIT has developed Val, which is a generative-text AI tool that won't share your information. Learn more about how to use Val effectively and appropriately in your studies in the Generative AI for students at RMIT module.","n":0.061},[{"v":"Media literacy","i":1,"n":0.707},{"v":"Artificial intelligence","i":0,"n":0.707}]]},{"i":143,"$":[{"v":"Understanding citations","n":0.707},{"v":"You know that you need to \u201ccite your sources\u201d, but what does that mean in practice? Learn what citations are and how to include them in different types of assessments.\r\n\r\nWhat is a citation?\r\nA citation is essentially an abbreviation with a special format. Each citation refers to a different source you are using in your work. It provides a couple of details about the source, and it points the audience to the reference list (sometimes called a bibliography) for further information.\r\n\r\nDepending on the referencing style you\u2019re using, citations can take different forms:\r\n\r\n \tParenthetical citations have information that appears between parentheses or brackets.\r\n \tFootnotes have two parts: a small number that appears above the text line, called a footnote number, and a note at the bottom of the page with more information.\r\n \tEndnotes have the same small numbers as footnotes, but the notes appear at the end of the document or section of writing.\r\n\r\nTips for inserting citations in written work\r\nParenthetical citations can appear at the beginning or end of a quote, paraphrase, or summary. You can also have more than one citation in a sentence if you are synthesising the ideas from multiple sources. When citations are placed at the end of a sentence, they generally appear inside the sentence punctuation.\r\nExamples\r\n\r\n\r\n \t\". . .\" (Author 2022:62).\r\n \tAccording to Author (2022), . . .\r\n \tIn contrast, Researcher\u2019s (2023) study suggested . . .\r\n \tWhile Author (2022) proposed . . ., this idea was rejected by Researcher (2023) due to . . .\r\n \tAuthor (2022) and Researcher (2023) agree that . . ., but others take the viewpoint that . . . (Writer 2022; Creator 2023).\r\n\r\n\r\n\r\nFootnote and endnote numbers appear at the end of the ideas you\u2019re referencing, but similar to parenthetical citations, you can integrate information about the source, such as the surnames of the authors, using reporting words. Unlike parenthetical citations, footnote and endnote numbers are placed outside the sentence punctuation.\r\nExamples\r\n\r\n\r\n \tAccording to Author, . . .1\r\n \tWhile Author proposed . . .,1 this idea was rejected by Researcher due to . . .2\r\n\r\n\r\nDon\u2019t forget that in general, footnote numbers need corresponding footnotes at the bottom of each page, and endnote numbers need endnotes at the end of the document or section of writing. For more examples and the complete guidelines for the referencing styles used at RMIT, visit Easy Cite.\r\nCitations in presentations and multimedia work\r\nIt is important to include citations in all of your work, including presentations and multimedia projects.\r\n\r\nWhen you are giving an oral presentation or creating a multimedia piece that references the ideas of others, make sure to incorporate the same information you would include in a written citation. Depending on the referencing style you\u2019re following, this might mean mentioning the surnames of authors and dates of publications when speaking. The same information should be included in your slides or other visual aids.\r\nExample\r\n\r\n\u201cThis chart summarises the findings from the 2023 study by Author and Researcher, who found that . . .\u201d\r\n\r\n\r\nAt the end of your work, be sure to include a list of references, just as you would for a written assessment.\r\nPlease note: The examples on this page use the RMIT Harvard and Chicago A referencing styles. Check your course handbook or speak to your instructor about the referencing style required in your area of study.","n":0.043},[{"v":"Referencing","i":1,"n":1},{"v":"Citation","i":0,"n":1}]]},{"i":144,"$":[{"v":"Understanding Turnitin","n":0.707},{"v":"If you are submitting written assignments for your course, your instructor will use a tool called Turnitin to make sure you haven\u2019t plagiarised in your work. Learn more about the feedback provided by Turnitin.\r\n\r\n\r\n\r\n\r\nWhen you submit your work to Turnitin, a similarity report will be generated. Your instructor will review this report to help them determine whether you have referenced everything correctly, or whether you might have plagiarised. \r\nUnderstanding the similarity report\r\nThe similarity report shows the amount of text that matches other texts, including websites, academic journals, and material previously uploaded to Turnitin. This may include work that you've submitted to RMIT in the past, as well as assignments from other institutions around the world.\r\n\r\nThe similarity report will tell you the percentage of your text that matches other materials. The match overview section highlights parts of text that match or are very similar to sources already on Turnitin. How the text matches are spread out is more important than the percentage. If you have large chunks of text which are highlighted, this could mean that more careful paraphrasing is needed. In some cases, you can make changes and upload your work again to produce a new similarity report. Check your course guide on Canvas to see if you can do this.\r\n\r\nIf there are small matches, look at these carefully. Correctly cited and formatted quotes may be highlighted, depending on the Turnitin settings. Additionally, items such as course titles, teacher names, and assignment titles may be highlighted in Turnitin.\r\nReducing the similarity score\r\nThese tips can help you reduce your similarity score:\r\n\r\n \tIf chunks of text are highlighted, then you need to paraphrase further as the wording is still too close to the original source.\r\n \tCheck that your quotations have double quotation marks. Single marks will always be ignored by Turnitin.\r\n \tCheck the settings. You can reduce the percentage immediately by excluding quotations and your reference list. Note that changing the settings does not impact what your instructor sees.\r\n\r\n\r\n\r\n\r\n \r\n\r\nTurnitin is text-matching software. It\u2019s used to detect plagiarism in written assignments submitted online. It scans everything that\u2019s been uploaded for text matches. If something is copied from another source and it\u2019s not cited correctly, it will be detected. The reason RMIT uses Turnitin is to ensure academic integrity. \r\n\r\nIt\u2019s important that any written work is your own or that you acknowledge that it has come from another source. Turnitin works by matching the text in assignments against electronic, print and other sources. This includes books and journals, websites, archived materials, and other students\u2019 assignments, from global universities. It detects incorrect citations and both accidental and intentional plagiarism. Turnitin makes sure that intellectual property is respected. \r\n\r\nTurnitin is constantly being updated with new entries. Assignments submitted by you and your classmates are scanned and added to the database. The similarity score is the percentage of text from your assignment that exactly matches other sources on the Turnitin database. \r\n\r\nThere are five possible similarity ranges, from 0\u2013100%. The colour of the report icon represents the score. It\u2019s very unlikely that your report icon will be blue, but don\u2019t panic. The score is just an indication of matching text and will be interpreted by your teacher depending on your discipline and the assignment. In some courses you may be able to redraft, using the report to see which areas need adjusting. \r\n\r\nOne way to improve your similarity score is to paraphrase. Avoid direct quotes unless they are important, for example when quoting literature, law or religious texts. Citations are critical to the success of your assessment. They validate your work and support your point of view. \r\n\r\nThere are some great resources on the Learning Lab to help you practice paraphrasing and citing. For more help on using Turnitin go to their website.\r\n","n":0.04},[{"v":"Plagiarism","i":1,"n":1},{"v":"Academic integrity","i":0,"n":0.707}]]},{"i":145,"$":[{"v":"Exam conduct","n":0.707},{"v":"You might be required to sit exams during your time at university. Ensure you are acting with academic integrity by following the instructions for what you can and cannot do during exams.\r\nMany university courses require students to sit exams at an appointed time. You might need to bring certain materials to your exams, and be prohibited from using other materials. It\u2019s important to follow the directions provided by your instructor in order to avoid breaching academic integrity.\r\n\r\nExam misconduct occurs when a student:\r\n\r\n \thas someone else sit the exam for them\r\n \tshares current exam questions and answers online, or searches for and copies exam answers online\r\n \tbrings technology or other items into the exam which are not allowed\r\n \treceives assistance or gives assistance to other people during the exam\r\n \tuses additional, unapproved software or technology in an online exam.\r\n\r\nThese behaviours breach academic integrity and are not acceptable.\r\nCheck your understanding\r\n \r\n\r\nImages in the activity by RMIT, licensed under CC BY-NC 4.0","n":0.079},[{"v":"Exam preparation","i":1,"n":0.707},{"v":"Academic integrity","i":0,"n":0.707}]]},{"i":146,"$":[{"v":"Ghostwriting and contract cheating","n":0.5},{"v":"If someone completes an assignment for you without payment, it is called ghostwriting. If they do it for payment, it is called contract cheating. Both are considered academic misconduct. Let's find out more.\r\nAssessments are your opportunity to show off your skills and prove what you have learned. It is crucial that you complete your assessments yourself so that your academic qualifications are legitimate.\r\n\r\nIf someone else completes an assessment on your behalf so that you can submit it as your own work, it means that you are missing out on the chance to learn and that you will not gain the skills you need for your future career. It also means that the person who actually wrote the paper is not credited for their work.\r\n\r\nFurthermore, there have been cases where the providers of contract cheating services have collected information about students and then used this information for exploitation and blackmail.\r\n\r\nGhostwriting and artificial intelligence tools\r\nYour instructor might permit you to use artificial intelligence tools that generate text, such as Val or ChatGPT, in your assessments. Make sure that you are following your assessment instructions carefully.  For example, when you do use an AI tool, it must be properly cited. Using an AI tool when it\u2019s not permitted could be considered ghostwriting and a breach of academic integrity.\r\nCheck your understanding\r\nRead the following scenarios and decide whether or not the students have acted with integrity.\r\n \r\nImages in the activity by RMIT, licensed under CC BY-NC 4.0","n":0.064},[{"v":"Artificial intelligence","i":1,"n":0.707},{"v":"Academic integrity","i":0,"n":0.707}]]},{"i":147,"$":[{"v":"Collusion","n":1},{"v":"Working with others is often a part of higher education, but there are times when group work is not permitted. Learn the difference between ethical collaboration and collusion, which is a breach of academic integrity.\r\n\r\nWhat is collusion?\r\nThere is a clear distinction between collaboration and collusion when working on various forms of assignments. As a university student, you will be asked to participate in both group and individual assignments. When you are doing an individual assignment and get help completing it, or help another student complete their individual assignment, that is collusion.\r\n\r\nCollusion can include:\r\n\r\n \tworking with someone else on an assignment that you are supposed to write individually\r\n \tcopying another student\u2019s work, or letting another student copy your work\r\n \tletting someone write part of an assignment for you.\r\n\r\nThis does not mean that you cannot work with others at all. Ethical collaboration includes:\r\n\r\n \tworking together on a group assignment with your group members (just remember to credit all group members)\r\n \thelping each other understand the assignment question\r\n \tstudying together.\r\n\r\nBe sure to ask your instructor if you are unsure about what kind of collaboration is permitted in your course.\r\nCheck your understanding\r\nRead the following scenarios and select the correct answer.\r\n\r\n\r\nImages in the activity by RMIT, licensed under CC BY-NC 4.0\r\nFile sharing\r\nMaterials like assessment and exam content, lecture slides, and study guides are protected by copyright. Uploading and sharing them online is prohibited and regarded as academic misconduct.\r\n\r\nYou might find websites and services offering payment for such materials. Be aware that this is not permitted and can result in serious consequences. It is your responsibility to keep your assessments secure.\r\n\r\nAlso, websites that accept notes and assignments from students lack quality control, so using work from these sites is unethical and may involve poorly written or inaccurate content. If you're struggling, it's best to contact your tutor or Study Support for additional assistance.\r\nCheck your understanding\r\nRead the following scenarios and select the correct answer.\r\n","n":0.056},[{"v":"Group work","i":1,"n":0.707},{"v":"Academic integrity","i":0,"n":0.707}]]},{"i":148,"$":[{"v":"Group work and academic integrity","n":0.447},{"v":"It is important to be conscious of academic integrity while working in a group. While breaches of academic integrity may be intentional or unintentional, the consequences can be extremely serious for all group members.\r\nGroup work is common in higher education and helps you learn to work well as part of a team. Here are a few things to keep in mind when working in a group.\r\nAlways refer to the assignment brief\r\nGroup work can take several different forms, so reading the assignment brief carefully and seeking clarification from your instructor where there is any doubt is essential. Where work is to be submitted in sections by students, ensure that you never take credit for the work done by another member of your group, as this is considered plagiarism.\r\nBe aware of group accountability\r\nEach team member is responsible for the work produced by the group. If one member plagiarises or compromises the academic integrity of the project in any way, there may be consequences for each of the group\u2019s members. One way to avoid this is to have team members revise each other\u2019s work, paying close attention to their references and citations.\r\nDo not accept outside help\r\nThe ideas and data expressed in your group\u2019s work should only come from the members of the group. Allowing someone who is not in your group to contribute to your work could be viewed as breach of the academic integrity standards of your university.","n":0.065},[{"v":"Group work","i":1,"n":0.707},{"v":"Academic integrity","i":0,"n":0.707}]]},{"i":149,"$":[{"v":"Plagiarism","n":1},{"v":"You wouldn\u2019t want someone else taking credit for your hard work\u2014which is why it\u2019s important to always give credit when using someone else\u2019s ideas. Not only is it the right thing to do, but it also shows that you have integrity and respect the work of others.\r\n\r\nWhat is plagiarism?\r\nPlagiarism means using someone else\u2019s work or ideas without giving them proper credit. It's considered a form of theft, because the work doesn't belong to you. It's also fraud, because it means you are claiming someone else\u2019s work or ideas as your own.\r\n\r\nIn short, any time you use the ideas of others in your own work, ensure you reference them properly.\r\nTypes of plagiarism\r\n\r\nIntentional plagiarism\r\nIntentional plagiarism is deliberately using someone else\u2019s ideas, words or images and presenting them as your own. This includes using paraphrasing or translation software to change materials without acknowledging the sources. Intentional plagiarism is theft and fraud, and has serious consequences.\r\n\r\n\r\nAccidental plagiarism\r\nAccidental plagiarism is when you do not reference sources correctly in an assignment. This usually happens when students haven't carefully noted their sources as they work. While accidental plagiarism might not be intended, it can still carry negative academic consequences. Visit the Referencing section to learn more.\r\n\r\n\r\nSelf plagiarism\r\nUsing parts of your old assignments in new assignments without referencing is self-plagiarism. Many courses do not allow self-referencing, but if you are permitted to use parts of your old assignments, make sure you reference them the same way you would reference any other source.\r\nAvoiding plagiarism\r\nFollow the steps below to help avoid accidental plagiarism:\r\n\r\n \tAccurately write down details of authors when taking notes.\r\n \tParaphrase carefully.\r\n \tBe accurate with citations and quotes.\r\n \tCheck your reference list is complete.\r\n \tProofread your work for citation and referencing errors.\r\n\r\nQuiz: Is it plagiarism?\r\nTake this quiz to test your knowledge.\r\n\r\n\r\n","n":0.058},[{"v":"Referencing","i":2,"n":1},{"v":"Plagiarism","i":1,"n":1},{"v":"Academic integrity","i":0,"n":0.707}]]},{"i":150,"$":[{"v":"Academic integrity","n":0.707},{"v":"Just as all communities have accepted standards for behaviour, academic communities expect members to act with academic integrity. As a student at RMIT, you are responsible for upholding the academic integrity of the university.\r\n\r\nWhat is academic integrity?\r\nUnderstanding academic integrity will help you succeed as a member of the RMIT academic community.\r\n\"Academic integrity means acting with the values of honesty, trust, fairness, respect and responsibility in learning, teaching and research. It is important for students, teachers, researchers and professional staff to act in an honest way, be responsible for their actions, and show fairness in every part of their work. All students and staff should be an example to others of how to act with integrity in their study and work.\u201d (Exemplary Academic Integrity Project, 2013).\r\nPlay the video for a brief overview of academic integrity:\r\n\r\nUnderstanding academic misconduct\r\nIt is important to understand what academic misconduct is in order to avoid it. The four main forms of academic misconduct are:\r\n\r\n \tCollusion \u2013 working with others to do an assessment that was meant to be completed individually\r\n \tImproper exam conduct \u2013 giving or receiving help during exams, including using prohibited technology or tools.\r\n \tGhostwriting and contract cheating \u2013 having someone else complete an assessment for you, or completing someone else\u2019s assessment\r\n \tPlagiarism \u2013 using someone else\u2019s work or ideas without giving them proper credit\r\n\r\nAcademic integrity in Australia\r\nAcademic integrity is taken very seriously in Australia, and there have even been people who have had their degrees withdrawn, after graduation, when they were found to have cheated.\r\n\r\nTertiary Education Quality and Standards Agency (TEQSA) is the federal government agency responsible for standards in tertiary education. TEQSA enforces regulations relating to universities, including academic integrity. In fact, contract cheating services\u2014companies that charge students money to write their assessments for them\u2014are illegal thanks to TEQSA. Not only is it illegal to take money to do someone else\u2019s assessments, it is illegal to buy an essay or other assignment from anyone.\r\n\r\nTEQSA has resources for students, and they also provide information in languages other than English.\r\n\r\nStudents need to be aware of the requirements for their studies. This includes knowing what is considered cheating or colluding, and what is acceptable practice. When in doubt, your instructor can answer questions about what is expected in your course.\r\nReferences\r\nExemplary Academic Integrity Project, (2023). Resources on academic integrity.\r\n \r\n\r\nAcademic integrity means acting with honesty, trust, fairness, respect and responsibility in learning, teaching and research. This includes doing your own original work, respecting and acknowledging the work of other people, and clearly stating when you use someone else's ideas, words or images by citing correctly in the body of your work and referencing your sources accurately.\r\n\r\nAcademic integrity involves recognising people and publications that have influenced your thinking and taking personal responsibility for the way you behave in an academic context. It's important to complete individual assignments on your own or acknowledge who you work with in group assignments. You should always behave with integrity in exams and assessments.\r\n\r\nFailing to recognise someone else\u2019s ideas by: not referencing, dishonestly working with another student on an individual assignment, or cheating on an exam or assignment, can result in a charge of Academic Misconduct. Other examples include paying or asking someone to write an assignment for you or attempting to bribe someone in the academic community.\r\n\r\nAcademic Misconduct can have serious consequences for both your academic and professional career. In some academic cultures, copying content may be seen as honouring the creator. However, at RMIT you must clearly state where ideas, words and images originated because they belong to the person or people who created them.\r\n\r\nEveryone has a responsibility for maintaining academic integrity, but don't worry, there's lots of support available. You can ask your tutor or lecturer to clarify the task. You can also take your assignment to the study support hub for extra help or use the Ask the Library online chat service. While they can\u2019t write the content for you, they can help you with structure and grammar.\r\n\r\nBy strengthening your understanding of academic integrity, you can develop skills which will be valuable throughout your work and life.","n":0.038},[{"v":"Academic integrity","i":0,"n":0.707}]]},{"i":151,"$":[{"v":"Synthesising","n":1},{"v":"To make a strong argument, you\u2019ll want to present evidence from a variety of different sources. This page explains the process of synthesising.\r\nSynthesising involves combining ideas from a range of sources in order to group together and present common ideas or positions on a topic. It\u2019s a necessary part of developing a sophisticated argument, and is used in many forms of academic communication.\r\n\r\nUnlike summarising and paraphrasing, which only use ideas from one source at a time, synthesising combines ideas from multiple sources.\r\n\r\nSynthesising allows you to:\r\n\r\n \tdevelop and strengthen your arguments with a variety of evidence\r\n \tshow that you have read widely on the topic\r\n \tdemonstrate analytical and critical thinking skills.\r\n\r\nHow to synthesise\r\nUse the following steps to synthesise information from different sources.\r\n\r\n \tRead materials relevant to your topic, and make brief notes on the key points in each source that are connected to your argument. This makes it easier to compare and contrast information across different sources.\r\n \tIdentify common ideas or themes in your notes.\r\n \tUse reporting words and linking words to explain common ideas, making sure to cite all the sources you have used.\r\n\r\nSynthesis example\r\nThe following example uses four fictional sources to show how you might take notes and then synthesise the ideas in a piece of writing.\r\nNote-taking for synthesis\r\nRead each source and take notes on the main ideas in dot point format.\r\n\r\nTopic: English is the dominant world language.\r\n\r\n\r\nWriting a synthesis from notes\r\nLook at your notes to identify similar and contradictory arguments. Group these together to develop and support your arguments, and cite your sources appropriately.\r\n\r\n\r\nPlease note: The examples on this page use the RMIT Harvard referencing style. Check your course handbook or speak to your instructor about the referencing style required in your area of study.","n":0.059},[{"v":"Synthesising","i":1,"n":1},{"v":"Referencing","i":0,"n":1}]]},{"i":152,"$":[{"v":"Paraphrasing","n":1},{"v":"Paraphrasing is one of the key skills students need at university. It means expressing information, ideas, or work done by someone else in your own words.\r\n\r\nTo paraphrase successfully, we need to do more than just change a few of the words. We can also:\r\n\r\nchange a sentence from active to passive or passive to active\r\nchange the order of the sentence\r\nbreak longer sentences into shorter sentences\r\nwrite in your own particular style.\r\n\r\nLet's use the following sentence as an example. It comes from a fictional book from 2024 called Emotion intelligence at home and at work by a fictional author called Lars Bergstr\u00f6m:\r\nOriginal text:\r\nEmotional intelligence is crucial for navigating workplace dynamics and fostering effective collaboration.\r\nFirst, let's look at an unsuccessful paraphrase of this sentence where the author has simply used a few synonyms:\r\nUnsuccessful paraphrase:\r\nEmotional intelligence is vital for managing office dynamics and encouraging effective collaboration.\r\n\r\nAnd now, here is a successful attempt at paraphrasing the sentence.\r\nIt is successful because:\r\n\r\n\tSynonyms or near synonyms replace words from the original:\r\n\t\t\r\n\t\t\tworkplace interactions replaces workplace dynamics\r\n\t\t\tmanage replaces navigating\r\n\t\t\tteam work replaces collaboration\r\n\t\t\tnecessary replaces crucial\r\n\t\t\tenhance replaces fostering\r\n\t\t\r\n\tThe sentence has a different structure: The original sentence begins with emotional intelligence, but in the paraphrase, emotional intelligence is mentioned toward the end of the sentence.\r\n\tThe citation Bergstr\u00f6m (2022) is included.\r\nSuccessful paraphrase:\r\nBergstr\u00f6m (2022) highlights that to effectively manage workplace interactions and enhance teamwork, developing the crucial skill of emotional intelligence is necessary.\r\n\r\nSteps to paraphrasing\r\nAn effective way to create paraphrases is to:\r\n\r\n\tread the information that you would like to paraphrase a few times.\r\n\twithout looking at the information, rewrite it in your own words. This way you will not be influenced by the original text.\r\n\treread your paraphrase and the original text, checking that the original meaning has been kept and that your paraphrase is not too similar to the original.\r\n\tcompare your language to the original text to make sure your version is different. A thesaurus or dictionary can help you with this. A good goal is to aim for no more than 4 words in a row that are the same as in the original source.\r\n\tDon\u2019t forget that you need to include a citation every time you paraphrase.\r\n\r\n\r\nAvoiding plagiarism\r\nCorrect paraphrasing is crucial to avoid plagiarism. Simply changing a few words or rearranging sentences from another source still constitutes plagiarism. A proper paraphrase must significantly differ in vocabulary and structure from the original. \r\n\r\nPlagiarism-detection tools like Turnitin will flag content that's too similar to existing works, even with minor word changes.\r\n\r\nNew AI tools, such as ChatGPT, can paraphrase text, but content from these tools also requires proper referencing. Their use may not always be allowed, so it's important to understand your course requirements and consult your instructor to avoid academic misconduct.\r\n\r\nPractice paraphrasing\r\nRead the text below, and then select the best paraphrase from the two options given.\r\n\r\n\r\n\r\nPlease note: The examples on this page use the APA 7th edition referencing style. Check your course handbook or speak to your instructor about the referencing style required in your area of study.\r\n\r\n ","n":0.045},[{"v":"Referencing","i":2,"n":1},{"v":"Paraphrasing","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":153,"$":[{"v":"Easy Cite and referencing help","n":0.447},{"v":"Have questions about referencing? Check out the links below for self-guided resources and personal support options.\r\n\r\nEasy Cite\r\nLooking for the ins and outs of referencing styles? Easy Cite explains the rules of common styles used at RMIT, along with referencing tips and examples.\r\n\r\nEasy Cite covers the following styles: RMIT Harvard, AGLC4, APA, Chicago A: footnotes and bibliography, Chicago B: author-date, IEEE, and Vancouver.\r\n\r\nWith Easy Cite, you can:\r\n\r\n \tlearn how to cite in each style\r\n \tunderstand how to correctly format reference list entries for different kinds of source material\r\n \tbookmark your commonly used sections\r\n \tdownload style guides.\r\n\r\nEasy Cite is intended only as a guide and some styles are open to interpretation. You should always check with your instructor to ensure you are using the correct style for your assignments and assessment tasks.\r\nSelf-guided resources\r\nThe resources below can help you learn more about why referencing is important, how to reference properly, and using reference management tools to track your sources.\r\n\r\n \tReferencing on Learning Lab\r\nLearning Lab is the home of self-guided study support at RMIT. Browse our resources on referencing, or use the keyword index to search for specific topics.\r\n \tEasy Cite\r\nEasy Cite explains the rules of common referencing styles at RMIT, along with referencing tips and examples.\r\n \tReference management tools\r\nLearn how tools such as EndNote, Mendeley, and Zotero can help you keep track of your sources while completing your assignments.\r\n\r\nMore information about reference management tools can also be found in the Research and writing skills for masters level students book.\r\n \tReferencing content generated by artificial intelligence (AI) tools\r\nIf you\u2019ve been told you can use an AI tool like ChatGPT in your assessment, this guide explains how to reference it, with examples from different referencing styles used at RMIT.\r\n\r\nPersonal support for RMIT students\r\nWant to talk to someone about referencing? These additional resources are available for RMIT students.\r\n\r\n \tAsk the Library\r\nChat with library staff online and get answers to your referencing and research questions.\r\n \tOne-on-one consultations\r\nArrange a consultation for expert help with referencing and assignment writing, as well as effective study strategies to support maths, statistics, chemistry and physics.\r\n \tVal\u2019s Writing Help persona\r\n  Get free, 24/7 AI-generated help with your assignments and coursework. After Val opens, use the persona menu to select Writing Help for feedback on your writing.\r\n","n":0.052},[{"v":"Student support","i":2,"n":0.707},{"v":"Referencing","i":1,"n":1},{"v":"Citation","i":0,"n":1}]]},{"i":154,"$":[{"v":"Summarising","n":1},{"v":"When you want to refer to the general ideas of a longer text, speech, or other creative work, without going into all the details, you can use a summary.\r\nA summary is an overview that contains the main points of a work. Summaries are much shorter than the works themselves, but still need to contain the author\u2019s original ideas and the degree of certainty with which those ideas are expressed.\r\n\r\nRemember that a summary does not need to contain every idea in the original work. You can focus on the ideas that are relevant to your argument, as long as you don\u2019t leave out an important aspect of the author\u2019s position that might mislead your audience.\r\n\r\nAs always, when you are referring to the ideas of others, you\u2019ll need to include a reference any time you summarise another work.\r\nHow to write a summary\r\nTo create a good summary of a text, you should identify:\r\n\r\n \tits main ideas and intentions\r\n \trelevant details\r\n \tthe tone used to express the ideas.\r\n\r\nUse the following steps to write a summary:\r\n\r\n \tLocate the topic and main idea of the work, and identify its purpose. Select the information that is relevant to the argument you are presenting in your own work, and take short notes using dot points or a mind map.\r\n \tWrite the summary directly from your notes. Make sure to use complete sentences.\r\n \tRefer back to the original text to ensure your summary reflects its main ideas.\r\n\r\nExample\r\nOriginal sample text\r\nTime management is a critical skill for students to develop. Weekly and semester timetables are an excellent way to plan a study program. Students can use them to manage their most important study, work and social commitments and to set themselves study goals. Blocks of time can then be set aside for study, reading, researching and writing. The most urgent tasks can be addressed, whilst work continues on preparing for lectures, tutorials and assignments. If time has been allocated for specific purposes, it is easier to avoid unexpected demands like phone calls, visitors and invitations. Assignments can be completed and submitted on time and to a satisfactory standard.\r\nSummary\r\nEffective time management allows students to utilise their time in order to successfully complete their study requirements (Author 2003).\r\nPractice summarising\r\nRead the fictional text below, and then choose the best summary from the three options given.\r\n\r\n\r\nPlease note: The examples on this page use the RMIT Harvard referencing style. Check your course handbook or speak to your instructor about the referencing style required in your area of study.","n":0.049},[{"v":"Essays","i":2,"n":1},{"v":"Academic writing","i":1,"n":0.707},{"v":"Summarising","i":0,"n":1}]]},{"i":155,"$":[{"v":"Integrating ideas with reporting words","n":0.447},{"v":"Whether you are paraphrasing or quoting, integrating the ideas of others with your own argument is an important part of clear, cohesive communication. It\u2019s easy to become repetitive, using \u2018this author states\u2019 and \u2018that author states\u2019 over and over. Keep reading to learn about how to use a variety of reporting verbs in your writing.\r\n\r\nWhen using reporting verbs, remember to use a variety of them. If you constantly use 'reports' or 'states', your writing will become repetitive. Also, reporting verbs reflect your attitude towards the information you have included in your writing, and generally they are sorted into three categories: tentative, neutral and strong.\r\n\r\nCompare the following sentences- can you tell which reporting verb is tentative, which is neutral and which is strong?\r\n\r\n\r\nCommon reporting verbs\r\nTentative\r\n\r\n \tadmit\r\n \tcaution\r\n \tconcede\r\n \tconsider\r\n \thypothesise\r\n \timply\r\n \tpostulate\r\n \tpropose\r\n \tquestion\r\n \trecommend\r\n \trestate\r\n \tspeculate\r\n \tsuggest\r\n\r\nNeutral\r\n\r\n \tassume\r\n \tcomment\r\n \tcompare\r\n \tcontrast\r\n \tdefine\r\n \tdemonstrate\r\n \tdiscuss\r\n \texamine\r\n \texplain\r\n \texplore\r\n \tfocus on\r\n \tfind\r\n \tindicate\r\n \tmention\r\n \tnote\r\n \tobserve\r\n \tpoint out\r\n \tpresent\r\n \treflect\r\n \tregard\r\n \tshow\r\n \tstate\r\n \ttake into consideration\r\n \tview\r\n\r\nStrong\r\n\r\n \tacknowledge\r\n \tadvocate\r\n \taffirm\r\n \targue\r\n \tassert\r\n \tchallenge\r\n \tconfirm\r\n \tconclude\r\n \tcontend\r\n \tdeny\r\n \tdetermine\r\n \tdispute\r\n \tdoubt\r\n \temphasise\r\n \tendorse\r\n \testablish\r\n \thighlight\r\n \tinfer\r\n \tinsist\r\n \tmaintain\r\n \tobject to\r\n \toppose\r\n \trecommend\r\n \treject\r\n \trefute\r\n \tstress\r\n\r\nUsing tenses when reporting\r\nNote, the tenses used when reporting published research can indicate much more than simply when something happened; the tense shows if the ideas are current. Tense use can vary a little with subject areas, so check the literature in your field for guidance. Here are some general patterns of tense use.\r\nPresent tense\r\nUse present tense to discuss findings from the literature that are generally considered factual and current.\r\nFor example: Trang (2020) emphasises that this is not the case.\r\nPresent perfect tense\r\nUse present tense to give the general background of past research and to show that previous research still influences current thinking.\r\nFor example: Research has indicated that economic factors are a major influence on the outcomes of planning strategies (Kendall, 2021; Steadman, 2022).\r\nPast tense\r\nUse past tense:\r\n\r\n \tWhen describing how a previous study was conducted.\r\nFor example: Smith (2021) conducted a review of data from 2022.\r\n \tWhen referring to specific findings of previous studies.\r\nFor example: Lee (2023) found that resource provision was problematic in this case.\r\n \tWhen referring to information or theories that were considered true in the past, but outdated today.\r\nFor example: Caldez (1932) claimed that the use of lead in paint created a faster-drying product.\r\n\r\nTest your knowledge\r\n\r\n","n":0.051},[{"v":"Referencing","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":156,"$":[{"v":"Quoting","n":1},{"v":"When you are researching a topic, sometimes you want to include the exact words of another author or speaker in your own work — this is called quoting, or including a direct quote.\r\n\r\nWhen to quote\r\nQuoting is the way to include the exact words of others in our own work. \r\n\r\nHere are some points to note about quoting:\r\n\r\n\tUse them sparingly. The person marking your work would much rather that you use paraphrases where possible because paraphrases are made up of your own words.\r\n\tSave direct quotes for particularly clear, powerful, or eloquent writing from the source text, for data and for when a paraphrase could possibly change the meaning of the original text.\r\n\r\n\r\nHow to insert a direct quote\r\nFollow these steps to use direct quotes in your written assignments.\r\n\r\n \tCopy the exact words from the original source. (If your quote is 40 words or longer, see the section below on using a block quote).\r\n \tUse quotation marks \" \" at the beginning and end of the copied text.\r\n \tCite the quote following the citation format for your referencing style. This often means including information such as the author\u2019s surname, year of publication, and the page number where the quote appears.\r\n\r\nWritten example\r\nWhile no single description exists for the concept of procrastination, it has been defined as \"delaying tasks to the point where personal discomfort is felt\" (Chin & Grey, 2023, p. 81).\r\nIf you are reading a direct quote as part of an oral or multimedia presentation, you'll want to include the same citation information. You might say something like the example below.\r\nSpoken example\r\nWhile no single description exists for the concept of procrastination, Solomon and Rothblum defined the concept in their 1984 study, page 503, as \"the act of needlessly delaying tasks to the point of experiencing subjective discomfort\".\r\n\r\nTips for quoting\r\n\r\n\tWhere possible, keep direct quotes short rather than using full sentences. You can do this by using key words from the original sentence as the direct quote. For example:\r\nAccording to Ivanov (2024), introducing continuing with the current manufacturing process would be a \"one-way ticket to failure\" (p. 11).\r\n\r\n\tDirect quotes need to be integrated into a sentence; they should not be a sentence on their own. \r\nFor example, change:\r\n\"Green spaces are essential for the mental and physical wellbeing of residents\" (Moretti, 2025, p. 6).\r\nTo:\r\nMoretti (2025) points out that \"[g]reen spaces are essential for the mental and physical wellbeing of residents\" (p.6).\r\nIf you need to make any changes to the original text, put the change inside square brackets, as can be seen in the previous example where the \"g\" in green has been changed to a lowercase letter.\r\n\tIf you need to make any changes to the original text, put the change inside square brackets, as can be seen in the previous example where the \"g\" in \"green\" has been changed to a lowercase letter.\r\n\r\n\r\n\r\n\r\n\r\nPlease note: The examples on this page use the APA referencing style. Check your course handbook or speak to your instructor about the referencing style required in your area of study.\r\n\r\nReferences\r\nSolomon, L.J., & Rothblum, E.D. (1984). Academic procrastination: Frequency and cognitive-behavioral correlates. Journal of Counseling Psychology, 31(4), 503-509. https://doi.org/10.1037/0022-0167.31.4.503\r\n\r\n","n":0.044},[{"v":"Quotation","i":2,"n":1},{"v":"Referencing","i":1,"n":1},{"v":"Citation","i":0,"n":1}]]},{"i":157,"$":[{"v":"When referencing isn't needed","n":0.5},{"v":"You know that you need to provide the source of ideas and language that are not your own, but what about facts and general knowledge? Does this information \u2018belong\u2019 to someone who should be referenced? Read on to understand when citations are and are not necessary.\r\nKnowing when to provide a reference can take some practice. In general, you will always include citations for work or ideas that are not your own. There are a couple of key exceptions, however.\r\nGeneral knowledge\r\nAs a guideline, you do not need to reference statements that are general knowledge, which are things that most people would know.\r\nExamples\r\nAustralia is between the Pacific Ocean to the east, and the Indian Ocean to the west.\r\nThis statement does not need a reference, particularly when writing in Australia.\r\nThe Pacific Ocean has a bigger surface area than all the land on earth combined (Encyclopedia Britannica, 2022).\r\nThis statement needs a reference, as it is not something that most people would know. Here a citation is provided in APA format and the full details appear in the reference list at the bottom of the page.\r\nHistorical overview\r\nBroad statements of historical overview also do not need to be referenced. A statement such as 'There were two world wars in the twentieth century\u2019 does not require a cited source.\r\nYour own ideas\r\nYou do not need to cite your own ideas, unless you have already used them in a previous assignment. If your ideas have been included in your prior work, check with your instructor to ensure that self-citation is permitted in your course.\r\nShould I cite it?\r\nThe following diagram can help you decide whether you need a reference. If in doubt, cite it. You will not be penalised for including a citation, and if you are citing too much, your instructor will let you know.\r\n\r\n\r\n\r\n\r\n\r\nPlease note: The examples on this page use the APA referencing style. Please check your course handbook or speak to your instructor about the referencing style required in your area of study.\r\n \r\nShould I cite it?\r\nIf it is your own idea, words or image, have you used it in previously submitted work?\r\n\r\n\tIf yes, you need to cite yourself. Check with your lecturer if this is allowed for your course.\r\n\tIf no, you don't need to cite it.\r\n\r\nIf it is not your idea, word or image, is it common or historical knowledge?\r\n\r\n\tIf yes, you don't need to cite it.\r\n\tIf no, you need to cite it.\r\n\r\n","n":0.05},[{"v":"Referencing","i":1,"n":1},{"v":"Citation","i":0,"n":1}]]},{"i":158,"$":[{"v":"What is referencing?","n":0.577},{"v":"We often use the ideas and work of others to support our own thinkingand make a convincing argument. If the examples or ideas we use come from other people, it is important that we give them appropriate credit for their work. This is called referencing.\r\n\r\nUsing other people's ideas\r\nIn academics and professional settings, referencing means acknowledging others' work and ideas. \r\n\r\nFor academic tasks, you often need to research topics, and collect data, images, and other information to support your point. Whenever you use someone else's ideas, you must inform the reader where the ideas came from.\r\n\r\n\r\nReferencing has two parts: citations, and the reference list.\r\n\r\n \tIn the body of your work, any time you want to use a direct quote from someone else, talk about their ideas more generally, or incorporate a piece of their creative work into your own, you\u2019ll include a citation. Citations offer your audience key pieces of information about the source of an idea, such as the author\u2019s surname and the year of publication.\r\n \tAt the end of your work, you\u2019ll include a reference list or bibliography. These provide the complete publication information for the sources you\u2019ve cited. Every citation will refer to an entry in the reference list or bibliography.\r\n\r\n\r\nEach professional discipline or area of study uses a specific set of referencing rules, called style guides. These style guides explain how to format citations and create a reference list. APA, Chicago, and RMIT Harvard are the names of some popular referencing styles. You can find the rules for these styles by visiting Easy Cite.\r\nWhy is referencing important?\r\nReferencing strengthens your arguments\r\n\r\nReferencing enables others to understand how you came to your conclusions\r\nReferences help others find the information you used to build your arguments. For example, if you use information from the Australian Bureau of Statistics, someone can find this information by checking your reference list and visiting the source. This allows others to check your statements and see how you reached your conclusions.\r\nReferencing shows you are acting with academic integrity\r\nWhen you state where you found your information, you are showing respect for the work of others. This is a key aspect of academic integrity. It shows your audience what knowledge or work already existed, and what ideas are your own. All publications are the intellectual property of the authors, so taking someone\u2019s ideas without acknowledgement and representing them as your own is a form of theft and fraud, known as plagiarism.","n":0.05},[{"v":"Referencing","i":2,"n":1},{"v":"Citation","i":1,"n":1},{"v":"Academic integrity","i":0,"n":0.707}]]},{"i":159,"$":[{"v":"Integration of trigonometric functions","n":0.5},{"v":"\r\nHow do you integrate a trigonometic function? This skill is important for solving problems involving wave patterns in physics, analysing alternating current circuits in engineering and determining structural loads with curved geomoetries in architecture. Use this resource to learn how.\r\n\r\nYou already learnt about the general forms of Circular functions. Examples include \\(\\sin(2x+3)\\), \\(\\cos(5x)\\) and \\(\\sec^{2}(x-2)\\).\r\nRemember:\r\n\\[\\dfrac{d}{dx}(\\cos(x))dx=-\\sin(x)\\] \\[\\dfrac{d}{dx}(\\sin(x))dx=\\cos(x)\\] \\[\\dfrac{d}{dx}(\\tan(x))dx=\\sec^{2}(x)\\]\r\nThis means that the antiderivatives are:\r\n\r\n\\[\\int\\sin(x)dx=-\\cos(x)+c\\] \\[\\int\\cos(x)dx=\\sin(x)+c\\] \\[\\int\\sec^{2}(x)dx=\\tan(x)+c\\]\r\n\r\nwhere \\(c\\) is a constant.\r\nThe more general forms are:\r\n\r\n\\[\\begin{align*} \\int\\sin\\left(ax+b\\right)dx & =-\\frac{1}{a}\\cos\\left(ax+b\\right)+c\\\\ \\int\\cos\\left(ax+b\\right)dx & =\\frac{1}{a}\\sin\\left(ax+b\\right)+c\\\\ \\int\\sec^{2}\\left(ax+b\\right)dx & =\\frac{1}{a}\\tan\\left(ax+b\\right)+c \\end{align*}\\]\r\nwhere \\(a\\), \\(b\\) and \\(c\\) are constants.\r\n\r\nExample 1 \u2013 integrating trigonometric functions\r\nFind \\(\\int5\\cos(x)dx\\).\r\nHere, \\(a=1\\) and \\(b=0\\).\r\n\\[\\int5\\cos(x)dx=5\\sin(x)+c\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercises\r\n\r\n \tCalculate the following.\r\n\t\r\n\t\t\\(\\int\\sec^{2}\\left(\\dfrac{x}{2}\\right)dx\\)\r\n\t\t\\(\\int\\sec^{2}(4x)dx\\)\r\n\t\t\\(\\int2\\cos(1-x)dx\\)\r\n\t\t\\(2\\sin\\left(\\dfrac{5-3x}{4}\\right)dx\\)\r\n\t\r\n\tEvaluate the following.\r\n\t\r\n\t\t\\(\\int_{0}^{\\frac{\\pi}{2}}3\\cos(2x+\\pi)dx\\)\r\n\t\t\\(\\int_{-\\pi}^{0}5\\sin\\left(\\dfrac{x}{2}\\right)dx\\)\r\n\t\t\\(\\int_{0}^{\\frac{\\pi}{2}}2\\sec^{2}\\left(\\dfrac{x}{3}\\right)dx\\)\r\n\t\r\n\tEvaluate the folowing.\r\n\t\r\n\t\t\\(\\int_{0}^{\\frac{\\pi}{6}}3\\sin(\\pi-2x)dx\\)\r\n\t\t\\(\\int_{0}^{\\frac{2\\pi}{3}}\\sec^{2}\\left(\\dfrac{x}{2}\\right)dx\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.098},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":160,"$":[{"v":"Integration of exponential functions","n":0.5},{"v":"\r\nHow do you integrate an exponential function? This skill is important for calculating compound interest over time in finance, determining population growth models in biology, and analysing radioactive decay in physics. Use this resource to learn how.\r\n\r\nExponential functions have the form:\r\n\\[y=ae^{kx}\\]\r\nwhere \\(a\\) and \\(k\\) are constants. Examples include \\(y=e^{2x+3}\\), \\(f(x)=e^{3x}\\) and \\(y=e^{x-1}\\).\r\nThe antiderivative or indefinite integral of \\(e^{x}\\) is \\(e^{x}\\). Therefore:\r\n\\[\\int e^{x}dx=e^{x}+c\\]\r\nwhere \\(c\\) is a constant.\r\nMore generally:\r\n\\[\\int e^{ax+b}dx = \\frac{1}{a}e^{ax+b}+c\\]\r\nwhere \\(a\\), \\(b\\) and \\(c\\) are constants.\r\n\r\nExample 1 \u2013 integrating exponential functions\r\nIntegrate \\(2e^{x}\\) with respect to \\(x\\).\r\n\\[\\int2e^{x} = 2e^{x}+c\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 integrating exponential functions\r\n\r\n \tCalculate the following.\r\n\t\r\n\t\t\\(\\int e^{3x}dx\\)\r\n\t\t\\(\\int e^{2-5x}dx\\)\r\n\t\t\\(\\int \\dfrac{9e^{3x}+5}{e^{2x}}dx\\)\r\n\t\tHint: Divide through first.\r\n\t\r\n\tEvaluate the following.\r\n\t\r\n\t\t\\(\\int_{0}^{2}e^{3x}dx\\)\r\n\t\t\\(\\int_{-1}^{3}e^{2-5x}dx\\)\r\n\t\t\\(\\int_{-1}^{1}\\dfrac{9e^{3x}+5}{e^{2x}}dx\\)\r\n\t\r\n\r\n\r\n\r\n\r\n","n":0.099},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":161,"$":[{"v":"Who will help me?","n":0.5},{"v":"There are many support services available to help you succeed in tertiary study. In this module we'll introduce you to some services that can help you achieve your goals, find your way and support you if things don't go as expected.\r\nDiscover the different services you will need during teritiary studies, including enrolment, IT support, study support, library services, health, wellbeing and safety.\r\n\r\n\r\n\r\nUniversity administration\r\nIf you're unsure where to turn for help with IT issues or administrative matters such as enrolment, or you're looking to connect with other students outside of class, the following resources can help.\r\n\r\n\r\n\r\nEnrolment and other admin\r\nThe enrolment process can be overwhelming and confusing. Support staff will be available to help you with the enrolment process, from accepting your offer to study, planning and enrolling in your subjects, to organizing your timetable and more.\r\n\r\nTo find access to these services, go to your institution's homepage and look for the 'Students' section on the website.\r\n\r\nThe college, school or faculty that manages your course can also help with administrative matters. Contact them online via the Learning Management System (LMS), or call or chat online with your course coordinator.\r\nInformation technology support\r\nIT support, sometimes called ITS, can help with wifi access, passwords, software, and access to university systems.\r\nStudent groups\r\nTertiary institutions have a range of student groups that can support you in your study, connect you with other students, and provide opportunities to engage in activities outside of your coursework.\r\n\r\nThere will be sporting, cultural and social clubs, as well as academic societies you can join to expand your social network and enrich your university experience.\r\n\r\nLook for your institution's student union online which will have information about student groups, as well as services which support your rights as a student.\r\n\r\n\r\n\r\n","n":0.059},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":162,"$":[{"v":"What will I need?","n":0.5},{"v":"Find out the technology and skills you will need in your studies, as well as why and how to use them.\r\nAt tertiary level, you will use many digital resources to help navigate your way through your program. The ability to understand and use various tools and platforms will support your studies and make your experience more rewarding.\r\n\r\n\r\n\r\nTechnology you'll need\r\nAccess to a laptop or PC\r\nYou'll need a laptop or PC to access most of your course material, assignments, and information. The school won't provide them, but you can use computers at computer labs or the library if they're available.\r\n\r\nWiFi access\r\nYou can log in to university WiFi using your student credentials. Having WiFi at home will let you access online services that can assist with your studies.\r\n\r\nA mobile phone\r\nUniversities often offer apps that help you find your way around campus, contact security, manage your timetable, and access your personalized communications. Find essential student resources in the RMIT App (not yet available in Vietnam).\r\n\r\nSoftware\r\nYou'll need to use various systems and tools throughout your studies, from enrolment to graduation. Depending on your program or descipline, you might need specific hardware or software. Ask your course coordinator or IT support for guidance.\r\n\r\nHow you'll use it\r\nYou will need to perform various tasks using digital technology. These might include:\r\n\r\n \tLearning\r\nAccess course information and interact with classmates and teachers online.\r\n \tParticipation\r\nJoin in and contribute to online classes, discussions and projects.\r\n \tCollaboration and communication\r\nShare information and ideas, network, and complete group assignments.\r\n \tCreation\r\nUse your computer to create videos, essays, reports, literature reviews, presentations, infographics, and spreadsheets for assessments.\r\n \tFile management\r\nManage and store\u00a0information and data,\u00a0assignments, required reading and research.\r\n \tPortfolios\r\nYou may consider collecting your work to present to prospective employers. It may also be a requirement of your studies.\r\n\r\nMany of these skills form the basis of digital capabilities considered essential in your studies, life and work.\r\n","n":0.057},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":163,"$":[{"v":"What will I do?","n":0.5},{"v":"Transitioning to higher education, whether from high school or work, is a significant step. Students from high school may feel underprepared, while those returning to study from work might may take some time adjusting to study expectations.\r\n\r\nPeople who have moved to Australia or Vietnam for the first time may feel overwhelmed by cultural adjustments.\r\n\r\nAll of these feelings are valid and normal. Let's explore some tips to help you get started on this journey and find out about RMIT's support services.\r\n\r\n\r\n\r\nOrientation programs\r\nTertiary institutions aim to ensure you get the most out of your time in your study. Your institution will run an orientation program to help you get started, including events, activities and resources.\r\nThe first few months of university study will be a time of significant change. You're expected to navigate a new campus, meet new people and develop a new way of learning. While you may work out many of these changes yourself, there's a wide range of support to help you transition to this new environment.\r\n\r\nOften your university orientation program begins before the start of classes, so you have the time to explore the campus, meet new people, and develop new habits. Your university email will provide important communication to help you get the most out of the orientation program. \r\n\r\nThere'll be events and activities run by clubs and societies and offers from partner organisations. Many of these will be promoted on campus, so travelling in to campus if you can during this time may help you get the most out of orientation.\r\n\r\nThe people who will help you through your studies will be your teachers, professional and academic support services, and your fellow students. This period is perfect to get to know them better. Once classes start, so does the coursework.\r\n\r\nIf you enrol late or miss the orientation program, you might have to work a little harder to adapt to university life. Keep an eye out for support information that will be available all year round.\r\n\r\n\r\n","n":0.055},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":164,"$":[{"v":"About the institution","n":0.577},{"v":"This section will introduce you to the people and places that make up a tertiary institution and help unpack some of the concepts so they feel a little more familiar.\r\nTertiary institutions are large places with lots of different spaces and staff members who support these spaces and students. Like any community, everyone contributes to keep things running. We might only view these educational organisations as places to learn. However, they do so much more to contribute to society, so having an understanding of how they operate will help you get the most out of your time there.\r\nSpaces and resources\r\nA university is much more than a collection of classrooms. Universities provide students with a variety of study, collaboration, social, recreational and creative spaces.\r\n\r\n\r\n\r\nThe campus\r\nUniversity campuses are vibrant places and can be crowded with people walking to various buildings and social spaces. The campus community is generally made up of students, teachers, professional staff, and visitors.\r\n\r\nSome of your learning may take place on campus where spaces are set up for a variety of different types of learning including:\r\n\r\n \tlectures\r\n \tlabs\r\n \tclassrooms for tutorials\r\n \tmedia studios\r\n \tcreative spaces\r\n \tstudy spaces\r\n \tthe library\r\n \tcommon areas\r\n \tcafes\r\n\r\nWifi allows you to connect your PC, laptop, mobile phone or tablet device without using a cable. Wifi access points are across the campus, making everywhere a study space.\r\n\r\nUniversity campuses provide recreational and social spaces for students such as sports facilities, like gyms, basketball courts and running tracks. Social events are held on campus for students to hang out with classmates and get to know their peers.\r\n\r\nUniversity libraries as community and cultural spaces\r\nUniversity libraries are centrally located on campus where the university community can come together to meet, collaborate and build a sense of place. Libraries are often positioned near other services where you may seek information and advice, like the enrolment centre and student union.\r\n\r\nLibraries help build community in several ways. They offer study support through drop-ins or consultations, workshops and training, as well as cultural events like exhibitions and seminars. At the same time, libraries also provide a quiet environment for students to study on their own.\r\n\r\nSome Libraries manage makerspaces, where students gather to work on projects while sharing ideas, equipment, and knowledge. A makerspace is often equipped with 3D printers and laser cutters.\r\n\r\n\r\n\r\nStudy spaces\r\nThe space in which you work plays an important role in productivity as well as improving focus and efficiency.\r\n\r\nWhen setting up a study space, it's a good idea to have all the materials you need within reach. That could be physical items like a notebook, pens, highlighters, and course material like course notes, textbooks, and calculators.\r\n\r\nWhen setting up a digital study space consider bookmarking study resources frequently used to improve organisation. It's also helpful to set up some folders for online documents for your courses.\r\n\r\nWhen working in a group space on campus, choose a space that will enable you and your group members to talk freely, exchange ideas and plan your work. Booking a group room on campus is one option and another is to arrange an online meeting with your group.\r\n\r\n","n":0.044},[{"v":"Student support","i":1,"n":0.707},{"v":"New students","i":0,"n":0.707}]]},{"i":165,"$":[{"v":"Higher order derivatives","n":0.577},{"v":"\r\nHigher order derivatives reveal even deeper insights into the behaviour of functions, such as jerk in motion or the curvature of graphs. These concepts are important in fields like physics, engineering and economics, where understanding complex dynamics is crucial.\r\n\r\nHigher order derivatives are derivatives of derivatives, providing insights into how the rate of change itself changes. You may have already encountered second derivatives when looking at Maxima and minima.\r\nOnce you reach a derivative that is \\(0\\), any further derivatives will also be \\(0\\).\r\nThere are a few ways to denote higher order derivatives for a function \\(y=f(x)\\), as summarised in the table.\r\n\r\n\r\n\r\nPoints of inflection\r\nWhen graphing a function, the second derivative is the rate of change of the derivative function. It describes how much the function curves inwards (its concavity) and indicates whether the slope of the function is increasing or decreasing.\r\n\r\n\r\n \tIf the graph of the function is concave upward, the second derivative is positive. This indicates that the slope is increasing.\r\n \tIf the graph of the function is concave downward, the second derivative is negative. This indicates that the slope is decreasing.\r\n \tAt a point of inflection, where the curve changes concavity, the second derivative is \\(0\\).\r\n\r\n\r\nExample 1 \u2013 finding higher order derivatives\r\nFind the \\(4^{\\textrm{th}}\\) derivative of \\(f(x)=\\sin(2x)\\).\r\n\\[\\begin{align*} f'(x) & = 2\\cos(2x)\\\\\r\nf''(x) & = -4\\sin(2x)\\\\\r\nf'''(x) & = -8\\cos(2x)\\\\\r\nf''''(x) & = 16\\sin(2x)\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding higher order derivatives\r\n\r\n\tFind the third derivative of \\(y=x^{5}-2x^{2}+8x\\).\r\n\tIf \\(f(x)=e^{x^{3}}\\), find \\(f''(x)\\).\r\n\tGiven \\(y=t^{4}+\\dfrac{1}{t^{4}}\\), find \\(\\dfrac{d^{3}y}{dt^{3}}\\).\r\n\tThe displacement of a particle at time \\(t\\) is described by the function \\(s(t)=t^{2}-\\cos(3t)\\). What is the initial acceleration of the particle?\r\n\tFind any points of inflection for the graph of the function \\(y=\\dfrac{x^{4}}{4}-2x^{3}+6x^{2}+x-10\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.061},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":166,"$":[{"v":"Partial differentiation","n":0.707},{"v":"\r\nPartial derivatives help us analyse the influence of each individual variable on a system. This is crucial for understanding surfaces and optimisation problems in fields like engineering, physics and economics. Use this resource to learn how to find partial derivatives.\r\n\r\nPartial derivatives reveal how a function with many variables changes when you adjust just one of the variables in the input. Consider the following equation: \r\n\\[z=x^{2}+4y^{2}\\]\r\nHere, we have two independent variables, so the function is \\(f(x,y)\\). The graph exists in a three-dimensional space.\r\nWhen we differentiate such a function, we need to be able to differentiate \\(z\\) with respect to either \\(x\\) or \\(z\\). This involves treating the other variable as a constant. We calculate the partial derivatives:\r\n\r\n\t\\(\\dfrac{\\partial z}{\\partial x}\\), the partial derivative of \\(z\\) with respect to \\(x\\) where we treat \\(y\\) as a constant \r\n\t\\(\\dfrac{\\partial z}{\\partial y}\\), the partial derivative of \\(z\\) with respect to \\(y\\) where we treat \\(x\\) as a constant.\r\n\r\nThe symbol we use to denote a partial differential is the curve d, \\(\\partial\\), often called \"del\".\r\nAn alternative notation for partial derivatives is \\(f_{x}\\) for \\(\\dfrac{\\partial f}{\\partial x}\\)and \\(f_{y}\\) for \\(\\dfrac{\\partial f}{\\partial y}\\).\r\n\r\nExample 1 \u2013 evaluating partial derivatives\r\nIf \\(z=x^{2}+4y^{2}\\), find \\(\\dfrac{\\partial z}{\\partial x}\\) and \\(\\dfrac{\\partial z}{\\partial y}\\).\r\n\\(\\dfrac{\\partial z}{\\partial x}\\) is the partial derivative of \\(z\\) with respect to \\(x\\), treating \\(y\\) as a constant.\r\n\\[\\begin{align*} \\frac{\\partial z}{\\partial x} & = 2x+0\\\\\r\n& = 2x\r\n\\end{align*}\\]\r\n\\(\\dfrac{\\partial z}{\\partial y}\\) is the partial derivative of \\(z\\) with respect to \\(y\\), treating \\(x\\) as a constant.\r\n\\[\\begin{align*} \\frac{\\partial z}{\\partial y} & = 0+8y\\\\\r\n& = 8y\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nEvaluating partial derivatives at a point\r\nWe can find partial derivatives at particular points. For example, \\(f_{x}(2,1)\\) refers to the value of the partial derivative of \\(f\\) with respect to \\(x\\) at the point where \\(x=2\\) and \\(y=1\\).\r\n\r\nExample \u2013 evaluating partial derivatives at a point\r\nIf \\(f(x,y)=2x^{3}y+3y^{2}\\), find \\(f_{y}(1,3)\\).\r\n\\(f_{y}\\) indicates that we are looking for the derivative of \\(f\\) with respect to \\(y\\), treating \\(x\\) as a constant. \r\n\\[f_{y} = 2x^{3}+6y\\]\r\nWe can then substitute \\(x=1\\) and \\(y=3\\) into \\(f_{y}\\) to find the partial derivative at that point.\r\n\\[\\begin{align*} f_{y}(1,3) & = 2(1)^{3}+6(3)\\\\\r\n& = 2+18\\\\\r\n& = 20\r\n\\end{align*}\\]\r\n\r\nHigher order partial derivatives\r\nA function \\(z=f(x,y)\\) has two partial derivatives: \\(\\dfrac{\\partial f}{\\partial x}\\) and \\(\\dfrac{\\partial f}{\\partial y}\\). This kind of function will have four second order partial derivatives.\r\n\r\n \tDifferentiating with respect to \\(x\\), then with respect to \\(x\\) gives us:\r\n\t\\[\\frac{\\partial}{\\partial x}\\left(\\frac{\\partial f}{\\partial x}\\right)=\\frac{\\partial^{2}f}{\\partial x^{2}}=f_{xx}\\]\r\n \tDifferentiating with respect to \\(y\\), then with respect to \\(y\\) gives us:\r\n\t\\[\\frac{\\partial}{\\partial y}\\left(\\frac{\\partial f}{\\partial y}\\right)=\\frac{\\partial^{2}f}{\\partial y^{2}}=f_{yy}\\]\r\n \tDifferentiating with respect to \\(x\\), then with respect to \\(y\\) gives us:\r\n\t\\[\\frac{\\partial}{\\partial y}\\left(\\frac{\\partial f}{\\partial x}\\right)=\\frac{\\partial^{2}f}{\\partial x\\partial y}=f_{xy}\\]\r\n \tDifferentiating with respect to \\(y\\), then with respect to \\(x\\) gives us:\r\n\t\\[\\frac{\\partial}{\\partial x}\\left(\\frac{\\partial f}{\\partial y}\\right)=\\frac{\\partial^{2}f}{\\partial y\\partial x}=f_{yx}\\]\r\n\r\n\\(f_{xy}\\) is always equal to \\(f_{yx}\\).\r\n\r\nExample \u2013 evaluating higher order partial derivatives\r\nFind all second order partial derivatives of \\(f(x,y)=x^{2}y^{3}+2x\\cos(y)\\).\r\nFirst, we need to find the first order partial derivative with respect to \\(x\\) and \\(y\\). Remember, when we derive \\(f\\) with respect to \\(x\\), we treat all the \\(y\\) terms as constants. In this case, it would be \\(y^{3}\\) and \\(2\\cos(y)\\).\r\n\\[f_{x}=2xy^{3}+2\\cos(y)\\]\r\nFor \\(f_{y}\\), we derive \\(f\\) with respect to \\(y\\), treating all the \\(x\\) terms as constants, i.e. \\(x^{2}\\) and \\(2x\\).\r\n\\[f_{y}=3x^{2}y^{2}-2x\\sin(y)\\]\r\nThe second order partial derivatives of \\(f(x,y)\\) are \\(f_{xx}\\), \\(f_{yy}\\) and \\(f_{xy}=f_{yx}\\).\r\nFor \\(f_{xx}\\), we find the derivative of \\(f_{x}=2xy^{3}+2\\cos(y)\\) with respect to \\(x\\) and treat the \\(y\\) terms as a constant, i.e. \\(y^{3}\\) and \\(2\\cos(y)\\).\r\n\\[f_{xx}=2y^{3}\\]\r\nFor \\(f_{xy}\\), we find the derivative of \\(f_{x}=2xy^{3}+2\\cos(y)\\) with respect to \\(y\\) and treat \\(x\\) as a constant, i.e. \\(2x\\).\r\n\\[f_{xy}=6xy^{2}-2\\sin(y)\\]\r\n\\(f_{xy}\\) is the same as \\(f_{yx}\\), so \\(f_{yx}=6xy^{2}-2\\sin(y)\\). We can prove this by deriving \\(f_{y}=3x^{2}y^{2}-2x\\sin(y)\\) with respect to \\(x\\) while treating the \\(y\\) terms, \\(3y^{2}\\) and \\(\\sin(y)\\), as constants.\r\n\\[f_{yx}=6xy^{2}-2\\sin(y)\\]\r\nFinally, for \\(f_{yy}\\), we find the derivative of \\(f_{y}=3x^{2}y^{2}-2x\\sin(y)\\) with respect to \\(y\\) while treating the \\(x\\) term, \\(x^{2}\\), as a constant.\r\n\\[f_{yy}=6x^{2}y-2x\\cos(y)\\]\r\n\r\nExercise \u2013 finding partial derivatives\r\n\r\n\tFind the partial derivatives \\(\\dfrac{\\partial z }{\\partial x}\\) and \\(\\dfrac{\\partial z }{\\partial y}\\) for each of the following.\r\n\t\r\n\t\t\\(z=3x^{4}+2y^{3}\\)\r\n\t\t\\(z=x^{2}y\\)\r\n\t\t\\(z=3xe^{2y}\\)\r\n\t\t\\(z=\\ln(x^{3}y^{5}-2)\\)\r\n\t\r\n\tFind the value of the indicated partial derivative at the given point.\r\n\t\r\n\t\t\\(f_{x}(2,3)\\) where \\(f(x,y)=x^{4}-4y^{2}\\)\r\n\t\t\\(f_{y}(-1,1)\\) where \\(f(x,y)=\\ln(x^{2}+y^{3})\\)\r\n\t\r\n\tFind the first and second order partial derivatives of the following.\r\n\t\r\n\t\t\\(f(x,y)=x\\ln(y)\\)\r\n\t\t\\(f(x,y)=x^{3}+x^{2}y-3xy^{2}+y^{3}\\)\r\n\t\t\\(f(x,y)=\\sin(xy)\\)\r\n\t\t\\(f(x,y)=x\\cos(y)+ye^{x}\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.039},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":167,"$":[{"v":"Implicit differentiation","n":0.707},{"v":"\r\n\r\nImplicit differentiation enables you to find the derivative of y with respect to x without having to solve the original equation for y.\r\n\r\nImplicit functions\r\nSometimes, equations involve variables that are intertwined and not easily separated. For example, \\(y^{5}+3xy+x^{2}-5=0\\) or \\(y=\\sin(xy))\\).\r\nIn such expressions, \\(y\\) is said to be an implicit function of \\(x\\) as we cannot rearrange the expression to the form \\(y=f(x)\\). We can use implicit differentiation techniques to find \\(\\dfrac{dy}{dx}\\) without having to solve the given equation for \\(y\\).\r\nTo do this, we need to use implicit differentiation. This process uses the chain rule because it is assumed that \\(y\\) is an unknown function of \\(x\\).\r\n\r\nUsing implicit differentiation\r\nConsider a function \\(f(y)\\) which has intertwined variables \\(x\\) and \\(y\\). To differentiate the function with respect to \\(x\\), we use the chain rule:\r\n\\[\\frac{d}{dx}(f(y))=\\frac{d}{dy}(f(y))\\times\\frac{dy}{dx}(f(y))\\]\r\nIn other words, to differentiate \\(f(y)\\) with respect to \\(x\\), we take the derivative with respect to \\(y\\), then multiply by the derivative with respect to \\(x\\) \\(\\left(\\dfrac{dy}{dx}\\right)\\).\r\n\r\nExample 1 \u2013 using implicit differentiation\r\nFind \\(\\dfrac{d}{dx}(y^{2}x)\\).\r\nFor this function, we can use the product rule first.\r\n\\[\\begin{align*} \\frac{d}{dx}(y^{2}x) & = \\left(y^{2}\\cdot\\frac{d}{dx}(x)\\right)+\\left(x\\cdot\\frac{d}{dx}(y^{2})\\right)\\\\\r\n& = (y^{2}\\cdot1)+\\left(x\\cdot\\frac{d}{dx}(y^{2})\\right)\r\n\\end{align*}\\]\r\nTo find \\(\\dfrac{d}{dx}(y^{2})\\), we need to use the chain rule. For every expression we differentiate that contains an \\(x\\) variable, we have to multiply by \\(\\dfrac{dy}{dx}\\).\r\n\\[\\begin{align*} \\frac{d}{dx}(y^{2}) & = \\frac{d}{dy}(y^{2})\\times\\frac{dy}{dx}(y^{2})\\\\\r\n& = 2y\\times\\frac{dy}{dx}(y^{2})\r\n\\end{align*}\\]\r\nPutting this all together:\r\n\\[\\begin{align*} \\frac{d}{dx}(y^{2}x) & = y^{2}+x\\cdot2y\\frac{dy}{dx}\\\\\r\n& = y^{2}+2xy\\frac{dy}{dx}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using implicit differentiation\r\n\r\n\tFind \\(\\dfrac{d}{dx}\\left(\\dfrac{x}{y}\\right)\\). Hint: use quotient rule.\r\n\tFind \\(\\dfrac{d}{dx}\\left(\\dfrac{x+y}{x-y}\\right)\\).\r\n\tFind the value of \\(\\dfrac{dy}{dx}\\) at the point \\((1, 2)\\) if \\(x^{2}+y=7-2xy\\).\r\n\tFind \\(\\dfrac{dy}{dx}\\) if e\\(^{x}\\sin(y)=x\\).\r\n\tFind the equation of the tangent line to the circle \\(x^{2}+y^{2}=4\\) at the point \\((1,\\sqrt{3})\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.063},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":168,"$":[{"v":"Linear approximation and differentials","n":0.5},{"v":"\r\n\r\nWe can use derivatives to estimate how a function behaves around a certain point. With linear approximation, we can analyse these small variations and predict values. These methods are useful in fields like physics and engineering, where precise measurements and predictions are crucial, but often challenging to obtain directly.\r\n\r\nSometimes a small change in one variable can cause a big change in another value. For example, a small increase (or error) in the radius of a sphere means a lot more volume is added. If you estimate the small error in one variable, you can calculate the significant change in a larger variable by using derivatives. This involves linear approximation using tangents.\r\n\r\nApproximating functions using the tangent\r\nThe curve of any function that we can differentiate will appear linear if we look at it over a small enough interval.\r\nIf \\(a\\) is a point within that interval, we can use the tangent line through \\((a,f(a))\\) to get an approximation of values of \\(f(x)\\), as long as \\(x\\) remains near point \\(a\\). This forms the basis of linear approximation.\r\nThe tangent line is defined by the equation:\r\n\\[y=f(a)+f'(a)(x-a)\\]\r\nWhen we use this to approximate a function, we get the linear approximation formula:\r\n\\[ L(x)=f(a)+f'(a)(x-a) \\]\r\nYou might be wondering: why bother using the tangent to approximate a function if you could just substitute the value into the original function and get the exact answer? In some cases, the original function might be too complex or intensive to evaluate. It can be much easier and quicker to use a linear tangent. Linear approximation is also handy when you don't know the original function or it is difficult to work with. \r\nRemember, the tangent method is used for approximation.\r\n\r\nExample 1 \u2013 approximating functions\r\nConsider the graph of the function \\(f(x)=e^{2-x}\\) near \\(x=2\\). Approximate the value of the function at \\(x=1.9\\) and \\(x=2.05\\).\r\nFirst, find \\(f'(x)\\).\r\n\\[f'(x)=-e^{2-x}\\]\r\nTo find the equation for the tangent:\r\n\\[\\begin{align*} y & = f(a)+f'(a)(x-a)\\\\\r\n& = f(2)+f'(2)(x-2)\\\\\r\n& = e^{2-2}+(-e^{2-2})(x-2)\\\\\r\n& = 1+(-1)(x-2)\\\\\r\n& = 1-x+2\\\\\r\n& = 3-x\r\n\\end{align*}\\]\r\nThus, the equation of the tangent at \\(x=2\\) is \\(y=3-x\\).\r\n\r\nWe can then estimate \\(L(1.9)\\) and \\(L(2.05)\\) by substituting them into the equation for the tangent.\r\n\\[\\begin{align*} L(1.9) & = 3-1.9\\\\\r\n& = 1.1\r\n\\end{align*}\\]\r\nCompare this to \\(f(1.9)=e^{2-1.9}\\approx1.105\\). It's a pretty good approximation.\r\nFor \\(L(2.05)\\):\r\n\\[\\begin{align*} L(2.05) & = 3-2.05\\\\\r\n& = 0.95\r\n\\end{align*}\\]\r\nCompare this to \\(f(2.05)=e^{2-2.05}\\approx0.951\\). \r\n\r\n\r\n\r\nDifferentials\r\nDifferentials can also be used to estimate the amount a function will change as a result of a small change in \\(x\\) or a small change in \\(y\\). These are denoted by \\(\\triangle x\\) and \\(\\triangle y\\), respectively.\r\nYou would use differentials instead of the linear approximation formula when you are looking at change, rather than to estimate specific values at a certain point.\r\nConsider this graph of function \\(y=f(x)\\). Point \\(P\\) is at \\(x=1\\), and another point \\(Q\\) is shown. \r\n\r\nThe tangent at \\(P\\) can be used to estimate the change in \\(x\\) and \\(y\\) values, based on the basic gradient formula: \\(y=mx\\).\r\nThe gradient \\(m\\) is the same as rise over run or the derivative of \\(y\\) with respect to \\(x\\). In other words, \\(\\dfrac{dy}{dx}\\).\r\nGenerally, as \\(x\\) approaches \\(0\\):\r\n\\[\\frac{\\triangle y}{\\triangle x}\\rightarrow \\frac{dy}{dx}\\]\r\nFor small values of \\(x\\), we can approximate \\(y\\):\r\n\\[\\triangle y\\approx\\frac{dy}{dx}\\triangle x\\]\r\n\r\nExample 1 \u2013 approximating small changes\r\nThe side of a square has length \\(5\\textrm{ cm}\\). How much will the area of the square increase when the side length is increased by \\(0.01\\textrm{ cm}\\)?\r\nLet the area of the square be \\(A\\) and the length of a side be \\(x\\textrm{ cm}\\). This means that \\(A=x^{2}\\) and \\(\\dfrac{dA}{dx}=2x\\). \r\n\\[\\begin{align*} \\triangle A & \\approx\\frac{dA}{dx}\\triangle x\\\\ \r\n& = 2x\\triangle x\\\\ \r\n& = 2\\times5\\times0.01\\\\\r\n& = 0.1 \r\n\\end{align*}\\]\r\nThe area of the square will increase by approximately \\(0.1\\textrm{ cm}^{2}\\).\r\n\r\n\r\n\r\nExercise \u2013 making approximations\r\nThe following formulas may be helpful for completing this exercise:\r\n\r\n\tSurface area of a sphere: \\(A=4\\pi r^{2}\\)\r\n\tVolume of a cylinder: \\(V=\\pi r^{2}h\\)\r\n\tVolume of a sphere: \\(V=\\dfrac{4}{3}\\pi r^{3}\\)\r\n\r\n\r\n \tUse linear approximation of the function \\(f(x)=\\sqrt{x}\\) to find the approximate value of \\(\\sqrt{16.1}\\).\r\n \tUse linear approximation to find an approximate value of \\(\\cos(59^{\\circ})\\). Hint: Convert degrees to radians using \\(1^{\\circ}=\\dfrac{\\pi}{180}\\textrm{ radians}\\).\r\n \tIf the radius of a sphere is increased from \\(10\\textrm{ cm}\\) to \\(10.1\\textrm{ cm}\\), what is the approximate increase in surface area?\r\n \tThe height of a cylinder is \\(10\\textrm{ cm}\\) and its radius is \\(4\\textrm{ cm}\\). Find the approximate increase in volume when the radius increases to \\(4.02\\textrm{ cm}\\).\r\n \tAn error of \\(3\\%\\) is made in measuring the radius of a sphere. Find the percentage error in volume.\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.037},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":169,"$":[{"v":"Rates of change","n":0.577},{"v":"\r\nIf there is a relationship between two or more variables, then there will also be a relationship between how these variables change. You may need to find how fast one variable changes in relation to another variable that is changing. This is called the rate of change.\r\n\r\nWe often study the relationship between two or more variables, like:\r\n\r\n\tthe area and radius of a circle (\\(A=\\pi r^{2}\\))\r\n\tthe length of a side and volume of a cube (\\(V=l^{3}\\))\r\n\tthe days since first case and number of people with an infectious disease.\r\n\r\nWhere there is a relationship between variables, there is also be a relationship between the rates at which the variables change.\r\n\r\nFinding the rate of change\r\nWe can use differentiation to find the function that defines the rate of change between variables. This involves differentiating with respect to time \\(t\\).\r\nTo do this, we use the chain rule.\r\n\\[\\frac{dy}{dt}=\\frac{dy}{dx}\\times\\frac{dx}{dt}\\]\r\nFor the example of area and radius of a circle:\r\n\\[\\begin{align*} A & = \\pi r^{2}\\\\\r\n\\frac{dA}{dr} & = 2\\pi r\r\n\\end{align*}\\]\r\nTo find the rate the area changes with respect to time \\(t\\), we would use:\r\n\\[\\begin{align*} \\frac{dA}{dt} & = \\frac{dA}{dr}\\times\\frac{dr}{dt}\\\\\r\n& = 2\\pi r\\times\\frac{dr}{dt}\r\n\\end{align*}\\]\r\nFor the example of length of a side and volume of a cube:\r\n\\[\\begin{align*} V & = l^{3}\\\\\r\n\\frac{dV}{dl} & = 3l^{2}\r\n\\end{align*}\\]\r\nTo find the rate the volume changes with respect to time \\(t\\), we would use\r\n\\[\\begin{align*} \\frac{dV}{dt} & = \\frac{dV}{dl}\\times\\frac{dl}{dt}\\\\\r\n& = 3l^{2}\\times\\frac{dl}{dt}\r\n\\end{align*}\\]\r\n\r\nExample 1 \u2013 finding the rate of change\r\nA balloon has a small hole and its volume \\(V\\) in cubic centimeters after \\(t\\) seconds is \\(V=66-10t-0.01t^{2}\\) for \\(t>0\\). Find the rate of change of volume after \\(10\\) seconds.\r\nFirst, we find the derivative.\r\n\\[\\begin{align*} V & = 66-10t-0.01t^{2}\\\\\r\n\\frac{dV}{dt} & = -10-0.02t\r\n\\end{align*}\\]\r\nNow, we can substitute \\(t=10\\) into the derivative to find the rate of change of volume.\r\n\\[\\begin{align*} \\frac{dV}{dt} & = -10-0.02(10)\\\\\r\n& = -10.2\\textrm{ cm}^{3}/{\\textrm{s}}\r\n\\end{align*}\\] \r\nThe volume of the balloon is decreasing at a rate of \\(10.2\\textrm{ cm}^{3}/{\\textrm{s}}\\).\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\nExercise \u2013 finding the rate of change\r\n\r\n \tThe radius of a spherical balloon is increasing at a rate of \\(3\\textrm{ cm/min}\\). At what rate is the volume increasing when the radius is \\(5\\textrm{ cm}\\)?\r\n \tIf the displacement of an object from a starting point is given by \\(s(t)=\\sin(t)-2\\cos(t)\\), find the velocity when \\(t=1\\). Hint: \\(v(t)=s'(t)=\\dfrac{ds}{dt}\\).\r\n \tThe function \\(n(t)=200t-100\\sqrt{t}\\) describes the spread of a virus where \\(t\\) is the number of days since the initial infection and \\(n\\) is the number of people infected. Find the rate at which \\(n\\) is increasing at the instant when \\(t=4\\).\r\n \tIf \\(y=\\left(x-\\dfrac{1}{x}\\right)^{2}\\) find \\(\\dfrac{dx}{dt}\\) when \\(x=2\\), given \\(\\dfrac{dy}{dt}=1\\).\r\n \tA hollow right circular cone is held point downwards under a tap leaking at the rate of \\(2\\textrm{ cm}^{3}/\\textrm{s}\\). Find the rate of rise of water level when the depth is \\(6\\textrm{ cm}\\) given that the height of the cone is \\(18\\textrm{ cm}\\) and its radius is \\(12\\textrm{ cm}\\). Hint: Use the properties of similar triangles to find a relationship between radius and height.\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.046},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":170,"$":[{"v":"Curve sketching","n":0.707},{"v":"\r\nYou can sketch an accurate graph of a function if you know some of its key characteristics. Use this resource to learn how to bring together everything to sketch a curve.\r\n\r\nTo sketch a curve, it is helpful to know the key points of the function, like:\r\n\r\n\tmaxima, minima, or turning points\r\n\t\\(x\\)- and \\(y\\)-intercepts\r\n\tregions where the gradient is positive or negative.\r\n\r\n\r\nStationary points\r\nA stationary point is a point on a graph of a function \\(y=f(x)\\) where the tangent to the curve is horizontal. From Maxima and minima, you should recognise that this is where \\(y'=f'(x)=0\\).\r\nThese are often also called turning points.\r\n\r\nMaximum stationary point\r\nA maximum stationary point occurs at \\(x=a\\) if \\(f'(a)=0\\) and \\(f'(x)>0\\) for \\(x<a\\) and \\(f'(x)<0\\) for \\(x>a\\).\r\n\r\n\r\nMinimum stationary point\r\nA minimum stationary point occurs at \\(x=a\\) if \\(f'(a)=0\\) and \\(f'(x)<0\\) for \\(x<a\\) and \\(f'(x)>0\\) for \\(x>a\\).\r\n\r\n\r\nSketching a curve\r\nTo sketch a curve, we can follow these steps:\r\n\r\n \tFind the \\(x\\)- and \\(y\\)-intercepts.\r\n \tFind the stationary point/s.\r\n \tDetermine whether the stationary points are maximum or minimum values.\r\n \tPlot the intercepts and stationary points and join them with a smooth curve.\r\n\r\n\r\nExample \u2013 sketching a curveSketch the graph of \\(y=x^{3}-x\\).\r\n\r\n\tFind the \\(x\\)-intercepts by letting \\(y=0\\) and solving for \\(x\\).\r\n\t\\[\\begin{align*} 0 & = x^{3}-x\\\\\r\n\t0 & = x(x^{2}-1)\\\\\r\n\tx=0 & \\textrm{ or }0=x^{2}-1\\\\\r\n\tx=0 & \\textrm{ or }x=\\pm1\r\n\t\\end{align*}\\]\r\n\tThis means the \\(x\\)-intercepts are at \\((-1,0)\\), \\((0,0)\\) and \\((1,0)\\). For this function, the \\(y\\)-intercept occurs at the origin.\r\n\tFind the stationary point/s by letting \\(y'=0\\) and solving for \\(x\\).\r\n\t\\[\\begin{align*} y' & = 3x^{2}-1\\\\\r\n\t& = 0\\\\\r\n\t1 & = 3x^{2}\\\\\r\n\t\\frac{1}{3} & = x^{2}\\\\\r\n\tx & = \\pm \\sqrt{\\frac{1}{3}}\\\\\r\n\t& = \\pm \\frac{1}{\\sqrt{3}}\\\\\r\n\t& \\approx \\pm0.58\r\n\t\\end{align*}\\]\r\n\tWe need to substitute \\(x=-0.58\\) and \\(x=0.58\\) back into \\(y\\) to find the coordinates for the stationary points.\r\n\t\\[\\begin{align*} y & = x^{3}-x\\\\\r\n\t& = (-0.58)^{3}-(-0.58)\\\\\r\n\t& = 0.38\r\n\t\\end{align*}\\] \\[\\begin{align*} y & = x^{3}-x\\\\\r\n\t& = (0.58)^{3}-(0.58)\\\\\r\n\t& = -0.39\r\n\t\\end{align*}\\]\r\n\tSo, the stationary points occur at around \\((-0.58, 0.38)\\) and \\((0.58,-0.39)\\).\r\n\tDetermine whether the stationary points are maximum or minimum values by taking the second derivative and evaluating whether it is \\(<0\\), \\(>0\\) or \\(=0\\).\r\n\t\\[y''=6x\\]\r\n\tFor \\(x=-0.58\\):\r\n\t\\[\\begin{align*} y'' & = 6(-0.58)\\\\\r\n\t& = -3.46\\\\\r\n\t& <0\\quad\\textrm{local maximum}\r\n\t\\end{align*}\\]\r\n\tFor \\(x=0.58\\):\r\n\t\\[\\begin{align*} y'' & = 6(0.58)\\\\\r\n\t& = 3.46\\\\\r\n\t& >0\\quad\\textrm{local minimum}\r\n\t\\end{align*}\\]\r\n\tWe can also look at values on either side of the stationary point to decide whether it is a maximum or minimum. For \\(x=-0.58\\):\r\n\t\r\n\tFor \\(x=0.58\\):\r\n\t\r\n\t\r\n\tPut it all together by first plotting the \\(x\\)-intercepts and stationary points.\r\n\t\r\n\tWe know that point \\(E\\) is a maximum and point \\(D\\) is a minimum, so we can graph the curve.\r\n\t\r\n\r\n\r\nExercise \u2013 sketching a curve\r\nSketch the graphs of the following functions, showing all intercepts and turning points.\r\n\r\n \t\\(y=x^{2}-4x\\)\r\n \t\\(y=x^{3}-2x^{2}+x\\)\r\n \t\\(y=6-x-x^{2}\\)\r\n \t\\(y=(x+1)^{4}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.049},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":171,"$":[{"v":"Sustainability","n":1},{"v":"Sustainability means being able to do the same thing, in the same way, over time. Sustainability is woven throughout our lives, so understanding how it impacts life and work is important for all of us, no matter our career paths or where we live.\r\nWe can already see changes towards a green and fair future being made by people and organisations in industries and workplaces around Australia and the world. A sustainable future is one where we can safely live on our planet Earth for a very long time.\r\n\r\n\r\n\r\nCreating systems, processes and products that minimise the negative impact humans have on the environment and our own wellbeing is a focus of sustainability. The climate crisis, waste, fast fashion, species loss, access to safe water, reducing natural disasters and preventing pollution are just some of the issues that are part of thinking about sustainability.\r\n\r\nThe United Nations (UN) is an international organisation with 193 member states where the world's nations can come together to discuss common problems and find shared solutions. The UN has, in recent years, set up working groups to develop guidelines and goals for sustainability. They call for \"action by all countries, poor, rich and middle-income to promote prosperity while protecting the planet\" (UN, n.d.). Individuals, organisations and governments can participate in these Sustainable Development Goals, which will be covered here.\r\n\r\nWhat you will learn:\r\n\r\n \twhat sustainability is and why is it important\r\n \twhere sustainability fits in the United Nations (UN) charter\r\n \twhat the UN Sustainable Development Goals (SDGs) are and where they are used\r\n \twhat systems thinking and a circular economy are\r\n \thow sustainability can be part of your life.\r\n\r\nIn this section\r\n\r\n\r\nUnited Nations Sustainable Development Goals (SDGs)\r\nUnderstanding the United Nations Sustainable Development Goals and how they were developed. You can even track their progress.\r\n\r\nSystems thinking\r\nExplore \"systems thinking\", which is a way of looking at the complex nature of our world and how things are interconnected. Systems thinking can be used as a way of understanding and dealing with environmental problems. This section examines the relationship between people, the environment and business, and gives an example of how to apply systems thinking to tricky problems.\r\n\r\nSustainable Development Goals in your life\r\nLearn about examples of the SDGs in industry, fashion, construction, events, creative practice, at RMIT and in other parts of your life. Learn how to create a green event or\u00a0about sustainable fashion and construction. This section also includes a learning activity that will help you to make your own sustainable development goal.\r\n\r\n\r\n\r\n\r\nReferences\r\nUnited Nations. (n.d.). Climate action.","n":0.049},[{"v":"Sustainability","i":0,"n":1}]]},{"i":172,"$":[{"v":"Colour theory","n":0.707},{"v":"\r\n\r\nThis open educational resource on Pressbooks explains the history of colour theory, how we see colour, and how to use colour systems to mix colour and create colour relationships. It's useful for anyone working with colour in creative and scientific fields.\r\n\r\nWhat is colour theory?\r\nColour theory covers a long history from antiquity to modern times. It includes academic and scientific investigations into how we see and understand colour. It also includes practical applications for using colour in creative work.\r\n\r\nIn this open educational resource, you can find information on colour theory divided into three chapters. This resource also includes quizzes to test your knowledge, and practical learning activities to guide users in applications of colour theory for art and design.\r\n\r\nVisit the home page of this resource (opens Pressbooks), or go directly to the chapters:\r\n\r\n\r\n\r\n\r\n\r\nChapter 1. Colour theory: history and culture\r\n(opens Pressbooks)An introduction to colour theory history with a summary of selected key scientific developments. Also covered are short introductions to cultural interpretations of colour and colour aesthetics.\r\n\r\n \t- Scientific discoveries relating to colour\r\n \t- Cultural interpretations of colour and colour symbolism\r\n \t- Colour aesthetics in art and design\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nChapter 2. Colour theory: the visible spectrum\r\n(opens Pressbooks)\r\nA scientific approach to the visible spectrum and how humans and other species perceive colour including:\r\n\r\n \t- Introduction to current scientific theory of the visible spectrum, physics, optics and chemistry\r\n \t- The human eye: how we see and interpret colour information in the brain\r\n \t- Colour blindness and other vision differences\r\n \t- Colour vision in animals\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nChapter 3. Colour theory: working with colour\r\n(opens Pressbooks)\r\nDetailed explanations of different colour systems for the practical applications of colour in physical materials and digital media including:\r\n\r\n \t- Understanding additive and subtractive colour\r\n \t- Understanding colour systems for working with:\r\n\r\n \t- digital media,\r\n \t- printing, and\r\n \t- colour pigments and dyes\r\n\r\n\r\n \t- Using colour wheels and creating colour relationships\r\n\r\n\r\n\r\n\r\n","n":0.057},[{"v":"Design","i":2,"n":1},{"v":"Colour Theory","i":1,"n":0.707},{"v":"Art","i":0,"n":1}]]},{"i":173,"$":[{"v":"Using starter phrases","n":0.577},{"v":"If you feel stuck for finding the right phrase to start your sentence, you can use a strategy called \u2018Syntactic borrowing\u2019 (Kamler & Thomson, 2006; Swales & Feak, 2004) in which you borrow the phrases used by others.\r\nTo do this, look at some sentences in various sections of a research journal in your discipline and remove all the content, leaving a sentence skeleton. This should give you some ideas for phrases that are often used.\r\nExamples\r\nTo help you along with the process, here are some lists of common phrases you might want to use in particular sections of your thesis or exegesis.\r\n\r\nNote: these are general academic phrases - there may be phrases in the lists that are not used in your discipline.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nReferences\r\nKamler, Barbara & Thomson, Pat. (2006). Helping Doctoral Students Write: Pedagogies for Supervision. 10.4324/9781315813639.\r\n\r\nSwales, J. M., & Feak, C. B. (2004). Academic writing for graduate students: Essential tasks and skills (2nd ed.). Ann Arbor, MI: The University of Michigan Press.","n":0.079},[{"v":"Academic writing","i":0,"n":0.707}]]},{"i":174,"$":[{"v":"Meanings of instructional words","n":0.5},{"v":"The first step of an assignment is to read the instructions carefully to understand what is required. However, some words in the instructions can be difficult to understand. This page provides examples of common instructional words and explains their meanings.\r\nSample assignment instructions and word meanings\r\nRead the sample assignment instructions and the definitions of the instructional words used in them.\r\nExample 1: reflective assignment\r\nReflect on your experiences working in a team during this semester. Describe specific instances where you faced challenges and how you overcame them. Analyse the skills you developed and assess how these experiences have contributed to your personal growth. Summarise your reflections in a 1,000-word essay.\r\n\r\n\r\nReflect: Think deeply about a topic, experience, or situation and describe the circumstances and importance of it. Visit the Reflective writing section of the Learning Lab for more information.\r\nDescribe: Explain what something is like in detail. For example, in Example 1, you have been asked to explain your challenges and solutions to these challenges.\r\nAnalyse: Break down the issue into parts, discuss them, and show their connections.\r\nSummarise: Provide the main points of an article, issue or theme without deep analysis.\r\n\r\nExample 2: presentation\r\nPrepare a 10-minute presentation on emerging technologies in healthcare. Explain how these technologies are transforming patient care and medical practices. Illustrate your points with case studies and statistical data. Evaluate the potential benefits and challenges associated with these technologies.\r\n\r\n\r\nExplain: Make something clear by describing it in detail.\r\nIllustrate: Clarify or explain using specific examples, diagrams, or charts.\r\nEvaluate: Critically assess the strengths and weaknesses of an argument, theme or study.\r\n\r\n\r\nExample 3: essay\r\nDiscuss the role of governments in economic development. Critique historical examples and current policies and outline different economic theories relevant to the topic. Review literature on the potential advantages and disadvantages of government intervention in the economy. Write a 2,000-word essay addressing these points.\r\n\r\nDiscuss: Explore a topic in detail and present different viewpoints.\r\nCritique: Evaluate in detail, describing both strengths and weaknesses.\r\nOutline: Provide a structured summary, highlighting the main points without in-depth analysis.\r\nReview: Critically examine and summarise a topic. This often includes your evaluation of different views or theories that you have summarised.\r\n\r\nExample 4: case study analysis\r\nExamine the economic impact of climate change on the agricultural sector in a specific region. Quantify the changes in crop yields, production costs, and market prices over the past decade. Interpret the data to identify trends and potential future impacts. Justify your analysis with evidence from credible sources and relevant economic theories.\r\n\r\n\r\nExamine: Explore a topic in detail.\r\nQuantify: Express in numerical terms, providing data or statistical evidence.\r\nInterpret: Explain the meaning or significance of something. If you interpret data, you look at it to uncover and explain patterns or trends.\r\nJustify: Provide reasons or evidence to support an idea, theory, or argument.\r\n\r\n","n":0.047},[{"v":"Understanding assignments","i":2,"n":0.707},{"v":"Assignments","i":1,"n":1},{"v":"Study skills","i":0,"n":0.707}]]},{"i":175,"$":[{"v":"Reading academic articles","n":0.577},{"v":"Reading academic articles can be tricky and time-consuming. However, there are some strategies that can help you find the information you need efficiently.\r\nFirst, let\u2019s learn about the different sections of articles, before seeing how we can apply active reading to the process.\r\nAcademic article sections\r\nHere are some of the sections commonly found in academic articles and a description of the information that they contain:\r\n\r\nTitle: Provides information about the contents of the article. \r\nAbstract: Contains a complete overview of what the research is about, what was done, how it was done, what was found and what the results mean.\r\nIntroduction: Often provides a brief review of previous research, a rationale (reason) for the research, and an outline of the research aims. Knowledge about this structure will help you read it efficiently.\r\nMethodology: Describes the methods used to carry out the research.\r\nResults: This section describes the findings. (Sometimes the results and discussion sections are combined.)\r\nDiscussion: Contains an interpretation of what the results mean in terms of the original research question or hypothesis, and in the field.\r\nConclusion: This is a discussion of the meaning and implications of the research. It may also detail the implications of the research and make recommendations for changes to policy and practice directions or directions for further research.\r\nReferences: A list of the literature on which the research was based.\r\n\r\nReading strategies for academic articles\r\nThe principles of active reading can be applied to academic articles.\r\nThink about why you are reading the article. \r\nKey texts\r\nIf the article is a key text that you need to understand thoroughly, then you might start by surveying the text — reading the article title, finding out about the author, looking at the heading titles and checking out any visual elements such as graphs and charts. \r\nWhile you read, remember to think critically. Just because you are reading an academic article, it does not mean that you must agree with everything theory or idea expressed in it. Consider whether the information is credible, whether the research has been carried out in an effective way and if the conclusions drawn from the research are logical.\r\nSearching for specific information\r\nThis is where skimming and scanning will come in handy. \r\nBefore you start reading, decide what information you are looking for — list down any key words, terms or numerical data that you would like to search for.\r\nNext, quickly read through the articles content, stopping when you find a key word or anything that you feel may be relevant. Then, read the text surrounding the key word or information carefully while taking notes.\r\nSearching for general information\r\nWhen searching for general information, many students start by reading the title of the article and the abstract. From this, they are often able to tell if the article will contain information that is useful to them. \r\nIf it doesn't, then they can move on to the next article without having lost time reading something that is not relevant. \r\nIf they believe the article does contain relevant information for their assignment, then they move on to reading the introduction for more information on the purpose of the article, followed by the discussion and/or conclusion sections. These sections will provide an interpretation of the results and summarise the main points.","n":0.043},[{"v":"Reading skills","i":2,"n":0.707},{"v":"Active reading","i":1,"n":0.707},{"v":"Academic articles","i":0,"n":0.707}]]},{"i":176,"$":[{"v":"Note-taking in classes and lectures","n":0.447},{"v":"Actively taking notes in class can help you understand the topic better. Below are some tips that can help you take good notes in class and revise them after class.\r\n\r\nIn class time\r\nDuring class teachers use various strategies to indicate to students what is important. Being aware of these helps you identify the main ideas and take useful notes. When taking notes, it\u2019s a good idea to pay attention to:\r\nIntroductions and conclusions\r\nIn the introduction, the lecturer usually outlines the important points, connects them with the previous and next lectures and puts the topic in context. The conclusion summarises the main points. Therefore, paying attention to the beginning and ending of lectures is very important.\r\n\r\nFor example: Today we\u2019ll be talking about the effect of information overload on the short-term memory. There are three main points \u2026\r\nVoice emphasis\r\nIn lectures and tutorials, important ideas, words and phrases are often emphasised. A lecturer or teacher may do this by saying them more slowly, more loudly, or by pausing. Gestures and movement may also be used for emphasis.\r\nLinking expressions\r\nListen for words and phrases that direct you to new and important ideas or show the relationships between the points being raised. Examples include:\r\n\r\n \tAddition words \u2013 also, too, in addition, furthermore, another\r\n \tEmphasis words \u2013 importantly, specifically, especially, obviously\r\n \tSequencing words \u2013 firstly, secondly, next, finally\r\n \tContrast words \u2013 however, but, despite, on the other hand, conversely\r\n \tConcession words \u2013 although, even though, in the light of, given that\r\n \tIllustration words \u2013 for example, to illustrate, that is\r\n \tReason and result words \u2013 consequently, because, therefore, leads to\r\n \tSummary words \u2013 in conclusion, to sum up, in summary\r\n\r\nRepetition and elaboration\r\nImportant points are repeated or illustrated with examples by your lecturer.\r\nAfter class\r\nYour independent study time after class is some of your most valuable time as a study; use it wisely. Revise notes within 24 hours of the class. Don\u2019t just file them away!\r\n\r\nSome different ways to revise notes after class:\r\n\r\n \tDiscuss ideas with a study group\r\n \tDraw a mind map of the key ideas\r\n \tDo some follow up reading about the key ideas\r\n \tThink about how ideas relate to your assignments, exams or future career\r\n","n":0.053},[{"v":"Study skills","i":1,"n":0.707},{"v":"Note-taking","i":0,"n":1}]]},{"i":177,"$":[{"v":"Risk assessment matrix","n":0.577},{"v":"A risk matrix is a matrix that is used during risk assessment to define the level of risk by considering the category of probability or likelihood against the category of consequence severity. This is a simple mechanism to increase the visibility of risks and assist management decision making.\r\nSelect each item of the interactive risk matrix below to examine the risk of each scenario.\r\nInteractive risk assessment matrix - zoo scenarios\r\n\r\n\r\nThe hierarchy of control should be used in conjunction with the risk assessment matrix. \r\n\r\nRisk assessment matrix interactive \u2013 zoo scenarios (interactive overlays transcript)\r\nRisk assessment matrix overlays text - Likelihood x Consequence Severity = Risk Impact\r\n\r\nIn order of lowest to highest risk\r\n\r\n 01. Highly unlikely x Negligible = Low risk\r\n\r\nAn orangutan learns to use sign language, insulting and frightening visitors as they pass by, gesticulating rudely and banging on the glass enclosure.\r\n\r\n02. Unlikely x Negligible = Low risk\r\n\r\nA butterfly from the insect enclosure gets caught in a visitor\u2019s hair resulting in some damage to the butterfly but only causing minor distress to the visitor.\r\n\r\n 03. Possible x Negligible = Low risk\r\n\r\nZoo kiosk coffee is too hot burning several visitors' tongues.\r\n\r\n 04. Likely x Negligible = Medium risk\r\n\r\nA visitor forgets to apply adequate sunscreen for a day at the zoo and sustains a moderate sunburn.\r\n\r\n 05. Very likely x Negligible = Medium risk\r\n\r\nWhile on school excursion, a child is separated from the group and gets lost, is found in a distressed state.\r\n\r\n 06. Highly unlikely x First aid = Low risk\r\n\r\nA zookeeper forgets to wear protective gloves and is bitten while handling a lizard.\r\n\r\n 07. Unlikely x First aid = Low risk\r\n\r\nA severe hailstorm injures several visitors and some animals, all requiring treatment for bruising and concussion.\r\n\r\n 08. Possible x First aid = Medium risk\r\n\r\nA child falls into an unfenced pond, swallows dirty pond water, and is covered in mud.\r\n\r\n 09. Likely x First aid = Medium risk\r\n\r\nA visitor trips on uneven ground caused by exposed tree roots, sustaining a grazed knee.\r\n\r\n 10. Very likely x First aid = Medium risk\r\n\r\nA free-roaming swan attacks a child and steals a sandwich, biting a finger which causes bleeding.\r\n\r\n 11. Highly unlikely x Minor injury = Medium risk\r\n\r\nDrinking fountains are contaminated with bacteria causing illness in some visitors.\r\n\r\n 12. Unlikely x Minor injury = Medium risk\r\n\r\nA visitor is knocked down while crossing a walkway in the zoo car park by a vehicle travelling at low speed.\r\n\r\n 13. Possible x Minor injury = Medium risk\r\n\r\nA photographer roams into a restricted area of the open lemur enclosure to take a better photograph of a young lemur and is attacked by several female lemurs defending their young, resulting in scratches and bites to the face and arms.\r\n\r\n 14. Likely x Minor injury = Medium risk\r\n\r\nA child climbs on to a dinosaur display and falls, breaking an arm.\r\n\r\n 15. Very likely x Minor injury = High risk\r\n\r\nA zookeeper crashes an electric cart into a tree at low speed, sustaining a leg injury and damaging the tree.\r\n\r\n 16. Highly unlikely x Major injury = Medium risk\r\n\r\nA zookeeper contracts a serious virus from a macaque monkey, is admitted to a quarantine ward, and placed on a ventilator.\r\n\r\n 17. Unlikely x Major injury = Medium risk\r\n\r\nA poacher sneaks into a safari zoo enclosure to steal rhinoceros horns. He shoots and seriously injures one rhinoceros but gets attacked by the others, causing multiple broken bones and internal injuries.\r\n\r\n 18. Possible x Major injury = High risk\r\n\r\nA gorilla escapes a damaged enclosure, attacking multiple visitors, resulting in severe injury.\r\n\r\n 19. Likely x Major injury = High risk\r\n\r\nStaff and visitors contract COVID19 from a source at the zoo resulting in hospital quarantine.\r\n\r\n 20. Very likely x Major injury = High risk\r\n\r\nAn animal handler fails to lock the gate properly during feeding time and is attacked by hungry tigers resulting in serious cuts and internal injuries.\r\n\r\n 21. Highly unlikely x Fatality = Medium risk\r\n\r\nExtinct dinosaurs are recreated from DNA trapped in fossilised mosquito blood. They break free from enclosures and rampage throughout the zoo killing staff, animals, and visitors.\r\n\r\n 22. Unlikely x Fatality = High risk\r\n\r\nGlass breaks in an aquatic display area causing a flood of glass and water, killing visitors, staff and animals.\r\n\r\n 23. Possible x Fatality = High risk\r\n\r\nA visitor falls into a dangerous animal enclosure and is mauled by lions.\r\n\r\n 24. Likely x Fatality = High risk\r\n\r\nA visitor or staff member is killed due to anaphylaxis from a wasp or bee sting while walking in zoo gardens.\r\n\r\n 25. Very likely x Fatality = Extreme risk\r\n\r\nDirectly caused by climate change, an extreme bush fire sweeps through a safari zoo resulting in the death of animals and staff.\r\n\r\n","n":0.036},[{"v":"Risk","i":2,"n":1},{"v":"Occupational health and safety (OH&S)","i":1,"n":0.447},{"v":"Assessment","i":0,"n":1}]]},{"i":178,"$":[{"v":"Sustainable development goals: life below water","n":0.408},{"v":"The United Nations (UN) Sustainable Development Goals (SDGs) cover many different areas of life. To understand the goals and what they achieve, we will look at one goal in particular, \u00a0Goal 14 \u2014 Life below water.\r\n\r\n\r\nTargets\r\nHaving a goal of preserving life below water sounds great, but how do we achieve it? How do we know if we are meeting the goal?\r\n\r\nThere are targets that break the goal down into smaller parts. Parts that individuals, countries, and businesses can work on. These targets include dates, so there is a deadline for completing them. They are strategic; they help to create positive change and achieve the goal of preserving \"Life below water.\"\r\nExamples\r\nRMIT is working on projects that will help to meet the SDG 14 targets set by the UN. These projects are only a few of many projects happening across the world. The power of an SDG is that it is global and targeted.\r\nThe Litter Trackers\r\nRMIT has been working with other organisations to create an education programme about litter in Melbourne. Litter left on the ground is washed into stormwater drains and from there into Port Phillip Bay. Rubbish and litter has a negative effect on the environment, the creatures living in the bay, birds and more. By tracking litter, the program can educate people about the importance of disposing of waste properly.\r\n\r\nFind out more about The Litter Trackers: Reducing Littering by Education.\r\n\r\n\r\n\r\nAquatic Pollution Prevention Partnership\r\nRMIT is not just working on educating people about litter in Port Phillip Bay, it is also preventing it from reaching the water.\r\n\r\nIn this partnership with Melbourne Water, RMIT is working on solutions to prevent litter and other waste from reaching Port Phillip Bay and other Victorian waterways.\r\n\r\nLearn more about the Aquatic Pollution prevention Partnership\r\n\r\nThese examples show that sustainability projects often address more than one SDG. For example, a project that educates, reduce litter and works to improve water quality, can have impacts on Goal 6 \u2014 Clean water and sanitation and Goal 4 \u2014 Quality education. The best goals are the ones that you can, and do, complete. Even better than completing a single goal is when that goal has a positive impact on many things.","n":0.053},[{"v":"UN Sustainable Development Goals","i":1,"n":0.5},{"v":"Sustainability","i":0,"n":1}]]},{"i":179,"$":[{"v":"United Nations Sustainable Development Goals (SDGs)","n":0.408},{"v":"Your future career will benefit from understanding the principles and practices that form the basis of the United Nation's (UN) approach to building a sustainable future with Sustainable Development Goals (SDGs).\r\nThere are many ways to think about and act on sustainability issues. Sustainability can mean different things to different people and at different times. What seems like a sustainable solution in one area of life, may not be sustainable when applied to another part of life.\r\n\r\nThe UN have ambassadors from every nation-state. Together, these ambassadors make international decisions. They work on many topics that affect people across the world, Human rights, Peace, Waste, Weapons and much more. The UN was formed in 1945, joined by 170 member countries. That number has increased over the years to the current 193 nation-states.\r\nIts purposes and principles are to maintain international peace and security, develop friendly relations among nations based on equal rights and self-determination, and to achieve international co-operation in solving economic, social, cultural, or humanitarian challenges without distinction for race, sex, language, or religion. (UN, n.d.)\r\nInitially, there were eight United Nations sustainable development goals, called the Millenium Development Goals, which started in 2000 and finished in 2015. The Millenium Development Goals were redeveloped into the Sustainable Development Goals (SDGs). The SDGs have a timelilne for completion by 2030.\r\n\r\nThere are currently 17 goals, 169 targets and 232 indicators. Providing the details of targets and indicators helps people measure the progress of each goal. Creating goals that are Strategic, Measurable, Achievable, Realistic and Timely (SMART goals) is a great way to make sure any goal will be completed and measured.\r\n\r\nRMIT University, like other institutions, has chosen to use the UN SDGs to guide its sustainability plans.\r\n\r\n\r\nThe 17 UN SDGs\r\n\r\n\r\n\r\n\r\n\r\n\r\nUnited Nations Sustainable Development Goals timeline\r\nExplore the information in the accordions folders to see a timeline of key moments in the UN's history, and the origins and future of the Sustainable Development Goals.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nQuiz\r\nTake the quiz to test your knowledge of the United Nations Sustainable Development Goals.\r\n\r\n\r\nImage by Derek Sutton via Unsplash, licensed under\u00a0CC0\r\n\r\n\r\nReferences\r\nSustainable Development Goals Fund. (n.d.). From MDGs to SDGs.\r\nUnited Nations. (n.d.) About us.\r\nUnited Nations. (n.d.). United Nations Charter, chapter I: Purposes and principles.\r\n\r\n\r\n\r\n\r\n\r\n","n":0.053},[{"v":"UN Sustainable Development Goals","i":1,"n":0.5},{"v":"Sustainability","i":0,"n":1}]]},{"i":180,"$":[{"v":"Systems thinking: The circular economy","n":0.447},{"v":"Systems thinking is a way of thinking about our products, businesses and services not just in terms of the final product or service, but also the ways that product or service impacts people and planet.\r\nAs we discussed briefly, a system is a group of objects and forces that influence each other. We live in many systems:\r\n\r\n \tsolar system\r\n \tjustice systems\r\n \tIT systems\r\n \teconomic systems\r\n\r\nSystems thinking means considering the effects on all parts of a system rather than just one. We live on the planet earth, which is a closed system. Waste we generate stays on earth. Damaging the environment in one area can have effects on other areas.\r\nCircular economy\r\nThe circular economy uses systems thinking to create products that are meant to be recycled or reused. You can see the difference between a traditional linear economy and a circular economy in engineering, in the image below. Circular economies do not end in waste; products are reused or recycled, or even better designed with the next purpose in mind.\r\n\r\n\r\nThese circular principles show how to recycle and feed energy and actions back into the lifecycle so that nothing is wasted.\r\nRepair, reuse, recycle\r\nA circular economy requires that products are made in a way that does not create waste. Making products that have little or no waste is much better than relying on recycling. Products that are built to be reusable or easily repaired are the best.\r\n\r\nThere are many examples from the past when people owned their clothes, furniture, and other items for life or even longer.\r\n\r\n\r\nTechnological waste\r\nTechnology produces a lot of waste, some of it is toxic. The materials used to make phones, laptops, games consoles and other technology are hard or impossible to recycle, and therefore create waste when they are created and waste at the end of their life.\r\n\r\nThere is a push for tech companies to make products that can be easily repaired rather than thrown away. Allowing people to fix their phones and replace broken parts will save a lot of money and stop a lot of waste.\r\nReusable bottles and containers\r\nIn many places, milk used to be sold in reusable glass bottles. The empty bottles were left out after use and taken back to be cleaned and refilled with fresh milk. There are some grocery stores that sell their food without any packaging today, where customers bring their own containers to fill with the products they want to buy.\r\nJars as drinking glasses\r\nFor many years, Vegemite and other products were sold in jars that were designed to be drinking glasses when empty. Jars were even printed with cartoon characters and pictures. When glass recycling became more available, the jars were not made anymore. But it is much better to reuse glass. Glass recycling uses a lot of energy \u2013 to melt the glass and reshape it \u2013 and putting jars on a truck makes air pollution. Putting Vegemite and Nutella in jars that were reused was a circular economy.\r\nRepair\r\nPeople did not always buy new things like we do today. In the past, people repaired things when they broke or reused them in a new way. There are many cities with 'repair cafes' where you can take broken items to be fixed, saving you money, and saving the environment.\r\n\r\nWhile there are many ideas from the past that we could bring back, there are also ways to create modern items that make less waste. \r\n\r\nLinear timeline:\r\n\r\n \tInnovate\r\n \tSelect\r\n \tSpecify\r\n \tRefine/define\r\n \tValidate/certify\r\n \tProduce\r\n \tOperate\r\n \tDispose\r\n\r\nCircular principals\r\n\r\nCore engineering principles that drive circularity:\r\n\r\n \tMaintainability/testability:\u00a0links Produce back to Innovate\r\n \tReliability: links Produce back to Specify\r\n \tAvailability: links Operate back to Operate\r\n \tMaintain/repair/overhaul: links Operate back to Operate\r\n \tSafety and approval: links Validate/Certify\u00a0back to Specify\r\n \tLife extension: links Operate back to Innovate\r\n\r\nNew circular principles that build on existing frameworks:\r\n\r\n \tDeliver capabilities: links Produce back to Innovate\r\n \tReusability/standardisation/modularity: links Operate back to Innovate\r\n \tEnergy from renewable sources: links Dispose and Operate back to Innovate\r\n \tRecycle/reuse: links Operate back to Operate\r\n \tWaste to energy: links Dispose back to Operate and Innovate\r\n\r\n","n":0.039},[{"v":"Systems thinking","i":1,"n":0.707},{"v":"Sustainability","i":0,"n":1}]]},{"i":181,"$":[{"v":"Understanding your audience","n":0.577},{"v":"You may not be aware of it, but we all present information in different ways at different times. Maybe you give your close friends more details when telling a story, or explain an idea more slowly and with simpler examples when talking to a child. You might even switch languages or dialects when communicating with certain people in your life.\r\nBecoming aware of the ways that we use language and structure our ideas gives us the power to be more intentional with our communication. And when we understand our audience, we can choose ways of communicating that will best convey our ideas to our listeners.\r\n\r\nRead on to learn more about audience characteristics and explore examples of information prepared for different audiences.\r\nAudience characteristics\r\nThere are several audience characteristics to consider when planning your spoken and written communication. These attributes can help you decide the type of language and vocabulary you might use, which examples you provide, the level of formality, and even whether the information is better conveyed in a spoken or written format.\r\n\r\n\r\n\r\n\r\n\r\n\r\nExplore examples\r\nThe slides below show how the same information about climate change and the Southern Ocean might be adapted for a different audience. The first text has been designed for adults who are interested in and have some knowledge of scientific principles, but are not scientists themselves. In the second text, the language, explanations and examples have been adjusted to suit an audience of primary school students.\r\n\r\n \r\n\r\nAttribution: This resource has been adapted from: Southern Ocean Sentinel. Authored by: Australian Broadcasting Corporation. License: CC BY: Attribution 4.0\r\n\r\n","n":0.062},[{"v":"Writing structures","i":2,"n":0.707},{"v":"Style","i":1,"n":1},{"v":"Presentations","i":0,"n":1}]]},{"i":182,"$":[{"v":"Parallel circuits","n":0.707},{"v":"Unlike the fairy lights from the previous page, the appliances in your home are wired in parallel. This is why your refrigerator keeps running when your toaster fails. Each component sits on its own independent path, so a break in one branch doesn't affect the others. This resource builds on series circuits to explore how current, voltage and resistance behave when components are connected in parallel.\r\n\r\nConnecting circuits in parallel\r\nIn parallel, components are connected side by side, each with its own path from the supply. Unlike a series circuit, the current splits across the branches \u2013 each component draws its own current independently, so the total current from the supply is the sum of all the branch currents. One important example is an electrical switch box. In a household, individual circuit breakers in the box control a separate parallel branch, such as lighting or your kitchen.\r\n\r\nThis means that adding more parallel branches gives the current more paths to flow through, which lowers the overall resistance. The effective resistance of resistors in parallel is therefore always less than the smallest individual resistor.\r\nConsider two resistors \\(R_{1}\\) and \\(R_{2}\\) connected in parallel across a battery.\r\n\r\nThe total current leaving the battery \\(I\\) is the sum of the two currents \\(I_{1}\\) and \\(I_{2}\\).\r\n\\[I = I_{1} + I_{2}\\]\r\nThe voltage across each parallel branch \\(V_{1}\\) and \\(V_{2}\\) equals the supply voltage \\(V\\).\r\nFor two resistors \\(R_1\\) and \\(R_2\\) in parallel, the equivalent resistance \\(R_{EQ}\\) is given by:\r\n\\[\\frac{1}{R_{\\text{EQ}}} = \\frac{1}{R_{1}} + \\frac{1}{R_{2}}\\]\r\nHere, \\(R_{\\text{EQ}}\\) is a single resistance equivalent to the two resistors in parallel.\r\n\r\nExample \u2013 measuring electricity in parallel circuits\r\nTwo pieces of nichrome wire are found to have resistances of \\(\\mathbf{10\\,\\Omega}\\) and \\(\\mathbf{20\\,\\Omega}\\).\r\n\r\n\tIf they are connected in parallel, what is their effective resistance?\r\n\r\n\tThe effective resistance can be found using \\(\\dfrac{1}{R_{EQ}}=\\dfrac{1}{R_{1}}+\\dfrac{1}{R_{2}}\\).\r\n\t\\[\\begin{align*} \\frac{1}{R_{EQ}} & = \\frac{1}{R_{1}}+\\frac{1}{R_{2}} \\\\[6pt]\r\n\t& = \\frac{1}{10}+\\frac{1}{20}\\\\[6pt]\r\n\t& = \\frac{3}{20}\\end{align*}\\]\r\n\tThis means that \\(R_{EQ}\\) is the reciprocal, \\(\\dfrac{20}{3}\\approx6.7\\,\\Omega\\). See how \\(R_{EQ}\\) is smaller than the two resistances \\(R_{1}\\) and \\(R_{2}\\)?\r\n\r\n\tWhat total current will flow through them and what power will be produced if the combination is placed across a \\(\\mathbf{12}\\,\\textbf{V}\\) battery?\r\n\r\n\tThe total current is given by \\(I=\\dfrac{V}{R}\\).\r\n\t\\[\\begin{align*} I & = \\frac{V}{R} \\\\[6pt]\r\n\t& = \\frac{12}{6.7}\\\\[6pt]\r\n\t& \\approx 1.791\\,\\text{A}\\end{align*}\\]\r\n\tTo then calculate the power, we need to use \\(P=V\\times I\\).\r\n\t\\[\\begin{align*} P & = V\\times I \\\\[6pt]\r\n\t& = 12\\times1.791\\\\[6pt]\r\n\t& = 21.49\\,\\text{W}\\end{align*}\\]\r\n\r\nExercise \u2013 measuring electricity in parallel circuits\r\n\r\n\tTwo torch bulbs are placed in parallel with each other across a \\(3.0\\,\\text{V}\\) battery. The current through the battery is \\(0.55\\,\\text{A}\\). The current through one of the bulbs is \\(0.25\\,\\text{A}\\).\r\n\t\t\r\n\t\t\tWhat is the current through the other bulb?\r\n\t\t\tWhat is the voltage across the other bulb?\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tWhat is the effective resistance of two \\(10\\,\\Omega\\) resistors:\r\n\t\t\r\n\t\t\tin series?\r\n\t\t\tin parallel?\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tTwo equal resistors are placed in parallel and found to have a combined resistance of \\(34\\,\\Omega\\). What is the resistance of each one?\r\n\r\n\r\n\r\n\r\n\tA \\(10\\,\\text{V}\\) power supply is used across two separate resistors. The current through one is found to be \\(0.4\\,\\text{A}\\), and through the other \\(0.5\\,\\text{A}\\). When they are combined in parallel:\r\n\t\t\r\n\t\t\twhat current will flow through them?\r\n\t\t\twhat is their effective resistance?\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tA current of \\(3\\,\\text{A}\\) is found to be flowing through two resistors of \\(20\\,\\Omega\\) and \\(10\\,\\Omega\\) in parallel.\r\n\t\t\r\n\t\t\tWhat is the effective resistance of the combination?\r\n\t\t\tWhat is the voltage across the pair of resistors?\r\n\t\t\tHow much current will be flowing in each resistor?\r\n\t\t\tHow much power will be dissipated (used up) in each of the two resistors?\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tThree resistors of \\(900\\,\\Omega\\), \\(1.5\\,\\text{k}\\Omega\\) and \\(2.0\\,\\text{k}\\Omega\\) are to be used in a circuit. What is their effective resistance if they are all placed:\r\n\t\t\r\n\t\t\tin series?\r\n\t\t\tin parallel?\r\n\t\t\r\n\t\r\n\r\n","n":0.042},[{"v":"Maths","i":3,"n":1},{"v":"Electricity","i":2,"n":1},{"v":"Electrical circuits","i":1,"n":0.707},{"v":"Physics","i":0,"n":1}]]},{"i":183,"$":[{"v":"Series circuits","n":0.707},{"v":"A single blown globe can knock out an entire string of fairy lights\u2014because in a series circuit, all components share one continuous path, and a break anywhere stops current flowing everywhere. This resource builds on current, voltage and resistance to explore how series circuits behave, including how voltage is shared across components and how Kirchhoff's laws help us analyse them.\r\n\r\nKirchhoff's laws\r\nThe principles of conservation of charge and conservation of energy can be used to establish two very important rules that apply to all electric circuits. These are called Kirchhoff's laws.\r\n\r\nKirchhoff's current law\r\nKirchhoff's current law states that in any electrical circuit, the sum of all currents flowing into any point is equal to the sum of the currents flowing out of it.\r\nFor example, if at any junction of three wires, there is \\(2\\,\\text{A}\\) flowing in on one wire and \\(3\\,\\text{A}\\) flowing in on another, then there must be a current of \\(5\\,\\text{A}\\) flowing out on the third.\r\nCurrents flowing into the point will be positive (\\(+\\)) and those flowing away from the point negative (\\(-\\)). As a result, Kirchhoff's current law is sometimes abbreviated to 'the sum of all currents at a point is zero.'\r\n\r\nKirchhoff's voltage law\r\nKirchhoff's voltage law states that the total potential drop around a closed circuit must be equal to the total voltage (EMF) in the circuit.\r\nFor example, if we know that a torch battery supplies a voltage (EMF) of \\(3.0\\,\\text{V}\\) and we measure a \\(2.8\\,\\text{V}\\) potential difference across the bulb, there must be a \\(0.2\\,\\text{V}\\) drop somewhere else in the circuit \u2013 possibly across the switch contacts if they are a little dirty.\r\n\r\nConnecting circuits in series\r\nNo matter how complex a circuit, it can always be broken up into sections in which circuit components are combined either in series or in parallel. In series, components are connected one after another in a single path so the same current flows through each component.\r\n\r\nIn this circuit, the sum of all the potential differences across the resistors \\(V_{1}\\) and \\(V_{2}\\) will equal the potential difference, or voltage of the supply \\(V\\). The voltage of the battery is therefore given by:\r\n\\[\\text{EMF} = V = V_1 + V_2\\]\r\nFor resistors in series, the equivalent resistance \\(R_{\\text{EQ}}\\) is equal to the sum of the individual resistances.\r\n\\[R_{\\text{EQ}} = R_1 + R_2\\]\r\nThe current \\(I\\) leaving the battery must flow through both \\(R_1\\) and \\(R_2\\), so the current in a series circuit is the same at every point. For resistors connected in series, the same current must pass through each resistor.\r\nIn addition to fairy lights, other examples of series circuits include a simple torch and a smoke detector circuit.\r\n\r\n\r\n\r\nExample \u2013 measuring electricity in series circuits\r\nTwo pieces of nichrome wire (as used in heater elements) have resistances of \\(\\mathbf{10}\\,\\Omega\\) and \\(\\mathbf{20}\\,\\Omega\\).\r\n\r\n\tWhat current would flow through them, and what power will be produced in them, if they are separately connected to a \\(\\mathbf{12}\\,\\textbf{V}\\) battery?\r\n\r\n\tThe current will be given by \\(I=\\dfrac{V}{R}\\). For the two wires separately, the currents will be \\(\\dfrac{12}{10}=1.2\\,\\text{A}\\) and \\(\\dfrac{12}{20}=0.6\\,\\text{A}\\).\r\n\tThe power is found using \\(P=VI\\). For the two wires separately, the powers will be \\(12\\times1.2=14.4\\,\\text{W}\\) and \\(12\\times0.6=7.2\\,\\text{W}\\).\r\n\r\n\tIf they are connected in series, what is their total resistance?\r\n\r\n\tWhen connected in series, the total resistance will be:\r\n\t\\[\\begin{align*} R_{EQ} & = R_{1} + R_{2} \\\\[6pt]\r\n\t& = 10+20\\\\[6pt]\r\n\t& = 30\\,\\Omega\\end{align*}\\]\r\n\r\n\tWhen placed in series across the \\(\\mathbf{12}\\,\\textbf{V}\\) battery, what current will flow through them and what power will be produced?\r\n\r\n\tThe current that flows from the \\(12\\,\\text{V}\\) battery will be:\r\n\t\\[\\begin{align*} I & = \\frac{V}{R} \\\\[6pt]\r\n\t& = \\frac{12}{30}\\\\[6pt]\r\n\t& = 0.4\\,\\text{A}\\end{align*}\\]\r\n\tThe total power will be:\r\n\t\\[\\begin{align*} P & = VI \\\\[6pt]\r\n\t& = 12\\times0.4\\\\[6pt]\r\n\t& = 4.8\\,\\text{W}\\end{align*}\\]\r\n\r\nExercise \u2013 measuring electricity in series circuits\r\n\r\n\tShiori has found a point in her car where four wires are attached together. She finds currents of \\(+2.5\\,\\text{A}\\) and \\(+1.0\\,\\text{A}\\) in two of the wires, and \\(-4.2\\,\\text{A}\\) in a third. \\(+\\) indicates current travelling into the point and \\(-\\) indicates current travelling out of the point. What is the current in the fourth wire?\r\n\r\n\r\n\r\n\r\n\tTwo torch bulbs are placed in series with each other and a \\(4.5\\,\\text{V}\\) battery. The current through one is found to be \\(0.25\\,\\text{A}\\) and the voltage across it is \\(2.1\\,\\text{V}\\).\t\r\n\t\t\r\n\t\t\tWhat is the current through the other bulb? Give your answer in \\(\\text{mA}\\).\r\n\t\t\tWhat is the voltage across the other bulb?\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tMarcos has bought two \\(12\\,\\text{V}\\) headlights for his truck but finds that it has a \\(24\\,\\text{V}\\) battery. He decides that the simplest way to overcome the problem is to wire them in series.\r\n\t\t\r\n\t\t\tWill the headlights work correctly?\r\n\t\t\tDo you see any problems with this scheme?\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tTwo equal resistors are placed in series and found to have a combined resistance of \\(34\\,\\Omega\\). What is the resistance of each one?\r\n\r\n\r\n\r\n\r\n\tA \\(10\\,\\text{V}\\) power supply is used across two separate resistors. The current through one is found to be \\(0.4\\,\\text{A}\\), and through the other \\(0.5\\,\\text{A}\\). When they are combined in series:\r\n\t\t\r\n\t\t\twhat current will flow through them? Give your answer in \\(\\text{mA}\\).\r\n\t\t\twhat is their effective resistance?\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tA \\(400\\,\\Omega\\) resistor and a \\(100\\,\\Omega\\) resistor are placed in series across a battery with a voltage (EMF) of \\(5\\,\\text{V}\\).\r\n\t\t\r\n\t\t\tHow much current will flow from the battery? Give your answer in \\(\\text{mA}\\).\r\n\t\t\tWhat will be the voltage across each resistor?\r\n\t\t\r\n\t\r\n\r\n","n":0.034},[{"v":"Maths","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Electricity","i":1,"n":1},{"v":"Electrical circuits","i":0,"n":0.707}]]},{"i":184,"$":[{"v":"Fashion: the Higg materials sustainability index","n":0.408},{"v":"The Higg Materials Sustainability Index was developed by the Sustainable Apparel Coalition, which works to reduce the impact of fashion and clothing to the environment. Keep reading to learn about this index.\r\nThe Higg Index looks at how materials are made, recycled, how much water they use and other things that have an impact on the environment. It also considers where the materials are made. Some countries will have different rules that can impact how much waste a company generates making materials.\r\n\r\nThe Higg Index developers applied five principles to understand how sustainable different fabrics are:\r\n\r\n \tclimate impact (how much heat this will add to the world's climate)\r\n \twater pollution (how much and what kind of pollution will go into waterways while making this fabric?)\r\n \twater use (some fabrics use a lot more water than others in production)\r\n \tuse of fossil fuels (this includes products made of fossil fuels and ones that use fossil fuels for transport or when they are made)\r\n \tchemistry (what is the fabric made of? How will those chemicals breakdown and react after they are not wanted any more?)\r\n\r\nUnderstanding the impact of different fabrics gives designers and companies a way to make more sustainable clothing. The Higg index has expanded since it was first developed. Now, companies can look at their impact from \"cradle-to-grave\" or from the raw materials they use to the waste after the product has finished. This lifecycle view looks at transportation, office spaces, suppliers, materials and much more to help companies find ways to be more sustainable.\r\n\r\nNote: the Higg Index has been criticised for scoring fabric from fossil fuels too highly (nylon and other plastic based fabrics) and rating them more highly than natural fibres (leather, silk and others). In response to this criticism, the Index is not being used as a promotional or marketing tool by fashion brands, anymore.\r\n\r\n\r\nBeyond fashion\r\nThe Higg Index suite of tools measures the environmental impact of different materials in other industries too. Home textiles, shoes, fashion and other uses of fabric and materials can have the Higg Index applied to their products. \"Materials\" in this context means the parts that a product is made of. A shoe might be made from leather, plastic and rubber \u2014 these are the shoe's materials.\r\n\r\nDesigners and developers can look at the Greenhouse Gas Emissions (GHG) of materials to find options that are lower in carbon emissions.\r\n\r\nAnyone can create an account and use the Higg index tools to green their business and make it more sustainable. There are clear instructions on how to use the index and how to compare materials.","n":0.048},[{"v":"Sustainable industries","i":2,"n":0.707},{"v":"Sustainability","i":1,"n":1},{"v":"Fashion","i":0,"n":1}]]},{"i":185,"$":[{"v":"Make your own sustainability goal","n":0.447},{"v":"This learning activity will help you to create your own sustainability goal using the UN Sustainable Development Goals (SDGs) as inspiration.\r\nUse this guide to creating SMART goals and read The lazy person's guide to saving the world for some simple and practical goals you can set. You can create as many goals as you like.\r\n\r\n\r\n\r\nNow that you have completed the activity, you can download your goal and put a reminder in your phone, so you can celebrate when you reach your target.","n":0.11},[{"v":"Sustainability","i":1,"n":1},{"v":"UN Sustainable Development Goals","i":0,"n":0.5}]]},{"i":186,"$":[{"v":"Sustainable industries: construction","n":0.577},{"v":"Construction, factories, railway lines, roads and many other things we rely on, are a big part of the modern world. Construction creates a lot of waste, greenhouse gases and uses a lot of valuable materials.\r\nConcrete uses a lot of natural resources \u2013 sand, gravel and water \u2013 and it relies on cement as a main ingredient. Cement, an ingredient of concrete, creates 8% of the world's greenhouse gas emissions.\r\n\r\nConcrete is also the second most used substance on earth (for construction) \u2014 the first is water.\r\n\r\nConstruction companies lose a lot of money because of climate change. Bad weather, worsening storms, and other changes stop construction and cause expensive delays. Climate change also means we will use more concrete to make things, as concrete is heavier, harder and better at making buildings that will last in big storms and other extreme weather conditions.\r\n\r\nFinding new materials or new ways to make things like concrete is key to creating sustainable construction.\r\n\r\nDangerous waste\r\nAnother big environmental problem is what to do with old tyres. Tyres:\r\n\r\n \tcontain toxic chemicals\r\n \tlast for many hundreds of years without breaking down\r\n \tare also very flammable \u2013 they catch on fire easily and the fires are hard to put out\r\n \tare dangerous in landfill because they are a fire risk.\r\n\r\nSustainable solution\r\nRMIT researchers have found a way to use recycled tyres in concrete. By adding recycled tyres and other waste materials to concrete, they have made stronger and longer lasting concrete. This means the concrete is not only a better quality product, they're also removing waste and recycling a dangerous substance, removing it from landfill. Finding methods like this to recycle, reuse, upcycle and reduce waste in construction is extremely important for sustainable construction.\r\n","n":0.06},[{"v":"Sustainability","i":2,"n":1},{"v":"Sustainable industries","i":1,"n":0.707},{"v":"Construction","i":0,"n":1}]]},{"i":187,"$":[{"v":"Sustainable industries: festivals and events","n":0.447},{"v":"There are many kinds of festivals and events: music, art, balloons, food and wine, and lots more. They're great fun, but are they bad for the environment? These days, many people try to organise green festivals and events that are planned with sustainability in mind.\r\n\r\nPlanning an event?\r\nThere are lots of organisations thinking about how to hold events in a sustainable way. From companies putting on 'green' festivals to guides for creating a more sustainable event.\r\n\r\nThere are some easy ways to make your event greener.\r\nWater\r\nEncourage people to bring their own water bottle to the event. Provide people with places they can refill water bottles. Consider selling or giving away reusable water bottles so everyone can use one.\r\n\r\nDid you know you can hire composting or chemical toilets? Using water to flush toilets uses anywhere between 6 to 15 litres of water per flush.\r\nWaste\r\nSorting waste on the site and making sure it goes to the right place is the first step. Where is the right place? Where the waste is recycled or reused rather than going to landfill.\r\n\r\nHaving clear signs will make sure the waste goes to the right bin, saving recycling from being contaminated with rubbish.\r\nFood waste\r\nSell tickets or ask people to register to come. Then you know how many people will be there and you can order enough food for everyone and not waste extra food. Any leftover food and green waste can be collected and turned to compost or soil.\r\nEnergy\r\nThink about where your power will come from. Can you use solar panels and batteries instead of electricity from fossil fuels?\r\nTransportation\r\nConsider how people will get to the festival. Will public transport be available? Make it easy for people to leave their cars at home by:\r\n\r\n \tchoosing\u00a0a site close to public transportation\r\n \tproviding bike parking\r\n \tproviding a bus from the train station\r\n \treducing\u00a0the cost of car parking for people who ride share\r\n \tcreating or using walking trails\r\n\r\n\r\n","n":0.056},[{"v":"Sustainable industries","i":2,"n":0.707},{"v":"Sustainability","i":1,"n":1},{"v":"Festivals and events","i":0,"n":0.577}]]},{"i":188,"$":[{"v":"Sustainable industries: fashion","n":0.577},{"v":"The clothing industry is global, extremely large and prolific because we all need clothes to wear. However, the fashion industry is creating a lot of waste due to the high turnover of clothes, which have short lifespans as they are discarded due to poor quality manufacture, or quickly changing fashion styles.\r\nBig companies can sell cheap clothes by manufacturing them in countries where workers may not be paid a living wage. In more affluent societies, these garments are considered disposable and not kept as long as they could be.\r\n\r\nA 'living wage' means paying workers enough to take care of themselves and their families, which includes affording things like housing, food, health care and education for their children.\r\n\r\nPressure is rising for designers and clothing manufacturers to reduce the impact of fashion. There are many areas of manufacture in the fashion and textile industry where changes can be made to make this industry more sustainable.\r\n\r\nTextile recycling \u2013 the textiles (fabric and other materials used to make clothing) can be broken down and the fibres (threads that make up the fabric) can be recycled into new clothes.\r\n\r\nRepair \u2013 creating garments that are easy to repair ensures that they can last longer.\r\n\r\nWages \u2013 garment workers (people who make clothes) are paid a living wage.\r\n\r\nReuse \u2013 high quality, well made clothes can be resold many times. Cheap, badly made clothes cannot.\r\nWhat is fast fashion?\r\nFast fashion is growing around the world. Fast fashion are clothes made very cheaply and quickly that are only worn a few times before being thrown away.\r\n\r\nPeople today own far more clothes than they used to in the past, and some of these clothes are cheap and not made to last. The actual cost of the clothes is much higher than the price people are paying to buy them if we think about the cost of waste produced.\r\n\r\nSome experts have suggested that by making clothes more expensive to cover the cost of dealing with clothing waste, we might stop people from buying so many cheap clothes.\r\n\r\nThe rise of Internet shopping, propelled further by the recent pandemic, has made it very easy to buy new clothes. Many of these clothes are very cheap. No one needs to buy new clothes every week or every month. Making clothes more expensive to cover the cost of waste is one way to change fast fashion.\r\n\r\nThere are companies recycling textiles that break down the clothes and create new threads and materials. There are many drop-off points in Australia that will accept old clothes for recycling. Buying recycled clothing can help to make the fashion industry more sustainable. If you are choosing new clothes, look for recycled fabrics and support companies that recycle, or are 'fair trade' and pay their workers a fair price for their labour.\r\n","n":0.047},[{"v":"Sustainability","i":2,"n":1},{"v":"Sustainable industries","i":1,"n":0.707},{"v":"Fashion","i":0,"n":1}]]},{"i":189,"$":[{"v":"Sustainable creative practice","n":0.577},{"v":"This page contains a reflective activity you can do at any time. The table below suggests ways to think about your creative practice. It outlines how you can support and inspire your work no matter what style or mediums you use. This activity can be applied to any creative arts or design practices.\r\nRemember that 'sustainable' means being able to do something for a long time. In this context, sustainable creative practice means developing methods and processes that help you to keep making creative work, while developing your skills and forging relationships and communities to help others develop their practices too. Aspects of creative practice can link to many of the United Nations Sustainable Development Goals (SGDs) including:\r\n\r\n \tGoal 3.\u00a0Good health and wellbeing\r\n \tGoal 4.\u00a0Quality education\r\n \tGoal 8.\u00a0Decent work and economic growth\r\n \tGoal 11.\u00a0Sustainable cities and communities\r\n\r\nReflective activity\r\nFor this activity, read the Developing a sustainable creative practice table.\r\nYou might recognise yourself at one of the stages shown in the table rows, such as 'Introductory practice' or 'Emerging practice'. \u00a0You can work through the columns and check if your methods and processes incorporate elements suggested here. Even if you don't identify yourself at the bottom row of 'Sustainable practice', you are still working towards a practice that you can take forward into the future.\r\n\r\nStructuring this information in a 'rubric' styled table can show you what's ahead and motivate you to work towards a more\u00a0established and sustainable creative practice.\r\n\r\nWhat you can do:\r\n\r\n \tMake a list of things you can do now to develop your practice based upon the information in the table.\r\n \tMake a wish-list of where you'd like your practice to be in 5 years.\r\n \tSave this work and remember to check in and review it regularly\r\n\tRefer to the Higg materials sustainability index to choose the most environmentally friendly materials for your creative works.\r\n\r\nDeveloping a sustainable creative practice\r\n\r\n\r\n\r\n\r\n\r\nCreate\r\nWork together\r\nReflect and\u00a0document\r\nBuild communities\r\n\r\n\r\n\r\n\r\nIntroductory practice\r\nExplore\u00a0emotions, thoughts and ideas. Experiment\u00a0with different mediums,\u00a0tools, concepts and creative modes.\r\nFind a mentor who you can work with on creative projects.\r\nStart a diary, a sketchbook, blog or other space to record your thoughts and reactions. Speak to other artists and designers\u00a0about creative projects.\r\nDiscuss your projects with your closest peers and other supporters.\r\n\r\n\r\nEmerging practice\r\nExpand your set of skills to include methods and processes that bring out your creativity and build a body of work.\r\nWork with a mentor and use their projects in your own way and for your needs.\r\nFind other artists, source their work. Credit the authors and makers, communicate progress and celebrate completion of your projects.\r\nTest and perfect lots of ways to share your work, and your process in novel places and to new people.\r\n\r\n\r\nEstablished practice\r\nDiscover\u00a0meaning and fulfillment in your practice. Give\u00a0voice your own authentic style.\r\nWork with other artists and creatives on projects. Contribute to the plans and execution of projects.\r\nDescribe your work and your practice.\u00a0Write an artist statement and\u00a0regularly review and update it. Record, rework, revise and release your current work.\r\nEstablish a relationship with a mentee who you can guide and support. You will be inspired watching them develop their practice.\r\n\r\n\r\nSustainable practice\r\nStart your own creative projects where you can experiment with different processes, methods and meanings.\r\nBring creatives from different practices together. Forge partnerships and work with other artists on new projects.\r\nInterrogate your own creative practice. Share your experiences and reflect on\u00a0feedback. Offer insight to other artists, where they request it.\r\nCampaign to increase access and equity for all who want to receive an education in the arts.\r\n\r\n\r\n\r\n\r\nThis table inspired by Tyler Denmead's Sustainable Creative Practice Rubric.","n":0.042},[{"v":"Sustainability","i":2,"n":1},{"v":"Design","i":1,"n":1},{"v":"Art","i":0,"n":1}]]},{"i":190,"$":[{"v":"Sustainable development goals in your life","n":0.408},{"v":"The United Nation's (UN) Sustainable Development Goals (SDG) can be found in many parts of your everyday life. RMIT is a UN SDG partner, and there are numerous projects running on campus, in classrooms and through industry partnerships.\r\n\r\nIn this section, we explore the SDGs in your everyday life. You can design your own sustainability goal by applying the principles we have explored in previous sections, and take action to make the world a little more sustainable.\r\n\r\nTopics\r\n\r\n \tSustainable industries: construction - What single product produces 8% of the world's greenhouse gas emissions? And how is RMIT helping?\r\nTime to complete 10 minutes.\r\n \tSustainable industries: fashion - Learn what the problem is with fast fashion and how we can fix it.\r\nTime to complete 25 minutes.\r\n \tSustainable industries: festivals and events - What can you do to make your festival or other events more sustainable?\r\nTime to complete 10 minutes.\r\n \tSustainable creative practise - A reflective activity that suggests ways you can sustain your practice as an artist or designer, including resources that explain how the arts are important for the SDGs.\r\nTime to complete 30 minutes.\r\n \tMake your own sustainability goal - Want to make the world and your life a little greener and more pleasant? Create a sustainability goal you can stick to.\r\nTime to complete 45 minutes\r\n\r\n\r\n\r\n","n":0.069},[{"v":"Sustainable industries","i":2,"n":0.707},{"v":"UN Sustainable Development Goals","i":1,"n":0.5},{"v":"Sustainability","i":0,"n":1}]]},{"i":191,"$":[{"v":"Systems thinking: unexpected consequences","n":0.5},{"v":"Sometimes we try and do the right thing but by mistake do the wrong thing. We might try to fix a problem and create more problems instead.\r\nHere is an example of creating more problems while trying to fix something.\r\n\r\nKalitha and Jain want to help kids at the local school. They know some of the children do not eat breakfast before school, so they decide to provide a free breakfast. After speaking to businesses, they find a company that will provide a discounted breakfast, so they raise money and buy breakfasts for the kids.\r\n\r\nThe children are happy to receive the breakfast and teachers report that the children are more focussed and answering more questions in the morning classes.\r\n\r\nAfter a week, the school asks Kalitha and Jain to stop providing the breakfasts. When Kalitha talks to the school principal, she learns that the breakfasts have created a waste problem. The breakfasts come in a plastic container and the school bins are overflowing with rubbish. The rubbish also costs the school money becasue they must pay a company to take away their waste and the extra waste costs more to remove.\r\n\r\nThe school, the children, parents and Kalitha and Jain want the program to continue but they do not want to create a waste problem and cost the school money.\r\n\r\nLuckily, they talk to some parents and one of the mothers works for a recycling company. The recycling company takes plastic and turns it into pellets that are sold to be made into new products. The company is happy to take the plastic waste, and they even pay the school for the plastic.\r\n\r\nThe company will only take the waste if it is properly sorted and there are no other kinds of rubbish mixed in. The company sends special plastic bins for collecting the extra waste. Kalitha and Jain create signs for the school to show the children which bin their rubbish needs to go in.\r\n\r\nFrom one bad consequence to lots of good outcomes!\r\n","n":0.055},[{"v":"Systems thinking","i":1,"n":0.707},{"v":"Sustainability","i":0,"n":1}]]},{"i":192,"$":[{"v":"Systems thinking: triple bottom line","n":0.447},{"v":"The triple bottom line means thinking about people, planet, and profit when making decisions in business.\r\nThe triple bottom line is not perfect, but it is well known. It is an easy way to remember the things that make a business sustainable and profitable. Finding ways to pay people properly, care for the environment and make a profit encourages business to be sustainable.\r\n\r\nSaving the planet shouldn't cost the earth. You might have heard people say this before. It means that business can find a way to create their products sustainably and still make them affordable.\r\n\r\nWhat do we mean by triple bottom line?\r\nPeople\r\nSustainability is not just about saving the natural environment; it is also about people. It is not sustainable for people to live in poverty to keep the costs of products low.\r\n\r\nBusinesses need to pay people a 'living wage'. A living wage is enough money to fed, clothe and house the worker. The worker needs to earn a living wage in a reasonable number of working hours. Working 40 hours a week and not being able to feed and home yourself, is not sustainable.\r\nPlanet\r\nPlanet here is talking about the natural environment, or nature. This includes trees, oceans, wild animals and birds. Taking care of the planet should be a priority for everyone.\r\n\r\nThe natural environment is damaged by many human activities: cutting down trees, mining, pollution and so much more. To make products sustainable, we need to stop damaging the environment. We need business to consider better ways to make their products.\r\nProfit\r\nBusinesses need to make money. Without profit, businesses fail. Sometimes sustainable options are more expensive. Asking people to pay a fair price and allowing the business to continue, is the sustainable choice.\r\n\r\nBusinesses might need to increase prices when they find better but more expensive ways to work. If people support businesses that are committed to the triple bottom line, everyone is better off.","n":0.057},[{"v":"Systems thinking","i":1,"n":0.707},{"v":"Sustainability","i":0,"n":1}]]},{"i":193,"$":[{"v":"Systems thinking","n":0.707},{"v":"We can think about the world we live in as a series of connected systems, from the solar system down to your local transport systems.\r\nSystems thinking is used in science to describe complicated relationships between parts of a whole. We talk about the 'solar system' to describe the group of planets and objects that orbit our sun. The solar system is a whole and each planet and the sun are the parts. There are also gravity, asteroids, heat, moons and more in this system.\r\nWhat is a system?\r\n\r\nSystems thinking in sustainability is a way of thinking about how we make goods and deliver services. It considers the environmental and social impacts for different businesses.\r\n\r\nIn the solar system, we know that changes in the sun or other objects in the system, have effects on Earth. In business, we know that changing how much workers are paid or what materials we use affects social and economic systems. We also know that what happens in business affects the system we all live in \u2014 the Earth.\r\n\r\nThe environment is a planet-wide system that affects us all. The economy is another system, a way of deciding value and exchanging goods and services. People live in societies in which there are also systems though these might not be as obvious as in a physical system.\r\n\r\nAll these different systems interact with each other. A change in the solar system can cause a change on Earth. A big solar storm causes blackouts. A change in the environment causes a change in society. Increased temperatures make people spend more time in their homes and less time socialising.\r\n\r\nThe United Nations Sustainable Development Goals are a way of making changes that have positive outcomes in many systems.\r\n\r\nWe have economic, social and environmental problems that are related to each other. The United Nations includes poverty, equality and justice in the Sustainable Development Goals because we cannot fix one system and cause a problem in another.\r\n\r\n\r\n\r\n\r\n","n":0.056},[{"v":"Systems thinking","i":1,"n":0.707},{"v":"Sustainability","i":0,"n":1}]]},{"i":194,"$":[{"v":"Scalar or dot product","n":0.5},{"v":"\r\nThere are two ways to multiply two vectors. Here, we will learn about the scalar product. It has many applications in STEM. For example, scalar products are used to calculate the work done by a system, in computer graphics to calculate the amount of light hitting surfaces, and in engineering to determine the stress applied to materials.\r\n\r\nThe scalar product is the result of multiplying the magnitudes of the components of two or more vectors. This does not result in a vector, but a scalar; a scalar or dot product does not have a direction.\r\nFor two vectors \\(\\vec{a}\\left(a_{1},a_{2},a_{3}\\right)\\) and \\(\\vec{b}\\left(b_{1},b_{2},b_{3}\\right)\\). The scalar product is defined by: \r\n\r\n\\[ \\vec{a}\\cdot\\vec{b}=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} \\]\r\nor\r\n\r\n\\[ \\vec{a}\\cdot\\vec{b}=\\left|\\vec{a}\\right|\\left|\\vec{b}\\right|\\cos\\theta \\] where \\(\\theta\\) is the angle between \\(\\vec{a}\\) and \\(\\vec{b}\\).\r\n\r\n\r\nIt is also commonly called the dot product as a dot is used between the two vectors being multiplied instead of a multiplication symbol.\r\n\r\nProperties of the scalar or dot product\r\nThere are some key characteristics of scalar products that are handy to remember.\r\n\r\n \tIf \\(\\vec{a}\\) and \\(\\vec{b}\\) are non-zero vectors and \\(\\vec{a}\\) is perpendicular to \\(\\vec{b}\\) (i.e. there is \\(90^{\\circ}\\) or \\(\\dfrac{\\pi}{2}\\) radians between them), then \\(\\vec{a}\\cdot\\vec{b}=0\\), since \\(\\cos\\left(90^{\\circ}\\right)=\\cos\\left(\\dfrac{\\pi}{2}\\right)=0\\).\r\n \tIf \\(\\vec{a}\\) is parallel to \\(\\vec{b}\\), then the angle between the vectors is \\(0\\) and \\(\\vec{a}\\cdot\\vec{b}=\\left|\\vec{a}\\right|\\left|\\vec{b}\\right|\\) as \\(\\cos\\left(0\\right)=1\\).\r\n \tThe dot product does not depend on the order of multiplication, so \\( \\vec{a}\\cdot\\vec{b}=\\vec{b}\\cdot\\vec{a} \\).\r\n \tIn three dimensions, with \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\) unit vectors along the \\(x\\)-, \\(y\\)- and \\(z\\)-axes, respectively: \\[\\begin{align*} \\vec{i}\\cdot\\vec{j} & = \\vec{j}\\cdot\\vec{k}=\\vec{k}\\cdot\\vec{i}=0\\\\ \\vec{i}\\cdot\\vec{i} & = \\vec{j}\\cdot\\vec{j}=\\vec{k}\\cdot\\vec{k}=1\r\n\t\\end{align*}\\]\r\n\r\n\r\nExample 1 \u2013 calculating the scalar product\r\n\r\n\\[\\begin{align*} \\left(2\\vec{i}+3\\vec{j}+4\\vec{k}\\right)\\cdot\\left(-\\vec{i}-2\\vec{j}+\\vec{k}\\right) & = \\left(2\\times\\left(-1\\right)\\right)+\\left(3\\times\\left(-2\\right)\\right)+\\left(4\\times1\\right)\\\\\r\n& = -4\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 calculating the scalar product\r\n\r\n\tCalculate the dot products of the following pairs of vectors.\r\n\t\r\n\t\t\\((2,5,-1)\\) and \\((4,1,1)\\)\r\n\t\t\\(3\\vec{i}\\) and \\(5\\vec{j}\\)\r\n\t\t\\(5\\vec{k}\\) and \\((\\vec{j}+2\\vec{k})\\)\r\n\t\r\n\tFind:\r\n\t\r\n\t\t\\((2,0,4)\\cdot(-3,1,3)\\)\r\n\t\t\\((0,5,1)\\cdot(4,0,0)\\)\r\n\t\t\\((2\\vec{i}+3\\vec{k})\\cdot(7\\vec{i}+2\\vec{j}+4\\vec{k})\\)\r\n\t\r\n\tDetermine which of the following vectors are perpendicular.\r\n\t\r\n\t\t\\((5,2,3)\\)\r\n\t\t\\((0,1,-1)\\)\r\n\t\t\\((-2,2,2)\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\nAngle between two vectors\r\nThe angle \\(\\theta\\) where \\(\\left(0\\leq\\theta\\leq\\pi\\right)\\), between two vectors can be found using same equations used to calculate the scalar product: \r\n\\[\\vec{a}\\cdot\\vec{b}=\\mid\\vec{a}\\mid\\mid\\vec{b}\\mid\\cos\\theta\\]\r\nWe can rearrange this to make \\(\\theta\\) the subject:\r\n\\[\\begin{align*} \\cos\\theta & = \\frac{\\vec{a}\\cdot\\vec{b}}{\\left|\\vec{a}\\right|\\left|\\vec{b}\\right|}\\\\\r\n\\theta & = \\cos^{-1}\\left(\\frac{\\vec{a}\\cdot\\vec{b}}{\\left|\\vec{a}\\right|\\left|\\vec{b}\\right|}\\right)\r\n\\end{align*}\\]\r\n\r\nExample 1 \u2013 finding the angle between two vectors\r\nIf \\(\\vec{a}=\\left(2,3,1\\right)\\) and \\(\\vec{b}=\\left(5,-2,2\\right)\\), find the angle \\(\\theta\\) between \\(\\vec{a}\\) and \\(\\vec{b}\\).\r\nWe can start by finding \\(\\vec{a}\\cdot\\vec{b}\\).\r\n\\[\\begin{align*} \\vec{a}\\cdot\\vec{b} & = \\left(2,3,1\\right)\\cdot\\left(5,-2,2\\right)\\\\\r\n& = 6\r\n\\end{align*}\\]\r\nNow, let's find \\(\\left|\\vec{a}\\right|\\) and \\(\\left|\\vec{b}\\right|\\).\r\n\\[\\begin{align*} \\left|\\vec{a}\\right| & = \\sqrt{2^{2}+3^{2}+1^{2}}\\\\\r\n& = \\sqrt{14}\r\n\\end{align*}\\]\r\n\\[\\begin{align*} \\left|\\vec{b}\\right| & = \\sqrt{25+4+4}\\\\\r\n& = \\sqrt{33}\r\n\\end{align*}\\]\r\nLet's substitute these values into the rearranged scalar product equation.\r\n\\[\\begin{align*} \\theta & = \\cos^{-1}\\left(\\frac{6}{\\sqrt{33}\\times\\sqrt{14}}\\right)\\\\\r\n& = \\cos^{-1}\\left(0.2791\\right)\\\\\r\n& = 73.8^{\\circ}\r\n\\end{align*}\\] \r\nTherefore, the angle between \\(\\vec{a}\\) and \\(\\vec{b}\\) is \\(73.8^{\\circ}\\).\r\n\r\n\r\n\r\nExercise \u2013 finding the angle between two vectors\r\n\r\n\tFind the angle between the following pairs of vectors.\r\n\t\r\n\t\t\\((1,2,3)\\) and \\((4,-1,0)\\)\r\n\t\t\\((2,1,-2)\\) and \\((1,5,-1)\\)\r\n\t\t\\((0,5,1)\\) and \\((2,0,0)\\)\r\n\t\t\\((1,-2,3)\\) and \\((-4,1,-3)\\)\r\n\t\t\\((2,1,-2)\\) and \\((0,4,0)\\)\r\n\t\t\\((0,3,0)\\) and \\((0,1,0)\\)\r\n\t\r\n\tConsider the vectors \\(\\vec{a}\\), \\(\\vec{b}\\) and \\(\\vec{c}\\) where \\(\\vec{a}=(2,2,2)\\), \\(\\vec{b}=(3,2,-1)\\) and \\(\\vec{c}=(-1,4,1)\\).\r\n\t\r\n\t\tShow that \\(\\vec{a}\\cdot\\vec{b}=\\vec{a}\\cdot\\vec{c}\\).\r\n\t\tRearranging \\(\\vec{a}\\cdot\\vec{b}=\\vec{a}\\cdot\\vec{c}\\) gives \\(\\vec{a}\\cdot\\left(\\vec{b}-\\vec{c}\\right)=0\\). Assuming that \\(\\vec{b}\\neq\\vec{c}\\), determine the relationship between \\(\\vec{a}\\) and \\(\\left(\\vec{b}-\\vec{c}\\right)\\).\r\n\r\n\r\n\r\n\r\n","n":0.047},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":195,"$":[{"v":"Vector or cross product","n":0.5},{"v":"The vector product is another way to multiply two vectors. Just like scalar products, vector products have many broad applications, such as in electrical engineering, quantum physics, software development, game programming and statistics. Use this resource to learn more.\r\n\r\nThe vector product is a vector resulting from multiplying the magnitudes (size) of two vectors. Compare this to the scalar or dot product, which gives a scalar. \r\nThe magnitudes are found using matrices and determinants, so make sure you are confident with these before you continue.\r\nLet's consider the unit vectors \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\) which are in the \\(x\\), \\(y\\) and \\(z\\) directions, respectively. We can write these as:\r\n\\[\\begin{align*} \\hat{i} & = \\hat{i}+0\\hat{j}+0\\hat{k}\\\\\r\n& = (1,0,0)\r\n\\end{align*}\\] \\[\\begin{align*} \\hat{j} & = 0\\hat{i}+\\hat{j}+0\\hat{k}\\\\\r\n& = (0,1,0)\r\n\\end{align*}\\] \\[\\begin{align*} \\hat{k} & = 0\\hat{i}+0\\hat{j}+\\hat{k}\\\\\r\n& = (0,0,1)\r\n\\end{align*}\\] They give us two vectors:\r\n\\[\\vec{a} = a_{1}\\hat{i}+a_{2}\\hat{j}+a_{3}\\hat{k}\\] \\[\\vec{b} = b_{1}\\hat{i}+b_{2}\\hat{j}+b_{3}\\hat{k}\\] Geometrically, the vector product is found using: \\[\\vec{a}\\times\\vec{b}=\\left|\\vec{a}\\right|\\left|\\vec{b}\\right|\\sin(\\theta)\\hat{n}\\] where \\(\\hat{n}\\) is a unit vector perpendicular to both \\(\\vec{a}\\) and \\(\\vec{b}\\) and \\(\\theta\\) is the angle between the vectors.\r\n\r\nThis helps us understand the magnitude and direction of the vector, but in practice, we use the component form:\r\n\\[\\begin{align*} \\vec{a}\\times\\vec{b} & = \\left| \\begin{array}{ccc} \\hat{i} & \\hat{j} & \\hat{k}\\\\\r\na_{1} & a_{2} & a_{3}\\\\\r\nb_{1} & b_{2} & b_{3}\r\n\\end{array} \\right| \\\\\r\n& = \\hat{i}\\left| \\begin{array}{cc} a_{2} & a_{3}\\\\\r\nb_{3} & b_{3}\r\n\\end{array} \\right| - \\hat{j}\\left| \\begin{array}{cc} a_{1} & a_{3}\\\\\r\nb_{1} & b_{3}\r\n\\end{array} \\right| + \\hat{k}\\left| \\begin{array}{cc} a_{1} & a_{2}\\\\\r\nb_{1} & b_{2}\r\n\\end{array} \\right| \\\\\r\n& = \\hat{i}(a_{2}b_{3}-b_{2}a_{3})-\\hat{j}(a_{1}b_{3}-b_{1}a_{3})+\\hat{k}(a_{1}b_{2}-b_{1}a_{2})\r\n\\end{align*}\\]\r\nThe vector product is also commonly referred to as the cross product as a cross (or multiplication symbol) is used between the two vectors being multiplied.\r\nKeep in mind that the vertical delimiters (\\(\\left| \\right|)\\) here denote the determinant of the matrix, and not the magnitude of the vector.\r\n\r\nProperties of the vector or cross product\r\nThere are some key characteristics of vector products that are handy to remember.\r\n\r\n\tIf \\(\\vec{a}\\) is parallel to \\(\\vec{b}\\), then \\(\\vec{a}\\times\\vec{b}=\\vec{0}\\) as \r\n\t\\[\\begin{align*} \\vec{a}\\times\\vec{b} & = \\left|\\vec{a}\\right|\\left|\\vec{b}\\right|\\sin(0)\\hat{n}\\\\\r\n\t& = \\vec{0} \\end{align*}\\]\r\n\tas \\(\\sin(0)=0\\). Since the vector product is a vector, we should write the answer as the zero vector, \\(\\vec{0}\\), instead of the number \\(0\\).\r\n \tThe order in which you take the cross product is important. In fact, \\(\\vec{a}\\times\\vec{b}=-\\vec{b}\\times\\vec{a}\\).\r\n \tThe direction of \\(\\vec{a}\\times\\vec{b}\\) is perpendicular (\\(90^{\\circ})\\) to the plane in which both \\(\\vec{a}\\) and \\(\\vec{b}\\) sit on. We can visualise this as the direction in which your thumb would point if the fingers of your right hand are curled from \\(\\vec{a}\\) to \\(\\vec{b}\\).\r\n\t\r\n\tThis is called the right hand rule.\r\n\r\n\r\nExample 1 \u2013 calculating the vector product\r\nFind \\(\\vec{a}\\times\\vec{b}\\) if \\(\\vec{a}=2\\vec{i}+3\\vec{j}+\\vec{k}\\) and \\(\\vec{b}=5\\vec{j}+3\\vec{k}\\).\r\nVectors must first be written in matrix form. We place the variables\u2014in this case, the unit vectors\u2014in the first row. In the second and third rows of the matrix, we put the corresponding coefficients for the vectors \\(\\vec{a}\\) and \\(\\vec{b}\\), respectively. \r\nThis gives us the matrix:\r\n\\[\\begin{align*} \\vec{a}\\times\\vec{b} & = \\left| \\begin{array}{ccc} \\hat{i} & \\hat{j} & \\hat{k}\\\\ \r\n2 & 3 & 1\\\\ \r\n0 & 5 & 3 \r\n\\end{array} \\right|\r\n\\end{align*}\\]\r\nTo find the vector or cross product, we simply find the determinant of the matrix.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nCartesian unit vectors\r\nIf we just look at the cartesian unit vectors\u2014that is, the unit vectors that lie along the \\(x\\)-, \\(y\\)- and \\(z\\)-axes\u2014there are a few rules that are helpful to remember. \r\nAgain, we will let \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\) be unit vectors in the \\(x\\), \\(y\\) and \\(z\\) directions, respectively. The rules are:\r\n\r\n\t\\(\\hat{i}\\times\\hat{j}=\\hat{k}\\)\r\n\t\\(\\hat{j}\\times\\hat{k}=\\hat{i}\\)\r\n\t\\(\\hat{k}\\times\\hat{i}=\\hat{j}\\)\r\n\t\\(\\hat{i}\\times\\hat{k}=-\\hat{j}\\)\r\n\t\\(\\hat{k}\\times\\hat{j}=-\\hat{i}\\)\r\n\t\\(\\hat{j}\\times\\hat{i}=-\\hat{k}\\)\r\n\r\nThese rules make finding the vector product more efficient as we can derive answers without having to complete calculations.\r\n\r\nExample 4 \u2013 calculating the vector product\r\nFind \\(\\hat{i}\\times\\hat{k}\\).\r\nWe can write these are vectors first. \r\n\\[\\hat{i}=\\hat{i}+0\\hat{j}+0\\hat{k}\\] \\[\\hat{k}=0\\hat{i}+0\\hat{j}+\\hat{k}\\]\r\nNow, let's find the vector product.\r\n\r\n\r\nExercise \u2013 calculating the vector product\r\n\r\n\tCalculate the following.\r\n\t\r\n\t\t\\(\\hat{j}\\times\\hat{k}\\)\r\n\t\t\\(\\hat{i}\\times4\\hat{i}\\)\r\n\t\t\\((2\\hat{i}+3\\hat{j}-\\hat{k})\\times(3\\hat{j}+2\\hat{k})\\)\r\n\t\t\\(3\\hat{j}\\times5\\hat{i}\\)\r\n\t\t\\((\\hat{i}-3\\hat{j}+\\hat{k})\\times(2\\hat{i}+\\hat{j}-\\hat{k})\\)\r\n\t\r\n\tFind a unit vector perpendicular to both \\((\\vec{i}-\\vec{k})\\) and \\((\\vec{i}+3\\vec{j}-2\\vec{k})\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.041},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":196,"$":[{"v":"Solving trigonometric equations","n":0.577},{"v":"\r\nA calculator may only give you one answer to an inverse trigonometry question between \\(0\\) and \\(90^{\\circ}\\). When you use the unit circle, you can visualise the many different solutions for the angle. Understanding this concept will help you in many STEM fields, such as satellite navigation in physics, and measurement of distances in architecture.\r\n\r\nDomain and range\r\nIf you know the value of a trigonometric function, you can find all of the possible angles that satisfy this expression using the unit circle.\r\nIn circular functions:\r\n\r\n\tthe domain is all of the possible values for the angles where the function is defined\r\n\tthe range is the range of possible values corresponding to the domain.\r\n\r\n\r\nFor example, if the question is: Solve \\(\\cos(\\theta)\\) if \\(-\\pi \\le \\theta \\le 3\\pi\\), then \\(-\\pi \\le \\theta \\le 3\\pi\\) is the domain and the range would be the relevant solutions for \\(\\cos(\\theta)\\).\r\nSolving these types of problems is easier if you can confidently identify the values of trigonometric functions on a unit circle. You also need to understand angles in degrees and radians. Go back and review circular functions and angle measurement before you proceed.\r\n\r\nExample 1 \u2013 solving trigonometric equations\r\nGiven that \\(\\sin\\theta=0.3\\), find all values of \\(\\theta\\) in the domain \\(\\left\\{ \\theta:0^{\\circ}\\leq\\theta\\leq360^{\\circ}\\right\\}\\).\r\nFirst, sketch a unit circle and locate the points on the unit circle for which \\(\\sin(\\theta)=0.3\\)\u2014that is, where \\(y=0.3\\).\r\n\r\nThere are at least two solutions shown in red and blue. Let's call the red one \\(\\theta_{r}\\) and the blue one \\(\\theta_{b}\\).\r\n\\[\\begin{align*} \\theta_{r} & = \\sin^{-1}\\left(0.3\\right)\\\\\r\n& = 17.46^{\\circ}\\quad\\textrm{from calculator}\r\n\\end{align*}\\]\r\nThe other angle can be found by symmetry of the unit circle.\r\n\\[\\begin{align*} \\theta_{b} & = 180^{\\circ}-\\theta_{r}\\\\\r\n& = 180^{\\circ}-17.46^{\\circ}\\\\ \r\n& = 162.54^{\\circ}\r\n\\end{align*}\\]\r\nHence, the solutions are \\(\\theta=17.46^{\\circ}\\) and \\(162.54^{\\circ}\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 solving trigonometric equations\r\nSolve for \\(\\phi\\) in each of the following.\r\n\r\n\tIf \\(\\sin(\\phi)=0.25\\), find \\(\\phi\\) for \\(0^{\\circ} \\le \\phi \\le 180^{\\circ}\\).\r\n\tIf \\(\\tan(\\phi)=0.8\\), find \\(\\phi\\) for \\(0^{\\circ} \\le \\phi \\le 360^{\\circ}\\).\r\n\tIf \\(\\cos(\\phi)=0.4\\), find \\(\\phi\\) for \\(0^{\\circ} \\le \\phi \\le 360^{\\circ}\\).\r\n\tIf \\(\\cos(\\phi)=-0.4\\), find \\(\\phi\\) for \\(-180^{\\circ} \\le \\phi \\le 360^{\\circ}\\).\r\n\tIf \\(\\tan(\\phi)=-1.5\\), find \\(\\phi\\) for \\(-180^{\\circ} \\le \\phi \\le 360^{\\circ}\\).\r\n\tIf \\(\\cos(\\phi)=-0.3\\), find \\(\\phi\\) for \\(0^{\\circ} \\le \\phi \\le 360^{\\circ}\\).\r\n\r\n\r\n\r\n\r\n\r\n","n":0.054},[{"v":"Trigonometry","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Functions","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":197,"$":[{"v":"Special matrices","n":0.707},{"v":"\r\nThere are a number of important special matrices to become familiar with if you are studying a STEM discipline. The transpose of a matrix is important for solving problems in structural mechanics, symmetric matrices can be applied to motion, dynamics and force calculations in physics, and orthogonal matrices have many applications in data analysis, signal processing and even medicine.\r\n\r\nTranspose of a matrix\r\nThe transpose of a matrix \\(A\\) is denoted by \\(A^{T}\\) and is found by interchanging the rows and the columns. The first row becomes the first column, the second row becomes the second column, and so on.\r\nIf \\(A\\) is an \\(m\\times n\\) matrix, then \\(A^{T}\\) is an \\(n\\times m\\) matrix.\r\n\r\nExample 1 \u2013 finding the transpose of matrices\r\nFind the transpose of \\(A= \\left[ \\begin{array}{cc} 1 & 2\\\\\r\n3 & 4\\\\\r\n5 & 6\r\n\\end{array} \\right] \\).\r\n\\[A^{T} = \\left[ \\begin{array}{ccc} 1 & 3 & 5\\\\\r\n2 & 4 & 6\r\n\\end{array} \\right] \\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the transpose of matrices\r\nFind the transpose of the following matrices.\r\n\r\n\t\\(A = \\left[ \\begin{array}{cc} 3 & -2 \\end{array} \\right] \\)\r\n\t\\(B = \\left[ \\begin{array}{cc} 1 & -3\\\\\r\n\t-3 & 0 \r\n\t\\end{array} \\right]\\)\r\n\t\\(C = \\left[ \\begin{array}{ccc} 2 & 0 & 5\\\\\r\n\t0 & -2 & 2\\\\\r\n\t5 & 2 & 1\r\n\t\\end{array} \\right]\\)\r\n\t\\(D = \\left[ \\begin{array}{ccc} 2 & -1 & 5\\\\ \r\n\t5 & -1 & 2\r\n\t\\end{array} \\right]\\)\r\n\t\\(E = \\left[ \\begin{array}{ccc} 0 & 1 & -1\\\\\r\n\t1 & 0 & 1\\\\\r\n\t1 & -1 & 0\r\n\t\\end{array} \\right]\\)\r\n\r\n\r\n\r\n\r\n\r\nSymmetric matrix\r\nA symmetric matrix is a square matrix which is equal to its transpose. It is also symmetric about its leading diagonal (top left to bottom right).\r\n\r\nExample 3 for finding the transpose of matrices is an example of a symmetric matrix. Let's look at some more examples.\r\n\r\nExample 1 \u2013 determining whether matrices are symmetric\r\nDetermine whether matrix \\(D\\) is symmetric.\r\n\\[D = \\left[ \\begin{array}{cc} 1 & 3\\\\\r\n3 & 4\r\n\\end{array} \\right] \\] \\[\\begin{align*} D^{T} & = \\left[ \\begin{array}{cc} 1 & 3\\\\\r\n3 & 4\\\\\r\n\\end{array} \\right]\\\\\r\n& = D\r\n\\end{align*} \\]\r\nTherefore, \\(D\\) is a symmetric matrix.\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 determining whether matrices are symmetric\r\nDetermine which of the following matrices are symmetric.\r\n\\[\\begin{align*} A & = \\left[ \\begin{array}{cc} 3 & -2 \\end{array} \\right] \\quad\r\nB = \\left[ \\begin{array}{cc} 1 & -3\\\\\r\n-3 & 0 \r\n\\end{array} \\right]\\quad\r\nC = \\left[ \\begin{array}{ccc} 2 & 0 & 5\\\\\r\n0 & -2 & 2\\\\\r\n5 & 2 & 1\r\n\\end{array} \\right]\\\\\r\nD & = \\left[ \\begin{array}{ccc} 2 & -1 & 5\\\\\r\n5 & -1 & 2\r\n\\end{array} \\right]\\quad\r\nE = \\left[ \\begin{array}{ccc} 0 & 1 & -1\\\\\r\n1 & 0 & 1\\\\\r\n1 & -1 & 0\r\n\\end{array} \\right] \r\n\\end{align*}\\]\r\n\r\n\r\nOrthogonal matrix\r\nA square matrix is orthogonal if \\(A^{T}A=AA^{T}=I\\), where \\(I\\) is the unit or identity matrix.\r\nOrthogonal matrices can be further classified as rotation or reflection matrices. If the determinant of the orthogonal matrix is \\(+1\\), it is a rotation matrix. If the determinant is \\(-1\\), it is a reflection matrix.\r\nExample \u2013 determining whether matrices are orthogonalDetermine whether matrix \\(B\\) is orthogonal.\r\n\\[B = \\left[ \\begin{array}{cc} 1 & 0\\\\\r\n0 & -1 \r\n\\end{array} \\right] \\]\r\nTo determine whether matrix \\(B\\) is orthogonal, we need to see whether \\(BB^{T}=I\\).\r\n\\[\\begin{align*} B^{T} & = \\left[ \\begin{array}{cc} 1 & 0\\\\\r\n0 & -1 \r\n\\end{array} \\right]\\\\\r\n& = B\r\n\\end{align*}\\] \\[\\begin{align*} BB^{T} & = \\left[ \\begin{array}{cc} 1 & 0\\\\\r\n0 & -1 \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{cc} 1 & 0\\\\\r\n0 & -1 \r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} (1\\times1)+(0\\times0) & (1\\times0)+(0\\times(-1))\\\\\r\n(0\\times1)+((-1)\\times0) & (0\\times0)+((-1)\\times(-1))\r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} 1 & 0\\\\\r\n0 & 1\r\n\\end{array} \\right]\\\\\r\n& = I\r\n\\end{align*}\\]\r\nTherefore, \\(B\\) is both a symmetric and orthogonal matrix.\r\n\r\nExercise \u2013 determining whether matrices are orthogonal\r\nConsider the following matrices.\r\n\r\n\r\n\tDetermine which of the matrices are orthogonal.\r\n\tDetermine which of the matrices is a rotation matrix. Hint: find the determinant.\r\n\r\n\r\n","n":0.042},[{"v":"Matrices","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":198,"$":[{"v":"Inverse of a 3\u00d73 matrix","n":0.447},{"v":"\r\nBeing able to find the inverse of a \\(3\\times3\\) matrix will help to simplify complex problems and enhances your ability to perform matrix operations efficiently. This is crucial in fields like engineering, physics and computer science.\r\n\r\nBefore you read further, make sure that you are familiar with augmented matrices and elementary row operations.\r\n\r\nInverting \\(3\\times3\\) matrices\r\nUnlike \\(2\\times2\\) matrices, finding the inverse of \\(3\\times3\\) matrices is a bit more complicated.\r\nLet's consider matrix \\(A\\):\r\n\\[\\left[ \\begin{array}{ccc} a_{11} & a_{12} & a_{13}\\\\\r\na_{21} & a_{22} & a_{23}\\\\\r\na_{31} & a_{32} & a_{33}\r\n\\end{array} \\right] \\]\r\nThere are a few methods to find the inverse of a \\(3\\times3\\) matrix, but here, we will use the following method:\r\n\r\n\tRewrite the matrix in the form of an augmented matrix with matrix \\(A\\) as the coefficient matrix and the \\(3\\times3\\) identity matrix as the constant matrix: \\(\\left[A|I\\right]\\).\r\n\t\\[\\left[ \\begin{array}{ccc} a_{11} & a_{12} & a_{13}\\\\\r\n\ta_{21} & a_{22} & a_{23}\\\\\r\n\ta_{31} & a_{32} & a_{33}\r\n\t\\end{array}\r\n\t\\left|\r\n\t\\begin{array}{ccc} 1 & 0 & 0 \\\\\r\n\t0 & 1 & 0\\\\\r\n\t0 & 0 & 1\r\n\t\\end{array} \\right. \\right] \\]\r\n\tUse elementary row operations to reduce \\(A\\) to \\(I\\), or in other words, make \\(A\\) look like \\(I\\).\r\n\t\\[\\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\\r\n\t0 & 1 & 0\\\\\r\n\t0 & 0 & 1\r\n\t\\end{array}\r\n\t\\left|\r\n\t\\begin{array}{ccc} b_{11} & b_{12} & b_{13}\\\\\r\n\tb_{21} & b_{22} & b_{23}\\\\\r\n\tb_{31} & b_{32} & b_{33}\r\n\t\\end{array} \\right. \\right] \\]Remember that because it is an augmented matrix, any operations applied to matrix \\(A\\) on the left-hand side must be applied to the identity matrix on the right-hand side.\r\n\tOnce \\(A\\) is transformed into the identity matrix, the final form on the right-hand side is the inverse of \\(A\\).\r\n\t\\[A^{-1} = \\left[ \\begin{array}{ccc} b_{11} & b_{12} & b_{13}\\\\\r\n\tb_{21} & b_{22} & b_{23}\\\\\r\n\tb_{31} & b_{32} & b_{33}\r\n\t\\end{array} \\right] \\]\r\n\t\r\n\r\nA square matrix ONLY has an inverse if its determinant is not equal to \\(0\\).\r\n\r\nElementary row operations\r\nRemember the elementary row operations:\r\n\r\n \tmultiplication of a row by a constant\r\n \taddition and subtraction of a multiple of a row to another row\r\n \tinterchanging two rows.\r\n\r\nWhen using elementary row operations to find the inverse, there is a preferred order to achieve the identity matrix:\r\n\r\n\\[\\begin{align*} \\underbrace{ \\left[ \\begin{array}{ccc} a_{11} & a_{12} & a_{13}\\\\ \r\na_{21} & a_{22} & a_{23}\\\\\r\na_{31} & a_{32} & a_{33} \r\n\\end{array} \\right]}_{\\textrm{Original matrix}} \\rightarrow \\underbrace{ \\left[ \\begin{array}{ccc} 1 & a_{12} & a_{13}\\\\\r\n0 & a_{22} & a_{23}\\\\\r\n0 & a_{32} & a_{33} \r\n\\end{array}\\right]}_{a_{11}=1,\\,a_{21}=a_{23}=0} \\rightarrow \\underbrace{ \\left[ \\begin{array}{ccc} 1 & a_{12} & a_{13}\\\\ \r\n0 & 1 & a_{23}\\\\ \r\n0 & a_{32} & a_{33} \r\n\\end{array} \\right]}_{a_{22}=1}\\\\\r\n\\rightarrow \\underbrace{ \\left[ \\begin{array}{ccc} 1 & 0 & a_{13}\\\\\r\n0 & 1 & a_{23}\\\\\r\n0 & 0 & a_{33} \r\n\\end{array} \\right]}_{a_{12}=a_{32}=0} \\rightarrow \\underbrace{ \\left[ \\begin{array}{ccc} 1 & 0 & a_{13}\\\\\r\n0 & 1 & a_{23}\\\\\r\n0 & 0 & 1 \r\n\\end{array} \\right]}_{a_{31}=1} \\rightarrow \\underbrace{ \\left[ \\begin{array}{ccc} 1 & 0 & 0\\\\ \r\n0 & 1 & 0\\\\ \r\n0 & 0 & 1 \r\n\\end{array} \\right]}_{a_{13}=a_{23}=0} \r\n\\end{align*}\\] If this looks confusing and a bit scary, don\u2019t worry. It's easier to understand how this works with examples containing actual numbers.\r\n\r\nExample 1 \u2013 finding the inverse of \\(3\\times3\\) matrices\r\nFind the inverse of the matrix \\(A = \\left[ \\begin{array}{ccc} 1 & 0 & 1\\\\ \r\n-1 & 2 & 2\\\\ \r\n1 & 1 & 2 \r\n\\end{array} \\right]\\).\r\nFirst, set up the augmented matrix \\(\\left[A|I\\right]\\):\r\n\\[\\left[ A|I \\right] = \\left[ \\begin{array}{ccc} 1 & 0 & 1\\\\\r\n-1 & 2 & 2\\\\ \r\n1 & 1 & 2 \r\n\\end{array} \r\n\\left|\r\n\\begin{array}{ccc} 1 & 0 & 0\\\\\r\n0 & 1 & 0\\\\\r\n0 & 0 & 1 \r\n\\end{array} \\right. \\right] \\]\r\nWe now use elementary row operations to reduce the matrix on the left-hand side (LHS) to the identity matrix.\r\nWe already have a \\(1\\) in the \\(a_{11}\\) position of the LHS matrix, so we first make \\(a_{21}=a_{31}=0\\). We can show this by writing the original augmented matrix, followed by the equations in the middle for the row operations, then the new augmented matrix.\r\n\r\nThe next step is to make \\(a_{22}=1\\) in the LHS matrix. There are two ways to do this. We could divide \\(R_{2}\\) by \\(2\\); however, this introduces a fraction that may be awkward. Fractions are often required, but we will avoid using them until absolutely necessary.\r\nThe other way is to interchange \\(R_{2}\\) and \\(R_{3}\\).\r\n\\[\\begin{align*} \\left[ \\begin{array}{ccc} 1 & 0 & 1\\\\ \r\n0 & 2 & 3\\\\ \r\n0 & 1 & 1 \r\n\\end{array}\r\n\\left|\r\n\\begin{array}{ccc} 1 & 0 & 0\\\\ \r\n1 & 1 & 0\\\\\r\n-1 & 0 & 1 \r\n\\end{array} \\right. \\right] & \\begin{array}{c} \\\\ \r\nR_{2}\\rightarrow R_{3}\\\\\r\nR_{3}\\rightarrow R_{2}\r\n\\end{array}\r\n\\left[ \\begin{array}{ccc} 1 & 0 & 1\\\\ \r\n0 & 1 & 1\\\\\r\n0 & 2 & 3 \r\n\\end{array}\r\n\\left|\r\n\\begin{array}{ccc} 1 & 0 & 0\\\\ \r\n-1 & 0 & 1\\\\\r\n1 & 1 & 0 \r\n\\end{array} \\right. \\right]\r\n\\end{align*}\\]\r\nNext, we make \\(a_{32}=0\\) in the LHS matrix by subtracting \\(2R_{2}\\) from \\(R_{3}\\). By completing this operation, we have also achieved \\(a_{33}=1\\) in the LHS matrix.\r\n\r\nFinally, we make \\(a_{13}=a_{23}=0\\) using \\(R_{3}\\) as follows:\r\n\r\nThe LHS matrix is the identity matrix and so the RHS is the inverse.\r\n\\[A^{-1} = \\left[\\begin{array}{ccc} -2 & -1 & 2\\\\ \r\n-4 & -1 & 3\\\\\r\n3 & 1 & -2 \r\n\\end{array} \\right] \\]\r\nIn general, for matrices \\(A\\) and \\(B\\), \\(AB\\ne BA\\), you should always check that \\(AA^{-1}=A^{-1}A=I\\).\r\n\\[\\begin{align*} AA^{-1} & = \\left[ \\begin{array}{ccc} 1 & 0 & 1\\\\ \r\n-1 & 2 & 2\\\\\r\n1 & 1 & 2\r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{ccc} -2 & -1 & 2\\\\ \r\n-4 & -1 & 3\\\\\r\n3 & 1 & -2 \r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{ccc} -2+3 & -1+1 & 2-2\\\\\r\n2-8+6 & 1-2+2 & -2+6-4\\\\ \r\n-2-4+6 & -1-1+2 & 2+3-4 \r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{ccc} 1 & 0 & 0\\\\ \r\n0 & 1 & 0\\\\ \r\n0 & 0 & 1\r\n\\end{array} \\right]\\\\\r\n& = I\r\n\\end{align*}\\] \\[\\begin{align*} A^{-1}A & = \\left[\\begin{array}{ccc} -2 & -1 & 2\\\\\r\n-4 & -1 & 3\\\\\r\n3 & 1 & -2 \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{ccc} 1 & 0 & 1\\\\\r\n-1 & 2 & 2\\\\ \r\n1 & 1 & 2 \r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{ccc} -2+1+2 & -2+2 & -2-2+4\\\\ \r\n-4+1+3 & -2+3 & -4-2+6\\\\ \r\n3-1-2 & 2-2 & 3+2-4 \r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{ccc} 1 & 0 & 0\\\\ \r\n0 & 1 & 0\\\\ \r\n0 & 0 & 1 \r\n\\end{array} \\right]\\\\\r\n& = I \r\n\\end{align*}\\]\r\nHence, \\(A^{-1} = \\left[ \\begin{array}{ccc} -2 & -1 & 2\\\\\r\n-4 & -1 & 3\\\\\r\n3 & 1 & -2 \r\n\\end{array} \\right] \\)\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the inverse of \\(3\\times3\\) matrices\r\nFind the inverse, if it exists, of the following matrices.\r\n\r\n\t\\(\\left[ \\begin{array}{ccc} 0 & 0 & -1\\\\\r\n\t2 & 1 & 2\\\\\r\n\t5 & 3 & 0\r\n\t\\end{array} \\right]\\)\r\n\t\\(\\left[ \\begin{array}{ccc} 1 & 2 & 0\\\\\r\n\t-1 & 3 & 1\\\\\r\n\t2 & 5 & 2\r\n\t\\end{array} \\right] \\)\r\n\t\\(\\left[ \\begin{array}{ccc} 1 & 2 & 0\\\\\r\n\t1 & 2 & 1\\\\\r\n\t1 & 2 & 4\r\n\t\\end{array} \\right] \\)\r\n\t\\(\\left[ \\begin{array}{ccc} -2 & 2 & 0\\\\\r\n\t1 & 0 & 1\\\\\r\n\t1 & -1 & -2\r\n\t\\end{array} \\right] \\)\r\n\r\n\r\n\r\n","n":0.03},[{"v":"Matrices","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":199,"$":[{"v":"Integration of polynomials","n":0.577},{"v":"\r\nHow do you integrate polynomial functions, where different terms have different powers and coefficients? This skill is important for calculating projectile trajectories in physics, beam deflections in engineering and determining cost functions and profit maximisation in economics. Use this resource to learn how.\r\n\r\nIn Antidifferentiation, you learned how to find the antiderivative of functions consisting of single algebraic terms. Here, we will look at polynomials which have two or more terms.\r\nPolynomial functions have the general form:\r\n\\[ f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\\cdots+a_{n}x^{n} \\]\r\nwhere \\(a_{i},\\,i=1,2,3,\\ldots,n\\) are constants.\r\n\r\nLinearity rules\r\nThere are two properties that make it easier for us to find the integrals of more complex functions. They are called the linearity rules.\r\nA constant that multiplies a function can be taken out of the integral. For a constant \\(a\\):\r\n\\[\\int af(x)dx =  a\\int f(x)dx\\]\r\nThe other rule states that when we differentiate the sum or difference of two functions \\(f(x)\\) and \\(g(x)\\), we can differentiate each function separately then add or substract the results.\r\n\\[\\int (f(x)+g(x))dx = \\int f(x)dx + \\int g(x)dx\\] \\[\\int (f(x)-g(x))dx = \\int f(x)dx - \\int g(x)dx\\]\r\nYou can have any number of terms in the integral. For example, if \\(a_{1},a_{2},a_{3}\\ldots a_{n}\\) are constant:\r\n\r\n\r\nExample 1 \u2013 using the linearity rules\r\nFind \\(\\int(5x^{2}-3x+4)dx\\).\r\nWe can use the linearity rules to separate these into two separate integrals.\r\n\\[\\int(5x^{2}-3x+4)dx = \\int5x^{2}dx+\\int(-3x)dx+\\int4dx\\]\r\nWe can now use the linearity rules to take the constants out of the integrals.\r\n\\[= 5\\int x^{2}dx - 3\\int xdx+4\\int dx\\]\r\nNow, we are ready to antidifferentiate.\r\n\\[= \\frac{5}{3}x^{3}-\\frac{3}{2}x^{2}+4x+c\\]\r\nThere is a constant for each of the integrals on the right hand side. However, a constant plus a constant still gives you a constant so we just use a single constant \\(c\\in\\mathbb{R}\\). \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using the linearity rules\r\nIntegrate the following with respect to \\(x\\).\r\n\r\n\t\\(x^{3}-2x\\)\r\n\t\\(2x^{4}+5x^{2}\\)\r\n\t\\(3x^{2}-7x^{5}\\)\r\n\t\\(x^{2}-2+x\\)\r\n\t\\(2x+2\\)\r\n\t\\(-6x^{3}+5x+2\\)\r\n\t\\(9x^{6}-3x-4\\)\r\n\r\n\r\n\r\n\r\nDefinite integrals\r\nSo far, we have looked at functions of \\(x\\) where the antiderivative involved a constant \\(c\\). These integrals are called indefinite integrals. They do not have a specific value.\r\nIf we define a definite integral for a function \\(f(x)\\), from \\(x=a\\) to \\(x=b\\) with respect to \\(x\\), we use the following notation:\r\n\\[ \\int_{a}^{b}f(x)dx=\\left[F(x)\\right]_{a}^{b}=F(b)-F(a) \\]\r\nwhere we assume \\(f(x)\\) is defined for all \\(x\\) in the interval \\([a,b]\\). \\(a\\) the lower limit of integration and \\(b\\) the upper limit of integration. \\([a,b]\\) is called the limits of integration.\r\nThe notation \\(\\left[F(x)\\right]_{a}^{b}\\) means to substitute \\(x=b\\) and \\(x=a\\) into \\(F(x)\\) and then subtract the values. This gets rid of the constant \\(c\\).\r\nTo find the definite integral:\r\n\r\n\tFind the antiderivative or indefinite integral \\(\\int xdx+c\\).\r\n\tSubstitute \\(x=b\\) into the integral, i.e. evaluate this at the upper limit of integration.\r\n\tSubstitute \\(x=a\\) into the integral, i.e. evaluate this at the lower limit of integration.\r\n\tMinus the integral at the lower limit from the integral at the upper limit. This lets us cancel out the constant \\(c\\).\r\n\r\nThis is known as the first fundamental theorem of calculus. Let's look at some examples of how this is applied.\r\n\r\nExample 1 \u2013 finding definite integrals\r\nFind \\(\\intop_{1}^{3}xdx\\).\r\nFirst, we find an antiderivative or indefinite integral \\(\\int xdx+c\\).\r\nAn antiderivative is:\r\n\\[\\frac{1}{2}x^{2}+c\\]\r\nNow, we can evaluate this at the limits of integration.\r\n\\[\\begin{align*} \\int_{1}^{3}xdx & = \\left[\\frac{1}{2}x^{2}+c\\right]_{x=1}^{x=3},\\,c\\in\\mathbb{R}\\\\ \r\n& = \\left(\\frac{1}{2}3^{2}+c\\right)-\\left(\\frac{1}{2}1^{2}+c\\right)\\\\ \r\n& = \\left(\\frac{9}{2}+c\\right)-\\left(\\frac{1}{2}+c\\right)\\\\ \r\n& = \\frac{8}{2}\\\\ \r\n& = 4\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding definite integrals\r\nEvaluate the following indefinite integrals:\r\n\r\n\t\\(\\intop_{1}^{2}(6x-2)dx\\)\r\n\t\\(\\intop_{2}^{3}(8x^{3}-9x^{2}-2x)dx\\)\r\n\t\\(\\intop_{0}^{2}(7x^{2}+2)dx\\)\r\n\t\\(\\intop_{-1}^{2}(7x^{3}-6x+3)dx\\)\r\n\t\\(\\intop_{-1}^{1}(8x^{3}-x^{2}+6)dx\\)\r\n\t\\(\\intop_{-1}^{0}(8x^{3}-x^{2}+6)dx\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.044},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":200,"$":[{"v":"Integration of reciprocal functions","n":0.5},{"v":"\r\nHow do you integrate a reciprocal function? This skill is important for calculating electrical currents in circuits and modelling harmonic motion in mechanical systems. Use this resource to learn how.\r\n\r\nReciprocal functions have the form:\r\n\\[y=\\frac{k}{x}\\]\r\nwhere \\(k\\) is a constant. We often deal with reciprocal functions when looking at Inverse functions. Examples include \\(y=\\dfrac{1}{x}\\) and \\(f(x)=\\dfrac{1}{3x-1}\\).\r\n\r\nThe power rule\r\nOne of the most important rules for integration is the power rule. It states that:\r\n\\[ \\int x^{n}dx=\\frac{1}{n+1}x^{n+1}+c,\\,n\\neq-1 \\] \r\nwhere \\(c\\) is a constant.\r\nThe reason that \\(n\\neq-1\\) is that if \\(n=-1\\), \\(\\dfrac{1}{n+1}\\) would be \\(\\dfrac{1}{0}\\), which is undefined. \\(n\\) can be any number (other than \\(-1\\)) \u2014 positive, negative or a fraction. Often, instead of saying \\(c\\) is a constant, we write \\(c\\in\\mathbb{R}\\) which means \\(c\\) is a real number.\r\nWhen we encounter a case when \\(n=-1\\), such as integrating \\(\\int\\left(\\dfrac{1}{x}\\right)dx\\), we define a new function:\r\n\\[ \\int\\frac{1}{x}dx=\\log_{e}\\left|x\\right|+c\\] \r\nwhere \\(c\\) is a constant. \r\nThe function \\(\\log_{e}\\left|x\\right|\\) is called the natural logarithm. It is sometimes written as \\(\\ln\\left|x\\right|\\). The absolute value sign is used as the natural logarithm is not defined for negative arguments.\r\nWe can now integrate \\(f(x)=\\dfrac{m}{x}\\) where \\(m\\) is any constant:\r\n\\[\\begin{align*} \\int\\frac{m}{x}dx & = m\\int\\frac{1}{x}dx\\\\\r\n& = m\\log_{e}\\left|x\\right|+c \\end{align*}\\]\r\nThere is a more general rule:\r\n \\[ \\int\\frac{m}{ax+b}dx=\\frac{m}{a}\\log_{e}\\left|ax+b\\right|+c \\]\r\nwhere \\(a\\), \\(b\\) and \\(m\\) be constants with \\(a\\neq0\\).\r\n\r\nExample 1 \u2013 integrating reciprocal functions\r\nIntegrate \\(f(x)=\\dfrac{1}{2x}\\) with respect to \\(x\\).\r\n\r\nHere, \\(m=1\\), \\(a=2\\) and \\(b=0\\).\r\n\\[\\int\\frac{1}{2x}dx=\\frac{1}{2}\\log_{e}\\left|2x\\right|+c,\\, c\\in\\mathbb{R}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 integrating reciprocal functions\r\n\r\n\tCalculate the following integrals.\r\n\t\r\n\t\t\\(\\int\\dfrac{1}{5x}dx\\)\r\n\t\t\\(\\int\\dfrac{3}{2x}dx\\)\r\n\t\t\\(\\int\\dfrac{3}{3x-5}dx\\)\r\n\t\t\\(\\int\\dfrac{6}{2-7x}dx\\)\r\n\t\r\n\tIntegrate the following functions with respect to \\(x\\).\r\n\t\r\n\t\t\\(f(x)=x^{2}-\\dfrac{2}{x}\\)\r\n\t\t\\(f(x)=\\dfrac{1}{3x-2}+\\dfrac{3}{1-x}\\)\r\n\t\t\\(f(x)=\\dfrac{3}{3x-7}-x^{2}\\)\r\n\t\t\\(f(x)=\\dfrac{2}{5x-4}-\\dfrac{3}{2-5x}\\)\r\n\t\r\n\tFind the following intervals.\r\n\t\r\n\t\t\\(\\int_{1}^{5}\\dfrac{2}{3x}dx\\)\r\n\t\t\\(\\int_{1}^{2}\\dfrac{2}{3x-1}dx\\)\r\n\t\t\\(\\int_{1}^{3}\\left(x^{2}-\\dfrac{2}{x}\\right)dx\\)\r\n\t\t\\(\\int_{2}^{4}\\left(\\dfrac{2}{5x-4}-\\dfrac{3}{2-5x}\\right)dx\\)\r\n\t\r\n\r\nHint: \\(\\log_{e}(1)=0\\).\r\n\r\n\r\n\r\n\r\n\r\n","n":0.065},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":201,"$":[{"v":"Area under a curve","n":0.5},{"v":"\r\nIntegrals can be used to find the area between a curve and the \\(x\\)-axis. We can use this to calculate the total accumulation of a quantity in physics, economics and engineering. Use this resource to learn how to calculate area under the curve.\r\n\r\nFinding the area under a curve\r\nConsider a curve defined by \\(y=f(x)\\). We can find the area under the curve between \\(x=a\\) and \\(x=b\\) using a definite integral:\r\n\\[ \\int_{a}^{b}f(x)dx \\]\r\nThis is the area bordered by the lines \\(x=a\\), \\(x=b\\), \\(y=0\\) (the \\(x\\)-axis) and the graph of \\(f(x)\\).\r\n\r\nThe area under a curve might be above the axis (and therefore, positive) or below the axis (and therefore, negative).\r\n\r\n\tIf \\(f(x)\\geq0\\) between \\(x=a\\) and \\(x=b\\), you would get a positive value for the definite integral. \r\n\t\r\n\tIf, \\(f(x)\\leq0\\) between \\(x=a\\) and \\(x=b\\), you would get a negative value for the definite integral.\r\n\t\r\n\r\nRegardless of whether the area is above or below the curve, the area under a curve is always given as a positive number.\r\n\\[ \\textrm{Area }=\\left|\\int_{a}^{b}f(x)dx\\right|\\]\r\nWhere \\(f(x)\\) crosses the \\(x\\)-axis between \\(a\\) and \\(b\\), the value of the definite integral may be positive or negative. However, the area under the curve is always taken as the sum of the modulus of the signed areas. \r\nFor example, the function \\(f(x)=x^{3}-x^{2}-2x\\) is plotted. The graph cuts the \\(x\\)-axis at \\(-1\\), \\(0\\) and \\(2\\). The integral for \\(-1\\geq x\\geq0\\) is positive and the integral for \\(0\\geq x\\geq2\\) is negative. The overall integral from \\(-1\\geq x\\geq2\\) is \\(-2.25\\), but the area under the curve is positive. See Example 1 for how to work this out.\r\n\r\n\r\nExample 1 \u2013 finding the area under the curve\r\nGiven the function \\(f(x)=x^{3}-x^{2}-2x\\), find the area under the curve between \\(x=-1\\) and \\(x=2\\).\r\nTo calculate the area, we need to evaluate two integrals. One from \\(-1\\) to \\(0\\) the other from \\(0\\) to \\(2\\). The actual area would then be the sum of the absolute value of these integrals. That is:\r\n\\[\\begin{align*} \\textrm{Area} & = \\left|\\int_{-1}^{0}\\left(x^{3}-x^{2}-2x\\right)dx\\right|+\\left|\\int_{0}^{2}\\left(x^{3}-x^{2}-2x\\right)dx\\right|\\\\\r\n& = \\left|\\left[\\frac{1}{4}x^{4}-\\frac{1}{3}x^{3}-x^{2}\\right]_{x=-1}^{x=0}\\right|+\\left|\\left[\\frac{1}{4}x^{4}-\\frac{1}{3}x^{3}-x^{2}\\right]_{x=0}^{x=2}\\right|\\\\\r\n& = \\left|0-\\left(\\frac{1}{4}\\left(-1\\right)^{4}-\\frac{1}{3}\\left(-1\\right)^{3}-\\left(-1\\right)^{2}\\right)\\right|+\\left|\\frac{1}{4}\\left(2\\right)^{4}-\\frac{1}{3}\\left(2\\right)^{3}-2^{2}\\right|\\\\\r\n& = \\left|-\\left(\\frac{1}{4}+\\frac{1}{3}-1\\right)\\right|+\\left|4-\\frac{8}{3}-4\\right|\\\\ \r\n& = \\left|\\frac{1}{4}+\\frac{1}{3}-1\\right|+\\frac{8}{3}\\\\\r\n& = \\frac{5}{12}+\\frac{8}{3}\\\\ \r\n& = \\frac{37}{12}\\textrm{ square units}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the area under the curve\r\n\r\n \tFind the area enclosed by the curve \\(y=6x-x^{2}\\) and the \\(x\\)-axis.\r\n \tFind the area bounded by the curve \\(y=\\sin(x)\\), the \\(x\\)\u2013 axis and the line \\(x=\\dfrac{\\pi}{2}\\).\r\n \tFind the area bounded by the curve \\(y=x^{2}-3x+2\\) and the \\(x\\)-axis between \\(x=0\\) and \\(x=2\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.051},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":202,"$":[{"v":"Integration by substitution","n":0.577},{"v":"\r\nIntegration by substitution simplifies the process of finding integrals by making a change of variables.  This is used for calculating areas and volumes in engineering, solving differential equations in physics, and optimising functions in economics. Use this resource to learn how to integrate by substitution.\r\n\r\nAn expression composed of two functions can be complicated to integrate. You can simplify this by substituting a single pronumeral (like \\(u\\)) to represent one of the functions. By substituting variables, we can transform the integral into a simpler form. \r\n\r\nThe substitution rule\r\nThe substitution rule for integration is like the chain rule for differentiation: it breaks a complex expression into manageable parts.\r\nThe basic rule is that if \\(u=g(x)\\) and \\(\\dfrac{du}{dx}=g'(x)\\), then:\r\n\\[ \\int f(g(x))g'(x)dx=\\int f(u)du \\]\r\nThis may also be written: \r\n\\[\\int f(u)\\frac{du}{dx}dx = \\int f(u)du\\]\r\n\r\nUsing the substitution rule\r\nA simple test may be used to see if substitution is useful. \r\nLook at the integrand (the function being integrated) and see if one part of it is the derivative of another part multiplied by a constant. If so, substitution may be helpful.\r\nFor example, consider \\(\\int2x\\cos(x^{2})dx\\). The derivative of \\(x^{2}\\) is \\(2x\\), and so one part is the derivative of another part and substitution can be used. \r\nAnother example is the integral \\(\\int x\\cos(x^{2})dx\\). The derivative of \\(x^{2}\\) is \\(2x\\). Since \\(x\\) is \\(\\dfrac{1}{2}\\) of the derivative of \\(x^{2}\\), substitution is possible.\r\n\r\nExample 1 \u2013 using the substitution rule\r\nCalculate \\(\\int2x\\cos(x^{2})dx\\).\r\nHere, we use the second form of the substitution rule.\r\n\\[ \\int f(u)\\frac{du}{dx}dx=\\int f(u)du \\]\r\nWe let \\(u=x^{2}\\) since the derivative is \\(2x\\), a part of the integrand. So, \\(\\cos(x^{2})=\\cos(u)=f(u)\\) and \\(\\dfrac{du}{dx}=2x\\). Substituting these in the integral gives:\r\n\\[\\begin{align*} \\int2x\\cos(x^{2})dx & = \\int\\frac{du}{dx}\\cos(u)dx\\\\ \r\n& = \\int f(u)\\frac{du}{dx}dx\\\\ \r\n& = \\int f(u)du\\\\\r\n& = \\int\\cos(u)du\\\\ \r\n& = \\sin(u)+c,\\,c\\in\\mathbb{R}\\\\ \r\n& = \\sin(x^{2})+c\r\n\\end{align*}\\]\r\nIn some other resources, you may encounter slightly different notation. For this example, \\(\\dfrac{du}{dx}=2x\\) may be rearranged to give \\(dx=\\dfrac{1}{2x}du\\). Then substituting for \\(x^{2}\\) and \\(dx\\) in the integral, you would get: \r\n\\[\\begin{align*} \\int2x\\cos(x^{2})dx & = \\int2x\\cos(u)\\frac{1}{2x}du\\\\ \r\n& =\\int\\cos(u)du\\\\ \r\n& =\\sin(u)+c,\\,c\\in\\mathbb{R}\\\\ \r\n& =\\sin(x^{2})+c\r\n\\end{align*}\\]\r\nThis method gives the same result and will always work. However, you should realise that \\(\\dfrac{du}{dx}\\) is not a fraction.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nSubstitution with definite integrals\r\nWe can also use the method of substitution to evaluate definite integrals, but we need to be careful with the limits of integration when we change the variable of integration.\r\nFor example, suppose we want to evaluate \\(\\int_{0}^{1}(2x+1)^{5}dx\\). We let \\(u=2x+1\\), which means \\(\\dfrac{du}{dx}=2\\) and so \\(1=\\dfrac{1}{2}\\dfrac{du}{dx}\\).\r\nNote that when \\(x=0\\), \\(u=1\\) and when \\(x=1\\), \\(u=3.\\)\r\nSubstituting into the integral, we get:\r\n\\[\\begin{align*} \\int_{0}^{1}(2x+1)^{5}dx & =\\int_{0}^{1}1\\times(2x+1)^{5}dx\\\\\r\n& =\\int_{0}^{1}\\frac{1}{2}\\frac{du}{dx}\\times u^{5}dx\\\\ \r\n& =\\frac{1}{2}\\int_{1}^{3}u^{5}du\\\\ \r\n& =\\frac{1}{2}\\frac{1}{6}\\left[u^{6}\\right]_{u=1}^{u=3}\\\\ \r\n& =\\frac{1}{12}(3^{6}-1^{6})\\\\ \r\n& =\\frac{1}{12}(729-1)\\\\ \r\n& =\\frac{728}{12}\\\\\r\n& =\\frac{182}{3}\r\n\\end{align*}\\]\r\nIn the third line of working, the limits have changed since we are integrating with respect to \\(u\\) instead of with respect to \\(x\\). This is a critical step! If you miss it, you will not get the correct answer.\r\n\r\nExercise \u2013 integrating by substitution\r\nFind the following:\r\n\r\n \t\\(\\int(5x+1)^{4}dx\\)\r\n \t\\(\\int3x^{2}(x^{3}+2)^{5}dx\\)\r\n \t\\(\\int(2x+1)(x^{2}+x+3)^{3}dx\\)\r\n \t\\(\\int6x^{2}\\sqrt{(x^{3}+3)}dx\\)\r\n \t\\(\\int\\sin(3x+2)dx\\)\r\n \t\\(\\int6x^{2}e^{x^{3}}dx\\)\r\n \t\\(\\int\\dfrac{2x}{2+x^{2}}dx\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.046},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":203,"$":[{"v":"Integration using partial fractions","n":0.5},{"v":"\r\nHow do you integrate an expression when there is an algebraic expression in the numerator and denominator of a fraction? Integrating using partial fractions helps you to solve this problem. Use this resource to learn how to integrate using partial fractions.\r\n\r\nSometimes a complex function may be integrated by breaking it up into partial fractions. The idea is that each partial fraction is an easier integral than the original. \r\nFor example:\r\n\\[\\int\\frac{1}{x^{2}+5x+6}dx=\\int\\frac{1}{x+2}dx-\\int\\frac{1}{x+3}dx\\]\r\nThe integrals on the right are much simpler than that on the left. \r\nBefore you dive into this resource, make sure you are confident with finding partial fractions.\r\n\r\nExample 1 \u2013 integrating using partial fractions\r\nFind the integral of \\(\\dfrac{x-5}{x^{2}+2x-3}\\) with respect to \\(x\\).\r\nFirst, find the partial fractions. Since \\(x^{2}+2x-3=(x+3)(x-1)\\), we are dealing with the case where the denominator has two distinct linear factors. Therefore, we factorise as follows:\r\n\\[\\begin{align*} \\frac{x-5}{x^{2}+2x-3} & = \\frac{x-5}{(x+3)(x-1)}\\\\\r\n& = \\frac{A}{x+3}+\\frac{B}{x-1}\r\n\\end{align*}\\]\r\nNow, we have to determine the constants \\(A\\) and \\(B\\). First, add the partial fractions by finding a common denominator. In this case, it is \\((x+3)(x-1)\\). \r\n\\[\\begin{align*} \\frac{A}{x+3}+\\frac{B}{x-1} & = \\frac{A}{(x+3)}\\times\\frac{(x-1)}{(x-1)}+\\frac{B}{(x-1)}\\times\\frac{(x+3)}{(x+3)}\\\\ \r\n& =\\frac{A(x-1)+B(x+3)}{(x+3)(x-1)}\\\\ \r\n& =\\frac{Ax-A+Bx+3B}{(x+3)(x-1)}\\\\ \r\n& =\\frac{Ax+Bx-A+3B}{x^{2}+2x-3}\\\\\r\n& =\\frac{(A+B)x-A+3B}{x^{2}+2x-3}\r\n\\end{align*}\\]\r\nEquating the numerators:\r\n\\[x-5=(A+B)x-A+3B\\]\r\nwhere \\(A+B\\) is the \\(x\\) term and \\(-A+3B\\) is the constant. On the left, the coefficient of \\(x\\) is \\(1\\) and on the right, the coefficient is \\(A+B\\). Similarly, the constant term on the left is \\(-5\\) and on the right, it is \\(-A+3B\\). \r\nThis gives us the following simultaneous equations:\r\n\\[\\begin{align*} A+B & = 1 \\\\ \r\n-A+3B & = -5 \r\n\\end{align*}\\]\r\nThese equations are easily solved by elimination. Adding the equations, we get:\r\n\\[\\begin{align*} 4B & = -4\\\\\r\nB & = -1\r\n\\end{align*}\\]\r\nSubstituting for \\(B\\) the first equation gives:\r\n\\[\\begin{align*} A-1 & = 1\\\\ \r\nA & = 2\r\n\\end{align*}\\]\r\nFinally, we have:\r\n\\[\\frac{x-5}{x^{2}+2x-3}=\\frac{2}{x+3}-\\frac{1}{x-1}\\] \r\nWe can now evaluate the integral:\r\n\\[\\begin{align*} \\int\\frac{x-5}{x^{2}+2x-3}dx & = \\int\\left(\\frac{2}{x+3}-\\frac{1}{x-1}\\right)dx\\\\\r\n& = \\int\\frac{2}{x+3}dx-\\int\\frac{1}{x-1}dx\\\\\r\n& = 2\\ln|x+3|-\\ln|x-1|+c,\\,c\\in\\mathbb{R}\\\\\r\n& = \\ln(x+3)^{2}-\\ln|x-1|+c\\\\ \r\n& = \\ln\\frac{(x+3)^{2}}{|x-1|}+c\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExercise \u2013 integrating using partial fractions\r\nPerform the following integrations.\r\n\r\n \t\\(\\int\\dfrac{4x+9}{x^{2}+x-12}dx\\)\r\n\t\\(\\int\\dfrac{21-8x}{x^{2}-x-6}dx\\)\r\n\t\\(\\int\\dfrac{5x^{2}-5x+2}{(x+1)(x-1)^{2}}dx\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.057},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":204,"$":[{"v":"Integration by parts","n":0.577},{"v":"\r\nIntegration by parts is used to solve integrals by expressing them in terms of simpler integrals. This is applied in areas like mechanical engineering for calculating work done by variable forces, in physics for solving complex integrals, and in economics for evaluating growth models. Use this resource to learn how to integrate using integration by parts.\r\n\r\nIf your expression consists of two functions multiplied together, you can integrate this by parts. This is like the product rule in differentiation and can be used on logarithmic, exponential, trigonometric and algebraic functions.\r\nIntegration by parts is particularly useful for integrating products of functions, such as when one function is easily differentiable and the other is easily integrable.\r\n\r\nThe product rule\r\nThe formula for integration by parts uses the product rule for differentiation. For any two differentiable functions \\(v(x)\\) and \\(u(x)\\), we have:\r\n\\[\\frac{d}{dx}(v(x)u(x))=v(x)\\frac{du(x)}{dx}+u(x)\\frac{dv(x)}{dx}\\]\r\nIntegrating both sides with respect to \\(x\\) gives:\r\n\\[\\begin{align*} \\int\\frac{d}{dx}(v(x)u(x))dx & = \\int\\left(v(x)\\frac{du(x)}{dx}+u(x)\\frac{dv(x)}{dx}\\right)dx\\\\ \r\nv(x)u(x) & = \\int v(x)\\frac{du(x)}{dx}dx+\\int u(x)\\frac{dv(x)}{dx}dx\r\n\\end{align*}\\]\r\nThis can be rearranged to give the product rule for integration:\r\n\\[\\int v(x)\\frac{du(x)}{dx}dx = v(x)u(x)-\\int u(x)\\frac{dv(x)}{dx}dx\\]\r\nThis is often abbreviated to \\(\\int vu'dx = vu-\\int uv'dx\\).\r\n\r\nUsing the product rule\r\nTo help you choose \\(v\\) and \\(u'\\), choose \\(v\\) to be the function that comes first in the following sequence of functions:\r\n\r\n \tlogarithmic function\r\n \tinverse trig function\r\n \talgebraic function\r\n \ttrigonometric function\r\n \texponential function.\r\n\r\nA common mnemonic to help you remember this is LIATE.\r\n\\(u'\\) is then the other function in the integrand. For example, the integrand in \\(\\int x\\sin(x)dx\\) is \\(x\\sin(x)\\). In this case, it is the product of an algebraic function \\(x\\) and a trigonometric function \\(\\sin(x)\\) so we would set \\(v=x\\), because algebraic functions are higher on the list than trigonometric functions. Then, \\(u'=\\sin(x)\\).\r\nThe hope is that by choosing functions \\(v\\) and \\(u'\\) carefully, the integral on the right-hand side is easier than the original integral. If the integral on the right is harder,  try swapping \\(v\\) and \\(u'\\).\r\n\r\nExample 1 \u2013 using the product rule\r\nFind \\(\\int x\\cos(x)dx\\).\r\nHere, we have an algebraic function \\(x\\) and a trigonometric function \\(\\cos(x)\\). Since algebraic functions are higher on the list than trigonometric functions, we let \\(v=x\\) and so \\(u'=\\dfrac{du}{dx}=\\cos(x)\\). \r\nThen, \\(v'=\\dfrac{dv}{dx}=1\\) and \\(u=\\sin(x)\\).\r\nUsing the formula, we have:\r\n\\[\\begin{align*} \\int x\\cos(x)dx & = vu-\\int uv'dx\\\\\r\n& = x\\sin(x)-\\int(\\sin(x))\\times1dx\\\\ \r\n& = x\\sin(x)+\\cos(x)+c,\\,c\\in\\mathbb{R}\r\n\\end{align*}\\]\r\nIf we chose the functions as \\(v=\\cos(x)\\) and \\(u'=x\\), then \\(v'=-\\sin(x)\\) and \\(u=\\dfrac{1}{2}x^{2}\\). Substituting in the formula will give:\r\n\\[\\begin{align*} \\int x\\cos(x)dx & =vu-\\int uv'dx\\\\\r\n& = -\\frac{1}{2}x^{2}\\sin(x)-\\int-\\frac{1}{2}x^{2}\\sin(x)dx\r\n\\end{align*}\\]\r\nThe integral on the right is harder than the one we began with and we make no progress. If this happens to you, change the selection.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nRepeated use of the product rule\r\nFor some integrals, you will need to use the integration by parts technique more than once. For example, consider \\(\\int x^{2}e^{x}dx\\).\r\nWe have an algebraic function \\(x^{2}\\) multiplying an exponential function \\(e^{x}\\). According to our guide, we set \\(v(x)=x^{2}\\) and \\(u'(x)=e^{x}\\). Then, \\(\\dfrac{dv}{dx}=2x\\) and \\(u(x)=e^{x}\\).\r\nSubstituting in the integration by parts formula, we get: \r\n\\[\\int x^{2}e^{x}dx = x^{2}e^{x}-\\int e^{x}(2x)dx\\]\r\nThe integral on the right-hand side requires another integration by parts. We set \\(v(x)=2x\\) and \\(u'(x)=e^{x}\\). Then \\(\\dfrac{dv}{dx}=2\\) and \\(u(x)=e^{x}\\).\r\nSubstituting in the integration by parts formula gives:\r\n\\[\\begin{align*} \\int e^{x}(2x)dx & = 2xe^{x}-\\int2e^{x}dx\\\\ \r\n& = 2xe^{x}-2e^{x}+c,\\,c\\in\\mathbb{R}\r\n\\end{align*}\\]\r\nNow, we substitute this result into the integration by parts formula to get the result:\r\n\\[\\begin{align*} \\int x^{2}e^{x}dx & = x^{2}e^{x}-\\int e^{x}(2x)dx\\\\ \r\n& = x^{2}e^{x}-(2xe^{x}-2e^{x}+c)\\\\\r\n& = x^{2}e^{x}-2xe^{x}+2e^{x}-c\\\\ \r\n& = e^{x}(x^{2}-2x+2)-c\r\n\\end{align*}\\]\r\nIt doesn\u2019t matter if we put \\(c\\) or \\(-c\\) as \\(c\\) is just any constant.\r\n\r\nExercise \u2013 integrating by parts\r\n\r\n\tFind the following.\r\n\t\r\n\t\t\\(\\int x\\sin(x)dx\\)\r\n\t\t\\(\\int xe^{x}dx\\)\r\n\t\t\\(\\int \\cos(4x)dx\\)\r\n\t\t\\(\\int x^{4}\\ln(x)dx\\)\r\n\t\r\n\tFind the following.\r\n\t\r\n\t\t\\(\\int x^{2}\\cos(x)dx\\)\r\n\t\t\\(\\int x^{3}e^{x^{2}}dx\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.042},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":205,"$":[{"v":"Double integrals","n":0.707},{"v":"What if you have an area that is bounded by limits on both the \\(x\\) and the \\(y\\)-axes? This is where double integrals come into play. We use double integrals in physics for finding mass in a region given a density function, engineering for determining properties of materials with varying thickness, and economics for analysing changes in multivariable contexts.\r\n\r\nA double integral is of the form:\r\n\\[\\underset{R}{\\iint}f(x,y)dA=\\underset{R}{\\iint}f(x,y)dxdy=\\underset{R}{\\iint}f(x,y)dydx\\]\r\nwhere:\r\n\r\n \t\\(z=f(x,y)\\) is a continuous function on the region \\(R\\)\r\n \t\\(R\\) is a region of integration in the \\(x\\)-\\(y\\) plane\r\n \t\\(dA\\) is an infinitesimal element of area in \\(R\\) and \\(dA=dxdy=dydx\\).\r\n\r\nIf \\(f(x,y)\\geq0\\), then the double integral is equal to the volume below the surface \\(f(x,y)\\) and above the region \\(R\\).\r\n\r\nWe can evaluate a double integral in two ways:\r\n\r\n \tIntegrating first with respect to \\(x\\), keeping \\(y\\) constant, with \\(a\\leq x\\leq b\\) and \\(c\\leq y\\leq d\\):\r\n\t\\[\\underset{R}{\\iint}f(x,y)dA=\\int_{c}^{d}\\left[\\int_{a}^{b}f(x,y)dx\\right]dy\\]\r\n\tThe \"inner\" integral is the integral with respect to \\(x\\) and the \"outer\" integral is the integral with respect to \\(y\\).\r\n \tIntegrating first with respect to \\(y\\), keeping \\(x\\) constant, with \\(a\\leq y\\leq b\\) and \\(c\\leq x\\leq d\\):\r\n\t\\[\\underset{R}{\\iint}f(x,y)dA=\\int_{c}^{d}\\left[\\int_{a}^{b}f(x,y)dy\\right]dx\\]\r\n\tThe \"inner\" integral is the integral with respect to \\(y\\) and the \"outer\" integral is the integral with respect to \\(x\\).\r\n\r\nIn practice, one may be easier than the other.\r\n\r\nEvaluating double integrals\r\nTo evaluate a double integral, it is helpful to:\r\n\r\n \tSketch the region of integration.\r\n \tDetermine the limits of integration.\r\n \tEvaluate the inner integral.\r\n \tEvaluate the outer integral.\r\n\r\nThe most straightforward example is when the region \\(R\\) is a rectangle. This is covered in Example 1. \r\nIn many cases, the region of integration is not bounded by limits of integration that are constants; sometimes, they are functions. It is important to be able to sketch the region to identify the limits of integration. See Example 2 and 3 for how to do this.\r\n\r\nExample 1 \u2013 evaluating double integrals\r\nEvaluate \\(\\underset{R}{\\iint}(3x^{2}-2xy)dxdy\\), where \\(R\\) is the rectangle with vertices \\((2,1),\\) \\((2,3),\\) \\((4,1),\\) \\((4,3)\\).\r\nFirst, sketch the region.\r\n\r\nLet's look at both ways of integrating. In this example, you could reverse the order of integration and still obtain the same answer.\r\n\r\nIntegrating with respect to \\(x\\) then \\(y\\)\r\nThe \"inner\" integral is the integral with respect to \\(x\\) and the \"outer\" integral is the integral with respect to \\(y\\). \r\n\r\n\tThe limits of integration for the inner integral are \\(x=2\\) to \\(x=4\\).\r\n\tThe limits of integration for the outer integral are \\(y=1\\) to \\(y=3\\).\r\n\r\nThe integral becomes:\r\n\\[\\int_{1}^{3}\\left[\\int_{2}^{4}(3x^{2}-2xy)dx\\right]dy\\]\r\nFirst, we work out the inner integral, integrating with respect to \\(x\\) while treating \\(y\\) as a constant. Then, we integrate with respect to \\(y\\) and evaluate.\r\n\\[\\begin{align*} \\int_{1}^{3}\\left[\\int_{2}^{4}(3x^{2}-2xy)dx\\right]dy & = \\int_{1}^{3}\\left[\\left(\\frac{3x^{3}}{3}-\\frac{2x^{2}y}{2}\\right)_{2}^{4}\\right]dy\\\\\r\n& = \\int_{1}^{3}\\left[\\left(x^{3}-x^{2}y\\right)_{2}^{4}\\right]dy\\\\ \r\n& = \\int_{1}^{3}\\left[\\left(4^{3}-4^{2}y\\right)-\\left(2^{3}-2^{2}y\\right)\\right]dy\\\\ \r\n& = \\int_{1}^{3}\\left[\\left(64-16y\\right)-\\left(8-4y\\right)\\right]dy\\\\\r\n& = \\int_{1}^{3}\\left[56-12y\\right]dy\\\\ \r\n& = \\left[56y-\\frac{12y^{2}}{2}\\right]_{1}^{3}\\\\\r\n& = \\left[56y-6y^{2}\\right]_{1}^{3}\\\\ \r\n& = \\left(56(3)-6(3)^{2}\\right)-\\left(56(1)-6(1)^{2}\\right)\\\\ \r\n& = 168-54-56+6\\\\\r\n& = 64 \r\n\\end{align*}\\]\r\n\r\nIntegrating with respect to \\(y\\) then \\(x\\)\r\nIn this case, the integral becomes:\r\n\\[\\int_{2}^{4}\\left[\\int_{1}^{3}(3x^{2}-2xy)dy\\right]dx\\]\r\nFirst, we work out the inner integral, integrating with respect to \\(y\\) while treating \\(x\\) as a constant. Then, we integrate with respect to \\(x\\) and evaluate.\r\n\\[\\begin{align*} \\int_{2}^{4}\\left[\\int_{1}^{3}(3x^{2}-2xy)dy\\right]dx & = \\int_{2}^{4}\\left[\\left(3x^{2}y-2x\\frac{y^{2}}{2}\\right)_{1}^{3}\\right]dx\\\\\r\n& = \\int_{2}^{4}\\left[\\left(3x^{2}y-xy^{2}\\right)_{1}^{3}\\right]dx\\\\ \r\n& = \\int_{2}^{4}\\left[\\left(3x^{2}(3)-x(3)^{2}\\right)-\\left(3x^{2}(1)-x(1)^{2}\\right)\\right]dx\\\\\r\n& = \\int_{2}^{4}\\left[\\left(9x^{2}-9x\\right)-\\left(3x^{2}-x\\right)\\right]dx\\\\ \r\n& = \\int_{2}^{4}\\left[6x^{2}-8x\\right]dx\\\\ \r\n& = \\left[\\frac{6x^{3}}{3}-\\frac{8x^{2}}{2}\\right]_{2}^{4}\\\\\r\n& = \\left[2x^{3}-4x^{2}\\right]_{2}^{4}\\\\\r\n& = \\left(2(4)^{3}-4(4)^{2}\\right)-\\left(2(2)^{3}-4(2)^{2}\\right)\\\\\r\n& = 128-64-16+16\\\\\r\n& = 64 \r\n\\end{align*}\\]\r\nAs expected, the same answer is obtained regardless of the order of integration.\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 evaluating double integrals\r\n\r\n\tEvaluate the following.\r\n\t\r\n\t\t\\(\\underset{R}{\\iint}(4xy)dxdy\\), where \\(R\\) is the rectangle with vertices \\((0,1)\\), \\((0,3)\\), \\((3,1)\\) and \\((3,3)\\)\r\n\t\t\\(\\underset{R}{\\iint}(6e^{x}+e^{y})dydx\\), where \\(R\\) is the region bounded by the lines \\(x=0\\), \\(x=4\\), \\(y=1\\) and \\(y=2\\)\r\n\t\t\\(\\int_{0}^{3}\\int_{1}^{4}(\\sin(\\pi x)+\\cos(\\pi y)dydx\\).\r\n\t\r\n\tFind \\(\\int_{0}^{b}\\int_{0}^{a}(5)dxdy\\). What is the physical significance of this result?\r\n\tEvaluate the following double integrals.\r\n\t\r\n\t\t\\(\\int_{0}^{1}\\int_{y=x^{2}}^{y=x}(x^{3}y)dydx\\)\r\n\t\t\\(\\int_{0}^{2}\\int_{y=x^{2}}^{y=4}(3+2y)dydx\\)\r\n\t\t\\(\\underset{R}{\\iint}(4xy)dA\\), where \\(R\\) is the region bounded by the graphs of \\(y=x\\) and \\(y=x^{2}-2x\\)\r\n\t\t\\(\\underset{R}{\\iint}(2x+y)dydx\\), where \\(R\\) is the region bounded by the graphs of \\(y+x=2\\) and \\(y=x^{2}\\).\r\n\t\r\n\tConsider the following:\r\n\t\\[\\int_{0}^{2}\\left[\\int_{y=0}^{y=x}(2x+2y)dy\\right]dx\\]\r\n\t\r\n\t\tEvaluate the double integral.\r\n\t\tSketch the region of integration defined by the integral.\r\n\t\tConfirm your answer by evaluating the double integral obtained by reversing the order of integration.\r\n\t\r\n\tConsider the following:\r\n\t\\[\\int_{0}^{1}\\left[\\int_{y=x}^{y=1}\\left(2x\\frac{e^{y}}{y^{2}}\\right)dy\\right]dx\\]\r\n\t\r\n\t\tSketch the region of integration defined by the integral.\r\n\t\tEvaluate the double integral.\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.04},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":206,"$":[{"v":"Fractional indices","n":0.707},{"v":" \r\nAn index can be an integer \u2013 a counting number - either positive or negative. It can also be a fraction such as \\(\\dfrac{1}{2}\\), \\(\\dfrac{3}{4}\\), or \\(2.5\\). Use this resource to learn how the laws of indices apply to fractional indices.\r\n\r\nIndex laws\r\nRemember the basic index laws:\r\n\r\n\t\\(a^{m}\\times a^{n}=a^{m+n}\\)\r\n\t\\(a^{m}\\div a^{n}=a^{m-n}\\)\r\n\t\\(\\left(a^{m}\\right)^{n}=a^{mn}\\)\r\n\t\\(a^{0}=1\\)\r\n\t\\(a^{-n}=\\dfrac{1}{a^{n}}\\)\r\n\r\nThese hold when \\(m\\) and \\(n\\) are whole numbers or fractions. From this point onwards, we will refer to them as the first index law, the second index law, and so on.\r\n\r\nBasic fractional indices\r\nLet's consider the fractional index, \\(3^{\\frac{1}{2}}\\).\r\nUsing the first index law:\r\n\\[\\begin{align*} 3^{\\frac{1}{2}}\\times3^{\\frac{1}{2}} & = 3^{\\frac{1}{2}+\\frac{1}{2}}\\\\ \r\n& = 3^{1}\\\\\r\n& = 3 \r\n\\end{align*}\\]\r\nWhen \\(3^{\\frac{1}{2}}\\) is multiplied by itself, it gives the base, \\(3\\). This is the same as:\r\n\\[\\sqrt{3}\\times\\sqrt{3}=3\\]\r\nTherefore, we can say that:\r\n\\[3^{\\frac{1}{2}}=\\sqrt{3}\\]\r\n\r\nNow, let's consider \\(2^{\\frac{1}{3}}\\).\r\nUsing the first index law:\r\n\\[\\begin{align*} 2^{\\frac{1}{3}}\\times2^{\\frac{1}{3}}\\times2^{\\frac{1}{3}} & = 2^{\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{3}}\\\\ \r\n& = 2^{1}\\\\ \r\n& =2\r\n\\end{align*}\\]\r\n\\(2^{\\frac{1}{3}}\\) therefore behaves like the cube root of \\(2\\):\r\n\\[2^{\\frac{1}{3}}=\\sqrt[3]{2}\\]\r\n\r\nIn general, we say that \\(a^{\\frac{1}{n}}\\) is the \\(n^{\\textrm{th}}\\) root of \\(a\\). That is:\r\n\\[a^{\\frac{1}{n}}=\\sqrt[n]{a}\\]\r\nfor any positive integer \\(n\\).\r\n\r\nExample 1 \u2013 evaluating basic fractional indices\r\nEvaluate \\(4^{\\frac{1}{2}}\\).\r\n\\[\\begin{align*} 4^{\\frac{1}{2}} & = \\sqrt{4}\\\\\r\n& = 2\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 evaluating basic fractional indices\r\nEvaluate the following expressions. If the answer is not exact, use a calculator to give a decimal approximation to two decimal places.\r\n\r\n\t\\(64^{\\frac{1}{2}}\\)\r\n\t\\(125^{\\frac{1}{3}}\\)\r\n\t\\(36^{\\frac{1}{4}}\\)\r\n\t\\(81^{-\\frac{1}{2}}\\)\r\n\t\\(128^{-\\frac{1}{7}}\\)\r\n\t\\(250^{\\frac{1}{5}}\\)\r\n\r\n\r\n\r\n\r\nMore complicated fractional indices\r\nConsider the fraction index, \\(a^{\\frac{2}{3}}\\).\r\nUsing the third index law:\r\n\\[\\begin{align*} a^{\\frac{2}{3}} & = \\left(a^{2}\\right)^{\\frac{1}{3}}\\\\ \r\n& = \\sqrt[3]{a^{2}}\r\n\\end{align*}\\]\r\n\r\nSince \\(2\\times \\dfrac{1}{3}\\) is the same as \\(\\dfrac{1}{3}\\times2\\), we can write:\r\n\\[\\begin{align*} a^{\\frac{2}{3}} & = a^{2\\times\\left(\\frac{1}{3}\\right)}\\\\ \r\n& = a^{\\left(\\frac{1}{3}\\right)\\times2}\\\\\r\n& = \\left(\\sqrt[3]{a}\\right)^{2}\r\n\\end{align*}\\]\r\n\r\nIn general,\r\n\\[a^{\\frac{p}{q}} = \\sqrt[q]{a^{p}}\\quad\\textrm{or}\\quad\\left(\\sqrt[q]{a}\\right)^{p}\\]\r\nwhere \\(p\\) and \\(q\\) are integers with \\(q\\neq0\\),\r\n\r\nExample 1 \u2013 evaluating more complex fractional indices\r\nEvaluate \\(5^{\\frac{3}{4}}\\).\r\n\\[\\begin{align*} 5^{\\frac{3}{4}} & = \\sqrt[4]{5^{3}}\\\\\r\n& = \\left(\\sqrt[4]{5}\\right)^{3}\\\\\r\n& = 3.34\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 evaluating more complex fractional indices\r\nSimplify the following expressions. Give your answers in index notation with positive indices.\r\n\r\n\t\\(\\left(\\frac{8}{27}\\right)^{\\frac{2}{3}}\\)\r\n\t\\(\\sqrt{5}\\times\\sqrt[3]{5}\\times\\sqrt[6]{5}\\)\r\n\t\\(2^{\\frac{5}{2}}-2^{\\frac{3}{2}}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.06},[{"v":"Surds","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Logarithms","i":1,"n":1},{"v":"Indices","i":0,"n":1}]]},{"i":207,"$":[{"v":"Surds","n":1},{"v":"\r\nHow do you find a square root of any number that isn\u2019t already a square? The square root of \\(16\\) is \\(4\\) and the square root of \\(25\\) is \\(5\\), but can you find the square root of a number between these, for example \\(\\sqrt{20}\\)? Learning how to identify and simplify surds to their simplest form makes complex expressions more manageable.\r\n\r\nA surd is an expression that cannot be simplified into a whole number or a simple fraction. They contain a square root, cube root or other root symbol. The number under the root is called the radicand.\r\nSurds are irrational numbers; in decimal form, they contain an infinite number of non-recurring digits. Expressions like \\(\\sqrt{2}\\), \\(\\sqrt{3}\\), \\(\\dfrac{1}{\\sqrt{8}}\\), \\(\\sqrt{20}\\), \\(\\sqrt[3]{10}\\) and \\(\\sqrt[4]{9}\\) are all surds.\r\n\\(\\sqrt{1}\\), \\(\\sqrt{9}\\), \\(\\dfrac{1}{\\sqrt{100}}\\), \\(\\sqrt{25}\\), \\(\\sqrt[3]{64}\\) and \\(\\sqrt[4]{16}\\) are not surds. They can be written as rational numbers or simple fractions: \\(1\\), \\(3\\), \\(\\dfrac{1}{10}\\), \\(5\\), \\(4\\), and \\(2\\), respectively.\r\nMore complex expressions such as \\(3\\sqrt{2}\\) and \\(\\sqrt{5}-\\sqrt{2}\\) are also surds.\r\n\r\nVideo tutorial \u2013 surds\r\nWatch this video to learn how to work with surds.\r\n\r\n\r\nSimplifying surds\r\nThere are two rules you can follow to simplify surds.\r\nRule 1 is:\r\n\\[\\sqrt{a\\times b}=\\sqrt{a}\\times\\sqrt{b}\\]\r\nRule 2 is:\r\n\\[\\sqrt{\\frac{a}{b}}=\\frac{\\sqrt{a}}{\\sqrt{b}}\\]\r\n\r\n\r\nExample 1 \u2013 simplifying surds\r\nSimplify \\(\\sqrt{200}\\).\r\n\\[\\begin{align*} & = \\sqrt{200}\\quad\\textrm{find any square factors}\\\\\r\n& = \\sqrt{100\\times2}\\quad\\textrm{use rule 1}\\\\\r\n& = \\sqrt{100}\\times\\sqrt{2}\\\\\r\n& = 10\\sqrt{2}\r\n\\end{align*}\\]\r\n\r\n\r\nAdding and subtracting surds\r\nThe rule for adding and subtracting surds is straightforward.\r\nOnly like surds can be added or subtracted.\r\n\r\nExample 1 \u2013 adding and subtracting surds\r\nSimplify \\(8\\sqrt{5}-3\\sqrt{5}\\).\r\n\\[8\\sqrt{5}-3\\sqrt{5}=5\\sqrt{5}\\]\r\n\r\n\r\n\r\n\r\n\r\nMultiplying surds\r\nTo multiply surds, multiply the coefficients outside of the root symbol and multiply the radicands.\r\n\\[a\\sqrt{b}\\times c\\sqrt{d} = ac\\sqrt{bd}\\]\r\n\r\nExample 1 \u2013 multiplying surds\r\nSimplify \\(\\sqrt{2}\\times\\sqrt{3}\\).\r\n\\[\\sqrt{2}\\times\\sqrt{3}=\\sqrt{6}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nDividing surds\r\nTo divide surds, combine the radicands under the same root.\r\n\\[\\frac{\\sqrt{a}}{\\sqrt{b}}=\\sqrt{\\frac{a}{b}}\\]\r\n\r\nExample 1 \u2013 dividing surds\r\nSimplify \\(\\dfrac{\\sqrt{18}}{\\sqrt{6}}\\).\r\n\\[\\begin{align*} \\frac{\\sqrt{18}}{\\sqrt{6}} & = \\sqrt{\\frac{18}{6}}\\\\\r\n& = \\sqrt{3}\r\n\\end{align*}\\]\r\n\r\n\r\nExpanding surds\r\nBrackets can be removed from surd expressions by multiplying everything in the brackets by everything outside of the brackets.\r\n\\[a\\left(b+c\\right)=ab+ac\\]\r\n\\[\\left(a+b\\right)\\left(c+d\\right)=ac+bc+ad+bd\\]\r\n\r\nExample 1 \u2013 expanding surds\r\nSimplify \\(\\sqrt{2}\\left(\\sqrt{2}+5\\right)\\).\r\n\\[\\begin{align*} \\sqrt{2}\\left(\\sqrt{2}+5\\right) & = \\left(\\sqrt{2}\\times\\sqrt{2}\\right)+\\left(\\sqrt{2}\\times5\\right)\\\\\r\n& = 2+5\\sqrt{2}\r\n\\end{align*}\\]\r\n\r\n\r\nRationalising surds\r\nSometimes, we need to express fractions containing surds with a rational denominator. We do this in a process called rationalising.\r\nTo rationalise a surd, we identify the surd in the denominator and multiply the top and bottom of the fraction by that surd.\r\n\r\nConjugate surds\r\nIt can get a bit more complicated when we deal with surd expressions that contain addition or subtraction. In these cases, we multiply by the conjugate surd. This is the surd with the addition or subtraction sign reversed.\r\n\\(\\sqrt{a}+\\sqrt{b}\\) and \\(\\sqrt{a}-\\sqrt{b}\\) are conjugate surds.\r\nWhen you multiply a pair of conjugate surds, the product does not contain any surd term.\r\n\\[ \\left(\\sqrt{a}+\\sqrt{b}\\right)\\left(\\sqrt{a}-\\sqrt{b}\\right)=(\\sqrt{a})^{2}-(\\sqrt{b})^{2}=a-b. \\]\r\n\r\nExample 1 \u2013 rationalising surds\r\nRationalise the denominator of \\(\\dfrac{2}{\\sqrt{5}}\\).\r\n\\[\\begin{align*} \\frac{2}{\\sqrt{5}} & = \\frac{2}{\\sqrt{5}} \\times \\frac{\\sqrt{5}}{\\sqrt{5}}\\\\\r\n& = \\frac{2\\sqrt{5}}{5}\r\n\\end{align*}\\]\r\nMultiplying by \\(\\dfrac{\\sqrt{5}}{\\sqrt{5}}\\) is the same as multiplying by \\(1\\). This keeps the fraction balanced.\r\n\r\n\r\n\r\n\r\n\r\nAdding and subtracting fractions involving surds\r\nWhen adding or subtracting fractions containing surds, you should  rationalise the denominators of each fraction first.\r\n\r\nExample \u2013 adding and subtracting fractions involving surds\r\nEvaluate \\(\\dfrac{2}{3\\sqrt{2}+1}+\\dfrac{1}{\\sqrt{3}-\\sqrt{2}}\\).\r\n\r\n\r\nExercise \u2013 simplifying surds\r\n\r\n \tExpand and simplify if possible.\r\n\t\r\n \t\\(\\sqrt{2}\\left(\\sqrt{2}-8\\right)\\)\r\n \t\\(\\left(\\sqrt{11}+3\\right)\\)\\(\\left(\\sqrt{11}-3\\right)\\)\r\n\t\r\n\tUse what you have learned so far to simplify the following.\r\n\t\r\n\t\t\\(\\sqrt{500}\\)\r\n\t\t\\(\\dfrac{2}{\\sqrt{20}}\\)\r\n\t\t\\(\\sqrt{2}+\\sqrt{7}+3\\sqrt{2}-4\\sqrt{7}\\)\r\n\t\t\\(\\sqrt{54}-\\sqrt{24}\\)\r\n\t\t\\(\\sqrt{10}\\times3\\sqrt{10}\\)\r\n\t\t\\(2\\sqrt{8}\\times2\\sqrt{50}\\times\\sqrt{2}\\)\r\n\t\t\\(\\dfrac{\\sqrt{32}}{\\sqrt{2}}\\)\r\n\t\t\\(\\dfrac{4\\sqrt{2}\\times3\\sqrt{3}}{6\\sqrt{6}}\\)\r\n\t\r\n\tExpress the following with a rational denominator.\r\n\t\r\n \t\\(\\dfrac{1}{\\sqrt{10}}\\)\r\n \t\\(\\dfrac{\\sqrt{3}+2}{\\sqrt{3}-2}\\)\r\n\t\r\n\tExpress \\(\\dfrac{2}{\\sqrt{3}-1}+\\dfrac{3}{2-\\sqrt{3}}\\) with a rational denominator and simplify.\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nI\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on surds. A surd is a number that cannot be written as a fraction.\r\n\r\nFor instance numbers like root two and root seven have got a series of decimal places after the decimal point if you type those into the calculator so you can\u2019t write those as fractions. Numbers that are not surds are written below. You can see that two can be written as a fraction two over one, one point three can be written as 13 over 10 and root four, which is two, can be written as two over one. In other words they can all be expressed as fractions.\r\n\r\nThe trick in simplifying surds is to look for square factors. Here are a number of square factors, four, which is two squared, nine, which is three squared and so on up to 100, which is 10 squared. Look at this example, we have the square root of 200 and what we do is we express it so that there\u2019s a square factor under the square root sign which in this case is 100, shown in red, 100 times two under the square root sign is the same as the square root of 100 times the square root of two. Notice the square root of 100 is 10 so our answer is 10 root two. Notice also the square root sign is sometimes called the radical. Similarly three root 48, 48 is not a square factor but 16 times three is, which gives us three the square root of 16 times three, the root 16 is four and root three is left on its own, in other words we now have two whole numbers, three and four, which gives us 12 root three, so our answer is 12 root three.\r\n\r\nWhen it comes to adding and subtracting surds one has to be mindful of the fact that only like surds can be added or subtracted. For instance, here\u2019s a question eight root five minus three root five. The root fives are the same in both terms, in other words we can subtract these two terms, eight root five minus three root five is five root five. However, if we have this question, where we have two root three plus five root seven minus 10 root three, notice there are two different surds here, a root three and a root seven, so we have to keep those separate in our answer, so that gives us five root seven minus eight root three.\r\n\r\nHere\u2019s the rule for multiplying surds and it\u2019s probably easier to illustrate this with an example. For instance root three times root two we can place both the root three and the root two under one square root sign, in other words that becomes the square root of three times two, three times two is six so the answer is root six.\r\n\r\nA second example, we have three root five times four root seven. Three and four are the whole numbers so you multiply those two separately, whereas the root five and the root seven can go under one square root sign, the square root of five times seven, so that gives us an answer of 12 root 35.\r\n\r\nFinally here\u2019s another one. We start with two root 10 times seven root six, which gives us initially 14 times the square root of 10 root six, which is root 60, but when you get a large square root sign in particular see if it will simplify into something a bit smaller, in other words look for a square factor which in this case is four, in other words root 60 is the square root of four times 15. Four times root 15, the square root of, gives us two times root 15, so we now have four times two which is 28 root 15.\r\n\r\nIn dividing surds we use a similar idea and again let\u2019s look at the example here, root 18 divided by root six is the same as the square root of 18 divided by six, which is three. In other words our answer is root three. Here\u2019s a number four divided by a surd, root two, so here we have four, which can be expressed as root 16 divided by root two. Notice I\u2019ve expressed both numerator and denominator as surds because we can now use the rule, which is the square root of 16 divided by two, which is root eight, and again notice that this simplifies because we can use a square factor here of four, so root four times two is two root two.\r\n\r\nRationalising a surd is a way of simplifying an expression where the surd is on the denominator. For instance if we have two over root five notice the surd is in the denominator here and to simplify that we can multiply two over root five by root five over root five, which is effectively one. In other words two over root five times root five over root five is two root five multiplying the top line, root five times root five is root 25 which is five. Notice the answer to root five over five has got no surd in the denominator.\r\n\r\nSimilarly here we have surds top and bottom but the purpose here is to try and get rid of the root two in the denominator, so we do as before by multiplying both top and bottom by root two over root two, which gives us root 10 on the top, root two times root two on the bottom is two times three which gives us six on the bottom. Again notice in both examples here that there are no surds in the denominator, that\u2019s rationalising surds.\r\n\r\nConjugate surds are just a slight extension of the rationalising surds problem we were doing in the previous slide. Here\u2019s an example; root three over five plus root two. What we do is we look at the denominator and then we write the conjugate of that, meaning we write the same terms five and root two but with a negative sign between them. In other words we\u2019re multiplying by five minus root two over five minus root two. Now we multiply out the problem, this gives us root three brackets five minus root two over five plus root two multiplied by five minus root two. If we multiply out the bottom line we\u2019ll get four terms, but notice in particular that when we multiply out this bottom line it\u2019ll be the surds that will cancel out. In other words we\u2019ll have no surd left in the denominator, so our answer is five root three minus root six divided by 23. Always make sure that the answer does not have a surd.\r\n\r\nNow try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.025},[{"v":"Surds","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Logarithms","i":1,"n":1},{"v":"Indices","i":0,"n":1}]]},{"i":208,"$":[{"v":"Relations and functions","n":0.577},{"v":"\r\nUnderstanding relations and functions is crucial for exploring more advanced topics in math and their applications in various scientific and engineering areas. Use this resouce to learn about what functions and relations are.\r\n\r\nRelations\r\nA relation is a set of ordered pairs. For example, \\((1,2),(2,6),(3,4),(x,y)\\) are ordered pairs. If the set is a relation, they are written inside curly brackets: \\(\\left\\{ (1,2),(2,6),(3,4),(x,y)\\right\\}\\).\r\n\r\n\tThe domain (dom) of a relation is the set of first elements or the \\(x\\)-values of the ordered pairs. For \\(\\left\\{ (1,2),(2,6),(3,4),(x,y)\\right\\}\\), \\(\\textrm{dom}=\\left\\{ 1,2,3,x\\right\\}\\).\r\n\tThe range (ran) of a relation is the set of second elements or the \\(y\\)-values of the ordered pairs. For \\(\\left\\{ (1,2),(2,6),(3,4),(x,y)\\right\\}\\), \\(\\textrm{ran}=\\left\\{ 2,4,6,y\\right\\}\\).\r\n\r\nIf the domain is not given then we assume the largest possible domain.\r\nThere is often a rule that links the domain and range. For example, consider a relation called \\(S\\) where:\r\n\\[ S = \\left\\{ (x,y):y>x,x\\in\\mathbb{R}\\right\\} \\]\r\nThe symbol \\(\\in\\) means \u201cis in\u201d or \u201cis an element of\u201d and \\(\\mathbb{R}\\) stands for the set of real numbers. The expression \\(x\\in\\mathbb{R}\\) means that \u201cx is an element of the set of real numbers\u201d. That is, \\(x\\) is a real number. \r\nTherefore, relation \\(S\\) is defined by the rule \\(y>x\\) and its domain \\(x\\in\\mathbb{R}\\). The relation consists of the set of all ordered pairs (\\(x\\) and \\(y\\)), where the \\(y\\) value is greater than the \\(x\\) value and where \\(x\\) must be a real number.\r\nThe rule of a relation may be thought of as: \\(\\textrm{DOMAIN}\\rightarrow\\textrm{RULE}\\rightarrow\\textrm{RANGE}\\). Values taken from the domain are applied to the rule, which then produces the values for the range.\r\n\r\nExample 1 \u2013 identifying the domain and range of relations\r\nConsider the graph of the following relation and state the domain and range.\r\n\\[ \\left\\{ (x,y):y=x^{2}\\right\\} \\] \r\nThe rule joining the set of ordered pairs \\((x,y)\\) is \\(y=x^{2}\\).\r\n\r\n\t\\(x\\) can be any real number, so \\(\\textrm{dom}=x\\in\\mathbb{R}\\).\r\n\t\\(y\\) must be greater than or equal to \\(0\\), so \\(\\textrm{ran}=\\left\\{ y:y\\geq0 \\right\\}\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 identifying the domain and range of relations\r\nState the domain and range of the following relations.\r\n\r\n\t\\(\\left\\{ (-2,1),(0,2),(2,5),(2,7),(3,9)\\right\\}\\)\r\n\t\\(\\left\\{ (4,1),5,2,6,3\\right\\}\\)\r\n\t\\(\\left\\{ x,y:x^{2}+y^{2}=25\\right\\}\\)\r\n\t\\(\\left\\{ x,y:2y=6-5x,x\\geq2\\right\\}\\)\r\n\r\n\r\n\r\n\r\nFunctions\r\nFor some relations, values in the domain (\\(x\\) values) may have many corresponding values in the range (\\(y\\) values), even an infinite number! When each point in the domain has a unique value in the range, the relation is called a function. In other words, every value of \\(x\\) has only one value of \\(y\\) in a function.\r\nFor example, the relation \\(\\left\\{ (-1,2),(-1,4),(1,6),(2,8),(3,10)\\right\\}\\) IS NOT a function because the value \\(x=-1\\) has two corresponding \\(y\\) values (\\(2\\) and \\(4\\)).\r\nHowever, \\(\\left\\{ (-1,1),(0,2),(1,3),(2,5),(3,7)\\right\\}\\) IS a function because for each \\(x\\) value, there is only one corresponding \\(y\\) value.\r\n\r\nMapping notation\r\nLet's consider the relation \\(F=\\left\\{ \\left(x,y\\right):y=\\sin(x),x\\in R\\right\\}\\).\r\n\r\nIf we choose any possible value of \\(x\\), there exists only one corresponding value of \\(y\\). Therefore, the relation \\(F\\) is a function. We can also write this function using mapping notation.\r\n\\[ f:X\\rightarrow Y,\\textrm{where }f(x)=\\sin (x)\\]\r\nThis means that the domain \\(X\\) is mapped onto the range \\(Y\\) using the rule \\(f(x)=\\sin(x)\\).\r\n\r\nVertical line test\r\nWhen you are given the graph of a relation, you can use a vertical line test to determine whether it is a function.\r\nIf you draw a vertical line through the \\(x\\)-axis, and it crosses the graph more than once, then it is NOT a function. This just means that there is an \\(x\\) value with more than one \\(y\\) value.\r\nConsider the graphs of \\(y=\\pm\\sqrt{x}\\) and \\(y=x^{2}\\).\r\n\r\nWhen a vertical line is drawn through \\(y=\\pm\\sqrt{x}\\), it crosses the graph more than once, so it is not a function. But when one is drawn through \\(y=x^{2}\\), the vertical line only crosses the graph once, so this relation is a function.\r\n\r\nExercise \u2013 identfying functions\r\nIdentify which of the following relations are functions.\r\n\r\n\t\\(\\left\\{ x,y:y=2x+4\\right\\}\\)\r\n\t\\(\\left\\{ x,y:y=4-x^{2}\\right\\}\\)\r\n\t\\(\\left\\{ x,y:x^{2}+y^{2}=36\\right\\}\\)\r\n\t\\(\\left\\{ x,y:y=7\\right\\}\\)\r\n\t\\(\\left\\{ x,y:x=-2\\right\\}\\)\r\n\t\\(\\left\\{ x,y:y=-\\sqrt{4-x^{2}}\\right\\}\\)\r\n\r\n\r\n\r\n\r\nImplied domain\r\nWhen only the rule of a function is given, we usually assume that the domain is all real numbers, \\(\\mathbb{R}\\), unless the function itself shows a different domain. There are a few examples of functions where the domain is restricted.\r\n\r\nFunctions with square roots\r\nIf a function involves a square root, the domain (in the real number system) is restricted to \\(x\\) values that give in a non-negative number under the square root sign. For example, the domain of the function \\(y=+\\sqrt{x-4}\\) would be restricted so that \\(x-4\\geq0\\). The domain is \\(\\left\\{ x:x\\geq4\\right\\}\\).\r\nSimilarly, the domain of the function \\(y=+\\sqrt{9-x^{2}}\\) would be restricted so that \\(9-x^{2}\\geq0\\), making the domain \\(\\left\\{ x:-3\\leq x\\leq3\\right\\}\\).\r\n\r\nFunctions with fractions\r\nIf the function involves a fraction, the value in the denominator must not equal \\(0\\). So, the domain of the function \\(y=\\dfrac{3}{x+5}\\) is restricted so that \\(x+5\\neq0\\). This means its domain is \\(\\left\\{ x:x\\neq-5\\right\\}\\). \r\nThe domain can also be written as \\(\\left\\{ x:x\\in\\mathbb{R}\\setminus\\left\\{ -5\\right\\} \\right\\}\\).  This means that the domain is the set of real numbers excluding \\(-5\\).\r\nConsider another example: \\(y=\\dfrac{3}{2x-8}\\). The domain of this function is restricted so that \\(x-8\\neq0\\); therefore, the domain is \\(\\left\\{ x:x\\neq4\\right\\}\\) or \\(\\left\\{ x:x\\in\\mathbb{R}\\setminus\\left\\{ 4\\right\\} \\right\\}\\).\r\n\r\nExercise \u2013 determining the domain of functions\r\nState the domain of the following functions.\r\n\r\n\t\\(\\left\\{ x,y:y=x+2\\right\\}\\)\r\n\t\\(\\left\\{ x,y:y=4-x^{2}\\right\\}\\)\r\n\t\\(\\left\\{ x,y:y=+\\sqrt{4-x}\\right\\}\\)\r\n\t\\(\\left\\{ x,y:y=\\dfrac{3}{x+2}\\right\\}\\)\r\n\t\\(\\left\\{ x,y:y=\\dfrac{5}{\\sqrt{x-7}}\\right\\}\\)\r\n\t\\(\\left\\{ x,y:y=\\dfrac{1}{x+2}-\\dfrac{3}{x-4}\\right\\}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.035},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":209,"$":[{"v":"Interval notation","n":0.707},{"v":"Interval notation helps to clearly express a continuous range of numbers in a simple, concise way. It is handy in many ways, such as to define acceptable weight limits for a machine, indicating a valid range of measurements in scientific experiments, and specifying temperature ranges in which a machine is safe to operate. Use this resource to learn about using interval notation.\r\n\r\nIntervals\r\nWhen the domain or range of a function is restricted to a subset of real numbers, \\(\\mathbb{R}\\), it is called an interval. The endpoints are indicated with \\(a\\) and \\(b\\) for the lower and upper limits, respectively. \r\nIntervals can be represented using:\r\n\r\n\tinterval notation, which uses brackets \\([\\,]\\) and parentheses \\(\\left(\\,\\right)\\)\r\n\tinequality notation, which uses inequality symbols: less than \\(\\), less than or equal to \\(\\leq\\) and greater than or equal to \\(\\geq\\).\r\n\r\nIntervals are always written with the smaller number on the left. For example, \\(\\left(-3,5\\right)\\) is correct and \\(\\left(5,-3\\right)\\) is not.\r\n\r\nClosed intervals\r\nA closed interval is a set of numbers that includes all the numbers between two endpoints, and the endpoints \\(a\\) and \\(b\\) themselves.\r\nConsider the interval on the line graph where the endpoints on the real number line are represented by solid circles.\r\n\r\n\\(a\\) and \\(b\\) are included in the interval. The subset is therefore a closed interval, so brackets \\([\\,]\\) are used. The interval notation would be \\([a,b]\\) and the inequality notation would be \\(a\\leq x \\leq b\\).\r\n\r\nOpen intervals\r\nIf the endpoints are not included in the interval, it is called an open interval. \r\nConsider the same interval with endpoints \\(a\\) and \\(b\\) represented by open circles on the line graph.\r\n\r\nCurved brackets are used to give the inverval notation \\(\\left(a,b\\right)\\). In inequality notation, the interval is \\(a<x<b\\). \r\n\r\nHalf-open and half-closed intervals\r\nIn some cases, one endpoint is included in the subset and the other is not. This is called a half-open or half-closed interval. Interval notation is written using one bracket and one parenthesis, and inequality notation is written using a mix of inequality symbols.\r\nFor some subsets, the range of values may continue indefinitely towards the right i.e. towards infinity (\\(\\infty\\)), or towards the left i.e. towards negative infinity (\\(-\\infty\\)). Infinity is not a numeral, so we should not write the inequality notation as \\([b,\\infty]\\), \\([-\\infty,a]\\) or \\(b\\leq x\\leq\\infty\\), etc. For the purposes of understanding the concept, this type of notation has been used in this resource, but elsewhere, we would typically write \\(x\\geq b\\) or \\(x\\leq a\\).\r\nThe table shows some examples of domains with endpoints \\(a\\) and \\(b\\), including their interval notation, inequality notation and line graph.\r\n\r\n\t\r\n\t\t\r\n\t\t\tInterval notation\r\n\t\t\tInequality notation\r\n\t\t\tLine graph\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\t\\([a,b]\\)\r\n\t\t\t\\(a\\leq x\\leq b\\)\r\n\t\t\t\r\n\t\t\r\n\t\t\r\n\t\t\t\\((a,b)\\)\r\n\t\t\t\\(a<x<b\\)\r\n\t\t\t\r\n\t\t\r\n\t\t\r\n\t\t\t\\([a,b)\\)\r\n\t\t\t\\(a\\leq x<b\\)\r\n\t\t\t\r\n\t\t\r\n\t\t\r\n\t\t\t\\((a,b]\\)\r\n\t\t\t\\(a<x\\leq b\\)\r\n\t\t\t\r\n\t\t\r\n\t\t\r\n\t\t\t\\([a,\\infty)\\)\r\n\t\t\t\\(x\\geq a\\)\r\n\t\t\t\r\n\t\t\r\n\t\t\r\n\t\t\t\\((a,\\infty)\\)\r\n\t\t\t\\(x>a\\)\r\n\t\t\t\r\n\t\t\r\n\t\t\r\n\t\t\t\\((-\\infty,b]\\)\r\n\t\t\t\\(x\\leq b\\)\r\n\t\t\t\r\n\t\t\r\n\t\t\r\n\t\t\t\\((-\\infty,b)\\)\r\n\t\t\t\\(x>b\\)\r\n\t\t\t\r\n\t\t\r\n\t\r\n\r\n\r\nExample 1 \u2013 using interval and inequality notation\r\nWrite \\([-2,3)\\) in inequality notation and graph it on the real number line.\r\nThe left bracket indicates that \\(-2\\) is included. The right parenthesis indicates that \\(3\\) is not included. Therefore, the inequality notation is \\(-2\\leq x\\leq 3\\).\r\nOn the number line, the \\(-2\\) endpoint uses a solid circle and the \\(3\\) endpoint uses an open circle.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using interval and inequality notation\r\n\r\n\tWrite the following in interval notation and graph them on a real number line.\r\n\t\r\n\t\t\\(1\\leq x<10\\)\r\n\t\t\\(-6\\leq x<-4\\)\r\n\t\t\\(x>5\\)\r\n\t\r\n\tWrite the following in interval notation and inequality notation.\r\n\t\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t\r\n\r\n\r\n\r\nMultiple intervals\r\nFunctions can have two or more sets of real values. Let's consider a function with one subset with endpoints \\(a\\) and \\(b\\), and a second subset with endpoints \\(c\\) and \\(d\\). \r\n\r\nTo indicate the intervals, we need to use the \\(\\cup\\) symbol, which means \"in union with\". In interval notation, this would be \\([a,b]\\cup[c,d]\\). In inequality notation, we write \\(a\\leq x\\leq b\\) with \\(c\\leq x\\leq d\\) OR \\(\\left\\{ x:a\\leq x\\leq b\\right\\} \\cup\\left\\{ x:c\\leq x\\leq d\\right\\} \\).\r\n\r\nExample \u2013 using interval and inequality notation for multiple intervals\r\nWrite the following interval and inequality notation.\r\n\r\nFor the left interval, the left arrow indicates the values extend indefinitely towards \\(-\\infty\\). The solid circle at \\(2\\) indicates that the subset includes \\(2\\). This interval would be \\((-\\infty,2]\\) or \\(x\\leq2\\).\r\nFor the right interval, the open circle at \\(5\\) indicates that the subset does not include \\(5\\). The solid circle at \\(12\\) indicates that the subset includes \\(12\\). This interval would be \\((5,12]\\) or \\(5<x\\leq12\\).\r\nPutting these together with the \\(\\cup\\) symbol, the interval notation would be \\((-\\infty,2]\\cup(5,12]\\). The inequality notation would be \\(x\\leq2\\) with \\(5<x\\leq12\\), or \\(\\left\\{ x:x\\leq2\\right\\}\\cup\\left\\{x:5<x\\leq12\\right\\}\\).\r\n\r\nExercise\r\nGraph the following on the real number line and write in inequality notation.\r\n\r\n\t\\((-\\infty,3)\\cup(8,13]\\)\r\n\t\\(\\left[-1,4\\right]\\cup\\left[6,9\\right]\\)\r\n\t\\((-\\infty,3]\\cup(6,\\infty)\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.038},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":210,"$":[{"v":"Inverse functions","n":0.707},{"v":"One-to-one functions can be inverted so that the inputs and outputs are reversed. They are crucial where we need effective reversal of processes and calculations, like undoing equations in physics, computer graphics transformations, or decrypting data. Use this resource to learn about inverse functions and how to graph them.\r\n\r\nIf a function \\(f\\) maps \\(x\\) to \\(y\\), then its inverse function, denoted by \\(f^{-1}\\), maps \\(y\\) back to \\(x\\). For a function to have an inverse, it must be one-to-one. This means that every \\(x\\) has a unique \\(y\\).\r\nTo inverse a function, we simply swap the \\(x\\) and \\(y\\) in each ordered pair. For example:\r\n\r\n\tCoordinate \\((a,b)\\) is an element of function \\(f\\) or \\((a,b)\\in f\\).\r\n\tCoordinate \\((b,a)\\) would be an element of function \\(f^{-1}\\) or \\((b,a)\\in f^{-1}\\).\r\n\r\nThe domain and range also swap. So:\r\n\r\n\tThe domain of \\(f\\) is the range of \\(f^{-1}\\).\r\n\tThe range of \\(f\\) is the domain of \\(f^{-1}\\).\r\n\r\n\r\nGraphs of inverse functions\r\nA function \\(f\\) and its inverse \\(f^{-1}\\) are symmetrical about the line \\(y=x\\).\r\n\r\n\r\nFinding inverse functions\r\nTo find the inverse of a function, we simply interchange \\(x\\) and \\(y\\), then rearrange the function so that it is in terms of \\(y\\).\r\n\r\nExample 1 \u2013 finding inverse functions\r\nFor \\(f:\\mathbb{R}\\rightarrow\\mathbb{R}\\), where \\(y=f(x)=\\dfrac{x-1}{2}\\), find the \\(f^{-1}\\).\r\nTo find the inverse function, we swap or interchange \\(x\\) and \\(y\\), then rearrange the function so that it is in terms of \\(y\\).\r\n\\[\\begin{align*} y & = \\frac{x-1}{2}\\\\\r\nx & = \\frac{y-1}{2}\\\\\r\n2x & = y-1\\\\\r\ny & = 2x+1\r\n\\end{align*}\\]\r\nTherefore, \\(f^{-1}(x)=2x+1\\).\r\n\r\n\r\n\r\nExercise\r\nFind the inverse of the following one-to-one functions.\r\n\r\n\t\\(y=x+5\\)\r\n\t\\(y=4x\\)\r\n\t\\(y=\\dfrac{2x+1}{3}\\)\r\n\t\\(y=\\sqrt{2x-1}\\) where \\(x\\geq\\dfrac{1}{2}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.064},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":211,"$":[{"v":"Absolute value functions","n":0.577},{"v":"\r\nAbsolute values turn negative values into positive ones, and leave positive ones unchanged. Absolute value functions are useful for expressing situations where only magnitude matters, without regard to direction, like measuring distance, temperature differences, profit or loss diffeerences, and margin of error.\r\n\r\nAbsolute values\r\nThe absolute value of any number \\(x\\) gives a measure of its size or magnitude, regardless of whether it is positive or negative. It is the distance from \\(0\\) on a number line.\r\nWe represent the absolute value of \\(x\\) using two vertical lines \\(\\left|\\,\\right|\\). For example, the absolute value of \\(x\\) would be \\(\\left|x\\right|\\).\r\n\r\nExample 1 \u2013 finding the absolute value\r\nFind \\(\\left|2\\right|\\).\r\nTo find the absolute value, we simply turn any negative value into a positive one. \\(2\\) is already positive, so \\(\\left|2\\right|=2\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the absolute value\r\nEvaluate the following.\r\n\r\n\t\\(\\left|-11\\right|\\)\r\n\t\\(\\left|-9+4\\right|\\)\r\n\t\\(-\\left|4\\right|-\\left|-5\\right|\\)\r\n\t\\(\\left|-12\\right|-\\left|3\\right|\\)\r\n\t\\(\\left|-30\\right|\\div\\left|5\\right|\\)\r\n\r\n\r\n\r\n\r\nSolving equations involving absolute values\r\nEquations involving absolute values often have two solutions because the absolute value measures distance from \\(0\\) and is always positive. This means the expression inside can be equal to either a positive or negative value.\r\nFor example, if \\(\\left|x\\right|=7\\), \\(x\\) can be \\(7\\) or \\(-7\\) since both give an answer of \\(7\\).\r\n\r\nExample 1 \u2013 solving equations involving absolute values\r\nSolve \\(\\left|x-2\\right|=3\\).\r\nWe consider two cases: \\(x-2=3\\) and \\(x-2=-3\\).\r\nFor \\(x-2=3\\):\r\n\\[\\begin{align*} x-2 & = 3\\\\\r\nx & = 5\r\n\\end{align*}\\]\r\nFor \\(x-2=-3\\):\r\n\\[\\begin{align*} x-2=-3\\\\\r\nx & = -1\r\n\\end{align*}\\]\r\nTherefore, \\(x=-1\\) and \\(x=5\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 solving equations involving absolute values\r\nFind the following, for \\(x\\in\\mathbb{R}\\).\r\n\r\n\t\\(\\left\\{x:\\left|x\\right|=6\\right\\}\\)\r\n\t\\(\\left\\{x:\\left|x-1\\right|<3\\right\\}\\)\r\n\t\\(\\left\\{x:\\left|\\dfrac{x-3}{2}\\right|\\geq1\\right\\}\\)\r\n\t\\(\\left\\{x:\\left|\\dfrac{x}{2}\\right|=\\left|x+2\\right|\\right\\}\\)\r\n\r\n\r\n\r\n\r\nGraphs of absolute value functions\r\nAbsolute value functions are hybrid functions. They have the general form:\r\n\\[f:\\mathbb{R}\\rightarrow\\mathbb{R},\\textrm{where }f(x)=\\left|x\\right|=\\begin{cases} -x & x<0\\\\ x & x\\geq0 \\end{cases}\\]\r\nThe graph of \\(y=\\left|x\\right|\\) is shown.\r\n\r\n\r\n\tThe domain of \\(y=\\left|x\\right|\\) is \\(\\mathbb{R}\\), the set of all real numbers.\r\n\tThe range of \\(y=\\left|x\\right|\\) is \\(\\mathbb{R}^{+}\\cup\\left\\{ 0\\right\\}\\), the set of all positive real numbers and \\(0\\).\r\n\r\nIn general, to sketch the graph of \\(y=\\left|f(x)\\right|\\), we need to sketch the graph of \\(y=f(x)\\) first, then reflect the \\(x\\)-axis portion of the graph below the \\(x\\)-axis.\r\n\r\nTransforming graphs of absolute value functions\r\nAbsolute value functions can be transformed like other graphs and functions. Let's look at the basic form \\(y=\\left|x\\right|\\) as an example.\r\n\r\nExample 1 \u2013 transforming graphs of absolute value functions\r\nGraph \\(y=\\left|x-2\\right|\\).\r\nThe \\(x-2\\) indicates that the graph is being translated \\(2\\) units to the right.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 transforming graphs of absolute value functions\r\nSketch the following graphs.\r\n\r\n\t\\(y=\\left|x+4\\right|\\)\r\n\t\\(y=\\left|x-1\\right|-3\\)\r\n\t\\(y=\\left|3-x^{2}\\right|\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.053},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":212,"$":[{"v":"Hybrid functions","n":0.707},{"v":"Hybrid functions combine different types of functions. They are useful for modelling real-world scenarios that have different behaviours under different conditions, like tax brackets or shipping rates. Other examples include analysing frictional forces in physics, designing profiles and shapes for vehicles, and modelling free-fall acceleration. Use this resource to learn about hybrid functions.\r\n\r\nFunctions which have different rules for different intervals of their domain are called hybrid functions. Sometimes, they are called piecewise functions.\r\n\r\nAn example of a hybrid function is: \r\n\\[y=f(x)=\\begin{cases} -x & x\\leq-1\\\\\r\n1 & -1<x<1\\\\\r\nx & x\\geq1\r\n\\end{cases}\\]\r\nThis hybrid function has three rules. Each depends on the value of \\(x\\) in its domain. A function is considered hybrid if it has two or more rules.\r\n\r\nGraphs of hybrid functions\r\nTo graph a hybrid function, we look at each rule and its specified interval separately. We have to play close attention to the endpoints and correctly use open or solid circles to show them.\r\nLet's see how to graph the example function.\r\n\r\nExample 1 \u2013 graphing hybrid functions\r\nGraph the following hybrid function.\r\n\\[y=f(x)=\\begin{cases} -x & x\\leq-1\\\\\r\n1 & -1<x<1\\\\\r\nx & x\\geq1\r\n\\end{cases}\\]\r\nThis is a hybrid function with three rules. Let's consider the graph of each of the rules, and pay attention to the restricted domains.\r\nRule \\(1\\) states that \\(y=-x\\) for \\(x\\leq-1\\). Since the less than or equal to sign is used, the endpoint \\(-1\\) has a solid circle. \\(-1\\) is included in the domain for this function.\r\n\r\nRule \\(2\\) states that \\(y=1\\) for \\(-1<x<1\\). Since the less than sign is used, the endpoints \\(-1\\) and \\(1\\) have an open circle. \\(-1\\) and \\(1\\) are not included in the domain for this function.\r\n\r\nRule \\(3\\) states that \\(y=x\\) for \\(x\\geq1\\). Since the greater than or equal to sign is used, the endpoint \\(1\\) has a solid circle. \\(1\\) is included in the domain for this function.\r\n\r\nThe \"graphical pieces\" from rules \\(1\\) to \\(3\\) can be put together to form the graph of the hybrid function.\r\n\r\n\r\n\r\n\r\nExercise \u2013 graphing hybrid functions\r\n\r\n\tGraph the following hybrid function.\r\n\t\\[y = f(x) = \\begin{cases} x+1 & x<0\\\\\r\n\tx-1 & x\\geq0\r\n\t\\end{cases}\\]\r\n\tGraph the following hybrid function.\r\n\t\\[y = f(x) = \\begin{cases} x^{2} & x<0\\\\\r\n\t-x^{2} & x\\geq0\r\n\t\\end{cases}\\]\r\n\tGraph the following hybrid function.\r\n\t\\[y = f(x) = \\begin{cases} -1 & x<-2\\\\\r\n\t0 & -2\\leq x\\leq2\\\\ \r\n\t1 & x>2\r\n\t\\end{cases}\\]\r\n\tGraph the following hybrid function.\r\n\t\\[y = f(x) = \\begin{cases} x+2 & x<-1\\\\\r\n\t1 & -1\\leq x\\leq1\\\\\r\n\tx & x>1\r\n\t\\end{cases}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.052},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":213,"$":[{"v":"Circular functions","n":0.707},{"v":"\r\n\r\nCircular functions produce periodic wave-like graphs and are foundational in trigonometry. They have broad applications across physics, engineering, and signal processing, where modelling repetitive phenomena and wave patterns is essential. Use this resource to explore circular functions.\r\n\r\nCircular functions include sine, cosine, and tangent. They describe relationships involving angles and lengths in circles.\r\nHere, we will just look at sine and cosine functions, and their graphs.\r\nThese functions have the general forms:\r\n\\[y=a\\sin(n(x-\\phi))+k\\] \\[y=a\\cos(n(x-\\phi)+k\\]\r\nYou will learn what \\(n\\), \\(k\\) and \\(\\phi\\) are when we discuss transformations.\r\n\r\nGraphs of circular functions\r\nGraphs of circular functions are unique because they involve a repeating waves\u2014that is, they are periodic. They are characterised by two key features:\r\n\r\n\tamplitude which is the height from the midline of the graph to the peak of the wave (or how tall a wave is)\r\n\tperiod which is the length of one complete cycle of the wave (or how frequently the wave repeats).\r\n\r\n\r\nHere is the graph of \\(y=\\sin(x)\\). It intercepts the \\(y\\)-axis at the origin \\((0,0)\\), has an amplitude of \\(1\\) and a period of \\(2\\pi\\).\r\n\r\nAnd here is the graph of \\(y=\\cos(x)\\). It intercepts the \\(y\\)-axis at \\((0,1)\\), has an amplitude of \\(1\\) and a period of \\(2\\pi\\).\r\n\r\nBoth functions have a domain of \\(\\mathbb{R}\\) and a range of \\([-1,1]\\).\r\nWhen looking at the graphs, remember \\(\\pi\\approx3.142\\), so \\(2\\pi\\approx6.284\\).\r\n\r\nTransforming graphs of circular functions\r\nWhen we have a graph of function \\(y=a\\sin(nx)\\) or \\(y=a\\cos(nx)\\):\r\n\r\n\tamplitude is \\(\\left|a\\right|\\)\r\n\tperiod is \\(\\dfrac{2\\pi}{n}\\).\r\n\r\nWe can transform circular functions just as we do with other graphs. Go back to Graphs and transformations if you need a refresher.\r\n\r\nVertical translation\r\nRemember that a translation is a shift of the whole graph up, down, to the left or right.\r\nApplying a vertical translation gives us the forms \\(y=a\\sin(nx)+k\\) and \\(y=a\\cos(nx)+k\\), where \\(k\\) is the number of units the graph is shifted. If \\(k\\) is positive, the graph is shifted up \\(k\\) units. If \\(k\\) is negative, the graph is shifted down \\(k\\) units.\r\nFor example, consider the graphs of \\(y=\\sin(x)+2\\) in red and \\(y=\\sin(x)\\) in blue. \\(k=2\\), so the \\(y=sin(x)\\) graph is shifted \\(2\\) units up so that it intercepts the \\(y\\)-axis at \\((0,2)\\).\r\n\r\n\r\nHorizontal translation\r\nReplacing the \\(x\\) with \\((x-\\phi)\\) shifts the graphs of \\(y=\\sin(x)\\) and \\(y=\\cos(x)\\) horizontally \\(\\phi\\) units to the right.\r\nReplacing the \\(x\\) with \\((x+\\phi)\\) shifts the graphs of \\(y=\\sin(x)\\) and \\(y=\\cos(x)\\) horizontally \\(\\phi\\) units to the left.\r\nConsider the graph of \\(y=\\cos(x+\\pi)\\) in red and \\(y=\\cos(x)\\) in blue. \\(\\phi=-\\pi\\), so the \\(y=\\cos(x)\\) graph is shifted \\(\\pi\\) units to the left. The new graph has a \\(y\\)-intercept of \\(-1\\). \r\n\r\n\r\nDilation\r\nA graph dilates when we change the \\(n\\) factor for \\(y=a\\sin(nx)\\) and \\(y=a\\cos(nx)\\). When \\(n>1\\), the graph is squashed by a factor of \\(n\\). When \\(0<n<1\\), the graph is stretched by a factor of \\(\\dfrac{1}{n}\\).\r\nConsider the graph of \\(y=\\sin(x)\\) in blue and \\(y=\\sin(2x)\\) in red as examples. \\(y=\\sin(x)\\) has a period of \\(2\\pi\\). \\(y=\\sin(2x)\\) has a period of \\(\\dfrac{2\\pi}{n}=\\dfrac{2\\pi}{2}=\\pi\\).\r\nThis means that we squash \\(y=\\sin(x)\\) by a factor of \\(2\\) parallel to the \\(x\\)-axis. The dilated graph has a period that is half that of the original graph.\r\n\r\n\r\n\r\nReflection\r\nChanging the sign of \\(a\\) in the equations \\(y=a\\sin(nx)\\) and \\(y=a\\cos(nx)\\) results in reflection about the \\(x\\)-axis.\r\n\r\nExample 1 \u2013 graphing circular functions\r\nGraph \\(y=3\\sin(x)\\).\r\nHere, \\(a=3\\) and \\(n=1\\), so the graph has an amplitude of \\(\\left|3\\right|=3\\) and a period of \\(\\dfrac{2\\pi}{n}=\\dfrac{2\\pi}{1}=2\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise\r\n\r\n\tSketch the graphs of the following functions for one complete cycle, and state the amplitude and period.\r\n\t\r\n\t\t\\(y=2\\cos(x)\\)\r\n\t\t\\(y=2\\sin(3x)\\)\r\n\t\t\\(y=\\dfrac{1}{2}\\sin(2x)\\)\r\n\t\t\\(y=3\\cos\\left(\\dfrac{x}{2}\\right)\\)\r\n\t\t\\(y=-2\\sin(3x)\\)\r\n\t\r\n\tSketch the graphs of the following functions for one complete cycle, and state the amplitude and period.\r\n\t\r\n\t\t\\(y=2\\sin(x-\\pi)\\)\r\n \t\t\\(y=3\\cos\\left(x+\\dfrac{\\pi}{2}\\right)\\)\r\n\t\r\n\tSketch the graphs of the following functions for one complete cycle, and state the amplitude and period.\r\n\t\r\n\t\t\\(y=2\\sin(3x-\\pi)\\)\r\n\t\t\\(y=3\\cos(4x-2\\pi)\\)\r\n\t\t\\(y=2\\sin\\left(2x+\\dfrac{\\pi}{3}\\right)\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.042},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":214,"$":[{"v":"Quadratic functions","n":0.707},{"v":"\r\nQuadratic functions have a distinct parabolic shape. Graphs of quadratic functions have applications in a wide range of fields, like modelling projectile motion, optimising business revenue, analysing population growth and designing bridges and arches. Use this resource to learn about parabolas and how to sketch them.\r\n\r\nA quadratic function has the form:\r\n\\[y=ax^{2}+bx+c\\] where \\(a\\neq0\\).\r\nThe graph of a linear function is called a parabola.\r\n\r\nParabolas\r\nTo sketch a parabola, there are three components you must label.\r\n\r\n \tThe \\(y\\)-intercept is where the graph cuts the \\(y\\)-axis (\\(x=0\\)).\r\n \tThe \\(x\\)-intercepts are where the graph cuts the \\(x\\)-axis (\\(y=0\\)).\r\n \tThe turning point is the vertex of the parabola.\r\n\r\nA parabola is a symmetrical about a vertical line that runs through the turning point.\r\n\r\nTurning point\r\nThe \\(a\\) in the quadratic equation tells you some information about the shape of the parabola.\r\n\r\n\tIf \\(a>0\\), the parabola opens upwards and has a minimum turning point.\r\n\tIf \\(a<0\\), the parabola opens downwards and has a maximum turning point. \r\n\r\n\r\nTo find the \\(x\\)-coordinate of the turning point, you substitute \\(b\\) and \\(a\\) into the equation:\r\n\\[x=-\\dfrac{b}{2a}\\]\r\nThen, you simply substitute the value of \\(x\\) into the quadratic equation to find the corresponding \\(y\\)-coordinate.\r\n\r\nYou can also write a quadratic function in turning point form:\r\n\\[y=a(x-h)^{2}+k\\] where \\((h,k)\\) is the turning point.\r\n\r\nExample 1 \u2013 finding the turning point\r\nFind the turning point of \\(y=(x-3)^{2}+4\\).\r\nThe quadratic function has been written in turning point form.\r\n\\[\\begin{align*} y & = a(x-h)^{2}+k\\\\\r\n& = 1(x-3)^{2}+4\r\n\\end{align*}\\]\r\n\\(h=3\\) and \\(k=4\\). The turning point is \\((h,k)=(3,4)\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the turning point\r\nFind the turning point of the graphs of the following functions.\r\n\r\n\t\\(y=(x-1)^{2}+5\\)\r\n\t\\(y=5(x-4)^{2}-12\\)\r\n\t\\(y=(x+2)^{2}+3\\)\r\n\t\\(y=-3(x+5)^{2}-3\\)\r\n\t\\(y=(x-6)^{2}\\)\r\n\t\\(y=-4x^{2}+3\\)\r\n\r\n\r\n\r\n\r\nExample 1 \u2013 graphing quadratic functions\r\nSketch \\(y=x^{2}\\).\r\nFirst, find the \\(x\\)- and \\(y\\)-intercepts. For \\(x\\), let \\(y=0\\).\r\n\\[\\begin{align*} y & = x^{2}\\\\\r\n0 & = x^{2}\\\\\r\nx & = 0\r\n\\end{align*}\\]\r\nFor \\(y\\), let \\(x=0\\).\r\n\\[\\begin{align*} y & = x^{2}\\\\\r\n& = 0^{0}\\\\\r\n& = 0\r\n\\end{align*}\\]\r\nThe graph intercepts the \\(x\\)-axis at \\(x=0\\), and the \\(y\\)-axis at \\(y=0\\). Next, we find the turning point. We can do this by rewriting the function in its turning point form.\r\n\\[\\begin{align*} y & = a(x-h)^{2}+k\\\\\r\n& = 1(x-0)^{2}+0\r\n\\end{align*}\\] \\(h=0\\) and \\(k=0\\), so the turning point occurs at \\((0,0)\\).\r\nSketch the graph, making sure to label the \\(x\\)- and \\(y\\)-intercepts, and the turning point.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 graphing quadratic functions\r\n\r\n\tSketch graphs of the following.\r\n\t\r\n\t\t\\(y=x^{2}-7\\)\r\n\t\t\\(y=(x-2)^{2}+1\\)\r\n\t\t\\(y=4-(x+3)^{2}\\)\r\n\t\t\\(y=(x-2)^{2}\\)\r\n\t\t\\(y=-(x-1)^{2}-1\\)\r\n\t\r\n\tSketch graphs of the following.\r\n\t\r\n\t\t\\(y=x^{2}-x-6\\)\r\n\t\t\\(y=-x^{2}-2x+8\\)\r\n\t\t\\(y=x^{2}-4x\\)\r\n\t\t\\(y=-2x^{2}-6x\\)\r\n\t\t\\(y=x^{2}-9\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.053},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":215,"$":[{"v":"Graphs and transformations","n":0.577},{"v":"Simple functions and relations can be transformed into more complicated functions. Seeing how changes in equations affect their visual representation will give you a deeper understanding of functions. This is useful in modelling real-world situations and solving complex maths problems. Use this resource to learn about transformations of graphs.\r\n\r\nSummary of basic graphs\r\nYou can always plot some points to determine the graph of a function, but it is very useful to know the graphs of some simple functions.\r\nGraphs of linear functions and quadratic functions have already been covered. Here is a brief overview of cubic, reciprocal, exponential and logarithmic functions.\r\n\r\nGraphs of cubic functions\r\nCubic functions have the general form:\r\n\\[y=ax^{3}+bx^{2}+cx+d\\]\r\nwhere \\(a\\), \\(b\\), \\(c\\) and \\(d\\) are constants.\r\nThe graph for \\(y=x^{3}\\) goes through the origin \\((0,0)\\). For positive \\(x\\) values, the graph increases. For negative \\(x\\) values, the graph decreases.\r\n\r\n\r\nGraphs of reciprocal functions\r\nReciprocal functions have the general form:\r\n\\[y=\\dfrac{a}{x-h}+k\\]\r\nwhere \\(a\\), \\(h\\) and \\(k\\) are constants.\r\nThe graph for \\(y=\\dfrac{1}{y}\\) has asymptotes at \\(x=0\\) and \\(y=0\\). This means that the graph approaches the \\(x\\)- and \\(y\\)-axes but never actually touches them. This goes on forever.\r\n\r\n\r\nGraphs of exponential functions\r\nExponential functions have the general form:\r\n\\[y=ab^{x}\\]\r\nwhere \\(a\\) is the \\(y\\)-intercept and \\(b\\) is the base that determines the rate of growth or decay. When \\(b>1\\), the function represents exponential growth. When \\(0<b<1\\), the function represents exponential decay.\r\nThe graph for \\(y=1.6^{x}\\) is shown. The negative \\(x\\)-axis is an asymptote; it approaches but never reaches \\(y=0\\).\r\n\r\n\r\nGraphs of logarithmic functions\r\nLogarithmic functions have the general form:\r\n\\[y=a\\log(b(x-h))+k\\]\r\nwhere \\(a\\), \\(b\\), \\(h\\) and \\(k\\) are constants.\r\nThe graph of \\(y=\\log(x)\\) has an asymptote at the \\(y\\)-axis; it approaches but never reaches \\(x=0\\). It intercepts the \\(x\\)-axis at \\(1\\) since \\(\\log(1)=0\\). \r\n\r\nThe graphs for \\(\\ln(x)=\\log_{e}(x)\\) and any other logarithmic function look similar to this.\r\n\r\nTransformations\r\nTransformations of graphs involves changing the position, size, or shape of a graph by applying various modifications. These transformations can include:\r\n\r\n\treflections, where the graph is flipped over an axis\r\n\ttranslations, where the graph is shifted up, down, left, or right\r\n\tdilations, where the graph is stretched or compressed (squashed).\r\n\r\nGraphs can also have a combination of these transformations applied.\r\n\r\nReflection\r\nA reflection is a mirror image of a graph. We can reflect a graph about any line but generally, this involves the \\(x\\)- or \\(y\\)-axes.\r\nConsider a graph given by \\(y=f(x)\\).\r\n\r\n\t\\(y=f(-x)\\) reflects the graph about the \\(y\\)-axis.\r\n\t\\(y=-f(x)\\) reflects the graph about the \\(x\\)-axis.\r\n\r\n\r\nExample 1 \u2013\u00a0reflecting graphs\r\nReflect the graph of \\(y=a^{x}\\):\r\n\r\n\tabout the \\(x\\)-axis\r\n\tabout the \\(y\\)-axis.\r\n\r\nThe graph of \\(y=a^{x}\\) is shown in blue. \r\nReflection about the \\(x\\)-axis is given by \\(y=a^{-x}\\) in red. Reflection about the \\(y\\)-axis is given by \\(y=-a^{x}\\) in green.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nTranslation\r\nTranslation involves shifting a graph horizontally or vertically along the \\(x-y\\) plane.\r\nFor horizontal translation:\r\n\r\n\tthe graph of \\(y=f(x-a)\\) is a shift of \\(y=f(x)\\) \\(a\\) units to the right\r\n\tthe graph of \\(y=f(x+a)\\) is a shift of \\(y=f(x)\\) \\(a\\) units to the left.\r\n\r\nFor vertical translation:\r\n\r\n\tthe graph of \\(y=f(x)+b\\) is a shift of \\(y=f(x)\\) \\(b\\) units up\r\n\tthe graph of \\(y=f(x)-b\\) is a shift of \\(y=f(x)\\) \\(b\\) units down.\r\n\r\n\r\nExample 1 \u2013 translating graphs\r\nTranslate \\(y=\\log(x)\\), \\(1\\) unit to the right.\r\nThe graph of \\(y=\\log(x)\\) is shown in blue. It has an asymptote at the \\(y\\)-axis and an \\(x\\)-intercept at \\(x=1\\).\r\nTranslation \\(1\\) unit to the right gives \\(y=\\log(x-1)\\) in red. This shifts the asymptote to \\(x=1\\) and the \\(x\\)-intercept to \\(x=2\\). When the asymptote is not at an axis, we show it using a dashed line.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nDilation\r\nDilations stretch or compress a graph of a function. The stretching or compression can occur in the directions of either the \\(x\\)-axis, \\(y\\)-axis, or both.\r\n\r\n\tThe graph of \\(y=af(x)\\) is a dilation of the graph of \\(y=f(x)\\) by a factor of \\(a\\) units in the direction parallel to the \\(y\\)-axis. If \\(a>1\\), the graph is stretched. If \\(a<1\\), the graph is compressed or squashed.\r\n\tThe graph of \\(y=f(bx)\\) is a dilation of the graph of \\(y=f(x)\\) by a factor of \\(\\dfrac{1}{b}\\) units in the direction parallel to the \\(x\\)-axis. If \\(b>1\\), the graph is compressed or squashed. If \\(b<1\\), the graph is stretched.\r\n\r\n\r\nExample \u2013 dilating graphs\r\nDilate \\(y=x^{2}\\) parallel to the \\(y\\)-axis by a factor of:\r\n\r\n\t\\(4\\)\r\n\t\\(\\dfrac{1}{3}\\).\r\n\r\nThe graph of \\(y=x^{2}\\) is shown in blue. It has a turning point at \\((0,0)\\).\r\nDilation parallel to the \\(y\\)-axis by a factor of \\(4\\) makes the parabola appear narrower and is given by \\(y=4x^{2}\\) in red. Dilation parallel to the \\(x\\)-axis by a factor of \\(\\dfrac{1}{3}\\) stretches the graph and is given by \\(y=\\dfrac{1}{3}x^{2}\\) in green.\r\nFor both dilated graphs, the turning point remains unchanged at \\((0,0)\\), since no translation has occurred.\r\n\r\n\r\nCombining transformations\r\nYou can get more complicated graphs by combining reflection, translation and dilation. You always start with the basic graph, then do the dilation first (if there is one), followed by translation or reflection (in any order).\r\n\r\nExample 1 \u2013 combining transformations\r\nGraph \\(y=2x^{3}+4\\).\r\nStart by identifying the basic graph. This one is a cubic function, so the basic form is \\(y=x^{3}\\).\r\n\r\nThen, we look to see what transformations have been applied. \r\n\r\n\tThe \\(2\\) in front of the \\(x^{3}\\) is a dilation by a factor of \\(2\\) parallel to the \\(y\\)-axis. It causes the graph to compress.\r\n\tThe \\(+4\\) is a translation of \\(4\\) units up. It causes the graph to shift up \\(4\\) units.\r\n\r\nRemember that we always do the dilation first. The graph of \\(y=2x^{3}\\) is shown with a red line.\r\n\r\nFinally, we translate the graph to get \\(y=2x^{3}+4\\). This is shown in green.\r\n\r\nThe only thing left to do is find the \\(x\\)-intercept for the new graph. Let \\(y=0\\).\r\n\\[\\begin{align*} 2x^{3}+4 & = 0\\\\\r\n2x^{3} & = -4\\\\ \r\nx^{3} & = -2\\\\\r\nx & = -1.26\r\n\\end{align*}\\]\r\nWe can add this information to the final graph.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 transforming graphs\r\nSketch the following graphs.\r\n\r\n\t\\(y=\\dfrac{2}{x+1}\\)\r\n\t\\(y=\\log(x+3)\\)\r\n\t\\(y=x^{2}-25\\)\r\n\t\\(y=2-(x+1)^{2}\\)\r\n\t\\(y=4^{x}\\)\r\n\t\\(y=2e^{x}+1\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.033},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":216,"$":[{"v":"Solving simultaneous equations","n":0.577},{"v":"\r\nDelve into solving simultaneous equations, where a solution works for multiple equations. This skill is essential for analysing systems with multiple variables and is widely used in fields such as engineering, economics, and science.\r\n\r\nSimultaneous equations\r\nSimultaneous equations are equations that share variables and must be solved at the same time. A pair of simultaneous equations therefore consists of two algebraic equations that share two variables, and must be solved together.\r\nThere are three methods for solving a pair of simultaneous equations using algebra:\r\n\r\n\telimination\r\n\tsubstitution\r\n\tgraphical.\r\n\r\nThis resource shows you how to solve the same pair of simultaneous equations using the three different methods.\r\n\r\nElimination method\r\nIn the elimination method, we manipulate the equations so that one of the variables has coefficients that are equal or opposite in sign. To do this:\r\n\r\n\r\n\tMultiply one or both of the equations by a constant to make the coefficients the same.\r\n\tIf the sign is opposite, add the equations to get rid of the variable.\r\n\tIf the sign is the same, subtract the equations to get rid of a variable.\r\n\tSolve for the remaining variable, then substitute the value into an original equation to solve for the other variable.\r\n\r\n\r\n\r\nExample 1 \u2013 solving simultaneous equations by elimination\r\nSolve the simultaneous equations \\(x+y=6\\) and \\(x-y=2\\).\r\nWe can write the equations on top of each other, and call them \\((1)\\) and \\((2)\\).\r\n\\[\\begin{align*} x+y & = 6\\quad(1)\\\\\r\nx-y & = 2\\quad(2)\r\n\\end{align*}\\]\r\nChoose one variable and multiply one or both equations by a constant to make the coefficient in front of the variable equal or opposite in sign. In this case, \\(y\\) has a coefficient of \\(1\\) in both equations, and is opposite in sign: \\(+\\) in \\((1)\\) and \\(-\\) in \\((2)\\).\r\nTo eliminate the variable, we need to add the equations. That is, we do \\((1)+(2)\\). Add the terms on the LHS together, and add the terms on the RHS together.\r\n\\[\\begin{align*} x+y+x-y & = 6+2\\quad(1)+(2)\\\\\r\n2x & = 8\\\\\r\nx & = 4\r\n\\end{align*}\\]\r\nTo solve for the other variable, we substitute \\(x=4\\) back into equation \\((1)\\) or \\((2)\\).\r\n\\[\\begin{align*} x+y & = 6\\quad\\textrm{equation (1)}\\\\\r\n(4)+y & = 6\\\\\r\ny & = 2\r\n\\end{align*}\\]\r\nThe final answer is \\(x=4\\) and \\(y=2\\). We can write this in coordinate or ordered pair form: \\((x,y)\\), which would be \\((4,2)\\).\r\nWe can check the solution by substituting in the original equations. For equation \\((1)\\):\r\n\\[\\begin{align*} x+y & = 6\\quad\\textrm{equation (1)}\\\\\r\n\\textrm{LHS} & = (4)+(2)\\\\\r\n& = 6\\\\\r\n& = \\textrm{RHS}\r\n\\end{align*}\\]\r\nFor equation \\((2)\\):\r\n\\[\\begin{align*} x-y & = 2\\quad\\textrm{equation (2)}\\\\\r\n\\textrm{LHS} & = (4)-(2)\\\\\r\n& = 2\\\\\r\n& = \\textrm{RHS}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nSubstitution method\r\nIn the substitution method, one equation is transposed to isolate one variable, and this is substituted into the other equation. To do this:\r\n\r\n\r\n\tSelect the simpler equation and transpose it to make one of the variables the subject.\r\n\tSubstitute the transposed equation into the second equation.\r\n\tSolve for the remaining variable, then substitute the value into an original equation to solve for the other variable.\r\n\r\n\r\n\r\nExample 1 \u2013 solving simultaneous equations by substitution\r\nSolve the simultaneous equations \\(x+y=6\\) and \\(x-y=2\\).\r\nWe can write the equations on top of each other, and call them \\((1)\\) and \\((2)\\).\r\n\\[\\begin{align*} x+y & =6\\quad(1)\\\\\r\nx-y & = 2\\quad(2)\r\n\\end{align*}\\]\r\nChoose the simpler equation and transpose it to make one of the variables the subject. These two are equally simple, so let's just go with \\(x\\) in equation \\((1)\\).\r\n\\[\\begin{align*} x+y & = 6\\quad(1)\\\\\r\nx & = 6-y\r\n\\end{align*}\\]\r\nSubstitute the transposed equation into the second equation. Solve for the remaining variable.\r\n\\[\\begin{align*} x-y & = 2\\quad(2)\\\\\r\n(6-y)-y & = 2\\\\\r\n6-2y & = 2\\\\\r\n-2y & = -4\\\\\r\ny & = 2\r\n\\end{align*}\\]\r\nSubstitute the value into an original equation to solve for the other variable. Here, we are substituting \\(y=2\\) into equation \\((1))\\).\r\n\\[\\begin{align*} x+y & = 6\\quad(1)\\\\\r\nx+(2) & = 6\\\\\r\nx & = 4\r\n\\end{align*}\\]\r\nThe final answer is \\(x=4\\) and \\(y=2\\). We can write this in coordinate or ordered pair form: \\((x,y)\\), which would be \\((4,2)\\).\r\nAs usual, you can substitute these values back into the original equations to check that your answers work.\r\n\r\n\r\n\r\n\r\nGraphical method\r\nIn the graphical method, the solution is the coordinates of the intersection between the two linear graphs. To find this:\r\n\r\n\r\n\tPlot the straight line graphs representing each equation on the same set of axes.\r\n\tFind the intersection of the two lines.\r\n\r\n\r\nThe graphs must be plotted accurately for this method to give an accurate answer. For this reason it is not usually used unless specifically asked for.\r\n\r\nExample \u2013 solving simultaneous equations using graphs\r\nSolve the simultaneous equations \\(x+y=6\\) and \\(x-y=2\\) graphically.\r\nGraphs representing each linear equation are drawn and the point of intersection is read from the graph.\r\n\r\nThe solution is \\((4,2)\\).\r\nAgain, you can check your answer by substituting the values into the equations.\r\n\r\nEquations without solutions\r\nIf the solution to a pair of simultaneous equations is the intersection of their graphs, it would make sense that some pairs of equations do not have a solution. Graphically, this is when the two lines do not intersect.\r\n\r\nExample \u2013 solving simultaneous equations\r\nSolve the simultaneous equations \\(2x-y=6\\) and \\(-4x+2y=-7\\).\r\nWe can write the equations on top of each other, and call them \\((1)\\) and \\((2)\\).\r\n\\[\\begin{align*} 2x-y & = 6\\quad(1)\\\\\r\n-4x+2y & = -7\\quad(2)\r\n\\end{align*}\\]\r\nYou can use any of the three methods above to try and solve the equations. We will use substitution. Choose the simpler equation and transpose it to make one of the variables the subject. Let's go with \\(y\\) in equation \\((1)\\).\r\n\\[\\begin{align*} 2x-y & = 6\\quad(1)\\\\\r\ny & = 2x-6\r\n\\end{align*}\\]\r\nSubstitute the transposed equation into the second equation. Solve for the remaining variable.\r\n\\[\\begin{align*} -4x+2y & = -7\\quad(2)\\\\\r\n-4x+2(2x-6) & = -7\\\\\r\n-4x+4x-12 & = -7\\\\\r\n-12 & = -7\r\n\\end{align*}\\]\r\nThe last line is not possible, so we can conclude that no solution exists.\r\nWe show see graphically that the two lines are parallel and never intersect, so no solution exists.\r\n\r\n\r\n\r\n\r\nEquations with multiple solutions\r\nSome pairs of simultaneous equations have multiple solutions. In this case, the equations actually have the same graph. That is, one line lies on the other line.\r\n\r\nExample \u2013 solving simultaneous equations\r\nSolve \\(y=1-3x\\) and \\(6x=2(1-y)\\).\r\nRearrange the equations so that \\(x\\) and \\(y\\) are on the same sides and number the equations.\r\n\\[\\begin{align*} 3x+y & = 1\\quad(1)\\\\\r\n6x+2y & = 2\\quad(2)\r\n\\end{align*}\\]\r\nThese two lines are the same because \\((2)=2\\times(1)\\). This means that they have the same graph and there are multiple values of \\(x\\) and \\(y\\) that are possible solutions.\r\nFor example, if we make \\(x=0\\):\r\n\\[\\begin{align*} 3x+y & = 1\\quad(1)\\\\\r\n3(0)+y & = 1\\\\\r\ny & = 1\r\n\\end{align*}\\]\r\nSo, \\((0,1)\\) is a solution.\r\nIf we make \\(y=0\\):\r\n\\[\\begin{align*} 3x+y & = 1\\quad(1)\\\\\r\n3x+(0) & = 1\\\\\r\nx & = \\frac{1}{3}\r\n\\end{align*}\\]\r\nSo, \\((\\dfrac{1}{3},0)\\) is also a solution.\r\nFor identical graphs, there are actually an infinite number of solutions, each corresponding to a point on the line \\(3x+y=1\\).\r\n\r\nExercise \u2013 solving simultaneous equations\r\n\r\n\tSolve the following simultaneous equations, using either the elimination or substitution method.\r\n\t\r\n\t\t\\(y=x-3\\) and \\(5x-2y=18\\)\r\n\t\t\\(y=3x+4\\) and \\(2x+3y=23\\)\r\n\t\t\\(4x+y=23\\) and \\(x-y=2\\)\r\n\t\t\\(-3x+2y=-4\\) and \\(5x-2y=8\\)\r\n\t\t\\(3x-5y=14\\) and \\(2x+y=5\\)\r\n\t\t\\(3x+2y=18\\) and \\(2x-5y=-7\\)\r\n\t\r\n\tSolve the following simultaneous equations graphically.\r\n\t\r\n\t\t\\(y=x-3\\) and \\(x+y=5\\)\r\n\t\t\\(5x-y=8\\) and \\(2y-3x=-2\\)\r\n\t\t\\(x+y=1\\) and \\(y-2x=7\\)\r\n\t\r\n\r\n\r\n\r\n","n":0.03},[{"v":"Maths equations","i":2,"n":0.707},{"v":"Maths","i":1,"n":1},{"v":"Equations","i":0,"n":1}]]},{"i":217,"$":[{"v":"Quadratic formula","n":0.707},{"v":"The quadratic formula provides a straightforward way to find the solutions of any quadratic equation, especially when factoring is difficult or impossible. By becoming familiar with this formula, you'll gain a reliable tool for solving quadratic equations and deepen your understanding of algebraic relationships.\r\n\r\nSolving quadratic equations\r\nThe solutions to any quadratic equation with the form \\(ax^{2}+bx+c=0\\) can be found by substituting the values \\(a\\), \\(b\\) and \\(c\\) into the quadratic formula:\r\n\\[ x=\\frac{-b\\pm\\sqrt{b^{2}-4ac}}{2a}\\]\r\n\r\nDiscriminant\r\nSolutions to quadratic equations may be real or complex numbers. Here, we will only consider real solutions.\r\nRemember from quadratic factorisation that the discriminant (\\(\\Delta)\\) can be used to determine the number and type of solutions.\r\n\r\n\tIf \\(\\Delta<0\\), there are no (real) solutions. The graph of \\(y=ax^{2}+bx+c\\) does not touch or cut the \\(x\\)-axis.\r\n\tIf \\(\\Delta=0\\), there is one solution. The graph of \\(y=ax^{2}+bx+c\\) touches the \\(x\\)-axis at the value of \\(x\\).\r\n\tIf \\(\\Delta>0\\), there are two solutions. The graph of \\(y=ax^{2}+bx+c\\) cuts the \\(x\\)-axis at these values of \\(x\\).\r\n\tIf \\(b^{2}-4ac\\) is a perfect square, the solutions will be rational. This means that the values of \\(x\\) can be expressed as a fraction \\(\\dfrac{p}{q}\\), where \\(p\\) and \\(q\\neq0\\) are integers.\r\n\r\n\r\nExample 1 \u2013 solving quadratic equations\r\nSolve the quadratic equation:\r\n\\[x^{2}-5x+4=0\\]\r\nIdentify \\(a\\), \\(b\\) and \\(c\\).\r\n\\[a=1,\\,b=-5,\\,c=4\\]\r\nSubstitute into the quadratic formula:\r\n\\[\\begin{align*} x & = \\frac{-b\\pm\\sqrt{b^{2}-4ac}}{2a}\\\\\r\n& = \\frac{-(-5)\\pm\\sqrt{(-5)^{2}-4(1)(4)}}{2(1)}\\\\\r\n& = \\frac{5\\pm\\sqrt{25-16}}{2}\\\\\r\n& = \\frac{5\\pm\\sqrt{9}}{2}\\\\ \r\n& = \\frac{5\\pm3}{2}\\\\ \r\n& = \\frac{5+3}{2}\\ \\textrm{ or }\\ \\frac{5-3}{2}\\\\ \r\n& = \\frac{8}{2}\\ \\textrm{ or }\\ \\frac{2}{2}\\\\\r\n& = 4\\ \\textrm{ or }\\ 1\r\n\\end{align*}\\]\r\nThe solution is \\(x=1\\)\u00a0or \\(x=4\\).\r\n\r\n\r\n\r\nExercise \u2013 solving quadratic equations\r\nFind the real solutions to the following quadratic equations.\r\n\r\n\t\\(x^{2}-4x-7=0\\)\r\n\t\\(x^{2}-2x-2=0\\)\r\n\t\\(x^{2}+6x-9=0\\)\r\n\t\\(x^{2}-x-7=0\\)\r\n\t\\(4x^{2}-12x+6=0\\)\r\n\t\\(2x^{2}+x-2=0\\)\r\n\t\\(x^{2}-4x+2=0\\)\r\n\t\\(2x^{2}-3x+2=0\\)\r\n\t\\(3x^{2}+5x-7=0\\)\r\n\t\\(3x^{2}=x+1\\)\r\n\t\\(4x^{2}+x+3=0\\)\r\n\t\\(\\dfrac{3x+1}{2}=\\dfrac{x+1}{x}\\)\r\n\r\n\r\n\r\n\r\n\r\n","n":0.063},[{"v":"Maths equations","i":2,"n":0.707},{"v":"Maths","i":1,"n":1},{"v":"Equations","i":0,"n":1}]]},{"i":218,"$":[{"v":"Solving cubic equations","n":0.577},{"v":"Learn how to solve cubic equations, where the highest power of the variable is three. These types of equations can model various real-world scenarios, from physics to engineering.\r\n\r\nThe factor theorem\r\nThe long division of polynomials is an important tool for solving cubic equations. However, the factor theorem is equally important. It is a trial-and-error approach to solving polynomial equations.\r\nThe factor theorem states that \\((x-a)\\) is a factor of the equation \\(f(x)\\) if \\(f(a)=0\\).\r\nIn other words, if we substitute \\(a\\) in place of the variable (e.g. \\(x\\)) in the polynomial and the answer is \\(0\\), then \\((x-a)\\) is a factor.\r\nWe can use the factor theorem to find one factor of a polynomial, and then use polynomial long division to find the remaining factor(s).\r\nThis will all make more sense once we look at some examples.\r\n\r\nExample \u2013 using the factor theorem\r\nSolve the equation \\(x^{3}-5x^{2}-2x+24=0\\).\r\nUse the factor theorem to find a linear factor of \\(x^{3}-5x^{2}-2x+24\\). We let:\r\n\\[ f(x)=x^{3}-5x^{2}-2x+24 \\]\r\nWe can try \\(x=1\\) first.\r\n\\[\\begin{align*} f(1) & = (1)^{3}-5(1)^{2}-2(1)+24\\\\\r\n& = 1-5-2+24\\\\\r\n& = 18\\\\\r\n& \\neq 0\r\n\\end{align*}\\]\r\nSince \\(f(1)\\neq0\\), \\((x-1)\\) is not a factor of the polynomial.\r\nWe can try \\(x=-1\\), too.\r\n\\[\\begin{align*} f(-1) & = (-1)^{3}-5(-1)^{2}-2(-1)+24\\\\\r\n& = -1-5+2+24\\\\\r\n& = 20\\\\\r\n& \\neq 0\r\n\\end{align*}\\]\r\nThus, \\((x+1)\\) is also not a factor of the polynomial.\r\nLet's try \\(x=2\\).\r\n\\[\\begin{align*} f(2) & = (2)^{3}-5(2)^{2}-2(2)+24\\\\\r\n& = 8-20-4+24\\\\\r\n& = 8\\\\\r\n& \\neq 0\r\n\\end{align*}\\]\r\n\\((x-2)\\) is not a factor of the polynomial.\r\nHow about \\(x=-2\\)?\r\n\\[\\begin{align*} f(-2) & = (-2)^{3}-5(-2)^{2}-2(-2)+24\\\\\r\n& = -8-20+4+24\\\\\r\n& = 0\r\n\\end{align*}\\]\r\nSince \\(f(-2)=0\\), this means that \\((x+2)\\) is a factor. We can then use polynomial long division to find the other factor. We divide \\(x^{3}-5x^{2}-2x+24\\) by \\(x+2\\).\r\n\\[\\begin{array}{r} x^2-7x+12\\phantom{)} \\\\ \r\nx+2{\\overline{\\smash{\\big)}\\,x^3-5x^2-2x+24\\phantom{)}}} \\\\\r\n\\underline{-(x^3+ 2x^2)\\phantom{bbbbbbbbbi)}} \\\\\r\n0- 7x^2 \\,-2x\\phantom{bbbbi)} \\\\\r\n\\underline{-~\\phantom{()}(-7x^2-\\!14x)}\\phantom{bbbi)} \\\\ \r\n12x+24 \\phantom{)} \\\\ \\underline{-~\\phantom{()}(12x+24)} \\\\ \r\n0+0\\phantom{)} \r\n\\end{array}\\]\r\nThis means that:\r\n\\[x^{3}-5x^{2}-2x+24=(x+2)(x^{2}-7x+12)\\]\r\nThe quadratic factor can also be further factorised:\r\n\\[x^{3}-5x^{2}-2x+24=(x+2)(x-3)(x-4)\\]\r\nAs \\(x^{3}-5x^{2}-2x+24=0\\), we can solve the cubic equation using the null factor law.\r\n\\[\\begin{align*} x+2=0 \\textrm{, } x-3 & = 0 \\textrm{ or } x-4=0\\\\\r\n\\therefore x=-2 \\textrm{, } x & = 3 \\textrm{ or } x=4\r\n\\end{align*}\\]\r\n\r\nExercise \u2013 using the factor theorem\r\nSolve the following cubic equations.\r\n\r\n\t\\(x^{3}+7x^{2}+11x+5=0\\)\r\n\t\\(4x^{3}+2x^{2}-2x=0\\)\r\n\t\\(-x^{3}-3x^{2}+x+3=0\\)\r\n\t\\(x^{3}-7x-6=0\\)\r\n\r\n\r\n\r\n\r\n","n":0.056},[{"v":"Maths equations","i":2,"n":0.707},{"v":"Maths","i":1,"n":1},{"v":"Equations","i":0,"n":1}]]},{"i":219,"$":[{"v":"Solving exponential equations","n":0.577},{"v":"\r\nExponential equations are common in fields such as science, finance, and population dynamics, where exponential growth or decay is observed. By learning how to isolate variables and find solutions to exponential equations, you will be able to handle a variety of real-world problems involving exponential relationships.\r\n\r\nExponential equations\r\nIndicial (or exponential) equations have a variable in the power or exponent. The general form of an exponential equation is: \\(a^{x}=b\\). Examples include:\r\n\r\n\t\\(5^{x}=17\\)\r\n\t\\(2^{1-x}=4^{3x}\\)\r\n\t\\(A=P(1+\\dfrac{r}{n})^{nt}\\)\r\n\r\nIf we can write \\(b\\) as a number with a base \\(a\\) and an index \\(x\\), then we can use a set of rules to find \\(x\\).\r\n\r\n\tIf two equal numbers have a common base \\((a)\\), then the indices \\((x)\\) must also be equal.\r\n\tThe logarithm of both sides can be taken to solve for \\(x\\).\r\n\r\nMathematically, these rules can be represented as:\r\n\r\n\\[\\begin{align*} a^{x} = a^{y} & \\Rightarrow x = y\\quad(1)\\\\\r\na^{x} = b \\Rightarrow & \\,\\log_{y}a^{x} = \\log_{y}b\\quad(2)\r\n\\end{align*}\\]\r\n\r\nThe rule you use to solve the exponential equation depends on whether it has integer or non-integer solutions.\r\n\r\nInteger solutions\r\nThe first rule comes in handy for exponential equations that have integer solutions. You can quickly tell whether to use  rule \\(1\\) by checking whether you can write the number on the other side of the equal sign, i.e. the \\(b\\) component, as an exponential with the same base. Let's look at some examples.\r\n\r\nExample 1 \u2013 solving exponential equations with integer solutions\r\nSolve \\(3^{x}=27\\).\r\nWe should first check whether \\(27\\) can be written to the same base as the exponential expression. In this case, the base is \\(3\\), so we need to check if \\(27\\) can be written as an exponential with the same base. It can be written as \\(27=3^{3}\\). Thus:\r\n\\[\\begin{align*} 3^{x} & = 27\\\\\r\n3^{x} & = 3^{3}\r\n\\end{align*}\\]\r\nThese expressions have the same base, so their indices must be equal.\r\n\\[x=3\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nNon-integer solutions\r\nNot all equations have integer solutions. That is, the solutions may not be whole numbers. For example: \\(3^{x}=10\\) has a solution between \\(2\\) and \\(3\\), since \\(3^{2}=9\\) and \\(3^{3}=27\\).\r\nHere, we must use rule \\(2\\) to solve the equation.\r\n\r\nRemember that for \\(a^{x}=b\\):\r\n\\[\\log_{y}a^{x} = \\log_{y}b\\Rightarrow x\\log_{y}a = \\log_{y}b\\]\r\n\r\nThese types of equations can be solved using a calculator that can do logarithms with base \\(10\\) or base \\(e\\). This means that \\(y=10\\) or \\(y=e\\).\r\nOn your calculator, the LOG button will calculate \\(\\log_{10}b\\) and the LN button will calculate \\(\\log_{e}b\\). Logarithms with base \\(e\\) are known as natural logarithms and sometimes the abbreviation \\(\\ln b\\) is used for \\(\\log_{e}b\\).\r\n\r\nExample 1 \u2013 solving exponential equations with non-integer solutions\r\nSolve \\(3^{x}=10\\) to three decimal places.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 solving exponential equations\r\nSolve for \\(x\\).\r\n\r\n\t\\(3^{1-x}=27\\)\r\n\t\\(2^{2x-1}=128\\)\r\n\t\\(9^{\\frac{1}{x}}=3^{-4}\\)\r\n\t\\(5^{x}=12\\)\r\n\t\\(2^{x-3}=9\\)\r\n\t\\(\\dfrac{1}{2^{x+1}}=4^{x+2}\\)\r\n\t\\(\\log_{10}x=\\log_{10}4-\\log_{10}2\\)\r\n\t\\(\\log_{2}2x-2\\log_{2}3=\\log_{2}6\\)\r\n\t\\(10^{x+1}=7\\)\r\n\r\n\r\n\r\n\r\nGrowth and decay\r\nGrowth and decay are often modelled using exponential and logarithmic expressions, across all branches of STEM.\r\n\r\nExample \u2013 solving problems involving growth and decay\r\nThe number of bacteria present in a sample is given by \\(N=800e^{0.2t}\\), where \\(t\\) is the time in seconds. Find:\r\n\r\n\tthe initial number of bacteria\r\n\tthe time taken for the bacteria count to reach \\(10000\\).\r\n\r\nFor part a, substitute \\(t=0\\) into the equation.\r\n\\[\\begin{align*} N & = 800e^{0.2t}\\\\\r\n& = 800e^{0.2\\times0}\\\\\r\n& = 800e^{0}\\\\\r\n& = 800\r\n\\end{align*}\\]\r\nThe initial bacteria count is 800.\r\nFor part b, substitute \\(N=10000\\) into the equation.\r\n\r\nIt takes \\(12.6\\) sec for the number of bacteria to reach \\(10000\\).\r\n\r\nExercise \u2013 solving problems involving growth and decay\r\n\r\n\tThe decay rate \\(R\\) for a radioactive element is given by \\(R=400e^{-0.03t}\\), where \\(t\\) is measured in seconds. Find:\r\n\t\r\n\t\tthe initial decay rate\r\n\t\tthe time for the decay rate to reduce to half of the initial decay rate.\r\n\t\r\n\tThe charge units \\(Q\\) on the plate of a condenser \\(t\\) seconds after it starts to discharge is given by \\(Q=Q_{0}10^{-kt}\\), where \\(Q_{0}\\) is the initial charge, \\(t\\) is the time in seconds, and \\(k\\) is a constant.\r\n\tIf the intial charge is \\(5076\\) units and \\(Q=1840\\) when \\(t=0.5\\) seconds, find:\r\n\t\r\n\t\tthe value of \\(k\\)\r\n\t\tthe time needed for the charge to fall to \\(1000\\) units\r\n\t\tthe charge after \\(2\\) sec.\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.04},[{"v":"Maths equations","i":2,"n":0.707},{"v":"Maths","i":1,"n":1},{"v":"Equations","i":0,"n":1}]]},{"i":220,"$":[{"v":"Partial fractions","n":0.707},{"v":"\r\nPartial fractions are used to simplify more complex algebraic fractions into simpler components. The ability to find partial fractions is used widely in calculus, like when solving problems involving Laplace transforms in engineering, analysing frequency response in electronics, and simplifying expressions in mathematical models. Use this resource to learn to find partial fractions.\r\n\r\nThe opposite of adding two fractions is to split one more complex fraction into partial fractions. Therefore, a partial fraction is an algebraic fraction broken into simpler parts that are added together.\r\nFor example:\r\n\\[\\frac{9x-7}{(3x+4)(x-5)}=\\frac{3}{3x+4}+\\dfrac{2}{x-5}\\]\r\nThe partial fractions are \\(\\dfrac{3}{3x+4}\\) and \\(\\dfrac{2}{x-5}\\).\r\nPartial fractions are particularly useful in calculus, like when the denominator consists of linear or quadratic factors. Understanding how to find partial fractions is important for Integration using partial fractions.\r\n\r\nFinding partial fractions\r\nTo express an algebraic fraction as partial fractions:\r\n\r\n \tFactorise the denominator. They may be:\r\n\t\r\n\t\tdistinct linear factors: \r\n\t\t\\[\\dfrac{mx+k}{(x+a)(x+b)}=\\dfrac{A}{x+a}+\\dfrac{B}{x+b}\\]\r\n\t\trepeated linear factors: \r\n\t\t\\[\\dfrac{mx+k}{(x+a)^{2}}=\\dfrac{A}{x+a}+\\dfrac{B}{(x+a)^{2}}\\]\r\n\t\tquadratic and linear factors: \r\n\t\t\\[\\dfrac{mx+k}{(ax^{2}+bx+c)(px+q)}=\\dfrac{Ax+B}{ax^{2}+bx+c}+\\dfrac{C}{px+q}\\]\r\n\t\r\n \tWrite the algebraic fraction as the sum of partial fractions with unknown constants (e.g. \\(A\\), \\(B\\), \\(C\\)).\r\n \tAdd the partial fractions by getting a common denominator.\r\n \tEquate the coefficients and use simultaneous equations, or substitute a value of \\(x\\) to determine the values of the constants.\r\n\r\n\r\nConditions for partial fractions\r\nYou can ONLY find partial fractions when the degree of the numerator is less than the degree of the denominator\u2014that is, the greatest power of \\(x\\) on the top must be less than the greatest power of \\(x\\) on the bottom.\r\nYou may need to divide the denominator into the numerator and then express the remaining fractional part as partial fractions.\r\nFor example:\r\n\\[\\begin{align*} \\frac{x^{2}+7x+7}{x^{2}+3x+2} & = 1+\\frac{4x+5}{x^{2}+3x+2}\\\\ \r\n& = 1+\\frac{4x+5}{(x+2)(x+1)}\\\\ \r\n& = 1+\\frac{3}{x+2}+\\frac{1}{x+1}\r\n\\end{align*}\\]\r\n\r\nExample 1 \u2013 finding partial fractions\r\nExpress \\(\\dfrac{x-5}{x^{2}+2x-3}\\) as a sum of partial fractions.\r\nFirst, we factorise the denominator.\r\n\\[\\frac{x-5}{x^{2}+2x-3}=\\frac{x-5}{(x-1)(x+3)}\\]\r\nHere, we have distinct linear factors. We can separate them out with constants \\(A\\) and \\(B\\) to find the generalised partial fraction form:\r\n\\[\\begin{align*} \\frac{x-5}{x^{2}+2x-3} & = \\frac{x-5}{(x-1)(x+3)}\\\\ \r\n& = \\frac{A}{x-1}+\\frac{B}{x+3}\r\n\\end{align*}\\]\r\nNext, we combine them again by cross-multiplying.\r\n\\[\\begin{align*} \\frac{A}{x-1}+\\frac{B}{x+3} & = \\frac{A(x+3)}{(x-1)(x+3)}+\\frac{B(x-1)}{(x-1)(x+3)}\\\\\r\n& = \\frac{A(x+3)+B(x-1)}{(x-1)(x+3)}\r\n\\end{align*}\\]\r\nWe can equate the numerator of the original fraction with our new numerator written in terms of \\(A\\) and \\(B\\).\r\n\\[\\begin{align*} x-5 & = A(x+3)+B(x-1)\\\\ \r\n& = Ax+3A+Bx-B\\\\\r\n& = (A+B)x+(3A-B)\r\n\\end{align*}\\]\r\nAgain, we can equate the terms involving \\(x\\) and the constants on the left- and right-hand side.\r\n\\[\\begin{align*} x & = (A+B)x\\\\ \r\n1 & = A+B\\textrm{assuming }x\\neq0\r\n\\end{align*}\\]\\[-5 = 3A-B\\]\r\nWe have two simultaneous equations that we can solve. We will look at just one method of solving them. \r\nFirst, let's add the two equations.\r\n\\[\\begin{align*} -4 & = 4A\\\\\r\n4A & = -4\\\\\r\nA & = -1\r\n\\end{align*}\\]\r\nSubstituting this result in the second equation:\r\n\\[\\begin{align*} 1 & = -1+B\\\\\r\nB & = 2\r\n\\end{align*}\\]\r\nWe can then substitute \\(A=-1\\) and \\(B=2\\) into the generalised partial fraction form:\r\n\\[\\frac{x-5}{x^{2}+2x-3}=\\frac{-1}{x-1}+\\frac{2}{x+3}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding partial fractions\r\n\r\n\tRewrite each of the following in the appropriate generalised partial fraction form. You do not need to calculate the constants.\r\n\t\r\n\t\t\\(\\dfrac{x+6}{2x^{2}+5x-12}\\)\r\n\t\t\\(\\dfrac{2x}{(x^{2}+3)(x+1)}\\)\r\n\t\t\\(\\dfrac{2x}{x^{2}+8x+16}\\)\r\n\t\t\\(\\dfrac{x^{2}}{x^{2}-1}\\)\r\n\t\r\n\tExpress the following as partial fractions.\r\n\t\r\n\t\t\\(\\dfrac{x+2}{x^{2}-5x+6}\\)\r\n\t\t\\(\\dfrac{3}{x^{2}-2x+1}\\)\r\n\t\t\\(\\dfrac{x^{2}-2x+2}{x^{3}+x^{2}+x}\\)\r\n\t\t\\(\\dfrac{x^{2}}{x^{2}-4}\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.046},[{"v":"Maths equations","i":3,"n":0.707},{"v":"Functions","i":2,"n":1},{"v":"Arithmetic","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":221,"$":[{"v":"Perfect squares","n":0.707},{"v":"\r\nUse this resource to explore perfect squares. Understanding perfect squares will help you with recognise patterns in algebraic expressions, making factorising and expanding algebraic expressions much easier.\r\n\r\nA perfect square is an algebraic term that is the square of an integer. In other words, it is a term multiplied by itself. Some examples of perfect squares are:\r\n\r\n\t\\(5^{2}\\)\r\n\t\\(x^{2}\\)\r\n\t\\(a^{2}b^{2}\\)\r\n\t\\((xy)^{2}\\)\r\n\t\\((x+1)^{2}\\)\r\n\r\nHere, we will deal with perfect squares that have the general form:\r\n\\[(x\\pm y)^{2}=x^{2}\\pm2xy+y^{2}\\]\r\nIt is useful to know this form because if a quadratic is a perfect square, it can be very neatly factorised.\r\n\r\nExpanding perfect squares\r\nThere are two rules you should remember when it comes to perfect squares. These will make your life easier when it comes to expanding and factorising algebraic expressions.\r\nThe first is \\((a+b)^{2}\\).\r\n\\[\\begin{align*} (a+b)^{2} & = (a+b)(a+b)\\\\\r\n& = a^{2}+ab+ba+b^{2}\\\\\r\n& = a^{2}+2ab+b^{2}\r\n\\end{align*}\\]\r\nThe equation you should remember is:\r\n\\[(a+b)^{2}=a^{2}+2ab+b^{2}\\]\r\nThe second rule is \\((a-b)^{2}\\).\r\n\\[\\begin{align*} (a-b)^{2} & = (a-b)(a-b)\\\\\r\n& = a^{2}-ab-ba+b^{2}\\\\\r\n& = a^{2}-2ab+b^{2}\r\n\\end{align*}\\]\r\nThe equation you should remember is:\r\n\\[(a-b)^{2}=a^{2}-2ab+b^{2}\\]\r\nIn these two equations:\r\n\r\n\tthe first and last terms must be positive and must be perfect squares\r\n \tthe middle term must be twice the product of the first and last terms and may be positive or negative.\r\n\r\n\r\nUsing perfect squares to factorise\r\nThe rules for expanding perfect squares can be used in reverse to factorise algebraic expressions. Let's look at some examples.\r\n\r\nExample 1 \u2013 using perfect squares to factorise quadratics\r\nFactorise \\(x^{2}+14x+49\\).\r\nTake the square root of \\(x^{2}\\) and \\(49\\). These are \\(x\\) and \\(7\\), respectively. We can confirm whether they are correct by checking that the coefficients \\(a\\) and \\(b\\) work with the equation.\r\n\\[2ab=2\\times x\\times7=14x\\]\r\nThis matches with the \\(2ab\\) term in the quadratic so it is a perfect square. Since the sign in front of the \\(2ab\\) term is positive, it must be the form \\((x+y)^{2}\\). Therefore:\r\n\\[x^{2}+14x+49=(x+7)^{2}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using perfect squares to factorise quadratics \r\nFactorise the following into perfect squares.\r\n\r\n\t\\(a^{2}+2a+1\\)\r\n\t\\(x^{2}-4x+4\\)\r\n\t\\(25x^{2}-10x+1\\)\r\n\t\\(4y^{2}-6y+9\\)\r\n\t\\(81x^{2}+108x+36\\)\r\n\t\\(9a^{2}-24a-16\\)\r\n\t\\(16x^{2}-40xy+25y^{2}\\)\r\n\t\\(121z^{2}+88z+64\\)\r\n\t\\(2x^{2}+8x+8\\)\r\n\r\n\r\n\r\n","n":0.058},[{"v":"Maths equations","i":1,"n":0.707},{"v":"Algebra","i":0,"n":1}]]},{"i":222,"$":[{"v":"Cubic factorisation","n":0.707},{"v":"\r\nCubic polynomials are more challenging to factorise and involve polynomial long division. This method helps break down complex polynomials into simpler components. You'll gain the ability to find factors and simplify cubic equations, which will help you solve higher-degree polynomial equations.\r\n\r\nPolynomials\r\nA polynomial is an algebraic expression that consists of the sum or difference of two or more algebraic terms, where the variables are raised to different powers.\r\nExamples of polynomials are:\r\n\r\n\t\\(x+7=0\\)\r\n\t\\(2x^{2}+9x+3=0\\)\r\n\t\\(7a^{2}-4=0\\)\r\n\t\\(n^{3}-3n+2=0\\)\t\r\n\r\nWhen a polynomial consists of two algebraic terms, it is called a binomial. When it consists of three terms, it is a trinomial, and so on.\r\nPolynomials can also be named according to the power to which the variables are raised.\r\n\r\n\tLinear polynomials are in the form \\(y=ax+b\\).\r\n\tQuadratic polynomials are in the form \\(y=ax^{2}+bx+c\\).\r\n\tCubic polynomials are in the form \\(y=ax^{3}+bx^{2}+cx+d\\).\r\n\r\nLinear and quadratic polynomials are far simpler to factorise that cubic polynomials.\r\n\r\nPolynomial long division\r\nIn the age of calculators, long division of numbers is not so important. However, it is still useful for factorising cubic (and higher degree) polynomials. Go back to the numbers page in arithmetic to review how to do long division.\r\nThe same algorithm (divide, multiply, subtract, bring down, repeat) is used to divide a polynomial by a polynomial of a lower degree.\r\nTo get familiar with the process, let's look at long division of a quadratic polynomial. Then, we can learn how to divide a cubic polynomial.\r\n\r\nExample \u2013 dividing quadratic polynomials using long division\r\nDivide \\(x^{2}-5x+6\\) by \\(x-3\\).\r\nFirst, we set up the long division by writing:\r\n\\[\\begin{array}{r} \\\\ x-3{\\overline{\\smash{\\big)}\\,x^2-5x+6\\phantom{)}}} \\end{array}\\]\r\nNow, we can start the algorithm.\r\nDivide the highest power of the polynomial \\(x^{2}-5x+6\\) by the highest power in the polynomial \\(x-3\\). That is, divide \\(x^{2}\\) by \\(x\\) to get \\(x\\). The \\(x\\) is placed above the \\(5x\\).\r\n\\[\\begin{array}{r} x\\phantom{)-2} \\\\\r\nx-3{\\overline{\\smash{\\big)}\\,x^2-5x+6\\phantom{)}}} \r\n\\end{array}\\]\r\nMultiply \\(x-3\\) by \\(x\\) to get \\(x^{2}-3x\\) and write the result below the quadratic polynomial.\r\n\\[\\begin{array}{r} x\\phantom{)-2} \\\\ \r\nx-3{\\overline{\\smash{\\big)}\\,x^2-5x+6\\phantom{)}}} \\\\ \r\n\\underline{-~\\phantom{(}(x^2-3x)\\phantom{-b)}} \r\n\\end{array}\\]\r\nSubtract \\(x^{2}-3x\\) from \\(x^{2}-5x\\) to get \\(-2x\\).\r\n\\[\\begin{array}{r} x\\phantom{)-2} \\\\ \r\nx-3{\\overline{\\smash{\\big)}\\,x^2-5x+6\\phantom{)}}} \\\\ \r\n\\underline{-~\\phantom{(}(x^2-3x)\\phantom{-b)}} \\\\ \r\n0\\,\\,\\,-2x\\phantom{)+6}\r\n\\end{array}\\]\r\nBring down the \\(6\\).\r\n\\[\\begin{array}{r} x\\phantom{)-2} \\\\ \r\nx-3{\\overline{\\smash{\\big)}\\,x^2-5x+6\\phantom{)}}}\r\n\\\\ \\underline{-~\\phantom{(}(x^2-3x)\\phantom{-b)}} \\\\ \r\n0-2x+6\\phantom{)}\r\n\\end{array}\\]\r\nRepeat as required until the remainder is \\(0\\).\r\nDivide \\(-2x\\) by the leading term in \\(x-3\\) to get \\(-2\\). That is, divide \\(x\\) into \\(-2x\\) to get \\(-2\\) and write the \\(-2\\) on the top line.\r\n\\[\\begin{array}{r} x-2\\phantom{)} \\\\ \r\nx-3{\\overline{\\smash{\\big)}\\,x^2-5x+6\\phantom{)}}} \r\n\\\\ \\underline{-~\\phantom{(}(x^2-3x)\\phantom{-b)}} \\\\ \r\n0-2x+6\\phantom{)} \r\n\\end{array}\\]\r\nMultiply \\(x-3\\) by \\(-2\\) to get \\(-2x+6\\) and write this below.\r\n\\[\\begin{array}{r} x-2\\phantom{)} \\\\\r\nx-3{\\overline{\\smash{\\big)}\\,x^2-5x+6\\phantom{)}}} \\\\ \r\n\\underline{-~\\phantom{(}(x^2-3x)\\phantom{-b)}} \\\\\r\n0-2x+6\\phantom{)} \\\\ \r\n\\underline{~\\phantom{()}(-2x+6)} \r\n\\end{array}\\]\r\nSubtract \\(-2x+6\\) from \\(-2x+6\\) to get \\(0\\). This is the end of the process and in this case, there is no remainder.\r\n\\[\\begin{array}{r} x-2\\phantom{)} \\\\ \r\nx-3{\\overline{\\smash{\\big)}\\,x^2-5x+6\\phantom{)}}} \\\\ \r\n\\underline{-~\\phantom{(}(x^2-3x)\\phantom{-b)}} \\\\ \r\n0-2x+6\\phantom{)} \\\\\r\n\\underline{-~\\phantom{()}(-2x+6)} \\\\ \r\n0+0\\phantom{)} \r\n\\end{array}\\]\r\n\\(x^{2}-5x+6\\) divides into \\(x-3\\) exactly and the answer is \\(x-2\\). In other words:\r\n\\[x^{2}-5x+6 = (x-3)(x-2)\\]\r\nIn this particular case, it would be much easier to just factorise the quadratic or use the quadratic formula to get the same result.\r\nHowever, these methods are not so helpful when the polynomials have higher degree.\r\n\r\nExample 1 \u2013 dividing cubic polynomials using long division\r\nDivide \\(x^{3}-6x^{2}+11x-6\\) by \\(x-1\\).\r\nFirst, we set up the long division by writing:\r\n\\[\\begin{array}{r} \\\\ \r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}} \r\n\\end{array}\\]\r\nNow, we can start the algorithm.\r\nDivide the highest power of the polynomial \\(x^{3}-6x^{2}+11x-6\\) by the highest power in the polynomial \\(x-1\\). That is, divide \\(x^{3}\\) by \\(x\\) to get \\(x^{2}\\). The \\(x\\) is placed above the \\(6x^{2}\\).\r\n\\[\\begin{array}{r} x^2\\phantom{\\,-5\\,x+6)x} \\\\\r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}} \r\n\\end{array}\\]\r\nMultiply \\(x-1\\) by \\(x^{2}\\) to get \\(x^{3}-x^{2}\\) and write the result below the cubic polynomial.\r\n\\[\\begin{array}{r} x^2\\phantom{\\,-5\\,x+6)x} \\\\ \r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}} \\\\\r\n\\underline{(x^3-\\,\\, x^2)\\phantom{bbbbbbbbbi)}} \r\n\\end{array}\\]\r\nSubtract \\(x^{3}-x^{2}\\) from \\(x^{3}-6x^{2}\\) to get \\(-5x^{2}\\).\r\n\\[\\begin{array}{r} x^2\\phantom{\\,-5\\,x+6)x} \\\\ \r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}} \\\\ \r\n\\underline{-(x^3-\\,\\, x^2)\\phantom{bbbbbbbbbi)}} \\\\\r\n0- \\, \\, 5x^2\\phantom{bbbi+11x)} \r\n\\end{array}\\]\r\nBring down the \\(11x\\).\r\n\\[\\begin{array}{r} x^2\\phantom{\\,-5\\,x+6)x} \\\\ \r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}} \\\\ \r\n\\underline{-(x^3-\\,\\, x^2)\\phantom{bbbbbbbbbi)}} \\\\ \r\n\\, \\, -5x^2+11x\\phantom{bbbi)} \r\n\\end{array}\\]\r\nDivide the highest power of the polynomial \\(-5x^{2}+11x-6\\) by the highest power in the polynomial \\(x-1\\). That is, divide \\(-5x^{2}\\) by \\(x\\) to get \\(-5x\\). The \\(-5x\\) is placed above the \\(11x\\).\r\n\\[\\begin{array}{r} x^2\\,\\,-5x\\phantom{)+6} \\\\ \r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}} \\\\ \r\n\\underline{-(x^3-\\,\\, x^2)\\phantom{bbbbbbbbbi)}} \\\\ \r\n0- \\, \\, 5x^2+11x\\phantom{bbbi)} \r\n\\end{array}\\]\r\nMultiply \\(x-1\\) by \\(-5x\\) to get \\(5x^{2}+5x\\) and write the result below the cubic polynomial.\r\n\\[\\begin{array}{r} x^2\\,-5\\,x\\phantom{)+6} \\\\\r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}} \\\\ \r\n\\underline{-(x^3-\\,\\, x^2)\\phantom{bbbbbbbbbi)}} \\\\ \r\n\\, \\, -5x^2+11x\\phantom{bbbi)} \\\\ \r\n\\underline{~\\phantom{()}(-5x^2+\\,\\,5x)}\\phantom{bbb)} \r\n\\end{array}\\]\r\nSubtract \\(-5x^{2}+5x\\) from \\(-5x^{2}+11x\\) to get \\(6x\\).\r\n\\[\\begin{array}{r} x^2\\,-5\\,x\\phantom{)+6} \\\\ \r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}}\\\\ \r\n\\underline{-(x^3-\\,\\, x^2)\\phantom{bbbbbbbbi)}}\\\\ \r\n- \\, \\, 5x^2+11x\\phantom{bbbi)}\\\\ \r\n\\underline{-~\\phantom{()}(-5x^2+\\,\\,5x)}\\phantom{bbb)}\\\\ \r\n6x \\phantom{)-6} \r\n\\end{array}\\]\r\nBring down the \\(-6\\).\r\n\\[\\begin{array}{r} x^2\\,-5\\,x\\phantom{)+6} \\\\\r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}} \\\\ \r\n\\underline{-(x^3-\\,\\, x^2)\\phantom{bbbbbbbbi)}}\\\\ \r\n0- \\, \\, 5x^2+11x\\phantom{bbbi)}\\\\\r\n\\underline{-~\\phantom{()}(-5x^2+\\,\\,5x)}\\phantom{bbb)}\\\\ \r\n6x-6 \\phantom{)} \r\n\\end{array}\\]\r\nDivide \\(6x-6\\) by the leading term in \\(x-1\\) to get \\(6x\\). That is, divide \\(x\\) into \\(6x\\) to get \\(6x\\) and write the \\(6x\\) on the top line.\r\n\\[\\begin{array}{r} x^2\\,-5\\,x+6\\phantom{)} \\\\\r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}}\\\\ \r\n\\underline{-(x^3-\\,\\, x^2)\\phantom{bbbbbbbbi)}}\\\\\r\n0- \\, \\, 5x^2+11x\\phantom{bbbi)} \\\\ \r\n\\underline{-~\\phantom{()}(-5x^2+\\,\\,5x)}\\phantom{bbb)}\\\\ \r\n6x-6 \\phantom{)} \r\n\\end{array}\\]\r\nMultiply \\(x-1\\) by \\(6x\\) and write this below.\r\n\\[\\begin{array}{r} x^2\\,-5\\,x+6\\phantom{)} \\\\\r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}} \\\\\r\n\\underline{-(x^3-\\,\\, x^2)\\phantom{bbbbbbbbi)}} \\\\\r\n0- \\, \\, 5x^2+11x\\phantom{bbbi)} \\\\\r\n\\underline{-~\\phantom{()}(-5x^2+\\,\\,5x)}\\phantom{bbb)} \\\\ \r\n6x-6 \\phantom{)} \\\\\r\n\\underline{\\phantom{()}(6x-6)} \r\n\\end{array}\\]\r\nSubtract \\(6x-6\\) from \\(6x-6\\) to get \\(0\\). This is the end of the process and this case, there is no remainder.\r\n\\[\\begin{array}{r} x^2\\,-5\\,x+6\\phantom{)} \\\\\r\nx-1{\\overline{\\smash{\\big)}\\,x^3-6x^2+11x-6\\phantom{)}}} \\\\ \r\n\\underline{-(x^3-\\,\\, x^2)\\phantom{bbbbbbbbi)}} \\\\\r\n0- \\, \\, 5x^2+11x\\phantom{bbbi)} \\\\ \r\n\\underline{-~\\phantom{()}(-5x^2+\\,\\,5x)}\\phantom{bbb)}\\\\ \r\n6x-6 \\phantom{)} \\\\ \r\n\\underline{-~\\phantom{()}(6x-6)} \\\\ \r\n0+0\\phantom{)} \r\n\\end{array}\\]\r\n\\(x^{3}-6x^{2}+11x-6\\) divides into \\(x-1\\) exactly and the answer is \\(x^{2}-5x+6\\). In other words: \r\n\\[x^{3}-6x^{2}+11x-6=(x-1)(x^{2}-5x+6)\\]\r\nThe quadratic factor can be further factorised, giving a final factorised cubic polynomial:\r\n\\[x^{3}-6x^{2}+11x-6=(x-1)(x-3)(x-2)\\]\r\n\r\n\r\n\r\nExercise \u2013 dividing polynomials using long division\r\nDivide the following.\r\n\r\n\t\\((2x^{3}-6x^{2}+5x+2)\\div (x-2)\\)\r\n\t\\((3x^{3}+13x^{2}+6x-12)\\div (3x+4)\\)\r\n\t\\((3x^{3}+x-1)\\div (x-1)\\)\r\n\t\\((x^{3}+2x-3)\\div (x-1)\\)\r\n\t\\((2x^{4}+5x^{3}+x+3)\\div (2x+1)\\)\r\n\t\\((2x^{3}+x^{2}+5x+12)\\div (2x+3)\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.035},[{"v":"Maths equations","i":1,"n":0.707},{"v":"Algebra","i":0,"n":1}]]},{"i":223,"$":[{"v":"Quadratics in algebraic fractions","n":0.5},{"v":"\r\nLet's take a look at algebraic fractions where the denominator is a quadratic expression. Such fractions are common in mathematics and engineering.\r\nBefore you continue with the content on this page, it is important that you are confident about identifying common factors, factorising quadratics using common factors and the difference of two squares rule.\r\nThe best way to grasp how to work with quadratics in algebraic fractions is through examples.\r\n\r\nExample \u2013 adding or subtracting algebraic fractions with quadratics\r\nSimplify \\(\\dfrac{2}{x^{2}+x-12}-\\dfrac{1}{x^{2}-9}\\).\r\nYou can multiply the denominators as is to form the lowest common denominator, but this would give you an \\(x^{4}\\). But since this can be tricky to work with, you can factorise the denominators instead.\r\n\\[\\begin{align*} x^{2}+x-12 & =(x+4)(x-3)\\\\\r\nx^{2}-9 & = (x+3)(x-3)\r\n\\end{align*}\\]\r\nWe can rewrite the original operation as:\r\n\r\nA common denominator is \\((x+4)(x-3)(x+3)\\).\r\nTo write the individual fractions with this common denominator, we need to multiply \\(2\\) by \\((x+3)\\), and \\(1\\) by \\((x+4)\\).\r\n\\[\\begin{align*} & = \\frac{2(x+3)}{(x+4)(x-3)(x+3)} - \\frac{1(x+4)}{(x+4)(x-3)(x+3)} \\\\\r\n& = \\frac{2(x+3)-1(x+4)}{(x+4)(x-3)(x+3)}\\\\\r\n& = \\frac{2x+6-x-4}{(x+4)(x-3)(x+3)}\\\\\r\n& = \\frac{x+2}{(x+4)(x-3)(x+3)}\\quad\\textrm{where \\(x\\neq-4,-3,3\\)}\r\n\\end{align*}\\]\r\n\r\nExercise \u2013 adding or subtracting algebraic fractions with quadratics\r\nSimplify \\(\\dfrac{2}{x^{2}+2x}+\\dfrac{1}{x^{2}-4}\\).\r\n\r\n\r\n\r\n\r\nExample \u2013 dividing algebraic fractions with quadratics\r\nSimplify \\(\\dfrac{1-\\frac{2}{x}}{1+\\frac{2}{x}}\\). \r\n\r\n\r\nExercise \u2013 dividing algebraic fractions with quadratics\r\nSimplify \\(\\dfrac{1-\\frac{6}{x}}{\\frac{x}{2}-3}\\).\r\n\r\n\r\nExample \u2013 simplifying algrebraic fractions with quadratics and mixed operations\r\nSimplify \\(\\dfrac{\\frac{-3}{x^{2}+2x-3}+\\frac{1}{x-1}}{\\frac{3}{x-1}+3}\\).\r\nThis expression involves nested fractions. We can step through it by looking at the numerator and denominator separately.\r\nLet's look at the numerator first. You can multiply the denominators as is to form the lowest common denominator, but this would give you an \\(x^{4}\\). This can be tricky to work with. Instead, you can factorise the denominators.\r\n\\[x^{2}+2x-3 = (x+3)(x-1)\\]\r\nWe can rewrite the original numerator as:\r\n\\[\\begin{align*} \\frac{-3}{x^{2}+2x-3}+\\frac{1}{x-1} & =\\frac{-3}{(x+3)(x-1)}+\\frac{1}{x-1}\\\\\r\n& =\\frac{-3}{(x+3)(x-1)}+\\frac{1}{x-1}\r\n\\end{align*}\\]\r\nA common denominator is \\((x+3)(x-1)\\).\r\nTo write the individual fractions with this common denominator, we need to multiply \\(1\\) by \\((x+3)\\).\r\n\\[\\begin{align*} & = \\frac{-3}{(x+3)(x-1)} + \\frac{(x+3)}{(x+3)(x-1)}\\\\\r\n& = \\frac{-3+x+3}{(x+3)(x-1)}\\\\\r\n& = \\frac{x}{(x+3)(x-1)}\r\n\\end{align*}\\]\r\nNow, we can look at the denominator. To make a common denominator, we need to multiply \\(3\\) by \\((x-1)\\).\r\n\\[\\begin{align*} \\frac{3}{x-1}+3 & = \\frac{3}{x-1}+\\frac{3(x-1)}{x-1}\\\\\r\n& = \\frac{3x}{x-1}\r\n\\end{align*}\\]\r\nPutting the numerator and denominator together, we get:\r\n\\[\\frac{\\frac{x}{(x+3)(x-1)}}{\\frac{3x}{x-1}}\\]\r\nWe can rewrite the fraction using a division symbol and complete the simplification.\r\n\r\n\r\nExercise \u2013 simplifying algebraic fractions with quadratics and mixed operations\r\nSimplify \\(\\dfrac{\\frac{1}{x^{2}-4}+\\frac{1}{2x+4}}{1+\\frac{2}{x-2}}\\).\r\n\r\n\r\n","n":0.054},[{"v":"Maths equations","i":3,"n":0.707},{"v":"Functions","i":2,"n":1},{"v":"Arithmetic","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":224,"$":[{"v":"Procrastination","n":1},{"v":"Procrastination means delaying or postponing something that must be done because it seems difficult or unpleasant. If you are someone who procrastinates, it's important to remember that it IS possible to teach ourselves new behaviours. Let's get started!\r\n\r\nReasons you might be procrastinating\r\nWhen procrastination becomes a problem, it can be helpful to explore reasons why you may find yourself avoiding certain tasks. These can include:\r\n\r\n \tlack of interest in the task\r\n \tfear of failure\r\n \tdoubting your ability to complete the task\r\n \thaving multiple tasks with looming deadlines\r\n\tfatigue\r\n \tdistractions such as technology.\r\n\r\nWhat can you do about it?\r\nEveryone is different, so some of the following tips may be perfect for you and others may not. This will depend on the cause of your procrastination and your preferred ways of studying.\r\nIdentify when you are procrastinating\r\nIt's never too early to start an assignment! You may be doing things that are keeping you busy instead of getting work done. This could include:\r\n\r\n \tpreparing your workspace rather than working at\u00a0it\r\n \tactively seeking distractions, such as checking your phone for updates\r\n \tif you're at home, you may prioritise housework over your studies.\r\n\r\nBreak procrastination cycles\r\n\r\n \tSet small goals by breaking a big task into small pieces.\r\n \tStart with a task that can easily be completed.\r\n \tStudy with your friends - seeing other people study can help you to get to work.\t\r\n \tWrite up a 'Done' list as an alternative as a 'To-Do' list to remind\u00a0yourself of your accomplishments.\r\n \tPlan regular times to focus on studying. Having a regular time to do something each day helps to form a good work habit.\r\n\r\nUnderstand your ideal workspace\r\nBeing aware of distraction is your workspace can help you minimise their effect. To remove distractions, you could:\r\n\r\nturn your mobile phone on silent or leave it in another room while you study\r\nYou can turn off notifications on your devices or set them to 'focus' mode.\r\ndeclutter your workspace by removing anything that isn't necessary for your studies\r\nif possible, adjust the lighting to avoid eye strain\r\nlisten to music or sounds that help you to focus.\r\n\r\nThe pomodoro technique\r\nThis technique involves choosing a task and working on it in 25-minute bursts followed by short breaks, as directed by a timer. After four of these bursts, you can take a longer break. An online pomodoro timer that might help you stick to your task is Pomofocus.\r\n\r\n\r\n\r\n\r\n","n":0.051},[{"v":"Time management","i":1,"n":0.707},{"v":"Study skills","i":0,"n":0.707}]]},{"i":225,"$":[{"v":"Quiz \u2013 Strength of acids and bases","n":0.378},{"v":"Understanding the strength of acids and bases is crucial for predicting their behaviour in chemical reactions and their impact in various applications. Test your ability to identify and differentiate between strong and weak acids and bases with this quiz.\r\n\r\nYour turn \u2013 strength of acids and bases\r\n\r\n \tThe \\(\\ce{HBr}\\) in a \\(0.1\\textrm{M}\\) \\(\\ce{HBr}\\) solution is \\(100\\%\\) dissociated. Calculate the molar concentrations of \\(\\ce{HBr}\\), \\(\\ce{H3O}^{+}\\) and \\(\\ce{Cl}^{-}\\) in the solution.\r\n\r\n\t\r\n\r\n\r\n \tDetermine the number of acidic and non-acidic hydrogens present in the following molecules.\r\n\t\r\n\t\t\\(\\ce{HNO3}\\)\r\n\t\r\n\t\t\r\n\t\r\n\t\t\\(\\ce{H2C4H4O4}\\) (succinic acid)\r\n\t\r\n\t\t\r\n\t\r\n\t\t\\(\\ce{NH3}\\)\r\n\t\r\n\t\t\r\n\t\r\n\t\t\\(\\ce{CH4}\\)\r\n\t\r\n\t\t\r\n\r\n\r\n\r\n \tUse the acid ionisation constants in the table to identify the stronger acid in each acid pair.\r\n\r\n\r\n\r\n\r\nFormula\r\n\\(K_{a}\\)\r\n\r\n\r\n\r\n\r\n\\(\\ce{H}_{3}\\ce{PO}_{4}\\)\r\n\\(7.5\\times10^{-3}\\)\r\n\r\n\r\n\\(\\ce{HF}\\)\r\n\\(6.8\\times10^{-4}\\)\r\n\r\n\r\n\\(\\ce{HNO}_{2}\\)\r\n\\(4.5\\times10^{-4}\\)\r\n\r\n\r\n\\(\\ce{HCN}\\)\r\n\\(4.9\\times10^{-10}\\)\r\n\r\n\r\n\\(\\ce{H}_{2}\\ce{CO}_{3}\\)\r\n\\(4.3\\times10^{-7}\\)\r\n\r\n\r\n\\(\\ce{HCO}_{3}^{-}\\)\r\n\\(5.6\\times10^{-11}\\)\r\n\r\n\r\n\r\n\r\n\t\r\n\t\t\\(\\ce{H3PO4}\\) and \\(\\ce{HNO2}\\)\r\n\t\r\n\t\t\r\n\t\r\n\t\t\\(\\ce{HCN}\\) and \\(\\ce{HF}\\)\r\n\t\r\n\t\t\r\n\t\r\n\t\t\\(\\ce{H2CO3}\\) and \\(\\ce{HCO}_{3}^{-}\\)\r\n\t\r\n\t\t\r\n\t\r\n\t\t\\(\\ce{HNO2}\\) and \\(\\ce{HCN}\\)\r\n\t\r\n\t\t\r\n\t\t\r\n\t\r\n\r\n\tA \\(0.0300\\textrm{M}\\) of a base is \\(7.5\\%\\) ionised. Calculate the base ionisation constant \\(K_{b}\\).\r\n\r\n\t\r\n\r\n\r\n\tA \\(0.0500\\textrm{M}\\) of an acid is \\(15\\%\\) ionised. Calculate the acid ionisation constant \\(K_{a}\\).\r\n\r\n\t","n":0.086},[{"v":"Salts","i":2,"n":1},{"v":"Chemistry exercises","i":1,"n":0.707},{"v":"Acids and bases","i":0,"n":0.577}]]},{"i":226,"$":[{"v":"Strength of acids and bases","n":0.447},{"v":"The strengths of acids and bases determine their reactivity and dissociation in solution. Strong acids and bases, like hydrochloric acid and sodium hydroxide, completely dissociate, while weak ones, like acetic acid and ammonia, only partially dissociate. Recognising these strengths is key to understanding their behaviour in chemical reactions and their diverse applications.\r\nAcids can be categorised as strong acids or weak acids depending on their ability to completely transfer protons in an aqueous solution. The extent of hydrogen transfer depends on the molecular structure of the acid. Similarly, bases can be categorised as strong bases or weak bases depending on their ability to completely accept donated protons in aqueous solution.\r\nExamples of strong and weak acids and bases are shown in the table.\r\n\r\n\r\n\r\n\r\nStrong acid\r\nWeak acid\r\nStrong base\r\nWeak base\r\n\r\n\r\n\r\n\r\nHydrochloric acid, \\(\\ce{HCl}\\)\r\nAcetic acid, \\(\\ce{CH3COOH}\\)\r\nAmmonia, \\(\\ce{NH3}\\)\r\nSodium hydroxide, \\(\\ce{NaOH}\\)\r\n\r\n\r\nHydrobromic acid, \\(\\ce{HBr}\\)\r\nFormic acid, \\(\\ce{HCOOH}\\)\r\nMethylamine, \\(\\ce{CH3NH2}\\)\r\nPotassium hydroxide, (\\(\\ce{KOH}\\)\r\n\r\n\r\nNitric acid, \\(\\ce{HNO3}\\)\r\nCarbonic acid, \\(\\ce{H2CO3}\\)\r\nAniline, \\(\\ce{C6H5NH2}\\)\r\nCalcium hydroxide, \\(\\ce{Ca(OH)2}\\)\r\n\r\n\r\nSulfuric acid, \\(\\ce{H2SO4}\\)\r\nPhosphoric acid, \\(\\ce{H3PO4}\\)\r\nPyridine, \\(\\ce{C5H5N}\\)\r\nBarium hydroxide, \\(\\ce{Ba(OH)2}\\)\r\n\r\n\r\nPerchloric acid, \\(\\ce{HClO4}\\)\r\nHydrofluoric acid, \\(\\ce{HF}\\)\r\nEthylamine, \\(\\ce{C2H5NH2}\\)\r\nLithium hydroxide, \\(\\ce{LiOH}\\)\r\n\r\n\r\n\r\n\r\n\r\nStrong acids\r\nStrong acids can transfer nearly all (\\(100\\%\\)) of their protons (hydrogen ions) to water. If the hydrogen is bound to a highly electronegative atom like oxygen or chlorine, the hydrogen is acidic and is able to undergo complete ionisation. The dominant species in solution would be \\(\\ce{H3O}^{+}\\) and the conjugate base \\(\\ce{A}^{-}\\) (the right side of the reaction arrow). An example is hydrochloric acid (\\(\\ce{HCl}\\)). Acid\u2013base reactions involving strong acids are written using reaction arrows (\\(\\rightarrow\\)).\r\n\\[\\ce{HCl}(\\textrm{g})+\\ce{H2O}(\\textrm{l})\\rightarrow\\ce{H3O}^{+}(\\textrm{aq})+\\ce{Cl}^{-}(\\textrm{aq})\\]\r\n\r\n\r\nWeak acids\r\nWeak acids can only transfer some of their protons to the water. If the hydrogen is bound to a less electronegative atom like carbon, the bond is not polar enough to release the hydrogen. The dominant species in solution would be the acid \\(\\ce{HA}\\) (the left side of the reaction arrow). An example is the hydrogen bound to carbon in acetic acid (\\(\\ce{CH3COOH}\\)), which makes it a weak acid. Acid\u2013base reactions involving weak acids are written using equilibrium arrows (\\(\\rightleftharpoons\\)).\r\n\\[\\ce{CH3COOH}(\\textrm{aq})+\\ce{H2O}(\\textrm{l})\\rightleftharpoons\\ce{H3O}^{+}(\\textrm{aq})+\\ce{CH3COO}^{-}(\\textrm{aq})\\]\r\n\r\n\r\n\r\n\r\nStrong bases\r\nStrong bases show a high affinity for protons in an aqueous solution. \\(\\ce{NaOH}\\), \\(\\ce{KOH}\\) and hydroxides of group \\(1\\) and \\(2\\) elements in the periodic table are strong bases. In a solution of a strong base, the dominant species in the solution would be \\(\\ce{OH}^{-}\\) and the conjugate acid \\(\\ce{BH}^{+}\\). Acid\u2013base reactions involving strong bases are written using reaction arrows (\\(\\rightarrow\\)).\r\n\\[\\ce{NaOH}(\\textrm{aq})+\\ce{H2O}(\\textrm{l})\\rightarrow\\ce{Na}^{+}(\\textrm{aq})+\\ce{OH}^{-}(\\textrm{aq})+\\ce{H2O}(\\textrm{l})\\]\r\n\r\nWeak bases\r\nWeak bases show less affinity for protons in aqueous solutions. \\(\\ce{NH3}\\) gas is a weak base. In a solution of a weak base, the dominant species in the solution is the base \\(\\ce{B}\\). Acid\u2013base reactions involving weak bases are written using equilibrium arrows (\\(\\rightleftharpoons\\)).\r\n\\[\\ce{NH3}(\\textrm{g})+\\ce{H2O}(\\textrm{l})\\rightleftharpoons\\ce{NH}_{4}^{+}(\\textrm{aq})+\\ce{OH}^{-}(\\textrm{aq})\\]\r\n\r\n\r\n\r\nRelationship between acid\u2013base strength\r\nThere is an inverse relationship between the strength of an acid and the strength of a base.\r\n\r\n\tThe conjugate base of a strong acid is always a weak base.\r\n\tThe conjugate base of a weak acid is always a strong base.\r\n\r\nThis relationship can be explained using the definition of a strong/weak acid and strong/weak base. The strong acid has the capacity to completely (nearly 100%) dissociate in water to donate protons, which means its conjugate base has less affinity to protons as the reaction favours the formation of products. A base showing less affinity for protons is a weak base.\r\nSimilarly, in water, the weak acid partially dissociates to produce protons, which means its conjugate base displays a high affinity for the protons. The reaction favours the formation of reactants.\r\nTherefore, the stronger the acid, the weaker its conjugate base, and the weaker the acid, the stronger its conjugate base.\r\n\r\nIonisation constants\r\nThe strengths of acids and bases can be quantified using ionisation constants. These constants are essentially the ratio of products (dissociated acid or base) compared to reactants (undissociated acid or base).\r\n\r\nAcid ionisation constants \u2013 \\(K_{a}\\)\r\nRecall that the dissociation of a weak acid in water can be written as:\r\n\\[\\ce{HA}(\\textrm{aq})+\\ce{H2O}(\\textrm{l})\\rightleftharpoons\\ce{H3O}^{+}(\\textrm{aq})+\\ce{A}^{-}(\\textrm{aq})\\]\r\nOnce the concentrations of \\(\\ce{H3O}^{+}\\), \\(\\ce{A}^{-}\\) and \\(\\ce{HA}\\) are experimentally determined, the acid ionisation constant (\\(K_{a}\\)) can be calculated using:\r\n\\[K_{a}=\\frac{ \\textrm{[}\\ce{H3O}^{+}\\textrm{]} \\textrm{[}\\ce{A}^{-}\\textrm{]} }{\\textrm{[}\\ce{HA}\\textrm{]}} \\]\r\nThe strength of an acid is directly proportional to the acid ionisation constant—that is, the higher the constant, the stronger the acid. Strong acids have larger \\(K_{a}\\) values (greater than \\(1\\)), as the formation of products is favoured; there is a larger concentration of \\(\\ce{H3O}^{+}\\) and \\(\\ce{A}^{-}\\) than \\(\\ce{HA}\\), Weak acids have \\(K_{a}\\) values less than \\(1\\), as ionisation is not favoured.\r\nWater is not included in the ionisation constant expression because it is the solvent and its concentration is effectively constant.\r\n\r\nBase ionisation constants \u2013 \\(K_{b}\\)\r\nRecall that the dissociation of a weak base in water can be written as:\r\n\\[\\ce{B}(\\textrm{aq})+\\ce{H2O}(\\textrm{l})\\rightleftharpoons\\ce{BH}^{+}(\\textrm{aq})+\\ce{OH}^{-}(\\textrm{aq})\\]\r\nThe strength of a base can also be predicted by an ionisation constant, the base ionisation constant (\\(K_{b}\\)). The general expression for \\(K_{b}\\) is:\r\n\\[K_{b}=\\frac{ \\textrm{[}\\ce{BH}^{+}\\textrm{]} \\textrm{[}\\ce{OH}^{-}\\textrm{]}  } {\\textrm{[}\\ce{B}\\textrm{]}} \\]\r\n\r\nExample \u2013 determining ionisation constants\r\nThe weak acid \\(\\ce{HA}\\) in \\(0.0200\\textrm{M}\\) solution is \\(10\\%\\) ionised. Determine the acid ionisation constant.\r\nStep 1: Write an equation for the ionisation of the acid.\r\n\\[\\ce{HA}(\\textrm{aq})+\\ce{H2O}(\\textrm{l})\\rightleftharpoons\\ce{H3O}^{+}(\\textrm{aq})+\\ce{A}^{-}(\\textrm{aq}) \\]\r\nStep 2: Recall the expression for \\(K_{a}\\).\r\n\\[K_{a}=\\frac{ \\textrm{[}\\ce{H3O}^{+}\\textrm{]} \\textrm{[}\\ce{A}^{-}\\textrm{]} }{\\textrm{[}\\ce{HA}\\textrm{]}} \\]\r\nStep 3: Calculate the concentrations of \\(\\ce{H3O}^{+}\\), \\(\\ce{A}^{-}\\) and \\(\\ce{HA}\\) using the percentage ionisation.\r\nThe concentration of \\(\\ce{H3O}^{+}\\) will be \\(10\\%\\) of the original concentration of \\(\\ce{HA}\\).\r\n\\[\\begin{align*}\r\n\\textrm{[}\\ce{H3O}^{+}\\textrm{]} & = 10\\% \\times 0.0200\\textrm{M}\\\\\r\n& = 0.0200\\textrm{M}\r\n\\end{align*} \\]\r\nUsing the molar ratio:\r\n\\[\\begin{align*}\r\n\\textrm{[}\\ce{A}^{-}\\textrm{]} & = \\frac{1}{1} \\times \\textrm{[}\\ce{H3O}^{+}\\textrm{]}\\\\\r\n& = 0.00200\\textrm{M}\r\n\\end{align*} \\]\r\nSince the concentrations are molar, \\(\\textrm{[}\\ce{H3O}^{+}\\textrm{]} + \\textrm{[}\\ce{HA}^{+}\\textrm{]} = \\textrm{[}\\ce{A}^{-}\\textrm{]}\\). We can calculate:\r\n\\[\\begin{align*}\r\n\\textrm{[}\\ce{HA}\\textrm{]} & = 0.0200\\textrm{M}-0.00200\\textrm{M}\\\\\r\n& =0.0180\\textrm{M}\r\n\\end{align*}\\]\r\nStep 4 Substitute the concentration values into the acid ionisation constant expression:\r\n\\[\\begin{align*}\r\nK_{a} & =\\frac{\\textrm{[}\\ce{H3O}^{+}\\textrm{]}\\textrm{[}\\ce{A}^{-}\\textrm{]}}{\\textrm{[}\\ce{HA}\\textrm{]}}\\\\\r\n& =\\frac{0.00200\\textrm{M}\\times0.00200\\textrm{M}}{0.0180\\textrm{M}}\\\\\r\n& =2.2\\times10^{-4}\r\n\\end{align*}\\]\r\nStep 5: Include the correct units.\r\nThe units for ionisation constants can vary. In this case, the numerator has \\(\\textrm{M}\\times\\textrm{M}\\) and the denominator has just \\(\\textrm{M}\\). \\(\\frac{\\textrm{M}^{2}}{\\textrm{M}}=\\textrm{M}\\), so the units for this example are just \\(\\textrm{M}\\). Thus, the correct answer is \\(2.2\\times10^{-4}\\textrm{M}\\).\r\n \r\n    Strong and weak acid reactions\r\n    \r\n        Strong acid\r\n        \r\n\t\tChemical equation: HA + H2O \u21cc H3O+ + A\u2013\r\n\t\tThe harpoon pointing towards the products (forward reaction) is larger than the one pointing to the reactants (backwards reaction), indicating the preference for formation of products.\r\n\t\t\r\n    \r\n    \r\n        Weak acid\r\n        \r\n\t\tChemical equation: HA + H2O \u21cc H3O+ + A\u2013\r\n\t\tThe harpoon pointing towards the products (forward reaction) is smaller than the one pointing to the reactants (backwards reaction), indicate the preference for formation of reactants.\r\n\t\t\r\n    \r\n \r\n    Strong and weak base reactions\r\n    \r\n        Strong base\r\n        \r\n\t\tChemical equation: B + H2O \u21cc BH+ + OH\u2013\r\n\t\tThe harpoon pointing towards the products (forward reaction) is larger than the one pointing to the reactants (backwards reaction), indicating the preference for formation of products.\r\n\t\t\r\n    \r\n    \r\n        Weak base\r\n        \r\n\t\tChemical equation: B + H2O \u21cc BH+ + OH\u2013\r\n\t\tThe harpoon pointing towards the products (forward reaction) is smaller than the one pointing to the reactants (backwards reaction), indicate the preference for formation of reactants.\r\n\t\t\r\n    \r\n","n":0.03},[{"v":"Salts","i":2,"n":1},{"v":"Chemistry","i":1,"n":1},{"v":"Acids and bases","i":0,"n":0.577}]]},{"i":227,"$":[{"v":"Quiz \u2013 Acids, bases and salts","n":0.408},{"v":"Can you confidently identify acids and bases? What about their conjugates? Test yourself with this quiz.\r\n\r\nYour turn \u2013 acids, bases and salts\r\nFor questions 1\u20135, identify the chemical species as a Br\u00f8nsted\u2013Lowry acid or base:\r\n\r\n\r\n\t\\(\\ce{HClO4}\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{HCOOH}\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{CN}^{-}\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{SO}_{3}^{2-}\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{H2S}\\)\r\n\r\n\t\r\n\r\nFor questions 6\u201310, write the chemical formulas for each of the following:\r\n\r\n\r\n\tthe conjugate acid of \\(\\ce{HS}^{-}\\)\r\n\r\n\t\r\n\r\n\r\n\tthe conjugate base of \\(\\ce{NH}_{4}^{+}\\)\r\n\r\n\t\r\n\r\n\r\n\tthe conjugate acid of \\(\\ce{PO}_{4}^{3-}\\)\r\n\r\n\t\r\n\r\n\r\n\tthe conjugate acid of \\(\\ce{HC2O4}^{-}\\)\r\n\r\n\t\r\n\r\n\r\n\tthe conjugate base of \\(\\ce{H2CO3}\\)\r\n\r\n\t\r\n\r\nFor questions 11\u201313, classify the acids as monoprotic, diprotic or triprotic:\r\n\r\n\r\n\t\\(\\ce{H2SO4}\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{HClO4}\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{H3PO4}\\)\r\n\r\n\t","n":0.115},[{"v":"Salts","i":2,"n":1},{"v":"Chemistry exercises","i":1,"n":0.707},{"v":"Acids and bases","i":0,"n":0.577}]]},{"i":228,"$":[{"v":"Acids, bases and salts","n":0.5},{"v":"Acids and bases are vital chemical compounds with significant roles in both science and daily life. Acids, with their sour taste and reactivity, are found in citrus fruits and vinegar, while bases, known for their bitter taste and slipperiness, appear in cleaning products and baking soda. When acids and bases react, they form salts, compounds that are essential in everything from seasoning food to preserving life through electrolytes. Use this resource to learn about acids and bases, including how Indigenous peoples use acid-base reactions in medicine.\r\n\r\nAcids and bases can be easily classified based on their useful properties, but how are they defined from a chemical perspective?\r\n\r\nArrhenius acids and bases\r\nArrhenius' theory of acids and bases defines acids and bases in terms of their behaviour in water.\r\n\r\n\tAn Arrhenius acid is any substance that increases the concentration of hydrogen ions (\\(\\ce{H}^{+}\\)) in water. They do this by dissociating or breaking down into their constituent ions when dissolved in water. Arrhenius acids contain \\(\\ce{H}^{+}\\) ions. Examples of Arrhenius acids are \\(\\ce{HCl}\\) and \\(\\ce{HNO3}\\). The equations for their dissociation are shown:\r\n\t\\[ \\ce{HNO3}(\\textrm{l})\\rightarrow\\ce{H}^{+}(\\textrm{aq})+\\ce{NO}_{3}^{-}(\\textrm{aq}) \\] \\[ \\ce{HCl}(\\textrm{g})\\rightarrow\\ce{H}^{+}(\\textrm{aq})+\\ce{Cl}^{-}(\\textrm{aq}) \\]\r\n\tAn Arrhenius base is any substance that increases the concentration of hydroxide ions (\\(\\ce{OH}_{-}\\)) in water. They achieve this by dissociating into their constituent ions when dissolved in water. Arrhenius bases contain \\(\\ce{OH}_{-}\\) groups. Examples of Arrhenius bases are \\(\\ce{NaOH}\\) and \\(\\ce{KOH}\\). The equations for their dissociation in water are:\r\n\t\\[ \\ce{NaOH}(\\textrm{s})\\rightarrow\\ce{Na}^{+}(\\textrm{aq})+\\ce{OH}^{-}(\\textrm{aq}) \\] \\[ \\ce{KOH}(\\textrm{s})\\rightarrow\\ce{K}^{+}(\\textrm{aq})+\\ce{OH}^{-}(\\textrm{aq}) \\]\r\n\r\n\r\nLimitations of Arrhenius' theory of acids and bases\r\nArrhenius's theory is limited to aqueous solutions, since it specifically defines acids and bases based on their behaviour in water. This theory cannot explain why compounds without hydroxyl groups behave as bases, such as ammonia generating a basic aqueous solution.\r\n\r\nBr\u00f8nsted\u2013Lowry acid\u2013base theory\r\nDue to the limitations of Arrhenius' theory of acids and bases, the Br\u00f8nsted\u2013Lowry acid\u2013base theory was introduced.\r\n\r\n\tA Br\u00f8nsted\u2013Lowry acid is any substance that can donate a hydrogen ion (or a proton) to another substance. It is called a proton because the hydrogen ion is a hydrogen atom that has lost its one electron, leaving behind a proton. Thus, a Br\u00f8nsted\u2013Lowry acid is a proton donor and is denoted by \\(\\ce{HA}\\), where \\(\\ce{H}\\) is the hydrogen atom and \\(\\ce{A}\\) is the rest of the acid.\r\n\tA Br\u00f8nsted\u2013Lowry base is any substance that can accept a hydrogen ion (or a proton) from another substance. To do this, it needs a lone pair of electrons. Br\u00f8nsted\u2013Lowry bases are proton acceptors. They are denoted by \\(\\ce{B}\\colon\\), where \\(\\ce{B}\\) is the base and the \\(\\colon\\) is the lone pair of electrons. In some instances, they are simply denoted by \\(\\ce{B}\\).\r\n\r\n\r\nPolyprotic acids\r\nBr\u00f8nsted\u2013Lowry acids such as \\(\\ce{HCl}\\) and \\(\\ce{HNO3}\\) can donate one hydrogen ion, so they are called monoprotic acids. Some Br\u00f8nsted\u2013Lowry acids can donate multiple hydrogen ions. For example, \\(\\ce{H2SO4}\\) is capable of donating two hydrogen ions (diprotic acid), and \\(\\ce{H3PO4}\\) can donate three hydrogen ions (triprotic acid).\r\n\r\nAcidity of hydrogens\r\nNot all hydrogens are acidic. Only the hydrogens bound to highly electronegative atoms such as oxygen and chlorine can be released as protons or hydrogen ions. Hydrogens in \\(\\ce{HCl}\\), \\(\\ce{HNO3}\\), \\(\\ce{H2SO4}\\), and \\(\\ce{H3PO4}\\) are all acidic as the hydrogens are bound to electronegative \\(\\ce{Cl}\\) and \\(\\ce{O}\\). However, if you consider acetic acid, \\(\\ce{CH3COOH}\\), only one of the four hydrogens is acidic. This is the hydrogen atom connected to the highly electronegative atom, \\(\\ce{O}\\). The others are bound to \\(\\ce{C}\\).\r\nExamine the structures below and determine which of the hydrogens in each are acidic.\r\n\r\n\r\n\r\nAcid\u2013base reactions\r\nThe presence of a species that can accept protons (a base) is vital for the existence of a Br\u00f8nsted\u2013Lowry acid, and vice versa. Proton donation and acceptance are complementary processes. The general equation of the Br\u00f8nsted\u2013Lowry acid\u2013base reaction is:\r\n\\[ \\ce{HA}+\\ce{B}\\colon\\rightarrow\\ce{A}^{-}+\\left[\\ce{B}\\colon\\ce{H}\\right]^{+} \\]\r\nLet's look at two examples written in this format.\r\n\r\n \tWhen \\(\\ce{HCl}\\) gas dissolves\u00a0in water, \\(\\ce{HCl}\\) is the Br\u00f8nsted\u2013Lowry acid as it donates the proton to the water. The polar water molecule is the Br\u00f8nsted\u2013Lowry base as it accepts protons to become\u00a0a hydronium (\\(\\ce{H3O}^{+}\\)) ion.\r\n\t\\[ \\ce{HCl}(\\textrm{g})+\\ce{H2O}(\\textrm{l})\\rightarrow\\ce{Cl}^{-}(\\textrm{aq})+\\ce{H3O}^{+}(\\textrm{aq}) \\]\r\n \tWhen \\(\\ce{NH3}\\) gas reacts with \\(\\ce{HCl}\\) gas, \\(\\ce{HCl}\\) is the Br\u00f8nsted\u2013Lowry that donates a proton to \\(\\ce{NH3}\\). The ammonia molecules are the Br\u00f8nsted\u2013Lowry bases that accept the protons and become ammonium ions (\\(\\ce{NH}_{4}^{+}\\)).\r\n\t\\[ \\ce{HCl}(\\textrm{g})+\\ce{NH3}(\\textrm{g})\\rightarrow\\ce{Cl}^{-}(\\textrm{aq})+\\ce{NH}_{4}^{+}(\\textrm{aq}) \\]\r\n\r\n\r\nConjugate acid\u2013base pairs\r\nMany Br\u00f8nsted\u2013Lowry acid\u2013base reactions are reversible. After donating protons, acids are able to act as bases and accept protons. In the same way, after accepting a proton, bases can act as acids and donate the proton.\r\nLet's consider the general equation for the Br\u00f8nsted-Lowry acid-base reaction. This reaction involves two acids and two bases:\r\n\\[ \\ce{HA}+\\ce{B}\\rightleftharpoons\\ce{HB}^{+}+\\ce{A}^{-} \\]\r\n\r\n\tIn the forward reaction, \\(\\ce{HA}\\) donates a proton to \\(\\ce{B}\\). So, \\(\\ce{HA}\\) functions as an acid and \\(\\ce{B}\\) functions as a base: \r\n\t\\[ \\ce{HA}+\\ce{B}\\rightarrow\\ce{HB}^{+}+\\ce{A}^{-} \\]\r\n\tIn the reverse reaction, \\(\\ce{HB}^{+}\\) donates a proton to \\(\\ce{A}^{-}\\). Therefore, \\(\\ce{HB}^{+}\\) functions as an acid and \\(\\ce{A}^{-}\\) functions as a base:\r\n\t\\[ \\ce{HB}^{+}+\\ce{A}^{-}\\rightarrow\\ce{HA}+\\ce{B} \\]\r\n\r\n\\(\\ce{HA}\\)/\\(\\ce{A}^{-}\\) and \\(\\ce{B}\\)/\\(\\ce{HB}^{+}\\) are referred to as conjugate acid\u2013base pairs. Conjugate acid\u2013base pairs are always present on opposite sides of the chemical equation. For instance, \\(\\ce{HA}\\) (acid) and \\(\\ce{A}^{-}\\) (conjugate base).\r\n\r\n\r\n\r\n\r\nIn a conjugate acid\u2013base pair, the acid always has one more hydrogen atom than the base. For example, the reaction for the dissociation of \\(\\ce{CH}_{3}\\ce{COOH}\\) (acetic acid) in water is:\r\n\\[ \\ce{CH3COOH}+\\ce{H2O}\\rightleftharpoons\\ce{CH3COO}^{-}+\\ce{H3O}^{+} \\]\r\n\\(\\ce{CH3COOH}/\\ce{CH3COO}^{-}\\) is a conjugate acid\u2013base pair. \\(\\ce{CH3COOH}\\) has one hydrogen atom more than the \\(\\ce{CH3COO}^{-}\\) ion. So, \\(\\ce{CH3COOH}\\) is the acid and \\(\\ce{CH3COO}^{-}\\) is base in the conjugate acid-base pair.\r\n\r\n\r\nReactions between acids and bases have been used for a very long time.\r\n\r\n\r\n\r\n\r\nExample 1 \u2013 identifying Br\u00f8nsted\u2013Lowry acids and bases\r\nIdentify the following substances as a Br\u00f8nsted-Lowry acid or base:\r\n\r\n\t\\(\\ce{HF}\\)\r\n\t\\(\\ce{PO}_{4}^{3-}\\).\r\n\r\nStep 1: Recall the definitions of a Br\u00f8nsted\u2013Lowry acid and a Br\u00f8nsted\u2013Lowry base.\r\nA Br\u00f8nsted\u2013Lowry acid is any substance that can donate a hydrogen ion (or a proton) to another substance; they typically have one or more hydrogen atoms attached to a highly electronegative atom. A Br\u00f8nsted\u2013Lowry base is any substance that can accept a hydrogen ion (or a proton) from another substance; they typically contain a lone pair.\r\nStep 2: Identify whether the substance can donate hydrogen ions or accept hydrogen ions.\r\n\r\n\t\\(\\ce{HF}\\) is a Br\u00f8nsted\u2013Lowry acid as it contains a hydrogen attached to an electronegative fluorine. This hydrogen can be donated as \\(\\ce{H}^{+}\\).\r\n\t\\(\\ce{PO}_{4}^{3-}\\) does not contain hydrogen. However, oxygen atoms present in the phosphate ion have lone pairs of electrons which can be shared with hydrogen ions. Therefore, \\(\\ce{PO}_{4}^{3-}\\) is a Br\u00f8nsted\u2013Lowry base.\r\n\r\n\r\nExample 2 \u2013 identifying acid\u2013base conjugate pairs\r\nWrite the formula for the:\r\n\r\n \tconjugate base of \\(\\ce{H3PO4}\\)\r\n \tconjugate acid of \\(\\ce{NO}_{3}^{-}\\).\r\n\r\nStep 1: Write the equations for the dissociation of each acid or base.\r\n\r\n\t\\( \\ce{H3PO4}\\rightarrow\\ce{H}^{+}+\\ce{H2PO}_{4}^{-} \\)\r\n\t\\( \\ce{NO}_{3}^{-}+\\ce{H}^{+}\\rightarrow\\ce{HNO3} \\)\r\n\r\nStep 2: In a conjugate acid\u2013base pair, the acid always has one more hydrogen atom than the base. Therefore, the conjugate base is the species with one less hydrogen atom, and the conjugate acid is the species with one more hydrogen atom.\r\n\\(\\ce{H3PO4}\\) is an acid that donates a hydrogen ion to become \\(\\ce{H2PO}_{4}^{-}\\). Therefore, the conjugate base of \\(\\ce{H3PO4}\\) is the \\(\\ce{H2PO}_{4}^{-}\\) ion.\r\nAs a base, \\(\\ce{NO}_{3}^{-}\\) accepts a hydrogen ion to become \\(\\ce{HNO3}\\). Therefore, the conjugate acid of \\(\\ce{NO}_{3}^{-}\\) is the \\(\\ce{HNO3}\\).\r\n\r\n \r\nChemical structures of acids\r\n\r\n\tThe single hydrogen in hydrochloric is acidic since it is bound to an electronegative chlorine.\r\n\tThe single hydrogen in nitric acid is acidic since it is bound to an electronegative oxygen.\r\n\tThe two hydrogens in sulfuric acid are acidic since they are bound to electronegative oxygens.\r\n\tThe three hydrogens in phosphoric acid are acidic since they are bound to electronegative oxygens.\r\n\tOnly one hydrogen in acetic acid is acidic since it is bound to an electronegative oxygen. The other three are bound to a carbon, which is not highly electronegative.\r\n\r\n","n":0.028},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Salts","i":2,"n":1},{"v":"Chemistry","i":1,"n":1},{"v":"Acids and bases","i":0,"n":0.577}]]},{"i":229,"$":[{"v":"Physical and chemical changes","n":0.5},{"v":"\r\n\r\nMatter is dynamic in nature. Explore how it undergoes physical and chemical changes, including transformations between different states such as melting, freezing, and evaporating, and how Aboriginal peoples apply knowledge of these changes to extract useful compounds.\r\n\r\nPhysical change\r\nA physical change is when the substance changes its physical appearance or state without affecting its chemical identity. Examples of physical changes include:\r\n\r\n \tthe grating of a piece of cheese\r\n \tthe boiling of water\r\n \tsugar dissolving in water.\r\n\r\n\r\nChange of state\r\nWhen a substance converts from one physical state to another, this is known as a change of state. Let's look at changes of state using water as an example.\r\n\r\nWater is one of the few substances that can exist in all three states: solid ice, liquid water, and gaseous steam or vapour. The process of state changes have different names depending on which states are involved.\r\n\r\n \tMelting is when a solid changes to liquid.\r\n \tFreezing is when a liquid changes to solid.\r\n \tCondensation is when a gas changes to liquid. \t\r\n\tVaporisation is when a liquid changes to gas.\r\n \tDeposition is when a gas changes to solid.\r\n \tSublimation is when a solid changes to gas.\r\n\r\n\r\n\r\n\r\nStart animationStop animation\r\n\r\n\r\nChemical change\r\nA chemical change is when a substance changes its chemical identity or composition. It involves the formation of new substances that have different properties compared to the original substance. Examples of chemical changes include:\r\n\r\n \tthe rusting of iron objects when exposed to moist air for a long period of time\r\n\tsugar heated in a saucepan to make caramel\r\n \tthe burning of potassium in water\r\n \tmilk turning sour.\r\n\r\n\r\nThese processes can be used to release substances like oils, colours and vapours from plant material, allowing these extracts to be used in medicines and other applications.\r\n\r\n\r\n\r\nYour turn \u2013 physical and chemical changes\r\nTest yourself on your understanding of physical and chemical changes.\r\n\r\n\r\n\r\n const manual = new Freezeframe('.freeze1', { trigger: false }); document .querySelector('.start') .addEventListener('click', () => { manual.start(); }); document .querySelector('.stop') .addEventListener('click', () => { manual.stop(); }); \r\nconst manual2 = new Freezeframe('.freeze2', { trigger: false }); document .querySelector('.start2') .addEventListener('click', () => { manual2.start(); }); document .querySelector('.stop2') .addEventListener('click', () => { manual2.stop(); }); \r\n\r\n","n":0.054},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Atoms","i":2,"n":1},{"v":"Matter","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":230,"$":[{"v":"Units and conversions","n":0.577},{"v":"Units are used to describe physical quantities. Being able to express units in the most appropriate way is crucial for scientific research, engineering, everyday tasks, and effective communication across measurement systems.\r\n\r\nA unit is a standard measurement used to express and quantify physical quantities. They provide a reference point that lets you measure and compare different quantities consistently. \r\nIf you expressed a quantity without a unit, it wouldn't mean much. Let's say that you tell a friend that you have gained \\(5.68\\textrm{ kg}\\) over the past three months. If you only say \\(5.68\\) without kilograms, it won't make much sense to your friend. The number \\(5.68\\) can be in any units we use to measure mass, such as grams, milligrams, kilograms, pounds, etc. Your friend might even think you gained \\(5.68\\) centimetres in height!\r\nAdding a unit next to the quantity is essential to communicate the correct information. We usually write the symbol for the unit, such as kg for kilograms.\r\n\r\n\r\n\r\nSI units\r\nA physical quantity can be described with several different units. For instance, weight can be expressed in kilograms, grams, pounds, etc. To avoid confusion, scientists have introduced standard units for each physical quantity. This system is known as the international system of units (SI units). Some SI units and their symbols are shown.\r\n\r\n\r\n\r\n\r\nQuantity\r\nUnit\r\nSymbol\r\n\r\n\r\n\r\n\r\nmass\r\nkilogram\r\n\\(kg\\)\r\n\r\n\r\nlength\r\nmetre\r\n\\(m\\)\r\n\r\n\r\nvolume\r\ncubic metre\r\n\\(m^{3}\\)\r\n\r\n\r\ntemperature\r\nKelvin\r\n\\(K\\)\r\n\r\n\r\ntime\r\nsecond\r\n\\(s\\)\r\n\r\n\r\nelectric current\r\nampere\r\n\\(A\\)\r\n\r\n\r\namount of substance\r\nmole\r\n\\(mol\\)\r\n\r\n\r\nluminous intensity\r\ncandela\r\n\\(cd\\)\r\n\r\n\r\n\r\n\r\nSI prefixes\r\nA common issue associated with any measuring system is that depending on the quantity, the unit we use might be inconvenient to express. For example, expressing the diameter of an atom \\(\\left(0.0000000002\\textrm{ m}\\right)\\) in metres is challenging. Therefore, prefixes have been introduced to express such measurements.\r\n\r\n\r\n\r\nPrefix\r\nSymbol\r\nMultiplied by\r\n\r\n\r\n\r\n\r\ngiga\r\n\\(G\\)\r\n\\(10^{9}\\)\r\n\r\n\r\nmega\r\n\\(M\\)\r\n\\(10^{6}\\)\r\n\r\n\r\nkilo\r\n\\(k\\)\r\n\\(10^{3}\\)\r\n\r\n\r\ndeci\r\n\\(d\\)\r\n\\(10^{-1}\\)\r\n\r\n\r\ncenti\r\n\\(c\\)\r\n\\(10^{-2}\\)\r\n\r\n\r\nmilli\r\n\\(m\\)\r\n\\(10^{-3}\\)\r\n\r\n\r\nmicro\r\n\\(\\mu\\)\r\n\\(10^{-6}\\)\r\n\r\n\r\nnano\r\n\\(n\\)\r\n\\(10^{-9}\\)\r\n\r\n\r\npico\r\n\\(p\\)\r\n\\(10^{-12}\\)\r\n\r\n\r\n\r\n\r\nPrefixes allow us to express the diameter of an atom in a more sensible way: \\(2 \\times 10^{\u201310}\\textrm{ m}\\). This representation of a physical quantity is called scientific notation.\r\n\r\n\r\nExample \u2013 converting between units\r\nConvert the following quantities into the required units:\r\n\r\n\\(150.0\\textrm{ mg}\\) to grams\r\n\\(0.275\\textrm{ g}\\) to milligrams\r\n\r\n\r\nWrite a conversion factor. In this case, we know from the prefixes that:\r\n\\[ 1000\\textrm{ mg} = 1\\textrm{ g}\\] Both \\(1000\\textrm{ mg}\\) and \\(1\\textrm{ g}\\) refer to the same amount, but use different units. Based on this relationship, we can create pair of conversion factors:\r\n\\[ \\frac{1000\\textrm{ mg}}{1\\textrm{ g}}\\:\\:\\textrm{and}\\:\\:\\frac{1\\textrm{ g}}{1000\\textrm{ mg}} \\]\r\nMultiply the original quantity by the correct conversion factor. This is the conversion factor with the units we want in the numerator (top of the fraction) and the units we have in the denominator (bottom of the fraction).\r\n\r\nFor part a, we want to convert from mg to g. So, we need to use the conversion factor with g in the numerator: \\(\\dfrac{1\\textrm { g}}{1000\\textrm{ mg}}\\). This gives us:\r\n\\[ 150.0\\textrm{ mg}\\times\\frac{1\\textrm{ g}}{1000\\textrm{ mg}}=0.1500\\textrm{ g}\\]\r\n\r\n\r\nFor part b, we want to convert from g to mg. So, we need to use the conversion factor with mg in the numerator: \\(\\dfrac{1000\\textrm{ mg}}{1\\textrm{ g}}\\). This gives us: \r\n\\[ 0.275\\textrm{ g}\\times\\frac{1000\\textrm{ mg}}{1\\textrm{ g}}=275\\textrm{ mg}\\]\r\n\r\nWhen you look at these two calculations, you can see that the conversion factor used from the pair depends on the required units.\r\n\r\n\r\nExercise \u2013 converting between units\r\n\r\n\tWrite the conversion factors for the following.\r\n\t\r\n\t\tcentimetres and metres\r\n\t\tmicrolitres and millilitres\r\n\t\tseconds and hours\r\n\t\r\n\tConvert the following quantities into the required units.\r\n\t\r\n\t\t\\(250.0\\textrm{ mg}\\) to grams\r\n\t\t\\(0.567\\textrm{ mL}\\) to microlitres\r\n\t\t\\(14.8\\textrm{ mL}\\) to litres\r\n\t\t\\(0^{\\circ}\\textrm{C}\\) to Kelvin (\\(\\textrm{K}=^{\\circ}\\textrm{C}+273\\))\r\n\t\r\n\r\n\r\n\r\n\r\n","n":0.044},[{"v":"Units","i":1,"n":1},{"v":"Measurement","i":0,"n":1}]]},{"i":231,"$":{"0":{"v":"Subject support","n":0.707},"2":[{"v":"Life science","i":0,"n":0.707}]}},{"i":232,"$":[{"v":"Argument analysis exercises","n":0.577},{"v":"The exercise on this page gives you the chance to practise and sharpen your argument analysis skills.\r\nIn this article, note how the writer has structured the paragraphs by focusing on one key argument that has its claim as the topic sentence. Each argument is supported by evidence and the use of language to persuade the reader. There are also assumptions used to support the writer's argument and overall contention.\r\n\r\nPractise argument analysis\r\nSteps for analysing the argument:\r\n\r\n1) Read the Mandatory Conscription: A necessity for national security.\r\n\r\n2) Identify the contention, the arguments, the claims, the evidence and underlying assumptions as well as the use of language to persuade you as the reader.\r\n\r\n3) Consider the quality of the overall argument in this text.\r\n\r\n \tThink of what specific additional evidence might weaken or lend support to the claims.\r\n \tAsk yourself what changes in the argument would make the reasoning more sound.\r\n\r\n\r\n\r\nMandatory conscription:A necessity for national security\r\nBy Major Kendra Horan\r\n\r\nThe reintroduction of mandatory conscription in Australia is a necessary step to bolster our national security and instil a sense of civic duty among young Australians. After all, what is more important than the safety of our families? \r\n\r\nCritics, generally made up of people who have never served their country, argue that conscription is an outdated and coercive practice, but they overlook the significant benefits it can bring to both individuals and the nation. Countries with mandatory conscription have stronger, more resilient defence forces. By requiring all eligible young adults to serve, Australia would ensure a steady influx of trained personnel, ready to respond to any threats. Furthermore, conscription fosters discipline, teamwork, and a sense of responsibility, qualities that are invaluable in both military and civilian life.\r\n\r\nMandatory conscription can serve as a powerful equaliser in society, providing opportunities for personal growth and development that might otherwise be inaccessible. Research published in the Journal of Military and Strategic Studies indicates that conscription programmes often lead to increased social cohesion and a greater understanding of national values. A common consensus across a number of researchers suggests that young Australians from diverse backgrounds would come together, breaking down social barriers and building lasting bonds.\r\n\r\nCritics who claim that conscription disrupts education and career plans fail to recognise the long-term advantages of the skills and experiences gained through military service. In fact, I would say that the vast majority of former conscripts would say that their time in the military was instrumental in shaping their future success.\r\n\r\n\r\n\r\n","n":0.05},[{"v":"Media literacy","i":1,"n":0.707},{"v":"Critical thinking","i":0,"n":0.707}]]},{"i":233,"$":[{"v":"Staying focused in online classes","n":0.447},{"v":"Are you new to online study? Want to discover some strategies for effective online learning? Here are some tips for maximising your online class time.\r\n\r\n   \r\n\r\nReflection\r\n\r\nReflect on how you learn best and apply it to your study habits.\r\n\r\n    Do you like to work with background noise?Or do you need silence?\r\n    Do you like to work in a room with others?Or do you prefer to work alone?\r\n    Are you most productive in the morning?Or do you prefer to work in the evening?\r\n    Do you like to take notes on pen and paper?Or do you take notes on a computer or laptop?\r\n\r\n\r\nMaking a good impression online\r\nCommunicating clearly and having positive interactions with your classmates and teachers is just as important in online classes as it is in person.\r\nMost people are used to talking to their friends and family online, but they might have less experience interacting in an online academic space. Do you know how to make a good impression in online classes?\r\n\r\nPlay the video to get some tips on:\r\n\r\n \tpositioning your camera\r\n \tengaging with group members\r\n \tcommunicating with your peers\r\n \twriting posts and chat messages\r\n \tfinding online support.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nG'day, I'm Tull! In this video I'm going to give you some tips on how to make a good impression online. It's important to remember that university is an academic and professional space. You could be studying among future employers or colleagues.\r\n\r\nSo how can you put your best foot forward online? When you're speaking, it's nice to turn your camera on if you can, so that your classmates and teacher can get to know you. Position the camera so that you are facing straight on, it might help to prop your laptop or computer up with a stand or some books. Try sitting in front of a window so that the natural sunlight lights your face. Think about your background. What's behind you? Can you angle yourself so that you have a neutral background? Of course, it's not always possible to turn your camera on or sit in front of a window or have a nice background but just do your best and don't let this stop you from contributing in class.\r\n\r\nIt's also helpful to mute your microphone when you aren't speaking. This minimises the background noise for everyone else. If your teacher puts you in a breakout room, it can help you to engage with your group members if you switch your camera on. If you can't switch your camera on, that's OK! Just let your group members know that you're still there, you're listening, and you're ready to talk.\r\n\r\nAlways try to be culturally sensitive and use inclusive language when you speak to your classmates and teachers. Be aware of making jokes and using sarcasm as humour can easily be misinterpreted online. Avoid using slang words as not everyone may understand these. And never use offensive language! Proofread what you type before you press send. And remember, the main class chat is not a place to have private discussions! Whether you are typing a message in the class chat, giving a presentation, responding on a discussion board, or speaking aloud in an online class, remember to be polite and respect others' points of view. Show people the same courtesy online that you would in person.\r\n\r\nSo, use these tips. Make the most of your online classes and join in! Contribute your ideas and listen to others. Ask questions and remember there are lots of places in the university where you can ask for support such as the Library, IT, and Student Connect.\r\n\r\nI hope these tips have been helpful. See you later!\r\n","n":0.041},[{"v":"Online learning","i":1,"n":0.707},{"v":"Online classes","i":0,"n":0.707}]]},{"i":234,"$":[{"v":"Mastering online class participation","n":0.5},{"v":"Maximise your university experience by learning how to put your best foot forward online and get the most out of your online classes.\r\nStaying focused and motivated while participating in online learning is a lifelong skill that will help you beyond university and in the workplace. These tutorials will help you prepare for your online classes as well as empower you to contribute in a respectful and meaningful way.\r\nPreparing for online learning\r\n\r\nBefore you attend your first online lecture or class, give yourself time to get familiar with Canvas and Collaborate Ultra. Learn about your technical settings, including your audio and your video.\r\nCanvas is RMIT's online Learning Management System. On Canvas, you will find all of your course content, including assessment tasks, and past lecture recordings.\r\nOn Collaborate Ultra, you can connect to your live online lectures and classes via the Collaborate Ultra link in Canvas.\r\nFamiliarise yourself with Canvas and Collaborate Ultra.\r\nNavigating Canvas\r\nLearn more about Canvas, like how to set up your notifications and announcements, find your assessment tasks, grades, feedback, and more in this video.\r\n\r\n\r\nCollaborate Ultra functions\r\nMake sure you understand the functions of Collaborate Ultra.\r\nDo you know how to:\r\n\r\nlog on to Collaborate Ultra?\r\nturn your audio on and off?\r\nturn your video on and off?\r\nraise and lower your hand?\r\nsend a message in the chat?\r\nfind your audio and video settings?\r\n\r\nRemember, before each online lecture or class:\r\n\r\n \tlog in early to allow time for any technical issues check your battery, power supply, and internet connection\r\n \tcheck your audio and video settings\r\n \tturn your sound up.\r\n\r\n\r\n\r\n\r\n \r\n\r\n[Start transcript]\r\n[Music]\r\n[Nhuan Pham]\r\nHi, my name is Nhuan and I\u2019m a student here at RMIT studying a Bachelor of Art.\r\n\r\nToday, I\u2019m here to walk you through how to log in and navigate Canvas, RMIT\u2019s Learning Management System.\r\n\r\nThe first thing you will need to do is open up your preferred web browser. Visit RMIT\u2019s website, which is www.rmit.edu.au. Select the student link at the top of your screen in the navigation bar.\r\n\r\n[Music continues]\r\n\r\nOnce you click on the Canvas link, you will be redirected to a login portal to access your personal RMIT section of the website. Login using your student number and password. When you are ready, click login.\r\n\r\nThe dashboard is the first think you will see when you log in to Canvas. The dashboard helps you see what is happening in all of your current courses. On the right side, you will see any work you need to complete, feedback, grades you have received and any upcoming events.\r\n\r\nThe global navigation menu is located on the left side of every page in Canvas. The global navigation links provide quick access to frequently used Canvas features. You can return to your dashboard at any time by clicking the dashboard link in the global navigation.\r\n\r\nThe first link in the global navigation is where you will find your Canvas account settings. In here, you can upload your profile picture and adjust your notification preferences. Canvas allows you to customise how you receive notifications from all of your courses. You can add different contact methods to receive notifications. So, it\u2019s just announcements and grades to your mobile number, the Canvas app or an alternative email address.\r\n\r\nIt\u2019s important that you spend time reviewing your notifications. And, in particular, make sure you have announcements turned on. This ensures you receive any information from your teachers straight away. For more information on setting up your notifications, see the links to the guide below the video.\r\n\r\nWhen you are ready, select your dashboard, and then click into the course you want to view. First is your homepage, which will be the first think you seen in your course. Next up, you\u2019ll have announcements which will have any messages from your teachers. After that it\u2019s your course syllabus. This contains the course summary, including topics, readings and assessment dates.\r\n\r\nIn a course, all your learning materials will be located in modules. Each module can contain files, discussions, assignments and other resources. You can select modules from the course navigation. And once you are there, you will be able to see a welcome module with introduction information and course resources organised according to the course delivery.\r\n\r\nIn Canvas, all your assessments information can be found in the assignment link. By clicking on the link, you get a list of all your assessment tasks for the course. When you click on the name of an assessment, it will display all information from the teachers about the task, including the due date and the rubric. It will also allow you to submit your assessment when the time comes. And finally, this is where you will be able to see your grades and view your feedback.\r\n\r\nIf you ever run into any trouble, you can always reach out to Canvas support. They are available 24 hours a day, seven days a week. You can find all your help options in the help link in the Canvas global navigation. Plus, you can always speak to your teacher.\r\n\r\nCanvas also has a student app, which you can download for free from the App Store. Canvas mobile apps for iOS and android gives you access to your course whenever you need it and wherever you are. You\u2019ll see the dashboard where you can access information in each of your courses and information about any groups that you are in.\r\n\r\nUsing the menu at the bottom of the screen, you can access your dashboard, calendar, task list, notifications and inbox. The app is really easy to use, and it means you can get updates, announcements and notifications instantly. You can participate in discussions when you are out and about. And it\u2019s a great way to get all your reading done on the train. You can also access all of your assignments easily and view the calendar or task list to check what assignments are due soon.\r\n\r\nGood luck and welcome to Canvas.\r\n\r\n[Music continues]\r\n\r\n[End transcript]\r\n","n":0.032},[{"v":"Online learning","i":1,"n":0.707},{"v":"Online classes","i":0,"n":0.707}]]},{"i":235,"$":[{"v":"Writing academic emails","n":0.577},{"v":"There are times at university when you need to email your lecturers, tutors, or other university staff members to communicate personal or non-public matters.\r\n\r\n\r\nUniversity emails are generally expected to be concise pieces of formal or semi-formal written communication. Learning how to structure an academic email, and the content to include, can help you get your message or request across clearly and effectively.\r\nEmails generally include the following sections:\r\n\r\n \tsubject line\r\n \tterms of address or greetings\r\n \tcontent\r\n \tsign off\r\n\r\nLet's explore these sections in more detail.\r\n\r\n\r\n\r\n\r\n\r\n","n":0.11},[{"v":"Study skills","i":3,"n":0.707},{"v":"Online learning","i":2,"n":0.707},{"v":"Online communication","i":1,"n":0.707},{"v":"Email communication","i":0,"n":0.707}]]},{"i":236,"$":[{"v":"Interacting on a discussion board","n":0.447},{"v":"Posting on discussion boards will give you the opportunity to improve your communication and collaboration skills by interacting with your peers and engaging more deeply with your course.\r\nHere are some tips for getting the most out of the experience and for interacting with your classmates in a thoughtful and respectful way.\r\nAcknowledge and respond to other people's posts\r\nRemember that discussion boards are not meant to be a series of unrelated posts on a topic. Therefore, it is important to read what other students have written, think about their view and respond respectfully. Then, other students will consider your post, and add their ideas to the discussion.\r\n\r\nHopefully, through this process, students will learn from each other, consider issues from other points of view and to develop deeper insights into the issue.\r\nAsking for further information\r\nTo help the discussion develop, you can ask your classmates for more information about their posts. The following are examples of how this can be done respectfully:\r\n\r\nCould you please explain your point a little more?\r\nCould you elaborate on that?\r\nI'm not entirely clear on this. Could you explain it further?\r\nCould you please clarify your last statement?\r\nThat\u2019s an interesting way to see the issue. Would you mind providing some more detail?\r\n\r\nBe open to new ideas\r\nTreat discussion board activities as a chance to learn from your peers. Even if you have strong ideas on the topic of discussion, there is always a chance to see the issue through someone else\u2019s eyes.\r\nAgreeing and disagreeing\r\nWhen interacting on discussion boards, you will often be required to agree or disagree with something a classmate has written. When doing this, you are required to write more than \u2018I agree\u2019 or \u2018I disagree\u2019. You will also need to provide a justification for your stance. By giving examples and explanations, you will help to develop a richer discussion.\r\nAvoid emotional responses\r\nIf you are passionate about the topic which is being discussed, it is possible that someone may write something that triggers an emotional response.\r\nInstead of posting immediately, take a moment to compose yourself before responding. Remember that this is an academic discussion and that our posts should be professional.","n":0.054},[{"v":"Discussion boards","i":1,"n":0.707},{"v":"Assignments","i":0,"n":1}]]},{"i":237,"$":[{"v":"Content and structure","n":0.577},{"v":"Typically, there will be two types of discussion boards that you may be asked to participate in: general discussions and task-based discussions. Let's find out how they are different and what is expected in each.\r\n\r\nWhen writing a post on a discussion board it's important to consider the nature of your post. While most discussion posts follow the same general rules, there are some differences in terms of the content you need to cover and the structure you need to follow.\r\n\r\nThe following information will help you understand the differences between these two types of discussion board posts.\r\n\r\nWriting a general discussion post\r\nHere are some simple tips can help you carry out meaningful discussions online. For a general discussion post:\r\n\r\n \tPost early and before you post, make sure you read all other posts to avoid repeating what others have already said.\r\n \tWrite clearly and concisely. As you can't rely on eye contact and body language, it's better to make sure you write clearly so that everyone can understand what you mean.\r\n \tRefer to points others have made in their posts when you respond. This will help them understand how your ideas relate to the discussion.\r\n \tExplain yourself clearly by elaborating on your answers or by sharing personal experiences and examples.\r\n \tCheck your grammar and punctuation and avoid using slang or abbreviations.\r\n \tAlways be respectful and remember that what you write may come across differently when it's read by others.\r\n\r\nWriting a task-based discussion post\r\nHere are some tups for writing a task-based discussion post:\r\n\r\n \tRead the task instructions carefully.\r\n \tMake sure you follow the required number of words, and the number of sources you need to have.\r\n \tCheck out the referencing requirements, for example the number of references to include in your post, and follow the referencing guidelines in your posts.\r\n \tRead the task rubric to understand what you are being marked on.\r\n \tEdit and proofread before you submit your post.\r\n","n":0.056},[{"v":"Online learning","i":1,"n":0.707},{"v":"Discussion boards","i":0,"n":0.707}]]},{"i":238,"$":[{"v":"Using discussion boards","n":0.577},{"v":"This tutorial explains online discussion board, which are commonly used in university courses to get students to engage with each other and share ideas.\r\n\r\nParticipating effectively in online discussions\r\nDiscussion boards are platforms on which you can communicate asynchronously* with your peers and tutors during an online course. The purpose of a discussion board is to provide you with the opportunity to exchange ideas, share content, and ask questions. Discussion boards are also used for giving and receiving feedback.\r\n\r\n*Not happening at the same time.\r\nWriting a discussion board post\r\nLearn how to start a thread and write a post on an online discussion board by exploring some tips and activities.\r\n\r\nStart and name a thread\r\nWhen starting a new discussion, the first step is to name your thread. These tips will help you title threads effectively:\r\n\r\n \tinclude keywords or only the most important information\r\n \tkeep thread titles short\r\n \tadd information like the unit number or unit name to make it easy for others to find your thread.\r\n\r\nDiscussion board title activity\r\nChoose the most suitable title for a discussion thread in the activity below.\r\n\r\n ","n":0.075},[{"v":"Online learning","i":1,"n":0.707},{"v":"Discussion boards","i":0,"n":0.707}]]},{"i":239,"$":[{"v":"Elements of an argument","n":0.5},{"v":"Let's unpack some of the key terms involved in argument analysis. On this page, you'll find some of the terminology you'll need to get started, along with explanations and examples.\r\n\r\nContention\r\n\r\n\r\n\r\nA contention is the main point of view or opinion the writer is wanting to convey to the reader, usually supported by the claims.\r\nAn author's contention is the main claim they're trying to prove throughout their article.\r\n\r\n \tPlacing their contention at the beginning is the most direct method, and has the effect of positioning the reader towards the author's beliefs from the outset.\r\n \tA contention placed at the end of an article can have the effect of seeming like a valid, logical conclusion to a well-thought through discussion.\r\n \tThe contention can also be repeated throughout the article. The author may have chosen to present it in this way in order to continue reiterating their main point in the audience's minds, aligning them to their views.\r\n\r\n\r\nArgument\r\nThis is a key message that a writer wants to convey to the readers to support their contention. The writer will use of a range of key arguments to support their contention. The writer will use language and evidence to try to persuade you to a particular point of view.\r\n\r\nHaving determined the contention of the overall argument you are making, you will need to identify key arguments (can be called sub-arguments) to support the overall contention of the article. This argument should take a strong stance so readers understand exactly what your claim is.\r\n\r\n\r\nClaims\r\nThese are the key points of the argument. Each key argument should have a central claim. This is often presented as the topic sentence for the key argument.\r\nA claim is a statement asserted within the argument that can be assessed to be true or false.\r\n\r\n \tEach claim can be supported by a number of sub claims, arguments or evidence.\r\n \tThe claim may contain additional contextual material, for example explaining terms used and scope.\r\n\r\n\r\n\r\nEvidence\r\nEvidence is information that an author weaves into the writing in order to substantiate claims. It supports the claims and arguments and can include support for the claim, such as examples, statistics, quotes etc. In academic writing, evidence used from other sources must be referenced.\r\nEvidence is presented in support of the claim and argument. All evidence provides some level of relevance. Always analyse the quality of each piece of evidence being presented to you as the reader. You need to be able to determine the quality of evidence as being strong or weak. Also consider whether the writer adequately presents evidence that contradicts their argument. This can indicate a writer that is presenting a balanced perspective.\r\n\r\n \tStrong evidence: Sufficient evidence, high quality evidence.\r\n \tWeak evidence: Insufficient evidence, poor quality evidence.\r\n\r\nA contention needs to be supported by relevant, reliable, credible, sufficient and current evidence.\r\n\r\nIf there is no evidence, it isn't an argument - it is only an opinion, a description, or an explanation.\r\n\r\nQuestion the evidence the author has used.\r\n\r\n \tDoes the evidence provided support the argument?\r\n \tHow credible (trustworthy) is the evidence?\r\n \tHas enough evidence been presented?\r\n \tDoes it come from different sources?\r\n \tAre there any missing links between the evidence and the author's contention?\r\n \tIs the evidence current?\r\n \tIs there other evidence available that might challenge the evidence used?\r\n\r\nTake notice of linking words and phrases which demonstrate cause/reason, effect/results, to help identify the evidence used to support the argument.\r\n\r\n\r\nAssumption\r\nAssumptions are information the writer believes the readers will already know. Assumptions are embedded in the arguments. They are seldom explicit and therefore can be difficult to identify.\r\nSometimes we assume the answer or reason we have given is \u2018obvious' or \u2018common sense'. This is because we believe (often incorrectly) that other people think and feel the same as we do. Assumptions come from our personal beliefs and values; they may be justified, or not. Often assumptions are also unstated or \u2018hidden' because the writer feels it is unnecessary to explain.\r\n\r\nWeak or false arguments are based on incorrect assumptions and are not well-supported by credible evidence.\r\n\r\n\r\nPersuasive language\r\nUsing persuasive language is a deliberate choice of words which are intended to influence or to trigger a response.\r\nThis refers to a range of different persuasive language techniques used by writers in their texts to trigger a particular effect on the reader.\r\n\r\nPersuasive language includes:\r\n\r\nRidicule, attacks: The writer attacks an opponent or idea. This is an assertion of the writer's viewpoint, and is used to prove the writer's point by using attacking and putting down persuasion techniques against the opponent or idea. These methods are often effective in placing the reader in a position of agreement with the writer.\r\n\r\nEmotional language:: Words used to deliberately create an emotional impact or response from the audience. The effect of using emotive language as a persuasive technique is to appeal to people's emotions and trigger an emotional response.\r\n\r\nHyperbole: A hyperbole is an extreme exaggeration used to make a point for emphasis. They are not meant to be taken literally. By using a hyperbole, a story can quickly become interesting or sound more convincing. Hyperboles are comparisons, but are exaggerated and often ridiculous.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.035},[{"v":"Academic writing","i":0,"n":0.707}]]},{"i":240,"$":[{"v":"Online presentations","n":0.707},{"v":"There are many similarities between presenting online and face-to-face. However, there are some differences that you should take into account. Here are some tips for delivering a successful online presentation.\r\n\r\nAvoid reading from a script\r\nIt can be tempting to read from a script in online presentations, but it is best to avoid this. When you read from a script, your voice can become robotic and lack theOnline presentations\r\n\r\nIt can also be distracting for your audience if your eyes are not focused on the camera and you are clearly reading line for line.\r\n\r\nA better approach is to learn your content well and rely on notes that contain key words. This way, you will appear more confident and authoratative.\r\n\r\nPositioning your camera\r\nYour audience does not want an extreme close-up of your nose — they want to be able to see your facial expressions and hand gestures because they convey meaning. Before you start your presentation, make sure that your camera is positioned appropriately.\r\n\r\nIf you are presenting at home, move things from the background that may distract your audience, such as drying clothes.\r\nReduce background noise\r\n\r\nKeep in mind that microphones can pick up a lot of background noise and that this can be distracting for your audience. Some tips for reducing background noise are:\r\n\r\n\tif you live with housemates, let them know ahead of time that you have an important presentation and ask them if they can be quiet during this time\r\n\tclose doors and windows that may allow street noise to be captured\r\n\tif you have pets, give them something to eat before the presentation or ask a friend to take them for a walk\r\n\tif possible, use a headset rather than the microphone on your device.\r\n\r\nTest your technology\r\nWhile you are preparing for your assignment, make sure that you are familiar with the platform you will be using for the presentation. If you haven't used it before, take some time to practice with it and learn how to  mute/unmute your microphone, turn your camera on/off and share your screen.\r\nOther considerations are:\r\n\r\n\tdoes the platform have permission to access your camera and microphone?\r\n\t is your Wi-Fi working well? If not, what is your plan b? (i.e. booking a study room in the library)\r\n\tis your device charged or plugged in?\r\n\tare there any pending updates on your device?\r\n\r\nMistakes and lessons learnt\r\nNow, let's find out how you can avoid some common presentation issues. Select the arrow button to move through the slides and read about how you can avoid some common presentation issues.\r\n","n":0.049},[{"v":"Study skills","i":3,"n":0.707},{"v":"Presentations","i":2,"n":1},{"v":"Online learning","i":1,"n":0.707},{"v":"Digital tools","i":0,"n":0.707}]]},{"i":241,"$":[{"v":"Using collaboration tools","n":0.577},{"v":"When your group has decided how to keep in touch, it\u2019s time to consider how you will work together from separate locations.\r\nOnline collaboration tools and platforms have become the preferred method of group work in many educational and professional settings. Understanding the basics of using these online tools to share, edit, and save documents will be extremely useful to your future working life and make group assignments easier and more efficient.\r\nWhat is an online collaboration platform?\r\nAn online collaboration platform is a digital workspace, which allows people to collaborate no matter where they are. Online platforms often include cloud storage and office applications, and sometimes chat, email, and online forums. Some examples are Microsoft 365 and Google Drive.\r\nWhat is a co-authoring tool?\r\nCo-authoring files can be stored in the cloud and shared with other people so that real-time reviewing and editing can take place, like it would in an office or classroom. These tools generally include features such as autosave, comments, track changes, user tagging, version history, and notifications. For example, Microsoft Word and Google Docs co-authoring tools allow multiple people to work on the same file at the same time.\r\nKey terms\r\nWhile collaborating online, you will likely come across the following key terms:\r\nLocal file: A digital file stored on your personal computer.\r\nCloud file: A digital file stored on internet servers (the cloud) rather than on your computer, which can be kept private or shared with others.\r\nCo-authoring: Collaborative authoring (working on a document together at the same time).\r\nReal-time editing: Working together on a document at the same time rather than working individually and sending updates.\r\nCollaboration steps\r\nBy following these three steps, you will be able to set your group up for online collaboration success.\r\nStep 1: Decide as a group which collaboration platform and tools you are going to use.\r\nStep 2: Create your online collaboration space.\r\nStep 3: Start creating, sharing, and co-authoring your work as a group.\r\nChoosing your collaboration tools\r\nRemember that the best option for a collaboration tool is the one which suits the whole group. Keep in mind that not everyone has used these tools before, so you might need to assist some group members in setting up collaboration spaces and documents, or you might need to ask them to help you - that's all part of group work!\r\nHere are some of the available options:\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.051},[{"v":"Study skills","i":3,"n":0.707},{"v":"Online learning","i":2,"n":0.707},{"v":"Group work","i":1,"n":0.707},{"v":"Digital tools","i":0,"n":0.707}]]},{"i":242,"$":[{"v":"Working well together online: Empathy","n":0.447},{"v":"When working on a group task, it's important to remember that we are all have our strengths and weaknesses, and that external factors can sometimes affect our university life.\r\nFor example, it can be easy to consider that team members are lazy when they are falling behind on group work. However, there may be reasons behind why they are struggling to keep up with a task.\r\nHere are some tips for getting to know your team members and understanding their situation:\r\n\r\n \tTake some time in the beginning to get to know each other. This will make it easier to communicate and work together.\r\n\r\n \tYou may like to start with a getting-to-know-you activity.\r\n\r\n\r\n \tAsk team members what their strengths are and what they have difficulties with. This can help your group to assign tasks to the best person.\r\n \tBe mindful that your group members may have family or work commitments, so they may not be able to respond to messages or emails straightaway.\r\n \tSometimes you will encounter a group member who is not contributing. Maybe they are shy or maybe they don't understand the task. Reach out and see if they are ok.\r\n \tIf you can't contact a group member, let your lecturer or tutor know as they may need extra support.\r\n\r\nTest your knowledge\r\nRead the scenario in the activity and decide what Stephen should do.\r\n\r\n\r\n","n":0.067},[{"v":"Study skills","i":3,"n":0.707},{"v":"Online learning","i":2,"n":0.707},{"v":"Online classes","i":1,"n":0.707},{"v":"Group work","i":0,"n":0.707}]]},{"i":243,"$":[{"v":"Working well together online: Respect","n":0.447},{"v":"This page outlines some useful tips on how to work well with your classmates online.\r\nAlthough it may be tempting to share a screenshot of something funny from your group meetings or post a cool infographic your team member has made on your social media, remember that it's important to respect everyone's digital privacy. You will need to ask your group members for permission before sharing anything they have created.\r\n\r\nHere are some examples of things you'll need permission to share:\r\n\r\n \ta recording of a virtual meeting or an audio conversation\r\n \ta screenshot of discussions between group members\r\n \ta screenshot of group members in virtual meetings\r\n \tacademic or creativework which has been shared by a group member.\r\n\r\nBe empathetic, inclusive, and listen to all group members. Remember that different opinions, expressed respectfully, should lead to a valuable discussion - not conflict.\r\n\r\nGive feedback to your group members but remember to remain respectful. Don't make it personal. See Interacting on a discussion board for more information on how to give and respond to feedback.\r\n\r\nIf you cannot resolve a conflict and it's getting in the way of your work, get in touch with your tutor or lecturer.\r\nTest your knowledge\r\nRead the scenario in the activity and decide what Stephen should do.\r\n\r\n","n":0.07},[{"v":"Study skills","i":3,"n":0.707},{"v":"Online learning","i":2,"n":0.707},{"v":"Online classes","i":1,"n":0.707},{"v":"Group work","i":0,"n":0.707}]]},{"i":244,"$":[{"v":"Working well together online: Organisation","n":0.447},{"v":"This section outlines some useful tips on how to work well with your classmates online.\r\nTaking time to plan at the beginning of your group project will set you up to succeed. Here are some tips for getting started:\r\n\r\n \tAs a group, analyse your task. Read the assignment instructions and the marking rubric. Agree on a goal. See the Planning page for more ideas.\r\n \tEnsure that everyone is involved and assign group roles.\r\n \tAcknowledge any differences in culture, skills, and abilities that your group members bring.\r\n \tSomeone may be more comfortable as a leader while another member might like to be a researcher or an editor. Create your own roles that will suit your project or assignment.\u00a0See the Starting page for ideas.\r\n \tCreate a group timeline and schedule any virtual meetings.\r\n \tBe mindful of everyone\u2019s time zones.\r\n \tIf you need to have a virtual meeting, agree on an appropriate time for all group members.\r\n\r\nFor more advice on how to succeed at group assignments, visit the Group work section of the Learning Lab.\r\nTest your knowledge\r\nRead the scenario in the activity and decide what Stephen should do.\r\n\r\n","n":0.074},[{"v":"Study skills","i":3,"n":0.707},{"v":"Online learning","i":2,"n":0.707},{"v":"Online classes","i":1,"n":0.707},{"v":"Group work","i":0,"n":0.707}]]},{"i":245,"$":[{"v":"Working well together online: Communication","n":0.447},{"v":"Clear communication is the key to success in online group work, while poor communication can lead to disastrous results.\r\nHere are some tips for communicating effectively online.\r\n\r\n\r\n \tAgree on how you will communicate.\r\n\r\n \tWill you use instant messaging applications, email, or discussion boards?\r\n \tWill you hold virtual meetings? See Choosing communication tools for more information.\r\n\r\n\r\n \tTake a moment to proofread your emails and posts before you hit send. Check your spelling and punctuation, use complete sentences and capital letters where appropriate.\r\n \tAvoid slang or jargon as it might not be understood by your group members.\r\n \tJokes and sarcasm are often dependent on tone of voice and body language to convey the message. It is best to avoid jokes and sarcasm online, unless you know your group members well.\r\n\r\nTest your knowledge\r\nRead the scenario in the activity and decide which strategies Stephen could use.\r\n\r\n\r\n","n":0.084},[{"v":"Study skills","i":3,"n":0.707},{"v":"Online learning","i":2,"n":0.707},{"v":"Online classes","i":1,"n":0.707},{"v":"Group work","i":0,"n":0.707}]]},{"i":246,"$":[{"v":"Choosing communication tools","n":0.577},{"v":"Once you and your group members have introduced yourselves, it's time to select the tools your group will use to stay connected.\r\nThere are many different tools to choose from. Your group might decide to use just one or multiple communication tools during your project. This section will take you through some of the tools you might like to use.\r\nInstant messaging applications\r\ne.g. WhatsApp, WeChat, Telegram, SMS\r\nBenefits:\r\n\r\n \tInstant messaging allows for efficient real-time communication.\r\n \tMessages can be sent from anywhere at any time - only a smartphone is needed.\r\n \tMessaging apps offer multiple forms of communication, such as text messages, voice and video chat, and image sharing.\r\n\r\n\r\nVirtual meeting tools\r\ne.g. Collaborate Ultra (RMIT), Teams, Zoom, Skype\r\nBenefits:\r\n\r\n \tReal-time collaboration often feels more social and collaborative.\r\n \tScreen-sharing, and file sharing functions are easy to use.\r\n \tMeetings can be recorded for those who can't attend.\r\n \tFace-to-face communication can help avoid miscommunication.\r\n\r\n\r\nEmail\r\ne.g. Outlook, Gmail, Yahoo!Mail\r\nBenefits:\r\n\r\n \tEmails can be written and sent at any time, so there is no need for everyone to be online.\r\n \tEmail accounts are free, and all students have an RMIT student email.\r\n \tEmail does not use a large amount of data or bandwidth.\r\n \tInformation is kept in one location and can be found later if necessary.\r\n\r\n\r\nUnit discussion board\r\nRMIT: Canvas\r\nBenefits:\r\n\r\n \tPosts can be written at any time, so there is no need for everyone to be online.\r\n \tRMIT students have access to the unit discussion boards on Canvas.\r\n \tEverything is in one place and can be found and referred to easily by all group members.\r\n \tFiles and images can be linked or embedded.\r\n\r\n\r\nSocial media\r\ne.g. Facebook, LinkedIn\r\nBenefits:\r\n\r\n \tMany students are frequently on social media applications, which will allow for prompt responses.\r\n \tSocial media can be accessed from both personal computers and smartphones.\r\n \tSocial media applications allow the sharing of multimedia and the option of video calls.\r\n \tCommunicating over social media can break down barriers and help group members become more comfortable with each other.\r\n\r\n\r\nTest your knowledge\r\nEach slide includes a scenario and an interactive question. Don't forget to check your answer before moving on to the next scenario.\r\n\r\n ","n":0.054},[{"v":"Study skills","i":3,"n":0.707},{"v":"Online learning","i":2,"n":0.707},{"v":"Group work","i":1,"n":0.707},{"v":"Digital skills","i":0,"n":0.707}]]},{"i":247,"$":[{"v":"Working well together online","n":0.5},{"v":"The ability to communicate and collaborate with team members online is an essential skill, in both academic and professional spaces. Learning how to collaborate effectively when doing group work online will help you get the most out of your units and assessments. This is an important digital literacy skill which is highly valued by employers.\r\n\r\nThese tutorials will help you:\r\n\r\n \tunderstand some of the issues which may occur when working together online and the best practices for resolving them\r\n \tlearn about the wide range of tools available for group communication and collaboration and what to consider when choosing which ones to use\r\n \tlearn about digital tools available for creating online presentations, and some useful strategies to help you present in a digital learning environment.\r\n\r\nTo be able to work well together online and become fully engaged, it is important to know how to:\r\n\r\n \tget organised for virtual environments\r\n \tcommunicate clearly and respectfully\r\n \tshow empathy to your group members.\r\n\r\n\r\n\r\n","n":0.08},[{"v":"Study skills","i":3,"n":0.707},{"v":"Online learning","i":2,"n":0.707},{"v":"Online communication","i":1,"n":0.707},{"v":"Group work","i":0,"n":0.707}]]},{"i":248,"$":[{"v":"Online learning skills","n":0.577},{"v":"Mastering online learning skills is essential for success in today's digital world. This section offers resources to enhance your abilities in collaboration, discussion boards, professional email writing, and online class participation, improving your digital literacy and academic experience.\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.162},[{"v":"Online learning","i":1,"n":0.707},{"v":"Digital skills","i":0,"n":0.707}]]},{"i":249,"$":[{"v":"Quiz \u2013 Concentration and dilutions","n":0.447},{"v":"If you are handling solutions, being able to complete calculations related to concentrations and dilutions is critical. Test your ability to do these with a quiz.\r\n\r\nYour turn \u2013 concentration and dilutions\r\n\r\n \tDetermine the volume (in \\(\\textrm{mL}\\)) of ethanol required to prepare \\(80\\textrm{ mL}\\) of a \\(70.0\\%(v/v)\\) solution.\r\n\r\n\t\r\n\r\n\r\n \tA solution contains \\(3.56\\textrm{ g}\\) of sodium chloride dissolved in a final volume of \\(25.0\\textrm{ mL}\\). Calculate the percent concentration of sodium chloride in \\(\\%(m/v)\\).\r\n\r\n\t\r\n\t\r\n\r\n \tWhen \\(2.50\\textrm{ g}\\) of \\(\\ce{LiBr}\\) is dissolved in \\(25.0\\textrm{ g}\\) of water, it gives a \\(10.0\\%(m/m)\\) \\(\\ce{LiBr}\\) solution. Calculate the mass of \\(\\ce{LiBr}\\) present in \\(15.0\\textrm{ g}\\) of the same solution.\r\n\r\n\t\r\n\t\r\n\r\n \tCalculate the mass in grams of sodium chloride required to prepare \\(500\\textrm{ mL}\\) of a \\(2.0\\%(m/v)\\) sodium chloride solution.\r\n\r\n\t\r\n\r\n\r\n \tBrass is a metal alloy composed of \\(\\ce{Cu}\\) and \\(\\ce{Zn}\\). Usually, \\(\\ce{Cu}\\) accounts for \\(60.0\\%(m/m)\\) of the composition. Determine the mass of brass that would contain \\(48.3\\textrm{ g}\\) of \\(\\ce{Cu}\\).\r\n\r\n\t\r\n\r\n\r\n \tCalculate the quantities of solute or solvent required to prepare the following solutions:\r\n\t\r\n\t\tthe mass of \\(\\ce{K2SO4}\\) required to prepare \\(48.0\\textrm{ g}\\) of a \\(3.00\\%(m/m)\\) \\(\\ce{K2SO4}\\) solution\r\n\t\r\n\t\t\r\n\t\r\n\t\tthe volume of acetone required to prepare \\(1.0\\textrm{ L}\\) of a \\(70.0\\%(v/v)\\) acetone solution\r\n\t\r\n\t\t\r\n\t\r\n\t\tthe mass of sodium azide required to prepare \\(500\\textrm{ mL}\\) of \\(1.50\\%(m/v)\\) sodium azide solution.\r\n\t\r\n\t\t\r\n\r\n\r\n\r\n \tThe maximum chlorine concentration allowed in drinking water is \\(4.0\\textrm{ mg}\\) per \\(1.0\\textrm{ L}\\) (\\(1.0\\textrm{ kg}\\)).\r\n\t\r\n\t \tExpress the maximum chlorine concentration in drinking water in ppm.\r\n\t\r\n\t\t\r\n\t\r\n\t \tCalculate the maximum mass of chlorine allowed in \\(240\\textrm{ mL}\\) or \\(240\\textrm{ g}\\) of drinking water.\r\n\t \r\n\t \t\r\n\r\n\r\n\r\n \tCalculate the molarity of the following solutions:\r\n\t\r\n\t\t\\(3.0\\textrm{ mol}\\) of silver nitrate in \\(0.50\\textrm{ L}\\) of solution\r\n\t\r\n\t\t\r\n\t\r\n \t\t\\(0.0625\\textrm{ mol}\\) of sodium bicarbonate in \\(250.0\\textrm{ mL}\\) of solution.\r\n\t\r\n\t\t\r\n\r\n\r\n\r\n \tCalculate the mass in grams of the solute in each of the following solutions:\r\n\t\r\n\t\t\\(750.0\\textrm{ mL}\\) of a \\(1.50\\textrm{M}\\) \\(\\ce{NaOH}\\) solution (\\(M(\\ce{NaOH})=39.997\\textrm{ g mol}^{-1}\\))\r\n\t\r\n\t\t\r\n \t\r\n\t\t\\(0.200\\textrm{ L}\\) of a \\(150.0\\textrm{mM}\\) \\(\\ce{NaCl}\\) solution (\\(M(\\ce{NaCl})=58.44\\textrm{ g mol}^{-1}\\)).\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n \tCalculate the volume (in litres) of \\(0.200\\textrm{M}\\) solution that can be prepared from \\(48.00\\textrm{ g}\\) of each of the following solutes:\r\n\t\r\n\t \t\\(\\ce{KCl}\\), \\(M=74.551\\textrm{ g mol}^{-1}\\)\r\n\t\r\n\t\r\n\t\r\n \t\t\\(\\ce{NaNO3}\\), \\(M=84.99\\textrm{ g mol}^{-1}\\).\r\n\t\r\n\t\r\n\r\n\r\n\r\n \tCalculate the molar concentration of a \\(5.0\\%(m/v)\\) \\(\\ce{KOH}\\) solution (\\(M=56.1056\\textrm{ g mol}^{-1}\\).\r\n\r\n\t\r\n\r\n\r\n \tCalculate the volume (in millilitres) of water that should be added to a \\(3.00\\textrm{M}\\) stock solution of \\(\\ce{HCl}\\) to make \\(100.0\\textrm{ mL}\\) of \\(1.00\\textrm{M}\\) \\(\\ce{HCl}\\) solution?\r\n\r\n\t\r\n\r\n\r\n \tA dilute \\(\\ce{NaOH}\\) solution was prepared by adding \\(100.0\\textrm{ mL}\\) of water to \\(400.0\\textrm{ mL}\\) of \\(0.1\\textrm{M}\\) stock solution. Determine the molar concentration of the diluted \\(\\ce{NaOH}\\) solution.\r\n\r\n\t\r\n\r\n\r\n \tCalculate the final volume of an \\(\\ce{H2SO4}\\) solution prepared from \\(100.0\\textrm{ mL}\\) of \\(0.5000\\textrm{M}\\) \\(\\ce{H2SO4}\\) if the final concentration is \\(0.1500\\textrm{M}\\).\r\n\r\n\t\r\n\t\r\n\r\n \tWater is added to \\(50.0\\textrm{ mL}\\) of \\(\\ce{NaOH}\\) stock solution with a concentration of \\(10.0\\%(m/v)\\) to make a final volume of \\(250.0\\textrm{ mL}\\). Determine the final concentration of the solution in \\(\\%(m/v)\\).\r\n\r\n\t\r\n","n":0.047},[{"v":"Dilutions","i":2,"n":1},{"v":"Concentrations","i":1,"n":1},{"v":"Chemistry exercises","i":0,"n":0.707}]]},{"i":250,"$":[{"v":"Quiz \u2013 The mole","n":0.5},{"v":"The ability to complete calculations using moles is very important in chemistry. Test yourself with a quiz.\r\n\r\nYour turn \u2013 mole calculations\r\n\r\n\r\n \tAspirin has the formula \\(\\ce{C}_{9}\\ce{H}_{8}\\ce{O}_{4}\\). Calculate the mass of aspirin in grams present in \\(0.650\\textrm{ mol}\\) of aspirin. The molar masses of \\(\\ce{C}\\), \\(\\ce{H}\\) and \\(\\ce{O}\\) are \\(12.01\\textrm{ g mol}^{\u20131}\\), \\(1.01\\textrm{ g mol}^{\u20131}\\), and \\(16.00\\textrm{ g mol}^{\u20131}\\), respectively.\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n\tCalculate the amount of \\(\\ce{C}\\), \\(\\ce{H}\\) and \\(\\ce{O}\\), in moles, present in \\(1.5\\textrm{ mol}\\) of glucose. The chemical formula of glucose is \\(\\ce{C}_{6}\\ce{H}_{12}\\ce{O}_{6}\\).\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n\tCalculate the mass of carbon in grams present in \\(1.8\\textrm{ g}\\) of glucose. The chemical formula of glucose is \\(\\ce{C}_{6}\\ce{H}_{12}\\ce{O}_{6}\\). The molar masses of \\(\\ce{C}\\), \\(\\ce{H}\\) and \\(\\ce{O}\\) are \\(12.01\\textrm{ g mol}^{\u20131}\\), \\(1.01\\textrm{ g mol}^{\u20131}\\), and \\(16.00\\textrm{ g mol}^{\u20131}\\), respectively.\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n \tCalculate the number of hydrogen atoms present in \\(1.5\\textrm{ mol}\\) of hydrogen.\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n \tCaffeine is an ingredient in coffee and tea that acts as a central nervous system stimulant. The chemical formula of caffeine is \\(\\ce{C}_{8}\\ce{H}_{10}\\ce{N}_{4}\\ce{O}_{2}\\).\r\n\r\n\r\n\t\r\n\t\tCalculate the molar mass of caffeine. The atomic masses are: \\(\\ce{C}=12.01\\textrm{ amu}\\), \\(\\ce{H}=1.01\\textrm{ amu}\\), \\(\\ce{O}=16.00\\textrm{ amu}\\), and \\(\\ce{N}=14.01\\textrm{ amu}\\).\r\n\t\r\n\t\r\n\t\r\n\t\tCalculate the number of molecules of caffeine present in a \\(970.95\\textrm{ g}\\) sample of caffeine.\r\n\t\r\n\t\r\n\r\n\r\n\t\r\n\r\n \tCalculate the mass in grams of a \\(1.2\\textrm{ mol}\\) sample of \\(\\ce{CO}_{2}\\). The molar masses of \\(\\ce{C}\\) and \\(\\ce{O}\\) are \\(12.01\\textrm{ g mol}^{\u20131}\\), \\(16.00\\textrm{ g mol}^{\u20131}\\), respectively.\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n \tA balloon filled with helium gas contains \\(1.505\\times10^{26}\\) atoms of helium. Determine the mass of helium present inside the balloon. The molar mass of \\(\\ce{He}=4.00\\textrm{ g mol}^{\u20131}\\).\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n \tRibose is a simple sugar present in RNA, the molecule involved in protein synthesis in living organisms. Ribose has the chemical formula \\(\\ce{C}_{5}\\ce{H}_{10}\\ce{O}_{5}\\). Calculate the amount of \\(\\ce{C}\\), \\(\\ce{H}\\) and \\(\\ce{O}\\) present in \\(0.8\\textrm{ mol}\\) of ribose sugar.\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n \tCalcium phosphate, \\(\\ce{Ca}_{3}\\left(\\ce{PO}_{4}\\right)_{2}\\), is an essential mineral present in the bones of many living organisms. Calculate the number of formula units of calcium phosphate present in \\(31.02\\textrm{ g}\\) of calcium phosphate sample. The molar masses of \\(\\ce{Ca}=40.08\\textrm{ g mol}^{\u20131}\\), \\(\\ce{P}=30.97\\textrm{ g mol}^{\u20131}\\), and \\(\\ce{O}=16.00\\textrm{ g mol}^{\u20131}\\).\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n \tChlorofluorocarbons have been used as refrigerants for many years. \\(\\ce{CFC}\\textrm{-}12\\) is one chlorofluorocarbon with the chemical formula \\(\\ce{CCl}_{2}\\ce{F}_{2}\\). Determine the mass of chlorine present in \\(12.9\\textrm{ g}\\) of a \\(\\ce{CFC}\\textrm{-}12\\) sample. \\(M(\\ce{CFC}\\textrm{-}12)=120.91\\textrm{ g mol}^{\u20131}\\) and \\(M(\\ce{Cl})=35.45\\textrm{ g mol}^{\u20131}\\).\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n \tKeratin is a fibrous protein found in our skin, hair and nails. It has the chemical formula \\(\\ce{C}_{28}\\ce{H}_{48}\\ce{N}_{2}\\ce{O}_{32}\\ce{S}_{4}\\). A sample of keratin contains \\(0.71\\textrm{ g}\\) of \\(\\ce{N}\\). Determine the mass of \\(\\ce{O}\\) present in the sample if the molar mass of \\(\\ce{N}\\) is \\(14.01\\textrm{ g mol}^{\u20131}\\) and \\(\\ce{O}\\) is \\(16.00\\textrm{ g mol}^{\u20131}\\).\r\n\r\n\t","n":0.049},[{"v":"The mole","i":3,"n":0.707},{"v":"Measurement","i":2,"n":1},{"v":"Chemistry exercises","i":1,"n":0.707},{"v":"Chemistry","i":0,"n":1}]]},{"i":251,"$":[{"v":"Quiz \u2013 Stoichiometry","n":0.577},{"v":"Have you mastered calculating reactants and products? Put your stoichiometry skills to the test with a quiz.\r\n\r\nYour turn \u2013 stoichiometry calculations\r\n\r\n\r\n \t\\(\\ce{CO}\\) reacts with \\(\\ce{O}_{2}\\) to form \\(\\ce{CO}_{2}\\) gas as shown by the equation:\r\n \t\\[2\\ce{CO}+\\ce{O}_{2}\\rightarrow2\\ce{CO}_{2}\\]\r\n\t\r\n\t\tCalculate the amount in moles of \\(\\ce{O2}\\) required to react with \\(0.58\\textrm{ mol}\\) of \\(\\ce{CO}\\).\r\n\t\r\n\t\r\n\t\r\n\t\tCalculate the amount in moles of \\(\\ce{CO2}\\) produced when \\(0.35\\textrm{ mol}\\) or \\(\\ce{CO}\\) reacts with \\(\\ce{O2}\\).\r\n\t\r\n\t\r\n\t\r\n\t\tCalculate the amount in moles of \\(\\ce{O2}\\) that must react with \\(\\ce{CO}\\) to produce \\(0.49\\textrm{ mol}\\) or \\(\\ce{CO2}\\).\r\n\t\r\n\t\r\n\r\n\r\n\r\n \tThe reaction between hydrazine and oxygen gas produces nitric oxide and water vapour according to the following balanced chemical equation:\r\n\t\\[\\ce{N}_{2}\\ce{H}_{4}\\left(\\textrm{l}\\right)+3\\ce{O}_{2}\\left(\\textrm{g}\\right)\\rightarrow2\\ce{NO}_{2}\\left(\\textrm{g}\\right)+2\\ce{H}_{2}\\ce{O}\\left(\\textrm{g}\\right)\\]\r\n\t\r\n\t \tCalculate the amount in moles of \\(\\ce{O}_{2}\\) required to react with \\(0.6\\textrm{ mol}\\) of \\(\\ce{N}_{2}\\ce{H}_{4}\\).\r\n\t\r\n\t\r\n\t\r\n \t\tCalculate the amount in moles of \\(\\ce{N}_{2}\\ce{H}_{2}\\) that must react with \\(\\ce{O}_{2}\\) to form \\(0.4\\textrm{ mol}\\) of \\(\\ce{NO}_{2}\\).\r\n\t\r\n\t\r\n\r\n\r\n\r\n \tCalculate the mass of \\(\\ce{O}_{2}\\) required to form \\(10.50\\textrm{ g}\\) of \\(\\ce{CO}_{2}\\) from the reaction between methane and oxygen:\r\n\t\\[\\ce{CH}_{4}+2\\ce{O}_{2}\\rightarrow\\ce{CO}_{2}+2\\ce{H}_{2}\\ce{O}\\]\r\n\tThe molar masses of \\(\\ce{CO}_{2}\\) and \\(\\ce{O}_{2}\\) are \\(44.01\\textrm{ g mol}^{-1}\\) and \\(32.00\\textrm{ g mol}^{-1}\\).\r\n\r\n\r\n\r\n\r\n \tConsider the equation for the following chemical reaction:\r\n\t\\[3\\ce{H}_{2}\\left(\\textrm{g}\\right)+\\ce{N}_{2}\\left(\\textrm{g}\\right)\\rightarrow2\\ce{NH}_{3}\\left(\\textrm{g}\\right)\\]\r\n\tCalculate the mass of \\(\\ce{N}_{2}\\) required to produce \\(48.5\\textrm{ g}\\) of \\(\\ce{NH}_{3}\\). The molar masses of \\(\\ce{NH}_{3}\\) and \\(\\ce{N}_{2}\\) are \\(17.03\\textrm{ g mol}^{-1}\\) and \\(28.02\\textrm{ g mol}^{-1}\\).\r\n\r\n\r\n\r\n\r\n \tSulfur dioxide gas is produced by the reaction between sulfur and oxygen: \r\n\t\\[\\ce{S}+\\ce{O}_{2}\\rightarrow\\ce{SO}_{2}\\]\r\n\t\\(12.8\\textrm{ g}\\) of \\(\\ce{S}\\) reacts with \\(18.5\\textrm{ g}\\) of \\(\\ce{O}_{2}\\). The molar masses of \\(\\ce{S}\\) and \\(\\ce{O}_{2}\\) are \\(32.07\\textrm{ g mol}^{-1}\\) and \\(32.00\\textrm{ g mol}^{-1}\\), respectively. The molar mass of \\(\\ce{SO2}\\) is \\(64.07\\textrm{ g mol}^{-1}\\).\r\n\t\r\n\t \tIdentify the limiting reagent.\r\n\t\r\n\t\r\n \t\r\n\t\tCalculate the maximum mass of \\(\\ce{SO}_{2}\\) that can be formed under the given conditions.\r\n\t\r\n\t\r\n\r\n\r\n\r\n \t\\(\\ce{HCl}\\) is formed according to the following reaction.\r\n\t\\[\\ce{H}_{2}+\\ce{Cl}_{2}\\rightarrow2\\ce{HCl}\\]\r\n\tCalculate the maximum mass of \\(\\ce{HCl}\\) that can be produced when \\(0.85\\textrm{ g}\\) of \\(\\ce{H}_{2}\\) reacts with \\(42.58\\textrm{ g}\\) of \\(\\ce{Cl}_{2}\\). The atomic molar masses of \\(\\ce{H}\\) and \\(\\ce{Cl}\\) are \\(1.01\\textrm{ g mol}^{-1}\\) and \\(35.45\\textrm{g mol}^{-1}\\).\r\n\r\n\r\n\r\n\r\n \tCalcium nitrite (\\(\\ce{Ca}_{3}\\ce{N}_{2}\\)) is produced using \\(\\ce{Ca}\\) and \\(\\ce{N}_{2}\\) as follows:\r\n\t\\[3\\ce{Ca}+\\ce{N}_{2}\\rightarrow\\ce{Ca}_{3}\\ce{N}_{2}\\]\r\n\tCalculate the theoretical yield of \\(\\ce{Ca}_{3}\\ce{N}_{2}\\) when \\(120.56\\textrm{ g}\\) of \\(\\ce{Ca}\\) is reacted with \\(42.87\\textrm{ g}\\) of \\(\\ce{N}_{2}\\). The atomic molar mass of \\(\\ce{Ca}\\) is \\(40.08\\textrm{g mol}^{-1}\\), and the atomic molar mass of \\(\\ce{N}\\) is \\(14.01\\textrm{g mol}^{-1}\\). The molar mass of \\(\\ce{Ca}_{3}\\ce{N}_{2}\\) is \\(148.26\\textrm{g mol}^{-1}\\).\r\n\r\n\r\n\r\n\r\n \tSalicylic acid (\\(\\ce{C}_{7}\\ce{H}_{6}\\ce{O}_{3}\\)) reacts with acetic anhydride (\\(\\ce{C}_{4}\\ce{H}_{6}\\ce{O}_{3}\\)) to form acetylsalicylic acid (\\(\\ce{C}_{9}\\ce{H}_{8}\\ce{O}_{4}\\)), which we call aspirin.\r\n\t\\[2\\ce{C}_{7}\\ce{H}_{6}\\ce{O}_{3}+\\ce{C}_{4}\\ce{H}_{6}\\ce{O}_{3}\\rightarrow2\\ce{C}_{9}\\ce{H}_{8}\\ce{O}_{4}+\\ce{H}_{2}\\ce{O}\\] \r\n\tThe percentage yield of aspirin from the reaction is \\(88.48\\%\\). Determine the mass of aspirin that was actually formed by the reaction of \\(250.82\\textrm{ g}\\) of salicylic acid. The molar masses of salicylic acid and acetylsalicylate acid are \\(138.21\\textrm{ g mol}^{-1}\\) and \\(180.16\\textrm{ g}\\), respectively.\r\n\r\n","n":0.048},[{"v":"The mole","i":2,"n":0.707},{"v":"Stoichiometry","i":1,"n":1},{"v":"Chemistry exercises","i":0,"n":0.707}]]},{"i":252,"$":[{"v":"Quiz \u2013 Chemical equations","n":0.5},{"v":"Do you feel confident about writing balanced equations for chemical reactions? Test yourself with a quiz.\r\n\r\nYour turn \u2013 chemical equations\r\nFor questions 1—5, identify the reactants and products in the chemical equations.\r\n\r\n\t\\(\\ce{2H}_{2}+\\ce{O}_{2}\\rightarrow2\\ce{H}_{2}\\ce{O}\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{CH}_{4}+2\\ce{O}_{2}\\rightarrow\\ce{CO}_{2}+\\ce{H}_{2}\\ce{O}\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(2\\ce{Fe}_{2}\\ce{O}_{3}+3\\ce{C}\\rightarrow4\\ce{Fe}+3\\ce{CO}_{2}\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\left(\\ce{NH}_{4}\\right)_{2}\\ce{Cr}_{2}\\ce{O}_{7}\\rightarrow\\ce{Cr}_{2}\\ce{O}_{3}+4\\ce{H}_{2}\\ce{O}+\\ce{N}_{2}\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{CaCO}_{3}\\rightarrow\\ce{CaO}+\\ce{CO}_{2}\\)\r\n\r\n\t\r\n\r\nFor questions 6—10, write balanced equations for the chemical reactions.\r\n\r\n\r\n\tAmmonia gas reacts with oxygen gas to produce nitrogen gas and water.\r\n\r\n\t\r\n\r\n\r\n\tSolid sodium sulfate reacts with solid elemental carbon to produce solid sodium sulfite and carbon dioxide gas.\r\n\r\n\t\r\n\r\n\r\n\tSolid elemental iron reacts with oxygen gas to produce solid ferric oxide.\r\n\r\n\t\r\n\r\n\r\n\tSolid ammonium chloride separates into ammonia and hydrogen chloride gas.\r\n\r\n\t\r\n\r\n\r\n\tGaseous hydrogen iodide decomposes into iodine gas and hydrogen gas.\r\n\r\n\t\r\n\t\r\nFor questions 11—22, balance the equations.\r\n\r\n\t\\(\\ce{FeI}_{2}(\\textrm{s})+\\ce{Cl}_{2}(\\textrm{g})\\rightarrow\\ce{FeCl}_{3}(\\textrm{s})+\\ce{I}_{2}(\\textrm{s})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{Fe}_{2}\\ce{O}_{3}(\\textrm{s})+\\ce{C}(\\textrm{s})\\rightarrow\\ce{Fe}(\\textrm{s})+\\ce{CO}_{2}(\\textrm{g})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{HNO}_{3}(\\textrm{l})+\\ce{H}_{2}\\ce{S}(\\textrm{g})\\rightarrow\\ce{NO}(\\textrm{g})+\\ce{S}(\\textrm{s})+\\ce{H}_{2}\\ce{O}(\\textrm{l})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{SO}_{2}(\\textrm{g})+\\ce{H}_{2}\\ce{O}(\\textrm{l})\\rightarrow\\ce{H}_{2}\\ce{SO}_{3}(\\textrm{aq})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{AgNO}_{3}(\\textrm{aq})+\\ce{KCl}(\\textrm{aq})\\rightarrow\\ce{KNO}_{3}(\\textrm{aq})+\\ce{AgCl}(\\textrm{s})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{PCl}_{3}(\\textrm{l})+\\ce{H}_{2}(\\textrm{g})\\rightarrow\\ce{PH}_{3}(\\textrm{g})+\\ce{HCl}(\\textrm{g})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{PbO}(\\textrm{s})+\\ce{NH}_{3}(\\textrm{g})\\rightarrow\\ce{Pb}(\\textrm{s})+\\ce{N}_{2}(\\textrm{g})+\\ce{H}_{2}\\ce{O}(\\textrm{l})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{NaClO}_{3}(\\textrm{s})\\rightarrow\\ce{NaCl}(\\textrm{s})+\\ce{O}_{2}(\\textrm{g})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{N}_{2}\\ce{O}(\\textrm{g})\\rightarrow\\ce{N}_{2}(\\textrm{g})+\\ce{O}_{2}(\\textrm{g})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{NH}_{4}\\ce{NO}_{3}(\\textrm{s})\\rightarrow\\ce{N}_{2}(\\textrm{g})+\\ce{O}_{2}(\\textrm{g})+\\ce{H}_{2}\\ce{O}(\\textrm{g})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{Mn}_{2}\\left(\\ce{CO}_{3}\\right)_{3}(\\textrm{s})\\rightarrow\\) \\(\\ce{Mn}_{2}\\ce{O}_{3}(\\textrm{s})+\\ce{CO}_{2}(\\textrm{g})\\)\r\n\r\n\t\r\n\r\n\r\n\t\\(\\ce{Na2CO3}(\\textrm{aq})+\\ce{Mg(NO3)2}(\\textrm{aq})\\rightarrow\\ce{MgCO3}(\\textrm{s}) +\\ce{NaNO3}(\\textrm{aq})\\)\r\n\r\n\t","n":0.099},[{"v":"Reactions","i":2,"n":1},{"v":"Equations","i":1,"n":1},{"v":"Chemical equations","i":0,"n":0.707}]]},{"i":253,"$":[{"v":"Quiz \u2013 Covalent bonding and Lewis structures","n":0.378},{"v":"Covalent bonding is a fundamental concept in chemistry that explains how atoms share electrons to form molecules. Molecular substances can be represented using Lewis structures. Test your understanding of Lewis structures and covalent compounds with this quiz.\r\n\r\nYour turn \u2013 covalent bonding and Lewis structures\r\n\r\n\r\n\tDraw the Lewis structures of the following molecules:\r\n\r\n\r\n\r\n\t\\(\\ce{SO}_{3}\\)\r\n\t\\(\\ce{COCl}_{2}\\)\r\n\t\\(\\ce{HF}\\)\r\n\t\\(\\ce{SF}_{6}\\)\r\n\t\\(\\ce{CS}_{2}\\)\r\n\t\\(\\ce{SeF}_{4}\\)\r\n\t\\(\\ce{AlCl}_{3}\\)\r\n\t\\(\\ce{Cl}_{2}\\ce{O}\\)\r\n\t\\(\\ce{SiBr}_{4}\\)\r\n\t\\(\\ce{N}_{2}\\ce{O}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\tDraw the Lewis structures of the following polyatomic ions:\r\n\r\n\r\n\r\n\t\\(\\ce{SO}_{3}^{2-}\\)\r\n\t\\(\\ce{PO}_{4}^{3-}\\)\r\n\t\\(\\ce{NO}_{2}^{-}\\)\r\n\t\\(\\ce{PF}_{4}^{+}\\)\r\n\t\\(\\ce{AlCl}_{4}^{-}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n \tDraw the Lewis structures for the following compounds:\r\n\t\r\n\t\t\\(\\ce{Cl}_{2}\\)\r\n\t\t\\(\\ce{HI}\\)\r\n\t\t\\(\\ce{PCl}_{3}\\).\r\n\t\r\n\r\n\r\n\r\n\r\n \tDetermine the molecular formulas for the following compounds by identifying the values for the subscripts \\(v,\\,w,\\,x\\) and \\(z\\):\r\n\t\r\n\t\t\\(\\ce{H}_{x}\\ce{S}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{CCl}_{z}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{PF}_{w}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{HBr}_{v}\\)\r\n\t\r\n\t\r\n\r\n\r\n\r\n \tDraw the structural formulas for the following compounds:\r\n\t\r\n\t\t\\(\\ce{C}_{2}\\ce{H}_{4}\\)\r\n\t\t\\(\\ce{N}_{2}\\)\r\n\t\t\\(\\ce{SO}_{3}\\)\r\n\t\t\\(\\ce{HCN}\\).\r\n\t\r\n\r\n\r\n\r\n\r\n \tUse your understanding of coordinate covalent bonds to explain the formation of the following polyatomic ions:\r\n\t\r\n\t\t\\(\\ce{NH}_{4}^{+}\\) from \\(\\ce{NH}_{3}\\) and \\(\\ce{H}^{+}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{H}_{3}\\ce{O}^{+}\\) from \\(\\ce{H}_{2}\\ce{O}\\) and \\(\\ce{H}^{+}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{BF}_{4}^{-}\\) from \\(\\ce{BF}_{3}\\) and \\(\\ce{F}^{-}\\).\r\n\t\r\n\t\r\n\r\n \r\n\r\n\r\n\r\n \tPredict the geometry of the following molecules:\r\n\t\r\n\t\t\\(\\ce{PF}_{3}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{SCl}_{2}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{SiBr}_{4}\\).\r\n\t\r\n\t\r\n\r\n\r\n\r\n \tPredict the geometry around each of the carbon atoms in \\(\\ce{C}_{2}\\ce{Cl}_{2}\\).\r\n\r\n\r\n\r\n\r\n \tPredict the geometry of the following polyatomic ions:\r\n\t\r\n\t\t\\(\\ce{NH}_{2}^{-}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{CO}_{3}^{2\u2013}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{SO}_{4}^{2-}\\).\r\n\t\r\n\t\r\n\r\n\r\n\r\n \tThe structure and 3D view of the amino acid aspartic acid is shown below. \r\n\t\r\n\tPredict the geometry around the indicated carbon atoms.\r\n\r\n\r\n\r\n\r\n \tDetermine whether bonds formed between the following pairs of elements are ionic, polar covalent or non-polar covalent:\r\n\t\r\n\t\t\\(\\ce{S}\\) and \\(\\ce{Cl}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{N}\\) and \\(\\ce{H}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{C}\\) and \\(\\ce{Br}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{Li}\\) and \\(\\ce{F}\\).\r\n\t\r\n\t\r\n\r\n\r\n\r\n \tPredict the polarity (polar or non-polar) of each of the following molecules:\r\n\t\r\n\t\t\\(\\ce{HCN}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{NH}_{3}\\)\r\n\t\r\n\t\r\n\t\r\n\t\t\\(\\ce{CH}_{3}\\ce{Br}\\).\r\n\t\r\n\t\r\n\r\n\r\n\r\n","n":0.07},[{"v":"Electron","i":2,"n":1},{"v":"Chemical bonding","i":1,"n":0.707},{"v":"Chemistry","i":0,"n":1}]]},{"i":254,"$":[{"v":"Quiz \u2013 Redox reactions","n":0.5},{"v":"Redox reactions, which involve the transfer of electrons between substances, are fundamental to understanding processes like combustion, respiration, and corrosion. Test your understanding of redox reactions with a quiz.\r\nYour turn \u2013 redox reactions\r\nFor questions 1\u201410, determine the oxidation numbers for the elements indicated.\r\n\r\n\r\n\t\\(\\ce{F}\\) in \\(\\ce{F}_{2}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(\\ce{Ca}\\) in \\(\\ce{Ca}^{2+}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(\\ce{S}\\) in \\(\\ce{SO}_{3}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(\\ce{P}\\) in \\(\\ce{PO}_{4}^{3-}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(\\ce{Cr}\\) in \\(\\ce{Cr}_{2}\\ce{O}_{3}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(\\ce{Al}\\) in \\(\\ce{AlCl}_{4}^{-}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(\\ce{N}\\) in \\(\\ce{NH}_{4}^{+}\\)\r\n\r\n\t\r\n\t\r\n\r\n \t\\(\\ce{Mn}\\) in \\(\\ce{KMnO}_{4}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(\\ce{Fe}\\) in \\(\\ce{Fe}_{2}\\ce{O}_{3}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(\\ce{C}\\) in \\(\\ce{CO}\\)\r\n\r\n\t\r\n\t\r\nFor questions 11\u201316, identify the oxidising agent, reducing agent, atom oxidised and atom reduced.\r\n\r\n \t\\(\\ce{FeO}+\\ce{CO}\\rightarrow\\ce{Fe}+\\ce{CO}_{2}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(3\\ce{MnO}_{2}+4\\ce{Al}\\rightarrow2\\ce{Al}_{2}\\ce{O}_{3}+3\\ce{Mn}\\)\r\n\r\n\t\r\n\t\r\n\r\n \t\\(\\ce{Zn}+\\ce{CuCl}_{2}\\rightarrow\\ce{ZnCl}_{2}+\\ce{Cu}\\)\r\n\r\n\t\r\n\t\r\n\r\n \t\\(\\ce{Cl}_{2}+2\\ce{KI}\\rightarrow2\\ce{KCl}+\\ce{I}_{2}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(2\\ce{Mg}+\\ce{CO}_{2}\\rightarrow2\\ce{MgO}+\\ce{C}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(\\ce{C}_{6}\\ce{H}_{12}\\ce{O}_{6}+6\\ce{O}_{2}\\rightarrow6\\ce{CO}_{2}+6\\ce{H}_{2}\\ce{O}\\)\r\n\r\n\t\r\n\r\nFor questions 17\u201318, balance the redox reactions, including states.\r\n\r\n \t\\(\\ce{H}^{+}+\\ce{Al}\\rightarrow\\ce{Al}^{3+}+\\ce{H}_{2}\\)\r\n\r\n\t\r\n\r\n\r\n \t\\(\\ce{Cr}_{2}\\ce{O}_{7}^{2-}+\\ce{Fe}^{2+}\\rightarrow\\ce{Cr}^{3+}+\\ce{Fe}^{3+}\\)\r\n\r\n\t","n":0.1},[{"v":"Redox reactions","i":3,"n":0.707},{"v":"Chemistry exercises","i":2,"n":0.707},{"v":"Chemistry","i":1,"n":1},{"v":"Chemical equations","i":0,"n":0.707}]]},{"i":255,"$":[{"v":"Concentration and dilution","n":0.577},{"v":"Concentration and dilution play vital roles in everyday life, from fine-tuning the strength of your morning coffee to ensuring the correct potency of medications. When mixing paints, artists rely on these concepts to achieve the perfect hue, while in the kitchen, chefs adjust the concentration of flavours to create culinary masterpieces. Understanding these principles is key to navigating and manipulating the mixtures that we encounter in our daily routines.\r\n\r\nExpressing concentrations\r\nConcentration (\\(c\\)) refers to the amount of solute present in a given quantity (known volume) of a solution. It is common to encounter concentration of a solute expressed using \\(c(\\textrm{chemical formula})\\) or \\(\\textrm{[chemical formula]}\\). For example, the concentration of \\(\\ce{NaCl}\\) can be expressed as \\(c(\\ce{NaCl}\\)) or \\([\\ce{NaCl}]\\).\r\n\r\nConcentration can be represented using a range of units:\r\n\r\n \tpercent concentration (%)\r\n \tparts per million (ppm)\r\n \tparts per billion (ppb)\r\n \tmolarity (M).\r\n\r\nPercent concentration\r\nConcentration can be represented as percentage. There are three ways to do this, depending on how the solution is prepared.\r\n\r\nWhen you dissolve a mass of solute in a mass of solvent, concentration can be expressed using percentage by mass or mass/mass, \\(\\%(\\textrm{m/m}\\)). This is calculated using:\r\n\r\n\\[ \\frac{\\textrm{Mass of solute (g)}}{\\textrm{Mass of solution (g)}}\\times100\\% \\]\r\n\r\nWhen you dissolve a volume of solute in a volume of solvent, concentration can be expressed using percentage by volume or volume/volume, \\(\\%(\\textrm{v/v}\\)). This is calculated using:\r\n\r\n\\[ \\frac{\\textrm{Volume of solute (mL)}}{\\textrm{Volume of solution (mL)}}\\times100\\% \\]\r\n\r\nWhen you dissolve a mass of solute in a volume of solvent, concentration can be expressed using percentage mass/volume, \\(\\%(\\textrm{m/v}\\)). This is calculated using:\r\n\r\n\\[ \\frac{\\textrm{Mass of solute (g)}}{\\textrm{Volume of solution (mL)}}\\times100\\% \\]\r\nParts per million and parts per billion\r\nParts per million is the number of parts (mass or volume) present in one million parts (mass or volume). Similarly, parts per billion is the number of parts (mass or volume) present in one billion parts (mass or volume). For example, \\(1\\textrm{ ppm}\\) solute could indicate that \\(1\\textrm{ g}\\) of solute is present in \\(1000000\\textrm{ mL}\\) of solution. This is the same as \\(1\\textrm{ mg}\\) of solute in \\(1\\textrm{ L}\\) of solvent.\r\n\r\nConcentration in \\(\\textrm{ppm}\\) is calculated for solutes and solvents expressed in the same units using:\r\n\r\n\\[ \\frac{\\textrm{Parts of solute (g)}}{\\textrm{Parts of solvent (g)}}\\times1000000 \\]\r\n\r\nIn the same way, concentration in \\(\\textrm{ppb}\\) is calculated using:\r\n\r\n\\[ \\frac{\\textrm{Parts of solute (g)}}{\\textrm{Parts of solvent (g)}}\\times1000000000 \\]\r\nMolarity\r\nMolarity or molar concentration (\\(M\\)) is the amount of solute in moles present in one litre of a solution. It is calculated using:\r\n\r\n\\[ \\textrm{Molarity (M)}=\\frac{\\textrm{Amount of solute (mol)}}{\\textrm{Volume of solution (L)}} \\]\r\n\r\nFrom this, we can derive a formula that relates molar concentration (\\(c\\)) with amount of solute in moles (\\(n\\)) nand volume of solute in litres (\\(V\\)):\r\n\r\n\\[ \\textrm{c}=\\frac{n}{V} \\]\r\nDilutions\r\nDilution is a process used to lower the concentration of the original solution by adding more solvent. The concentrated solution is known as the stock solution. During this process, the amount of solute present in the solution remains constant before and after the dilution; only the volume changes.\r\n\r\nThe following equation shows the relationship between concentration and volume before and after dilution. This formula is built upon the fact that the same amount of solute is present in the solution before (stock solution) and after (diluted solution) the dilution.\r\n\r\n\\[ c_{1}V_{1}=c_{2}V_{2} \\]\r\n\r\nIn this formula,\r\n\r\n \t\\(c_{1}\\)= concentration of the stock solution\r\n \t\\(V_{1}\\)= volume of the stock solution\r\n \t\\(c_{2}\\)= concentration of the diluted solution\r\n \t\\(V_{2}\\)= volume of the diluted solution\r\n\r\nExample 1 \u2013 representing concentration using percent concentration\r\nCalculate the concentrations of the following solutions and express them in the units indicated.\r\n\r\n \tA sodium chloride solution is prepared by dissolving \\(3.25\\textrm{ g}\\) of sodium chloride in \\(42.50\\textrm{ g}\\) of water. Express the concentration of the solution using percentage mass/mass.\r\n \t\\(8.00\\textrm{ mL}\\) of ethanol is dissolved in enough water to give \\(80.00\\textrm{ mL}\\) of\u00a0solution. Calculate the percent by volume concentration of ethanol in the resulting solution.\r\n \tIf \\(5.0\\textrm{ g}\\) of \\(\\ce{MgCl2}\\) is dissolved in enough water to give \\(200\\textrm{ mL}\\) of solution, calculate the concentration of \\(\\ce{MgCl2}\\) in \\(\\%(\\textrm{m/v}\\)).\r\n\r\nStep 1: For part a, calculate the concentration using the formula for \\(\\%(\\textrm{m/m}\\)).\r\n\\[\\begin{align*}\r\nc(\\ce{NaCl}) & = \\frac{\\textrm{Mass of solute (g)}}{\\textrm{Mass of solution (g)}}\\times100\\textrm{%}\\\\\r\n& = \\frac{3.25\\textrm{ g}}{42.50\\textrm{ g}}\\times100\\%\\\\\r\n& = 7.65\\%(\\textrm{m/m})\r\n\\end{align*}\\]\r\n\r\nStep 2: For part b, calculate the concentration using the formula for \\(\\%(\\textrm{v/v}\\)).\r\n\\[\\begin{align*}\r\nc(\\textrm{Ethanol}) & = \\frac{\\textrm{Volume of solute (mL)}}{\\textrm{Volume of solution (mL)}}\\times100\\textrm{%}\\\\\r\n& = \\frac{8.00\\textrm{ mL}}{80.00\\textrm{ mL}}\\times100\\%\\\\\r\n& = 10.0\\%(\\textrm{v/v})\r\n\\end{align*}\\]\r\n\r\nStep 3: For part c, calculate the concentration using the formula for \\(\\%(\\textrm{m/v}\\)).\r\n\\[\\begin{align*}\r\nc(\\ce{MgCl2}) & = \\frac{\\textrm{Mass of solute (g)}}{\\textrm{Volume of solution (mL)}}\\times100\\%\\\\\r\n& = \\frac{5.0\\textrm{ g}}{200\\textrm{ mL}}\\times100\\%\\\\\r\n& = 2.5\\%(\\textrm{m/v})\r\n\\end{align*}\\]\r\nExample 2 \u2013 representing concentration using parts per million\r\nDrinking water contains \\(0.002\\textrm{ g}\\) of \\(\\ce{Mg}^{+}\\) ions in \\(100\\textrm{ g}\\) of water. Calculate the concentration in parts per million.\r\n\r\nStep 1: Calculate the concentration using the formula for \\(\\textrm{ppm}\\).\r\n\r\n\\[\\begin{align*}\r\nc(\\ce{Mg}^{+}) & = \\frac{0.002\\textrm{ g}}{100\\textrm{ g}}\\times1000000\\\\\r\n& = 20\\textrm{ ppm}\r\n\\end{align*}\\]\r\nExample 3 \u2013 representing concentration using molarity\r\n\\(2.50\\textrm{ mol}\\) of \\(\\ce{CuSO4}\\) is dissolved in enough water to give a \\(500.0\\textrm{ mL}\\) solution. Calculate the molar concentration of the resulting solution.\r\n\r\nStep 1: Calculate the concentration using the formula for \\(M\\). Remember that the volume needs to be converted to litres.\r\n\r\n\\[\\begin{align*}\r\nc(\\ce{CuSO4}) & = \\frac{2.50\\textrm{ mol}}{500.0\\textrm{ml}\\times\\frac{1\\textrm{ L}}{1000\\textrm{ mL}}}\\\\\r\n& = 5.00\\textrm{ M}\r\n\\end{align*}\\]\r\nExample 4 \u2013 calculating the quantity of solute using concentration\r\nCalculate the mass of glucose present in \\(250\\textrm{ mL}\\) of \\(1.50\\textrm{M}\\) glucose solution. The molar mass of glucose is \\(180.15\\textrm{ g mol}^{-1}\\).\r\n\r\nStep 1: Calculate the amount in moles using the formula for \\(M\\). Remember that the volume needs to be converted to litres.\r\n\r\n\\[\\begin{align*}\r\nc & = \\frac{n}{V}\\\\\r\nn & = c\\times V\\\\\r\n& = 1.50\\textrm{M} \\times 250\\textrm{ mL} \\times \\frac{1\\textrm{ L}}{1000\\textrm{ mL}}\\\\\r\n& = 0.375 \\textrm{ mol}\r\n\\end{align*} \\]\r\n\r\nStep 2: Calculate the mass in grams using \\(M=\\frac{m}{n}\\).\r\n\r\n\\[\\begin{align*}\r\nM & = \\frac{m}{n}\\\\\r\nm & = n \\times M\\\\\r\n& = 0.375 \\textrm{ mol} \\times 180.15\\textrm{ g mol}^{-1}\\\\\r\n& = 67.6 \\textrm{ g}\r\n\\end{align*} \\]\r\n\r\n\r\nExample 6 \u2013 calculating dilutions\r\nA chemist prepared \\(10\\textrm{ mL}\\) of a \\(1.00\\textrm{M}\\) \\(\\ce{HCl}\\) solution by adding water to the \\(3.0\\textrm{M}\\) stock solution of \\(\\ce{HCl}\\). Determine the volume of stock solution that should be taken for the dilution.\r\n\r\nStep 1: Identify the known variables and the unknown variable.\r\n\r\nWe are given the concentration of the diluted solution (\\(c_{2}=1.00\\textrm{M}\\)), the volume of the diluted solution (\\(V_{2}=10\\textrm{ mL}\\)), and the concentration of the stock solution (\\(c_{1}=3.0\\textrm{M}\\)). The unknown variable is the volume of the undiluted stock solution (\\(V_{1}\\)).\r\n\r\nStep 2: Calculate the volume of the stock solution using the dilution formula. It is helpful to rearrange the formula first. Make sure all concentration values are in the same units, and all volume values are in the same units.\r\n\r\n\\[\\begin{align*}\r\nc_{1}V_{1} & = c_{2}V_{2}\\\\\r\nV_{1} & = \\frac{c_{2}V_{2}}{c_{1}}\\\\\r\n& = \\frac{1.00\\textrm{M} \\times 10\\textrm{ mL}}{3.0\\textrm{M}}\\\\\r\n& = 3.3\\textrm{ mL}\r\n\\end{align*}\\]","n":0.031},[{"v":"Solutions","i":2,"n":1},{"v":"Concentrations","i":1,"n":1},{"v":"Chemistry exercises","i":0,"n":0.707}]]},{"i":256,"$":[{"v":"Solubility","n":1},{"v":"Solubility is the measure of how well a solute can dissolve in a solvent to form a solution. It dictates how substances dissolve and interact, impacting fields like pharmaceuticals, environmental science, cooking, and chemical reactions. Understanding solubility allows for better control and optimisation of processes in these areas. Use this resource to learn the basics of solubility and learn about a real world example of detoxifying cycad seeds.\r\n\r\nSolutes, solvents and solutions\r\nA solute is a substance that is dissolved to form a solution. A solvent is the substance that the solute dissolves in. Solutions are therefore the resulting mixture. The formation of a mixture involves attractive intermolecular forces between solutes and solvent molecules, which allow them to mix together.\r\n\r\nIn a solution, the solvent can be liquid water or something else such as alcohol-based solvents, acetone and ethyl ether. When liquid water is the solvent, the solution is an aqueous solution. If the solvent is not liquid water, we call it a non-aqueous solution.\r\n\r\nTo prepare a solution, a known mass of solute (e.g. \\(5.85\\textrm{ g}\\) of \\(\\ce{NaCl}\\)) is dissolved in a solvent, like water, to give a known volume of solution (e.g. \\(250.0\\textrm{ mL}\\) aqueous solution). If the solid solute readily dissolves in the solvent, it is soluble. If it does not readily dissolve and remains as a solid, it is considered insoluble.\r\nSaturation\r\nThe level of saturation describes the extent to which a solution contains dissolved solute relative to its maximum solubility at a given temperature and pressure. There are three ways to classify the level of saturation:\r\n\r\n \tUnsaturated solutions contain less than the maximum amount of solute that can be dissolved under given conditions.\r\n \tSaturated solutions contain the maximum amount of solute that can be dissolved under the given conditions.\r\n \tSupersaturated solutions contain more dissolved solute molecules than a saturated solution. They exist in an unstable, temporary condition. A supersaturated solution always tries to revert to a saturated state with time, through the formation of crystals.\r\n\r\n\r\n\r\n\r\n\r\nThe solubility rule\r\nThe solubility of a solute depends on whether it can form attractive forces with the solvent molecules, which in turn, depends on the types of forces it can form.\r\n\r\n \tPolar solutes dissolve in polar solvents. Polar substances can interact via electrostatic forces, due to the presence of partial or complete charges.\r\n \tNon-polar solutes dissolve in non-polar solvents. Non-polar substances are unlikely to form electrostatic forces. Instead, they interact with each other via dispersion forces.\r\n\r\nThus, when you are predicting the solubility of a solute in a solvent, you should check whether the solute is an ionic or molecular compound, and whether the solvent liquid contains polar or non-polar molecules. This concept is summarised by the solubility rule:\r\n\"Like dissolves like.\"\r\nFactors affecting solubility\r\nAs well as the polarity of the solute and solvent, there are other external factors that affect solubility. One key example is temperature. Many solids show increased solubility in water at higher temperatures. In contrast, gases generally have a lower solubility in water with increased temperature.\r\nExamples \u2013 predicting solubility\r\n\r\n\r\n\r\n\r\n","n":0.045},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Solvents","i":2,"n":1},{"v":"Solutions","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":257,"$":[{"v":"Solutions and concentration","n":0.577},{"v":"Solutions and concentration are fundamental concepts in chemistry that describe how substances mix and interact. From the sugar dissolving in your morning coffee to the saline solution used in medical treatments, understanding how solutes and solvents work together is essential for grasping how mixtures are formed and how their concentrations can affect chemical reactions. Use this resource to learn about different types and properties of solutions.\r\n\r\nOverview\r\n\r\n \tSolubility \u2013 how do substances form solutions? What kind of solutions exist and what is solubility? 25 minutes to complete\r\n \tConcentration and dilution \u2013 how is the concentration of a solution expressed? 25 minutes to complete\r\n \tQuiz \u2013 Concentration and dilution \u2013 can you complete calculations involving concentration and dilutions? 30 minutes to complete\r\n \tAcids, bases and salts \u2013 what makes a substance acid or basic? How do salts fit in? 40 minutes to complete\r\n \tQuiz \u2013 Acids, bases and salts \u2013 can you correctly classify substances as acids or bases? 20 minutes to complete\r\n \tStrength of acids and bases \u2013 how do we determine how acidic or basic a substance is? 30 minutes to complete\r\n \tQuiz \u2013 Strength of acids and bases \u2013 how well do you understand the strengths of acids and bases? 40 minutes to complete\r\n\r\n","n":0.07},[{"v":"Solutions","i":2,"n":1},{"v":"Concentrations","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":258,"$":[{"v":"Redox reactions","n":0.707},{"v":"A redox reaction, short for reduction\u2013oxidation reaction, is a chemical process in which one substance loses electrons (oxidation) while another gains electrons (reduction). Keep reading to learn more, including Aboriginal research and applications of combustion reactions.\r\n\r\nOxidation and reduction\r\nRedox reactions involve the net movement of electrons from one reactant to the other. This causes some reactants to gain electrons (reduction), while others lose electrons (oxidation). An easy way to remember which is which is by using the acronym OIL RIG:\r\n\r\n \tOxidation is the loss of electron(s) by an atom or ion.\r\n \tReduction is the gain of electron(s) by an atom or ion.\r\n\r\nRedox reactions can be described using full redox equations, such as:\r\n\r\n\\[ \\ce{CuO}(\\textrm{s})+\\ce{H2}(\\textrm{g})\\rightarrow\\ce{Cu}(\\textrm{s})+\\ce{H2O}(\\textrm{g}) \\]\r\n\r\nThis can be broken into half-equations (sometimes called \"half-reactions\"). Half-equations show the loss of electrons (oxidation) or gain of electrons (reduction):\r\n\r\n \tThe reduction half-equation shows \\(\\ce{Cu}^{2+}\\) being reduced: \\(\\ce{Cu}^{2+}+2e\\rightarrow\\ce{Cu}\\). In this half-equation, the redox couple is \\(\\ce{Cu}^{2+}\\)/\\(\\ce{Cu}\\).\r\n \tThe oxidation half-equation shows \\(\\ce{H2}\\) being oxidised: \\(\\ce{H}_{2}\\rightarrow2\\ce{H}^{+}+2e\\). In this half-equation, the redox couple is \\(\\ce{H2}\\)/\\(\\ce{H}^{+}\\).\r\n\r\nAnother example is the full redox equation: \\(2\\ce{Ag}^{+}+\\ce{Zn}\\rightarrow2\\ce{Ag}+\\ce{Zn}^{2+} \\)\r\n\r\n \tThe reduction half-equation shows \\(\\ce{Ag}^{+}\\) being reduced: \\(2\\ce{Ag}^{+}+2e\\rightarrow2\\ce{Ag}\\). The redox couple is \\(\\ce{Ag}^{+}\\)/\\(\\ce{Ag}\\).\r\n \tThe oxidation half-equation shows \\(\\ce{Zn}\\) being oxidised: \\(\\ce{Zn}\\rightarrow\\ce{Zn}^{2+}+2e\\). The redox couple is \\(\\ce{Zn}\\)/\\(\\ce{Zn}^{2+}\\).\r\n\r\nRedox equations are usually written as balanced net ionic equations. These equations exclude spectator ions (ions that do not take part in the overall redox process).\r\n\r\n\r\nOne of the most useful redox reactions is combustion. You see this reaction on your gas stovetop, where carbon-based fossil fuels are burned to heat your saucepan.\r\n\r\n\r\n\r\nOxidising agents and reducing agents\r\nOxidising agents (or oxidants) can become reduced (gains electrons) while causing another reactant to become oxidised (lose electrons). Reducing agents (or reductants) reduce another reactant by offering electrons to the other reactant. Thus, the reducing agent becomes oxidised (by losing electrons) while reducing another reactant (causing it to gain electrons).\r\nOxidation number\r\nThe oxidation number is the apparent electrical charge on the atom. It helps you keep track of the electrons during a reaction.\r\n\r\n \tWhen a chemical species is oxidised, its oxidation number increases.\r\n \tWhen a chemical species is reduced, its oxidation number decreases.\r\n\r\nYou can assign an oxidation number to any element in a given compound using the rules in the table.\r\n\r\n\r\n\r\nRule\r\nExample\r\n\r\n\r\n\r\n\r\n1. The oxidation number of an atom as an element is zero.\r\nThe \\(\\ce{O}\\) in \\(\\ce{O2}\\) has an oxidation number of \\(0\\). The oxidation numbers of \\(\\ce{N}\\) and \\(\\ce{Br}\\) are \\(0\\) in \\(\\ce{N2}\\) and \\(\\ce{Br2}\\), respectively. The oxidation number of \\(\\ce{Fe}\\) in \\(\\ce{Fe}\\) metal is also \\(0\\).\r\n\r\n\r\n2. The oxidation number of \\(\\ce{O}\\) is usually \\(-2\\), except when it is bonded to \\(\\ce{F}\\) or in a peroxide (like \\(\\ce{H2O2}\\)).\r\nThe \\(\\ce{O}\\) in \\(\\ce{H2O}\\) has an oxidation number of \\(-2\\). In \\(\\ce{OF2}\\), \\(\\ce{O}\\) has an oxidation number of \\(+2\\). In \\(\\ce{H2O2}\\), it is \\(-1\\).\r\n\r\n\r\n3. The oxidation number of a monoatomic ion is equal to the ion's charge.\r\nThe oxidation number of \\(\\ce{Na}^{+}\\) is \\(+1\\). The oxidation number of \\(\\ce{Cl}^{-}\\) is \\(-1\\). The oxidation number of \\(\\ce{S}^{2-}\\) is \\(-2\\).\r\n\r\n\r\n4. The algebraic sum of the oxidation numbers in a neutral polyatomic compound is zero.\r\n\\(\\ce{NaCl}\\) has an oxidation number of \\(0\\). The oxidation numbers of \\(\\ce{Na}\\) and \\(\\ce{Cl}\\) are \\(+1\\) and \\(-1\\), respectively. \\(+1-1=0\\)\r\n\r\n\r\n5. In a polyatomic ion, the sum of the oxidation numbers is equal to the ion's overall charge.\r\n\\(\\ce{CO}_{3}^{2-}\\) has an overall charge of \\(-2\\). The oxidation numbers of its constituent atoms are \\((+4)+(-2)+(-2)+(-2)=-2\\).\r\n\r\n\r\n6. In a compound containing multiple elements, the more electronegative element has its characteristic negative oxidation number and the less electronegative element has a positive oxidation number.\r\n\\(\\ce{N}\\) has an oxidation number of \\(-3\\), \\(\\ce{O}\\) has an oxidation number of \\(-2\\) and \\(\\ce{Cl}\\) has an oxidation number of \\(-1\\). \\(\\ce{H}\\), \\(\\ce{Ca}\\) and \\(\\ce{Al}\\) have the oxidation numbers of \\(+1\\), \\(+2\\) and \\(+3\\), respectively.\r\n\r\n\r\n7. The oxidation number of \\(\\ce{H}\\) is usually \\(+1\\) because it is less electronegative than that of the main group elements to which it is normally bonded. With more electropositive elements, \\(\\ce{H}\\) has an oxidation number of \\(-1\\).\r\nIn \\(\\ce{HCl}\\), \\(\\ce{H}\\) has an oxidation number of \\(+1\\). However, in \\(\\ce{NaH}\\), it has an oxidation number of \\(-1\\).\r\n\r\n\r\n\r\nStandard reduction potential\r\nRedox reactions involve the movement of electrons, so there is also a voltage associated with it. This is called the standard reduction potential, \\(E^{0}\\). It is the tendency for a reactant to attract electrons, cause oxidation, and be reduced itself. It is measured in volts (\\(\\textrm{V}\\)) and is defined for an oxidising agent at a concentration of \\(1\\textrm{M}\\) for solutions, \\(1\\textrm{ atm}\\) pressure for gases, and \\(25\u00baC\\) temperature.\r\n\r\nThe greater (more positive) the value of \\(E^{0}\\), the greater the tendency for the oxidising agent to spontaneously become reduced itself and cause another reactant to undergo oxidation. It will remove electrons from a redox couple with a less positive \\(E^{0}\\). Thus, there would be a spontaneous redox reaction.\r\nExample 1 \u2013 determining oxidation numbers\r\nFind the oxidation number of \\(\\ce{S}\\) in \\(\\ce{H2SO4}(\\textrm{aq})\\).\r\n\r\nStep 1: Identify the relevant oxidation number rule/s.\r\n\r\nRule 4 states that the algebraic sum of the oxidation numbers in a neutral polyatomic compound is zero, so:\r\n\r\n\\[(2\\times\\ce{H})+\\ce{S}+(4\\times\\ce{O})=0\\]\r\n\r\nRule \\(2\\) states that the oxidation number of \\(\\ce{O}\\) is \\(-2\\) and rule \\(6\\) states the oxidation number of \\(\\ce{H}\\) is \\(+1\\).\r\n\r\nStep 2: Use algebra to calculate the missing oxidation number.\r\n\r\nLet the oxidation number of \\(\\ce{S}=x\\).\r\n\r\n\\[\\begin{align*} (2\\times\\left(+1\\right))+x+(4\\times\\left(-2\\right)) & =0\\\\\r\n2+x-8 & =0\\\\\r\nx-6 & =0\\\\\r\nx & =+6\r\n\\end{align*}\\]\r\n\r\nTherefore, the oxidation number of \\(\\ce{S}\\) in \\(\\ce{H2SO4}\\) is \\(+6\\).\r\n\r\n\r\nExample 3 \u2013 predicting the products of redox reactions\r\nPredict the products formed from the spontaneous redox reaction between the two redox couples: \\(\\ce{MnO}_{4}^{-}/\\ce{Mn}^{2+}\\) and \\(\\ce{H}_{2}\\ce{SO}_{3}/\\ce{SO}_{4}^{2-}\\), given the data in the table.\r\n\r\nStep 1: Identify the redox couple with the greater/more positive \\(E^{0}\\).\r\n\r\n\\(\\ce{MnO}_{4}^{-}/\\ce{Mn}^{2+}\\) has a \\(E^{0}\\) of \\(+1.51\\textrm{ V}\\).\r\n\r\nStep 2: Identify the oxidising agent in this couple. The oxidising agent becomes reduced itself, so its oxidation number decreases.\r\n\r\nWe can use the worked example above to find the oxidation numbers of each atom in the redox pair.\r\n\r\nIn \\(\\ce{MnO}_{4}^{-}\\), the oxidation number of \\(\\ce{Mn}\\) is \\(+7\\) and the oxidation number of \\(\\ce{O}\\) is \\(-2\\). In \\(\\ce{Mn}^{2+}\\), the oxidation number of \\(\\ce{Mn}^{2+}\\) is \\(+2\\).\r\n\r\nThe oxidising agent becomes reduced, so it is the species that contains the atom with the higher oxidation number. Here, it would be \\(\\ce{MnO}_{4}^{-}\\). It has the greater affinity (liking) for electrons. Therefore, \\(\\ce{MnO}_{4}^{-}\\) will be the electron acceptor (i.e. oxidising agent).\r\n\r\nStep 3: Identify the reducing agent in the other couple. The reducing agent becomes oxidised itself, so its oxidation number increases.\r\n\r\nThe sulfur system must donate electrons; hence, \\(\\ce{H}_{2}\\ce{SO}_{3}\\) will be the electron donor (i.e. reducing agent). We can check the oxidation numbers to confirm this.\r\n\r\nIn \\(\\ce{SO}_{4}^{2-}\\), the oxidation numbers of \\(\\ce{S}\\) and \\(\\ce{O}\\) are \\(+6\\) and \\(-2\\), respectively. In \\(\\ce{H2SO3}\\), the oxidation numbers of \\(\\ce{H}\\), \\(\\ce{S}\\) and \\(\\ce{O}\\) are \\(+1\\), \\(+4\\), and \\(-2\\), respectively.\r\n\r\nThe reducing agent would contain the atom with the lower oxidation number, i.e. \\(\\ce{H2SO3}\\).\r\n\r\nStep 4: Write the half-equations for the redox reaction.\r\n\r\n\\(\\ce{MnO}_{4}^{-}\\) is reduced to \\(\\ce{Mn}^{2+}\\), so the reduction half-equation is:\r\n\r\n\\[\\ce{MnO}_{4}^{-}\\rightarrow\\ce{Mn}^{2+}\\]\r\n\r\n\\(\\ce{H2SO3}\\) is oxidised to \\(\\ce{SO}_{4}^{2-}\\), so the oxidation half-equation is:\r\n\r\n\\[\\ce{H}_{2}\\ce{SO}_{3}\\rightarrow\\ce{SO}_{4}^{2-}\\]\r\nExample 4 \u2013 writing balanced half-equations for redox equations\r\nBalance the following full redox equation for the reaction occurring between the reactants in the previous example:\r\n\r\nStep 1: Assign oxidation numbers to each element in the equation to determine which reactants are being oxidised and reduced.\r\n\r\nThe oxidation number of \\(\\ce{Mn}\\) in \\(\\ce{MnO}_{4}^{-}\\) is \\(+7\\) and the \\(\\ce{Mn}\\) oxidation number in \\(\\ce{Mn}^{2+}\\) is \\(+2\\). Thus, \\(\\ce{Mn}^{7+}\\)is being reduced to \\(\\ce{Mn}^{2+}\\).\r\n\r\nThe oxidation number of \\(\\ce{S}\\) in \\(\\ce{H}_{2}\\ce{SO}_{3}\\) is \\(+4\\), and the oxidation number of \\(\\ce{S}\\) in \\(\\ce{SO}_{4}^{2-}\\) ion is \\(+6\\). So, \\(\\ce{S}\\) is being oxidised to \\(\\ce{SO}_{4}^{2-}\\).\r\n\r\nStep 2: Divide the equation into the appropriate half-reactions which can be balanced separately.\r\n\r\nThe reduction half-reaction is: \\(\\ce{MnO}_{4}^{-}\\rightarrow\\ce{Mn}^{2+}\\)\r\n\r\nThe oxidation half-reaction is: \\[ \\ce{H}_{2}\\ce{SO}_{3}\\rightarrow\\ce{SO}_{4}^{2-}\\]\r\n\r\nStep 3: Write separate balanced chemical equations for reduction and oxidation. Use the acronym KOHES:\r\n\r\n \tKey elements (elements aside from oxygen and hydrogen)\r\n \tOxygen \u2013 add water molecules to balance the oxygen atoms\r\n \tHydrogen \u2013 add hydrogen ions (\\(\\ce{H}^{+}\\)) to balance the hydrogen atoms\r\n \tElectrons \u2013 add electrons (\\(e^{-}\\)) to balance the charges\r\n \tStates \u2013 add states for all species except electrons, which have no state\r\n\r\nStarting with the reduction half-equation, the key elements are already balanced as there is \\(1\\textrm{ Mg}\\) on either side of the reaction arrow. There are \\(4\\textrm{ O}\\) atoms on the reactant side and none on the product side, so to balance the \\(\\textrm{ O}\\) atoms, we need to add \\(4\\textrm{ H}_{2}\\textrm{O}\\) to the product side.\r\n\r\n\\[\\ce{MnO}_{4}^{-}\\rightarrow\\ce{Mn}^{2+}+4\\ce{H2O}\\]\r\n\r\nNext, we need to balance the \\(\\textrm{H}\\) by adding \\(8\\textrm{ H}^{+}\\) ions to the reactant side. This will give us \\(8\\textrm{ H}\\) atoms on each side.\r\n\r\n\\[\\ce{MnO}_{4}^{-}+8\\ce{H}^{+}\\rightarrow\\ce{Mn}^{2+}+4\\ce{H2O}\\]\r\n\r\nTo balance the charge, add \\(e^{-}\\). On the reactant side, the charge is \\((-1)+(+8)=(+7)\\) and on the product side, the charge is \\(+2\\). We need to add \\(5e^{-}\\) to the reactant side to make the charge \\(+2\\) on both sides.\r\n\r\n\\[\\ce{MnO}_{4}^{-}+8\\ce{H}^{+}+5e^{-}\\rightarrow\\ce{Mn}^{2+}+4\\ce{H2O}\\]\r\n\r\nFinally, we can add states.\r\n\r\nWe can confirm that this is a reduction half-equation by looking at which side of the arrow the electrons are on. For reduction half-equations, the electrons are always on the reactant (left-hand) side. This is indeed the case.\r\n\r\nStep 4: Repeat step 3 for the oxidation half-equation.\r\n\r\nThere is one key element, \\(\\ce{S}\\), and it is already balanced, so we can move on to \\(ce{O}\\). There are \\(3\\textrm{ O}\\) atoms on the reactant side and \\(4\\textrm{ O}\\) atoms on the product side. We need to add \\(1\\textrm{ H}_{2}\\textrm{O}\\) to the reactant side.\r\n\r\n\\[\\ce{H2SO3}+\\ce{H2O}\\rightarrow\\ce{SO}_{4}^{2-}\\]\r\n\r\nTo balance the \\(\\ce{H}\\) atoms, we add \\(4\\textrm{ H}^{+}\\) to the product side.\r\n\r\n\\[\\ce{H2SO3}+\\ce{H2O}\\rightarrow\\ce{SO}_{4}^{2-}+4\\ce{H}^{+}\\]\r\n\r\nTo balance the charge, we add \\(2e^{-}\\) to the product side.\r\n\r\n\\[\\ce{H2SO3}+\\ce{H2O}\\rightarrow\\ce{SO}_{4}^{2-}+4\\ce{H}^{+}+2e^{-}\\]\r\n\r\nAnd finally, states.\r\n\r\n\r\n\r\nWe can confirm that this is an oxidation question as the electrons are on the product (right-hand) side.\r\n\r\nStep 5: To get the full equation, the electrons need to be eliminated. We do this by first making sure that the number of electrons in both equations is equal.\r\n\r\nThe reduction half-equation has \\(5e^{-}\\), whereas the oxidation half-equation has \\(2e^{-}\\). To make them equal, we need to find the least common multiple. In this case, it is 10, so all of the coefficients in the reduction half-equation need to be multiplied by \\(2\\) and all of the coefficients in the oxidation half-equation need to be multiplied by \\(5\\).\r\n\r\n\r\n\r\nAdding these together gives:\r\n\r\n\r\n\r\nStep 6: Cancel out the equal number of electrons on each side of the equation and simplify the equation by eliminating any chemical species common to each side.\r\n","n":0.024},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Chemical equations","i":2,"n":0.707},{"v":"Redox reactions","i":1,"n":0.707},{"v":"Chemical bonding","i":0,"n":0.707}]]},{"i":259,"$":[{"v":"Chemical equations and reactions","n":0.5},{"v":"A chemical reaction is represented by a chemical equation that involves ions, atoms and molecular formulas. Use this resource to will learn about the main components of a chemical equation, and how to write and balance chemical equations. Also read about how Aboriginal peoples apply their understanding of chemical reactions.\r\nThe main components of a chemical equation are:\r\n\r\n \tReactants \u2013 the starting materials of the reaction. Reactants are written on the left side of the equation.\r\n \tProducts \u2013 the materials obtained at the end of the reaction. Products are written on the right side of the equation.\r\n \tAn arrow \u2013 drawn between the reactants and products pointing towards the products to indicate a chemical change\r\n \tConditions required for the reaction to occur—such as pressure and temperature—written above the arrow\r\n \tPhysical state of the substances involved \u2013 written next to the chemical formulas using brackets and abbreviations\r\n\t\r\n\t\tsolid \u2013 (s)\r\n\t\tliquid \u2013 (l)\r\n\t\tgas \u2013 (g)\r\n\t\taqueous solution (a substance dissolved in water) \u2013 (aq)\r\n\t\r\n \tCoefficients \u2013 the numbers written before the chemical formulas to balance the chemical equation.\r\n\r\nAccording to the law of conservation of mass, matter cannot be created nor destroyed. None of the atoms disappear during a chemical reaction. Instead, atoms rearrange to form new chemical compounds. Therefore, the total number of atoms of each type on the reactant side of the arrow must be equal to that on the product side.\r\n\r\nExample 1 \u2013 writing balanced equations for chemical reactions\r\nWrite the balanced chemical equation for the reaction between nitrogen gas and hydrogen gas to produce ammonia gas.\r\nStep 1: Write the reactants and products of the chemical equation using correct symbols and formulas. The reactants are written on the left side of the arrow and the products are written on the right. Use a plus (+) sign to separate each reactant or product from others.\r\n\\[\\ce{N}_{2}+\\ce{H}_{2}\\ce{\\rightarrow}\\ce{NH}_{3}\\]\r\nStep 2: Balance the chemical equation by writing the correct coefficients in front of each chemical formula. In a balanced chemical equation, an equal number of atoms of each element involved must be present on either side  of the arrow. Choose one atom to balance at a time. Let's start with the compound that is composed of the greatest number of atoms.\r\nThe compound with the greatest number of atoms is ammonia. The element in ammonia with the highest number of atoms is \\(\\ce{H}\\). The reactant side contains \\(3\\textrm{ H}\\) atoms. On the product side, \\(\\ce{H}\\) contains \\(2\\textrm{ H}\\) atoms.\r\nTo balance the number of \\(\\ce{H}\\) atoms, \\(6\\textrm{ H}\\) atoms are needed on either side, as \\(6\\) is the least common multiple. To achieve \\(6\\textrm{ H}\\) atoms on either side, we need \\(2\\ce{NH3}\\) molecules and \\(3\\ce{H2}\\) molecules. \r\n\\[\\ce{N}_{2}+3\\ce{H}_{2}\\rightarrow2\\ce{NH}_{3}\\]\r\nStep 3: Repeat step 2 with each atom, as required.\r\nThe next atom will be nitrogen. The reactant side has \\(2\\textrm{ N}\\) atoms and the product side also has \\(2\\textrm{ N}\\) atoms. Therefore, nitrogen is already balanced.\r\nStep 4: Do a final check to make sure all types of atoms in the equation are balanced.\r\n\\[\\ce{N}_{2}+3\\ce{H}_{2}\\rightarrow2\\ce{NH}_{3}\\]\r\nOn the reactant side, there are \\(2\\ce{N}\\) atoms and \\(6\\ce{H}\\) atoms. On the product side, there are \\(2\\ce{N}\\) atoms and \\(6\\ce{H}\\) atoms. The equation is balanced. This is summarised in the table.\r\n\r\n\r\n\r\nElement\r\nReactant side\r\nProduct side\r\n\r\n\r\n\r\n\r\n\\(\\ce{N}\\)\r\n\\(2\\)\r\n\\(2\\)\r\n\r\n\r\n\\(\\ce{H}\\)\r\n\\(3\\times2=6\\)\r\n\\(2\\times3=6\\)\r\n\r\n\r\n\r\nStep 5: Check whether the coefficients are in their lowest possible whole numbers.\r\n\\[1\\ce{N}_{2}+3\\ce{H}_{2}\\rightarrow2\\ce{NH}_{3}\\]\r\nThe ratio of \\(1:3:2\\) is the lowest possible in whole numbers. In the final equation, any coefficients of \\(1\\) do not need to be written. That is, \\(1\\ce{N2}\\) is simply \\(\\ce{N2}\\).\r\nStep 6: Add the states for each compound.\r\n\\[\\ce{N}_{2}(\\textrm{g})+3\\ce{H}_{2}(\\textrm{g})\\rightarrow2\\ce{NH}_{3}(\\textrm{g})\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nClick each of the following to reveal Indigenous applications of chemical reactions.\r\n\r\n\r\n\r\n\r\n","n":0.042},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Reactions","i":2,"n":1},{"v":"Equations","i":1,"n":1},{"v":"Chemical equations","i":0,"n":0.707}]]},{"i":260,"$":[{"v":"Chemical equations","n":0.707},{"v":"Chemical equations represent chemical reactions using symbols and letters. This tutorial will teach you what the properties of chemical equations are, how to read an equation as well as how to balance an equation.\r\nOverview\r\n\r\n\tChemical equations and reactions \u2013 how do equations represent the differences in substances before and after a chemical reaction? What do equations tell us? 45 minutes to complete\r\n\tQuiz \u2013 Chemical equations \u2013 how well do you understand chemical equations? Can you confidently balance equations? 25 minutes to complete\r\n\tRedox reactions \u2013 what is a redox reaction? 45 minutes to complete\r\n\tQuiz \u2013 Redox reactions \u2013 how well do you understand redox reactions 15 minutes to complete\r\n\r\n","n":0.097},[{"v":"Chemistry","i":1,"n":1},{"v":"Chemical equations","i":0,"n":0.707}]]},{"i":261,"$":[{"v":"Stoichiometry","n":1},{"v":"In our daily lives, from the food we eat to the fuel that powers our vehicles, stoichiometry plays a crucial role in ensuring that chemical processes are efficient and effective. Improve your knowledge of stoichiometry by using this resource to learn about the stoichiometric coefficient, the molar ratio and chemical equations.\r\nStoichiometry describes the quantitative relationship between relative quantities of reactants and products in a balanced chemical equation. Stoichiometric coefficients are the numbers written immediately before the atoms, ions and molecules in a balanced chemical equation. These numbers indicate the ratio of any reactant or product that reacts or forms in a reaction.\r\nAs an example, let's consider the reaction between hydrogen gas and nitrogen gas to form ammonia. The balanced chemical equation can be written as:\r\n\\[3\\ce{H}_{2}\\left(\\textrm{g}\\right)+\\ce{N}_{2}\\left(\\textrm{g}\\right)\\rightarrow2\\ce{NH}_{3}\\left(\\textrm{g}\\right)\\]\r\nThe numbers three, one and two are written in front of \\(\\ce{H}\\), \\(\\ce{N}\\) and \\(\\ce{NH}_{3}\\), respectively. These are the stoichiometric coefficients. They show that three molecules of hydrogen gas react with one molecule of nitrogen gas to form two ammonia molecules. When there is no coefficient before an atom, ion or molecule, we interpret it as one. For example, the coefficient of molecular nitrogen in this reaction is one.\r\n\r\nMolar ratios\r\nThe stoichiometric coefficients in a balanced chemical equation can also be interpreted on the molar scale. Molar scale interpretation relates to the quantities of substances involved in the reaction. Therefore, if we know the amount of one substance involved in the reaction, we can use the molar ratio to determine the amounts of others involved in the reaction. Let's consider the same example:\r\n\\[3\\ce{H}_{2}\\left(\\textrm{g}\\right)+\\ce{N}_{2}\\left(\\textrm{g}\\right)\\rightarrow2\\ce{NH}_{3}\\left(\\textrm{g}\\right)\\]\r\nUsing the molar scale, we can say that three moles of hydrogen gas react with one mole of nitrogen gas to form two moles of ammonia gas.\r\n\r\nLimiting and excess reactants\r\nIn chemical reactions, limiting and excess reactants (or reagents) determine the extent of the reaction and the amount of products formed. The limiting reactant is the substance that is consumed completely first, thus limiting the amount of product that can be formed. Once the limiting reactant is used up, the reaction cannot proceed further, even if other reactants are still available. On the other hand, excess reactants are those that remain after the reaction has reached completion.\r\nLet's consider the reaction between hydrogen gas and nitrogen gas to form ammonia gas again: \r\n\\[3\\ce{H}_{2}\\left(\\textrm{g}\\right)+\\ce{N}_{2}\\left(\\textrm{g}\\right)\\rightarrow2\\ce{NH}_{3}\\left(\\textrm{g}\\right)\\]\r\nBased on stoichiometry, we know that if we mix three moles of \\(\\ce{H}_{2}\\) with one mole of \\(\\ce{N}_{2}\\), we can obtain two moles of \\(\\ce{NH}_{3}\\). In other words, \\(\\ce{H}_{2}\\) reacts with \\(\\ce{N}_{2}\\) in a molar ratio of \\(3:1\\).\r\nIf we mix three moles of hydrogen gas and two moles of nitrogen gas, two moles of \\(\\ce{NH}_{3}\\) will be produced. However, the entire quantity of \\(\\ce{H}_{2}\\) will be used before the \\(\\ce{N}_{2}\\) as we start with the exact amount of \\(\\ce{H}_{2}\\) required for the reaction. The product mixture will contain one unreacted mole of \\(\\ce{N}_{2}\\) as we only required one mole of \\(\\ce{N}_{2}\\) for the reaction. If we had more \\(\\ce{H}_{2}\\), we could have produced more \\(\\ce{NH}_{3}\\) as we already have excess \\(\\ce{N}_{2}\\). Therefore, in this reaction, we call \\(\\ce{H}_{2}\\) the limiting reagent since it limits the production of \\(\\ce{NH}_{3}\\).\r\nHence, the limiting reagent is the reactant that runs out first, limiting the formation of products. Nitrogen gas is called the excess reagent because we have a greater amount than required for the reaction.\r\n\r\nReaction yield\r\nThe theoretical yield is the maximum amount of product that can form under the given reaction conditions. This is determined by the amount of limiting reagent available. Theoretical yield can be calculated using the stoichiometric coefficients (or molar ratios) in the balanced chemical equation. We calculate theoretical yield assuming no losses occur during the reaction, and all the reactants are converted into products. However, in reality, losses and inefficiencies occur during chemical reactions. Therefore, the actual amount you isolate and measure from a chemical reaction is less than what you expect in theory.\r\nThe actual yield of a reaction is the amount of the product you isolate from the chemical reaction. Actual yield must be determined experimentally, so it often referred to as experimental yield. You need to isolate the product and weigh it to get the actual yield.\r\nThe percentage yield indicates how efficient the reaction was in converting reactants into products. If the value is closer to \\(100\\textrm{%}\\), most of the reactants were converted into products; the reaction was closer to completion. Percentage yield can be calculated using:\r\n\\[\\textrm{Percentage yield}=\\frac{\\textrm{Actual yield}}{\\textrm{Theoretical yield}}\\times100\\textrm{%}\\]\r\n\r\nExample 1 \u2013 calculating the amount of product formed from a chemical reaction\r\nCalculate the amount in moles of ammonia gas that can be produced from \\(0.18\\textrm{ mol}\\) of hydrogen gas. \\[3\\ce{H}_{2}\\left(\\textrm{g}\\right)+\\ce{N}_{2}\\left(\\textrm{g}\\right)\\rightarrow2\\ce{NH}_{3}\\left(\\textrm{g}\\right)\\]\r\nStep 1: Determine the molar ratio of the compounds and write it as a fraction. The stoichiometric coefficient for the compound we are interested in (the \"unknown\") is written as the numerator at the top, and the coefficient for the compound we have the amount for (the \"known\") is written as the denominator at the bottom.\r\n\\[\\begin{align*}\r\n\\textrm{Stoichiometric ratio} & =\\frac{\\textrm{coefficient of unknown}}{\\textrm{coefficient of known}}\\\\\r\n& = \\frac{2}{3}\r\n\\end{align*}\\]\r\nStep 2: Calculate the amount of product formed by multiplying the amount of reactant by the molar ratio.\r\n\\[\\begin{align*}\r\nn(\\ce{NH3}) & = \\frac{2}{3}\\times0.18\\textrm{ mol}\\\\\r\n& =0.12\\textrm{ mol}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExample 3 \u2013 calculating the theoretical yield of a reaction\r\nPhosphoric acid is produced by treating phosphate rock, which contains calcium phosphate, with sulfuric acid:\r\n\\[\\ce{Ca}_{3}\\left(\\ce{PO}_{4}\\right)_{2}+3\\ce{H}_{2}\\ce{SO}_{4}\\rightarrow3\\ce{CaSO}_{4}+2\\ce{H}_{3}\\ce{PO}_{4}\\]\r\nA chemist in a fertiliser production plant mixes \\(144.5\\textrm{ g}\\) of calcium phosphate with \\(160.5\\textrm{ g}\\) of sulfuric acid. Determine the maximum mass of phosphoric acid that can be formed. The molar masses are given in the table.\r\n\r\n\r\n\r\nCompound\r\nMolar mass (\\(\\textrm{g mol}^{-1}\\))\r\n\r\n\r\n\r\n\r\n\\(\\ce{Ca}_{3}\\left(\\ce{PO}_{4}\\right)_{2}\\)\r\n\\(310.2\\)\r\n\r\n\r\n\\(\\ce{H}_{2}\\ce{SO}_{4}\\)\r\n\\(98.1\\)\r\n\r\n\r\n\\(\\ce{H}_{3}\\ce{PO}_{4}\\)\r\n\\(97.99\\)\r\n\r\n\r\n\r\n\r\nStep 1: Identify the limiting reagent in this reaction by first, finding the number of moles of each reactant.\r\n\\[\\begin{align*}\r\nn(\\ce{Ca}_{3}\\left(\\ce{PO}_{4}\\right)_{2}) & = \\frac{m}{M}\\\\\r\n& =\\frac{144.5\\textrm{ g}}{310.2\\textrm{ g mol}^{-1}}\\\\\r\n& =0.4658\\textrm{ mol}\r\n\\end{align*}\\]\r\n\r\n\\[\\begin{align*}\r\nn(\\ce{H}_{2}\\ce{SO}_{4}) & = \\frac{m}{M}\\\\\r\n& =\\frac{160.5\\textrm{ g}}{98.1\\textrm{ g mol}^{-1}}\\\\\r\n& =1.64\\textrm{ mol}\r\n\\end{align*}\\]\r\n\r\nStep 2: Compare the calculated amounts of two reactants. The easiest way to find the limiting reactant is to use the formula:\r\n\\[\\frac{n}{\\textrm{coefficient}}\\]\r\nThe reactant with the smallest \\(\\dfrac{n}{\\textrm{coefficient}}\\) is the limiting reactant.\r\nFor \\(\\ce{Ca}_{3}\\left(\\ce{PO}_{4}\\right)_{2}\\):\r\n\\[\\begin{align*}\r\n\\frac{n}{\\textrm{coefficient}} & = \\frac{0.4659\\textrm{ mol}}{1}\\\\\r\n& = 0.4659\\textrm{ mol}\r\n\\end{align*}\\]\r\nFor \\(\\ce{H}_{2}\\ce{SO}_{4}\\):\r\n\\[\\begin{align*}\r\n\\frac{n}{\\textrm{coefficient}} & = \\frac{1.64\\textrm{ mol}}{2}\\\\\r\n& = 0.82\\textrm{ mol}\r\n\\end{align*}\\]\r\nAs the \\(\\dfrac{n}{\\textrm{coefficient}}\\) for \\(\\ce{Ca}_{3}\\left(\\ce{PO}_{4}\\right)_{2}\\) is smaller, it is the limiting reactant.\r\nStep 3: Determine the molar ratio of the compounds and write it as a fraction. The stoichiometric coefficient for the compound we are interested in (the \"unknown\") is written as the numerator at the top, and the coefficient for the limiting reactant (the \"known\") is written as the denominator at the bottom.\r\n\\[\\begin{align*}\r\n\\textrm{Stoichiometric ratio} & =\\frac{\\textrm{coefficient of unknown}}{\\textrm{coefficient of known}}\\\\\r\n& = \\frac{2}{1}\r\n\\end{align*}\\] \r\nStep 4: Calculate the amount of product formed by multiplying the amount of reactant by the molar ratio.\r\n\\[\\begin{align*}\r\nn(\\ce{H3PO4}) & = \\frac{2}{1}\\times0.4659\\textrm{ mol}\\\\\r\n& = 0.9317\\textrm{ mol}\r\n\\end{align*}\\]\r\nStep 5: Calculate the mass of product formed. This is the theoretical yield.\r\n\\[\\begin{align*}\r\nm(\\ce{H3PO4}) & = n\\times{M}\\\\\r\n& = 0.9317\\textrm{ mol}\\times 97.99\\textrm{ g mol}^{-1}\\\\\r\n& = 91.29\\textrm{ g}\r\n\\end{align*}\\]\r\n\r\nExample 4 \u2013 calculating percentage yield\r\nWhen \\(64.5\\textrm{ g}\\) of methane is burnt in the presence of excess oxygen, it forms \\(136.2\\textrm{ g}\\) of carbon dioxide and water vapour. \r\n\\[\\ce{CH}_{4}\\left(\\textrm{g}\\right)+2\\ce{O}_{2}\\left(\\textrm{g}\\right)\\rightarrow\\ce{CO}_{2}\\left(\\textrm{g}\\right)+2\\ce{H}_{2}\\ce{O}\\left(\\textrm{g}\\right)\\]\r\nHowever, the theoretical yield of carbon dioxide for this reaction is \\(176.4\\textrm{ g}\\). Calculate the percentage yield of carbon dioxide for this reaction.\r\n\r\nStep 1: Recall the formula for calculating percentage yield.\r\n\\[\\textrm{Percentage yield}=\\frac{\\textrm{Actual yield}}{\\textrm{Theoretical yield}}\\times100\\textrm{%}\\]\r\nStep 2: Substitute the values into the formula to calculate percentage yield.\r\n\\[\\begin{align*}\r\n\\textrm{Percentage yield} & =\\frac{136.2\\textrm{ g}}{176.4\\textrm{ g}}\\times100\\textrm{%}\\\\\r\n& = 77.21\\textrm{%}\r\n\\end{align*}\\]\r\n\r\n","n":0.029},[{"v":"The mole","i":2,"n":0.707},{"v":"Stoichiometry","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":262,"$":[{"v":"The mole","n":0.707},{"v":"The mole (\\(n\\)) is an important unit of measurement in chemistry because it allows scientists to count and quantify atoms, molecules, and ions in a given sample. This then allows them to perform calculations relating to chemical reactions.\r\nIn everyday language, we use \\(1\\) dozen \\(=12\\) object, \\(1\\) dozen \\(=10\\) years, and \\(1\\) ream \\(=500\\) sheets of paper. In chemistry, \\[ 1\\textrm{ mole}=6.022\\times10^{23}\\textrm{ objects} \\]\r\nThe objects can be anything such as houses, sand, eggs, or oranges. In chemistry, objects are mainly particles like atoms, ions or molecules. The symbol for moles is  \\(mol\\) .\r\n\r\nOne mol of anything contains \\(6.022\\times10^{23}\\) objects. This means that:\r\n\r\n\r\n\t\\(1\\textrm{ mol}\\) of eggs contains \\(6.022\\times10^{23}\\) eggs\r\n\t\\(1\\textrm{ mol}\\) of carbon contains \\(6.022\\times10^{23}\\) carbon atoms\r\n\t\\(1\\textrm{ mol}\\) of carbon dioxide contains \\(6.022\\times10^{23}\\) carbon dioxide molecules\r\n \t\\(3\\textrm{ mol}\\) of particles is \\(3 \\times 6.022\\times10^{23}\\) particles \r\n\t\\(10\\textrm{ mol}\\) of particles is \\(10 \\times 6.022\\times10^{23}\\) particles\r\n \r\n If we had \\(3.011\\times10^{24}\\) atoms of gold (Au), the amount of gold in moles would be:\r\n \r\n \r\nThe mole and chemical formulas\r\nChemical formulas indicate the number of atoms of each element present in the compound. The ratio of atoms is also consistent with the relative number of moles of each atom. \r\n\r\nUsing sodium carbonate, \\(\\ce{Na}_{2}\\ce{CO}_{3}\\), as an example, \\(1 \\textrm{ mol}\\) of sodium carbonate units contains \\(2 \\textrm{ mol}\\) of sodium atoms, \\(1 \\textrm{ mol}\\) of carbon atoms and \\(3 \\textrm{ mol}\\) of oxygen atoms.\r\n \r\nAvogadro's number (\\(N_\\textrm{A}\\))\r\nAvogadro's number is the term used for \\(6.022\\times10^{23}\\) objects.\r\n\\[ N_\\textrm{A}=6.022\\times10^{23}\\textrm{ mol}^{\u20131} \\]\r\n\r\nAvogadro's number links the number of atoms, molecules or ions of a substance and the amount in moles of the substance:\r\n\\[ N=n \\times N_\\textrm{A} \\]\r\n\r\nMolar mass (\\(M\\))\r\nMolar mass is the mass of one mole of a substance. For example, the molar mass of \\(^{12}\\ce{C}\\) is \\(12.0000\\textrm{ g mol}^{\u20131}\\). This is calculated from atomic mass, the average mass of an atom of an element measured in atomic mass units (amu). \r\n\r\nOne atomic mass unit is equal to \\(1.66054\\times10^{\u201324}\\textrm{ g}\\), which is equal to \\(\\dfrac{1}{12}\\) of the mass of a \\(^{12}\\ce{C}\\) atom.\r\n\r\nTherefore, the mass of one \\(^{12}\\ce{C}\\) atom is: \r\n \r\n\r\nThe mass of \\(1 \\textrm{ mol}\\) of \\(^{12}\\ce{C}\\) atoms can then be calculated:\r\n\r\n\r\n\r\nAlthough the number of objects in \\(1 \\textrm{ mol}\\) is fixed, molar mass can vary based on the composition of atoms in a substance. \r\n\r\nMolar mass is related to the amount in moles of substance and the mass of the substance (\\(m\\),  in grams) by the formula:\r\n\\[ M=\\frac{m}{n} \\]\r\n\r\nExample 1 \u2013 calculating the number of molecules\r\nCalculate the number of acetaminophen molecules present in \\(0.25\\textrm{ mol}\\) of acetaminophen.\r\nStep 1: Recall the formula relating Avogadro's number, the amount in moles and the number of molecules.\r\n\\[ N=n \\times N_\\textrm{A} \\]\r\n\r\nStep 2: Substitute the given values for \\(N_\\textrm{A}\\) and \\(n\\). It is good practice to write the units as well. \r\n\r\n\r\n\r\n\r\n\r\n\r\nExample 3 \u2013 calculating molar mass\r\nDetermine the molar mass of \\(\\ce{Na}_{2}\\ce{CO}_{3}\\).\r\n\r\nStep 1: Find the atomic mass of each atom in \\(\\ce{Na}_{2}\\ce{CO}_{3}\\) from the periodic table.\r\n\r\n\\(M\\textrm{(Na)}=23.0\\textrm{ g mol}^{\u20131}\\)\r\n\\(M\\textrm{(C)}=12.0\\textrm{ g mol}^{\u20131}\\)\r\n\\(M\\textrm{(O)}=16.0\\textrm{ g mol}^{\u20131}\\)\r\n\r\n\r\nStep 2: Multiply each atomic mass by the number of each type of atom in \\(\\ce{Na}_{2}\\ce{CO}_{3}\\) .\r\n\r\n\\(2 \\times 23.0\\textrm{ g mol}^{\u20131}=46.0\\textrm{ g mol}^{\u20131}\\)\r\n\\(1 \\times 12.0\\textrm{ g mol}^{\u20131}=12.0\\textrm{ g mol}^{\u20131}\\)\r\n\\(3 \\times 16.0\\textrm{ g mol}^{\u20131}=48.0\\textrm{ g mol}^{\u20131}\\)\r\n\r\n\r\nStep 3: Take the sum of the atomic masses calculated in Step 2.\r\n\\[\\begin{align*}\r\nM(\\ce{Na}_{2}\\ce{CO}_{3}) & =46.0+12.0+48.0\\\\ & =106\\textrm{ g mol}^{\u20131}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExample 5 \u2013 calculating amount in moles\r\nCalculate the amount of silver present in \\(50.00\\textrm{ g}\\) of silver (\\(M(\\ce{Ag})=107.87\\textrm{ g mol}^{\u20131}\\)).\r\n\r\nStep 1: When a question asks you to calculate the \"amount\", it is simply asking for the number of moles. As the mass and molar mass of silver are given in the question, you can use \\(M=\\frac{m}{n}\\) to find the number of moles \\(\\left(n\\right)\\).\r\nFirst, rearrange the formula to make \\(n\\) the subject.\r\n\\[ n = \\frac{m}{M} \\]\r\n\r\nStep 2: Substitute the given values for \\(M\\) and \\(m\\). It is good practice to write the units as well. \r\n\\[\\begin{align*}\r\nn(\\ce{Ag}) & =\\frac{m}{M}\\\\\r\n& =\\frac{50.00\\textrm{ g}}{107.87\\textrm{ g mol}^{\u20131}}\\\\\r\n& =0.4635\\textrm{ mol}\r\n\\end{align*}\\]\r\n\r\n","n":0.039},[{"v":"The mole","i":2,"n":0.707},{"v":"Measurement","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":263,"$":[{"v":"Covalent bonding","n":0.707},{"v":"Covalent bonding is responsible for creating many substances we encounter in our daily lives. Water (\\(\\ce{H2O}\\)) is essential for life, carbon dioxide (\\(\\ce{CO2}\\)) is a key component of the Earth's atmosphere, and methane (\\(\\ce{CH4}\\)) is a common fuel source. Use this resource to learn more about how covalent bonds form, including how Aboriginal peoples take advantage of the covalent bonds in spinifex grass.\r\n\r\nFormation of molecular compounds\r\nCovalent bonding primarily occurs between non-metal atoms or elements located in the \\(p\\)-block. Whereas ionic bonding involves the transfer of valence electrons, covalent bonding involves the sharing of valence shell electrons to achieve a complete outer shell. Atoms held together by covalent bonds are called molecular compounds. The figure below shows the formation of three covalent molecular compounds: \\(\\ce{H}_{2}\\), \\(\\ce{F}_{2}\\) and \\(\\ce{H}_{2}\\ce{O}\\).\r\n\r\n\r\n \tHydrogen gas, \\(\\ce{H2}\\), is formed when two hydrogen atoms share one electron in their valence electron shell (\\(1s\\)) with each other. This allows each hydrogen atom to achieve a full electron shell with the electron configuration of a noble gas, \\(\\ce{He}\\): \\(1s^{2}\\)\r\n \tFluorine gas, \\(\\ce{F2}\\), is formed when two fluorine atoms share a pair of electrons in their valence electron shell (\\(2p\\)). This allows each fluorine atom to achieve a full electron shell with the electron configuration of a noble gas, \\(\\ce{Ne}\\): \\(1s^{2}\\,2s^{2}\\,2p^{6}\\)\r\n \tWater, \\(\\ce{H2O}\\), is formed when two hydrogen atoms share one electron in their valence electron shell (\\(1s\\)) with an oxygen atom. The oxygen atom shares one electron in their valence electron shell (\\(2p\\)) with each hydrogen atom. This allows each hydrogen atom to achieve a full electron shell with the electron configuration of a noble gas, \\(\\ce{He}\\): \\(1s^{2}\\). It also allows the oxygen atom to achieve a full electron shell with the electron configuration of a noble gas, \\(\\ce{Ne}\\): \\(1s^{2}\\,2s^{2}\\,2p^{6}\\)\r\n\r\nNumber of covalent bonds\r\nGenerally, the number of covalent bonds an atom forms depends on how many electrons an atom requires to achieve an octet. For instance, \\(\\ce{C}\\) forms four covalent bonds as \\(\\ce{C}\\) \\(\\left(1s^{2}\\,2s^{2}\\,2p^{2}\\right)\\) needs four more electrons to acquire an octet in their valence shell. A few more examples are listed in the table.\r\n\r\nHowever, the number of covalent bonds formed by an atom is not always guided by the octet rule. Let's take \\(\\ce{S}\\) as an example. \\(\\ce{S}\\) can form four or six covalent bonds, such as \\(\\ce{SF}_{4}\\) and \\(\\ce{SF}_{6}\\). Some other exceptions to the octet rule are \\(\\ce{BF}_{3}\\), \\(\\ce{PCl}_{5}\\) and some of the \\(p\\) block elements in period \\(4\\), as they have empty \\(d\\) orbitals.\r\n\r\nTypes of covalent bonds\r\nWhen two electrons are shared between two atoms, a single bond forms.\r\n\r\nMultiple covalent bonds are formed between two atoms when they share more than one pair of electrons to acquire valence shell octet. For instance, \\(\\ce{CO}_{2}\\): \\(\\ce{C}\\) forms two covalent bonds (two pairs of electrons or four electrons) with each oxygen atom involved. These are called double bonds. Double bonds are stronger than single bonds.\r\n\r\n\r\n\r\n\r\n\r\n\r\nIn the same way, when six electrons are shared between two atoms, a triple bond forms. Ethyne is an example of a molecular compound that forms a triple bond. One carbon atom shares six electrons with another carbon atom to complete the octet. These types of bonds are stronger than double bonds.\r\n\r\n\r\n\r\n\r\n\r\nCoordinate covalent bonds\r\nTypically, each atom contributes one or more electrons to the covalent bond. When the electrons shared between two atoms are donated by only one atom, the resulting bond is known as a coordinate covalent bond. In a coordinate covalent bond, an atom with a filled orbital shares its two electrons with an empty orbital of another atom.\r\n\r\n\r\nExamples of molecular compounds with coordinate covalent bonds are carbon monoxide (\\(\\ce{CO}\\)) and ozone (\\(\\ce{O3}\\)).\r\n\r\nThe shape of molecular compounds\r\nThe three-dimensional arrangement of a molecular compound gives it a molecular shape. Shape is primarily determined by valence shell electron repulsion (VSEPR) theory. Being able to draw Lewis structures also helps with determining molecular shape.\r\n\r\nValence shell electron pair repulsion (VSEPR) theory\r\nVSEPR theory states that the shape of a molecule is determined by the number of electron clouds around the central atom. Non-bonding (lone pair) and bonding valence electrons are considered electron clouds. Due to the repulsion forces between electron clouds, they attempt to stay as far as possible.\r\n\r\nFor instance, a carbon dioxide molecule has two electron clouds (two \\(\\ce{C}-\\ce{O}\\) bonds) around the central atom \\(\\ce{C}\\). As a result of repulsion forces between these two electron clouds, the molecule is arranged into linear geometry with a bond angle of \\(180^{\\circ}\\). Multiple bonds, such as double and triple bonds, are considered one electron cloud.\r\n\r\nDepending on the number of electron clouds around a central atom, the number of bonds formed and the number of lone pairs, the molecular shape will vary. The different molecular shapes are summarised in the table below.\r\n\r\nWhen more than one central atom is present in the molecule, the molecular geometry of each central atom is predicted and the results are combined. For example, \\(\\ce{C}_{2}\\ce{H}_{4}\\) has two central \\(\\ce{C}\\) atoms. Each \\(\\ce{C}\\) atom is attached to three electron clouds. This means each \\(\\ce{C}\\) atom has a trigonal planar geometry. In this case, the whole molecule is also planar, as the bond angles \\(\\ce{H}-\\ce{C}-\\ce{H}\\) and \\(\\ce{H}-\\ce{C}-\\ce{C}\\) are \\(120^{\\circ}\\) each.\r\n\r\n\r\nThe polarity of molecular compounds\r\nIn chemistry, polarity refers to the distribution of electric charge around atoms or molecules. The concept of polarity is crucial in understanding how covalent bonds and entire molecules interact with each other, influencing their physical and chemical properties.\r\nBond polarity\r\nIn the context of covalent bonds, polarity arises when there is an unequal sharing of electrons between atoms due to differences in their electronegativities. This results in one end of the bond having a partial negative charge (designated \\(\\delta-\\)) and the other a partial positive charge (designated \\(\\delta+\\)). The partially negative end is electron-rich, while the partially positive end is electron-poor. This creates something called a dipole moment.\r\n\r\nA covalent bond in which this occurs is called a polar covalent bond. One example is \\(\\ce{HF}\\). The electronegativity (EN) of the \\(\\ce{H}\\) atom is \\(2.20\\). The EN of the \\(\\ce{F}\\) atom is \\(3.98\\). The difference in EN is \\(3.98-2.20=1.78\\). Valence electrons shared by \\(\\ce{H}\\) and \\(\\ce{F}\\) are more attracted to the electronegative \\(\\ce{F}\\) atom. The \\(\\ce{F}\\) atom therefore has a partial negative charge and is labelled \\(\\delta-\\), and the \\(\\ce{H}\\) atom has a partial positive charge and is labelled \\(\\delta+\\).\r\n\r\nA bond in which electrons are shared equally is simply called a non-polar covalent bond. An example is the bond between the two \\(\\ce{H}\\) atoms that make \\(\\ce{H2}\\). The difference in EN is \\(0\\).\r\n\r\nThe difference between the EN of participating atoms in a bond also indicates whether the bond is ionic, non-polar covalent or polar covalent. Compared to ionic bonds, the difference between the EN of atoms participating in covalent bonding is much smaller.\r\n\r\n\r\n\r\nElectronegativity difference\r\nType of bond\r\nExamples\r\n\r\n\r\n\r\n\r\n\\(\\geqslant2.0\\)\r\nIonic\r\n\\(\\ce{NaCl}\\), \\(\\ce{CsF}\\)\r\n\r\n\r\n\\(0.5-1.9\\)\r\nPolar covalent\r\n\\(\\ce{HCl}\\), \\(\\ce{H}_{2}\\ce{O}\\)\r\n\r\n\r\n\\(0-0.4\\)\r\nCovalent\r\n\\(\\ce{CH}_{4}\\), \\(\\ce{O}_{2}\\)\r\n\r\n\r\n\r\n\r\nMolecular polarity\r\nWhen discussing molecules, polarity depends on the polarity of the individual bonds, the contribution from lone pair electrons, and the molecular shape. In a polar molecule, electrons are more strongly attracted to one end of the molecule. For example, in the \\(\\ce{H2O}\\) molecule, due to the high EN of the \\(\\ce{O}\\) atom, electrons are strongly attracted to the \\(\\ce{O}\\) atom. This creates a net polarity point between two \\(\\ce{O}-\\ce{H}\\) bonds. The polarity of the molecule is indicated using a crossed arrow that points in the direction of electron attraction.\r\n\r\n\r\n\r\nIn some cases, although the individual covalent bonds are polar, overall the molecule becomes non-polar due to the shape of the molecule. For instance, a carbon dioxide molecule has polar \\(\\ce{C}-\\ce{O}\\) bonds. However, due to the linear symmetrical shape of the molecule, polar bonds (dipoles) cancel each other, creating a zero net polarity.\r\n\r\n\r\nNaming molecular compounds\r\nTo name molecular compounds, you can follow the steps below. Let's use \\(\\ce{CCl4}\\) as an example.\r\n\r\n \tName the less electronegative atom first. Non-metals located on the left side of the periodic table are written before non-metals located on the right side. For \\(\\ce{CCl4}\\), carbon is less electronegative than the chlorine, so \"carbon\" is written first.\r\n \tName the more electronegative atom with the suffix \"\u2013ide\", as used for anions. For \\(\\ce{CCl4}\\), this would be \"chloride\".\r\n \tIdentify the number of atoms of each element involved in the compound and use the prefixes di\u2013, tri\u2013, tetra\u2013, penta\u2013, etc, to show how many of each atom is present. No prefix is used if there is only one atom of the element. \\(\\ce{CCl4}\\) has one carabon atom and four chlorine atoms, so is named \"carbon tetrachloride\".\r\n\r\nSome other examples are \"carbon monoxide\" for \\(\\ce{CO}\\), \"carbon dioxide\" for \\(\\ce{CO2}\\), \"boron trichloride\" for \\(\\ce{BCl3}\\) and \"dinitrogen pentoxide\" for \\(\\ce{N2O5}\\).\r\n\r\n\r\nThe ability for molecules to form covalent bonds is important in many fields, particularly materials science.\r\n\r\n\r\n \r\n\r\nHydrogen molecule (H2):\r\nTwo hydrogen atoms (H), each with 1s electron, combine to form an H2 molecule with a shared electron pair.\r\n\r\nFluorine molecule (F2):\r\nTwo fluorine atoms (F), each with 2p electrons, combine to form an F2 molecule with a shared electron pair.\r\n\r\nWater molecule (H2O):\r\nTwo hydrogen atoms and one oxygen atom combine. The oxygen atom (with 2p electrons) shares electrons with two hydrogen atoms to form an H2O molecule, showing shared electron pairs.\r\n\r\n \r\nTypical covalent bonds:\r\nShows two atoms, X and Y, with partially filled orbitals. They combine to form a shared electron pair.\r\n\r\nCoordinate covalent bonds:\r\nIllustrates atom X with an entirely filled orbital and atom Y with an empty orbital. They combine to form a shared electron pair.\r\n \r\nTwo molecular geometry diagrams depicting central carbon atoms (labeled as central atom 1 and central atom 2), each bonded to hydrogen atoms, showing three separate orbitals. On the right, a simplified structure with two carbon atoms double-bonded, displaying bond angles of 120 degrees, and labeled with positions 1 and 2.\r\n \r\nWater polarity diagram showing a central atom bonded to two hydrogen atoms (H). The electron cloud is depicted with grey lobes, indicating areas of higher electron density. The symbols \u03b4- and \u03b4+ indicate partial negative and positive charges, respectively, with \u03b4- on the grey lobes and \u03b4+ by the hydrogen atoms. An arrow points upwards, indicating the direction of polarity.\r\n\r\n","n":0.025},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Chemical bonding","i":2,"n":0.707},{"v":"Covalent bonds","i":1,"n":0.707},{"v":"Chemistry","i":0,"n":1}]]},{"i":264,"$":[{"v":"Working with physical quantities","n":0.5},{"v":"Chemistry often deals with quantities (amounts) of chemicals. What do moles have to do with chemistry? How do we measure the mass of different components in a chemical reaction? The mole and stoichiometry are explained here.\r\nOverview \r\n \r\n\tThe mole \u2013 what number does a mole represent and how is it used in chemistry? 25 minutes to complete\r\n\tQuiz \u2013 The mole \u2013 can you complete calculations involving moles? 30 minutes to complete\r\n\tStoichiometry \u2013 what is the relationship between reactants and products in chemistry? 40 minutes to complete\r\n\tQuiz \u2013 Stoichiometry \u2013 can you complete calculations involving stoichiometry? 25 minutes to complete\r\n\r\n","n":0.101},[{"v":"Units","i":3,"n":1},{"v":"The mole","i":2,"n":0.707},{"v":"Stoichiometry","i":1,"n":1},{"v":"Measurement","i":0,"n":1}]]},{"i":265,"$":[{"v":"Lewis structures","n":0.707},{"v":"Lewis electron dot structures are crucial because they visually represent the valence electrons of atoms, helping to predict molecular bonding, structure, and reactivity in chemistry. This resource contains all you need to know to get started with these structures.\r\nLewis structures, also known as electron dot structures, are diagrams that represent the valence electrons of atoms within a molecule. These structures are used to show how the electrons are arranged around individual atoms in a molecule. The main purpose of Lewis structures is to predict the number and type of bonds that may form around an atom.\r\n\r\nExample 1 \u2013 drawing Lewis structures of molecular compounds\r\nDraw the Lewis structure of sulfur dioxide, \\(\\ce{SO}_{2}\\).\r\nStep 1: Count the total number of valence shell electrons of all the atoms involved in the molecule.\r\nSulfur dioxide has two sulfur atoms and one oxygen atom.\r\n\r\nThe total number of valence shell electrons is:\r\n\\[6+\\left(6\\times2\\right)=18\\]\r\n\r\nStep 2: Predict the central atom. This is the atom that can form most bonds. If two different atoms can form the same number of bonds, then the least electronegative atom should be the central atom.\r\nAmong sulfur and oxygen, sulfur is the least electronegative atom. Therefore, it is the central atom.\r\n\r\nStep 3: Draw single bonds connected to the central atom.\r\n\r\n\r\nStep 4: Add the lone pairs of electrons to the atoms connected to the central atom. Calculate the remaining valence electrons after drawing bonds.\r\nFour valence electrons are used for two bonds. The remaining valence electrons are:\r\n\\[=18-4=14\\]\r\nThese valence electrons can be distributed among atoms connected to the central atom as lone pairs to complete the octet.\r\n\r\n\r\nStep 5: Assign lone pairs of electrons to the central atom if any valence electrons remain.\r\nTo assign lone pairs to the connected atoms, we have used 12 electrons. Therefore, we can calculate the electrons remaining:\r\n\\[14-12=2\\]\r\nThese electrons can be allocated to the central atom as a lone pair.\r\n\r\n\r\nStep 6: If the central atom still lacks the octet, create multiple bonds with connected atoms using lone pairs of electrons.\r\n\r\nSince sulfur has not achieved an octet, a double bond must be created with one of the oxygen atoms using a lone pair. With the double bond, all three atoms have acquired an octet.\r\n\r\n\r\nStep 7: Calculate the formal charges for each atom. Formal charges should be closer to zero for each atom. The formula to calculate the formal charges is:\r\n\\[\\textrm{Formal charge}=\\textrm{Valence electrons}-\\textrm{Lone pairs}-\\textrm{Bonding electrons}\\]\r\nFor \\(\\ce{S}\\), formal charge \\(=6-2-3=+1\\).\r\nFor the single-bonded \\(\\ce{O}\\), formal charge \\(=6-6-1=-1\\).\r\nFor the double-bonded \\(\\ce{O}\\), formal charge \\(=6-4-2=0\\).\r\n\r\nExample 2 \u2013 drawing Lewis structures of polyatomic ions\r\nDraw the Lewis structure of the carbonate ion, \\(\\ce{CO}_{3}^{2-}\\).\r\nStep 1: Count the total number of valence shell electrons of all the atoms involved in the molecule, including the charge of the ion.\r\nThe carbonate ion is composed of three oxygen atoms and one carbon atom. It also has a \\(2-\\) charge, so two extra electrons.\r\n\r\nThe total number of valence shell electrons is:\r\n\\[=4+\\left(6\\times3\\right)+2=24\\]\r\n\r\nStep 2: Predict the central atom.\r\nAmong carbon and oxygen, carbon is the least electronegative atom. Therefore, it is the central atom.\r\n\r\nStep 3: Draw single bonds connected to the central atom.\r\n\r\n\r\nStep 4: Add the lone pairs of electrons to the atoms connected to the central atom. Calculate the remaining valence electrons after drawing bonds.\r\nSix valence electrons are used for three bonds. The remaining valence electrons are:\r\n\\[=24-6=18\\]\r\nThese valence electrons can be distributed among atoms connected to the central atom as lone pairs to complete the octet.\r\n\r\n\r\nStep 5: Assign lone pairs of electrons to the central atom if any valence electrons remain.\r\nTo assign lone pairs to the connected atoms, we have used 18 electrons. We have used all the valence electrons.\r\n\r\nStep 6: If the central atom still lacks the octet, create multiple bonds with connected atoms using lone pairs of electrons.\r\n\r\nSince carbon has not achieved an octet, a double bond must be created with one of the oxygen atoms using a lone pair. With the double bond, all four atoms have acquired an octet.\r\n\r\n\r\nStep 7: Calculate the formal charges for each atom. Allocate the two negative charges.\r\nFor \\(\\ce{C}\\), formal charge \\(=4-0-4=0\\).\r\nFor the double-bonded \\(\\ce{O}\\), formal charge \\(=6-4-2=0\\).\r\nFor the single-bonded \\(\\ce{O}\\)s, formal charge \\(=6-6-1=-1\\).\r\n\r\n\r\n\r\nExample 4 \u2013 using Lewis structures to determine molecular shape\r\nPredict the molecular shape of \\(\\ce{NH3}\\).\r\nStep 1: Draw the Lewis structure of the molecule and identify the central atom.\r\n\r\nThe central atom is \\(\\ce{N}\\) as it forms most bonds.\r\n\r\nStep 2: Sum up the number of bonds and lone pairs connected to the central atom to get the total number of electron clouds around the central atom.\r\nThe central atom \\(\\ce{N}\\) forms three bonds\u2014one with each \\(\\ce{H}\\) atom\u2014and one lone pair. Therefore there are four electron clouds (three bonds and one lone pair) around the central atom.\r\n\r\n\r\nStep 3: Recall the VSEPR theory to determine how the electron clouds will be arranged.\r\nAccording to VSEPR theory, the four electron clouds around the \\(\\ce{N}\\) will arrange in such a way that each cloud stays as far as possible from the others to minimise the repulsion forces between them. This will result in a tetrahedral shape. Molecules with a tetrahedral shape have a bond angle of \\(109^{\\circ}\\). However, due to the presence of a lone pair on the central atom, the overall shape of ammonia becomes pyramidal rather than tetrahedral with a bond angle of \\(107^{\\circ}\\).\r\n\r\n\r\n\r\n \r\n    Molecular geometry diagram\r\n\r\n    \r\n        Tetrahedron\r\n        \r\n            A diagram of a tetrahedral shape with a central atom labeled \"C\" (in red) and surrounding atoms labeled \"A\" (in green). The legend indicates that \"C\" is the central atom and \"A\" represents other atoms.\r\n        \r\n    \r\n    \r\n    \r\n        Ammonia \u2013 pyramid shape\r\n\t\t\r\n        \r\n            The chemical structure of ammonia (NH\u2083), showing a nitrogen atom bonded to three hydrogen atoms in a pyramidal shape. The nitrogen has a lone pair represented by a grey lobe.\r\n        \r\n        \r\n            A 3D model of the ammonia molecule with a blue nitrogen atom and grey hydrogen atoms arranged in a pyramidal shape.\r\n        \r\n\t\t\r\n    \r\n","n":0.032},[{"v":"Chemical bonding","i":2,"n":0.707},{"v":"Electron","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":266,"$":[{"v":"Elements of critical thinking","n":0.5},{"v":"When it comes to critically analysing articles at university, there are generally four steps that you need to work through: identifying the topic, analysis, synthesis and evaluation. Let's take a look at what's involved at each stage.\r\n\r\n\r\n\r\nIdentifying the topic\r\n\r\nThe first step in the critical analysis of a text is identifying the topic.\r\n\r\n'Topic' refers to the word or sentence which states the main subject of the work. The 'work' is the author's explanation of the topic. The topic is often identified in the main title and it should be clear from the abstract or opening paragraph what the topic is. Topics answer the questions of who, what and why through data and facts.\r\n\r\nOnce you have identified the topic, it's time to start the critical thinking process, which has three key elements:\r\n\r\n \tanalyse\r\n \tevaluate\r\n \tsynthesise.\r\n\r\nAnalyse\r\nWhen you analyse, you identify the parts of an argument. This includes:\r\n\r\n \tidentifying the main elements of a text, particularly the key ideas, the argument and the evidence\r\n \tdrawing out inferences and implications\r\n \tidentifying any persuasive tactics that have been used.\r\n\r\nExamples of analytical language\r\nHere are some examples of language that can be used when analysing information:\r\n\r\nWright's (2022) claim that ... is supported by...\r\nThe evidence is comprised of three main components, which are...\r\nThe research demonstrates that...\r\n\r\nEvaluate\r\nWhen you evaluate, you make judgements about arguments and theories.\r\nThis includes:\r\n\r\n \tidentifying the strengths and weaknesses in an argument (its credibility)\r\n \tweighing up the value of evidence (its validity)\r\n \tidentifying and evaluating the assumptions underlying the argument (its integrity).\r\n\r\nExamples of evaluative language\r\nHere are some examples of language that can be used when evaluating arguments:\r\n\r\nNguyen's (2021) argument is based on quantitative research conducted by...\r\nThe evidence used to support this theory is credible because...\r\nThis research is based on unreliable theories because...\r\nThe results do not support the conclusions of the report due to...\r\nThis argument has been adequately supported by evidence such as...\r\n\r\nSynthesise\r\nSynthesising means bringing together different sources of information and considering them together. \r\nIt includes:\r\n\r\n \tputting information together in a new pattern\r\n \tproviding a new point of view\r\n \tshowing how the relationship between the parts, and between the parts and the whole, producing a unique communication.\r\n\r\nExamples of language used for synthesising\r\nHere are some examples of language that can be used when synthesising information:\r\n\r\nThe study by Akbar (2024) contributes to the research on the topic of...\r\nOverall, the research demonstrates that...\r\nFrom these studies, it can be concluded that...\r\n\r\nCritical writing example\r\nIn this example paragraph, note how the writer has structured their paragraph to support their contention stated in the topic sentence.\r\n\r\n\r\n\r\n","n":0.049},[{"v":"Critical thinking","i":1,"n":0.707},{"v":"Analysis","i":0,"n":1}]]},{"i":267,"$":[{"v":"Maxima and minima","n":0.577},{"v":"\r\nTaking the derivative of the function lets us find critical points where the slope is \\(0\\) or undefined. These points tell us where the function might have a maximum or a minimum value. This approach is useful in many real-world situations, like optimising resources or finding the best design for a project. Learn how to use derivatives to find maxima and minima in functions.\r\n\r\nFinding maxima and minima\r\nThe maximum or minimum values of a function occur where the derivative is \\(0\\). This is where the graph of the function has a horizontal tangent.\r\nThis means that if we find where a function has a derivative of \\(0\\), we can locate the maxima (highest point of a curve) and minima (lowest point of a curve).\r\nMathematically, we are solving for \\(x\\) when:\r\n\\[f'(x)=0\\]\r\n\r\nSecond derivative test\r\nThe second derivative is the derivative of the derivative. For \\(f(x)\\), the derivative is \\(f'(x)\\) and the second derivative is \\(f''(x)\\).\r\nDepending on the size of the second derivative, we can tell whether the \\(x\\) value we have calculated is an inflection point (where the curve changes from concave to convex or vice versa), a local maximum or local minimum.\r\nFor \\(x\\) where \\(f'(x)=0\\), if:\r\n\\[f''(x) \\begin{cases} = 0,x\\textrm{ is an inflection point}\\\\ \r\n<0,f(x)\\textrm{ is a local maximum}\\\\\r\n>0,f(x)\\textrm{ is a local minimum}\r\n\\end{cases}\\]\r\n\r\n\r\n\r\nThe second derivative for a function \\(y\\) in terms of \\(x\\) can also be written as \\(\\dfrac{d^{2}y}{dx^{2}}\\).\r\n\r\nExample 1 \u2013 finding maxima and minima\r\nThe distance \\(s\\textrm{ km}\\) of a fishing boat from port at any time \\(t\\) hours, to the nearest kilometre, is given by the formula:\r\n\\[s=2+8t-2.5t^{2}\\]\r\nWhen is the boat furthest from port and what is its distance from the port at that time?\r\nThe furthest distance from the port is the \\(s\\) value when \\(s'=0\\). We need to solve for \\(t\\).\r\nFirst, we find \\(s'\\).\r\n\\[s'=8-5t\\]\r\nNow, we can solve \\(s'=0\\) for \\(t\\).\r\n\\[\\begin{align*} s' & = 0\\\\\r\n8-5t & = 0\\\\\r\n8 & = 5t\\\\\r\nt & = \\frac{8}{5}\\\\\r\n& = 1.6\\textrm{ h}\r\n\\end{align*}\\]\r\nWe can check that this is a maximum distance using the second derivative test.\r\n\\[\\begin{align*} s'' & = -5\\\\\r\n& < 0\r\n\\end{align*}\\]\r\n\\(t=1.6\\) is therefore a maximum.\r\nTo find the distance from the port at that time, we substitute \\(t=1.6\\) into \\(s\\).\r\n\\[\\begin{align*} s & = 2+8t-2.5t^{2}\\\\\r\n& = 2+8(1.6)-2.5(1.6)^{2}\\\\\r\n& = 8.4\\textrm{ km}\r\n\\end{align*}\\]\r\nTo summarise, the boat is furthest from port after \\(1.6\\textrm{ h}\\), where it is \\(8.4\\textrm{ km}\\) away.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding maxima and minima\r\n\r\n \tFind the turning point of the parabola defined by \\(y=f(x)=5x^{2}-30x+17\\).\r\n \tFind two positive numbers whose sum is \\(18\\) such that the sum of their squares is a minimum. \r\n\tWhat is the maximum area that can be enclosed if a rectangle is created with a piece of wire \\(48\\textrm{ cm}\\) long?\r\n \tThe annual profit \\(P\\) made on a garment is related to the number \\(n\\) that are produced by the formula \\(P(n)=300n-7200-0.2n^{2}\\). How many garments should be produced to maximise profit?\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.047},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":268,"$":[{"v":"Gradients, tangents and derivatives","n":0.5},{"v":"\r\nGradients and tangents help us understand how functions change. We use derivatives to find these. Learn how these concepts work together to analyse and interpret the behaviour of functions.\r\n\r\nTangent at a point along a curve\r\nA tangent is a line that touches a curve at only one point. Where that point sits along the function curve determines the slope (i.e. the gradient) of the tangent to that point.\r\nWe can find the gradient of a tangent at any point on a curve \\(f(x)\\) using the derivative of the function, \\(f'(x)\\). If we plug an \\(x\\) value into the derivative function, we will get the slope of the tangent at that point \\(x\\).\r\nHere, the derivative functions \\(f'(x)\\) have been given to you. To learn how to find derivatives, go to Differentiation from first principles.\r\n\r\nGradient of a tangent\r\nYou should already be familiar with how to find the gradient of a straight line. If not, you can go to Linear functions for a recap. For straight lines, the gradient is the same for the entire function.\r\nWhen we are concerned with curves, the gradient can differ depending on where we are on the curve. Let's consider the curve \\(f(x)\\) with point \\(P\\).\r\n\r\nWe can approximate the tangent at point \\(P(x,f(x))\\) by using another point \\(Q\\) on the curve, which has coordinate \\(Q(x+h,f(x+h))\\). The approximate gradient would be:\r\n\\[\\textrm{Gradient of }PQ = \\frac{f(x+h)-f(x)}{h}\\]\r\nAs \\(h\\) decreases and \\(Q\\) gets closer to \\(P\\), the approximate of the gradient becomes more accurate. In other words, the gradient becomes more accurate as \\(h\\) approaches \\(0\\). So, we can determine a more accurate gradient by finding the limit:\r\n\\[f'(x) = \\lim_{h\\rightarrow0}\\left(\\frac{f(x+h)-f(x)}{h}\\right)\\quad h\\neq0\\]\r\nThe gradient formula for the curve \\(y=f(x)\\) is defined as the derivative function, which gives the slope of the tangent to the curve \\(f(x)\\) at any point \\(x\\).\r\n\r\nExample \u2013 finding the gradient of a tangent\r\nThe derivative of the function \\(f(x)=\\dfrac{3}{x}\\) is \\(f'(x)=-\\dfrac{3}{x^{2}}\\). Find the slope of the tangent to the curve at \\(x=4\\).\r\nAt \\(x=4\\):\r\n\\[\\begin{align*} f'(x) & =-\\frac{3}{x^{2}}\\\\\r\nf'(4) & = -\\frac{3}{4^{2}}\\\\ \r\n& = -\\frac{3}{16}\r\n\\end{align*}\\]\r\nThe slope of the tangent at \\(x=4\\) is \\(-\\dfrac{3}{16}\\).\r\n\r\nExercise \u2013 finding the gradient of a tangent\r\n\r\n \tIf the derivative function for \\(f(x)=x^{3}-x\\) is \\(f'(x)=3x^{2}-1\\), find the slope of the tangent to this curve at:\r\n\t\r\n\t\t\\(x=2\\)\r\n\t\t\\(x=0\\)\r\n\t\t\\(x=-9\\)\r\n\t\r\n\tIf the derivative function of \\(f(x)=\\sin\\left(x\\right)\\) is \\(f'(x)=\\cos\\left(x\\right)\\), find the gradient of \\(f(x)=\\sin\\left(x\\right)\\) at:\r\n\t\r\n\t\t\\(x=0\\)\r\n\t\t\\(x=\\dfrac{\\pi}{2}\\)\r\n\t\t\\(x=3.5\\)\r\n\t\r\n\tDetermine \\(\\lim_{h\\rightarrow0}\\left(\\frac{(x+h)^{2}-x^{2}}{h}\\right)\\) and hence find the slope of the tangent to the curve \\(y=x^{2}\\) at:\r\n\t\r\n\t\t\\(x=2\\)\r\n\t\t\\(x=0\\)\r\n\t\t\\(x=-9\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.051},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":269,"$":[{"v":"Directional derivatives","n":0.707},{"v":"Imagine navigating a hilly landscape, searching for the steepest ascent or smoothest descent\u2014this is the essence of directional derivatives. These mathematical tools are vital in meteorology for predicting temperature changes along wind paths, in finance for analysing portfolio shifts, and in machine learning for optimising algorithms. Use this resource to learn how to calculate directional derivatives.\r\n\r\nThis concept combines your understanding of vectors and calculus. Before you read any further, make sure that you are confident with differentiation.\r\n\r\nRemember that the partial derivative of a function gives us the gradient or rate of change of the function along each axis. For example, let's consider a function with two variables: \\(z=f(x,y)\\).\r\n\r\n\tFor the rate of change of the function \\(f\\) with respect to \\(x\\), we find \\(\\dfrac{\\partial f}{\\partial x}\\). This is the gradient of the graph as we move in the \\(x\\) direction.\r\n\tFor the rate of change of the function \\(f\\) with respect to \\(y\\), we find \\(\\dfrac{\\partial f}{\\partial y}\\). This is the gradient of the graph as we move in the \\(y\\) direction.\r\n\r\n\r\nDirectional derivatives build on this and allow us to find the rate at which a function changes as we move in any direction defined by a vector.\r\n\r\nImagine you're hiking on a mountain, and the landscape is a complex surface. The directional derivative is like asking, \"If I walk in a specific direction, how steep will the path be?\"\r\n\r\n\r\nJust as the steepness of your path changes depending on the direction you choose to walk (uphill, downhill, or along the side of the mountain), the directional derivative measures how steep or fast a function changes in a particular direction at a given point.\r\n\r\n\tIf the directional derivative is positive, the path goes uphill.\r\n\tIf the directional derivative is negative, the path goes downhill.\r\n\tIf the directional derivative is \\(0\\), the path is flat in that direction.\r\n\r\n\r\nFinding the directional derivative\r\nYou can find the directional derivative of a function containing two variables \\(f(x,y)\\) or three variables \\(g(x,y,z)\\) at a point and in a given direction. \r\nThe directional derivative of function \\(f\\) in the direction of a vector \\(\\vec{u}\\) is denoted by \\(D_{u}\\) and is given by:\r\n\\[D_{u} = \\nabla f\\cdot\\hat{u}\\]\r\nwhere \\(\\nabla f\\) (\"grad \\(f\\)\") is the gradient of the function \\(f\\) in the direction of the vector \\(\\vec{u}\\) and \\(\\hat{u}\\) is a unit vector in the direction of \\(\\vec{u}\\).\r\n\\(\\nabla f\\) is found by partial differentiation according to the following:\r\n\\[\\nabla f = \\frac{\\partial f}{\\partial x}\\hat{i} + \\frac{\\partial f}{\\partial y}\\hat{j} + \\frac{\\partial f}{\\partial z}\\hat{k}\\]\r\nAnd of course, the unit vector for vector \\(\\vec{u}\\) is just the vector divided by its magnitude.\r\nLet's look at the mountain example again. \r\n\r\nImagine you are standing at the point indicated by the red dot in the figure. \r\n\r\n\tThe tangent of the slope at that point is shown by the solid line.\r\n\tThe gradient \\(\\nabla f\\) is perpendicular to the tangent.\r\n\tWe want to know how steep the path is in the direction of vector \\(\\vec{u}\\); this is given by \\(D_{u}\\).\r\n\t\\(D_{u}\\) is the projection of the gradient onto the direction vector.\r\n\r\nSince the directional derivative relies on a dot product where \\(\\vec{a}\\cdot\\vec{b}=\\left|\\vec{a}\\right|\\left|\\vec{b}\\right|\\cos\\theta\\):\r\n\r\n \t\\(D_{u}\\) will be maximum when \\(\\vec{u}\\) and \\(\\nabla f\\) are in the same direction, i.e. when \\(\\theta=0\\) and \\(\\cos\\theta=1\\).\r\n \t\\(D_{u}\\) will have its greatest negative value when \\(\\vec{u}\\) and \\(\\nabla f\\) are in opposite directions, i.e. when \\(\\theta=\\pi\\).\r\n \t\\(D_{u}\\) will be \\(0\\) when \\(\\vec{u}\\) and \\(\\nabla f\\) are at right angles, i.e. perpendicular to each other or when \\(\\theta=\\dfrac{\\pi}{2}\\).\r\n\r\n\r\nExample 1 \u2013 finding the directional derivative\r\nFind the directional derivative of \\(f(x,y)=x^{2}+y^{2}\\) at the point \\((2,3)\\) in the direction of the vector \\(\\vec{u}=4\\hat{i}-3\\hat{j}\\).\r\nThe first step is to find the vector \\(\\nabla f\\) at the point \\(x=2\\), \\(y=3\\).\r\n\\[\\begin{align*} \\nabla f & = \\frac{\\partial f}{\\partial x}\\hat{i} +\\frac{\\partial f}{\\partial y}\\hat{j}\\\\\r\n& = \\frac{\\partial}{\\partial x}(x^{2}+y^{2})\\hat{i} + \\frac{\\partial}{\\partial y}(x^{2}+y^{2})\\hat{j}\\\\\r\n& = 2x\\hat{i}+2y\\hat{j}\\\\\r\n& = 2(2)\\hat{i}+2(3)\\hat{j}\\\\\r\n& = 4\\hat{i}+6\\hat{j}\r\n\\end{align*}\\]\r\nNext, we need to find the dot product of \\(\\nabla f\\) and \\(\\hat{u}\\), the unit vector in the direction of \\(\\vec{u}\\). Let's find the \\(\\hat{u}\\) first.\r\n\r\nNow, we can substitute \\(\\nabla f\\) and \\(\\hat{u}\\) into the formula to find the directional derivative.\r\n\\[\\begin{align*} D_{u} & = (4\\hat{i}+6\\hat{j})\\cdot(\\frac{4}{5}\\hat{i}-\\frac{3}{5}\\hat{j})\\\\\r\n& = (4\\times\\frac{4}{5})+(6\\times-\\frac{3}{5})\\\\\r\n& = \\frac{16}{5}-\\frac{18}{5}\\\\\r\n& =-\\frac{2}{5}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExercise \u2013 finding the directional derivative\r\n\r\n \tFind the directional derivative of the given function \\(f\\) at the given point \\((x,y)\\) in the direction of the indicated vector \\(\\vec{u}\\).\r\n\t\r\n\t\t\\(f(x,y)=xy^{2}\\) at point \\((3,2)\\) and in the direction of \\(\\vec{u}=4\\hat{i}+3\\hat{j}\\)\r\n\t\t\\(f(x,y)=e^{xy}\\) at point \\((0,2)\\) and in the direction of \\(\\vec{u}=\\hat{i}\\)\r\n\t\t\\(f(x,y,z)=x^{2}y^{3}z\\) at point \\((2,-1,3)\\) and in the direction of \\(\\vec{u}=\\hat{i}-2\\hat{j}-2\\hat{k}\\)\r\n\t\r\n\tFind the unit vector in the direction in which \\(f\\) increases most rapidly at \\(P\\left(1,\\dfrac{\\pi}{2}\\right)\\) for \\(f(x,y)=x^{2}+\\cos(xy)\\).\r\n\r\n\r\n \r\n\r\n\r\n\r\n","n":0.037},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":270,"$":[{"v":"Distance from a point to a plane","n":0.378},{"v":"\r\nThe distance between a point in 3D space and a plane can be determined using vectors and trigonometry. This is handy in many situations, like making sure components are specific distances from certain planes in construction, determining distances between a airplane and a mountain or building in safe flight planning, and analysing how far away you need to place a light source or camera from a surface to create realistic shading. Use this resource to learn more.\r\n\r\nIn the figure, the blue plane extends in all directions. Points \\(P_{1}\\) and \\(P_{2}\\) lie on the plane. Point \\(A\\) is a point in the 3D space that does not lie on the plane.\r\n\r\nWe can have any number of distances from a point \\(A\\) to the blue plane. For example, the distance from point \\(A\\) to point \\(P_{1}\\) is one possible answer. The distance from point \\(A\\) to point \\(P_{2}\\) is another. \r\nBut when you are asked to find the distance between a point and a plane, you must provide the shortest distance. This is found by drawing a line at a right angle to the plane. Hence, the distance is also called the perpendicular distance between the point and the plane.\r\nIn this example, it would be the distance between point \\(A\\) and \\(P_{3}\\) in the following diagram.\r\n\r\n\r\nPerpendicular distance\r\nConsider the following figure. \r\n\r\nTo find the shortest distance from point \\(A\\) to the blue plane:\r\n\r\n\tChoose any point \\(P\\) on the plane.\r\n\tDefine a vector from \\(P\\) to \\(A\\), \\(\\overrightarrow{PA}\\).\r\n\tFind the vector that is normal to \\(\\overrightarrow{PA}\\).\r\n\tFind a unit vector that is normal to the plane. This is sometimes called the unit normal or scalar resolute of \\(\\overrightarrow{PA}\\) in the direction of \\(\\hat{n}\\).\r\n\t\\[\\hat{n}=\\frac{\\vec{n}}{\\left|\\vec{n}\\right|}\\]\r\n\tGo back to review how to calculate magnitude of a vector if you need to.\r\n\tFind the perpendicular distance using:\t\r\n\t \\[\\begin{align*} d & = \\overrightarrow{PA}\\cdot\\hat{n}\\\\\r\n\t & = \\overrightarrow{PA}\\cdot\\frac{\\vec{n}}{\\left|\\vec{n}\\right|}\r\n\t \\end{align*}\\]\r\n\tThis may be easier to understand if we rotate the figure so that we are looking at the blue plane from its edge.\r\n\t\r\n\r\n\r\nExample \u2013 finding the distance between a point and a plane\r\nFind the distance from the point \\((2,5,4)\\) to the plane \\(x+2y+2z=2\\).\r\nLet point \\(A=(2,5,4)\\). We need to then find a point \\(P\\) on the plane. It can be any point but to make it easy, we can take \\(y=z=0\\). \r\nSubstituting in the equation for the plane, we get \\(x=2\\). So, the point on the plane is \\(P=(2,0,0)\\). \r\nThe vector from \\(P\\) to \\(A\\) is:\r\n\\[\\begin{align*} \\overrightarrow{PA} & = \r\n(2-2)\\hat{i}+(5-0)\\hat{j}+(4-0)\\hat{k}\\\\\r\n& = 5\\hat{j}+4\\hat{k}\r\n\\end{align*}\\]\r\nNow, we need to find a unit vector that is normal to the plane. A normal vector is:\r\n\\[\\vec{n} = \\hat{i}+2\\hat{j}+2\\hat{k}\\] \r\nWe can then find a unit normal vector:\r\n\\[\\begin{align*} \\hat{n} & = \\frac{\\vec{n}}{\\left|\\vec{n}\\right|}\\\\\r\n& = \\frac{\\hat{i}+2\\hat{j}+2\\hat{k}}{\\sqrt{1^{2}+2^{2}+2^{2}}}\\\\\r\n& = \\frac{\\hat{i}+2\\hat{j}+2\\hat{k}}{\\sqrt{9}}\\\\ \r\n& = \\frac{1}{3}\\left(\\hat{i}+2\\hat{j}+2\\hat{k}\\right)\r\n\\end{align*}\\] \r\nFinally, we substitute the values into the formula to find the distance, \\(d\\).\r\n\\[\\begin{align*} d & = \\overrightarrow{PA}\\cdot\\hat{n}\\\\ \r\n& = \\left(5\\hat{j}+4\\hat{k}\\right)\\cdot\\frac{1}{3}\\left(\\hat{i}+2\\hat{j}+2\\hat{k}\\right)\\\\ \r\n& = \\frac{1}{3}\\left(10+8\\right)\\\\ \r\n& = 6\r\n\\end{align*}\\]\r\n\r\nExercise \u2013 finding the distance between a point and a plane\r\nFind the distance from the given point to the given plane.\r\n\r\n\t\\((0,0,0)\\) and \\(2x+3y-z=6\\)\r\n\t\\((-1,1,2)\\) and \\(x+y=2\\)\r\n\t\\((1,2,3)\\) and \\(3x+4y-z=1\\)\r\n\r\n\r\n \r\n\r\n\r\n\r\n","n":0.045},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":271,"$":[{"v":"Intersecting planes in 3D","n":0.5},{"v":"Two planes in 3D can intersect. Finding this intersection has many real-life applications, including the design of buildings in architecture, 3D rendering and modelling in computer graphics, and determining paths of movement in robotics. Use this resource to learn how to determine the angle between two intersecting planes and the equation of the line of intersection.\r\n\r\n\r\nAs long as they are not parallel, two lines in 3D can intersect at a point. In the same way, two 2D planes can intersect or slice through each other given they are not parallel. The set of points where they interact forms a line. There will also be an angle between the two planes. \r\n\r\n\r\nAngle of intersection between two planes\r\nThe angle between two intersecting planes is the same as the angle between the normals of the planes.\r\nIf \\(N_{1}\\) and \\(N_{2}\\) are the normals of two intersecting planes, we know that:\r\n\\[\\overrightarrow{N_{1}}\\cdot\\overrightarrow{N_{2}} = \\left| \\overrightarrow{N_{1}}\\right| \\left|\\overrightarrow{N_{2}} \\right| \\cos(\\theta)\\] where \\(\\theta\\) is the angle between them. This follows from the definition of the scalar or dot product.\r\nWe can rearrange this to make \\(\\theta\\) the subject:\r\n\\[\\theta = \\cos^{-1}\\left( \\frac{\\overrightarrow{N_{1}}\\cdot\\overrightarrow{N_{2}}}{\\left|\\overrightarrow{N_{1}}\\right|\\left|\\overrightarrow{N_{2}}\\right|} \\right) \\] \r\nIn general, there are two possible angles. One is obtuse and the other is acute. Together they add up to \\(180^{\\circ}\\).\r\nIf we want to find only the acute angle between the planes, then we just take the absolute value of the numerator:\r\n\\[\\theta = \\cos^{-1}\\left(\\frac{\\left|\\overrightarrow{N_{1}}\\cdot\\overrightarrow{N_{2}}\\right|}{\\left|\\overrightarrow{N_{1}}\\right|\\left|\\overrightarrow{N_{2}}\\right|} \\right) \\]\r\n\r\nExample \u2013 finding the angle of intersection between two planes\r\nFind the acute angle of intersection of the planes \\(x+y+z=0\\) and \\(x-3y+z=1\\).\r\nThe plane \\(x+y+z=0\\) has the normal vector \\(\\overrightarrow{N_{1}}=\\overrightarrow{i}+\\overrightarrow{j}+\\overrightarrow{k}\\).\r\nThe plane \\(x-3y+z=1\\) has the normal vector \\(\\overrightarrow{N_{2}}=\\overrightarrow{i}-3\\overrightarrow{j}+\\overrightarrow{k}\\).\r\nWe can start with the numerator of the formula: \\(\\left| \\overrightarrow{N_{1}}\\cdot\\overrightarrow{N_{2}} \\right|\\) first.\r\n\\[\\begin{align*} \\left| \\overrightarrow{N_{1}}\\cdot\\overrightarrow{N_{2}} \\right| & = \\left|(1\\times1)+(1\\times(-3))+(1\\times1)\\right| \\\\\r\n& = \\left|-1\\right|\\\\\r\n& = 1 \r\n\\end{align*}\\]\r\nNow, the denominator: \\(\\left| \\overrightarrow{N_{1}} \\right|\\) and \\(\\left| \\overrightarrow{N_{2}} \\right|\\).\r\n\\[\\begin{align*} \\left|\\overrightarrow{N_{1}}\\right| & =\\sqrt{1^{2}+1^{2}+1^{2}}\\\\ & =\\sqrt{3} \\end{align*}\\] \\[\\begin{align*} \\left|\\overrightarrow{N_{2}}\\right| & =\\sqrt{1^{2}+3^{2}+1^{2}}\\\\ & =\\sqrt{11} \\end{align*}\\]\r\nWe can substitute these values into the formula:\r\n\\[\\begin{align*} \\theta & = \\cos^{-1} \\left(\\frac{1}{\\sqrt{3}\\sqrt{11}} \\right) \\\\\r\n& = \\cos^{-1} \\left( \\frac{\\sqrt{33}}{33} \\right)\\\\\r\n& \\approx \\cos^{-1} \\left( 0.1741 \\right)\\\\\r\n& \\approx 80^{\\circ}\r\n\\end{align*}\\]\r\nThe angle of intersection of the two planes is approximately \\(80^{\\circ}\\).\r\n\r\nExercise - finding the angle of intersection between two planes\r\n\r\n\tFind the angle of intersection between the following planes.\r\n\t\r\n\t\t\\(x-y+z=1\\) and \\(2x+y-z=3\\)\r\n\t\t\\(2x+y-z=2\\) and \\(3x+y-z=3\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\nLine of intersection between two planes\r\nConsider planes \\(P_{1}\\) and \\(P_{2}\\) which intersect along the line \\(L\\),\r\n\r\nTo find the equation of the line of intersection, \\(L\\), we solve the simultaneous equations.\r\n\r\nExample \u2013 finding the line of intersection between two planes\r\nFind the parametric equations for the line of intersection between \\(P_{1}:\\,2x+4y-z=4\\) and \\(P_{2}:\\,x-2y+z=3\\).\r\nWe have two equations but three unknowns. This means we need to introduce a parameter. We can pick any of \\(x\\), \\(y\\) or \\(z\\) but here, let's set \\(z=t\\), where \\(t\\in\\mathbb{R}\\). We can also shift \\(t\\) over so that it sits with the constant. This would give us the equations:\r\n\\[2x+4y=4+t\\] \\[x-2y=3-t\\]\r\nWe can multiply the second equation by \\(2\\) so that they become:\r\n\\[2x+4y=4+t\\] \\[2x-4y=6-2t\\]\r\nAdding these two equations, we can get rid of the \\(y\\) term:\r\n\\[\\begin{align*} (2x+4y)+(2x-4y) & = (4+t)+(6-2t)\\\\\r\n4x & = 10-t\\\\\r\nx & = \\frac{10-t}{4}\\\\\r\n& = \\frac{5}{2}-\\frac{1}{4}t\r\n\\end{align*}\\]\r\nNow that we have \\(x\\), we can substitute it into the second equation to find \\(y\\) in terms of \\(t\\):\r\n\\[\\begin{align*} \\left(\\frac{5}{2}-\\frac{1}{4}t\\right)-2y & = 3-t\\\\\r\n-2y & = 3-t-\\frac{5}{2}+\\frac{1}{4}t\\\\\r\n& = \\left(3-\\frac{5}{2}\\right)+\\left(-1+\\frac{1}{2}\\right)t\\\\\r\n& = \\frac{6-5}{2}+\\frac{-4+1}{4}t\\\\\r\n& = \\frac{1}{2}-\\frac{3}{4}t\\\\\r\ny & = -\\frac{1}{4}+\\frac{3}{8}t\r\n\\end{align*}\\]\r\nThis gives us the parametric equations for the line of the intersection of the two planes:\r\n\\[x=\\frac{5}{2}-\\frac{1}{4}t\\] \\[y=-\\frac{1}{4}+\\frac{3}{8}t\\] \\[z=t\\] where \\(t\\in\\mathbb{R}\\).\r\n\r\nExercise \u2013 finding the line of intersection between two planes\r\nFind the parametric equations for the line of intersection and the angle of intersection of the planes \\(x+y+2z=0\\) and \\(2x-y+z=5\\).\r\n\r\n \r\n\r\n","n":0.041},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":272,"$":[{"v":"Equation of a plane in 3D","n":0.408},{"v":"Just as we can define an equation for a line in three dimensions, we can do the same for an entire plane\u2013that is, we can define an equation that represents a two-dimensional space within a 3D space.\r\n\r\nA plane is a subset of three dimensional space. You can think of it as a flat surface that extends infinitely in two directions. It may be defined as:\r\n\r\n \tthe surface that goes through three points; or\r\n \tthe surface containing a point and having a fixed normal vector.\r\n\r\n\r\n\r\nCartesian equation of a plane\r\nA Cartesian equation is a way to describe shapes using the \\(x\\)\u2014\\(y\\) coordinate system. It is handy for analysing the properties and positions of geometric 2D shapes in 3D space.\r\nConsider the following diagram, which shows a part of a plane.\r\n\r\nOn the plane are two points, \\(P(x,y,z)\\) and \\(P_{0}(x_{0},y_{0},z_{0})\\). The equation for the vector extending from \\(P_{0}\\) to \\(P\\) is:\r\n\\[\\overrightarrow{P_{0}P} = (x-x_{0})\\hat{i}+(y-y_{0})\\hat{j}+(z-z_{0})\\hat{k}\\]\r\nThe diagram also shows a normal vector to the plane\u2014that is, a vector at right angles to the plane\u2014represented by:\r\n\\[\\overrightarrow{N}=a\\hat{i}+b\\hat{j}+c\\hat{k}\\]\r\nwhere \\(a\\), \\(b\\) and \\(c\\) are constants.\r\nAnother normal is given by \\(-\\overrightarrow{N}=-a\\hat{i}-b\\hat{j}-c\\hat{k}\\); this just a vector in the opposite direction of \\(\\overrightarrow{N}\\) and is still at right angles to the plane.\r\nAs \\(\\overrightarrow{N}\\) is perpendicular to \\(\\overrightarrow{P_{0}P}\\), their scalar or dot product must equal \\(0\\). Go back to the page on scalar or dot product to review this concept.\r\nThis means \\(\\overrightarrow{N}\\cdot\\overrightarrow{P_{0}P}=0\\). Written out in full, this is:\r\n\\[(a\\hat{i}+b\\hat{j}+c\\hat{k})\\cdot((x-x_{0})\\hat{i}+(y-y_{0})\\hat{j}+(z-z_{0})\\hat{k})\\]\r\nSo:\r\n\\[a(x-x_{0})+b(y-y_{0})+c(z-z_{0}) = 0\\]\r\nThis equation defines the plane in Cartesian coordinates.\r\nWe can rearrange the equation by expanding the brackets and moving all of the \\(P_{0}\\) coordinates to one side and the \\(P\\) coordinates to the other:\r\n\\[\\begin{align*} a(x-x_{0})+b(y-y_{0})+c(z-z_{0}) & = 0\\\\\r\nax+by+cz & = ax_{0}+by_{0}+cz_{0}\\\\ \r\n& = d\r\n\\end{align*}\\]\r\nwhere \\(d\\) is a constant. \r\nIn fact, the general equation of a plane with the normal vector \\(\\overrightarrow{N}=a\\hat{i}+b\\hat{j}+c\\hat{k}\\) is: \\[ax+by+cz = d\\]\r\nSome examples of equations representing planes are:\r\n\r\n\t\\(2x-3y+z=6\\)\r\n\t\\(x+y=4\\)\r\n\t\\(z=x-2y+3\\).\r\n\r\n\r\nParallel planes\r\nThere are an infinite number of parallel planes that are perpendicular to a given vector. For instance \\(2x-3y+z=6\\), \\(2x-3y+z=0\\), \\(2x-3y+z=-1\\) and \\(2x-3y+z=11\\) are all perpendicular to the vector \\(2\\hat{i}-3\\hat{j}+\\hat{k}\\). However, only one of these will pass through a given point.\r\n\r\nExample 1 \u2013 finding the equation of a plane\r\nIf a plane has the normal vector \\(\\overrightarrow{N}=\\hat{i}+2\\hat{j}-5\\hat{k}\\) and contains the point \\((3,4,1)\\), determine the equation that represents it.\r\nSubstitute the values into the correct formula to find the equation. The normal vector is in the form \\(a\\hat{i}+b\\hat{j}+c\\hat{k}\\) and the point is \\((x_{0},y_{0},z_{0})\\).\r\n\\[\\begin{align*} a(x-x_{0})+b(y-y_{0})+c(z-z_{0}) & = 0\\\\\r\n1(x-3)+2(y-4)-5(z-1) & = 0\\\\\r\nx-3+2y-8-5z+5 & = 0\\\\\r\nx+2y-5z & = 3+8-5\\\\\r\nx+2y-5z & = 6\r\n\\end{align*}\\]\r\nYou may notice that the coefficients for \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\) for the normal vector \\(\\overrightarrow{N}\\) are the same as the coefficients for \\(x\\), \\(y\\) and \\(z\\) in the equation for the plane.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the equation of a plane\r\n\r\n \tDetermine the vectors normal to the following planes.\r\n\t\r\n\t\t\\(2x+3y+7z=13\\)\r\n\t\t\\(z=x+3y-9\\)\r\n\t\r\n\tFind the equation of the plane that contains the point \\((2,3,1)\\) and has the normal vector \\(4\\hat{i}+3\\hat{j}-2\\hat{k}\\).\r\n\tFind the equation of the plane that contains the point \\((5,-3,2)\\) and has the normal vector \\(\\hat{i}-9\\hat{j}-4\\hat{k}\\).\r\n\tFind the equation of the plane that contains the point \\((1,-1,0)\\) and is parallel to the plane \\(x-3y+2z=0\\).\r\n\tFind the equation of the plane that contains the points \\((1,2,3)\\), \\((2,1,1)\\) and \\((-3,0,4)\\).\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n","n":0.044},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":273,"$":[{"v":"Intersecting lines in 3D","n":0.5},{"v":"\r\nVectors can be used to determine whether two lines in 3D cross each other (or intersect), and identify the point at which they intersect. This is used in a variety of STEM disciplines, including detection of collisions between objects in robotics and modelling complex structures in computer graphics and game development. Use this resource to learn more.\r\nTo find the point of intersection of two lines in 3D space, it is best to have both equations in parametric form, with different parameters for each line. For example, the set of equations for the first line may be written in terms of \\(t\\) while the other for second line is written in terms of \\(s\\).\r\nIf the lines intersect, there must be values of \\(s\\) and \\(t\\) that give the same point on each of the lines. If this is not the case, the lines do not intersect.\r\n\r\nThe basic approach is:\r\n\r\n \tWrite each line in parametric form.\r\n \tSolve the simultaneous equations for the parameters.\r\n \tIf a solution exists, the lines intersect. If not, they don\u2019t intersect.\r\n\r\nLet's look at some examples.\r\n\r\nExample 1 \u2013 determining the point of intersection between two lines\r\nFind the point of intersection of the two lines \\(L_{1}\\) and \\(L_{2}\\) where \\(L_{1}\\) is the line represented by the equation \\(\\dfrac{x-5}{1}=\\dfrac{y-1}{-1}=\\dfrac{z-8}{-3}\\) and \\(L_{2}\\) is the line represented by \\(x=1+2s\\), \\(y=-4+s\\) and \\(z=8-2s\\).\r\nThe first step is to write each line in parametric form. Line \\(L_{2}\\) is already in parametric form:\r\n\\[x=1+2s\\] \\[y=-4+s\\] \\[z=8-2s\\]\r\nFor line \\(L_{1}\\), we let each component equal to another parameter that is not \\(s\\). Let's go with \\(t\\).\r\n\\[\\frac{x-5}{1}=t\\] \\[\\frac{y-1}{-1}=t\\] \\[\\frac{z-8}{-3}=t\\]\r\nRearranging them so that they are written with \\(x\\), \\(y\\) and \\(z\\) as the subjects:\r\n\\[x=5+t\\] \\[y=1-t\\] \\[z=8-3t\\]\r\nTo find the point of intersection of lines \\(L_{1}\\) and \\(L_{2}\\), we must find a point \\(P\\) that lies on both lines.\r\n\r\nAt the point of intersection, the coordinates for \\(L_{1}\\) will equal the coordinates for \\(L_{2}\\). This means there exist values of \\(s\\) and \\(t\\) such that the \\(x,\\) \\(y\\) and \\(z\\) coordinates of the two lines are equal. To work this out, we equate the \\(x\\), \\(y\\) and \\(z\\) coordinates.\r\nFor \\(x\\):\r\n\\[\\begin{align*} 5+t & = 1+2s\\\\\r\n4 & = 2s-t\r\n\\end{align*}\\]\r\nFor \\(y\\):\r\n\\[\\begin{align*} 1-t & = -4+s\\\\\r\n5 & = s+t\r\n\\end{align*}\\]\r\nFor \\(z\\):\r\n\\[\\begin{align*} 8-3t & = 8-2s\\\\\r\n0 & = 3t-2s\r\n\\end{align*}\\]\r\nTo solve for \\(s\\) and \\(t\\), we can add the \\(x\\) and \\(y\\) equations:\r\n\\[\\begin{align*} 4+5 & = 2s-t+s+t\\\\\r\n9 & = 3s\\\\\r\n\\frac{9}{3} & = s\\\\\r\n3 & = s\r\n\\end{align*}\\]\r\nWe can substitute \\(s=3\\) into another equation to find \\(t\\). Let's start with the equation for \\(x\\):\r\n\\[\\begin{align*} 4 & = 2s-t\\\\\r\n4 & = 2(3)-t\\\\\r\n4 & = 6-t\\\\\r\nt & = 2\r\n\\end{align*}\\]\r\nThe solution \\(s=3\\) and \\(t=2\\) may not necessary satisfy all of the equations, so you must check all of them. For the \\(y\\) equation:\r\n\\[\\begin{align*} \\textrm{RHS} & = s+t\\\\\r\n& = 3+2\\\\\r\n& = 5\\\\\r\n& = \\textrm{LHS}\r\n\\end{align*}\\]\r\nAnd finally, the \\(z\\) equation:\r\n\\[\\begin{align*} \\textrm{RHS} & = 3t-2s\\\\\r\n& = 3(2)-2(3)\\\\\r\n& = 0\\\\\r\n& = \\textrm{LHS}\r\n\\end{align*}\\]\r\n\\(s=3\\) and \\(t=2\\) satisfy all three equations, so it is the solution. To find the point of intersection, we substitute the \\(s\\) value into the equations for \\(L_{2}\\) and the \\(t\\) value into the equations for \\(L_{1}\\).\r\nFor \\(L_{1}\\):\r\n\\[\\begin{align*} x & = 5+t\\\\\r\n& = 5+2\\\\\r\n& = 7\r\n\\end{align*}\\] \\[\\begin{align*} y & = 1-t\\\\\r\n& = 1-2\\\\\r\n& = -1\r\n\\end{align*}\\] \\[\\begin{align*} z & = 8-3t\\\\\r\n& = 8-3(2)\\\\\r\n& = 2\r\n\\end{align*}\\]\r\nFor \\(L_{2}\\):\r\n\\[\\begin{align*} x & = 1+2s\\\\\r\n& = 1+2(3)\\\\\r\n& = 7\r\n\\end{align*}\\] \\[\\begin{align*} y & = -4+s\\\\\r\n& = -4+3\\\\\r\n& = -1\r\n\\end{align*}\\] \\[\\begin{align*} z & = 8-2s\\\\\r\n& = 8-2(3)\\\\\r\n& = 2\r\n\\end{align*}\\]\r\nYou can see that these give the same coordinates: \\((7,-1,2)\\). Here, it doesn't matter which set of equations you substitute the values into, as it is the point of intersection that both lines share.\r\n\r\n\r\n\r\nExercise \u2013 determining the point of intersection between two lines\r\nFind the point of intersection of the lines \\(x=2t+2\\), \\(y=2t+3\\), \\(z=t+3\\) and \\(\\dfrac{x+2}{2}=\\dfrac{y-2}{-1}=\\dfrac{z}{2}\\).\r\n\r\n \r\n\r\n\r\n\r\n","n":0.04},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":274,"$":[{"v":"Equation of a line in 3D","n":0.408},{"v":"\r\nUniquely defining a line for a vector in three-dimensional space is useful in a range of scenarios. In mechanical and civil engineering, they are needed to create models and analyse alignment. In computer graphics, they can be used to define paths for animations. In physics, they represent trajectories of particles and forces.\r\n\r\nYou should already know that a line in the \\(x\\)\u2014\\(y\\) plane can be represented in the form \\(y=mx+c\\) where \\(m\\) is the gradient (or slope) of the line and \\(c\\) is the \\(y\\)-intercept. But how do we uniquely define a line in three dimensions (3D), where we need a third coordinate?\r\n\r\nWe can define lines in 3D in three different ways:\r\n\r\n\tthe vector equation\r\n\tthe parametric equation\r\n\tthe symmetric equation.\r\n\r\n\r\nVector equation of a line in 3D\r\nThe vector equation of a line is expressed using a position vector and a direction vector.\r\nConsider the following figure.\r\n\r\nHere:\r\n\r\n\tTwo vectors \\(\\vec{r_{1}}\\) and \\(\\vec{r_{2}}\\) point from the origin \\(O\\) to the points \\(R_{1}\\) and \\(R_{2}\\), respectively.\r\n\tThe line is drawn such that it intersects points \\(R_{1}\\) and \\(R_{2}\\).\r\n\t\\(\\vec{V}\\) is a vector along the line. It is the direction vector.\r\n\t\\(P_{0}\\) is a point on the line, with \\(\\vec{P_{0}}\\) being its position vector\u2014that is, \\(\\vec{P_{0}}\\) tells us where point \\(P_{0}\\) is relative to the origin \\(O\\).\r\n\r\nThe equation of the line, written in vector form, is:\r\n\\[\\vec{r}=\\vec{P_{0}}+t\\vec{V}\\]\r\nwhere \\(t\\) is a constant, or \\(t\\in\\mathbb{R}\\).\r\nWhat this means is that to get to any point \\(\\vec{r}\\) along the line, we need to first get to the line by travelling along \\(\\vec{P_{0}}\\). Then, we travel a distance \\(t\\) along the line in the direction of \\(\\vec{V}\\). \r\n\r\n\tTo get from \\(O\\) to \\(R_{1}\\), we travel along \\(\\vec{P_{0}}\\), then we travel a distance along \\(\\vec{V}\\) given by \\(\\vec{r_{1}}=\\vec{P_{0}}+t\\vec{V}\\) where \\(t\\) is a positive constant.\r\n\tTo get from \\(O\\) to \\(R_{2}\\), we travel along \\(\\vec{P_{0}}\\), then we travel a distance along \\(\\vec{V}\\) given by \\(\\vec{r_{2}}=\\vec{P_{0}}+t\\vec{V}\\) where \\(t\\) is a negative constant.\r\n\r\nThe positive or negative \\(t\\) indicates the direction we are travelling in, relative to \\(\\vec{V}\\).\r\n\r\nParametric equation of a line in 3D\r\nLines can broken down into their components and expressed using parametric equations. These equations indicate the coordinates of any point on the line as functions of the parameter \\(t\\), hence, \"parametric\".\r\nSince there are three axes in 3D, we define three separate parametric equations.\r\nLet's look at the following figure.\r\n\r\nSuppose we wish to find the line that goes through the point \\(P_{0}(x_{0},y_{0},z_{0})\\) in the direction of the vector \\(\\vec{V}=a\\hat{i}+b\\hat{j}+c\\hat{k}\\). A general point \\(P(x,y,z)\\) lies on the line.\r\nWe can define vector \\(\\overrightarrow{P_{0}P}\\) using:\r\n\\[\\overrightarrow{P_{0}P} = (x-x_{0})\\hat{i}+(y-y_{0})\\hat{j}+(z-z_{0})\\hat{k}\\]\r\nThis vector will be in the same direction as \\(\\vec{V}\\) but will likely have a different magnitude. In other words, \\(\\overrightarrow{P_{0}P}\\) will be a multiple of the direction vector \\(\\vec{V}\\). So \\(\\overrightarrow{P_{0}P}=t\\vec{V}\\) where \\(t\\in\\mathbb{R}\\).\r\nWe can substitute \\(\\vec{V}\\) into this equation to get:\r\n\\[\\begin{align*} \\overrightarrow{P_{0}P} & = t\\vec{V}\\\\\r\n& = t\\left(a\\hat{i}+b\\vec{j}+c\\vec{k}\\right)\\\\\r\n& = at\\hat{i}+bt\\hat{j}+ct\\hat{k}\r\n\\end{align*}\\] We can equate the components of this equation with the earlier one to get three parametric equations:\r\n\\[\\begin{align*} \\left(x-x_{0}\\right) & = at\\\\\r\nx & = x_{0}+at\r\n\\end{align*}\\] \\[\\begin{align*} \\left(y-y_{0}\\right) & = bt\\\\\r\ny & = y_{0}+bt\r\n\\end{align*}\\] \\[\\begin{align*} \\left(z-z_{0}\\right) & = ct\\\\\r\nz & = z_{0}+ct\r\n\\end{align*}\\]\r\nThese are parametric equations of the line that contains the point \\((x_{0},y_{0},z_{0})\\) and has the direction vector \\(\\vec{V}=a\\hat{i}+b\\hat{j}+c\\hat{k}\\). The variable \\(t\\in\\mathbb{R}\\) is called a parameter.\r\n\r\nSymmetric equation of a line in 3D\r\nThe symmetric equation eliminates the parameter \\(t\\) by solving the parametric equations for \\(t\\). This helps identifying points on a line without expressing it in terms of a parameter.\r\nWe start by rearranging the parametric equations so that \\(t\\) is the subject. \r\n\\[\\begin{align*} t = \\frac{x-x_{0}}{a}\\quad t=\\frac{y-y_{0}}{b}\\quad t=\\frac{z-z_{0}}{c} \\end{align*}\\]\r\nYou might like to review rearranging (or transposing) algebraic formulas.\r\nSince they are all equal, we can say that:\r\n\r\n\\[\\frac{x-x_{0}}{a} = \\frac{y-y_{0}}{b} = \\frac{z-z_{0}}{c}\\]\r\nThe equation is symmetric in the sense that each coordinate is treated in a similar manner, highlighting the proportional relationships along the line.\r\n\r\nExample 1 \u2013 finding equations of a line in 3D\r\nFind the parametric and symmetric equations of the line that goes through the point \\((3,2,3)\\) and is in the direction of the vector \\(2\\hat{i}+\\hat{j}-5\\hat{k}\\).\r\nFirst, we can define the variables. The point is written in the form \\((x_{0},y_{0},z_{0})\\), so \\(x=3\\), \\(y=2\\) and \\(z=3\\). The vector is written in the form \\(a\\hat{i}+b\\hat{j}+c\\hat{k}\\), so \\(a=2\\), \\(b=1\\) and \\(c=-5\\).\r\nWe can now substitute them into the equations to form the parametric equations.\r\n\\[\\begin{align*} x & = x_{0}+at\\\\\r\n& = 3+2t\r\n\\end{align*}\\] \\[\\begin{align*} y & = y_{0}+bt\\\\\r\n& = 2+t\r\n\\end{align*}\\] \\[\\begin{align*} z & = z_{0}+ct\\\\\r\n& = 3-5t\r\n\\end{align*}\\]\r\nTo get the symmetric equation, we substitute them into the relevant equation.\r\n\\[\\begin{align*} & \\frac{x-x_{0}}{a} = \\frac{y-y_{0}}{b} = \\frac{z-z_{0}}{c}\\\\\r\n\\Rightarrow & \\quad \\frac{x-3}{2} = \\frac{y-2}{1} = \\frac{z-3}{-5}\\\\\r\n\\Rightarrow & \\quad \\frac{x-3}{2} = y-2 = \\frac{z-3}{-5}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding equations of a line in 3D\r\nFind the symmetric and parametric equations of the lines that satisfy the given conditions.\r\n\r\n\tcontains the point \\((1,-1,2)\\) with the direction vector \\(2\\hat{i}-2\\hat{j}+3\\hat{k}\\)\r\n\tcontains the point \\((3,4,-1)\\) with the direction vector \\(\\hat{i}+\\hat{j}+5\\hat{k}\\)\r\n\tcontains the point \\((2,3,-1)\\) parallel to \\(\\dfrac{x-1}{2}=\\dfrac{y-3}{-1}=\\dfrac{x+1}{3}\\)\r\n\tcontains the points \\((2,2,1)\\) and \\((1,1,3)\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.036},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":275,"$":[{"v":"Projection of vectors","n":0.577},{"v":"\r\nYou will have learned that vectors can be resolved in two dimensions along the horizontal and vertical axes. It is also possible to resolve one vector along the line of another vector, instead of along the \\(x\\)- and \\(y\\)-axes.\r\n\r\nOften, in physics, engineering and mathematics courses, you are asked to resolve a vector into two component vectors that are perpendicular to one another. This typically involves projecting the vector in the horizontal and vertical directions.\r\nBut, you can project a vector in any direction, not only horizontally and vertically. Here, we will look at resolving a vector in the direction of a second vector.\r\nThis is called vector projection. There are two types of vector projection:\r\n\r\n \tscalar projection, which tells you the length of the vector pointing in a certain direction\r\n \tvector projection, which gives you a vector that represents how much of the first vector lies in the direction of the second vector.\r\n\r\nThis will make more sense when we look at some examples.\r\n\r\nScalar projection\r\nConsider the following diagram where \\(\\overrightarrow{PQ}=\\vec{a}\\) and \\(\\overrightarrow{PS}=\\vec{b}\\). \r\n\r\nThe scalar projection of the vector \\(\\vec{a}\\) in the direction of vector \\(\\vec{b}\\) is the length of the straight line \\(PR\\) or \\(\\left|\\overrightarrow{PR}\\right|\\).\r\nThis forms a right-angled triangle, where:\r\n\r\n\t\\(\\vec{a}\\) is the hypotenuse\r\n\t\\(\\theta\\) is the angle between \\(\\vec{a}\\) and the horizontal; and\r\n\tthe scalar projection of \\(\\vec{a}\\) in the direction of \\(\\vec{b}\\) is the adjacent side.\r\n\r\nThis means that the scalar projection can be calculated using:\r\n\\[\\begin{align*} \\cos\\left(\\theta\\right) & = \\frac{\\text{adjacent}}{\\text{hypotenuse}}\\\\\r\n& = \\frac{\\left|\\overrightarrow{PR}\\right|}{\\left|\\vec{a}\\right|}\r\n\\end{align*}\\]\r\nWe can rearrange this to find the scalar projection:\r\n\\[\\left|\\overrightarrow{PR}\\right| = \\left|\\vec{a}\\right|\\cos\\left(\\theta\\right)\\]\r\nPreviously, you learned how to find the angle between two vectors using a rearranged form of the scalar product equation. We can actually write the scalar projection equation in terms of this.\r\nRemember:\r\n\\[\\cos\\theta=\\frac{\\vec{a}\\cdot\\vec{b}}{\\left|\\vec{a}\\right|\\left|\\vec{b}\\right|}\\]\r\nSubstituting this into the scalar projection equation gives us:  Therefore, the scalar projection of a vector \\(\\vec{a}\\) in the direction of a second vector \\(\\vec{b}\\) is given by:\r\n\\[\\left|\\overrightarrow{PR}\\right|=\\vec{a}\\cdot\\hat{b}\\]\r\nCalculating the scalar projection of vectors requires you to be able to calculate the magnitude of a vector and the scalar product. You may like to review these concepts before diving into an example.\r\n\r\nExample \u2013 finding the scalar projection of a vector\r\nFind the scalar projection of the vector \\(\\vec{a}=(2,3,1)\\) in the direction of vector \\(\\vec{b}=(5,-2,2)\\).\r\nWe need to find \\(\\hat{b}\\) which is given by \\(\\dfrac{\\vec{b}}{\\left|\\vec{b}\\right|}\\).\r\n\\[\\begin{align*} \\left|\\vec{b}\\right| & =\\sqrt{5^{2}+(-2)^{2}+2^{2}}\\\\\r\n& = \\sqrt{25+4+4}\\\\\r\n& = \\sqrt{33} \r\n\\end{align*}\\] Therefore: \\[\\begin{align*} \\hat{b} & = \\frac{\\vec{b}}{\\left|\\vec{b}\\right|}\\\\\r\n& = \\frac{(5,-2,2)}{\\sqrt{33}}\\\\\r\n& = \\frac{1}{\\sqrt{33}}(5,-2,2) \r\n\\end{align*}\\] The scalar projection is calculated using \\(\\vec{a}\\cdot\\hat{b}\\): \\[\\begin{align*} \\vec{a}\\cdot\\hat{b} & = (2,3,1)\\cdot\\frac{(5,-2,2)}{\\sqrt{33}}\\\\ \r\n& = \\frac{\\left(2\\times5\\right)+\\left(3\\times\\left(-2\\right)\\right)+\\left(1\\times2\\right)}{\\sqrt{33}}\\\\\r\n& = \\frac{10-6+2}{\\sqrt{33}}\\\\ \r\n& = \\frac{6}{\\sqrt{33}}\r\n\\end{align*}\\]\r\n\r\nVector projection\r\nThe vector projection of a vector \\(\\vec{a}\\) in the direction of vector \\(\\vec{b}\\) is a vector pointing in the direction of \\(\\vec{b}\\) that has the same magnitude as the straight line \\(PR\\) or \\(\\left|\\overrightarrow{PR}\\right|\\). \r\nYou might notice that we are essentially looking at the same thing as scalar projection, but instead of finding the magnitude of the vector, we are looking at the vector itself. This gives us information about its direction.\r\n\r\n\r\nThe vector projection of \\(\\vec{a}\\) in the direction of \\(\\vec{b}\\) is the scalar projection multiplied by a unit vector in the direction of \\(\\vec{b}\\).\r\n\\[\\left(\\vec{a}\\cdot\\hat{b}\\right)\\hat{b} = \\frac{\\left(\\vec{a}\\cdot\\vec{b}\\right)\\vec{b}}{\\left|\\vec{b}\\right|^{2}} \\]\r\n\r\nExample 1  \u2013 finding the vector projection of a vector\r\nFind the vector projection of vector \\(\\vec{a}=(2,3,1)\\) in the direction of vector \\(\\vec{b}=(5,-2,2)\\).\r\nVectors \\(\\vec{a}\\) and \\(\\vec{b}\\) are the same as Example \u2013 finding the scalar projection of a vector, so this makes our lives easier since we already have \\(\\vec{a}\\cdot\\hat{b}\\) and \\(\\hat{b}\\).\r\n\\[\\begin{align*} \\left(\\vec{a}\\cdot\\hat{b}\\right)\\hat{b} & = \\frac{6}{\\sqrt{33}}\\times\\frac{1}{\\sqrt{33}}\\left(5,-2,2\\right)\\\\\r\n& = \\frac{6\\left(5,-2,2\\right)}{33}\r\n\\end{align*}\\] We can also calculate this using \\(\\left(\\vec{a}\\cdot\\hat{b}\\right)\\hat{b}=\\dfrac{\\left(\\vec{a}\\cdot\\vec{b}\\right)\\vec{b}}{\\left|\\vec{b}\\right|^{2}}\\).\r\n\\[\\begin{align*} \\left(\\vec{a}\\cdot\\hat{b}\\right)\\hat{b} & = \\frac{\\left(\\vec{a}\\cdot\\vec{b}\\right)\\vec{b}}{\\left|\\vec{b}\\right|^{2}}\\\\\r\n& = \\frac{\\left(\\left(2,3,1\\right)\\cdot\\left(5,-2,2\\right)\\right)\\left(5,-2,2\\right)}{\\left(\\sqrt{33}\\right)^{2}}\\\\\r\n& = \\frac{\\left(2\\times5+3\\times\\left(-2\\right)+1\\times2\\right)\\left(5,-2,2\\right)}{33}\\\\\r\n& = \\frac{\\left(10-6+2\\right)\\left(5,-2,2\\right)}{33}\\\\\r\n& = \\frac{6\\left(5,-2,2\\right)}{33} \r\n\\end{align*}\\]\r\nWe can further simplify this and write it in vector form:\r\n\\[\\begin{align*} \\frac{6\\left(5,-2,2\\right)}{33} & = \\frac{6}{33}\\left(5,-2,2\\right)\\\\ \r\n& = \\frac{2}{11}\\left(5\\hat{i}-2\\hat{j}+2\\hat{k}\\right)\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExercise \u2013 finding the scalar and vector projections of a vector\r\n\r\n\tFor \\(\\vec{a}=\\left(2,3,1\\right)\\), \\(\\vec{b}=\\left(5,0,3\\right)\\), \\(\\vec{c}=\\left(0,0,3\\right)\\) and \\(\\vec{d}=\\left(-2,2,-1\\right)\\), find:\r\n\t\r\n\t\tthe scalar projection of \\(\\vec{a}\\) in the direction of \\(\\vec{b}\\)\r\n\t\tthe vector projection of \\(\\vec{a}\\) in the direction of \\(\\vec{b}\\)\r\n\t\tthe scalar projection of \\(\\vec{c}\\) in the direction of \\(\\vec{b}\\)\r\n\t\tthe vector projection of \\(\\vec{c}\\) in the direction of \\(\\vec{a}\\)\r\n\t\tthe vector projection of \\(\\vec{d}\\) in the direction of \\(\\vec{a}\\)\r\n\t\tthe vector projection of \\(\\vec{b}\\) in the direction of \\(\\vec{d}\\).\r\n\t\r\n\tConsider \\(\\vec{a}=(2,0,-1)\\) and \\(\\vec{b}=(3,5,6)\\).\r\n\t\r\n\t\tFind the scalar projection of \\(\\vec{a}\\) in the direction of \\(\\vec{b}\\).\r\n\t\tDescribe the relationship between \\(\\vec{a}\\) and \\(\\vec{b}\\).\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.038},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":276,"$":[{"v":"Resolution of vectors","n":0.577},{"v":"\r\n\r\nBreaking vectors down into their components\u2014or resolving them\u2014makes it easier to add or subtract them, especially when dealing with vectors that don't act along the same line. You will encounter this in many areas of STEM, like when analysing forces involved in robotics, studying the projectile motion of objects launched into the air, and in fluid mechanics. Use this resource to learn how to find the resolution of vectors.\r\n\r\nTo help simplify calculations, vectors can be thought of as the sum of several other vectors. For example, imagine that a single force is made up of several forces. If we break the force into imaginary vertical and horizontal components that are at right angles to each other, it is much less scary to work with.\r\nOnce we break the forces down into their components, we can just group the components that are parallel to each other and complete the required operations. Let's see how this works with some examples.\r\n\r\nComponents of a vector\r\nIf you pull on an object with a force \\(\\vec{F}\\), the effect is the same as when two people pull with forces \\(\\vec{a}\\) and \\(\\vec{b}\\) at right angles to each other \u2013 one horizontal and the other vertical.\r\n\\(\\vec{a}\\) and \\(\\vec{b}\\) are the components of the force vector. The horizontal component is \\(\\vec{a}\\) and the vertical component is \\(\\vec{b}\\). They are often called rectangular components because they are at right angles to each other.\r\n\r\n\r\n\r\nFrom \\(\\vec{a}\\) and \\(\\vec{b}\\), we get a vector triangle, where \\(\\theta\\) is the angle, \\(\\vec{F}\\) is the hypotenuse, \\(\\vec{a}\\) is the adjacent and \\(\\vec{b}\\) is the opposite.\r\n\r\n\r\n\r\nFrom this, we can find the magnitude of the component vectors using trigonometry. You might like to go back and review right-angled triangle trigonometry first.\r\nWe can find the length of the horizontal component using:\r\n\\[\\begin{align*} \\cos\\left(\\theta\\right) & = \\frac{\\textrm{Adjacent}}{\\textrm{Hypotenuse}}\\\\\r\n& = \\frac{\\left|\\vec{a}\\right|}{\\left|\\vec{F}\\right|}\r\n\\end{align*}\\]\r\nRearranging this in terms of \\(\\left|\\vec{a}\\right|\\) gives:\r\n\\[\\vec{\\left|a\\right|} = \\vec{\\left|F\\right|}\\cos\\theta\\]\r\nWe can also find the length of the vertical component using:\r\n\\[\\begin{align*} \\sin\\left(\\theta\\right) & = \\frac{\\textrm{Opposite}}{\\textrm{Hypotenuse}}\\\\\r\n& = \\frac{\\left|\\vec{b}\\right|}{\\left|\\vec{F}\\right|}\r\n\\end{align*}\\]\r\nRearranging this in terms of \\(\\left|\\vec{b}\\right|\\) gives:\r\n\\[\\vec{\\left|b\\right|} = \\vec{\\left|F\\right|}\\sin\\theta\\]\r\n\r\nExample 1 \u2013 resolving components of force vectors\r\nCalculate the rectangular components of a force of \\(10\\textrm{ N}\\) that acts in a direction of North \\(30^{\\circ}\\) East.\r\nFirst, we should always draw the vector triangle. North \\(30^{\\circ}\\) East is the same as East \\(60^{\\circ}\\) North, or \\(60^{\\circ}\\) from the horizontal.\r\nWe can label the horizontal component of the vector triangle as \\(\\vec{a}\\) and the vertical component as \\(\\vec{b}\\). \\(\\vec{a}\\) is in the East direction and \\(\\vec{b}\\) is in the North direction.\r\n\r\n\r\n\r\nThe magnitudes of these forces can be calculated using trigonometry.\r\n\\[\\begin{align*} \\vec{\\left|a\\right|} & = \\vec{\\left|F\\right|}\\cos\\left(\\theta\\right)\\\\ \r\n& = 10\\cos\\left(60^{\\circ}\\right)\\\\ \r\n& = 5\\textrm{ N}\r\n\\end{align*}\\] \\[\\begin{align*} \\vec{\\left|b\\right|} & = \\vec{\\left|F\\right|}\\sin\\left(\\theta\\right)\\\\ \r\n& = 10\\sin\\left(60^{\\circ}\\right)\\\\\r\n& = 8.7\\textrm{ N} \r\n\\end{align*}\\]\r\n\r\nThus, the \\(10\\) newtons force can be resolved into two rectangular components: \\(5\\textrm{ N}\\) to the East and \\(8.7\\textrm{ N}\\) to the North.\r\n\r\n\r\n\r\nExercise \u2013 resolving components of force vectors\r\n\r\n \tDetermine the horizontal component of a force of \\(11\\textrm{ N}\\) acting at \\(\\theta=60^{\\circ}\\) to the horizontal.\r\n \tDetermine the northerly component of wind blowing at \\(15\\textrm{ m s}^{-1}\\) from the South East, to the nearest whole number.\r\n \tDefine the vector whose components are \\(3\\textrm{ N}\\) horizontally and \\(4\\textrm{ N}\\) vertically.\r\n \tDetermine the vertical and horizontal components of a force of \\(24\\textrm{ N}\\) acting at \\(\\theta=30^{\\circ}\\) to the horizontal, to the nearest whole number.\r\n \tA plane taking off leaves the runway at \\(\\theta=32^{\\circ}\\) to the horizontal, traveling at \\(180\\textrm{ km h}^{-1}\\). Determine how long it will take for the plane to reach an altitude of \\(1000\\textrm{ m}\\).\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\n","n":0.042},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":277,"$":[{"v":"Introduction to vectors","n":0.577},{"v":"Vectors are used to represent forces in physics, handle 2D and 3D manipulation with computer graphics, and to calculate the forces acting on materials in textiles. You also deal with vector quantities in your everyday life, from driving your car down the road to planning the shortest route to get to uni. Use this resource to gain a basic understanding of vectors.\r\n\r\nVectors and scalars\r\nA vector is a measurable quantity with direction. Examples of vectors are:\r\n\r\n\tvelocity \u2013 the speed at which an object is travelling in a given direction\r\n\tforce \u2013 an action that causes an object to change its velocity\r\n\tdisplacement \u2013 the distance of an object from its original position\r\n\tacceleration \u2013 the change in speed over time.\r\n\r\nOn the other hand, a scalar quantity has magnitude only, i.e. no direction. Examples include:\r\n\r\n\theat \u2013 the amount of thermal energy transferred or transformed within a system or between systems\r\n\tmass \u2013 the amount of matter in an object\r\n\ttime \u2013 the amount of time that has passed\r\n\tvolume \u2013 the amount of space occupied by a substance.\r\n\r\n\r\nRepresenting vectors\r\nVectors are shown using an arrow that points in specific direction. The length of the line indicates the magnitude of the vector\u2013in other words, the longer the line, the greater the magnitude. This means that a force vector of \\(10\\textrm{ N}\\) will be \\(5\\) times longer than a force vector of \\(2\\textrm{ N}\\).\r\nVectors in three-dimensional space are defined by three directions that are perpendicular to each other. They can be denoted as bold letters or by using a vector arrow: \\(\\vec{a}\\) or \\(\\vec{b}\\) or \\(\\vec{c}\\).\r\nIn the following figure, point P is the tail of the vector and point Q is the head of the vector. The vector is referred to as \\(\\overrightarrow{PQ}\\), as the arrow points from point P to point Q.\r\n\r\nVectors can also be defined along horizontal or vertical axes \u2014 that is, in two directions. A vector in the opposite direction from \\(\\vec{a}\\) is denoted by \\(-\\vec{a}\\).\r\n\r\n\r\nComponents of a vector\r\nIn the diagram, the vector \\(\\vec{r}\\) is represented by \\(\\overrightarrow{OP}\\) where \\(P\\) is the point \\((x,y,z)\\). This is called a position vector, which indicates where the vector is relative to the origin \\(O(0,0,0)\\).\r\n\r\nIf \\(\\vec{i}\\), \\(\\vec{j}\\), and \\(\\vec{k}\\) are vectors parallel to the positive directions of the \\(x\\)-, \\(y\\)- and \\(z\\)-axis, respectively, then:\r\n\r\n\t\\(x\\vec{i}\\) is a vector of length \\(x\\) in the direction of the \\(x\\)-axis\r\n\t\\(y\\vec{i}\\) is a vector of length \\(y\\) in the direction of the \\(y\\)-axis\r\n\t\\(z\\vec{k}\\) is a vector of length \\(z\\) in the direction of the \\(z\\)-axis.\r\n\r\n\\(\\overrightarrow{OP}\\) is then the position vector \\(x\\vec{i}+y\\vec{j}+z\\vec{k}\\), and \\(x\\), \\(y\\) and \\(z\\) are called the components of the vector.\r\nThe notation \\((x,y,z)\\) is used to denote the vector \\(\\vec{r}=(x\\vec{i}+y\\vec{j}+z\\vec{k})\\), as well as the coordinates of a point \\(P(x,y,z)\\).\r\n\r\nDirected line segment\r\nIn the previous diagram, we defined vector \\(P\\) which extended from the origin O to point \\(P(x,y,z)\\). We can also define vectors extending from two points in space, such as from point \\(P(x_{1},y_{1},z_{1})\\) to point \\(Q(x_{2},y_{2},z_{2})\\). \r\n\\(\\overrightarrow{PQ}\\) is called a directed line segment or geometric vector. This vector is found by subtracting the coordinates of \\(P\\) (the initial point) from the coordinates of \\(Q\\) (the final point).\r\n\\[\\overrightarrow{PQ}=(x_{2}-x_{1})\\vec{i}+(y_{2}-y_{1})\\vec{j}+(z_{2}-z_{1})\\vec{k}\\]\r\nLet's consider the points \\(P(3,4,1)\\) and \\(Q(5,6,-1)\\).\r\n\r\nWhen we apply the general formula:\r\n\\[\\begin{align*} PQ & = (5-3)\\vec{i}+(6-4)\\vec{j}+(-1-1)\\vec{k}\\\\\r\n& = 2\\vec{i}+2\\vec{j}-2\\vec{k}\r\n\\end{align*}\\]\r\nTherefore, the directed line segment \\(\\overrightarrow{PQ}\\) is represented by the vector \\(2\\vec{i}+2\\vec{j}-2\\vec{k}\\) or \\((2,2,-2)\\). This is also used to represent ANY directed line segment with the same length and direction as \\(\\overrightarrow{PQ}\\).\r\nThe directed line segment \\(\\overrightarrow{QP}\\) has the same length as \\(\\overrightarrow{PQ}\\) but is in the opposite direction.\r\n\\[\\begin{align*} \\overrightarrow{QP} & = -\\overrightarrow{PQ}\\\\\r\n& = -(2\\vec{i}+2\\vec{j}-2\\vec{k})\\\\ \r\n& = -2\\vec{i}-2\\vec{j}+2\\vec{k}\\quad\\textrm{or}\\quad(-2,-2,2) \r\n\\end{align*}\\]\r\n\r\nExercise \u2013 defining directed line segments and position vectors\r\n\r\n \tGiven the points \\(A(3,0,4)\\), \\(B(-2,4,3)\\), and \\(C(1,-5,0)\\), find:\r\n\t\r\n\t\t\\(\\overrightarrow{AB}\\)\r\n\t\t\\(\\overrightarrow{AC}\\)\r\n\t\t\\(\\overrightarrow{CB}\\)\r\n\t\t\\(\\overrightarrow{BC}\\)\r\n\t\t\\(\\overrightarrow{CA}\\)\r\n\t\r\n\tLook at your answers for questions b and e, and c and d. Describe what you notice.\r\n\tDefine the position vectors of the points \\(A\\), \\(B\\) and \\(C\\).\r\n\r\n\r\n\r\nAdding and subtracting vectors\r\nVectors may be added or subtracted by adding or subtracting their corresponding components. For vectors pointing in the same direction, we simply place them head to tail and add them algebraically. Here, \\(\\overrightarrow{AB}+\\overrightarrow{CD}=\\overrightarrow{AD}\\).\r\n\r\nWe can apply the same method to vectors pointing in opposite directions. \r\n\r\nFor vectors pointing in different directions, this is done graphically using the triangle law.\r\nLet's consider two vectors: \\(\\vec{a}\\) extends from point \\(P\\) to point \\(Q\\), and \\(\\vec{b}\\) extends from point \\(Q\\) to point \\(R\\).\r\nTo add vectors \\(\\vec{a}\\) and \\(\\vec{b}\\), we place the tail of vector \\(\\vec{b}\\) at the head of vector \\(\\vec{a}\\) (point Q). The vector sum, \\(\\vec{a}+\\vec{b}\\), is the vector \\(\\overrightarrow{PR}\\), from the tail of vector \\(\\vec{a}\\) to the head of \\(\\vec{b}\\).\r\n\r\n To subtract \\(\\vec{b}\\) from \\(\\vec{a}\\), we reverse the direction of \\(\\vec{b}\\) to give \\(-\\vec{b}\\), then place them head to tail like we did with addition. \\(\\vec{b}\\) is now extending from point \\(R\\) to point \\(Q\\). \r\n\\(\\vec{a}\\) and \\(-\\vec{b}\\) are added together. This means that \\(\\overrightarrow{PR}\\) is the same as \\(\\vec{a}-\\vec{b}\\).\r\n\r\n\r\nExample \u2013 adding and subtracting vectors\r\nIf \\(\\vec{a}=(-3,4,2)\\) and \\(\\vec{b}=(-1,-2,3)\\), find \\(\\vec{a}+\\vec{b}\\) and \\(\\vec{a}-\\vec{b}\\).\r\n\\[\\begin{align*} \\vec{a}+\\vec{b} & = (-3,4,2)+(-1,-2,3)\\\\ \r\n& = (-3+(-1),4+(-2),2+3)\\\\\r\n& = (-4,2,5) \r\n\\end{align*}\\] \\[\\begin{align*} \\vec{a}-\\vec{b} & = (-3,4,2)-(-1,-2,3)\\\\\r\n& = (-3-(-1),4-(-2),2-3)\\\\ \r\n& = (-2,6,-1)\r\n\\end{align*}\\]\r\n\r\nExercise \u2013 adding and subtracting vectors\r\n\r\n\tGiven \\(\\vec{a}=(2,1,1)\\), \\(\\vec{b}=(1,3,-3)\\) and \\(\\vec{c}=(0,3,-2)\\), find:\r\n\t\r\n\t\t\\(\\vec{a}+\\vec{b}\\)\r\n\t\t\\(\\vec{a}+\\vec{c}\\)\r\n\t\t\\(\\vec{c}-\\vec{b}\\)\r\n\t\t\\(\\vec{a}-\\vec{b}\\)\r\n\t\r\n\tAdd \\(\\vec{60}\\) kilometres East to \\(\\vec{25}\\) kilometres East.\r\n\tAdd \\(\\vec{60}\\) kilometres East to \\(\\vec{25}\\) kilometres West.\r\n\tAdd \\(\\vec{30}\\) kilometres West to \\(\\vec{40}\\) kilometres North.\r\n\tAdd \\(\\vec{30}\\) kilometres West to \\(\\vec{30}\\) kilometres South.\r\n\tA ship sails \\(\\vec{40}\\) kilometres North East, and then changes direction and sails \\(\\vec{40}\\) kilometres South East. Find the displacement of the ship.\r\n\tTwo forces \\(100\\textrm{ N}\\) North and \\(100\\textrm{ N}\\) East are acting away from the same point. Find the sum of the vectors, or the resultant force.\r\n\tA plane is flying due North at a speed of \\(300\\textrm{ km h}^{-1}\\) and a Westerly wind of \\(25\\textrm{ km h}^{-1}\\) is blowing it off course. Calculate the true speed and direction of the plane.\r\n\r\n\r\n\r\n\r\n\r\nMagnitude of a vector\r\nOnce again, let's consider vector \\(P(x,y,z)\\) where \\(\\vec{P}=x\\vec{i}+y\\vec{j}+z\\vec{k}\\). The length or magnitude of \\(\\vec{P}\\) is denoted by \\(\\left|\\vec{P}\\right|\\) or \\('\\vec{P}'\\), where:\r\n\\[\\left|\\vec{P}\\right|=\\sqrt{x^{2}+y^{2}+z^{2}}\\]\r\nTake, for example, the vector \\(2\\vec{i}+3\\vec{j}-5\\vec{k}\\). Its length is:\r\n\\[\\sqrt{2^{2}+3^{2}+(-5)^{2}}=\\sqrt{38}\\]\r\n\r\nUnit vector\r\nAny vector with a magnitude of \\(1\\) is called a unit vector. If \\(\\vec{P}\\) is any vector, then a unit vector parallel to \\(\\vec{P}\\) is written \\(\\hat{P}\\) (read as P \u201chat\u201d). The \u201chat\u201d symbolises a unit vector.\r\nThe unit vector \\(\\hat{P}\\) equals vector \\(\\vec{P}\\) divided by its magnitude \\(\\left|\\vec{P}\\right|\\).\r\n\\[\\hat{P} = \\frac{\\vec{P}}{\\left|\\vec{P}\\right|}\\]\r\nThis can be rearranged to give:\r\n\\[\\vec{P} = \\left|\\vec{P}\\right|\\hat{P}\\]\r\nThe rearranged equation shows that vector \\(P\\) is the magnitude of the vector multiplied by its unit vector.\r\nYou should remember unit vectors \\(\\vec{i}\\), \\(\\vec{j}\\) and \\(\\vec{k}\\), which are parallel to the \\(x\\)-, \\(y\\)- and \\(z\\)-axes, respectively. Sometimes, they are written as \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\), especially when emphasising that the vectors have a magnitude of \\(1\\) and point in the positive directions.\r\n\r\nExample 1 \u2013 finding unit vectors\r\nIf \\(\\overrightarrow{PQ}\\) is the line \\(2\\vec{i}-5\\vec{j}+\\vec{k}\\), find the unit vector parallel to \\(\\overrightarrow{PQ}\\).\r\nTo find the unit vector, divide the vector by its magnitude. Let's calculate the magnitude first.\r\n\\[\\begin{align*} \\overrightarrow{PQ} & = \\sqrt{2^{2}+\\left(-5\\right)^{2}+1^{2}}\\\\\r\n& = \\sqrt{30}\r\n\\end{align*}\\] The unit vector parallel to \\(\\overrightarrow{PQ}\\) is therefore:\r\n\\[\\begin{align*} \\overrightarrow{PQ} & = \\frac{\\overrightarrow{PQ}}{\\left|\\overrightarrow{PQ}\\right|}\\\\\r\n& = \\frac{(2\\vec{i}-5\\vec{j}+\\vec{k})}{\\sqrt{30}}\\\\\r\n& = \\frac{1}{\\sqrt{30}}(2\\vec{i}-5\\vec{j}+\\vec{k}) \r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExercise \u2013 finding unit vectors\r\n\r\n \tFind the length of the vectors.\r\n\t\r\n\t\t\\((3,-1,-1)\\)\r\n\t\t\\((0,2,4)\\)\r\n\t\t\\((0,-2,0)\\)\r\n\t\r\n\tGiven the points \\(A(3,0,4)\\), \\(B(0,4,3)\\) and \\(C(1,-5,0)\\), find the unit vectors parallel to:\r\n\t\r\n\t\t\\(\\overrightarrow{BA}\\)\r\n\t\t\\(\\overrightarrow{CB}\\)\r\n\t\t\\(\\overrightarrow{AC}\\)\r\n\t\r\n\r\n\r\n\r\nMultiplying vectors by a scalar\r\nA scalar is a number that you multiply by to change the magnitude of a value. To multiply a vector by a scalar, \\(m\\), you just multiply each component by \\(m\\).\r\n\\[m\\vec{P}=mx\\vec{i}+my\\vec{j}+mz\\vec{k}\\]\r\nThe result is a vector of length \\(m\\times\\left|\\vec{P}\\right|\\).\r\n\r\n\tIf \\(m>0\\), the resultant vector is in the same direction as \\(\\vec{P}\\).\r\n\tIf \\(m<0\\), the resultant vector is in the opposite direction as \\(\\vec{P}\\).\r\n\r\nTwo vectors, \\(\\vec{a}\\) and \\(\\vec{b}\\), are said to be parallel if and ONLY if \\(\\vec{a}=k\\vec{b}\\) where \\(k\\) is a real constant.\r\n\r\nExample \u2013 multiplying vectors by a scalar\r\nMultiply \\(\\vec{a}=(3\\vec{i}+\\vec{j}-2\\vec{k})\\) by \\(7\\) and show that \\(\\left|7\\vec{a}\\right|=7\\left|\\vec{a}\\right|\\).\r\nTo multiply by a scalar, multiply each component of the vector by the scalar.\r\n\\[\\begin{align*} 7\\vec{a} & = 7(3\\vec{i}+\\vec{j}-2\\vec{k})\\\\\r\n& = 21\\vec{i}+7\\vec{j}-14\\vec{k}\r\n\\end{align*}\\]\r\nNow, let's find \\(\\left|7\\vec{a}\\right|\\) and \\(7\\left|\\vec{a}\\right|\\).\r\n\\[\\begin{align*} \\left|7\\vec{a}\\right| & = \\sqrt{21^{2}+7^{2}+(-14)^{2}}\\\\\r\n& = \\sqrt{686}\\\\\r\n& = \\sqrt{49\\times14}\\\\ \r\n& = 7\\sqrt{14}\r\n\\end{align*}\\] \\[\\begin{align*} \\left|\\vec{a}\\right| & = \\sqrt{3^{2}+1^{2}+(-2)^{2}}\r\n& = \\sqrt{14}\\\\\r\n7\\left|\\vec{a}\\right| & = 7\\sqrt{14} \r\n\\end{align*}\\]\r\nTherefore, \\(\\left|7\\vec{a}\\right| = 7\\left|\\vec{a}\\right| = 7\\sqrt{14}\\).\r\n\r\nExercise \u2013 multiplying vectors by a scalar.\r\n\r\n\tExpand the following.\r\n\t\r\n\t\t\\(3(\\vec{i}+3\\vec{j}-5\\vec{k})\\)\r\n\t\t\\(-4(\\vec{j}-3\\vec{k})\\)\r\n\t\r\n\tIf \\(\\vec{a}=(2,-2,1)\\), \\(\\vec{b}=(0,1,1)\\) and \\(\\vec{c}=(-1,3,-2)\\), find:\r\n\t\r\n\t\t\\((2\\vec{a}+3\\vec{b})\\)\r\n\t\t\\((3\\vec{a}-2\\vec{b})\\)\r\n\t\t\\((2\\vec{a}-\\vec{b}+2\\vec{c})\\)\r\n\t\ta unit vector parallel to \\(2\\vec{a}-\\vec{b}\\)\r\n\t\r\n \tIdentify a vector three times the length of \\((6\\vec{i}+2\\vec{j}-5\\vec{k})\\) and in the opposite direction.\r\n\r\n\r\n\r\n","n":0.027},[{"v":"Vectors","i":3,"n":1},{"v":"Physics","i":2,"n":1},{"v":"Maths","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":278,"$":[{"v":"Fallacies using feelings instead of facts","n":0.408},{"v":"Arguments are claims that are backed up by evidence to support the claim. The logical fallacies on this page are claims that have feelings to support them instead of evidence.\r\n\r\n\r\n\r\nGenetic\r\nWhen someone judges an idea or argument based on its origin or source rather than its actual merit.\r\nFor example:\r\n\u201cDid you read that in the mainstream media? They cannot be trusted.\u201d\r\nLoaded question\r\nThis is where a question is asked that has a presumption built into it.\r\nFor example:\r\n\u201cAre you na\u00efve enough to believe people on the Internet?\u201d\r\nAppeal to emotion\r\nAn attempt to manipulate someone\u2019s emotions rather than state a fact or make a claim.\r\nFor example:\r\n\"You should donate to this charity. If you don't, so many children will continue to suffer.\"\r\nPersonal incredulity\r\nRefusing to believe something is true because you don\u2019t understand it.\r\nFor example:\r\n\"Evolution is so complicated.There's no way it could be true.\"\r\nTu quoque\r\nThe literal translation of tu quoque is \u201cyou too\u201d. It is the logical equivalent of saying \u201cI know you are, but what am I?\u201d\r\nFor example:\r\n\u201cOh you think my claim is fallacious, well you\u2019re using logical fallacies too.\u201d\r\n\r\nTest your knowledge\r\nThink you can sort the facts from the feelings? Give this quiz a try.\r\n\r\n\r\n\r\n","n":0.073},[{"v":"Reasoning","i":2,"n":1},{"v":"Logical fallacies","i":1,"n":0.707},{"v":"Critical thinking","i":0,"n":0.707}]]},{"i":279,"$":[{"v":"Misused evidence","n":0.707},{"v":"Logical fallacies are mistakes in reasoning. They make flawed claims. The examples below are using flawed evidence or evidence that is not relevant to make their case.\r\n\r\nHere are five examples of fallacies that involve the misuse of evidence.\r\n\r\n\r\n\r\nTexas sharpshooter\r\nWhere someone cherry-picks data or finds patterns in random data to falsely fit their argument. \r\nFor example:\r\n\u201cThree of the five countries with the highest vaccination rates also have the highest rates of brain cancer.\u201d\r\nBurden of proof\r\nYou make a claim without evidence and assert that it is up to other people to disprove it rather than for you to offer proof.\r\nFor example:\r\n\"If you can't prove that ghosts don't exist, they must be real.\"\r\nBegging the question\r\nPresenting an argument where the premise includes the conclusion, meaning the argument just repeats itself without proving anything. It can also be called 'circular reasoning'.\r\nFor example:\r\n\u201cAustralia is the best country to live in because it's better than all the other countries.\u201d\r\nAppeal to nature\r\nClaiming that because something is natural, it is good. And conversely, if something is not found in nature, it is bad.\r\nFor example:\r\n\u201cAntibiotics are man-made, so they are bad for you. I\u2019ll stick to green tea to stay healthy.\u201d\r\nAnecdotal\r\nUsing a personal experience or anecdote as evidence.\r\nFor example:\r\n\u201cI got the flu vaccine last year and it made me sick with the flu for three weeks.\u201d\r\nTake the quiz and test your logic skills\r\nCan you pick the fallacy?\r\n\r\n\r\n\r\n\r\n\r\n","n":0.066},[{"v":"Reasoning","i":2,"n":1},{"v":"Logical fallacies","i":1,"n":0.707},{"v":"Critical thinking","i":0,"n":0.707}]]},{"i":280,"$":[{"v":"History of the periodic table","n":0.447},{"v":"The periodic table contains all the elements that we know exist in the universe. It classifies elements into categories and makes predictions about elements that have not yet been found by humans. Let's take a look at some key milestones that led to it coming to its current form.\r\n\r\nFrom alchemical table to periodic table\r\nLong before the periodic table, the term \"elements\" referred to earth, air, water, and fire. At the time of ancient Greek philosophy (5th century BCE), the understanding of the chemical elements was represented by the alchemical table. This also ordered elements and substances and their corresponding symbols.\r\nAs more elements were discovered and a modern understanding of chemistry emerged, some components were removed from the table and others were added. Hennig Brand is the first person credited with discovering a modern element, phosphorus. Brand was an alchemist searching for the philosopher's stone. He distilled many substances including human urine, from which he obtained pure phosphorus.\r\n\r\nAdding oxygen to the mix\r\nMany consider Antoine Lavoisier the father of modern chemistry. His work was groundbreaking and instrumental to the current understanding of chemistry. Lavoisier discovered oxygen and hydrogen, and correctly explained the role of oxygen in combustion. He also predicted the existence of silicon. Many people built on the work of Lavoisier, contributing to the periodic table and its elements.\r\n\r\nOrganising the elements\r\nIn 1869, a Russian scientist named Dmitri Mendeleev organised the elements into the table structure we know today. He arranged the elements by atomic mass, starting with the lightest elements. By grouping elements into columns, he was able to correctly predict elements that had not yet been discovered. Mendeleev also knew that the proposed atomic masses of some elements were incorrect, and used his periodic table to correctly predict the actual masses.\r\n\r\nThe development of periodic law\r\nGrouping the elements by atomic mass also led to Mendeleev discovering the periodic law. This law states that by grouping elements by atomic mass, we see trends in their properties. Periodic law has since been used to reliably predict the properties of elements.\r\n\r\n","n":0.054},[{"v":"Periodic table","i":2,"n":0.707},{"v":"History","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":281,"$":[{"v":"Chemical bonding","n":0.707},{"v":"How do chemicals form bonds to create molecules, compounds and more? Learn about different forms of chemical bonding, and how those bonds are broken, using this resource.\r\n\r\nOverview\r\n\r\n \tIonic bonding \u2013 what are ionic bonds and how do they form? How are ionic compounds named? 80 minutes to complete\r\n \tCovalent bonding \u2013 what are covalent bonds and how do they form molecular compounds? How do you predict the molecular geometry of molecules? 60 minutes to complete\r\n  \tLewis structures \u2013 what are Lewis electron dot structures and how do you draw them? 40 minutes to complete\r\n \tQuiz \u2013 Covalent bonding and Lewis structures \u2013 how well do you understand covalent bonding? Do you feel confident drawing Lewis structures? 60 minutes to complete\r\n\r\n\r\n","n":0.091},[{"v":"Covalent bonds","i":3,"n":0.707},{"v":"Ionic bonds","i":2,"n":0.707},{"v":"Chemistry","i":1,"n":1},{"v":"Chemical bonding","i":0,"n":0.707}]]},{"i":282,"$":[{"v":"The periodic table","n":0.577},{"v":"The periodic table, also known as the periodic table of elements, elegantly arranges the building blocks of the universe. It reveals patterns and relationships that govern the elements and their interactions. Use this resource to explore how this essential tool has transformed our understanding of the natural world, including the history of the periodic table and how to read and understand it.\r\n\r\nOverview\r\n\r\n \tHistory of the periodic table \u2013 how have people been using the periodic table to categorise elements and predict new atoms? How do elements interact with the known elements? 15 minutes to complete\r\n \tThe structure of the periodic table \u2013 how is the periodic table organised? 65 minutes to complete\r\n \tTrends in the periodic table \u2013 what is the relationship between the position of an atom on the periodic table and its properties? What is the periodic law? 45 minutes to complete\r\n\r\n\r\n\r\n","n":0.083},[{"v":"Periodic table","i":2,"n":0.707},{"v":"History","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":283,"$":[{"v":"Matter","n":1},{"v":"Matter is anything that has mass and occupies space. It includes things that can be seen (all living and non-living, natural and synthetic things) and cannot be seen to the naked eye (e.g. air and microorganisms). Studying chemistry always begins with understanding how matter behaves and is classified.\r\n\r\nOverview\r\n\r\n \tStates and properties of matter \u2013 how is matter described? What is the difference between a liquid, a gas and a solid? 15 minutes to complete\r\n\tPhysical and chemical changes \u2013 how does matter change? 10 minutes to complete\r\n\r\n","n":0.108},[{"v":"Matter","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":284,"$":[{"v":"Atoms and atomic structures","n":0.5},{"v":"Everything in the universe is made of atoms. From the screen you are looking at, to the hand you are using to navigate this page, atoms are the building blocks of all matter. Let's step through what atoms are made of and how they make other molecules and compounds. This resource also covers electrons, how they are structure, how they behave, and the relationship between a quantum number and an electron.\r\n\r\nOverview\r\n\r\n\r\n\tAtoms, molecules and compounds \u2013 what is an atom, a molecule and a compound, and how are they related? 25 minutes to complete\r\n\tAtomic structure \u2013 what are atoms made of and how are they structured? 35 minutes to complete\r\n\tThe electron \u2013 what is an electron and how does it behave? 30 minutes to complete\r\n\tQuantum numbers \u2013 how do we know where an electron will be in an atom? 15 minutes to complete\r\n\tQuiz \u2013 Quantum numbers \u2013 how well do you understand quantum numbers? 20 minutes to complete\r\n\r\n\r\n\r\n","n":0.08},[{"v":"Chemistry","i":1,"n":1},{"v":"Atomic Structures","i":0,"n":0.707}]]},{"i":285,"$":[{"v":"Famous chemists and milestones","n":0.5},{"v":"Chemists have shaped the modern world in many important ways.\r\n\r\nNotable chemists\r\nSome notable chemists are:\r\n\r\n\r\n \tJoseph Priestley \u2013 discovered oxygen in 1774, helping to explain the combustion reaction\r\n \tAlfred Nobel \u2013 invented dynamite in 1866 and established the prestigious Nobel Prizes\r\n \tDmitri Mendeleev \u2013 created the periodic table of elements and proposed the periodic law (also known as Mendeleev's law)\r\n \tMarie Curie \u2013 was the pioneer in radioactivity research and discovered radium and polonium, winning two Nobel Prizes for her discoveries, one in Chemistry (1911) and the other in Physics (1903)\r\n \tRosalind Franklin \u2013 paved the way for the discovery of the molecular structure of DNA\r\n\tEllen Swallow Richards \u2013 was a pioneer in the field of sanitary engineering, helping to establish water-quality standards and the discipline of home economics in America.\r\n\r\n\r\n\r\n\r\nChemistry milestones\r\nThere are so many important milestones in the field of chemistry. Some examples, which have earned their founders a Nobel Prize are:\r\n\r\n \tthe discovery of fullerenes by Professor Robert F. Curl, Harold W. Kroto and Richard E. Smalley (Nobel Prize in Chemistry 1996). A fullerene is a compound composed entirely of the element carbon. It has applications in several fields such as medicine, electronics, and polymer industry.\r\n \tthe discovery of gene-editing tool CRISPR by Emmanuelle Charpentier and Jennifer A. Doudna (Nobel Prize in Chemistry 2020). CRISPR has numerous applications, specially in precision medicine, immunology, and agriculture.\r\n \tthe development of a process to synthesise ammonia using elements nitrogen and hydrogen by Fritz Haber (Nobel Prize in Chemistry 1918). Ammonia is used as a fertiliser to provide nitrogen to plants.\r\n \tthe discovery of macromolecules called polymers by Hermann Staudinger (Nobel Prize in Chemistry 1953). Polymers are giant molecules composed of smaller repeating units. They can be natural (such as cellulose, proteins, carbohydrates) or synthetic (nylon, polyethylene).\r\n \tthe identification of the molecular structure for vitamin B12 by Dorothy Crowfoot (Nobel Prize in Chemistry 1964). Deficiency of vitamin B12 causes anaemia (where the body lacks healthy red blood cells), leading to weakness in the body, nerve damage, and poor memory.\r\n","n":0.055},[{"v":"History","i":2,"n":1},{"v":"Chemists","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":286,"$":[{"v":"Introduction to chemistry","n":0.577},{"v":"What were the origins of chemistry? What has it done for us? Who are the stars in the field? The history of chemistry is fascinating. Let's take a quick tour through what chemistry is and some famous chemists.\r\n\r\nOverview\r\n\r\n \t\r\n\t\r\n\t\tFrom alchemy to chemistry \u2013 the start of a modern science \u2013 when did chemistry become a science? 15 minutes to complete\r\n\t\tChemistry today \u2013 what do we use chemistry for in the modern world? 10 minutes to complete\r\n\t\tFamous chemists and milestones \u2013 who are some famous chemists and what have they achieved? 8 minutes to complete\r\n\t\r\n\t\r\n\r\n","n":0.103},[{"v":"History","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":287,"$":[{"v":"Chemistry today","n":0.707},{"v":"Chemistry is a vital part of the modern world. From DNA to space flight, virtually all parts of science and technology have a chemistry component.\r\nStudying chemistry will help you to understand the building blocks of all matter in the universe, including how substances interact with each other and form everything we encounter day-to-day. For example, knowledge of chemistry will allow you to:\r\n\r\n \tpredict how an acid will react when you add a base and save your cakes in food technology\r\n \tkeep your cleaning products active in industrial chemistry; or\r\n \tget the pH of your plant fertiliser just right.\r\n\r\n\r\n\r\nChemistry in context \u2013 fertilisers in agriculture\r\nModern agriculture and farming could not exist without chemistry. All plants need sunlight and water; today, we know that plants also need a mix of potassium, nitrogen and phosphorus to grow. This understanding is vital to creating large farms that can feed today's populations.\r\nWe provide plants with compounds containing these elements through mixtures called fertilisers. We also need to know where to find natural sources of these elements, or how to synthesise the fertilisers that supply them. Modern life depends heavily on fertilisers to grow the plants we need to feed ourselves \u2013 not just the fruits and vegetables we enjoy, but also the feed for dairy, chickens, and other farm animals.\r\n\r\nCareers in chemistry\r\nAtoms and elements are the foundations of chemistry. Since everything we can see and touch is made of chemicals, it\u2019s not surprising that chemistry is important in many careers. A sound understanding of chemistry is essential for success in disciplines of health and medicine, all the different fields of engineering, food technology, textiles, construction, architecture, geology, aeronautics, design, archaeology and so much more. It is vital to many science subjects and STEM (science, technology, engineering and maths) careers.\r\n","n":0.058},[{"v":"Chemists","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":288,"$":[{"v":"Alchemy to chemistry \u2013 the start of a modern science","n":0.316},{"v":"Isaac Newton would not have said he was a scientist, but he would have said he was an alchemist, or someone who practised alchemy. In fact, it was not until 1833 that people began calling themselves \"scientists\". Before that time, people studied natural philosophy, the study of nature, causes, astrology and many more subjects. Astrology, medicine, magic, and experiments were used in alchemy to understand the world and to create new substances. What we now think of as science came from natural philosophy and alchemy.\r\n\r\n\r\n\r\nTurning lead to gold\r\nAlchemy was the art of attempting to transmute or to change metals, particularly lead, into gold. For many thousands of years, alchemists had tried to unlock the secrets of matter, to bring great wealth and create a cure for all diseases. Alchemists believed that by studying the stars, metals, herbs and other substances, they could find a way to create the philosopher's stone. The philosopher's stone was believed to cure all disease, turn metals to gold and make people immortal.\r\nFor over four thousand years and across many countries, alchemy was practised by people who believed that matter could be classified and understood through experiments. We can describe alchemy as the start of the system of modern science in many parts of the world. Of the modern sciences, chemistry is the closest to alchemy in name and in practice. Classifying matter into different elements, states and categories was an alchemical pursuit.\r\n\r\nThe alchemical table\r\n\r\n\r\nYou may recognise some of the elements included on the alchemical table. Some of the metals named here are still very much a part of modern science and our understanding of matter. Other components, such as the four elements, are no longer part of scientific beliefs.\r\n\r\nChemistry is born\r\nRobert Boyle published \"The Sceptical Chymist\" at the end of the 17th century. This book disproved the idea that matter was made of a mix of some or all of the four elements; earth, fire, air and wind. Instead, Boyle used new equipment and experiments to prove his theories about the world. This emphasis on experiments that could be repeated anywhere and give the same results was not new, but Boyle helped to set the standards for the modern chemist.\r\nIn the 19th century, the scientific method and the separation of the sciences into specific disciplines became the hallmarks of modern science. \r\nAlchemical symbols\r\nThe three principles:\r\n\r\n\tSulfur\r\n\tSalt\r\n\tMercury\r\n\r\nThe four elements\r\n\r\n\tFire\r\n\tAir\r\n\tWater\r\n\tEarth\r\n\tThe elements\r\n\r\nPlanets and metals\r\n\r\n\tVenus or copper\r\n\tJupiter or tin\r\n\tSun or gold\r\n\tMars or iron\r\n\tMoon or silver\r\n\tMercury or quicksilver\r\n\tSaturn or lead\r\n\r\n","n":0.049},[{"v":"History","i":1,"n":1},{"v":"Chemistry","i":0,"n":1}]]},{"i":289,"$":[{"v":"Logical fallacies","n":0.707},{"v":"Ever get the feeling that someone's argument is a bit off, but you can't put your finger on why? Feel like the evidence a politician uses is not enough for the claim they are making? Want to know why when your uncle says \"reading something online means it's true\" it is actually a bad argument? What's the name for these things? The answer is logical fallacies.\r\n\r\n\r\n\r\nLogical fallacies are used all the time: to make arguments that don't have the right evidence, to claim a relationship between two unrelated things and to attack people's characters instead of their claims. Logical fallacies are really common and generally reveal poor critical thinking.\r\n\r\nUnderstanding and identifying commonly used fallacies will help you to critically analyse claims people make- whether they come from a journal article, news item, opinion piece or an annoying relative. This will help you know a flawed argument when you hear one.\r\n\r\nFive common fallacies\r\nLet's start off by exploring five of the most common fallacies.\r\nSlippery slope\r\nClaiming that one action will lead to larger and unintended changes.\r\nFor example:\r\n\u201cIf we raise minimum wage, businesses will fail.\u201d\r\nAd hominem\r\nWhen you attack a person rather than their claim.\r\nFor example:\r\n\u201cYou are stupid and wrong.\u201d\r\nAppeal to authority\r\nUsing a figure of authority to prove your argument is correct.\r\nFor example:\r\n\"My Mum's friend who is a doctor, said you should eat 3 bananas a day.\"\r\nFalse cause\r\nSeeing a relationship between two things when they are probably not the cause of each other.\r\nFor example:\r\n\u201cI ate an apple every day in March, and I did not get sick.\u201d\r\nStrawman\r\nYou misrepresent someone's argument to make it easier to attack.\r\nFor example:\r\n\"You said petrol causes climate change, so you want us to live in caves?\"\r\nTake the quiz to test your knowledge\r\n\r\n","n":0.06},[{"v":"Media literacy","i":3,"n":0.707},{"v":"Reasoning","i":2,"n":1},{"v":"Logical fallacies","i":1,"n":0.707},{"v":"Critical thinking","i":0,"n":0.707}]]},{"i":290,"$":[{"v":"Introduction to critical thinking","n":0.5},{"v":"Being able to think and act critically is an important and necessary skill for your academic success as well as your future employment. Critical thinking is highly valued by employers in many careers, from law to teaching to business. Seeing the reasoning behind arguments and to find flaws and strengths in other people's ideas helps you succeed in the modern world.\r\n\r\nWhy think critically?\r\nCritical thinking can be defined as:\r\n\"being thoughtful, asking questions, not simply accepting as true, everything your read/hear\".\r\n\u2013 Williams cited by the University of Hull\r\n\r\n\r\nIt is not about being negative or trying to find something wrong.\r\n\r\nCritical thinking is about asking good questions, identifying bias, challenging ideas, finding credible evidence, reflecting on your own learning and transferring these skills to your professional practice.\r\n\r\nCritical thinking is like an onion \u2026 as you peel off one layer (begin to analyse the argument), there is another layer (more questions), and another layer (deeper questions) \u2026 and on \u2026 and on.\r\nWhat is critical thinking?\r\nCritical thinking is a key skill for higher education studies. It is necessary for every aspect of your study, including:\r\n\r\n \treading\r\n \tlistening\r\n \tnote-making\r\n \twriting\r\n \tpresentations\r\n \tprofessional practice\r\n\r\nAt university in Australia, lecturers expect you to do more than just describe a theory or concept in your assessment tasks. You are expected to show higher level thinking and reasoning: to analyse, evaluate or judge a concept, and apply a concept or theory to a practical situation. You may be asked to compare and contrast theories, to find flaws or gaps in evidence, look for new evidence or reflect on your own learning.\r\n\r\n","n":0.062},[{"v":"Media literacy","i":2,"n":0.707},{"v":"Critical thinking","i":1,"n":0.707},{"v":"Critical reading","i":0,"n":0.707}]]},{"i":291,"$":[{"v":"Critical thinking","n":0.707},{"v":"Critical thinking is an essential skill for succeeding in your studies, and life. These tutorials will take you from understanding the basics of critical thinking, refining your research skills and finally analysing your sources.\r\n","n":0.171},[{"v":"Critical thinking","i":0,"n":0.707}]]},{"i":292,"$":[{"v":"Reflective writing in design","n":0.5},{"v":"Reflecting on practice is an important means of developing expertise in design. Structured reflection involves identifying significant events that lead to learning, identifying their value and considering how they will contribute to your future projects and your practice as a whole.\r\nOne example of reflective writing in art and design is the Studio Knowledge Object (SKO), which is a record of learning and insights gained in a design studio course.\r\n\r\nThese insights can be anything from learning a new method for improving your workflow to a lesson that completely changed your perspective of art and design. Your specific learning moments are individual to you. By articulating them in words and demonstrating them through design, you can communicate the value of your learning to others.\r\n\r\nLet's explore how SKOs are structured.\r\n\r\nThe SKO text has three main sections:\r\n\r\n \tContextualising the field of inquiry that the studio investigates gives a non-expert reader an understanding of the key issues and approaches that your reflection relates to.\r\n \tReflecting on your learning identifies key insights and learning from the studio, describing specific activities, concepts or approaches and evaluating their impact. You will probably cover 2-4 insights in your SKO.\r\n \tProposing future application speculates about how you can transfer this new knowledge to other projects and how it might affect your future practice.\r\n\r\nThe way that you combine these elements will help to tell your story, so consider:\r\n\r\n \ttext (title, headings and paragraphs)\r\n \timages (of your work, diagrams or other content)\r\n \tdesign (type and graphic elements)\r\n \tlayout (arrangement)\r\n\r\nAll elements should reflect the content of your studio and the learning you have gained. \r\n","n":0.062},[{"v":"Assignments","i":2,"n":1},{"v":"Design","i":1,"n":1},{"v":"Art","i":0,"n":1}]]},{"i":293,"$":[{"v":"Case study report for nursing","n":0.447},{"v":"Do you need to write a case study report for nursing? The information on this page will help get you started.\r\n\r\nPart 1: What is it and why?\r\nThis video introduces the concept of the case study and identifies how case studies differ to other written assignments.\r\n\r\n\r\n\r\nPart 2: Research and writing\r\nThis video provides tips on how to research and structure paragraphs in order to keep a patient centred emphasis in the writing.\r\n\r\n \r\n\r\nThe most important thing in writing up a case study report is that first, second, and thirdly: It is about the patient, not the condition nor the treatment. Students often put the emphasis on the condition or the treatment because the question usually focuses on this. While you need to explore these things, you still have to bring it back to the patient, and keep the patient central to the whole assignment.\r\n\r\nWe can understand if busy medical professionals start to see their patients in terms of a condition to be fixed, or a procedure to be performed. But ideally, we should remember that a patient is a person with a complex array of needs and preferences. This means that a simple textbook answer is not always best for them.\r\n\r\nIn your career, you will need to negotiate adjustments and anticipate problems particular to the patient. It does not make sense to discuss a condition in isolation from the individual. Each patient has their own unique kaleidoscope of history, personality type, fears, medical challenges, needs during recovery, so forth and so on. You need to see their condition or their treatment overall, as part of the whole person. Writing a case study teaches you to do that, if you don\u2019t explore the condition or treatment as part of a holistic system, you case study becomes theoretical, it becomes another essay.\r\n\r\nLet\u2019s compare the case study to the essays you might have already written. The case study in nursing explores a particular patient\u2019s situation, while your essays will more likely have focused on disease, a treatment, a concept, or a system. A case study is very grounded in the real and the now, even if the patient is a hypothetical invention. It still provides some fairly authentic experience with dealing with an individual patient.\r\n\r\nEssays, on the other hand, tend to be literature-based. Of course, you need to do a lot of research and reading for your case study too, but the difference is for a case study you must always view and evaluate the information in the context of your patient. So in a case study, you need to show that you can find and apply information to you patient. For an essay you need to look at the information itself, pull it apart and evaluate it for its own sake.\r\n\r\nSo even if the question looks like an essay, don\u2019t be distracted. There is still an expectation to relate this to the patient. It is possible to explore the content and theory in a generalised way throughout the assignment, but the beginning of the assignment (and the beginning of each paragraph, in fact) should foreground you're patient and their relationship to this theory.\r\n\r\n \r\n\r\nAn essay, as opposed to a case study, usually has an question or a thesis which defines the boundaries of what you research, You aim to keep it within those parameters. and you delve as deeply as you can into the material. It loses focus if you include stuff outside of the topic.\r\n\r\nWith a case study, however, your focus is the patient so you need to cast your net widely and generously at first to find something that is relevant to them. And then when you find something relevant, you need to research deeply. So, doing a good case study report involves a lot more reading than you might think. You\u2019ll have to research both generally and deeply.\r\n\r\nThere can be a lot of research, and fairly unguided research involved in a case study. Don\u2019t assume it\u2019s like a reading comprehension exercise, the problems are not always clearly defined and the solutions have not necessarily been covered in lectures. Think of your case study as a mystery puzzle with a few clues scattered through it.\r\n\r\nIn reality, patients do not arrive with instructions, so it\u2019s up to the professional to notice that it is obvious and to think outside of the box. It is up to you to anticipate problems for the patient, it is up to you to follow through with the threads of any clues, and to see if they offer any valuable information, and these may only be hinted at indirectly. With years of nursing experience, the clues and danger signs get easier to spot, but it is very challenging for a beginner if you don\u2019t know what to look for, and that is why the broad reading is important.\r\n\r\nSo, every piece of information about your patient is valuable. When the information is given in the assignment, the facts are not random facts, they have usually been included for a reason, so use them and use them as much as you can. You may, on the other hand, have been told very little in the assignment and be expected to elicit the information yourself from the patient, through a blog, or a question-and-answer session. Asking for more details about their own condition just gives you more details about their condition, and unless you have some reasoning in mind for asking about that, more details per se is usually not that very helpful.\r\n\r\nInstead, ask about things that may affect the treatment or recovery; ask about things that may complicate matters. Smoking, relevant illnesses, lack of family support, strenuous lifestyles or jobs are all factors that may generate important decisions about how the patient is treated or cared for. The secret to keeping it about the patient is to keep them foremost into the discussion. You do this by making them the subject of each paragraph, and you do this by making them feature in the topic sentence: at the beginning and, if possible, a summary sentence at the end.\r\n\r\nIn the middle of a paragraph, you have more flexibility to talk about the procedure or the condition in a more general way. But, at the end of the day, you need to remind us of how this is important for your patient. If you don\u2019t, your writing becomes an essay rather than a case study. And if you feel like you\u2019re being asked to research and to discuss material that is not directly relevant to your patient, you can make it relevant by at least explaining why it is not relevant. Remember, everything you write about should be in relation to your patient.\r\n\r\nSo remember, it is all about the patient, keep everything relevant to the patient; keep the patient in mind throughout the research and the writing. Explore and discuss what the question requires, but theory for the sake of theory should be avoided. Your mission is to find the best solutions available for your patient, and develop the best understanding that you can of your patient.\r\n\r\nYou have now complete the Case Study module. We hope you have enjoyed the content. For more study skill resources, please visit the RMIT Learning Lab.\r\n","n":0.029},[{"v":"Reflective writing","i":2,"n":0.707},{"v":"Nursing","i":1,"n":1},{"v":"Case studies","i":0,"n":0.707}]]},{"i":294,"$":[{"v":"Mind mapping","n":0.707},{"v":"Mind mapping is a creative and logical means of note-taking and note-making that literally \"maps out\" your ideas. Keep reading to find out how to create an effective mind map.\r\nMind maps are a useful tool for making notes, revising and organising information. They are a visual representation of connections between concepts and ideas that are related to one central topic.\r\nWhy are mind maps useful?\r\nMind maps are an extremely useful study tool for several reasons:\r\n\r\n \tmind maps help to clarify the key concepts of a topic\r\n \tthey show links and relationships between ideas\r\n \tmind maps organise ideas and information by making it accessible on a single page\r\n \tthey facilitate memory through key words and phrases\r\n \tmind maps save time as they are very quick to do.\r\n\r\nMind maps are particularly useful for scoping and planning a topic for an assignment or essay and for revising and preparing for tests and exams.\r\nTest yourself\r\nLook at the following list of ideas about 'Forms of water'.\r\nForms of water\r\n\r\nNow, look at the mind map version of the same ideas. Which version is easier to remember?\r\n","n":0.075},[{"v":"Understanding the question","i":1,"n":0.577},{"v":"Mind mapping","i":0,"n":0.707}]]},{"i":295,"$":[{"v":"Writing paragraphs","n":0.707},{"v":"A paragraph is a group of connected sentences. It is generally made up of one main idea and a group of supporting sentences. Paragraphs are the basic unit for building longer pieces of writing.\r\n\r\nThe importance of paragraphs\r\nThis short video explains why writing good paragraphs in your assessment tasks is important. It also introduces the easy-to-follow TEEL structure for your paragraphs to improve your writing skills.\r\n\r\n\r\n\r\n\r\nUsing the TEEL structure\r\n\r\nWell structured paragraphs are important because if the information is well organised it is easier to read. The TEEL strategy is very helpful for knowing what should be included in a paragraph.\r\n\r\n\r\n\r\n \tTopic sentence (contains the main idea and is usually first in the paragraph)\r\n \tExplain\r\n \tEvidence and example (use references)\r\n \tLink (refer back or sum up main idea) - optional\r\n\r\nSample paragraph\r\n\r\n\r\n\r\n\r\nParagraphs activity\r\nTest your skills! Try this short quiz to see if you can identify the most appropriate topic sentence for the paragraph.\r\n\r\n\r\n\r\n\r\n \r\n\r\nIf information is organised and your main points are clear, you will get better marks. Making your ideas and arguments stand out is mostly done through having clear and well-structured paragraphs. A really good way to write paragraphs is to use the TEEL structure:\r\n\r\n \tTopic Sentence: The topic of the paragraph\r\n \tExplain: Why is the topic important?\r\n \tEvidence/Example: Support the topic (references)\r\n \tLink: Link back to the topic\r\n\r\nThis is an example:\r\n\r\n\"One factor within the team that seems to be important is the notion of team cohesiveness. Team cohesiveness enables a diverse group of individuals to work towards common goals. Addison (1996, p. 107) argues that highly cohesive teams 'are more effective at achieving goals they set for themselves, and have higher member satisfaction'. For example, cohesive teams are more likely to have high morale (Smith 1996) and the ability to cooperate, and so work effectively together (McAlfee & Champagne 1987). Therefore, team cohesiveness enhances team effectiveness.\"\r\n\r\nSee how the topic sentence tells you the main idea of the paragraph. The rest just supports this main idea. If it doesn't relate to the main idea, it doesn't belong in the paragraph. It is sort of like a formula for writing.\r\n\r\nLinking your ideas helps the paragraph to flow and it shows how your analysis has developed. Have a look at the examples. The paragraph with linking words flows and it easy to read.\r\n\r\nSeparate sentences: \"Many countries are in recession\", and \"China's economy is growing\".\u00a0 These are two separate ideas, so how do we link them?\r\n\r\nRevised sentence: \"Although many countries are in recession, China's economy is growing\". The linking word in the paragraph here is \"Although\".\r\n\r\nHere are a few more linking words: \"furthermore\u2026\", \"in addition\u2026\", \"however\u2026\", \"as a result\u2026\", \"although\u2026\". \"therefore\u2026\".\r\n\r\nThe paragraph with linking words flows and it easy to read. Without linking words, it is disjointed and more difficult to follow. By using linking words, it looks like you have thought more deeply about your research.\r\n\r\nFor more information about paragraphs, try the tutorial.\r\n\r\nThanks for watching.\r\n\r\n \r\n\r\nThe missing sentence will be wrapped in square brackets. Read the sentence and at the end, select the sentence you think is correct.\r\nExercise 1\r\n[ ]In China, tea is usually drunk without milk or sugar in small cups. In Arab countries, tea is usually flavoured with mint leaves and sugar. However, in India tea is taken with both milk and sugar. Tea is now a popular drink worldwide.\r\n\r\n \tTea is a healthy and refreshing drink.\r\n \tTea is taken in a variety of ways in different countries.\r\n \tTea is important part of the global economy.\r\n\r\nCorrect answer: 2\r\n\r\n[Tea is taken in a variety of ways in different countries.]In China, tea is usually drunk without milk or sugar in small cups. In Arab countries, tea is usually flavoured with mint leaves and sugar. However, in India tea is taken with both milk and sugar. Tea is now a popular drink worldwide.\r\nExercise 2\r\n[ ]Thirty years ago, most Australians ate a typical Northern European diet. Since the 1960's, immigrants have brought their food knowledge with them and shared it with Australians. Australian diets now strongly reflect those styles of cooking.\r\n\r\n \tThe Australian diet has changed dramatically to the influences of immigration.\r\n \tThe Australian diet is now more interesting and healthy than before.\r\n \tNew migrants have imported new foods to Australia.\r\n\r\nCorrect answer: 1\r\n\r\n[The Australian diet has changed dramatically to the influences of immigration.]Thirty years ago, most Australians ate a typical Northern European diet. Since the 1960's, immigrants have brought their food knowledge with them and shared it with Australians. Australian diets now strongly reflect those styles of cooking.\r\nExercise 3\r\n[ ]Good office systems ensure the work is done efficiently. They also enable clear procedures and communications. These help to maintain good relationships in the office. Most importantly, they enable each worker to clearly understand what is required.\r\n\r\n \tA well-run office is one where staff respect each other.\r\n \tGood organisation is the basis of a well-run office.\r\n \tHappy staff make an effecient office.\r\n\r\nCorrect answer: 1\r\n\r\n[A well-run office is one where staff respect each other.] Good office systems ensure the work is done efficiently. They also enable clear procedures and communications. These help to maintain good relationships in the office. Most importantly, they enable each worker to clearly understand what is required.\r\n\r\n","n":0.034},[{"v":"Topic sentences","i":3,"n":0.707},{"v":"TEEL","i":2,"n":1},{"v":"Paragraphs","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":296,"$":[{"v":"Artist statement: editing","n":0.577},{"v":"Edit and edit again...\r\nEditing is an essential step in the writing process. The first draft of your statement may be too long and disorganised as you elaborate on ideas from previous stages.\r\nEditing your work\r\nHere are some suggestions to make your writing more engaging:\r\n\r\n\r\n\r\nWrite your own short artist statement\r\n(practical writing activity - 30 minutes)\r\n\r\nWrite your own short artist statement that describes your creative practice. Use your mind map and other activity outcomes to start the process, using the learning resources provided here to help you.\r\n\r\nChallenge: to practise writing a concise statement with impact, try and limit your text to between 50-100 words!\r\n\r\nShare your statement with others and get some feedback. Make sure you save it somewhere with the date so you can return to it and reflect/review over time.\r\n\r\nMaterials \r\n\r\nPaper and pen/pencil or any text editing application.","n":0.086},[{"v":"Editing","i":3,"n":1},{"v":"Design","i":2,"n":1},{"v":"Artist statement","i":1,"n":0.707},{"v":"Art","i":0,"n":1}]]},{"i":297,"$":[{"v":"Group work","n":0.707},{"v":"This tutorial gives you strategies for being part of a successful group, which you can apply to your future academic and professional life.\r\nThis section will step you through a process to work together to create an effective product, as well as giving advice on how to manage each other and challenges as they arise.\r\nWhy group work?\r\nMany courses require you to work in groups, including virtual or online groups. Group work requires you to use your organisational and interpersonal skills.\r\n\r\nGroup work helps you to:\r\n\r\n \tdevelop your ability to communicate in a professional context\r\n \tuse interpersonal skills to work productively with others\r\n \tapply leadership and project management skills to solve complex problems\r\n \tthink critically and reflectively in your subject area.\r\n\r\nEffective groups usually think of themselves as 'teams' when working together to complete a project. An effective group assignment is one where the whole team is responsible for the outcome.\r\n\r\n\r\n\r\n\r\n","n":0.082},[{"v":"Group work","i":0,"n":0.707}]]},{"i":298,"$":[{"v":"Artist statement: mind mapping","n":0.5},{"v":"If you don't know how to start writing your artist statement, mind mapping words and ideas that relate to your creative practice is a good way to visualise concepts and create relationships between them.\r\n\r\nMind mapping activity\r\nThis activity is best done on paper, however, if you have a mobile device with a drawing app, you can create a digital version in the same way.\r\n\r\nFor your artist statement mind map, get your creative work together and look at it as a whole before you start. This can help you to identify key themes and the development path of your practice.\r\n\r\nRead other artist statements (see Artist statement: written examples), as these can help you to think about tone and vocabulary for your own writing.\r\n\r\nThink about the following topics and how they relate to your creative work.\r\n\r\n \tAudience/purpose \u2013 who are you writing for? What is the purpose of the statement?\r\n \tWhat: Think of words that describe the format of your creative work, materials used, ideas and themes explored.\r\n \tHow: Think of words that describe methods, process, and techniques you use in your creative practice.\r\n \tWhy: What is the reason for your creative practice? Why is it important to you and your community?\r\n \tContext: What is your place in a continuum of creative practice? What are your influences and where do your ideas come from? Think of words or phrases that describe where you are in the world and how this affects your work.\r\n\r\nMaterials\r\n\r\nPaper and coloured pens/pencils or any digital drawing application on computer or mobile device.\r\n\r\nWrite down words and ideas that relate to your practice.\r\n\r\nOrganise:\r\nOnce you have some words written down from the first part of this activity, you can organise related ideas. You may prefer to keep working on paper or drawing app for this stage or start grouping your ideas into a text document. Using colour pencils, highlighters etc. can help you to visualise the relationships between words and ideas.\r\n\r\n\r\n\r\nGroup the words and ideas that describe aspects of your practice.\r\n\r\nIdentify topics:\r\nThese will form the sentences or paragraphs for the next stage of the process.\r\n\r\n\r\n \r\n\r\nThe image shows handwritten text, \"Me as a Textile Designer\", in a box in the centre. Various branches of sub-topics are handwritten around that text.\r\n\r\nSome of the sub-topics include:\r\n\r\n \tDaily search\r\n \tBotanicals\r\n \tGuache\r\n \tPencil\r\n \tMethodical\r\n \tCollective goal\r\n \tCollage\r\n \tNew technologies\r\n \tCraftsmanship\r\n\r\n \r\n\r\nAs above, the image shows handwritten text, \"Me as a Textile Designer\", in a box in the centre. Various branches of sub-topics are handwritten around that text.\r\n\r\nIn this iteration, sub-topics are highlighted in different colours and linked together into groups.\r\n\r\nThe groups are:\r\n\r\n \tIdeas and inspiration\r\n \tFuture\r\n \tHow\r\n \tValue and beliefs\r\n \tNew approach\r\n\r\n \r\n\r\nAs above, the image shows handwritten text, \"Me as a Textile Designer\", in a box in the centre.\r\n\r\nThere are five branches leading from the center, each labeled with a question.\r\n\r\nThe branch labeled \"What is your approach to design?\" has the following sub-topics:\r\n\r\n \tColor combinations\r\n \tMotifs symbols (everyday things)\r\n \tPersonal creative dialogue\r\n \tBotanicals\r\n \tNature strips\r\n \tSuburban architecture\r\n \tCity/wildness connection\r\n\r\nThe branch labeled \"What inspired you?\" has the following sub-topics:\r\n\r\n \tSoft forms to convey emotions\r\n \tSoft structure\r\n \tMood/meaning associations\r\n \tCraftsmanship/hand processes\r\n\r\nThe following items under \"new technologies\" are labelled as \"processes\"\r\n\r\n \tProcesses:\r\n \tNatural fibers\r\n \tHand dyed yarns\r\n \tKnitting\r\n \tExperimental & fresh\r\n\r\nThe branch labeled \"What values or beliefs underpin your work?\" has the following sub-topics:\r\n\r\n \tIdeas energy flow\r\n \tCollective goal\r\n \tEnthusiasm\r\n \tFull drop collective\r\n \tHolistic: concept to product\r\n\r\nThe branch labeled \"Where do you see yourself in the future?\" has the following sub-topics:\r\n\r\n \tChange in industry\r\n \tSmaller considered manufacture of textiles\r\n \tAustralian designer/makers\r\n \tSustainable\r\n\r\nThe branch labeled \"How do you work?\" has two groups of sub-topics:\r\n\r\nThe first\r\n\r\n \tLino\r\n \tGouache\r\n \tPencil\r\n \tInk\r\n\r\nThe second\r\n\r\n \tStamp\r\n \tCut\r\n \tCollage\r\n \tMethodical\r\n \tConstructed\r\n\r\n \r\n\r\nThis image refines the previous image, it still has the handwritten text, \"Me as a Textile Designer\", in a box in the centre. There are five branches leading from the center, each labeled with a question. Each branch has a list of sub-topics and a label underneath.\r\n\r\nThe branch labeled \"What is your approach to design?\" has the label \"DAILY SEARCH & CREATIVITy\". The following sub-topics are also diaplayed:\r\n\r\n \tColor combinations\r\n \tMotifs symbols (everyday things)\r\n \tPersonal creative dialogue\r\n \tBotanicals\r\n \tNature strips\r\n \tSuburban architecture\r\n \tCity/wildness connection\r\n\r\nThe branch labeled \"What inspired you?\" has the label \"INTEREST IN PERSONAL NARRATIVE\". The following sub-topics are also diaplayed:\r\n\r\n \tSoft forms to convey emotions\r\n \tSoft structure\r\n \tMood/meaning associations\r\n \tCraftsmanship/hand processes\r\n\r\nThe following items under \"new technologies\" are labelled as \"processes\"\r\n\r\n \tProcesses:\r\n \tNatural fibers\r\n \tHand dyed yarns\r\n \tKnitting\r\n \tExperimental & fresh\r\n\r\nThe branch labeled \"What values or beliefs underpin your work?\" has the label \"COLLABORATION\". The following sub-topics are also diaplayed:\r\n\r\n \tIdeas energy flow\r\n \tCollective goal\r\n \tEnthusiasm\r\n \tFull drop collective\r\n \tHolistic: concept to product\r\n\r\nThe branch labeled \"Where do you see yourself in the future?\" has the label \"HUMAN & RELATABLE\". The following sub-topics are also diaplayed:\r\n\r\n \tChange in industry\r\n \tSmaller considered manufacture of textiles\r\n \tAustralian designer/makers\r\n \tSustainable\r\n\r\nThe branch labeled \"How do you work?\" has the label \"SMALL BUSINESS\".\r\n\r\nIt has two groups of sub-topics:\r\n\r\nThe first\r\n\r\n \tLino\r\n \tGouache\r\n \tPencil\r\n \tInk\r\n\r\nThe second\r\n\r\n \tStamp\r\n \tCut\r\n \tCollage\r\n \tMethodical\r\n \tConstructed\r\n\r\n","n":0.035},[{"v":"Mind mapping","i":3,"n":0.707},{"v":"Design","i":2,"n":1},{"v":"Artist statement","i":1,"n":0.707},{"v":"Art","i":0,"n":1}]]},{"i":299,"$":[{"v":"Artist statement: the writing process","n":0.447},{"v":"Writing an artist statement has several stages. Here you will learn how to go through the process of writing a clear and concise text.\r\n\r\nVideo: Writing an artist statement\r\nThis video explains an approach to writing an artist statement. It is based on textile design but can be easily adapted to suit other mediums. It describes the steps to identifying the key components of the statement and then shows how to write it. Specifically, it introduces the concepts of brainstorming, organising and ordering your ideas, and then the writing process. There are models and additional resources to help you draft your statement, and there is a short tutorial outlining the essentials of the artist biography.\r\n\r\nVocabulary tips\r\nHere are some vocabulary tips to help you through your writing process.\r\nVocabulary 'do's\r\n\r\nVocabulary 'don't's\r\n\r\n \r\n\r\nWriting an artist statement and biography\r\nSlide 1: Introduction\r\nSlide description:\r\nSlide title: Introduction: Artist statement\r\nDevelop\r\nA page with the title: Professional document: reflecting on these questions:\r\n\r\n \tWhat is your approach to design?\r\n \tWhat values or beliefs underpin your work?\r\n \tHow do you work?\r\n \tWhere do you see yourself in the future?\r\n \tWho inspired you?\r\n\r\nNarration: The skills you learn in this tutorial will help you in your professional life. Where you might use these statements as a whole, or you may take parts of it to use separately.\r\n\r\nAn artist statement is an important professional document that reflects on these following five questions:\r\n\r\n \tWhat is your approach to design?\r\n \tWhat values or beliefs underpin your work?\r\n \tHow do you work?\r\n \tWhere do you see yourself in the future?\r\n \tWho inspired you?\r\n\r\nSlide 2: Overview\r\nSlide description:\r\nSlide title: 1. Mind mapping\r\nA three-step process is mapped out in three blocks; in-between them are arrows pointing in the right direction. Demonstrating a step-by-step approach when developing a mind-map.\r\n\r\n \tBrainstorm: Ideas and words around the reflective questions and free write\r\n \tOrganise: Your brainstorm into order under each question to identify topics\r\n \tOrder: The topics and then individual points in your mind map\r\n\r\nNarration: To develop your artist statement, the following steps to a mind map are useful and will help you find suitable words and ideas. The first step is to brainstorm to generate ideas. Next you need to organise the ideas into categories and finally develop your ideas into a map or plan.\r\nSlide 3: Brainstorm\r\nSlide description: A mind map example. The topic \u2018Me as a textile designer\u2019 is written at the center of the page. Brainstorm ideas begin to appear around the topic.\r\n\r\nNarration: At this stage all ideas are acceptable and the aim is to get as many ideas as possible that could go towards writing about yourself as a textile designer.\r\nSlide 4: Organise\r\nSlide description: Building on top of the previous example. Connections are made between certain words by using different colours.\r\n\r\nNarration: Step two involves finding connections between the ideas in relation to the five reflective questions. Colour is a useful way to distinguish connections around the different questions.\r\nSlide 5: Order\r\nSlide description: A new mind-map is created with the topic \u2018Me as a textile designer\u2019 written in the center of the page. This time, the five reflective questions are written on the page:\r\n\r\n \tWhat is your approach to design?\r\n \tWhat values or beliefs underpin your work?\r\n \tHow do you work?\r\n \tWhere do you see yourself in the future?\r\n \tWho inspired you?\r\n\r\nThe connections that were identified from the previous mind-map are organised into these categories.\r\n\r\nNarration: The final stage is to order the ideas into groups around the questions and create a spidergram or something similar to complete your mind map you will use this as a plan for your writing.\r\nSlide 6: Overview\r\nSlide description:\r\nSlide title: The writing process\r\nA three-step process is mapped out in three blocks; in-between them are arrows pointing in the right direction demonstrating a step-by-step approach during the writing process. The steps are:\r\n\r\n \tIdentify: Themes and ideas using your mind-map\r\n \tBegin writing: By creating topic sentences for each of the paragraph\r\n \tDevelop your ideas: By structuring your paragraphs using TEEL\r\n\r\nNarration: Now, it's time to start writing your artist statement. First, look at each category in your mind-map and identify the main theme or idea that answers each question. Complete the writing processes by then structuring your paragraphs based on the mind map.\r\nSlide 7: Part 1 - Identify\r\nSlide description:\r\nSlide title: Using your mind-map\r\nLook at the bigger picture of what you are trying to say and identify the overall themes or ideas.\r\nE.g. What values or beliefs underpin your work?\r\n\r\n \tIdeas + energy flow\r\n \tCollective goal\r\n \tEnthusiasm eg. \u2018Full Drop Collective\u2019\r\n \tHolistic: concept to product\r\n\r\nTheme identified from these points is collaboration.\r\n\r\nNarration: Here is an example of how you would identify one main theme from each list that will become the topic sentence for each paragraph. In this case all the ideas are about working with other people and sharing ideas, so a useful word here is collaboration. This will become the topic of your paragraph.\r\nSlide 8: Part 2 - Identify\r\nSlide description: The main themes are written under their relevant sections on the mind map.\r\n\r\nNarration: Notice how the main themes have been identified for each part of the mind map.\r\nSlide 9: Begin writing\r\nSlide description: Start by writing a topic sentence for each paragraph. Use the themes you have identified for each question.\r\nFor example:\r\nQ. What beliefs and values underpin your work?\r\nStep 1: Start by directly answering the question even if it seems predictable.\r\nFor example: I believe that collaboration is the most important thing for my work.\r\n\r\nStep 2: Now change the topic to make it stronger or more interesting.\r\nFor example:\r\n\r\n \tCollaboration is key for me.\r\n \tCollaboration is essential to my work.\r\n\r\nNarration: To write a topic sentence for your paragraph, start by writing a direct answer to the question, then you can rewrite the sentence to improve it.\r\nSlide 10: Develop your ideas\r\nSlide description:\r\nSlide title: Paragraph writing\r\nComplete your paragraph by following the TEEL structure\r\nTEEL STRUCTURE:\r\n\r\n \tTOPIC: Write a topic sentence containing the main idea. This is your voice and your idea\r\n \tEXPLANATION: Write one or more sentences that explain or elaborate on the main idea. This shows your reflection and analysis of the topic.\r\n \tEVIDENCE: Illustrate using examples from your own work and your experience, references to other artists and designers you admire and from your knowledge of history and theory.\r\n \tLINK: sum up by referring back to the main idea\r\n\r\nNarration: Now you can use the rest of the information from the mind map to complete the paragraph by writing supporting sentences that explain and give examples. The TEEL structure is a very helpful framework for structuring your paragraphs and making sure that you cover everything that is important.\r\nSlide 11: Begin writing\r\nSlide description: Visual representation of the narration: a hamburger animated in layers:\r\n\r\n \tTop bun = topic sentence.\r\n \tTomato and cheese = explanation.\r\n \tMeat and lettuce = evidence and examples.\r\n \tBottom bun = linking sentence.\r\n\r\nNarration: The hamburger is a good metaphor for a paragraph. Notice how the bread roll which is the topic sentence at the beginning and the linking sentence at the end hold the ingredients together because without it you don't have a hamburger!\r\nSlide 12: Begin writing\r\nDescription: Using the TEEL paragraph (example):\r\n\r\n \tTopic sentence: Collaboration is key for me.\r\n \tExplanation: The ideas and energy which grow when a group of like-minded individuals are moving towards a collective goal fills me with enthusiasm and makes me feel like anything is possible.\r\n \tEvidence: At the end of 2011, I started working as a part of Full Drop Collective, a group of six females textile designers exploring the possibilities in designer-maker goods and collaborative projects.\r\n \tLink: Being part of a project from start to finish and thinking about design in a holistic sense from concept to product has been important part of my practice.\r\n\r\nNarration: No Narration.\r\nSlide 14: Introduction\r\nSlide description:\r\nDevelop an artist biography with the following:\r\n\r\n \tAn introduction to yourself\r\n \tA summary of your resume\r\n \tYour approach to your work\r\n\r\nNarration: An artist biography is an introduction to you and a summary to your resume and approach to your work.\r\nSlide 15: What is it for?\r\nSlide description:\r\n\r\n \tExhibition proposals / galleries / curators\r\n \tWebsites\r\n \tCatalogues prefacing your images\r\n \tIndustry / potential employers\r\n \tRecruitment agencies\r\n \tPublications of an kind / media.\r\n\r\nNarration: Your artist biography needs to be redesigned for each audience. Here are some examples of where your biography would be used.\r\nSlide 16: What is in it?\r\nSlide description: A pyramid diagram (base at the top then narrowing down), broken down into four different sections. This is a representation of importance.\r\nArtist name and work\r\n\r\n \tMaterials and techniques that are employed by the artist\r\n \tInfluences on your work\r\n\r\nBrief history\r\n\r\n \tEducation, training\r\n\r\nProfessional experience\r\n\r\n \tExperience in your field eg. Projects, commissions, own businesses\r\n\r\nPhilosophy / ethos\r\n\r\n \tShort description of what the artist would like to accomplish with their art\r\n\r\nNarration : The artist biography is structured with the most important point at the beginning, which is who the artist is and their name. Followed by a brief history experience and summary of their philosophy.\r\nSlide 17: Example: Artist biography?\r\nSlide description:\r\nArtist biography (example):\r\n[Artist name and work] Tania Swinn is a textile designer who creates softly textured fabrics in vibrant colours. Her work features botanical inspired designs using recycled materials. [Brief history]Born and raised in Tasmania, Tania received her Bachelor in Textile Design from RMIT. [Professional history] She has exhibited widely in Australia and recently worked on projects with DESSI. [Ethos/philosophy] She is currently completing a placement with Ilke. The emphasis is on the quality of workmanship and the respect for sustainability.\r\n\r\nNarration: No Narration.\r\n","n":0.025},[{"v":"Professional writing","i":3,"n":0.707},{"v":"Design","i":2,"n":1},{"v":"Artist statement","i":1,"n":0.707},{"v":"Art","i":0,"n":1}]]},{"i":300,"$":[{"v":"Artist statement: Written examples","n":0.5},{"v":"View the different structural parts of each example across a number of artists working in different mediums and media.\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\n \r\n\r\n\r\n \r\n \r\n \r\n\t\t\r\n \t \r\n\t\t\r\n","n":0.196},[{"v":"Design","i":2,"n":1},{"v":"Artist statement","i":1,"n":0.707},{"v":"Art","i":0,"n":1}]]},{"i":301,"$":[{"v":"Artist statement: formats","n":0.577},{"v":"Your artist statement will change over time as you evolve your creative practice \u2013 you will need to write more than one!\r\nYou will also need different types of statements for different purposes.\r\n\r\nAn artist statement is useful for a portfolio of work, as part of a creative project, or as part of an application process for a job, project, exhibition, performance, grant, or other professional opportunity.\r\n\r\nHere are some common formats for artist statements that you may need to write as a creative professional:\r\n\r\n\r\n\r\nArtist statement or biography: what's the difference?\r\n\tAn artist statement should be about your work \u2013 the what, why and how of your creative practice.\r\n\t\r\n\tAn artist biography is about the who \u2013 it tells the reader about your personal history, career development, and achievements - where relevant to your creative practice.\r\n\t\r\n\tYou may need one or more of these statement types when writing an application or proposal. An organisation may specify which kind of statement they want and the maximum word-length for each text.\r\n","n":0.078},[{"v":"Professional writing","i":3,"n":0.707},{"v":"Design","i":2,"n":1},{"v":"Artist statement","i":1,"n":0.707},{"v":"Art","i":0,"n":1}]]},{"i":302,"$":[{"v":"Quiz \u2013 Quantum numbers","n":0.5},{"v":"Quantum numbers provide a detailed framework for describing the arrangement and behavior of electrons in atoms. Test your understanding of quantum numbers with a quiz.\r\n\r\nYour turn \u2013 quantum numbers\r\n\r\n \tIf \\(n=5\\), what are the possible values for \\(\\ell\\)?\r\n\r\n\t\r\n\r\n\r\n \tFor a certain electron, \\(\\ell=0\\). What is the shape of the subshell electron cloud of that electron?\r\n\r\n\t\r\n\t\r\n\r\n \tWhat are the \\(n\\) and \\(\\ell\\) values for \\(4d\\) electrons?\r\n\r\n\t\r\n\r\n\r\n \tFor an electron, the principle quantum number, \\(n\\) is \\(3\\). What are the possible values for \\(\\ell\\) and \\(m_{\\ell}\\)?\r\n\r\n\t\r\n\r\n\r\n \tWhat are the names of the orbitals with the principle quantum number \\(n=3\\)?\r\n\r\n\t\r\n\r\n\r\n \tWhat are the possible values for \\(\\ell\\), \\(m_{\\ell}\\) and \\(m_{s}\\) for an electron in the \\(n=1\\) energy state?\r\n\r\n\t\r\n\r\n\r\n \tHow many valid combinations of quantum numbers exist for \\(2p\\) electrons?\r\n\r\n\t\r\n\r\n\r\n \tCan an electron exist with \\(n=4\\), \\(\\ell=2\\) and \\(m_{\\ell}=4\\) quantum numbers? Explain, using possible quantum number combinations.\r\n\r\n\t\r\n\r\n\r\n \tCan two electrons have the following combination of quantum numbers?\r\n\t\\(n=2\\), \\(\\ell=0\\) and \\(m_{\\ell}=0\\)\r\n\r\n\t\r\n\r\n\r\n \tWhich type of orbital has the following combination of quantum numbers?\r\n\t\\(n=4\\), \\(\\ell=2\\), and \\(m_{\\ell}=1\\)\r\n\r\n\t","n":0.077},[{"v":"Quantum","i":2,"n":1},{"v":"Chemistry","i":1,"n":1},{"v":"Atomic Structures","i":0,"n":0.707}]]},{"i":303,"$":{"0":{"v":"Search the Learning Lab","n":0.5},"2":[{"v":"Archive","i":0,"n":1}]}},{"i":304,"$":[{"v":"Quantum numbers","n":0.707},{"v":"Quantum numbers are used to describe orbitals. They serve as an \"address\" for locating an electron within an atom, specifying its energy level, shape, orientation, and spin. Therefore, they provide a detailed understanding of the regions in an atom where electrons are likely to be found, which is crucial for predicting and explaining chemical bonding and molecular geometry.\r\nThere are four quantum numbers:\r\n\r\n \tThe principal quantum number (\\(n)\\) describes the energy of the electron. This can only be positive integers, such as \\(n=1,\\,2,\\,3\\) and so on.\r\n \tThe orbital quantum number (also known as the Azimuthal quantum number) (\\(\\ell)\\) represents the shape of the subshell electron cloud (orbital shape). \\(\\ell=0,\\,1,\\,2,\\,.....n-1\\) (positive integers less than \\(n\\)).\r\n\r\n\r\n\t\\(\\ell=0=s\\) orbital: spherical shape\r\n\t\t\\(\\ell=0=s\\) orbital: spherical shape\r\n\t\t\\(\\ell=2=d\\) orbital: four-leaf clovers.\r\n\r\n\r\n\tThe magnetic quantum number (\\(m_{\\ell}\\)) indicates the number of orbitals in the subshell and their orientation in space. \\(m_{\\ell}=0,\\,\\pm1,\\,\\pm2,\\,......\\pm\\ell\\) (integers between \\(-\\ell\\) and \\(+\\ell\\)).\r\n\r\n\r\n\t\\(m_{\\ell}=0=\\) one \\(s\\) orbital\r\n\t\t\\(m_{\\ell}=0,\\,\\pm1=\\) three \\(p\\) orbitals\r\n\t\t\\(m_{\\ell}=0,\\,\\pm1,\\,\\pm2=\\) five \\(d\\) orbitals.\r\n\r\n\r\n \tThe spin quantum number (\\(m_{s}\\)) represents the direction of electron spin. These are often shown as arrows in orbital \"boxes\". \\(m_{s}=+\\dfrac{1}{2}\\) or \\(-\\dfrac{1}{2}\\), often represented as\u00a0\\(\\uparrow\\)\u00a0or\u00a0\\(\\downarrow\\), indicate spin up or down. \r\n\r\n\r\nExample \u2013 determining quantum numbers\r\nFor an electron with a quantum number of 2, \\(n=2\\), what are the possible values for \\(\\ell\\) and \\(m_{\\ell}\\)?\r\nStep 1: Recall what \\(\\ell\\) is and how it is determined.\r\n\\(\\ell\\) is the orbital quantum number. The possible values for \\(\\ell\\) are any positive integers less than \\(n\\). Therefore, if \\(n=2\\), then \\(\\ell=0,\\,1\\).\r\nStep 2: Recall what \\(m_{\\ell}\\) is and what orbitals they represent.\r\nThe \\(m_{\\ell}\\) is \\(\\ell\\). Since \\(\\ell=0,\\,1\\):\r\n\r\n\t\\(m_{\\ell}=0\\) for \\(\\ell=0\\). This indicates one orbital. Since \\(n=2\\), it is one \\(2s\\) orbital.\r\n\t\\(m_{\\ell}=0,\\pm1\\) for \\(\\ell=1\\). This indicates three separate orbitals (\\(m_{\\ell}=0,-1,+1\\)). Since \\(n=2\\), it is three \\(2p\\) orbitals.\r\n","n":0.059},[{"v":"Quantum","i":3,"n":1},{"v":"Electron","i":2,"n":1},{"v":"Chemistry","i":1,"n":1},{"v":"Atomic Structures","i":0,"n":0.707}]]},{"i":305,"$":[{"v":"Poisson distribution","n":0.707},{"v":"\r\nThe Poisson distribution deals with the number of random occurrences over a period of time or a given space, such as the number of people who enter a shop every hour or the number of flaws in a sheet of glass. It is very handy, well beyond applications in STEM. Use this resource to learn how to use the Poisson distribution.\r\n\r\nThe Poisson distribution is a discrete distribution like the binomial distribution. This means that it involves events that we can count. It measures the probability that a certain number of random independent events occur within a period of time (or distance or area, etc.) when the average rate of occurrence is known.\r\nIt was developed in the \\(1800\\)s when mathematicians were asked to theorise on such matters as the annual number of wrongful criminal convictions in a particular country or to work out the number of soldiers in the Prussian cavalry who were likely to be kicked to death by a horse (honest!)\r\nSome more modern examples of events described by a Poisson distribution distribution are:\r\n\r\n \tthe number of errors or crashes a software system experiences in a day\r\n\tnumber of online orders received for a particular item per hour during a sale\r\n \tthe numbers of cars that pass on a stretch of road over a certain time period\r\n\tthe number of stars in a given area of a telescope's field of view\r\n\tthe number of communication failures within a specific network area\r\n \tthe number of phone calls a customer support centre receives per minute.\r\n\r\n\r\nPoisson probabilities\r\nPoisson probabilities are calculated using:\r\n\\[\\Pr(X=x)=\\frac{e^{-\\lambda}\\lambda^{x}}{x!}\\]\r\nwhere \\(\\lambda\\) is the mean or expected number of occurrences per unit of time, distance, area or volume.\r\nThe Poisson distribution varies with \\(\\lambda\\).\r\n\r\n\r\n\tAs \\(\\lambda\\) increases, the shape of the Poisson distribution becomes more spread out and symmetric. The average rate of occurences is higher, so the distribution looks more like a normal distribution.\r\n\tFor smaller \\(\\lambda\\) values, most probabilities cluster around \\(x=0\\). In other words, when the average rate of occurence is small, the likelihood of observing a higher number of events is very small.\r\n\r\n\r\nExample \u2013 calculating Poisson probabilities\r\nSuppose that over the past years, a company receives an average of \\(3\\) defective returns from customers each month. What is the probability that next month there will be \\(4\\) defective returns?\r\nHere, \\(\\lambda=3\\) as it is the average or expected value. The number of successes is \\(x=4\\). We can use these values to calculate the probability.\r\n\\[\\begin{align*} \\Pr(X=x) & = \\frac{e^{-\\lambda}\\lambda^{x}}{x!}\\\\\r\n& = \\frac{e^{-3}3^{4}}{4!}\\\\\r\n& = \\frac{81e^{-3}}{4\\times3\\times2\\times1}\\\\\r\n& = 0.168\r\n\\end{align*}\\]\r\n\r\nApproximating the binomial distribution\r\nOften, the normal distribution can be used to get fairly good approximations of binomial probabilities. This works best when \\(p\\) is close to \\(0.5\\). The normal distribution becomes closer to the binomial probability as \\(n\\) increases.\r\nAs a general rule, both \\(np\\) and \\(n(1-p)\\) should be greater than or equal to \\(5\\) in order to use the normal approximation.\r\nBut, what if these rules are not satisfied? If \\(p<0.05\\) and \\(n\\) is large, we can use the Poisson distribution to approximate the binomial distribution. When we do this, we calculate \\(\\lambda\\):\r\n\\[\\lambda=np\\]\r\n\r\nExample \u2013 approximating binomial distributions\r\nA sample of \\(50\\) plants are taken from a field, where \\(1\\%\\) of the plants are expected to have a certain pest infection. What is the probability that \\(2\\) of these plants are infested? Compare the binomial probability and the probability approximated using the Poisson distribution.\r\nUsing the binomial distribution, we have \\(n=50\\), \\(p=0.01\\) and \\(1-p=0.99\\). To calculate the probability for a binomially distributed data, we use \\(\\Pr(X=x)=^{n}C_{x}\\times p^{x}\\times(1-p)^{n-x}\\).\r\n\\[\\begin{align*} ^{n}C_{x} & = \\frac{n!}{x!(n-x)!}\\\\\r\n& = \\frac{50!}{2!(50-2)!}\\\\\r\n& = 1225\r\n\\end{align*}\\] \\[\\begin{align*} \\Pr(X=2) & = ^{n}C_{x}\\times p^{x}\\times(1-p)^{n-x}\\\\\r\n& = ^{50}C_{2}\\times p^{2}\\times(1-50)^{50-2}\\\\\r\n& = 1225\\times0.01^{2}\\times0.99^{48}\\\\\r\n& = 0.0756\r\n\\end{align*}\\]\r\nTo approximate the binomial probability using the Poisson distribution, we first need to calculate \\(\\lambda\\).\r\n\\[\\begin{align*} \\lambda & = np\\\\\r\n& = 50\\times0.01\\\\\r\n& = 0.5\r\n\\end{align*}\\]\r\nThen, we can use the Poisson approximation equation.\r\n\\[\\begin{align*} \\Pr(X=2) & = \\frac{e^{-\\lambda}\\lambda^{x}}{x!}\\\\\r\n& = \\frac{e^{-0.5}0.5^{2}}{2!}\\\\ \r\n& = 0.0758\r\n\\end{align*}\\]\r\nThe binomial probability is \\(0.0756\\) and the approximation is \\(0.0758\\). You can see that the Poisson distribution gives a good approximation. It would be even more accurate if \\(n\\) was larger.\r\n\r\nExercise \u2013 approximating binomial distributions\r\n\r\n\tOn average, one in every \\(100\\) births produces twins. Use the Poisson distribution to find the probability that there is one set of twins in the next \\(250\\) births.\r\n\tA manufacturer of electric light globes knows from past experience that \\(2\\%\\) of globes produced are defective. Use the Poisson distribution to approximate the probability that, out of the next \\(200\\) globes, less than \\(2\\) are defective?\r\n\tThe chance of death by drowning is \\(\\dfrac{1}{30000}\\). Use the Poisson distribution to estimate the probability that, out of a random selection of \\(1500\\) people, there will be at least one death by drowning.\r\n\tInsurance companies have found that the probability that a residential property is damaged by fire at some stage over a \\(25\\) year period is \\(0.01\\). If a company insures \\(20\\) homes in a particular street, use the Poisson distribution to approximate the probability that \\(2\\) of them will be damaged by fire in the next \\(25\\) years.\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.035},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":306,"$":[{"v":"Tests of proportion","n":0.577},{"v":"\r\nRather than comparing a sample mean to a population mean, we can compare a sample proportion to a population proportion. These tests of proportion are used broadly, like for checking if a new fabric meets quality standards by having fewer defects or evaluating customer satisfaction levels for financial services. Use this resource to learn how to perform tests of proportions.\r\n\r\nA test of proportion is used to determine whether or not a sample from a population represents the true proportion from the entire population. \r\nTests of proportion are useful when we want to compare categorical data, like the percentage of defective items in a batch or the proportion of people preferring a new product over an old one. They help organisations make informed decisions based on sample data.\r\n\r\nConducting tests of proportions\r\nAgain, the process of conducting a test of proportions is similar to what you have already encountered for \\(z\\)- and t-tests.\r\n\r\n \tState the null and alternative hypotheses. We can let \\(\\hat{p}\\) (or \\(\\pi\\)) be the sample proportion and \\(p\\) be the null hypothesised population proportion. For a two-tailed test: \r\n\t\r\n\t\t\\(H_{0}:\\hat{p}=p\\) (the proportion of the population from which the sample is drawn is the same as the given population proportion)\r\n\t\t\\(H_{a}:\\hat{p}\\neq p\\) (the proportion of the population from which the sample is drawn is not the same as the given population proportion)\r\n\t\r\n\tChoose a significance level. Often, we choose \\(\\alpha=0.05\\).\r\n \tFind the critical values \\(z_{c}\\) using a \\(z\\)-table.\r\n\t\r\n\tIf \\(np>5\\) and \\(n(1-p)>5\\), we assume that the test statistic is normally distributed. If these are not satisfied, we should make note of this, or use a different and more exact test.\r\n\tCalculate the test statistic.\r\n\t\\[z = \\frac{ \\hat{p}-p}{\\sqrt{\\frac{p(1-p)}{n}}}\\]\r\n \tCompare the test statistic to the critical value to make a decision about the null hypothesis. If the test statistic is more extreme than the critical value, we reject the null hypothesis. If not, we do not reject the null hypothesis.\r\n \tDraw a conclusion from the hypothesis test.\r\n\t\r\n\t\tIf we have rejected the null hypothesis, this means that the proportion of the population from which the sample is drawn is not the same as the given population proportion after allowing for chance variation.\r\n\t\tIf we have not rejected the null hypothesis, this means that the proportion of the population from which the sample is drawn is the same as the given population proportion after allowing for chance variation.\r\n\t\r\n\r\n\r\nExample \u2013 conducting tests of proportion\r\nA lecturer knows that, historically, \\(60\\%\\) of students do not begin their statistics assignment until within seven days of the due date. As a result, they do not do well. To address this unfortunate reality, Maths support services are widely promoted during the early weeks of the semester. \r\nA survey of \\(80\\) randomly selected students revealed that \\(42\\) had started the assignment before the final week. Is there evidence to suggest the promotion had been effective at the \\(10\\%\\) significance level?\r\n\r\n\tIn this case, \\(\\hat{p}=\\dfrac{42}{80}=0.525\\). The hypotheses would be:\r\n\t\r\n\t\t\\(H_{0}:\\hat{p}=0.6\\)\r\n\t\t\\(H_{a}:\\hat{p}<0.6\\)\r\n\t\r\n\tThe significance level has already been set for us at \\(\\alpha=0.10\\).\r\n\tTo make sure that the test statistic is normally distribution, we need to check that \\(np>5\\) and \\(n(1-p)>5\\).\r\n\t\r\n\t\t\\(np=48\\)\r\n\t\t\\(n(1-p)=32\\)\r\n\t\r\n\tIt is normally distributed, so we can use the \\(z\\)-table to find the critical value, \\(z=-1.28\\).\r\n\t\r\n\tWe then find the test statistic.\r\n\t\\[\\begin{align*} z & = \\frac{ \\hat{p}-p}{\\sqrt{\\frac{p(1-p)}{n}}}\\\\\r\n\t& = \\frac{0.525-0.6}{\\sqrt{\\frac{0.6(1-0.6)}{80}}}\\\\\r\n\t& = -1.37\r\n\t\\end{align*}\\]\r\n\t\\(-1.37\\) is more extreme than \\(-1.28\\); therefore, we reject the null hypothesis.\r\n\tThere is evidence to suggest that the promotion has been effective and that students are beginning their assignments earlier.\r\n\r\n\r\nExercises \u2013 conducting tests of proportion\r\n\r\n\tA survey conducted in \\(2012\\) showed that \\(18\\%\\) of students were the first in their family to attend university. To evaluate widening participation programs, a randomly selected sample of \\(120\\) enrolment records indicated that in \\(2017\\), this number had increased to \\(24\\%\\). Does this data provide convincing evidence to suggest that the percentage of \"first in family\" students has changed over these five years? Use \\(\\alpha=0.10\\).\r\n\tA car manufacturer advertises that at least \\(80\\%\\) of vehicles will obtain fuel economy of \\(7\\textrm{ L}/100\\textrm{ km}\\) or better. To test the claim, a consumer interest group randomly selects \\(22\\) vehicles and finds that only \\(15\\) achieve this figure. Is there evidence that the company's claim is false at the \\(5\\%\\) level of significance?\r\n\tIn a rural community hospital, \\(6\\) out of \\(7\\) babies born in a particular week were boys. Assuming that in the general population, the probability of female and male babies is equal, test the claim that babies born at this hospital are more likely to be boys at the \\(10\\%\\) level of significance.\r\n\r\n\r\n\r\n\r\n\r\n","n":0.037},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":307,"$":[{"v":"One-tailed tests","n":0.707},{"v":"\r\n\r\nOne-tailed hypothesis tests are useful when you have a clear idea about the direction of an effect, such as testing if a new drug performs better than the standard treatment, whether a new fabric treatment increases the durability of a textile, or if a new alloy used for aircrafts is stronger than the minimum required strength for safety reasons. Use this resource to learn how to conduct one-tailed hypothesis tests.\r\n\r\nTesting two-sided claims\r\nIn the hypothesis tests we have looked at so far, we have looked for evidence that there has been a change in the population mean. The null and alternative hypotheses have been:\r\n\r\n\t\\(H_{o}:\\mu_{\\overline{x}}=\\mu\\) (the sample mean is the same as the population mean after allowing for chance variation)\r\n\t=\\(H_{a}:\\mu_{\\overline{x}}\\neq\\mu\\) (the sample mean is not the same as the population mean after allowing for chance variation).\r\n\r\nThese tests look for a change in either direction: has the population mean increased or has the population mean decreased? They involve two-sided claims.\r\nNon-directional tests are called two-sided (or two-tailed) tests. The two-sided approach helps you focus on detecting differences in a specific direction, providing more power to find a significant effect if it exists.\r\n\r\nTesting one-sided claims\r\nWhen claims use words like \"greater\", \"less than\", \"lower\", higher\", \"more\", or similar, they are one-sided claims. They involve changes in one direction. Examples are:\r\n\r\n \tQueenslanders have a greater chance of skin cancer.\r\n \tThe drop-out rate for Business students is lower than the university average.\r\n \tDrug A is more effective than drug B.\r\n \tBattery X lasts longer than \\(300\\) minutes.\r\n \tSuperannuation company Bib outperformed superannuation company Bob in \\(2017\\).\r\n\r\nWhen we test hypotheses related to one-sided claims, we use one-sided (or one-tailed) tests.\r\nThe null and alternative hypotheses for one-tailed tests may be:\r\n\r\n\t\\(H_{0}:\\mu_{\\overline{x}}\\leq\\mu\\) (the sample mean is not greater than the population mean after allowing for chance variation)\r\n\t\\(H_{a}:\\mu_{\\overline{x}}>\\mu\\) (the sample mean is greater than the population mean after allowing for chance variation)\r\n\r\nOR\r\n\r\n\t\\(H_{0}:\\mu_{\\overline{x}}\\geq\\mu\\) (the sample mean is not less than the population mean after allowing for chance variation)\r\n\t\\(H_{a}:\\mu_{\\overline{x}}<\\mu\\) (the sample mean is less than the population mean after allowing for chance variation)\r\n\r\n\r\nCritical values\r\nIn a two-sided test with a significance level of \\(\\alpha\\), there are two critical values and the area of each upper and lower rejection region is \\(\\dfrac{\\alpha}{2}\\).\r\n\r\n\tIf the test statistic is smaller than (or to the left of) the critical value in a one-sided test, the sample mean is smaller than the claimed mean. For these claims, we use a left-tailed test.\r\n\t\r\n\r\n\r\nExample \u2013 conducting one-sided tests\r\nA baking company claims that the amount of meat in its pies is at least \\(50\\textrm{ g}\\). However, a consumer group has had many complaints and decides to conduct a test to check if the pies are being underfilled. A random sample of \\(10\\) pies resulted in the following data:\r\n\r\nIs there evidence in this sample that the pies are being underfilled (\\(\\alpha=0.05\\))?\r\nFrom the data, we can calculate \\(\\overline{x}\\) and \\(s\\).\r\n\\[\\begin{align*} \\overline{x} & = \\frac{\\sum{x_{n}}}{n}\\\\\r\n& = \\frac{477}{10}\\\\\r\n& = 47.7\r\n\\end{align*}\\] \\[\\begin{align*} s & = \\sqrt{\\frac{\\sum(x-\\overline{x})^{2}}{n-1}}\\\\\r\n& = \\frac{30.6}{10-1}\\\\\r\n& = 3.4\r\n\\end{align*}\\]\r\nWe can then state the null and alternate hypotheses.\r\n\r\n\t\\(H_{o}:\\mu_{\\overline{x}}\\geq50\\)\r\n\t\\(H_{a}:\\mu_{\\overline{x}}<50\\)\r\n\r\nWe know the significance level is \\(\\alpha=0.05\\) and since we only have the sample standard deviation \\(s\\), we use a t-test. As we are dealing with a directional claim, it is a one-sided t-test. The degrees of freedom is \\(n-1=10-1=9\\).\r\nFrom the t-table, the critical value is \\(-1.833\\). We need to check whether the test statistic is less than the critical value; this indicates that the sample mean is significantly smaller than the claimed mean.\r\n\r\nLet's calculate the test statistic.\r\n\\[\\begin{align*} t & = \\frac{\\overline{x}-\\mu}{s_{\\overline{x}}}\\\\\r\n& = \\frac{47-50}{\\frac{3.4}{\\sqrt{10}}}\\\\\r\n& = -2.14\r\n\\end{align*}\\]\r\nThe test statistic is indeed less than the critical value, so we reject the null hypothesis. There is enough evidence to suggest that the pies are being filled with less than \\(50\\textrm{ g}\\) of meat.\r\n\r\nExercise \u2013 conducting one-sided tests\r\n\r\n\tNutrition guidelines for an aged care facility recommend that no more than \\(71.1\\textrm{ mg}\\) of a particular vitamin be consumed each day. A random sample of \\(7\\) residents revealed \\(\\overline{x}=71.3\\) and \\(s=0.214\\). Use a test with \\(\\alpha=0.05\\) to decide if there is evidence of a problem with the dietary intake at this facility.\r\n\tA long-term analysis of Rabbit University students\u2019 study habits revealed that, on average, students study for \\(18\\) hours per week with a standard deviation of \\(4.2\\) hours. However, a random sample of \\(45\\) students in \\(2017\\) had a mean weekly study time of \\(14.68\\) hours. At the \\(1\\%\\) significance level, does this data suggest that students at Rabbit University are now studying less?\r\n\tAn IVF process resulted in \\(50\\) successful live births of which \\(33\\) were females. Use the p-value method to decide, at the \\(1\\%\\) level of significance, whether female children are more likely to result from this procedure.\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.036},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":308,"$":[{"v":"P-values","n":1},{"v":"\r\nAs well as comparing the test statistic to a critical value, we can test hypotheses by comparing a p-value to the significance level, \\(\\alpha\\). Understanding p-values helps us decide if our results are meaningful or just due to random chance. Use this resource to learn how to use p-values.\r\n\r\nIn hypothesis testing, there are two ways for us to make a decision about the null hypothesis. So far, we have compared the test statistic with the critical value/s and asked the question: Is the test statistic more extreme than the critical value?\r\n\r\n\tIf yes, we reject the null hypothesis.\r\n\tIf no, we do not reject the null hypothesis.\r\n\r\nThe other approach is to calculate the p-value and compare it with \\(\\alpha\\).\r\nThe p-value is a probability. It is a measure of the likelihood of the test statistic being obtained if the null hypothesis is true. If the probability is lower than \\(\\alpha\\), we reject the null hypothesis. Our question becomes: Is \\(p<\\alpha\\)?\r\n\r\n\tIf yes, we reject the null hypothesis.\r\n\tIf no, we do not reject the null hypothesis.\r\n\r\n\r\nCalculating p-values\r\nThe p-value differs depending on whether we are looking at a two-tailed or one-tailed test.\r\n\r\np-values in two-tailed tests\r\nFor a two-tailed test where \\(x_{o}\\) is the test statistic, the p-value is the sum of the areas of the two regions beyond \\(-\\left|x_{o}\\right|\\) and \\(\\left|x_{o}\\right|\\).\r\n\r\n\r\np-values in one-tailed tests\r\nFor a one-tailed test where \\(x_{o}\\) is the test statistic, the p-value is either the area to the left of \\(x_{o}\\) for a left-sided test, or the area to the right of \\(x_{o}\\) for a right-sided test.\r\n\r\n\r\nUsing p-values\r\nThe steps for a hypothesis test using the p-value are:\r\n\r\n \tState the null and alternative hypotheses.\r\n \tChoose the level of significance, \\(\\alpha\\).\r\n \tCalculate the test statistic.\r\n \tFind the p-value of the test statistic. This number is obtained from a graphics calculator or an online tool like Stat Trek.\r\n \tDecide whether to reject the null hypothesis by comparing the p-value to \\(\\alpha\\).\r\n \tState your conclusion.\r\n\r\n\r\nExample \u2013 using p-values\r\nThe student queries service at a university claim that during orientation week, the average wait time to be seen is \\(23\\) minutes. However, this has been disputed, as feedback from student focus groups indicates the wait is much longer.\r\nTo test the hypothesis, the wait time for \\(40\\) randomly selected students is recorded during the next orientation week. The average wait time is found to be \\(25\\) minutes with a standard deviation of \\(5\\) minutes. Is there evidence to support the hypothesis that the average wait time exceeds \\(23\\) minutes at a \\(5\\)% level of significance?\r\n\r\n \tHypotheses:\r\n\t\r\n\t\t\\(H_{0}:\\mu=23\\)\r\n\t\t\\(H_{a}:\\mu>23\\)\r\n\t\r\n \tSignificance level: \\(\\alpha=0.05\\)\r\n \tThe population standard deviation is unknown, so we use a t-test. The test statistic, \\(t\\) is:\r\n\t\\[\\begin{align*} t & = \\frac{\\overline{x}-\\mu}{s_{\\overline{x}}}\\\\\r\n\t& = \\frac{25-23}{\\frac{5}{\\sqrt{40}}}\\\\\r\n\t& = 2.53\r\n\t\\end{align*}\\]\r\n\tThe p-value is the area to the right of \\(2.53\\). We are looking at a right-tailed test because we want to know if the sample mean is greater than the claimed mean.\r\n\t\r\n\tThe area is \\(0.0078\\).\r\n\tTo make a decision, we check if \\(p<\\alpha\\). \\(0.0078<0.05\\), so we reject the null hypothesis.\r\n\tThere is evidence to suggest that the average wait time for student queries during orientation week is more than \\(23\\) minutes.\r\n\r\n\r\nExercise \u2013 using p-values\r\n\r\n\tA sample of \\(10\\) is drawn from a normally distributed population and the mean is found to be \\(37.22\\). Use the p-value method to see if the sample mean is significantly less than the population mean of \\(40\\). The population standard deviation is known to be \\(3.27\\). Use \\(\\alpha=0.05\\).\r\n\tTraditionally, it is believed that \\(35\\%\\) of RMIT students are aware of the SLC Maths drop-in service. Management have recently implemented a new marketing strategy and results of a new survey show that out of the \\(100\\) randomly sampled students, \\(42\\) knew of the service. \r\n\tIs there evidence to suggest that the proportion of students at RMIT who are aware of the Maths drop-in service has changed after the implementation of the marketing initiative? Use \\(\\alpha=0.1\\) with the p-value method.\r\n\tThe manufacturer of Branbix cereal claims the mean weight of contents of each box is \\(450\\textrm{ g}\\). A consumer interest magazine weighs a sample of \\(100\\) boxes and finds that they have a mean of \\(444\\textrm{ g}\\) with a standard deviation of \\(28\\textrm{ g}\\).\r\n\tUse the p-value method to decide whether the sample data contradicts the manufacturers and the boxes are being underfilled? Use \\(\\alpha=0.01\\).\r\n\tThe university is considering opening an evening Maths drop-in service, but only on the condition that attendance would be at least \\(20\\%\\) of the weekly daytime average. A random survey of \\(400\\) attendees indicated that \\(92\\) would attend in the evening. Use the p-value method \\((\\alpha=0.1)\\) to decide what the drop-in service should do.\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.036},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":309,"$":[{"v":"T-tests","n":1},{"v":"\r\nT-tests help us understand if the differences we observe in data are due to random chance or if they're meaningful. They are commonly used in research to compare groups, like testing if a new teaching method improves scores more than the old one. Use this resource to learn how to use t-tests to test hypotheses.\r\n\r\nT-distribution\r\nHypothesis testing usually uses the population standard deviation (\\(\\sigma\\)) to calculate a \\(z\\)-score. If the population standard deviation is unknown, we approximate using the sample standard deviation (\\(s\\)):\r\n\\[s_{\\overline{x}}=\\frac{s}{\\sqrt{n}}\\]\r\nA sampling distribution that results from approximating \\(\\sigma\\) with \\(s\\) is called a t-distribution. The t-distribution curve is a bit distorted compared to the standard normal distribution. This is because \\(\\sigma\\) is a fixed value and \\(s\\) varies with each sample. The varied \\(s\\) creates a level of uncertainty. \r\nThe precise shape of the t-distribution depends on the size of the sample. It is always bell-shaped but varies slightly according to the size of the sample \\(n\\). As the sample size increases, a t-distribution becomes more like the standard normal distribution distribution. For small \\(n\\) values, it has heavier tails.\r\n\r\n\r\nT-tests\r\nLike the \\(z\\)-test, a t-test is a hypothesis test used to make decisions about the similarity of sample and population means.\r\nIt involves calculating a test statistic called the t-value:\r\n\\[t = \\frac{\\overline{x}-\\mu}{\\frac{s}{\\sqrt{n}}} \\]\r\nYou might notice that the equation is almost identical to the test statistic for a \\(z\\)-test, but \\(s\\) is used instead of \\(\\sigma\\).\r\nLike a \\(z\\)-test, we assume that the sample data is drawn randomly and that the distribution of sample means follows a normal distribution.\r\n\r\nConducting t-tests\r\nThe process for conducting a t-test is similar to that for a \\(z\\)-test, but using an additional parameter and a \\(t\\)-table.\r\n\r\n \tState the null and alternate hypotheses.\r\n\t\r\n\t\t\\(H_{0}:\\overline{x}=\\mu\\) (the sample mean is the same as the population mean after allowing for chance variation)\r\n\t\t\\(H_{a}:\\overline{x}\\neq\\mu\\) (the sample mean is not the same as the population mean after allowing for chance variation)\r\n\t\r\n\tChoose a significance level. Often, we choose \\(\\alpha=0.05\\). In other words, we define \"reasonable\" as what we can expect \\(95\\%\\) of the time.\r\n\tFind the critical values (\\(t_{c}\\)) that correspond to the chosen significance level, \\(\\alpha\\). Te critical t-value also depends on the degrees of freedom, \\(n-1\\). This is the number of separate pieces of information that we need to calculate a statistic. \r\n\tThese \\(t\\)-values are the cutoff points beyond which we reject the null hypothesis.\r\n\tCalculate the test statistic. This is the standardised difference between the sample mean (calculated from the data) and the known population mean:\r\n\t\\[t = \\frac{\\overline{x}-\\mu}{\\frac{s}{\\sqrt{n}}}\\]\r\n\tCompare the test statistic to the critical value to make a decision about the null hypothesis.\r\n\t\r\n\t\tIf the test statistic falls beyond the cutoff point/s of the critical value, we reject the null hypothesis with \\(5\\%\\) confidence, as per the chosen significance level.\r\n\t\tIf the test statistic falls within the cuttoff point/s of the critical value, we do not reject the null hypothesis with \\(5\\%\\) confidence, as per the chosen significance level.\r\n\t\r\n\tDraw a conclusion from the hypothesis test.\r\n\t\r\n\t\tIf we have rejected the null hypothesis, this means that the sample mean is not the same as the population mean after allowing for chance variation.\r\n\t\tIf we have not rejected the null hypothesis, this means that the sample mean is the same as the population mean after allowing for chance variation.\r\n\t\r\n\r\nJust like for \\(z\\)-tests, remember:\r\n\r\n\tThe decision about the null hypothesis is not made with certainty but with a level of confidence that the error in the decision is small; for example, \\(5\\%\\) if \\(\\alpha=0.05\\).\r\n\tTerminology is important! The decision relates only to rejecting or not rejecting the null hypothesis, \\(H_{0}\\). We never make a decision that references the alternate hypothesis, \\(H_{a}\\), nor do we say we \"accept\" either hypothesis.\r\n\r\n\r\nT-test or \\(z\\)-test?\r\nSometimes, it can be confusing whether to use a \\(z\\)-test or a t-test to test your hypothesis.\r\nGenerally:\r\n\r\n \tuse a \\(z\\)-test if you know \\(\\sigma\\)\r\n \tuse a t-test if you do not know \\(\\sigma\\).\r\n\r\nBut there may be some exceptions!\r\n\r\n \tMany textbooks only provide t-tables for \\(n\\leq30\\). For larger samples, the standard normal distribution is a reasonable approximation and a \\(z\\)-test may be used even though \\(\\sigma\\) is unknown.\r\n \tSome courses provide students with more comprehensive t-tables for \\(n\\) up to \\(50\\) or \\(100\\). If you have access to such tables, you should use a t-test whenever \\(\\sigma\\) is not known, no matter the size of the sample.\r\n \tIf you know that \\(s\\) is a good approximation for \\(\\sigma\\), then a \\(z\\)-test may be appropriate.\r\n \tIf a computer package such as SPSS, Minitab or Microsoft Excel is being used for the hypothesis test, then a t-test should be selected when \\(\\sigma\\) is unknown.\r\n\r\n\r\nExample \u2013 conducting t-tests\r\nA manufacturer of batteries claims that, on average, a battery lasts \\(200\\) hours. To test this claim, \\(7\\) batteries were randomly selected and the sample had an average of only \\(190\\) hours with a standard deviation of \\(9\\) hours. Does this sample provide evidence at the \\(1\\)% level of significance that the manufacturers claim is incorrect?\r\n\r\n\tWe start with the hypotheses.\r\n\t\r\n\t\t\\(H_{0}:\\mu_{\\overline{x}}=200\\)\r\n\t\t\\(H_{a}:\\mu_{\\overline{x}}\\neq200\\)\r\n\t\r\n\tThe significance level has been given to us in the question: \\(1\\%\\) means \\(\\alpha=0.01\\).\r\n\tWe use the \\(t\\)-table to find the critical values for \\(\\alpha=0.01\\) and degrees of freedom\\(=n-1=6\\).  You can go back to Confidence intervals if you need some help finding these.\r\n\r\n\tFor \\(\\alpha=0.01\\), \\(t=-3.7074\\) or \\(z=3.7074\\). The interval of interest would be \\(\\left[-3.7074,3.7074\\right]\\).\r\n\r\n\tWe use the formula to calculate the test statistic.\r\n\t\\[\\begin{align*} t & = \\frac{\\overline{x}-\\mu}{\\frac{s}{\\sqrt{n}}}\\\\\r\n\t& = \\frac{190-200}{\\frac{9}{\\sqrt{7}}}\\\\\r\n\t& = -2.94\r\n\t\\end{align*}\\]\r\n\tThe test statistic does not fall outside of the cutoffs \\(-3.7074\\) and \\(3.7074\\), so we do not reject the null hypothesis.\r\n\tFinally, we can draw a conclusion. Based on the hypothesis test, there is not enough evidence to suggest that the average battery life differs from \\(200\\) hours.\r\n\r\n\r\nExercise \u2013 conducting t-tests\r\n\r\n\tThe number of nic nac lollies in a small pack follows a normal distribution and the average is claimed to be \\(40\\). A suspicious customer buys \\(10\\) packs and counts the nic nacs in each pack. The results are: \\(38\\,45\\,36\\,40\\,42\\,44\\,35\\,41\\,36\\,40\\). Use a t-test with \\(\\alpha=0.05\\) to decide whether the customer's suspicion is justified.\r\n\tA hospital claims a new process will reduce the waiting time for surgery to treat a non-life threatening condition. To date, the average waiting time has been \\(15.7\\) months. A random sample of \\(33\\) patients diagnosed with the condition were observed and the time until surgery was recorded (\\(\\overline{x}=13.664\\) and \\(s=2.544\\)). Test the hospitals claim at the \\(1\\)% significance level.\r\n\tThe quality of a pre-natal program for at-risk mothers is to be assessed by comparing the birth weight of newborn babies to the historical average of \\(2800\\textrm{ g}\\). The babies of the \\(25\\) mothers in the program had a mean birth weight of \\(3075\\textrm{ g}\\) and a standard deviation of \\(300\\textrm{ g}\\). Comment on the success of the program using a t-test and \\(\\alpha=0.01\\).\r\n\r\n\r\n\r\n\r\n\r\n","n":0.03},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":310,"$":[{"v":"Hypothesis testing","n":0.707},{"v":"\r\nHypothesis testing is used anywhere we conduct research to determine whether there is a relationship between variables. These days, hypothesis testing forms the basis of machine learning and data science, where it is used to validate models and make sure they reliably detect patterns and make predictions. Use this resource to learn how to set up and test hypotheses.\r\n\r\nHypotheses and experiments\r\nA hypothesis is an informed prediction or explanation of a phenomenon. Often, it connects two or more variables, proposing the effect of one on another, or the relationship between them.\r\nHypothesis testing is a method we use to decide if there is enough evidence to support a specific claim about a population. It helps us determine whether our observations are due to chance or if there is a real effect, by looking at the maths.\r\nThis involves defining a null hypothesis (\\(H_{0}\\)), which says that the effect being studied does not exist. The claim that the effect exists is called the alternate hypothesis (\\(H_{a}\\)). The table gives some examples of null hypotheses and alternate hypotheses.\r\n\r\n\t\r\n\t\t\r\n\t\t\tNull hypothesis (\\(H_{0}\\))\r\n\t\t\tAlternate hypothesis (\\(H_{a}\\))\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tTeenagers aged \\(13-15\\) spend more than \\(10\\) hours a week on Instagram.\r\n\t\t\tTeenagers aged \\(13-15\\) spend no more than \\(10\\) hours a week on Instagram.\r\n\t\t\r\n\t\t\r\n\t\t\tThe average weight of Australian men is different from what it was in \\(1990\\).\r\n\t\t\tThe average weight of Australian men is the same as it was in \\(1990\\).\r\n\t\t\r\n\t\t\r\n\t\t\tStudents from private schools have a different mean ATAR score to the Victorian average.\r\n\t\t\tStudents from private schools have the same mean ATAR score as the Victorian average.\r\n\t\t\r\n\t\t\r\n\t\t\tThe mean winter rainfall for the last \\(10\\) years is the same as the historical mean.\r\n\t\t\tThe mean winter rainfall for the last \\(10\\) years is different from the historical mean.\r\n\t\t\t\t\r\n\t\r\n\r\n\r\nOur confidence about the probabilities of values drawn from normally distributed populations and sampling distributions lets us formally test hypotheses (or claims).\r\n\r\nChance variation\r\nWhen we perform an experiment to test a hypothesis, we know there will be chance variation. For example, if we toss a supposedly fair coin \\(100\\) times, we would not be surprised to obtain \\(48\\) or \\(45\\) or perhaps even \\(40\\) heads. However, we would be surprised to obtain only \\(5\\) heads. \r\nIf we were testing a coin for fairness, we might decide beforehand what we would consider a reasonable number of heads. In hypothesis testing, reasonable is defined as what we could expect \\(95\\%\\) (or \\(99\\%\\) or \\(90\\%\\), etc) of the time.\r\nIn a hypothesis test, we assess how unusual our result is:\r\n\r\n\twhether it is reasonable chance variation (like obtaining \\(45\\) heads in \\(100\\) tosses of a coin); or\r\n\twhether the result is too extreme to be considered chance variation (like obtaining \\(5\\) heads in \\(100\\) tosses of a coin).\r\n\r\nA hypothesis test formalises the process of deciding whether a result is reasonable. It helps us see if a difference is likely due to random chance or if it is significant enough to suggest a true effect, beyond what we'd expect from natural variation.\r\n\r\nStatistical significance\r\nIn hypothesis testing, statistical significance helps us determine if the observed relationship is likely to be real or just due to chance. We use a significance level (\\(\\alpha\\)), often set at \\(0.05\\) or \\(5\\%\\), as a threshold.\r\nThe significance level is how much chance variation we are willing to accept. If we find that the \\(z\\)-score falls outside of the range we consider reasonable, we say that the effect is statistically significant and not likely due to random chance.\r\nThis helps us make informed decisions about whether the findings support our hypothesis.\r\n\r\nConducting \\(z\\)-tests\r\nHypothesis testing involves several steps. Let's look at the steps for testing the null hypothesis that there is no significant difference between a sample mean and a population mean, also called a \\(z\\)-test.\r\n\r\n\tState the null and alternate hypotheses.\r\n\t\r\n\t\t\\(H_{0}:\\overline{x}=\\mu\\) (the sample mean is the same as the population mean after allowing for chance variation)\r\n\t\t\\(H_{a}:\\overline{x}\\neq\\mu\\) (the sample mean is not the same as the population mean after allowing for chance variation)\r\n\t\r\n\tChoose a significance level. Often, we choose \\(\\alpha=0.05\\). In other words, we define \"reasonable\" as what we can expect \\(95\\%\\) of the time.\r\n\tFind the critical values (\\(z_{c}\\)) that correspond to the chosen significance level. These \\(z\\)-scores are the cutoff points beyond which we reject the null hypothesis.\r\n\t\r\n\tCalculate the test statistic. This is the standardised difference between the sample mean (calculated from the data) and the known population mean:\r\n\t\\[z = \\frac{\\overline{x}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}\\]\r\n\tCompare the test statistic to the critical value to make a decision about the null hypothesis.\r\n\t\r\n\t\tIf the test statistic falls beyond the cutoff point/s of the critical value, we reject the null hypothesis with \\(5\\%\\) confidence, as per the chosen significance level.\r\n\t\tIf the test statistic falls within the cuttoff point/s of the critical value, we do not reject the null hypothesis with \\(5\\%\\) confidence, as per the chosen significance level.\r\n\t\r\n\tDraw a conclusion from the hypothesis test.\r\n\t\r\n\t\tIf we have rejected the null hypothesis, this means that the sample mean is not the same as the population mean after allowing for chance variation.\r\n\t\tIf we have not rejected the null hypothesis, this means that the sample mean is the same as the population mean after allowing for chance variation.\r\n\t\r\n\r\nRemember:\r\n\r\n\tThe decision about the null hypothesis is not made with certainty but with a level of confidence that the error in the decision is small; for example, \\(5\\%\\) if \\(\\alpha=0.05\\).\r\n\tTerminology is important! The decision relates only to rejecting or not rejecting the null hypothesis, \\(H_{0}\\). We never make a decision that references the alternate hypothesis, \\(H_{a}\\), nor do we say we \"accept\" either hypothesis.\r\n\r\nThis resource outlines just one way of testing a hypothesis. You can read about ,  and .\r\n\r\nExample \u2013 conducting \\(z\\)-tests\r\nStudents had previously found a statistics course very difficult, with the average score over many years was \\(48\\%\\) with a standard deviation of \\(12\\%\\). A bridging program was introduced to support students, and in the first year the program ran, the \\(120\\) students that attended achieved a mean score of\\(50\\%\\) in the final exam. Is there evidence that the scores of those who attended the bridging program have changed, at a \\(1\\%\\) level of significance?\r\n\r\n\tWe start with the hypotheses.\r\n\t\r\n\t\t\\(H_{0}:\\overline{x}=48\\)\r\n\t\t\\(H_{a}:\\overline{x}\\neq48\\)\r\n\t\r\n\tThe significance level has been given to us in the question: \\(1\\%\\) means \\(\\alpha=0.01\\).\r\n\tWe use the \\(z\\)-table to find the critical values. You can go back to Confidence intervals if you need some help finding these.\r\n\tFor \\(\\alpha=0.01\\), \\(z=-2.58\\) or \\(z=2.58\\). The interval of interest would be \\(\\left[-2.58,2.58\\right]\\).\r\n\t\r\n\tWe use the formula to calculate the test statistic.\r\n\t\\[\\begin{align*} z & = \\frac{\\overline{x}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}\\\\ \r\n\t& = \\frac{50-48}{\\frac{12}{\\sqrt{120}}}\\\\\r\n\t& = 1.83\r\n\t\\end{align*}\\]\r\n\tThe test statistic \\(1.83\\) does not fall outside the cutoffs of \\(-2.58\\) and \\(2.58\\), so we do not reject the null hypothesis.\r\n\tFinally, we can draw a conclusion. Based on the hypothesis test, there is not enough evidence to suggest that the scores of those who attended the bridging program have changed. It is reasonable that the apparent improvement in scores was due to chance variation.\r\n\r\n\r\nExercise \u2013 conducting \\(z\\)-tests\r\n\r\n\tStudents had previously found a statistics course very difficult, with the average score over many years was \\(48\\%\\) with a standard deviation of \\(12\\%\\). A bridging program was introduced to support students, and in the first year the program ran, the \\(120\\) students that attended achieved a mean score of\\(50\\%\\) in the final exam. Is there evidence that the scores of those who attended the bridging program have changed, at a \\(10\\%\\) level of significance?\r\n\tA random sample of \\(36\\) soft drinks from vending machines had an average content of \\(370\\textrm{ mL}\\) with a standard deviation of \\(20\\textrm{ mL}\\). Test the null hypothesis that \\(\\mu=375\\textrm{ mL}\\) against the alternative hypothesis \\(\\mu\\neq375\\textrm{ mL}\\) at the \\(1\\%\\) significance level.\r\n\tA bank manager has historical data that shows over lunchtime Mon\u2013Fri, the mean number of customers that come into the bank is \\(32\\). Accordingly, he believes he has no need to change the number of tellers. However, a branch survey conducted every lunchtime over eight weeks found that the mean number of customers was \\(36\\) with a standard deviation of \\(8.2\\).\r\n\tConduct a hypothesis test with a \\(5\\%\\) level of significance to test whether the mean number of lunchtime customers has changed.\r\n\tWhat recommendation would you make to the bank manager?\r\n\tThe manufacturer of LongLast batteries claims the mean lifetime of his batteries is \\(450\\) hours. A consumer interest magazine samples \\(100\\) batteries and finds that they have a mean of \\(444\\) hours with a standard deviation of \\(28\\) hours. Does the sample data contradict the manufacturer's claim? Use \\(\\alpha=0.02\\)).\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.027},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":311,"$":[{"v":"Confidence intervals","n":0.707},{"v":"\r\nWhen we use the mean of a sample to estimate the mean of an entire population, we use confidence intervals to help us understand the precision of our estimates and how confident we can be about them. This tool is useful for making informed decisions based on sample data. Use this resource to learn how to use confidence intervals.\r\n\r\nWe use the statistics we obtain from samples to make inferences or estimates about the population from which the sample was drawn. \r\n\r\n \tA batch may be selected in a factory production process to assess how the process is operating.\r\n \tSurveys of consumers are used to determine the preferred brands in the population.\r\n \tPolls are conducted on samples of the voting population before elections to predict the result of the election.\r\n\r\nBut these estimates can vary from one sample to another, so we use tools called confidence intervals. They give us a range of values that likely include the true population parameter, like a mean or proportion.\r\nFor example, if we calculate a \\(95\\%\\) confidence interval, it means we expect the true value to fall within that range \\(95\\%\\) of the time if we repeated the experiment many times.\r\n\r\nConstructing confidence intervals\r\nReporting a confidence interval together with our estimate of the population parameter lets us make statements like: \"we are \\(95\\)% confident that the true mean weight of boxes of cereal labelled \\(450\\textrm{ g}\\) is in the interval \\([449.5,453.8]\\)\".\r\nFor large samples (\\(n\\geq30\\)), we can use the mean of a sample, \\(\\overline{x}\\), to find an interval estimate for the mean of the population, \\(\\mu\\):\r\n\\[\\mu=\\overline{x}\\pm\\left(z\\times\\frac{\\sigma}{\\sqrt{n}}\\right)\\]\r\nIf the standard deviation for the population data \\(\\sigma\\) is not known, we use the standard deviation for the sample \\(s\\):\r\n\\[\\mu=\\overline{x}\\pm\\left(z\\times\\frac{s}{\\sqrt{n}}\\right)\\]\r\n\r\nProbability and confidence intervals\r\nThe \\(z\\)-score is determined by the level of confidence and can be found using \\(z\\)-tables, a graphics calculator or statistics software.\r\nAs shown in the standard normal distribution curve, for a \\(95\\%\\) confidence interval, \\(z=1.96\\).\r\n\r\nSimilarly:\r\n\r\n\tFor a \\(99\\%\\) confidence interval, \\(z=2.575\\).\r\n\tFor a \\(90\\%\\) confidence interval, \\(z=1.645\\).\r\n\r\n\r\nExample \u2013 constructing confidence intervals\r\n\\(36\\) of a certain type of fish were caught in Port Phillip Bay. This sample had a mean length of \\(30\\textrm{ cm}\\) and a standard deviation of \\(3\\textrm{ cm}\\).\r\n\r\n \tConstuct a \\(95\\%\\) confidence interval for the true mean length of this type of fish.\r\n \tConstruct a \\(98\\%\\) confidence interval for the true mean length of this type of fish.\r\n\r\nFor the \\(95\\%\\) confidence interval, \\(z=1.96\\). We only have the standard deviation for the sample, \\(s=3\\). We substitute these, along with \\(\\overline{x}=30\\) and \\(n=36\\), into the equation to estimate the mean of the population.\r\n\\[\\begin{align*} \\mu & = \\overline{x}\\pm\\left(z\\times\\frac{s}{\\sqrt{n}}\\right)\\\\\r\n& = 30\\pm\\left(1.96\\times\\frac{3}{\\sqrt{36}}\\right)\\\\\r\n& = 30\\pm0.98\\\\\r\n& = \\left[29.02,30.98\\right]\r\n\\end{align*}\\]\r\nWe can state with \\(95\\%\\) confidence that the mean of the entire population of fish will be between \\(29.02\\textrm{ cm}\\) and \\(30.98\\textrm{ cm}\\).\r\nWe can do the same for the \\(98\\%\\) confidence. Looking this up on a \\(z\\)-table, we find that \\(z=2.326\\).\r\n\\[\\begin{align*} \\mu & = \\overline{x}\\pm\\left(z\\times\\frac{s}{\\sqrt{n}}\\right)\\\\\r\n& = 30\\pm\\left(2.326\\times\\frac{3}{\\sqrt{36}}\\right)\\\\\r\n& = 30\\pm1.163\\\\\r\n& = \\left[28.84,31.16\\right]\r\n\\end{align*}\\]\r\nWe can state with \\(98\\%\\) confidence that the mean of the entire population of fish will be between \\(28.84\\textrm{ cm}\\) and \\(31.16\\textrm{ cm}\\).\r\n\r\nExercise \u2013 constructing confidence intervals\r\n\r\n\tIn an effort to improve appointment scheduling, a doctor agreed to estimate the average time spent with each patient. A random sample of \\(49\\) patients had a mean of \\(30\\) minutes and a standard deviation of \\(7\\) minutes.\r\n\t\r\n\t\tConstruct a \\(95\\%\\) confidence interval for the true mean.\r\n\t\tConstruct an \\(80\\%\\) confidence interval for the true mean.\r\n\t\r\n\tTo estimate the average weight of males in the town of Cityville, a random sample of \\(100\\) men was drawn from the population of \\(10000\\) men and weights were recorded. The mean weight was found to be \\(83\\textrm{ kg}\\) and the standard deviation \\(12\\textrm{ kg}\\).\r\n\t\r\n\t\tConstruct a \\(99\\%\\) confidence interval for the mean weight of the male population.\r\n\t\tIn two of the suburbs of Cityville, Subtown and Tubtown, the mean weights for males were found to be \\(80\\textrm{ kg}\\) and \\(88\\textrm{ kg}\\), repectively. Comment on these results.\r\n\t\r\n\tAfter randomly sampling \\(400\\) individuals and obtaining a sample mean of \\(56.5\\), a research company was able to claim they were \\(90\\%\\) certain that the true mean of the population was between \\(56.089\\) and \\(56.911\\). What was the standard deviation of the sample?\r\n\r\n\r\n\r\n\r\n\r\n","n":0.038},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":312,"$":[{"v":"Sampling distributions","n":0.707},{"v":"\r\nSampling distributions help us understand the behaviour of sample statistics, like means or proportions, from different samples of the same population. By examining these distributions, we can see how sample results might vary and how close they are likely to be to the actual population value. This concept is important for making predictions and decisions based on sample data, such as estimating averages or testing hypotheses.\r\n\r\nA sampling distribution is the probability distribution for the means of all samples of size \\(n\\) from a specific, given population. It gives us an idea of the range of possible statistical outcomes for a population.\r\n\r\nPopulation parameters\r\nThere are two key parameters that we look at when we deal with sampling distributions and population data.\r\n\r\n\tThe mean of the sampling distribution, or\u00a0the mean of all the sample means, is denoted by \\(\\mu_{\\overline{x}}\\):\r\n\t\\[\\mu_{\\overline{x}}=\\mu\\]\r\n\twhere \\(\\mu\\) is the population mean.\r\n\tThe standard deviation of the sampling distribution is denoted by \\(\\sigma_{\\overline{x}}\\):\r\n\t\\[\\sigma_{\\overline{x}}=\\dfrac{\\sigma}{\\sqrt{n}}\\]\r\n\twhere \\(\\sigma\\) is the population standard deviation and \\(n\\) is the sample size.\r\n\r\nThe sampling distribution with parameters \\(\\mu_{\\overline{x}}\\) and \\(\\sigma_{\\overline{x}}\\) tends to follow a normal distribution, if either:\r\n\r\n \tthe population from which the samples are drawn is normally distributed; or\r\n \tthe samples are large \\((n\\geq30)\\).\r\n\r\n\r\nProperties of a sampling distribution\r\n\r\n \tThe sampling distribution has the same centre as the population.\r\n \t\\(\\sigma_{\\overline{x}}\\) is the measure of variability of a sampling distribution or the standard error.\r\n \tThe distribution of means is not as spread out as the values in the population from which the sample was drawn.\r\n \tIf we do not know the population standard deviation, we approximate with the sample standard deviation: \\(s_{\\overline{x}}\\approx\\sigma_{\\overline{x}}\\) and \\(\\dfrac{s}{\\sqrt{n}}\\approx\\dfrac{\\sigma}{\\sqrt{n}}\\) if the sample is large.\r\n\r\n\r\nProbability and sampling distributions\r\nWhen sampling distributions follow a normal distribution, we can find probabilities (\\(\\Pr(\\overline{X})\\)). See Probability and the normal distribution if you need a refresher.\r\nLet's consider the small population of values \\(P=1\\,2\\,3\\,4\\,5\\). This population has \\(\\mu=3\\) and \\(\\sigma=1.41\\). If a sample of size \\(n=3\\) was drawn from this population, it could be any one of:\r\n\\[\\begin{array}{cccc} (1\\,2\\,3) & (1\\,2\\,4) & (1\\,2\\,5) & (1\\,3\\,4)\\\\\r\n(1\\,3\\,5) & (1\\,4\\,5) & (2\\,3\\,4) & (2\\,3\\,5)\\\\\r\n(2\\,4\\,5) & (3\\,4\\,5)\r\n\\end{array}\\]\r\nThe means of each of the samples are shown in the table.\r\n\r\nFrom this, we can calculate the population mean and standard deviation:\r\n\\[\\begin{align*} \\mu_{\\overline{x}} & = \\frac{\\sum \\overline{x}_{n}}{n}\\\\\r\n& = \\frac{30}{10}\\\\\r\n& = 3\r\n\\end{align*}\\] \\[\\begin{align*} \\sigma_{\\overline{x}} & = \\frac{\\sigma}{\\sqrt{n}}\\\\\r\n& = \\sqrt {\\frac {\\sum (x-\\overline{x})^{2} } {n-1} } \\times \\frac{1}{\\sqrt{n}}\\\\\r\n& = \\frac{1.92}{\\sqrt{10}}\\\\\r\n& = 0.61\r\n\\end{align*}\\]\r\nBy converting the values to \\(z\\)-scores, we can find the probabilities where \\(\\Pr(\\overline{X}=\\overline{x})\\), as shown in the table.\r\n\r\nPlotting the distribution of means on a histogram, we can see that even though the sample is small and the population is not normally distributed (though it is symmetric), the sampling distribution is reasonably normally distributed.\r\n\r\nHere, the mean of the sampling distribution (the mean of all the means) is the same as the population mean, \\(\\mu_{\\overline{x}}=\\mu=3\\). \r\nHowever, the variability in the sampling distribution is less than that of the population: \\(\\sigma_{\\overline{x}}=0.61\\) and \\(\\sigma=1.41\\). \r\nLarger samples, or those drawn from normally distributed populations will follow a normal distribution, so we can use the properties of normal distributions to find probabilities relating to samples:\r\n\\[\\begin{align*} z_{\\overline{x}} & = \\frac{\\overline{x}-\\mu}{\\sigma_{\\overline{x}}}\\\\\r\n& = \\frac{\\overline{x}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}\r\n\\end{align*}\\]\r\n\r\nExample \u2013 calculating probabilities for sampling distributions\r\nThe shire of Bondara has \\(1200\\) preschoolers. The mean weight of preschoolers is known to be \\(18\\textrm{ kg}\\) with a standard deviation of \\(3\\textrm{ kg}\\). What is the probability that a random sample of \\(50\\) preschoolers will have a mean weight more than \\(19\\textrm{ kg}\\)?\r\nHere, \\(n=50\\), \\(\\mu=18\\) and \\(\\sigma=3\\). We are looking for \\(\\Pr(\\overline{X}>19)\\).\r\nThe sampling distribution of the means for samples of size \\(50\\) will have \\(\\mu_{\\overline{x}}=\\mu=18\\) and standard error:\r\n\\[\\begin{align*} \\sigma_{\\overline{x}} & = \\frac{\\sigma}{\\sqrt{n}}\\\\\r\n& = \\frac{3}{\\sqrt{50}}\\\\\r\n& = 0.42\r\n\\end{align*}\\]\r\nFrom this, we can convert the value of interest into a \\(z\\)-score, which we can then look up on a \\(z\\)-table.\r\n\\[\\begin{align*} z_{\\overline{x}} & = \\frac{\\overline{x}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}\\\\\r\n& = \\frac{19-18}{\\frac{3}{\\sqrt{50}}}\\\\ \r\n& = 2.38\r\n\\end{align*}\\]\r\nFinally, we can calculate the probability.\r\n\\[\\begin{align*} \\Pr(\\overline{x}>19) & = \\Pr(z_{\\overline{x}}>2.38)\\\\\r\n& = 1-0.9913\\\\\r\n& = 0.0087 \r\n\\end{align*}\\]\r\n\r\nExercise \u2013 calculating probabilities for sampling distributions\r\n\r\n\tList all samples of size \\(2\\) for the population \\((1\\,2\\,3\\,4\\,5\\,6)\\), then find the probability of obtaining a sample mean of less than \\(3\\).\r\n\tSamples of size \\(40\\) are drawn from a population with \\(\\mu=50\\) and \\(\\sigma=5\\).\r\n\t\r\n\t\tWhat are the mean and standard error of the sampling distribution?\r\n\t\tWhat is the probability that a particular sample has a mean less than \\(48.5\\)?\r\n\t\r\n\tIf IQ in the general population of secondary students is known to follow a normal distribution with \\(\\mu=100\\) and \\(\\sigma=10\\):\r\n\t\r\n\t\tFind the mean and standard error for a random sample of size \\(100\\).\r\n\t\tTo test whether a secondary school is representative of the general population, a sample of \\(100\\) students from that school is chosen. What is the probability of the mean IQ being more than \\(105\\)? What would be your conclusion?\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n","n":0.036},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":313,"$":[{"v":"Probability and the normal distribution","n":0.447},{"v":"To find probabilities in normal distributions that aren't standard, we need to adjust the data to a common scale using \\(z\\)-scores. This allows us to find the likelihood of certain outcomes in a range of contexts, like identifying manufacturing tolerances, testing the tensile strength of new materials or calculating the probability of electronic failure. Use this resource to learn how!\r\n\r\nEven when data follows a normal distribution, different data sets will have their own mean and standard deviation and a different bell shaped curve. More often than not, they do not follow the standard normal distribution.\r\n\r\nFor any normal distribution, the mean and standard deviation can be used to convert it to a standard normal distribution. We can then find probabilities using a \\(z\\)-table.\r\nEvery score in a normally distributed dataset, regardless of the shape, has an equivalent score in the standard distribution. \r\n\r\n\tThe mean of the normal distribution would be standardised to \\(0\\).\r\n\tOne standard deviation from the mean \\(\\mu\\pm\\sigma\\) would shift to \\(\\pm1\\).\r\n\tTwo standard deviations from the mean \\(\\mu\\pm2\\sigma\\) would shift to \\(\\pm2\\).\r\n\tThree standard deviations from the mean \\(\\mu\\pm3\\sigma\\) would shift to \\(\\pm3\\).\r\n\r\n\r\nConverting \\(z\\)-scores\r\nFor \\(z\\)-scores aside from the mean and standard deviation, we can use a formula:\r\n\\[z = \\dfrac{x-\\mu}{\\sigma}\\]\r\nRemember that the \\(z\\)-score is the number of standard deviations away from the mean.\r\n\r\nExample \u2013 converting \\(z\\)-scores\r\nIf we have a normal distribution with a mean of \\(1.2\\) and a standard deviation of \\(0.4\\),  find the standardised score for \\(x=1.7\\).\r\n\r\n\\[\\begin{align*} z & = \\frac{(x-\\mu)}{\\sigma}\\\\\r\n& = \\frac{(1.7-1.2)}{0.4}\\\\\r\n& = 1.25 \r\n\\end{align*}\\]\r\nA score of \\(1.7\\) in the distribution with mean \\(1.2\\) and standard deviation \\(0.4\\) is equivalent to a standardised score of \\(1.25\\). Alternatively, we could say that the score \\(1.7\\) is \\(1.25\\) standard deviations above the mean for that distribution.\r\n\r\nCalculating probabilities using \\(z\\)-scores\r\nOnce we have converted the scores of our distribution into standard scores or \\(z\\)-scores, we can use \\(z\\)-tables to calculate precise percentages and probabilities.\r\nRemember that the probability is the area under the curve. Because the total area under the standardised curve is \\(1\\), \\(\\Pr(z<\\beta)\\) is the area to the left of \\(\\beta\\).\r\n\r\nAs the normal distribution is a continuous distribution, we can find the probability that \\(x\\) is greater than or less than a particular value, but not that \\(x\\) is equal to a particular value.\r\n\r\nExample 1 \u2013 calculating probabilities using \\(z\\)-scores\r\nIf the mean maximum temperature for Melbourne in January is \\(25.9^{\\circ}\\textrm{C}\\) with a standard deviation of \\(2.1^{\\circ}\\textrm{C}\\), what is the probability that the mean maximum temperature for January \\(2015\\) will be above \\(28^{\\circ}\\textrm{C}\\)?\r\nAlways start with a diagram.\r\n\r\nThe \\(x\\) value on interest is \\(28\\), so we can go ahead and convert it.\r\n\\[\\begin{align*} z & = \\frac{(x-\\mu)}{\\sigma}\\\\\r\n& = \\frac{(28-25.9)}{2.1}\\\\ \r\n& = 1\r\n\\end{align*}\\]\r\nThis means that \\(\\Pr(x>28)\\) for the normal distribution is equivalent to \\(\\Pr(z>1)\\) on the standard normal distribution curve.\r\n\\[\\begin{align*} \\Pr(x>28) & = \\Pr(z>1)\\\\ \r\n& = 1-\\Pr(z<1)\\\\\r\n& = 1-0.8413\\\\\r\n& = 0.1587\r\n\\end{align*}\\]\r\nThe probability that the mean maximum temperature for January \\(2015\\) will be above \\(28^{\\circ}\\textrm{C}\\) is \\(0.1587\\).\r\n\r\n\r\n\r\nExercise \u2013 calculating probabilities using \\(z\\)-scores\r\nUse the following \\(z\\)-table to answer the questions.\r\n\r\n\r\n\tIf a population has a mean IQ of \\(100\\) and a standard deviation of \\(15\\), find the probability that an individual chosen at random will have an IQ:\r\n\t\r\n\t\tbetween \\(110\\) and \\(130\\)\r\n\t\tgreater than \\(87\\).\r\n\t\r\n\tA coffee machine is regulated to deliver \\(200\\textrm{ mL}\\) per cup. In reality, the amount of coffee varies, following a normal distribution with a mean of \\(200\\textrm{ mL}\\) and a standard deviation of \\(10\\textrm{ mL}\\). What is the probability that a cup will contain:\r\n\t\r\n\t\tless than \\(195\\textrm{ mL}\\)?\r\n\t\tmore than \\(220\\textrm{ mL}\\)?\r\n\t\tbetween \\(195\\) and \\(215\\textrm{ mL}\\)?\r\n\t\r\n\tThe heights of a group of men follow a normal distribution with a mean of \\(180\\textrm{ cm}\\) and a standard deviation of \\(6\\textrm{ cm}\\).\r\n\t\r\n\t\tWhat is the probability that a man chosen from this group is less than \\(185\\textrm{ cm}\\) tall?\r\n\t\tIf the tallest \\(10\\%\\) of this group are automatically eligible for a basketball team, what is the qualifying height?\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.04},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":314,"$":[{"v":"Standard normal distribution","n":0.577},{"v":"A standard normal distribution helps us compare different datasets by converting them to a common scale using values called \\(z\\)-scores. Understanding standard normal distributions makes it easier to calculate probabilities and assess how individual data points relate to the average. Use this resource to learn how to analyse standard normal distributions and use \\(z\\)-tables.\r\n\r\n\\(z\\)-distributions\r\nA normal distribution with a mean of \\(\\mu=0\\) and a standard deviation of \\(\\sigma=1\\) is called the standard normal distribution. \r\n\r\n\r\n\\(z\\)-tables\r\nAreas under the standard normal distribution curve represent probabilities which can be found using a calculator or a \\(z\\)-table. \r\nThe \\(z\\)-score represents how many standard deviations a data point is from the mean in a standard normal distribution. For example, if \\(z=0.05\\), then the data point is \\(0.05\\) standard deviations above the mean. Typical \\(z\\)-tables only show positive standard deviations, but since the standard normal distribution is symmetrical, we can still find the answers we are looking for.\r\nWhen we work with the standard normal distribution, we are not restricted to \\(1\\), \\(2\\) and \\(3\\) standard deviations from the mean and the \\(68-95-99\\) rule for other normal distributions. \\(z\\)-tables let us find proportions, percentages or probabilities for any value in the distribution.\r\n\\(z\\)-tables come in different layouts, but here, we show a small section of a table that gives \\(z\\)-values up to \\(2\\) decimal places.\r\n\r\nInterpreting \\(z\\)-tables\r\nWhen using a \\(z\\)-table, you need to look at your \\(z\\)-score in two parts. Let's consider \\(z=1.32\\). You would need to break this up into:\r\n\\[1.30+0.02\\]\r\nThe larger value, \\(1.30\\), is found in the first column of the table. The smaller value \\(0.02\\) is found in the first row. The corresponding area value in the table is the proportion, percentage or probability that \\(z\\) is less than \\(1.32\\).\r\nSome other examples are:\r\n\r\n\t\\(\\Pr(z<0.00)=0.500\\) or \\(50\\%\\). This is \\(0.0\\) in the first column and \\(0.00\\) in the first row.\r\n\t\\(\\Pr(z<0.03)=0.512\\) or \\(51.2\\%\\). This is \\(0.0\\) in the first column and \\(0.03\\) in the first row.\r\n\t\\(\\Pr(z<0.75)=0.7734\\) or \\(77.34\\). This is \\(0.7\\) in the first column and \\(0.05\\) in the first row.\r\n\r\nYou can also use a graphics calculator or computer (instead of a \\(z\\)-table) to find areas, proportions and probabilities in a normal distribution.\r\n\r\nExample 1 \u2013 interpreting \\(z\\)-tables\r\nIn a standard normal distribution, what percentage of values will be less than \\(1.28\\)?\r\nIt is always good to start with a diagram of the standard normal distribution. It will also help you sense-check your answer.\r\n\r\nFrom this, it is clear that we are looking for the shaded area to the left of \\(z=1.28\\). We can use the following table to find that the area to the left is \\(0.8997\\).]Standard normal distribution: table values represent area to the left of the Z score.\r\n\r\nThis means that \\(89.97\\%\\) of values will be less than \\(1.28\\). This agrees with the shaded region in our diagram; a large proportion of the curve is shaded.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 interpreting \\(z\\)-tables\r\nUse the following \\(z\\)-table to answer the questions.\r\n\r\n \tIn a standard normal distribution, what percentage of values will be:\r\n\t\r\n\t\tless than \\(1.95\\)?\r\n\t\tless than \\(-1.95\\)?\r\n\t\tbetween \\(-1.95\\) and \\(0\\)?\r\n\t\tgreater than \\(1.95\\)?\r\n\t\tbetween \\(-1.95\\) and \\(1.95\\)?\r\n\t\r\n\tIn a standard normal distribution, what proportion of values lie between \\(-0.5\\) and \\(1.5\\)?\r\n\tIn a standard normal distribution, what proportion of values lie outside the interval \\(\\pm1.7\\)?\r\n\tGiven that a value in a standard normal distribution is greater than \\(-1\\), what is the probability that it will be less than \\(2\\)? (Hint: use the conditional probability formula.)\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.043},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":315,"$":[{"v":"Normal distributions","n":0.707},{"v":"\r\nIf data is normally distributed, it is represented by a bell-shaped curve. Normal distributions are common in nature and statistics, like heights or test scores. Understanding this distribution helps us make predictions and compare data from different sources easily.\r\n\r\nGraphical representations of data may look quite different.\r\n\r\nMany things that we can measure follow a bell-shaped curve, such as\r\n\r\n \theights of people\r\n \tblood pressure\r\n \terrors in measurement\r\n \tscores on a test\r\n\r\nSuch data produces a normal distribution. A normal distribution curve shows that most data points cluster around the middle, or mean, with fewer points appearing as you move away from the centre.\r\n\r\n\r\nProperties of a normal distribution\r\nIn a normal distribution curve:\r\n\r\n \tthere is symmetry about the mean\r\n \tthe mean = median = mode\r\n \t\\(50\\%\\) of values are greater than the mean and \\(50\\%\\) are less than the mean\r\n \t\\(68\\%\\) of values fall within one standard deviation either side of the mean (approximately)\r\n \t\\(95\\%\\) of values fall within two standard deviations either side of the mean (approximately)\r\n \t\\(99.7\\%\\) of values fall within three standard deviation either side of the mean (approximately).\r\n\r\n\r\nRemember the \\(68-95-99\\) rule.\r\nEven though most of the data will fall within three standard deviations of the mean, there is in theory, no upper or lower bound to a normal distribution. We are just less and less likely to find values beyond these points.\r\n\r\nExample \u2013 analysing normal distributions\r\nIf scores on an IQ test are normally distributed with a mean of \\(\\mu=100\\) and standard deviation of \\(\\sigma=10\\), what percentage of people would you expect to:\r\n\r\n \tscore between \\(90\\) and \\(110\\)?\r\n \tscore less than \\(80\\)?\r\n\r\nThe values \\(90\\) and \\(110\\) are \\(10\\) away from the mean. This is one standard deviation away. Therefore, you would expect \\(68\\%\\) of people to score between \\(90\\) and \\(110\\).\r\n\\(80\\) is \\(20\\) away from the mean. This is two standard deviations away. We know that \\(95\\%\\) of the scores fall between two standard deviations of the mean, so \\(5\\%\\) must fall outside of this range. This \\(5\\%\\) would include scores two standard deviations below the mean \\((<80)\\) and two standard deviations above the mean \\((>120)\\), so half of this percentage would score less than \\(80\\), i.e. \\(2.5\\%\\).\r\n\r\nExercise \u2013 analysing normal distributions\r\n\r\n\tScores on a general achievement test are normally distributed with a mean of \\(80\\) and a standard deviation of \\(15\\). Adam scored \\(95\\). What proportion of students had a higher score than Adam?\r\n\tThe actual weights of cereal boxes that are supposed to contain \\(500\\textrm{ g}\\) are normally distributed with mean of \\(510\\textrm{ g}\\) and a standard deviation of \\(5\\textrm{ g}\\). What proportion of boxes are under-filled?\r\n\tIn a Maths class, the bottom \\(16\\%\\) of students are given an F grade. If the class mean is \\(63\\) and the standard deviation is \\(18\\), what score must a student get to pass?\r\n\tIf newborn birth weights in a certain hospital are normally distributed with a mean of \\(3200\\textrm{ g}\\) and a standard deviation of \\(400\\textrm{ g}\\):\r\n\t\r\n\t\twhat percentage of babies weigh more than \\(3200\\textrm{ g}\\)?\r\n\t\twhat percentage of babies weigh between \\(2400\\textrm{ g}\\) and \\(4000\\textrm{ g}\\)?\r\n\t\twhat percentage of babies weigh less than \\(3600\\textrm{ g}\\)?\r\n\t\tif the \\(16\\%\\) of babies with the lowest birth weights are placed in the special care nursery, will a baby that weighs \\(2500\\textrm{ g}\\) need special care?\r\n\t\r\n\t\\(95\\%\\) of people in a clinical study had systolic blood pressure readings between \\(116\\) and \\(144\\). If the blood pressure measurements follow a normal distribution, what is the mean and standard deviation of the blood pressures for this group?\r\n\tA class of ten students get the following marks in a test: \\(13,23,41,55,66,78,49,33,35,67\\). If anyone who scored less than one standard deviation below the mean fails, how many students will fail?\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.041},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":316,"$":[{"v":"Trends in the periodic table","n":0.447},{"v":"The periodic table not only organises elements based on their atomic structure but also highlights recurring patterns in their properties. These trends, observed across periods and groups, provide insights into the behaviour of elements and their interactions with others.\r\nSome key trends emerge from the periodic table, including how atomic size (atomic radii) changes across periods and down groups, how much energy is required to remove the first electron from the valence shell (first ionisation energy), how readily an atom can accept an electron (electron affinity), and how strongly atoms attract electrons in chemical bonds (electronegativity).\r\n\r\nAtomic radii\r\nAtomic radii (or radius) increases down a group (from top to bottom) of the periodic table. As \\(n\\) increases when moving down a group, the number of electron shells become larger; the electron orbitals are therefore larger, leading to an increase in atomic radii.\r\nWhen moving across a period (from left to right), atomic radii decreases. As you move across a period, each successive element has an additional proton in the nucleus and an additional electron in the same valence shell. The increase in positive charge in the nucleus attracts the electrons more strongly, pulling them closer to the nucleus. Although the number of electrons increases, these electrons are added to the same energy level and do not significantly increase electron\u2013electron repulsion that would otherwise counteract the pull of the nucleus.\r\nThis stronger attraction between the increased positive charge of the nucleus and the electrons results in a smaller atomic radii as you move across a period.\r\nAn exception to the trend in atomic radii across a period is in the transition metals. As you move across the transition metals (\\(d\\)-block elements), the decrease in atomic radii is less pronounced. This means that elements in the same period within the \\(d\\)-block will have a similar atomic radii.\r\n\r\nFirst ionisation energy\r\nThe first ionisation energy is the minimum energy required to remove the first electron from a neutral atom. The second and third ionisation energies are the energies necessary to remove the second and third electrons from the atom, respectively.\r\nFirst ionisation energy increases when moving across a period as the attraction between the nucleus and the valence electrons increases. More energy is therefore required to overcome this strong attraction and remove the first valence electron.\r\nWhen moving down a group, the first ionisation energy decreases. This is because a valence electron in a higher energy level (greater \\(n\\)) is easier to remove.\r\n\r\nElectron affinity\r\nElectron affinity gives an indication of how easy it is for a neutral atom to accept an electron and form a negatively-charged ion (an anion). Quantitatively, it is the amount of energy released when an electron is added to an atom in the gaseous phase to form an anion. \r\nThe electron affinity increases when moving across a period. The interaction between the nucleus and the electrons increases, and more energy is released when an electron is added. Non-metals, particularly the halogens, have high electron affinities because they are close to achieving a full valence shell, making it easier for them to gain electrons.\r\n\r\n\r\nElectronegativity\r\nElectronegativity is the power of an atom in a molecule to attract electrons from another atom to form chemical bonds. The larger the value, the larger the electron-attracting ability.\r\nElectronegativity decreases from top to bottom and increases from left to right of the periodic table. Across a period, atoms have more protons and a higher effective nuclear charge as you move across a period, which increases the attraction for electrons. Down a group, the atomic radii increases and the addition of electron shells leads to a greater distance between the nucleus and the valence electrons, and a reduced attraction for electrons.\r\nNoble gases are generally not considered in electronegativity trends because they have full valence electron shells and are typically inert. They do not tend to form chemical bonds, so electronegativity is not so relevant\u2014although, some heavier noble gases like xenon can form chemical bonds under specific conditions. This means that fluorine is generally considered the most electronegative element.\r\n\r\n\r\n\r\nExample \u2013 predicting periodic trends\r\nPredict which atom is larger for the following pairs of atoms:\r\n\r\n\t\\(\\ce{C}\\) and \\(\\ce{F}\\)\r\n\t\\(\\ce{Na}\\) and \\(\\ce{Cs}\\)\r\n\t\\(\\ce{Sc}\\) and \\(\\ce{Ni}\\)\r\n\r\nStep 1: Recall the trend in atomic radii in the periodic table.\r\nWhen moving across a period (left to right), atomic radii decreases. As you move across a period, the number of valence increases, making the electron cloud more negative. However, the number of protons (atomic number) also increases, making the nucleus more positive. This allows the positively-charged nucleus to pull the electrons in closer, decreasing the atomic radius.\r\nWhen moving down a group (top to bottom), atomic radii increases. Going down a group, elements gain an electron shell. As there will be a greater distance between the valence electrons and the nucleus, the atomic radius increases.\r\nStep 2: Locate the elements on the periodic table and compare them.\r\n\r\n\t\\(\\ce{C}\\) and \\(\\ce{F}\\) are both located in period \\(2\\), but \\(\\ce{F}\\) is located further right in group \\(17\\) compared to \\(\\ce{C}\\) in group \\(14\\).\r\n\t\\(\\ce{Na}\\) and \\(\\ce{Cs}\\) are both located in group \\(1\\), but \\(\\ce{Cs}\\) is located further down in period \\(6\\) compared to \\(\\ce{Na}\\) in period \\(3\\).\r\n\t\\(\\ce{Sc}\\) and \\(\\ce{Ni}\\) are both located in period \\(4\\), but \\(\\ce{Ni}\\) is located further right in group \\(10\\) compared to \\(\\ce{Sc}\\) in group \\(3\\).\r\n\r\nStep 3: Apply your understanding of the relative locations of the elements on the periodic table with the trend in atomic radii to predict which atom is larger.\r\n\r\n\t\\(\\ce{C}\\) would be expected to have a bigger atomic radius than \\(\\ce{F}\\) and would therefore be larger.\r\n\t\\(\\ce{Cs}\\) would be expected to have a bigger atomic radius than \\(\\ce{Na}\\) and would therefore be larger.\r\n\t\\(\\ce{Sc}\\) would be expected to have a bigger atomic radius than \\(\\ce{Ni}\\) and would therefore be larger. However, across \\(d\\)-block and \\(f\\)-block, atomic radii change is not significant. Thus, in reality, \\(\\ce{Sc}\\) and \\(\\ce{Ni}\\) have approximately the same radius.\r\n\r\n\r\nYour turn \u2013 trends in the periodic table\r\nTest yourself on your understanding of trends in the periodic table.\r\n\r\n\r\n \r\nPeriodic table illustrating trends. Atomic radii decrease across periods and increase down groups. First ionisation energy increases across periods and decreases down groups. Electronegativity increases across periods and decreases down groups.\r\n","n":0.031},[{"v":"Periodic table","i":1,"n":0.707},{"v":"Chemistry","i":0,"n":1}]]},{"i":317,"$":[{"v":"Ionic bonding","n":0.707},{"v":"Chances are you sprinkle table salt on your meals, use baking soda in your cooking, or encounter calcium carbonate in chalk. Ionic bonding explains how these substances form. Understanding ionic bonds helps us see how different materials are made and why they have certain properties.\r\nWhen forming bonds, atoms attempt to achieve a stable noble gas configuration by sharing, losing or gaining electrons for each of the atoms associated. In the case of ionic bonds, atoms lose or gain electrons to form ions, which then form electrostatic interactions.\r\n\r\nIons\r\nIons are charged species formed when electrons are transferred from one atom to another. The charge of an ion depends on:\r\n\r\n\twhether the atom gains electrons to form negatively-charged ions (or anions), or loses electrons to form positively-charged ions (or cations)\r\n\tthe number of electrons gained or lost.\r\n\r\nFor example, \\(\\ce{Na}^{+}\\) is an \\(\\ce{Na}\\) atom that has lost one electron. \\(\\ce{O}^{2-}\\) is an \\(\\ce{O}\\) atom that has gained two electrons.\r\n\r\nMonatomic ions\r\nMonatomic (or monoatomic) ions are ions consisting of a single atom. Ions formed from main group elements are named using the name of the element, followed by the word \"ion\". The chemical symbol for monatomic ions is written using the chemical symbol for the element, with the charge of the ion written afterwards as a superscript. For instance, the lithium ion is written as \\(\\ce{Li}^{+}\\), the magnesium ion is \\(\\ce{Mg}^{2+}\\) and the aluminium ion is \\(\\ce{Al}^{3+}\\).\r\n\r\nTransition metals (\\(d\\)-block and \\(f\\)-block) and some \\(p\\)-block elements can form multiple cations such as \\(\\ce{Cu}^{+}\\) and \\(\\ce{Cu}^{2+}\\), and \\(\\ce{Pb}^{2+}\\) and \\(\\ce{Pb}^{4+}\\). When naming these ions, the ion with the smaller charge is written with an \"\u2013ous\" suffix, and the larger charge is written with an \"\u2013ic\". For example, copper would form the cuprous ion (\\(\\ce{Cu}^{+}\\)) and the cupric ion (\\(\\ce{Cu}^{2+}\\)). The charge of the ion can also be written in Roman numerals within parenthesis next to the element name. For example, \\(\\ce{Cu}^{+}\\) is copper(I) ion, and \\(\\ce{Cu}^{2+}\\) is copper(II). There is no space between the element name and the parenthesis.\r\n\r\nAnions are named by removing the ending of the element name and replacing it with \"\u2013ide\". Let's consider the anion formed when chlorine (\\(\\ce{Cl}\\)) gains an electron. The anion formed would be the chloride ion (\\(\\ce{Cl}^{-}\\)).\r\n\r\nPolyatomic ions\r\nPolyatomic ions consist of more than one atom. Some example are the \\(\\ce{SO}_{4}^{2-}\\) (sulfate), \\(\\ce{NO}_{3}^{-}\\) (nitrate), \\(\\ce{PO}_{4}^{3-}\\) (phosphate) and \\(\\ce{NH}_{4}^{+}\\) (ammonium) ions. Atoms in polyatomic ions are held together by covalent bonds.\r\n\r\nThe octet rule\r\nThe valence electron configuration of noble gases is considered the most stable. Except for helium, all noble gases have eight electrons in their valence shell (completely filled \\(s\\) and \\(p\\) subshells). Main group elements gain or lose electrons to achieve a full outer valence shell containing eight electrons. This is the octet rule. \r\nThe main group metals attempt to achieve the electron configuration of the noble gas just before them in the periodic table by losing electrons. For instance, potassium's electron configuration is \\(1s^{2}\\,2s^{2}\\,2p^{6}\\,3s^{2}\\,3p^{6}\\,4s^{1}\\). the potassium atom will lose one \\(4s\\) electron to become a  potassium cation (\\(\\ce{K}^{+}\\)) to achieve a noble gas electron configuration: \\(1s^{2}\\,2s^{2}\\,2p^{6}\\,3s^{2}\\,3p^{6}\\). This is the same as the electron configuration for argon.\r\nMain group non-metals acquire the electron configuration of the noble gas just after them in the periodic table by gaining electrons. Consider the electron configuration of chlorine: \\(1s^{2}\\,2s^{2}\\,2p^{6}\\,3s^{2}\\,3p^{5}\\). To achieve a noble gas electron configuration, the chlorine atom gains one \\(3p\\) electron to become a chloride anion (\\(\\ce{Cl}^{-}\\)). The electron configuration for the chloride ion is \\(1s^{2}\\,2s^{2}\\,2p^{6}\\,3s^{2}\\,3p^{6}\\), which is also the same as the electron configuration of argon.\r\nIn general, metals form cations and non-metals form anions.\r\nExceptions to the octet rule\r\nIn some molecules, atoms do not follow the octet rule. Atoms such as sulfur, phosphorus and chlorine can deviate from the octet rule by forming molecules in which there are more than eight electrons in the valence shell. This is due to the availability of a vacant \\(d\\) orbital in the third shell.\r\n\r\nIonic compounds\r\nIonic compounds are held together by electrostatic interactions between cations and anions involved. The cations and anions are arranged in such a way that gaps between ions are filled efficiently and permit maximum electrostatic interactions. Ionic compounds are polar materials.\r\n\r\nFormation of ionic compounds\r\nIonic bonds forms between metals and non-metals through the transfer of electrons from metals to non-metals. Metals provide the cation (by losing electrons), and non-metals provide the anion (by gaining electrons). \r\n\r\n\r\n\r\nGenerally, chemical bonds formed between atoms with larger electronegativity (e.g. elements in groups 16 and 17 of the periodic table) and smaller electronegativity (e.g. elements in groups 1 and 2 of the periodic table) are considered ionic. As an example, \\(\\ce{NaCl}\\) consists of \\(\\ce{Na}\\) which has an electronegativity value of \\(0.93\\) and \\(\\ce{Cl}\\) which has an electronegativity of 3.16. The difference between these values is \\(2.23\\).\r\nFor an ionic compound to be formed, the difference between electronegativities of two participating atoms must be larger than or equal to \\(2\\). This permits strong electrostatic interactions between the electrons of one atom and the nucleus of the other atom.\r\n\r\nWriting chemical formulas for ionic compounds\r\nIonic compounds are neutral species. Ions involved in an ionic compound combine in such a ratio that the net charge becomes zero. The ratio of ions in an ionic compound can be represented using a chemical formula, using subscripts after each chemical symbol. This displays the lowest possible ratio in which ions can combine to form the compound. Examples are \\(\\ce{MgO}\\) and \\(\\ce{CaCl2}\\).\r\nParentheses are used to enclose polyatomic ions when more than one polyatomic ion is involved in an ionic compound. In this case, the subscript is written outside the parentheses. Examples are \\(\\ce{Fe2(SO4)3}\\) and \\(\\ce{(NH4)2CO3}\\).\r\n\r\nNaming ionic compounds\r\nIonic compounds can be named by following some easy steps. Let's look at how we can name the compound \\(\\ce{CaCl2}\\).\r\n\r\n \tName the cation by writing the name of the metal element. The cation in \\(\\ce{CaCl2}\\) would simply be \"calcium\".\r\n\tName the anion by writing the name of the non-metal element. Remove the ending of the element name and replacing it with \"\u2013ide\". The anion in \\(\\ce{CaCl2}\\) would be \"chloride\".\r\n\r\nTransitional metals and some of the \\(p\\) block elements have more than one charge. Therefore, if the cation is from those groups, you must include the charge within parentheses next to its name in Roman numerals or use \"\u2013ous\" and \"\u2013ic\" to indicate smaller charge and larger charge, respectively. For instance, \\(\\ce{CuCl}\\) is written as copper(I) chloride or cuprous chloride, \\(\\ce{CuCl2}\\) is written as copper(II) chloride or cupric chloride.\r\n\r\n\r\n\r\nPhysical properties of ionic compounds\r\nIonic compounds exhibit a range of distinctive properties that make them essential in various applications. The table below lists some properties of ionic compounds and examples of how the properties are applied.\r\n\r\n\r\n\r\nProperty\r\nExample application\r\n\r\n\r\n\r\nDurable and beautiful as crystalline solids\r\nPotassium aluminium sulfate is used in traditional bookbinding to increase the toughness and lustre of decorative paste papers.\r\n\r\n\r\nConduct electricity in solution\r\nLithium salts in lithium\u2013ion batteries are used to conduct electricity by facilitating the flow of charge.\r\n\r\n\r\nHigh melting and boiling points\r\nMagnesium oxide is used in furnace linings because it can withstand extreme temperatures.\r\n\r\n\r\nSoluble in polar solvents\r\nPotassium chloride is a medication used in the management and treatment of hypokalemia. Its solubility in bodily fluids allows it to be absorbed more effectively.\r\n\r\n\r\n\r\n\r\n\r\nExample 1 \u2013 determining the number of electrons gained or lost\r\nDetermine how many electrons an aluminium atom has to lose or gain to obtain a noble gas electron configuration.\r\nStep 1: Write the electron configuration for aluminium.\r\n\t\\[1s^{2}\\,2s^{2}\\,2p^{6}\\,3s^{2}\\,3p^{1}\\]\r\nStep 2: Identify the nearest noble gas. Write its electron configuration.\r\n\tThe nearest noble gas is \\(\\ce{Ne}\\), which has an electron configuration of \\(1s^{2}\\,2s^{2}\\,2p^{6}\\).\r\nStep 3: The purpose of losing or gaining an electron is to acquire an octet in the valence shell. Identify the number of electrons in the valence shell and deduce whether aluminium will gain or lose electrons to achieve an octet.\r\n\tThe valence shell contains three electrons: two in the \\(3s\\) orbital and one in the \\(3p\\) orbital. It can gain the electron configuration of \\(\\ce{Ne}\\) by losing these three electrons.\r\n\r\n\r\n\r\n\r\nExample 3 \u2013 writing chemical formulas\r\nDetermine the chemical formula of the compound produced when \\(\\ce{Ca}^{2+}\\) and \\(\\ce{F}^{-}\\) ions become electrostatically attracted.\r\nStep 1: Write the symbol of the cation with its charge, followed by the anion with its charge.\r\n\t\\[\\ce{Ca}^{2+}\\ce{F}^{-}\\]\r\nStep 2: Cross each ion's charge (only the numerical value, not the sign) and write it down as subscripts. The calcium ion's charge will be written next to fluoride ion, and the fluoride ion's charge will be written next to calcium ion, both as subscripts. Where the subscript is a \\(1\\), simply exclude it.\r\n\t\\(\\ce{Ca}_{1}\\ce{F}_{2}\\), which becomes \\(\\ce{CaF2}\\)\r\n\r\n\r\n\r\n\r\n\t\r\n\r\nYour turn \u2013 ionic bonding\r\nUnderstanding ionic bonding is essential for grasping how elements interact to form compounds with unique properties. Test your understanding of ionic bonding with this quiz.\r\n\r\n\r\n\r\n \r\nThe diagram shows the transfer of electrons from Atom 1 (metal) to Atom 2 (non-metal). Electron transfer results in Atom A becoming positively charged and Atom B negatively charged. They interact via an ionic bond to form an ionic compound.\r\n","n":0.026},[{"v":"Ionic bonds","i":2,"n":0.707},{"v":"Chemistry","i":1,"n":1},{"v":"Chemical bonding","i":0,"n":0.707}]]},{"i":318,"$":[{"v":"Probability rules","n":0.707},{"v":"\r\nProbability rules help us understand how likely different outcomes are. They are applied in a range of ways, like risk assessment and modelling. Having a good understanding of these basics will help you to predict and make sense of different possibilities.\r\nWhen considering a set of all possible outcomes, an event is a particular outcome of interest. For example:\r\n\r\n\tWhen tossing a coin, the event of interest might be \"obtaining a head\".\r\n\tWhen considering the weather for Saturday, the event of interest might be \"it doesnt rain\".\r\n\tWhen throwing a die, the event of interest might be \"rolling a \\(3\\).\"\r\n\r\nThe probability of an event \\(E\\) can be found with the formula:\r\n\\[\\Pr(E)=\\frac{\\textrm{Number of ways $E$ can occur}}{\\textrm{Total number of possible outcomes}}\\]\r\nThis assumes all outcomes are equally likely.\r\n\\(\\Pr(E)\\) is expressed as a number between \\(0\\) and \\(1\\):\r\n\\[0\\leq Pr(E)\\leq1\\]\r\n\r\n\tIf the event is impossible, \\(\\Pr(E)=0\\).\r\n\tIf the event is certain, \\(\\Pr(E)=1\\).\r\n\r\n\r\nExample 1 \u2013 finding the probability of an outcome\r\nIf two coins are tossed, find the probability of obtaining two heads.\r\nLet \\(E\\) be the event \"two heads\". If H = heads and T = tails, the total number of possible outcomes is \\(4\\): HH HT TH TT.\r\n\\[\\begin{align*} \\Pr(E) & = \\frac{\\textrm{Number of ways E can occur}}{\\textrm{Total number of possible outcomes}}\\\\\r\n& = \\frac{1}{4}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nThe multiplication principle\r\nThe multiplication principle is used to find the probability of two events happening at the same time by multiplying their probabilities\u2014that is, the probability of \\(A\\) and \\(B\\) occurring.\r\nWe can only use the multiplication principle if the probability of event \\(A\\) does not affect the probability of event \\(B\\) occurring. If this is true, we say that \\(A\\) and \\(B\\) are independent events.\r\nThe probability of \\(A\\) and \\(B\\) occurring together is denoted by \\(\\Pr(A\\cap B)\\) where \\(\\cap\\) is the symbol for intersection.\r\n\\[\\Pr(A\\textrm{ and }B)=\\Pr(A\\cap B)=\\Pr(A)\\times(B)\\]\r\n\r\nExample \u2013 using the multiplication principle\r\nIf two coins are tossed, find the probability of obtaining two heads.\r\nThis is the same question as Example 1 \u2013 finding the probability of an outcome. Because tosses of a coin are independent events, an alternate way of calculating the probability would be to use the multiplication principle.\r\nTwo coins are tossed, each representing an event. In both, the possibility of a heads is \\(\\dfrac{1}{2}\\).\r\nThe probability of a head on the first toss (\\(H_{1}\\)) and a head on the second toss (\\(H_{2}\\)) is:\r\n\\[\\begin{align*} \\Pr(H_{1}\\cap H_{2}) & = \\Pr(H_{1})\\times \\Pr(H_{2})\\\\\r\n& = \\frac{1}{2}\\times\\frac{1}{2}\\\\\r\n& = \\frac{1}{4}\r\n\\end{align*}\\]\r\n\r\nThe addition principle\r\nThe addition principle tells us how to find the probability of one event or another happening by adding their individual probabilities\u2014that is, the probability of \\(A\\) or \\(B\\) occurring.\r\nWe denote this using the union symbol, \\(\\cup\\).\r\n\\[\\Pr(A\\textrm{ or }B)=\\Pr(A\\cup B)=\\Pr(A)+\\Pr(B)-\\Pr(A\\cap B)\\]\r\nIf \\(A\\) and \\(B\\) are mutually exclusive or cannot happen at the same time (i.e. \\(\\Pr(A\\cap B)=0\\), we use:\r\n\\[\\Pr(A\\textrm{ or }B)=\\Pr(A\\cup B)=\\Pr(A)+\\Pr(B)\\]\r\n\r\nExample \u2013 using the addition principle\r\nIf a die is tossed twice, find the probability that a \\(6\\) occurs.\r\nWe are looking at two events, the first die toss and the second die toss. The \\(6\\) could occur on the first toss (\\(S_{1}\\)) or the second toss (\\(S_{2}\\)).\r\nIn both events, the possibility of a \\(6\\) is \\(\\dfrac{1}{6}\\).\r\n\\[\\begin{align*} \\Pr(S_{1}\\cup S_{2}) & = \\Pr(S_{1})+\\Pr(S_{2})-\\Pr(S_{1}\\cap S_{2})\\\\\r\n& = \\frac{1}{6}+\\frac{1}{6}-\\frac{1}{36}\\\\\r\n& = \\frac{11}{36}\r\n\\end{align*}\\]\r\n\r\nComplementary events\r\nComplementary events show us what happens when an event doesn't occur, helping to see the whole picture.\r\nIf \\(E\\) is an event, then \"not \\(E\\)\" is the complement of \\(E\\). The complement of \\(E\\) is denoted by \\(\\overline{E}\\), \\(E'\\) or \\(E^{c}\\). \r\nSome examples of complementary events are:\r\n\r\n\t\"winning the grand final\" and \"not winning the grand final\"\r\n\t\"passing a test\" and \"failing a test\"\r\n\t\"being left handed\" and \"being right handed\".\r\n\r\nTo find the probability of a complementary event, we just minus the probability of the event occurring from \\(1\\).\r\n\\[\\Pr(\\overline{E})=1-\\Pr(E)\\]\r\n\r\nExample \u2013 finding the probability of a complementary event\r\nIf a die is tossed twice, find the probability of not getting a \\(6\\) on either toss.\r\nFrom Example \u2013 using the addition principle, we found that \\(\\Pr(S_{1}\\cup S_{2})=\\dfrac{11}{26}\\).\r\n\\[\\begin{align*} \\Pr(\\overline{S_{1}\\cup S_{2}}) & = 1-\\Pr(S_{1}\\cup S_{2})\\\\ \r\n& = 1-\\frac{11}{36}\\\\\r\n& = \\frac{25}{36}\r\n\\end{align*}\\]\r\n\r\nExercise \u2013 using the probability rules\r\n\r\n\tIf \\(1000\\) tickets are sold in a raffle and one winning ticket is chosen at random, what is the probability of winning the raffle if \\(5\\) tickets are purchased?\r\n\tIf a die is rolled, what is the probability that the number is greater than \\(4\\)?\r\n\tA bag contains \\(6\\) white marbles and \\(4\\) black marbles. A marble is chosen, the colour recorded and it is then returned to the bag. This is repeated three times. What is the probability that all three marbles are white?\r\n\tThe probability that phone A will continue to function in \\(10\\) years' time is \\(0.7\\). The probability that phone B will continue to function in \\(10\\) years' time is \\(0.4\\). Find the probability that:\r\n\t\r\n\t\tboth are still functioning in \\(10\\) years' time\r\n\t\tneither are still functioning in \\(10\\) years' time\r\n\t\tonly one is still functioning in \\(10\\) years' time\r\n\t\tat least one is still functioning in \\(10\\) years' time.\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.036},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":319,"$":[{"v":"Sample spaces","n":0.707},{"v":"\r\nA sample space lists of all the possible outcomes, giving a complete picture of the situation. Sample spaces form the foundation for calculating probabilities and making predictions about future events. Use this resource to learn some ways to represent a sample space.\r\n\r\nA sample space is a list or diagram showing all possible outcomes in a probability experiment. \r\nWe can rewrite the probability equation to:\r\n\\[\\begin{align*} \\Pr(E) & = \\frac{\\textrm{Number of ways $E$ can occur}}{\\textrm{Number of outcomes in the sample space}}\\\\\r\n& = \\frac{n(E)}{n(S)}\r\n\\end{align*}\\]\r\nAgain, this assumes all outcomes are equally likely.\r\nFor tossing a single die, the sample space is \\(1,2,3,4,5,6\\). We can immediately tell that \\(\\Pr(1)=\\Pr(2)=\\Pr(3)=\\Pr(4)=\\Pr(5)=\\Pr(6)=\\dfrac{1}{6}\\).\r\n\r\nFor a four-colour spinner with equal sectors, the sample space is Red, Green, Yellow, Blue. \\(\\Pr(R)=\\Pr(G)=\\Pr(Y)=\\Pr(B)=\\dfrac{1}{4}\\).\r\n\r\nThe sum of the probabilities of the distinct outcomes within a sample space is \\(1\\).\r\nLet's look into some ways to represent sample spaces.\r\n\r\nTree diagrams\r\nA tree diagram can be used to find the sample space. It uses a branching format, starting from a single point and splitting into paths that represent different possibilities. Each branch represents an event and the complete diagram shows all outcomes at the ends of the branches.\r\n\r\nExample 1 \u2013 finding probabilities using a tree diagram\r\nUse a tree diagram to find the sample space for two coins tosses, then determine the probability of getting at least one head.\r\nThere are two possible outcomes for each toss: H = Heads and T = Tails. We draw one branch to represent the outcomes of the first coin toss: H or T. Then, from each of these initial outcomes, we draw another branch: H or T. \r\n\r\nThe sample space for tossing the two coins is: HH HT TH TT. We can see that each has a probability of \\(\\dfrac{1}{4}\\).\r\nTo find the probability of at least one head, we need to add up the probabilities of getting HH, HT or TH:\r\n\\[\\begin{align*} \\Pr(HH\\cup HT\\cup TH) & = \\frac{1}{4}+\\frac{1}{4}+\\frac{1}{4}\\\\\r\n& = \\frac{3}{4}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nTables\r\nTables are helpful to organise information that makes finding probabilities easier. Let's see an example.\r\n\r\nExample \u2013 finding probabilities using a table\r\nThe following table shows the results of a study that looked at the association between smoking (S) and lung cancer (C).\r\n\r\n\t\r\n\t\t\r\n\t\t\t\r\n\t\t\tLung cancer\r\n\t\t\tNo lung cancer\r\n\t\t\tTotal\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\tSmoker\r\n\t\t\t\\(30\\)\r\n\t\t\t\\(70\\)\r\n\t\t\t\\(100\\)\r\n\t\t\r\n\t\t\r\n\t\t\tNon-smoker\r\n\t\t\t\\(10\\)\r\n\t\t\t\\(90\\)\r\n\t\t\t\\(100\\)\r\n\t\t\r\n\t\t\r\n\t\t\t\r\n\t\t\t\\(40\\)\r\n\t\t\t\\(160\\)\r\n\t\t\t\\(200\\)\r\n\t\t\r\n\t\r\n\r\nFind the probability that an individual:\r\n\r\n\tis a smoker \\((\\Pr(S))\\)\r\n\thas lung cancer \\((\\Pr(C'))\\)\r\n\tis a smoker with lung cancer \\((\\Pr(S\\cap C))\\).\r\n\r\nThere are \\(100\\) smokers out of the total \\(200\\) individuals, so:\r\n\\[\\begin{align*} \\Pr(S) & = \\frac{100}{200}\\\\\r\n& = \\frac{1}{2}\r\n\\end{align*}\\]\r\nThere are \\(160\\) individuals of the \\(200\\) who do not have lung cancer, so:\r\n\\[\\begin{align*} \\Pr(C') & = \\frac{160}{200}\\\\\r\n& = \\frac{4}{5}\r\n\\end{align*}\\]\r\nThere are \\(30\\) individuals of the \\(200\\) who are smokers and have lung cancer, so:\r\n\\[\\begin{align*} \\Pr(S\\cap C) & = \\frac{30}{200}\\\\\r\n& = \\frac{3}{20}\r\n\\end{align*}\\]\r\n\r\nVenn diagrams\r\nVenn diagrams are also handy for showing sample spaces. They consist of overlapping circles to show relationships between different sets of outcomes in a probability experiment.\r\nEach circle shows a group of outcomes. They overlap where outcomes are common to those groups.\r\nVenn diagrams are especially useful for showing how events relate to each other, such as mutually exclusive events.\r\n\r\nExample \u2013 finding probabilities using a Venn diagram\r\nThe Venn diagram shows the number of people in a survey who regularly ate KitKats (K), Mars bars (M) or Rocky Road (R).\r\n\r\nFind the probability that a person:\r\n\r\n\teats KitKats only \\((\\Pr(K))\\)\r\n\teats Mars bars and Rocky Road \\((\\Pr(M\\cap R))\\)\r\n\teats KitKats and Mars bars but not Rocky Road \\((\\Pr(K\\cap M\\cap R'))\\)\r\n\teats at least one of the chocolates \\((\\Pr(K\\cup M\\cup R))\\).\r\n\r\nFrom the Venn diagram, we can see that there are \\(256\\) people. This is the sum of all sections of the diagram: \\(26+46+30+38+40+24+32+20\\).\r\nPeople who eat KitKats include those who:\r\n\r\n\teat KitKats only: \\(26\\)\r\n\teat KitKats and Mars bars only: \\(30\\)\r\n\teat KitKats and Rocky Road only: \\(46\\)\r\n\teat all three: \\(38\\).\r\n\r\nThis is \\(26+30+46+38=140\\) of the \\(256\\) people, so:\r\n\\[\\begin{align*} \\Pr(K) & = \\frac{140}{256}\\\\\r\n& = \\frac{35}{64}\r\n\\end{align*}\\]\r\nThere are \\(40\\) people who eat Mars bars and Rocky Road, but also \\(38\\) people who eat all three, so:\r\n\\[\\begin{align*} \\Pr(M\\cap R) & = \\frac{40+38}{256}\\\\\r\n& = \\frac{78}{256}\\\\\r\n& = \\frac{39}{128}\r\n\\end{align*}\\]\r\nThere are \\(30\\) people who eat KitKats and Mars bars, but not Rocky Road, so:\r\n\\[\\begin{align*} \\Pr(K\\cap M\\cap R') & = \\frac{30}{256}\\\\\r\n& = \\frac{15}{128}\r\n\\end{align*}\\]\r\nThere are \\(20\\) people who do not eat any of these, so the rest of the people must eat at least one. Therefore:\r\n\\[\\begin{align*} \\Pr(K\\cup M\\cup R) & = 1-\\Pr(K\\cup M\\cup R)'\\\\\r\n& = 1-\\frac{20}{256}\\\\\r\n& = \\frac{236}{256}\\\\\r\n& = \\frac{59}{64}\r\n\\end{align*}\\]\r\nThis is the same as adding up all the numbers inside the Venn diagram:\r\n\\[\\begin{align*} \\Pr(K\\cup M\\cup R) & \\frac{26+46+38+30+32+40+24}{256}\\\\\r\n& = \\frac{236}{256}\\\\\r\n& = \\frac{59}{64}\r\n\\end{align*}\\]\r\n\r\nDiagrams\r\nDepending on the type of data, it may help to use a unique diagram to represent the sample space. \r\n\r\nExample \u2013 finding probabilities using a diagram\r\nThe sample space for a deck of \\(52\\) cards is shown. Find the probability of the event (E) where the card drawn is a diamond (D) or an ace (A).\r\n\r\n\\(\\dfrac{1}{4}\\) of the cards are diamonds. \\(\\dfrac{4}{52}\\) are aces. Therefore:\r\n\\[\\begin{align*} \\Pr(E) & = \\Pr(D\\cup A)\\\\\r\n& = \\Pr(D)+\\Pr(A)-\\Pr(D\\cap A)\\\\\r\n& = \\frac{13}{52}+\\frac{4}{52}-\\frac{1}{52}\\\\\r\n& = \\frac{16}{52}\\\\\r\n& = \\frac{4}{13}\r\n\\end{align*}\\]\r\n\r\nExercise \u2013 finding probabilities using sample spaces\r\n\r\n\tUse a tree diagram to find the sample space for a two child family. Then, find the probability that:\r\n\t\r\n\t\tboth children are girls\r\n\t\tthe oldest child is a girl\r\n\t\tat least one child is a girl.\r\n\t\r\n\tThe table shows the sample space for tossing a single die twice. Going down a column is the result of the first throw and going along a row is the result of the second throw.\r\n\t\r\n\t\t\r\n\t\t\t\r\n\t\t\t\t\r\n\t\t\t\t\\(\\mathbf{1}\\)\r\n\t\t\t\t\\(\\mathbf{2}\\)\r\n\t\t\t\t\\(\\mathbf{3}\\)\r\n\t\t\t\t\\(\\mathbf{4}\\)\r\n\t\t\t\t\\(\\mathbf{5}\\)\r\n\t\t\t\t\\(\\mathbf{6}\\)\r\n\t\t\t\r\n\t\t\r\n\t\t\r\n\t\t\t\r\n\t\t\t\t\\(\\mathbf{1}\\)\r\n\t\t\t\t\\((1,1)\\)\r\n\t\t\t\t\\((1,2)\\)\r\n\t\t\t\t\\((1,3)\\)\r\n\t\t\t\t\\((1,4)\\)\r\n\t\t\t\t\\((1,5)\\)\r\n\t\t\t\t\\((1,6)\\)\r\n\t\t\t\r\n\t\t\t\r\n\t\t\t\t\\(\\mathbf{2}\\)\r\n\t\t\t\t\\((2,1)\\)\r\n\t\t\t\t\\((2,2)\\)\r\n\t\t\t\t\\((2,3)\\)\r\n\t\t\t\t\\((2,4)\\)\r\n\t\t\t\t\\((2,5)\\)\r\n\t\t\t\t\\((2,6)\\)\r\n\t\t\t\r\n\t\t\t\r\n\t\t\t\t\\(\\mathbf{3}\\)\r\n\t\t\t\t\\((3,1)\\)\r\n\t\t\t\t\\((3,2)\\)\r\n\t\t\t\t\\((3,3)\\)\r\n\t\t\t\t\\((3,4)\\)\r\n\t\t\t\t\\((3,5)\\)\r\n\t\t\t\t\\((3,6)\\)\r\n\t\t\t\r\n\t\t\t\r\n\t\t\t\t\\(\\mathbf{4}\\)\r\n\t\t\t\t\\((4,1)\\)\r\n\t\t\t\t\\((4,2)\\)\r\n\t\t\t\t\\((4,3)\\)\r\n\t\t\t\t\\((4,4)\\)\r\n\t\t\t\t\\((4,5)\\)\r\n\t\t\t\t\\((4,6)\\)\r\n\t\t\t\r\n\t\t\t\r\n\t\t\t\t\\(\\mathbf{5}\\)\r\n\t\t\t\t\\((5,1)\\)\r\n\t\t\t\t\\((5,2)\\)\r\n\t\t\t\t\\((5,3)\\)\r\n\t\t\t\t\\((5,4)\\)\r\n\t\t\t\t\\((5,5)\\)\r\n\t\t\t\t\\((5,6)\\)\r\n\t\t\t\r\n\t\t\t\r\n\t\t\t\t\\(\\mathbf{6}\\)\r\n\t\t\t\t\\((6,1)\\)\r\n\t\t\t\t\\((6,2)\\)\r\n\t\t\t\t\\((6,3)\\)\r\n\t\t\t\t\\((6,4)\\)\r\n\t\t\t\t\\((6,5)\\)\r\n\t\t\t\t\\((6,6)\\)\r\n\t\t\t\r\n\t\t\r\n\tFind the probability that:\r\n\t\r\n\t\tthe first toss is a \\(4\\)\r\n\t\tthe sum of the tosses is \\(5\\)\r\n\t\tat least one toss is a \\(6\\)\r\n\t\tneither toss is a \\(6\\).\r\n\t\r\n\tIn a classroom of \\(20\\) Year 12 VCE students, \\(10\\) study Maths Methods, \\(7\\) study Specialist Maths and \\(5\\) study both. Organise the information in a Venn diagram and find the probability that a student chosen at random studies:\r\n\t\r\n\t \tneither of these Maths subjects\r\n \t\tMaths Methods but not Specialist Maths.\r\n\t\r\n\tFind the probability that a card drawn at random from a pack is:\r\n\t\r\n\t\ta red card\r\n\t\tlower than a \\(5\\).\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nTree diagram for two coin tosses\r\nThe image is a tree diagram showing all possible outcomes for two consecutive coin flips, starting from the \"Start\" point.\r\n\r\n\tH (Heads): Leads to a second flip with two possible outcomes:\r\n\t\r\n\t\tH (Heads) resulting in the outcome HH\r\n\t\tT (Tails) resulting in the outcome HT\r\n\t\r\n\tT (Tails): Leads to a second flip with two possible outcomes:\r\n\t\r\n\t\tH (Heads) resulting in the outcome TH\r\n\t\tT (Tails) resulting in the outcome TT\r\n\t\r\n\r\n \r\nTree diagram for a family with three children\r\nThe image is a tree diagram showing all possible outcomes for a family with three children.\r\n\r\n\tb (boy): Leads to a second child with two possible outcomes:\r\n\t\r\n\t\tb (boy): Leads to a third child with two possible outcomes\r\n\t\t\r\n\t\t\tb (boy) resulting in the outcome bbb\r\n\t\t\tg (girl) resulting in the outcome bbg\r\n\t\t\r\n\t\tg (girl): Leads to a third child with two possible outcomes\r\n\t\t\r\n\t\t\tb (boy) resulting in the outcome bgb\r\n\t\t\tg (girl) resulting in the outcome bgg\r\n\t\t\r\n\t\r\n\tg (girl): Leads to a second child with two possible outcomes:\r\n\t\r\n\t\tb (boy): Leads to a third child with two possible outcomes\r\n\t\t\r\n\t\t\tb (boy) resulting in the outcome gbb\r\n\t\t\tg (girl) resulting in the outcome gbg\r\n\t\t\r\n\t\tg (girl): Leads to a third child with two possible outcomes\r\n\t\t\r\n\t\t\tb (boy) resulting in the outcome ggb\r\n\t\t\tg (girl) resulting in the outcome ggg\r\n\t\t\r\n\t\r\n\r\n","n":0.029},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":320,"$":[{"v":"Conditional probability","n":0.707},{"v":"\r\nIf two events are not independent, then the outcome of one event can change the probability of the second event occurring. Understanding the connection between these helps us make better predictions and decisions based on specific situations. Use this resource to learn about the concept of conditional probability.\r\nConditional probability is the chance that something happens, given that another event has already occurred. \r\n\r\n\r\nDependent events\r\nConditional probability often involves dependent events. Two events are dependent if the outcome or occurrence of the first affects the probability of the second. \r\nFor example, if you draw a card from a deck and don't replace it, the probability of drawing a second card of a certain type depends on the first draw.\r\nIn the same way, if you know that it is raining, the chance of carrying an umbrella might be higher.\r\nWhen two events, \\(A\\) and \\(B\\), are dependent, the probability of both occurring is given by: \r\n\\[\\Pr(A\\textrm{ and }B)=\\Pr(A\\cap B)=\\Pr(A)\\times \\Pr(B|A)\\]\r\nWe can rearrange this rule to find conditional probability. The probability of \\(B\\) given that \\(A\\) has occurred already is given by \\(\\Pr(B|A)\\), where:\r\n\\[\\Pr(B|A)=\\dfrac{\\Pr(A\\cap B)}{\\Pr(A)}\\]\r\nThe probability of \\(A\\) given that \\(B\\) has occurred already is given by \\(\\Pr(A|B)\\), where:\r\n\\[\\Pr(A|B)=\\dfrac{\\Pr(A\\cap B)}{\\Pr(B)}\\]\r\n\r\nExample 1 \u2013 calculating conditional probability\r\nA card is chosen at random from a pack. If the first card chosen is the jack of diamonds and it is not replaced, what is the probability that the second card is:\r\n\r\n \ta diamond?\r\n\ta jack?\r\n\tthe queen of clubs?\r\n\r\nBecause the first card was the jack of diamonds and not replaced, there is one less diamond in the pack and one less card in the pack. There are \\(12\\) diamonds remaining in the deck of \\(51\\) cards. Here, the second event is dependent on the first event because there is no replacement of cards.\r\n\\[\\begin{align*} \\Pr(\\diamondsuit_{2}|J\\diamondsuit_{1}) & = \\frac{12}{51}\\\\\r\n& = \\frac{4}{17}\r\n\\end{align*}\\]\r\nAfter a jack of diamonds is drawn, there are \\(3\\) jacks remaining in the deck of \\(51\\) cards. Here, the second event is also dependent on the first event because there is no replacement of cards\r\n\\[\\begin{align*} \\Pr(J_{2}|J\\diamondsuit_{1}) & = \\frac{3}{51}\\\\\r\n& = \\frac{1}{17}\r\n\\end{align*}\\]\r\nThere is only \\(1\\) queen of clubs in the remaining deck of \\(51\\) cards. Here, the second event is independent of the first event.\r\n\\[\\Pr(Q\\clubsuit_{2}|J\\diamondsuit_{1}) = \\frac{1}{51}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 calculating conditional probability\r\n\r\n\r\n\tThe results of a survey of music preferences are shown in the Venn diagram. \r\n\tFind the probability that a student likes live music given that they like karaoke.\r\n\t\r\n\tThree cards are chosen at random from a pack without replacement. What is the probability of choosing \\(3\\) aces?\r\n\tIn a maths class of \\(20\\) students, \\(5\\) failed the final exam. If \\(2\\) students are chosen at random without replacement, what is the probability that the first passed but the second failed?\r\n\tIf \\(\\Pr(X)=0.5\\), \\(\\Pr(Y)=0.5\\) and \\(\\Pr(X\\cap Y)=0.2\\), find the probability of:\r\n\t\r\n\t\t\\(\\Pr(X|Y)\\)\r\n\t\t\\(\\Pr(X\\cup Y)\\)\r\n\t\t\\(\\Pr(X)\\times \\Pr(Y|X)\\)\r\n\t\r\n\tIn a family with three children, what is the probability that all three children will be girls given that the first child is a girl?\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.046},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":321,"$":[{"v":"Binomial probability","n":0.707},{"v":"Binomial probability is used in various scenarios, like quality control and genetics. By understanding binomial probability, we can make informed predictions and decisions based on repeated experiments. Use this resource to learn how to apply the concepts of binomial probability.\r\n\r\nBinomial probability (\"bi\" meaning \"two\" and \"nomos\" meaning \"parts\") helps us find the chance of getting a certain number of successes in a series of tries. It's like flipping a coin and wanting to know how often it lands on heads after several flips.\r\nThis helps us predict outcomes when we have repeated experiments or tests, making it useful for planning and decisions.\r\n\r\n\r\nBinomial experiments\r\nIn a statistical experiment, a random variable may be described as having a binomial distribution when there are:\r\n\r\n\ta number of repeated trials\r\n\tonly two possible outcomes on each trial: success or failure.\r\n\r\nIf it produces a binomial distribution or binomial data, it involves binomial probability.\r\nSome examples of such experiments include:\r\n\r\n \tfinding the probability of obtaining \\(5\\) heads in \\(6\\) tosses of a coin, with six repeated trials, each with two possible outcomes: head or tail\r\n \tfinding the probability that in a randomly selected group of \\(30\\) people, none of them will have a particular disease, with \\(30\\) repeated \"trials\", each with two possible outcomes: have disease or do not have disease\r\n \tfinding the probability that in a sample of \\(100\\) manufactured components, no more than \\(2\\) will be defective, with \\(100\\) repeated trials, each with two possible outcomes: defective or not defective.\r\n\r\n\r\nExample \u2013 finding probability in binomial experiments\r\nIn a particular family, the probability that a child will have red hair is \\(\\dfrac{1}{4}\\). If the parents have three children, find the probability that:\r\n\r\n\tall three will have red hair\r\n\tnone will have red hair\r\n\tat least one child will have red hair\r\n\tonly the first child will have red hair\r\n\texactly one child will have red hair.\r\n\r\nThis is a binomial probability experiment because it involves three \"events\" (having three children) and two possible outcomes for each event (having red hair and not having red hair). Let's let \\(R\\) be the outcome of having red hair and \\(R'\\) be the outcome of not having red hair.\r\nFor all three to have red hair, we are looking at \\(\\Pr(R_{1}\\cap R_{2}\\cap R_{3})\\).\r\n\\[\\begin{align*} \\Pr(R_{1}\\cap R_{2}\\cap R_{3}) & = \\frac{1}{4}\\times\\frac{1}{4}\\times\\frac{1}{4}\\\\\r\n& = \\frac{1}{64}\r\n\\end{align*}\\]\r\nFor none with red hair, we are looking at \\(\\Pr(R'_{1}\\cap R'_{2}\\cap R'_{3})\\).\r\n\\[\\begin{align*} \\Pr(R'_{1}\\cap R'_{2}\\cap R'_{3}) & = \\frac{3}{4}\\times\\frac{3}{4}\\times\\frac{3}{4}\\\\\r\n& = \\frac{27}{64}\r\n\\end{align*}\\]\r\nThe outcome where at least one child with red hair complements the event where no children have red hair, so we can use the formula:\r\n\\[\\begin{align*} 1-\\Pr(R'_{1}\\cap R'_{2}\\cap R'_{3}) & = 1-\\frac{27}{64}\\\\\r\n& = \\frac{37}{64}\r\n\\end{align*}\\]\r\nFor only the first child with red hair, we are looking at:\r\n\\[\\begin{align*} \\Pr(R_{1}\\cap R'_{2}\\cap R'_{3}) & = \\frac{1}{4}\\times\\frac{3}{4}\\times\\frac{3}{4}\\\\\r\n& = \\frac{9}{64}\r\n\\end{align*}\\]\r\nThere are many ways in which only one child will have red hair. It could be the first child only, the second child only, or the first child only. Because of this, we must sum the probabilities of each of these.\r\n\r\n\r\nFinding binomial probability\r\nIn Example \u2013 finding probability on binomial experiments, we used the basic probability rules but there are specific rules we can use when we know we are dealing with binomial probability.\r\nIf \"\\(n\\)\" is the number of trials and \"\\(p\\)\" is the probability of the outcome of interest, then the probability of \"\\(x\\)\" outcomes is given by the formula:\r\n\\[\\Pr(X=x) = ^{n}C_{x}\\times p^{x}\\times(1-p)^{n-x}\\]\r\nwhere \\(^{n}C_{x}\\) is the number of ways of selecting \\(x\\) items from a population of \\(n\\) items, without regard to order, and \\(^{n}C_{x}=\\dfrac{n!}{x!(n-x)!}\\).\r\nThe exclamation mark (\\(!\\)) means that we are looking at factorials. For \\(n!\\), we multiply \\(n\\) by all of the integers counting from \\(n\\) to \\(1\\). This means \\(3!\\) would be \\(3\\times2\\times1=6\\) and \\(6!\\) would be \\(6\\times5\\times4\\times3\\times2\\times1=720\\).\r\nThese formulas may look a bit scary, but looking at an example will help!\r\n\r\nExample \u2013 using the binomial probability formula\r\nOne in every hundred items a machine produces is defective. What is the probability that, in a sample of five items produced by this machine:\r\n\r\n\texactly \\(3\\) are defective?\r\n\tnone are defective?\r\n\tat least \\(1\\) is defective?\r\n\r\nThe binomial probability formula uses \\(n\\) which is the number of items, \\(p\\) is the probability of the outcome and \\(x\\) is the number of times the outcome is achieved.\r\n\r\n\t\\(n=5\\)\r\n\t\\(p=\\dfrac{1}{100}\\)\r\n\r\n\\(x\\) depends on the question being asked: a, b or c.\r\nFor a, \\(x=3\\). Let's find \\(^{n}C_{x}\\) first.\r\n\\[\\begin{align*} ^{n}C_{x} & = \\frac{n!}{x!(n-x)!}\\\\\r\n^{5}C_{3} & = \\frac{5\\times4\\times3\\times2\\times1}{3\\times2\\times1(2\\times1)}\\\\\r\n& = 10\r\n\\end{align*}\\]\r\nNow, we can substitute this into the probability formula:\r\n\\[\\begin{align*} \\Pr(X=3) & = ^{n}C_{x}\\times p^{x}\\times(1-p)^{n-x}\\\\\r\n& = 10\\times\\left(\\frac{1}{100}\\right)^{3}\\times\\left(1-\\frac{1}{100}\\right)^{5-3}\\\\\r\n& = 10\\times\\left(\\frac{1}{1000000}\\right)\\times\\left(\\frac{99}{100}\\right)^{2}\\\\\r\n& \\approx 0.00001\r\n\\end{align*}\\]\r\nFor b, \\(x=0\\).\r\n\\[\\begin{align*} ^{n}C_{x} & = \\frac{n!}{x!(n-x)!}\\\\\r\n^{5}C_{0} & = \\frac{5\\times4\\times3\\times2\\times1}{1(5\\times4\\times3\\times2\\times1)}\\\\\r\n& = 1\r\n\\end{align*}\\] \r\n\\[\\begin{align*} \\Pr(X=0) & = ^{n}C_{x}\\times p^{x}\\times(1-p)^{n-x}\\\\\r\n& = 1\\times\\left(\\frac{1}{100}\\right)^{0}\\times\\left(1-\\frac{1}{100}\\right)^{5}\\\\\r\n& = \\left(\\frac{99}{100}\\right)^{5}\\\\\\\r\n& = 0.95\r\n\\end{align*}\\]\r\nFor c, at least \\(1\\) defective is complementary to all defective, so we can calculate the probability easily.\r\n\\[\\begin{align*} \\Pr(X\\geq1) & = 1-\\Pr(X=0)\\\\\r\n& = 1-0.95\\\\\r\n& = 0.05\r\n\\end{align*}\\]\r\n\r\nBinomial probability distribution\r\nA list of all possible outcomes of an event and their associated probabilities is called a probability distribution.\r\nThis is useful for calculating values like:\r\n\r\n\tmean (\\(\\mu\\) or \\(E(X)\\)), which tells us the expected number of successes or expected result in a given number of trials\r\n\t\\[\\mu=E(X)=np\\]\r\n\tstandard deviation (\\(\\sigma\\) or \\(SD(x)\\), which tells us how much the actual number of successes might vary from the average.\r\n\t\\[\\sigma=SD(x)=\\sqrt{np(1-p)}\\]\r\n\r\n\r\nExample \u2013 summarising data from binomial probability experiments\r\nOne in every hundred items a machine produces is defective. The probability distribution table shown is for the event where the item is defective \\(X\\), to five decimal places.\r\n\r\n\t\r\n\t\t\r\n\t\t\t\\(\\mathbf{x}\\) (Number of defective items)\r\n\t\t\t\\(0\\)\r\n\t\t\t\\(1\\)\r\n\t\t\t\\(2\\)\r\n\t\t\t\\(3\\)\r\n\t\t\t\\(4\\)\r\n\t\t\t\\(5\\)\r\n\t\t\r\n\t\r\n\t\r\n\t\t\r\n\t\t\t\\(\\mathbf{\\Pr(X=x)}\\)\r\n\t\t\t\\(0.95099\\)\r\n\t\t\t\\(0.04803\\)\r\n\t\t\t\\(0.00097\\)\r\n\t\t\t\\(0.00001\\)\r\n\t\t\t\\(0.00000\\)\r\n\t\t\t\\(0.00000\\)\r\n\t\t\r\n\t\r\n\r\nFind the expected number of defectives in a batch of \\(1000\\) items, including the standard deviation.\r\n\\[\\begin{align*} \\mu & = np\\\\\r\n& = 1000\\times\\frac{1}{100}\\\\\r\n& = 10\r\n\\end{align*}\\] \\[\\begin{align*} \\sigma & = \\sqrt{np(1-p)}\\\\\r\n& = \\sqrt{1000\\times\\frac{1}{100}\\left(1-\\frac{99}{100}\\right)}\\\\\r\n& = 3.15\r\n\\end{align*}\\]\r\nThis means that in a batch of \\(1000\\), we would expect to get \\(10\\) defectives and the number of defectives will deviate from this amount by an \"average\" of \\(3.15\\). \r\nWe would expect most batches to have between: \\(10-3.15\\approx7\\) and \\(10+3.15\\approx13\\) defective items.\r\n\r\nExercise \u2013 finding binomial probability\r\n\r\n\tThe probability that an archer will hit a bullseye is \\(0.7\\). If he is allowed ten attempts, find the probability that he:\r\n\t\r\n\t\thits it every time\r\n\t\tmisses each time\r\n\t\tscores two bulls-eyes\r\n\t\tscores at least two bulls-eyes\r\n\t\r\n\tHow many bulls-eyes would you expect the archer from the previous question to score?\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.032},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":322,"$":[{"v":"Solving quadratic equations","n":0.577},{"v":"Just as linear equations can be solved to find the value of an unknown variable, we can do the same with quadratic equations. Here, we will learn how to solve quadratic equations using the null factor law. By applying this principle, you can factorise quadratic equations and determine their solutions efficiently.\r\n\r\nVideo tutorial \u2013 solving quadratic equations\r\nWatch this video to learn how to solve quadratic equations using the null factor law.\r\n\r\n\r\nSolving quadratic equations\r\nJust like how solving linear equations gives you the coordinates where the graph intercepts the axes, the solutions for a quadratic equation are also the intercepts.\r\n\r\nWhen \\(a\\neq0\\), the quadratic graph can intercept the \\(x\\)-axis at two points. Therefore, there can be \\(2\\) solutions for a quadratic.\r\n\r\nNull factor law\r\nIf the equation can be factorised, then the null factor law can be used to find the solutions.\r\nThe null factor law states that if the product of two terms is \\(0\\), then the terms themselves are \\(0\\). That is, if \\(m\\times n=0\\), then \\(m=0\\) and/or \\(n=0\\).\r\nIf the product of two or more factors is zero, then any one of the individual factors may be zero and can provide a solution for the equation.\r\nWe can apply this to quadratics. Suppose that the quadratic equation can be factorised into the form \\((a+b)(c+d)=0\\). Using the null factor law, we can say that \\(a+b=0\\) and \\(c+d=0\\).\r\n\r\n\r\nExample 1 \u2013 solving quadratic equations using the null factor law\r\nFind all possible values of \\(y\\) for the quadratic \\(y^{2}=5y\\).\r\nRearrange the equation to the form \\(ax^{2}+bx+c\\), then factorise.\r\n\\[\\begin{align*} y^{2} & =5y\\\\\r\ny^{2}-5y & = 0\\quad\\textrm{rearrange and factorise}\\\\\r\ny(y-5) & = 0\r\n\\end{align*}\\]\r\nAccording to the null factor law, if \\(m\\times n=0\\), then \\(m=0\\) and \\(n=0\\). Here:\r\n\\[\\begin{align*} y=0 &\\textrm{ or }y-5=0\\\\\r\n\\therefore y=0 &\\textrm{ or }y=5\r\n\\end{align*}\\]\r\nWe can check our answer by substituting the values back into the quadratic equation. If \\(y=0\\):\r\n\\[\\begin{align*} \\textrm{LHS} & = y^{2}\\\\\r\n& = (0)^{2}\\\\\r\n& = 0\\\\\r\n\\textrm{RHS} & = 5y\\\\\r\n& = 5(0)\\\\\r\n& = 0\\\\\r\n& = \\textrm{LHS}\r\n\\end{align*}\\]\r\nIf \\(y=5\\):\r\n\\[\\begin{align*} \\textrm{LHS} & = y^{2}\\\\\r\n& = (5)^{2}\\\\\r\n& = 25\\\\\r\n\\textrm{RHS} & = 5y\\\\\r\n& = 5(5)\\\\\r\n& = 25\\\\\r\n& = \\textrm{LHS}\r\n\\end{align*}\\]\r\nTherefore, \\(y=0\\) and \\(y=5\\) are both valid solutions for \\(y^{2}=5y\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 solving quadratic equations using the null factor law\r\nSolve the following quadratic equations using the null factor law.\r\n\r\n\t\\(x^{2}-6x+8=0\\)\r\n\t\\(x^{2}+2x-3=0\\)\r\n\t\\(2x^{2}-3x-2=0\\)\r\n\t\\(6-z-z^{2}=0\\)\r\n\t\\(2x^{2}+7x=15\\)\r\n\t\\(11p=3(2p^{2}+1)\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on solving quadratic equations.\r\n\r\nThe general form of a quadratic equation is A X squared plus B X plus C equals zero where A is not equal to one. For instance, five X squared minus three X plus nine equals zero. In this case A equals five, B equals minus three and C equals nine and if you have the equation X squared equals five X minus four you then have to rearrange it into the form A X squared plus B X plus C equals zero, in other words this becomes X squared minus five X plus four equals zero where A equals one, B equals minus five and C equals four.\r\n\r\nLet\u2019s start with Y squared equals five Y. As you can see this is not of the form A X squared plus B X plus C equals zero, so rearrange it to make it look like that, Y squared minus five Y equals zero. Notice that this will factorise into Y brackets Y minus five. Then we have the form M times N equals zero, Y times Y minus five equals zero, so we can use the null factor law where Y equals zero or Y minus five equals zero, from which Y equals zero or Y equals five. Notice when you get your answers it\u2019s a good idea to check that when substituting Y equals zero and Y equals five back into the original Y squared equals five Y that you get the left hand side equal to the right hand side. Do this for yourself.\r\n\r\nLet\u2019s look at this example, X squared minus five X plus four equals zero. Note how it\u2019s different to example one, there\u2019s a trinomial here, X squared term and X term and a number term, a constant. That will factorise into X minus four X minus one, again we have our null factor law X minus four times X minus one, so using that law we get X minus four equals zero or X minus one equals zero from which X equals four or X equals one. It\u2019s always a good idea to check to see that the answers are correct by substituting back into the original equation. You can do this for yourself.\r\n\r\nLet\u2019s look at another trinomial, P square plus 10 P plus 25 equals zero. This will factorise into P plus five P plus five. Notice it\u2019s a perfect square and thus, using the null factor law, we have just one P plus five equals zero from which P equals minus five. Again, do the check to make sure the left hand side of the original equation equals the right hand side, which is zero.\r\n\r\nHere\u2019s another example, four M squared minus 49 equals zero. Notice that this is the difference of a perfect square, in other words it becomes two M plus seven two M minus seven equals zero. Again using the null factor law two M plus seven equals zero two M minus seven equals zero from which M equals minus seven over two or M equals plus seven over two.\r\n\r\nAnd finally here\u2019s a more complicated quadratic equation to solve. Here we\u2019re given X equals minus six over one minus two X so we have to do some rearranging first to turn it into A X square plus B X plus C equals zero, and that\u2019s what I\u2019ve done in this second line, so we have two X squared minus X minus six equals zero which needs to be factorised and it does, into two X plus three X minus two equals zero, using the null factor law we arrive at X equals minus three over two or X equals two, so those are our solutions.\r\n\r\nNow do some problems for yourselves. The answers to these six questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.031},[{"v":"Maths equations","i":2,"n":0.707},{"v":"Maths","i":1,"n":1},{"v":"Equations","i":0,"n":1}]]},{"i":323,"$":[{"v":"The structure of the periodic table","n":0.408},{"v":"The periodic table is more than just a list of elements; it is a structured map that reveals patterns and relationships among them. By examining the arrangement of elements, we gain valuable insights into their properties and behaviours.\r\n\r\nThe periodic table arranges elements in order of increasing atomic number. The rows in the periodic table are known as periods. As an example, \\(\\ce{C}\\), \\(\\ce{N}\\) and \\(\\ce{O}\\) are in period \\(2\\). The columns in the periodic table are called groups. \\(\\ce{H}\\), \\(\\ce{Li}\\) and \\(\\ce{Na}\\) are group \\(1\\) elements.\r\n\r\nYou can select elements in the periodic table to learn more about their properties and history.\r\n\r\n\r\n\r\n\r\n\r\nClassification of the elements by properties\r\nElements can be classified based on their physical and chemical properties. \r\nPhysical properties\r\nBased primarily on selected physical properties, elements are classified as metals and non-metals. Metals are located on the left side of the periodic table, and non-metals are located on the right. Elements know as metalloids exhibit both metallic and non-metallic properties. Metalloids are located between metals and non-metals in the periodic table.\r\nThe physical properties of metals, non-metals and metalloids are shown in the table.\r\n\r\nChemical properties\r\nIf we focus on chemical properties, we can classify elements as alkali metals, alkaline earth metals, halogens and noble gases. The chemical properties of elements in these classes are shown in the table.\r\n\r\nClassification of the elements by electron configuration\r\nElements in the same group have similar chemical properties due to the similarity in their outermost electron shell (valence) configurations. Thus, the electrons in the outermost shell (largest and highest-energy) are known as valence electrons. Valence electrons are accessible to other chemical species. This allows them to participate in chemical reactions. Elements in the same group have the same number of valence electrons.\r\nThe rest of the electrons in the atom are known as core electrons. Core electrons are tightly bound to the nucleus and are considered inaccessible.\r\nAll of the elements in the table below have one \\(s\\) electron in their outermost electron shell. Group \\(1\\) elements, except hydrogen, are soft metals and are highly reactivity with water and oxygen.\r\n\r\n\\(s\\)-, \\(p\\)-, \\(d\\)-, and \\(f\\)-blocks\r\nThe elements in the periodic table are categorised into \\(s\\)-, \\(p\\)-, \\(d\\)- and \\(f\\)-blocks based on the location of the last electron in the valence shell. The \\(s\\)- block contains elements that have their last electron in \\(s\\) subshell. Similarly, \\(p\\)-, \\(d\\)-, and \\(f\\)-blocks have elements with their last electron located in \\(p\\), \\(d\\), and \\(f\\) subshells, respectively. The \\(s\\)-, \\(p\\)-, \\(d\\)- and \\(f\\)-blocks are shown in the figure.\r\n\r\n\r\n\r\n\r\nBased on the location of the last electron in an element, elements can also be classified as a noble gas, main group (or representative) element, transition element or inner transition element. This is summarised in the table and figure.\r\n\r\n\r\n\r\nExample \u2013 interpreting the periodic table\r\nDetermine the number of valence electrons in \\(\\ce{P}\\), \\(\\ce{F}\\), \\(\\ce{Ca}\\) and \\(\\ce{Fe}\\).\r\nStep 1: Recall how the periodic table is organised, in relation to the number of valence electrons.\r\nThe periodic table is organised such that elements in the same group have the same number of valence electrons. Thus, the group numbers of these elements can be used to determine the number of valence electrons.\r\nStep 2: For groups \\(1-10\\), group number is equal to the number of valence electrons. For groups \\(12-18\\), subtract \\(10\\) out of the group number to get the number of valence electrons. For group \\(11\\), we would expect the number of valence electrons to be \\(1\\) due to the \\(s^{1}\\) electron configuration. However, there are a few exceptions in this case.\r\n\r\nYour turn \u2013 the structure of the periodic table\r\nTest yourself on your understanding of the structure of the periodic table.\r\n\r\n\r\n\r\n \r\nInteractive periodic table\r\nInteractive periodic table displaying elements. Users can hover over or click on elements to view details such as atomic number, symbol, and properties.\r\n \r\nA colour-coded periodic table showing the s block in green, d block in blue, p block in pink, and f block in orange. Each block includes the corresponding elements grouped by their electron configurations.\r\n \r\nA periodic table colour-coded by categories: representative elements in pink, transition elements in blue, inner transition elements in green, and noble gases in orange. Each category groups elements with similar properties.\r\n","n":0.038},[{"v":"Periodic table","i":1,"n":0.707},{"v":"Chemistry","i":0,"n":1}]]},{"i":324,"$":[{"v":"The electron","n":0.707},{"v":"Electrons are fundamental subatomic particles with a negative electric charge. They play a crucial role in many aspects of science and technology.\r\n\r\nBohr planetary (particle) model and Schr\u00f6dinger's wave mechanical model\r\nElectrons have historically been described as tiny particles that revolve around the nucleus in circular orbits similar to the solar system. This is Bohr's particle model. However, this model overly simplifies the path of electrons within atoms and was eventually rejected.\r\n\r\nThe modern understanding of electron behaviour is based on Schr\u00f6dinger's wave mechanical model. \r\nThe wave model shows that electrons don't have exact paths. Instead, they exist in regions of space where they are likely to be found. These regions are called orbitals. In this model, electrons occupy a three-dimensional space around the nucleus.\r\n\r\n\r\nWhen an electron's wave fits perfectly within an orbital, it creates a stable energy level, known as a \"stationary state.\" This means the electron's energy is quantised, meaning that it can exist only in certain allowed states. These stable energy levels are crucial for understanding how electrons behave in atoms, influencing how atoms absorb and emit energy, and determining the chemical properties of elements. Orbitals, therefore, provide a framework for predicting where electrons are likely to be and how they will interact with other atoms in chemical reactions.\r\n\r\nElectron energy levels\r\nThe space where electrons move around the nucleus is divided into subspaces known as shells, subshells and orbitals.\r\n\r\nElectron shells\r\n\r\n\r\n\r\n\r\n\r\nElectron subshells\r\nWithin each electron shell are subshells. The number of subshells within a shell is the same as the shell number. For example, the \\(n=3\\) shell contains \\(3\\textrm{ subshells}\\) and the \\(n=6\\) shell contains \\(6\\textrm{ subshells}\\). Subshells are labeled as \\(s,p,d\\) and \\(f\\). Each subshell contains a certain number of electrons:\r\n\r\n\tAn \\(s\\) subshell contains up to \\(2\\) electrons.\r\n\tA \\(p\\) subshell contains up to \\(6\\) electrons.\r\n\tA \\(d\\) subshell contains up to \\(10\\) electrons.\r\n\tA \\(f\\) subshell contains up to \\(14\\) electrons.\r\n\r\nLet's consider the \\(n=4\\) shell. This shell contains \\(4\\) subshells, which are labeled as \\(4s\\), \\(4p\\), \\(4d\\), and \\(4f\\). Remember, the \\(4s\\) subshell can hold up to \\(2\\) electrons, the \\(4p\\) subshell can hold up to \\(6\\) electrons, the \\(4d\\) subshell can accommodate up to \\(10\\) electrons, and the \\(4f\\) subshell can contain up to \\(14\\) electrons. Altogether, the \\(n=4\\) shell can hold a maximum of \\(2+6+10+14=32\\) electrons.\r\nElectron subshells are filled in order of increasing energy. The figure shows the electron filling order. Generally, shells fill from \\(1\\) to \\(2\\) to \\(3\\) and subshells, from \\(s\\) to \\(p\\) to \\(d\\), but there is an exception. Subshell \\(4s\\) has lower energy than subshell \\(3d\\); therefore, \\(4s\\) fills before \\(3d\\).\r\n\r\n\r\nElectron orbitals\r\nWithin each electron subshell are electron orbitals. Each electron orbital can hold up to two electrons, where one spin upwards and the other spins downwards. This means that:\r\n\r\n\tAn \\(s\\) subshell, which contains up to \\(2\\) electrons, has one orbital.\r\n\tA \\(p\\) subshell, which contains up to \\(6\\) electrons, has three orbitals.\r\n\tA \\(d\\) subshell, which contains up to \\(10\\) electrons, has five orbitals.\r\n\tAn \\(f\\) subshell, which contains up to \\(14\\) electrons, has seven orbitals.\r\n\r\nYou can see that the number of orbitals is half of the number of maximum electrons.\r\nOrbitals have distinct shapes. For example, \\(s\\) orbitals are spherical and \\(p\\) orbitals are dumbbell-shaped. Orbitals within the same subshell that have the same shape are generally of equal energy but have a different orientation in space, such as \\(p_{x}\\), \\(p_{y}\\) and \\(p_{z}\\), where \\(x\\), \\(y\\), and \\(z\\) denote the orientation in space.\r\n\r\n\r\nLike shells and subshells, electron fill orbitals from lowest to highest energy level. This is called the Aufbau principle. Electrons also fill orbitals of a subshell such that each orbital acquires one electron before any orbital acquires a second electron. All single electrons must have the same spin.\r\n\r\nElectron configurations\r\nElectron configurations show how the electrons are distributed within an atom. They are written from lowest energy subshell to the highest energy subshell. Each subshell is written with the number of electrons in each subshell shown as a superscript. The superscripts in an electron configuration will add up to the total number of electrons for that atom. \r\nFull electron configurations\r\nFull electron configurations show the location of all electrons in an atom. The full electron configurations of the first 10 elements are shown in the table.\r\n\r\n\r\n\r\nElement\r\nAtomic number\r\nElectron configuration\r\n\r\n\r\n\r\n\r\n\\(\\ce{H}\\)\r\n\\(1\\)\r\n\\(1s^{1}\\)\r\n\r\n\r\n\\(\\ce{He}\\)\r\n\\(2\\)\r\n\\(1s^{2}\\)\r\n\r\n\r\n\\(\\ce{Li}\\)\r\n\\(3\\)\r\n\\(1s^{2}\\,2s^{1}\\)\r\n\r\n\r\n\\(\\ce{Be}\\)\r\n\\(4\\)\r\n\\(1s^{2}\\,2s^{2}\\)\r\n\r\n\r\n\\(\\ce{B}\\)\r\n\\(5\\)\r\n\\(1s^{2}\\,2s^{2}2p^{1}\\)\r\n\r\n\r\n\\(\\ce{C}\\)\r\n\\(6\\)\r\n\\(1s^{2}\\,2s^{2}2p^{2}\\)\r\n\r\n\r\n\\(\\ce{N}\\)\r\n\\(7\\)\r\n\\(1s^{2}\\,2s^{2}2p^{3}\\)\r\n\r\n\r\n\\(\\ce{O}\\)\r\n\\(8\\)\r\n\\(1s^{2}\\,2s^{2}2p^{4}\\)\r\n\r\n\r\n\\(\\ce{F}\\)\r\n\\(9\\)\r\n\\(1s^{2}\\,2s^{2}2p^{5}\\)\r\n\r\n\r\n\\(\\ce{Ne}\\)\r\n\\(10\\)\r\n\\(1s^{2}\\,2s^{2}2p^{6}\\)\r\n\r\n\r\n\r\n\r\n\r\nCondensed electron configuration\r\nCondensed electron configurations are a short way of showing electron configurations. They are often also called noble gas configurations and are handy to avoid writing out all of the subshells. This way of representing electron configurations states the previous noble gas in square brackets, followed by the remaining electrons.\r\nFor example, let's look at the magnesium atom, which has a full electron configuration of \\(1s^{2}\\,2s^{2}\\,2p^{6}\\,3s^{2}\\). The previous noble gas, as shown on the periodic table, is neon. Neon has an electron configuration of \\(1s^{2}\\,2s^{2}\\,2p^{6}\\). You can see that the location of the first 10 electrons is the same as in magnesium, so we can replace this part of the electron configuration with the chemical symbol for the noble gas in square brackets: \\(\\textrm{[Ne]}\\). This makes the condensed electron configuration for magnesium: \\(\\textrm{[Ne]}\\,3s^{2}\\).\r\n\r\nOrbital diagrams\r\nOrbital diagrams depict spin of individual electrons within each subshell. The orbital diagrams for the first 10 elements are also shown.\r\n\r\n\r\n\r\n\r\nExample 1 \u2013 writing full electron configurations\r\nWrite the full electron configuration for \\(\\ce{Na}\\).\r\nStep 1: Determine the number of electrons. This is the same as the atomic number (Z).\r\n\\[Z=11\\]\r\nTherefore, \\(\\ce{Na}\\) has \\(11\\textrm{ electrons}\\).\r\nStep 2: Place the electrons into orbitals according to the Aufbau principle. The first two electrons go into the \\(1s\\) orbital as a pair. The next two electrons are placed in the \\(2s\\) orbital. The next six electrons are placed in \\(2p\\) orbitals as three pairs. The remaining electron is placed in \\(3s\\) orbital.\r\n\r\n\r\n\r\nStep 3: Convert the orbital diagram into an electron configuration, listing the subshells in order of increasing energy. Write the total number of electrons in each subshell as a superscript.\r\n\\(\\ce{Na}\\) has the full electron configuration \\(1s^{2}\\,2s^{2}\\,2p^{6}\\,3s^{1}\\).\r\n\r\n\r\n\r\nExample 3 \u2013 writing condensed electron configurations\r\nWrite the condensed electron configuration for \\(\\ce{Ca}\\).\r\nStep 1: Use the steps in Example 1 \u2013 writing full electron configurations to write the full electron configuration for \\(\\ce{Ca}\\).\r\n\\(\\ce{Ca}\\) has the full electron configuration \\(1s^{2}\\,2s^{2}\\,2p^{6}\\,3s^{2}\\,3p^{6}\\,4s^{2}\\).\r\nStep 2: Determine the full electron configuration of the previous noble gas, according to the periodic table. For \\(\\ce{Ca}\\), the previous noble gas is \\(\\ce{Ar}\\).\r\n\\(\\ce{Ar}\\) has the full electron configuration \\(1s^{2}\\,2s^{2}\\,2p^{6}\\,3s^{2}\\,3p^{6}\\).\r\nStep 3: Write the condensed electron configuration by replacing the electron configuration of \\(\\ce{Ar}\\) within the electron configuration of \\(\\ce{Ca}\\) with \\(\\textrm{[Ar]}\\).\r\n\\(\\ce{Ca}\\) has the condensed electron configuration \\(\\textrm{[Ar]}\\,4s^{2}\\).\r\n\r\nYour turn \u2013 the electron\r\nTest yourself on your understanding of electrons.\r\n\r\n\r\n\r\n \r\n    Bohr and Schr\u00f6dinger models\r\n    \r\n        The Bohr model\r\n        \r\n            A central black dot labeled \"p+\" representing the nucleus.\r\n            A red oval path around the nucleus labeled \"e-\" representing an electron orbit.\r\n        \r\n    \r\n    \r\n        The Schr\u00f6dinger model\r\n        \r\n            A central black dot labeled \"p+\" representing the nucleus.\r\n            A complex, cloud-like blue shape around the nucleus labeled \"e-\" representing an electron cloud.\r\n        \r\n    \r\n \r\n    Electron shell diagram:\r\n    \r\n        A central black dot representing the nucleus.\r\n        Three concentric circles around the nucleus, labeled:\r\n            \r\n                n=1 (innermost circle)\r\n                n=2 (middle circle)\r\n                n=3 (outermost circle)\r\n            \r\n        \r\n        A red arrow pointing outward from the nucleus through the circles, indicating \"Increasing energy and distance from the nucleus\".\r\n    \r\n \r\n    Electron shell filling diagram\r\n    \r\n        Left side: A vertical arrow labeled \"Energy\" pointing upwards.\r\n        \r\n            Top labels:\r\n            \r\n                \"Shell number\" on the left\r\n                \"Filling order\" on the right\r\n            \r\n        \r\n        \r\n            Energy levels (from bottom to top):\r\n            \r\n                Shell 1: 1s (blue line)\r\n                Shell 2:\r\n                    \r\n                        2s (red line)\r\n                        2p (red line)\r\n                    \r\n                \r\n                Shell 3:\r\n                    \r\n                        3s (yellow line)\r\n                        3p (yellow line)\r\n                        3d (yellow line)\r\n                    \r\n                \r\n                Shell 4:\r\n                    \r\n                        4s (blue line)\r\n                        4p (blue line)\r\n                        4d (blue line)\r\n                    \r\n                \r\n                Shell 5:\r\n                    \r\n                        5s (red line)\r\n                        5p (red line)\r\n                    \r\n                \r\n            \r\n        \r\n        Connecting lines: Lines indicate the order of filling, showing connections between shells such as from 3p to 4s to 3d.\r\n    \r\n \r\n    Electron orbitals\r\n    \r\n        s orbital\r\n        \r\n            A red spherical shape labeled \"s orbital\".\r\n            Axes labeled x, y, and z around the sphere.\r\n        \r\n    \r\n    \r\n    \r\n        p orbitals\r\n        \r\n            \r\n                px orbital: \r\n                \r\n                    Two blue lobes along the x-axis labeled \"px orbital\".\r\n                    Axes labeled x, y, and z.\r\n                \r\n            \r\n            \r\n                py orbital: \r\n                \r\n                    Two blue lobes along the y-axis labeled \"py orbital\".\r\n                    Axes labeled x, y, and z.\r\n                \r\n            \r\n            \r\n                pz orbital: \r\n                \r\n                    Two blue lobes along the z-axis labeled \"pz orbital\".\r\n                    Axes labeled x, y, and z.\r\n                \r\n            \r\n        \r\n \r\n    Aufbau principle diagram\r\n    \r\n        \r\n            Energy levels labeled n=1 to n=7 on the left.\r\n        \r\n        \r\n            Diagonal lines indicate the filling order for orbitals:\r\n            \r\n                n=1: 1s\r\n                n=2: 2s, 2p\r\n                n=3: 3s, 3p, 3d\r\n                n=4: 4s, 4p, 4d, 4f\r\n                n=5: 5s, 5p, 5d, 5f\r\n                n=6: 6s, 6p, 6d, 6f\r\n                n=7: 7s, 7p, 7d, 7f\r\n            \r\n        \r\n        \r\n            Black and red arrows guide the sequence in which orbitals are filled.\r\n        \r\n    \r\n \r\n    Electron configurations and corresponding orbital diagrams\r\n    \r\n        \r\n            \r\n                Element\r\n                Electron configuration\r\n                Orbital diagram\r\n            \r\n        \r\n        \r\n            \r\n                H\r\n                1s1\r\n                1s: \u2191\r\n            \r\n            \r\n                He\r\n                1s2\r\n                1s: \u2191\u2193\r\n            \r\n            \r\n                Li\r\n                1s2 2s1\r\n                1s: \u2191\u2193 2s: \u2191\r\n            \r\n            \r\n                C\r\n                1s2 2s2 2p2\r\n                1s: \u2191\u2193 2s: \u2191\u2193 2p: \u2191 \u2191\r\n            \r\n            \r\n                N\r\n                1s2 2s2 2p3\r\n                1s: \u2191\u2193 2s: \u2191\u2193 2p: \u2191 \u2191 \u2191\r\n            \r\n            \r\n                O\r\n                1s2 2s2 2p4\r\n                1s: \u2191\u2193 2s: \u2191\u2193 2p: \u2191\u2193 \u2191 \u2191\r\n            \r\n        \r\n    \r\n \r\n\t\r\n\t\r\n    \r\n    Element\r\n    Electron configuration\r\n    Orbital diagram\r\n    \r\n    \r\n    \r\n    \r\n    Na\r\n    1s2 2s2 2p6 3s1\r\n    1s: \u2191\u2193 2s: \u2191\u2193 2p: \u2191\u2193 \u2191\u2193 \u2191\u2193 3s: \u2191\r\n    \r\n\t\r\n\t\r\n","n":0.025},[{"v":"Electron","i":2,"n":1},{"v":"Chemistry","i":1,"n":1},{"v":"Atomic Structures","i":0,"n":0.707}]]},{"i":325,"$":[{"v":"Eigenvalues and eigenvectors","n":0.577},{"v":"\r\nEigenvalues and eigenvectors are important parts of an engineer\u2019s mathematical toolbox. They give us an understanding of how buildings, structures, automobiles and materials behave in real life. But that's not all they are used for; you can see their use in many other areas of STEM, including colour theory, electric circuits, facial recognition systems, data science and quantum mechanics.\r\n\r\nAn eigenvalue is a special number, called a scalar, that is linked to a square matrix. It shows how much an eigenvector, which is a specific non-zero vector, is stretched or compressed by the matrix.\r\n\"Eigen\" comes from the German word for \"own\", so eigenvectors and eigenvalues are vectors and values owned by the matrix.\r\nLet's consider a square, \\(n\\times n\\) matrix called \\(A\\). If \\(X\\) is a non-zero vector of order \\(n\\times1\\) and \\(\\lambda\\) is a scalar, and they satisfy the equation:\r\n\\[AX = \\lambda X\\]\r\nthen \\(\\lambda\\) is the eigenvalue of matrix \\(A\\), and \\(X\\) is the eigenvector.\r\nIn simpler terms, when the matrix \\(A\\) acts on the vector \\(X\\), it transforms it, and the eigenvalue \\(\\lambda\\) tells us how much \\(X\\) is stretched or shrunk.\r\n\\(X\\) is called a column vector (or column matrix) and is generally written as:\r\n\\[X=\\left[ \\begin{array}{c} x_{1}\\\\\r\nx_{2}\\\\\r\n\\vdots\\\\\r\nx_{n}\r\n\\end{array} \\right] \\]\r\nThis means that if \\(A\\) is a \\(2\\times2\\) matrix:\r\n\\[X=\\left[ \\begin{array}{c} x\\\\\r\ny\r\n\\end{array} \\right] \\]\r\nSimilarly, if \\(A\\) is a \\(3\\times3\\) matrix:\r\n\\[X=\\left[ \\begin{array}{c} x\\\\\r\ny\\\\\r\nz\r\n\\end{array} \\right] \\]\r\n\r\nFinding eigenvalues for \\(2\\times2\\) matrices\r\nTo find the eigenvalue for a square matrix \\(A\\), we solve the characteristic equation.\r\n\\[\\det \\left[ A-\\lambda I \\right] = \\left| A-\\lambda I \\right| = 0 \\]\r\nIf \\(A\\) is an \\(n\\times n\\) matrix, there will be at most \\(n\\) distinct eigenvalues of \\(A\\).\r\n\r\nExample 1 \u2013 finding eigenvalues for \\(2\\times2\\) matrices\r\nFind the eigenvalues of the matrix:\r\n\\[A = \\left[ \\begin{array}{cc} -4 & -2\\\\\r\n3 & 3 \r\n\\end{array} \\right] \\]\r\nWe need to find the characteristic equation for matrix \\(A\\). It includes the \\(2\\times2\\) identity matrix.\r\n\\[I = \\left[ \\begin{array}{cc} 1 & 0\\\\ \r\n0 & 1 \r\n\\end{array} \\right] \\]\r\nWe can substitute this into \\(A-\\lambda I\\).\r\n\\[\\begin{align*} A-\\lambda I & = \\left[ \\begin{array}{cc} -4 & -2\\\\\r\n3 & 3\r\n\\end{array} \\right]-\\lambda \\left[ \\begin{array}{cc} 1 & 0\\\\ \r\n0 & 1\r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} -4 & -2\\\\\r\n3 & 3\r\n\\end{array} \\right]-\\left[ \\begin{array}{cc} \\lambda & 0\\\\ \r\n0 & \\lambda \r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} -4-\\lambda & -2\\\\\r\n3 & 3-\\lambda\r\n\\end{array} \\right]\r\n\\end{align*}\\]\r\nTo find the eigenvalues, we solve \\(\\det[A-\\lambda I]=\\left|A-\\lambda I\\right|=0\\).\r\n\\[\\begin{align*} \\det \\left[ A-\\lambda I\\right] & = 0\\\\ \\det \\left[ \\begin{array}{cc} -4-\\lambda & -2\\\\\r\n3 & 3-\\lambda \\end{array} \\right] & = 0\\\\\r\n\\left( -4-\\lambda \\right)\\left( 3-\\lambda \\right) -3 \\left( -2 \\right) & = 0\\\\ \r\n-12+4\\lambda-3\\lambda+\\lambda^{2}+6 & = 0\\\\ \r\n\\lambda^{2}+\\lambda-6 & = 0\r\n\\end{align*}\\]\r\nThe characteristic equation that forms is a quadratic equation, which we can factorise and use the null factor law or quadratic formula to solve.\r\n\\[\\begin{align*} 0 & = \\lambda^{2}-\\lambda-6\\\\\r\n& = \\left( \\lambda+3\\right) \\left( \\lambda-2\\right)\r\n\\end{align*}\\] \\[\\begin{align*} \\lambda+3 & = 0\\\\\r\n\\lambda & = -3\r\n\\end{align*}\\] \\[\\begin{align*} \\lambda-2 & = 0\\\\\r\n\\lambda & = 2\r\n\\end{align*}\\]\r\nThe \\(2\\times2\\) matrix \\(A\\) has two eigenvalues: \\(\\lambda_{1}=-3\\) and \\(\\lambda_{2}=2\\).\r\n\r\n\r\n\r\nFinding eigenvectors for \\(2\\times2\\) matrices\r\nEach eigenvalue of a matrix has a corresponding eigenvector. To find the eigenvectors, we solve the following equation for each eigenvalue:\r\n\\[\\begin{align*} AX & =\\lambda X\\\\\r\nAX-\\lambda X & = 0\\\\\r\n(A-\\lambda I)X & = 0\r\n\\end{align*} \\]\r\nIn the very last line, you might notice that we have not written \\((A-\\lambda)X=0\\). This wouldn't make sense, because \\(A\\) is an \\(n\\times n\\) matrix and \\(\\lambda\\) is a scalar (number). We cannot subtract a number from a matrix, so we need to complete a transformation in which we multiply by the identity matrix.\r\nFor each eigenvalue, \\(\\lambda\\), solve the following to obtain the eigenvector, \\(X\\):\r\n\\[(A-\\lambda I)X=0\\]\r\nA multiple of any eigenvector is an eigenvector. We do not consider \\(\\vec{X}=\\vec{0}\\) to be an eigenvector.\r\n\r\nExample \u2013 finding eigenvectors for \\(2\\times2\\) matrices\r\nFind the eigenvectors of the matrix:\r\n\\[A = \\left[ \\begin{array}{cc} -4 & -2\\\\ \r\n3 & 3 \r\n\\end{array} \\right]\\]\r\nThis is the same matrix as Example 1 \u2013 finding eigenvectors for \\(2\\times2\\) matrices, so we already know that the eigenvalues are \\(\\lambda_{1}=-3\\) and \\(\\lambda_{2}=2\\).\r\nTo find the corresponding eigenvectors, we solve \\((A-\\lambda I)X=0\\) for \\(X\\). We can write \\(X\\) in the form \\(\\left[ \\begin{array}{c} x\\\\ y \\end{array} \\right]\\) and substitute all of these into the equation.\r\n\\[ \\left[ \\begin{array}{cc} -4-\\lambda & -2\\\\\r\n3 & 3-\\lambda \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{c} x\\\\\r\ny\r\n\\end{array} \\right] = 0 \\]\r\nFor \\(\\lambda_{1}=-3\\):\r\n\\[\\begin{align*} \\left[ \\begin{array}{cc} -4-(-3) & -2\\\\\r\n3 & 3-(-3) \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{c} x\\\\\r\ny\r\n\\end{array} \\right] & = 0\\\\\r\n\\left[ \\begin{array}{cc} -1 & -2\\\\\r\n3 & 6\r\n\\end{array} \\right] \r\n\\left[ \\begin{array}{c} x\\\\\r\ny\r\n\\end{array} \\right] & = 0\r\n\\end{align*}\\]\r\nThis gives two linear equations:\r\n\\[\\begin{align*} -x-2y & = 0\\\\ \r\n3x+6y & = 0 \r\n\\end{align*}\\]\r\nThese cannot be solved to give a unique solution, as one is a multiple of the other. This means there are an infinite number of solutions.\r\nRemember that when you encounter this case, you have to let one variable be free. We can let \\(y=t\\) where \\(t\\in\\mathbb{R}\\). This gives us \\(x=-2t\\) and \\(y=t\\), which we can substitute into the general eigenvector form:\r\n\\[\\begin{align*} \\left[ \\begin{array}{c} x\\\\\r\ny \r\n\\end{array} \\right] & = \\left[ \\begin{array}{c} -2t\\\\\r\nt\r\n\\end{array} \\right]\\\\\r\n& = t \\left[ \\begin{array}{c} -2\\\\\r\n1\r\n\\end{array} \\right]\r\n\\end{align*}\\]\r\nAs \\(t\\) can be any number, there are an infinite number of eigenvectors. However, the convention is that we recognise any eigenvector multiplied by a scalar is still an eigenvector and we state the eigenvector in the lowest possible terms. So, the eigenvector that corresponds to \\(\\lambda_{1}=-3\\) is:\r\n\\[\\begin{align*} X_{1} & = \\left[ \\begin{array}{c} -2\\\\\r\n1 \r\n\\end{array} \\right]\r\n\\end{align*}\\]\r\nFor \\(\\lambda_{2}=2\\):\r\n\\[\\begin{align*} \\left[ \\begin{array}{cc} -4-2 & -2\\\\\r\n3 & 3-2 \r\n\\end{array} \\right] \\left[ \\begin{array}{c} x\\\\ \r\ny \\end{array} \\right] & = 0\\\\\r\n\\left[ \\begin{array}{cc} -6 & -2\\\\ \r\n3 & 1\r\n\\end{array} \\right] \\left[ \\begin{array}{c} x\\\\\r\ny \r\n\\end{array} \\right] & = 0\\end{align*}\\]\r\nThis gives two linear equations:\r\n\\[\\begin{align*} -6x-2y & = 0\\\\ \r\n3x+y & = 0 \r\n\\end{align*}\\]\r\nAgain, we see that the one equation is a multiple of the other. Let's let \\(x=t\\), which gives us \\(y=-3t\\) where \\(t\\in\\mathbb{R}\\). We can substitute \\(x\\) and \\(y\\) into the eigenvector :\r\n\\[\\begin{align*} \\left[ \\begin{array}{c} x\\\\\r\ny \\end{array} \\right] & = \\left[ \\begin{array}{c} t\\\\\r\n-3t\r\n\\end{array}\\right]\\\\\r\n& = t\\left[\\begin{array}{c} 1\\\\ \r\n-3 \\end{array} \\right]\r\n\\end{align*}\\]\r\nIn lowest terms, the eigenvalue corresponding to \\(\\lambda_{2}=2\\) is:\r\n\\[X_{2} = \\left[ \\begin{array}{c} 1\\\\\r\n-3 \r\n\\end{array} \\right] \\]\r\n\r\nExercise \u2013 finding eigenvalues and eigenvectors for \\(2\\times2\\) matrices\r\nFind the eigenvalues and eigenvectors for the following matrices.\r\n\r\n\t\\(\\left[ \\begin{array}{cc} 2 & 1\\\\\r\n\t1 & 2\r\n\t\\end{array} \\right] \\)\r\n\t\\(\\left[ \\begin{array}{cc} 4 & -1\\\\\r\n\t1 & 2\r\n\t\\end{array} \\right] \\)\r\n\t\\(\\left[ \\begin{array}{cc} -1 & -3\\\\\r\n\t1 & 3 \r\n\t\\end{array} \\right] \\)\r\n\r\n\r\n\r\n\r\n\r\n\r\nEigenvalues and eigenvectors for \\(3\\times3\\) matrices\r\nRegardless of the size of the matrix, the method for finding eigenvalues and eigenvectors is the same.\r\nThe working out is quite long for these types of questions, but as long as you work through them carefully and follow the correct method, you will reach the correct answer.\r\n\r\nExample 1 \u2013 finding eigenvalues and eigenvectors for \\(3\\times3\\) matrices\r\nFind the eigenvalues and eigenvectors of the matrix:\r\n\\[B = \\left[ \\begin{array}{ccc} 1 & 1 & -2\\\\\r\n-1 & 0 & 1\\\\\r\n-2 & 1 & 1 \r\n\\end{array} \\right]\\]\r\nWe can find the eigenvalues by solving \\(\\det\\left[B-\\lambda I\\right]\\).\r\n\r\nThis is the characteristic equation. It is a cubic polynomial. The degree of the polynomial is generally the same as \\(n\\) for a \\(n\\times n\\) matrix.\r\nWe factorise the characteristic equation to solve it.\r\n\\[\\begin{align*} -\\lambda^{3} + 2\\lambda^{2} + 3\\lambda & = 0\\\\\r\n-\\lambda(\\lambda^{2}-2\\lambda-3) & = 0\\\\\r\n-\\lambda(\\lambda-3)(\\lambda+1) & = 0\r\n\\end{align*}\\]\r\nHence, the eigenvalues of \\(B\\) are \\(\\lambda_{1}=0\\), \\(\\lambda_{2}=3\\) and \\(\\lambda_{3}=-1\\).\r\nTo find the eigenvectors for each eigenvalue, we solve \\((A-\\lambda I)X=0\\). In this case, since the order of the matrix is \\(3\\times3\\), \\(X = \\left[ \\begin{array}{c} x\\\\ y\\\\ z \\end{array} \\right]\\). Here, we haven't showed all of the steps.\r\nFor \\(\\lambda_{1}=0\\):\r\n\\[\\begin{align*} \\left(B-\\lambda I\\right)X & = 0\\\\ \\left[ \\begin{array}{ccc} 1-0 & 1 & -2\\\\\r\n-1 & 0-0 & 1\\\\ \r\n-2 & 1 & 1-0 \r\n\\end{array} \\right] \r\n\\left[ \\begin{array}{c} x\\\\\r\ny\\\\ \r\nz \r\n\\end{array} \\right] & = 0\\\\ \r\n\\left[ \\begin{array}{ccc} 1 & 1 & -2\\\\\r\n-1 & 0 & 1\\\\ \r\n-2 & 1 & 1 \r\n\\end{array} \\right] \r\n\\left[ \\begin{array}{c} x\\\\ \r\ny\\\\ \r\nz\r\n\\end{array} \\right] & = 0\\\\\r\nx+y-2z & = 0\\\\\r\n-x+z & = 0\\\\ \r\n-2x+y+z & = 0\r\n\\end{align*}\\]\r\nFrom the second equation, we get \\(z=x\\).\r\nWe can substitute this into the first equation:\r\n\\[\\begin{align*} x+y-2z & = 0\\\\\r\nx+y-2(x) & = 0\\\\\r\ny-x & = 0\\\\\r\ny & = x\r\n\\end{align*}\\]\r\nSince \\(x=y=z\\), we can let them \\(=t\\), where \\(t\\in\\mathbb{R}\\). The eigenvector is therefore:\r\n\\[\\begin{align*} X & = \\left[ \\begin{array}{c} t\\\\\r\nt\\\\\r\nt\r\n\\end{array} \\right]\\\\\r\n& = t \\left[ \\begin{array}{c} 1\\\\\r\n1\\\\\r\n1\r\n\\end{array} \\right]\r\n\\end{align*} \\]\r\nIn its lowest form, \\(X_{1} = \\left[ \\begin{array}{c} 1\\\\ 1\\\\ 1 \\end{array} \\right] \\).\r\nFor \\(\\lambda_{2}=3\\):\r\n\\[\\begin{align*} \\left(B-\\lambda I\\right)X & = 0\\\\\r\n\\left[ \\begin{array}{ccc} 1-3 & 1 & -2\\\\ \r\n-1 & 0-3 & 1\\\\ \r\n-2 & 1 & 1-3 \r\n\\end{array} \\right] \r\n\\left[ \\begin{array}{c} x\\\\\r\ny\\\\\r\nz \r\n\\end{array} \\right] & = 0\\\\ \r\n\\left[ \\begin{array}{ccc} -2 & 1 & -2\\\\\r\n-1 & -3 & 1\\\\\r\n-2 & 1 & -2 \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{c} x\\\\\r\ny\\\\ \r\nz \r\n\\end{array}\\right] & = 0\\\\ \r\n-2x+y-2z & = 0\\\\\r\n-x-3y+z & = 0 \\\\ \r\n-2x+y-2z & = 0\r\n\\end{align*}\\]\r\nThe first and third equations are identical, so we can pick just the first and second to solve. We can let \\(z=t\\), where \\(t\\in\\mathbb{R}\\).\r\n\\[\\begin{align*} -2x+y & = 2t\\\\\r\n-x-3y & = -t\r\n\\end{align*}\\]\r\nWe can multiply the second equation by \\(2\\) and subtract from the first equation to get:\r\n\\[\\begin{align*} -2x+y - 2(-x-3y) & = 2t-(2\\times(-t))\\\\\r\n-2x+y+2x+6y & = 2t+2t\\\\\r\n7y & = 4t\\\\\r\ny & = \\frac{4}{7}t\r\n\\end{align*}\\]\r\nWe can then multiply the first equation by \\(3\\) and add it to the second equation to get:\r\n\\[\\begin{align*} 3(-2x+y) + (-x-3y) & = 3(2t) + (-t)\\\\\r\n-6x+3y-x-3y & = 6t-t\\\\\r\n-7x & = 5t\\\\\r\nx & = -\\frac{5}{7}t\r\n\\end{align*}\\]\r\nThus, \\(y=\\dfrac{4}{7}t\\) and \\(x=-\\dfrac{5}{7}t\\).\r\nHence, the eigenvector has the form:\r\n\\[\\begin{align*} X & = \\left[ \\begin{array}{c} -\\frac{5}{7}t\\\\\r\n\\frac{4}{7}t\\\\\r\n-\\frac{5}{7}t\r\n\\end{array} \\right] \\\\\r\n& = t\\left[ \\begin{array}{c} -\\frac{5}{7}\\\\ \r\n\\frac{4}{7}\\\\\r\n1 \r\n\\end{array}\\right] \r\n\\end{align*}\\]\r\nTo get the lowest form, we can make \\(t=7\\) to get rid of the fraction. Hence, \\(X_{2} = \\left[ \\begin{array}{c} -5\\\\ 4\\\\ 7 \\end{array} \\right]\\) or \\(\\left[ \\begin{array}{c} 5\\\\ -4\\\\ -7 \\end{array} \\right]\\).\r\nFor \\(\\lambda_{3}=-1\\):\r\n\\[\\begin{align*} \\left(B-\\lambda I\\right)X & = 0\\\\ \r\n\\left[ \\begin{array}{ccc} 1-(-1) & 1 & -2\\\\ \r\n-1 & 0-(-1) & 1\\\\\r\n-2 & 1 & 1-(-1)\r\n\\end{array} \\right]\r\n\\left[\r\n\\begin{array}{c} x\\\\\r\ny\\\\\r\nz \r\n\\end{array} \\right] & = 0\\\\\r\n\\left[ \\begin{array}{ccc} 2 & 1 & -2\\\\ \r\n-1 & 1 & 1\\\\\r\n-2 & 1 & 2 \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{c} x\\\\ \r\ny\\\\ \r\nz \r\n\\end{array}\\right] & = 0\\\\\r\n2x+y-2z & = 0\\\\\r\n-x+y+z & = 0\\\\ \r\n-2x+y+2z & = 0\r\n\\end{align*}\\]\r\nWe can add the first equation to the third equation to get:\r\n\\[\\begin{align*} 2x+y-2z + (-2x+y+2z) & = 0+0\\\\\r\n2x+y-2z-2x+y+2z & = 0\\\\\r\n2y & = 0\\\\\r\ny & = 0\r\n\\end{align*}\\]\r\nWe can substitute \\(y=0\\) into the second equation:\r\n\\[\\begin{align*} -x+y+z & = 0\\\\\r\n-x+0+z & = 0\\\\\r\nx & = z\r\n\\end{align*}\\]\r\nWe can then let \\(z=t\\), where \\(t\\in\\mathbb{R}\\), so that the eigenvector has the form:\r\n\\[\\begin{align*} X & = \\left[ \\begin{array}{c} t\\\\ \r\n0\\\\\r\nt\r\n\\end{array} \\right]\\\\\r\n& = t \\left[ \\begin{array}{c} 1\\\\\r\n0\\\\\r\n1\r\n\\end{array} \\right] \r\n\\end{align*}\\]\r\nThis gives us the lowest form, \\(X_{3} = \\left[ \\begin{array}{c} 1\\\\ 0\\\\ 1 \\end{array} \\right] \\).\r\nIn summary, the eigenvalues of matrix \\(B\\) are \\(\\lambda_{1}=0\\), \\(\\lambda_{2}=3\\) and \\(\\lambda_{3}=-1\\). The corresponding eigenvectors are \\(X_{1}=\\left[ \\begin{array}{c} 1\\\\ 1\\\\ 1 \\end{array} \\right]\\), \\(X_{2}=\\left[ \\begin{array}{c} 5\\\\ -4\\\\ -7 \\end{array} \\right]\\) and \\(X_{3}=\\left[ \\begin{array}{c} 1\\\\ 0\\\\ 1 \\end{array} \\right]\\).\r\n\r\n\r\n\r\nExercise \u2013 finding eigenvalues and eigenvectors for \\(3\\times3\\) matrices\r\n\r\n\tDetermine which of the following are eigenvectors of the matrix \\(A = \\left[ \\begin{array}{ccc} 2 & 0 & -1\\\\\r\n\t0 & 1 & 0\\\\\r\n\t4 & 0 & 2\r\n\t\\end{array} \\right]\\)\r\n\t\r\n\t\t\\(\\left[ \\begin{array}{c} 0\\\\\r\n\t\t1\\\\\r\n\t\t0 \r\n\t\t\\end{array} \\right] \\)\r\n\t\t\\(\\left[ \\begin{array}{c} 0\\\\ \r\n\t\t0\\\\\r\n\t\t0 \r\n\t\t\\end{array} \\right] \\)\r\n\t\t\\(\\left[ \\begin{array}{c} 1\\\\ \r\n\t\t0\\\\ \r\n\t\t2 \r\n\t\t\\end{array} \\right] \\)\r\n\t\r\n\tFind the eigenvalues and eigenvectors of the following matrices.\r\n\t\r\n\t\t\\(\\left[ \\begin{array}{ccc} 1 & -1 & -1\\\\\r\n\t\t0 & 2 & 0\\\\\r\n\t\t0 & -1 & -1\r\n\t\t\\end{array} \\right] \\)\r\n\t\t\\(\\left[ \\begin{array}{ccc} -2 & 0 & -1\\\\\r\n\t\t0 & 5 & 2\\\\\r\n\t\t3 & 0 & 2\r\n\t\t\\end{array} \\right] \\)\r\n\t\r\n\r\n\r\n\r\n\r\nSummary\r\nSince eigenvalues and eigenvectors can involve some complex calculations, let's review the most important points:\r\n\r\n \tAn \\(n\\times n\\) matrix will have, at most, \\(n\\) distinct eigenvalues.\r\n \tAn eigenvalue may be \\(0\\).\r\n \tAn eigenvector multiplied by any number other than \\(0\\) is also an eigenvector. In particular, if \\(X\\) is an eigenvector, then so is \\(-X\\).\r\n \tAn eigenvector cannot be the zero vector, \\(\\vec{0}\\).\r\n","n":0.023},[{"v":"Matrices","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":326,"$":[{"v":"Summation notation","n":0.707},{"v":"\r\nYou may have seen the Greek symbol sigma (\\(\\Sigma\\)) used in maths, but what does it mean? It is a way to simply express long numbers, using something called summation notation. A good understanding of summation notation is important in statistics for calculating averages and sums. Use this resource to learn how to interpret and use summation notation.\r\n\r\nSummation (or sigma) notation is a shorthand way to write the sum or addition of a sequence of numbers or mathematical terms. It is a great way to express the sum of a large data set that has a mathematical relationship.\r\n\r\nThe general formula for summation is:\r\n\r\n\\[\\sum_{n=1}^{N}x_{n}=x_{1}+x_{2}+x_{3}+\\cdots+x_{n}\\]\r\nIt has a few parts:\r\n\r\n\tThe Greek symbol sigma \\(\\Sigma\\) is used to denote summation. It is called the summation sign.\r\n\t\\(n\\) is the index of summation. It is the value that you start at. For example, if \\(n=2\\), then first value you use is \\(x_{2}\\).\r\n\t\\(N\\) is the upper limit of summation. It is the very last value of \\(n\\) you include in your sum. For example, if \\(n=1\\) and \\(N=3\\), then your sum will only include \\(x_{1}+x_{2}+x_{3}\\).\r\n\t\\(x\\) tells you what to sum. It can be a single value or an algebraic expression.\r\n\t\\(x_{1}\\) is the first item in a data set, \\(x_{2}\\) is the second item, \\(x_{3}\\) is the third, and so on. They are added in increments of \\(1\\).\r\n\r\n\r\nUsing summation notation\r\nConsider summation notation of a data set where \\(x\\) is a value.\r\n\\[\\begin{align*} \\sum_{n=1}^{5}n & = 1+2+3+4+5\\\\\r\n& = 15\r\n\\end{align*}\\]\r\n\\(n=1\\), so we start adding from \\(x_{1}=1\\). \\(N=5\\), so we add in increments of \\(1\\) until we have added the value \\(x_{5}=5\\).\r\nThis is just a very simple summation; summation notation can involve more complex sequences.\r\nLet's consider the same notation, but where \\(x_{n}=2n\\).\r\n\\[\\begin{align*} \\sum_{n=1}^{5}2n & = \\underbrace{(2\\times1)}_{n=1}+\\underbrace{(2\\times2)}_{n=2}+\\underbrace{(2\\times3)}_{n=3}+\\underbrace{(2\\times4)}_{n=4}+\\underbrace{(2\\times5)}_{n=5}\\\\\r\n& = 30\r\n\\end{align*}\\]\r\nWe start from \\(x_{1}=2\\times1=2\\) and add in increments of \\(1\\) until we have added the value \\(x_{5}=2\\times5=10\\).\r\n\r\nExample 1 \u2013 using summation notation\r\nExpand and evaluate \\(\\sum_{n=0}^{3}(n^{2}-3)\\).\r\nWe start from \\(x_{0}=0^{2}-3=-3\\) and add in increments sof \\(1\\) until we have added the value \\(x_{3}=3^{2}-3=6\\).\r\n\\[\\begin{align*} \\sum_{n=0}^{3}(n^{2}-3) & = \\underbrace{(0^{2}-3)}_{n=0}+\\underbrace{(1^{2}-3)}_{n=1}+\\underbrace{(2^{2}-3)}_{n=2}+\\underbrace{(3^{2}-3)}_{n=3}\\\\\r\n& = (-3)+(-2)+1+6\\\\\r\n& = 2\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using summation notation\r\n\r\n\tEvaluate the following.\r\n\t\r\n\t\t\\(\\sum_{n=1}^{5}3n\\)\r\n\t\t\\(\\sum_{n=1}^{3}(5n-2)\\)\r\n\t\t\\(\\sum_{n=1}^{3}(5n)-2\\)\r\n\t\r\n\tGiven \\(x_{1}=-2\\), \\(x_{2}=0\\), \\(x_{3}=1\\), \\(x_{4}=3\\) and \\(x_{5}=3\\), evaluate the following.\r\n\t\r\n\t\t\\(\\sum_{n=1}^{5}10x_{n}\\)\r\n\t\t\\(10\\sum_{n=1}^{5}x_{n}\\)\r\n\t\t\\(\\sum_{n=1}^{5}(x_{n})^{2}\\)\r\n\t\t\\(\\left(\\sum_{n=1}^{5}x_{n}\\right)^{2}\\)\r\n\t\t\\(\\sum_{n=1}^{5}(N\\times x_{n})\\)\r\n\t\t\\(\\overline{x}=\\dfrac{\\sum_{n=1}^{5}x_{n}}{N}\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.053},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":327,"$":[{"v":"Data","n":1},{"v":"Data is everywhere. It drives many aspects of our day-to-day lives. Use this resource to learn about the different types of data and some ways of displaying it.\r\n\r\nCollecting and processing data\r\nTo study the relationships between variables (a characteristic, number, or quantity that can be measured or counted), you need to collect information or data. The relationship can apply to a whole population, which is the total group of individuals or items being considered. Usually, we only take a small subset of this population, called a sample.\r\n\r\nOnce we have the data, we can calculate numbers from the sample to get statistics. When we calculate number from population data, we get a parameter.\r\n\r\nTypes of data\r\nData can be classified as either:\r\n\r\n\tqualitative \u2013 categorical or non-numerical (e.g. blood type, occupation, colour) \r\n\tquantitative \u2013 numerical (e.g. mass, length, duration).\r\n\r\nThere are two types of quantitative data:\r\n\r\n\tdiscrete data \u2013 data that can only take on certain values or be counted (e.g. the number of children in a family)\r\n\tcontinuous data \u2013 data that can take on any value or be measured (e.g. height).\r\n\r\n\r\nExercise \u2013 classifying data\r\nClassify each of the following as qualitative or quantitative data, including whether they are discrete or continuous if they numerical.\r\n\r\n\tHair colour\r\n\tTemperature in the fridge throughout the day\r\n\tHeight\r\n\tMovie genre\r\n\tStudents in a class\r\n\tLanguage spoken\r\n\tWeights of babies born on a certain day\r\n\tNumber of eggs laid by a chicken\r\n\tDistance travelled to work by RMIT staff\r\n\tThe make of car driven by RMIT students\r\n\r\n\r\n\r\n\r\nDisplaying data\r\nA summary table of your data is one way of presenting results, but sometimes, it is important to turn it into a visual format like a chart or graph. \r\nVisualising data helps people to understand the information quickly by showing patterns and trends that might be hard to spot in a list of numbers. This makes it easier to share findings with others and helps in making smart decisions based on the data.\r\nThe way you chose to present data can depend on the type of data you have collected.\r\n\r\n\tQualitative data is often represented using a bar chart or a pie chart.\r\n\tQuantitative data is often represented using a frequency bar chart called a histogram.\r\n\r\n\r\nThere are many other ways of presenting data; these are just some basic forms.\r\n\r\nExample 1 \u2013 displaying data\r\nThe table shows the percentage of Australian imports from various countries. Represent the data with a pie chart.\r\n\r\n\r\n\r\nCountry\r\nImports \\((\\%)\\)\r\n\r\n\r\n\r\n\r\nChina\r\n\\(22\\)\r\n\r\n\r\nJapan\r\n\\(20\\)\r\n\r\n\r\nSouth Korea\r\n\\(8\\)\r\n\r\n\r\nIndia\r\n\\(8\\)\r\n\r\n\r\nUSA\r\n\\(5\\)\r\n\r\n\r\nUK\r\n\\(4\\)\r\n\r\n\r\nNew Zealand\r\n\\(4\\)\r\n\r\n\r\nOthers\r\n\\(29\\)\r\n\r\n\r\n\r\nFor a pie chart, the size of the wedge corresponds to the proportion of \\(100\\%\\). The data we have is already in percentages, so we just need to accurately represent them in the chart. \r\nThese days, it is uncommon to prepare pie charts by hand. We usually use graphing software like Microsoft Excel to help, but it can be done.\r\n\r\n\r\n\r\n\r\nExercise \u2013 displaying data\r\n\r\n\tRepresent the data from Example 1 \u2013 displaying data in a bar graph.\r\n\tA group of employees recorded the time that it took them to travel to work on a particular day in a table. Represent this data using a histogram.\r\n\t\r\n\t\r\n\t\r\n\tTime (min)\r\n\tFrequency\r\n\t\r\n\t\r\n\t\r\n\t\r\n\t\\(0-<15\\)\r\n\t\\(2\\)\r\n\t\r\n\t\r\n\t\\(15-<30\\)\r\n\t\\(12\\)\r\n\t\r\n\t\r\n\t\\(30-<45\\)\r\n\t\\(23\\)\r\n\t\r\n\t\r\n\t\\(45-<60\\)\r\n\t\\(9\\)\r\n\t\r\n\t\r\n\t\\(60-<75\\)\r\n\t\\(3\\)\r\n\t\r\n\t\r\n\t\\(75-<90\\)\r\n\t\\(1\\)\r\n\t\r\n\t\r\n\tTotal\r\n\t\\(50\\)\r\n\t\r\n\t\r\n\t\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.045},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":328,"$":[{"v":"Mean, mode, median","n":0.577},{"v":"The mean, mode, and median provide a single value that is typical of the data. These measures of central tendency help summarise large datasets, making it easier to understand the overall pattern. Understanding these concepts is essential for analysing and interpreting data effectively.\r\n\r\nMeasures of central tendency\r\nThe mean, median, and mode are important tools for understanding data. They each give a single value that represents the data as a whole. These are called measures of central tendency because they show the centre or most typical values in a dataset, helping us quickly see what is common or typical.\r\n\r\n \tThe mode is the value that occurs most often. It is the value with the highest frequency (f).\r\n \tThe median is the middle value when the data is arranged in order. You can quickly find it using:\r\n\t\\[\\dfrac{n+1}{2}\\]\r\n\twhere \\(n\\) is the total number of data points. If the value is \\(4\\), the median will be the fourth value in the ordered dataset.\r\n \tThe mean or average (denoted by \\(\\overline{x}\\)) is the sum of all the scores divided by the number of scores in the dataset: \r\n\t\\[\\overline{x} = \\dfrac{\\sum x_{n}}{n}\\]\r\n\twhere \\(\\sum x_{n}\\) is the sum of each score \\(x\\) multiplied by its frequency (f).\r\n\r\n\r\nExample 1 \u2013 finding the measures of central tendency\r\nConsider the data set \\(3,2,0,5,2\\). Find the mode, median and mean.\r\nTo find the mode, we find the value that occurs most frequently in the dataset. The mode is \\(2\\) because it occurs twice.\r\nTo find the median, we need to rearrange the data so that it is ordered from smallest to largest and find the middle number:\r\n\\[0,2,\\underline{2},3,5\\]\r\nThe median is \\(2\\).\r\nTo find the mean, we use \\(\\overline{x}=\\dfrac{\\sum x_{i}}{n}\\).\r\n\\[\\begin{align*} \\overline{x} & = \\frac{\\sum x_{i}}{n}\\\\\r\n& = \\frac{3+2+0+5+2}{5}\\\\\r\n& = 2.4 \r\n\\end{align*}\\]\r\nThe mean is \\(2.4\\).\r\n\r\n\r\n\r\nExercise \u2013 finding the measures of central tendency\r\n\r\n\tConsider the following scores: \\(12,12,13,14,14,15,15,15,16\\).\r\n\t\r\n\t\tFind the mean score.\r\n\t\tFind the median.\r\n\t\tFind the mode.\r\n\t\r\n\tDetermine the mean, mode and median for the data in the frequency table.\r\n\t\r\n\t\r\n\t\r\n\tScore\r\n\tFrequency\r\n\t\r\n\t\r\n\t\r\n\t\r\n\t\\(40\\)\r\n\t\\(1\\)\r\n\t\r\n\t\r\n\t\\(50\\)\r\n\t\\(4\\)\r\n\t\r\n\t\r\n\t\\(60\\)\r\n\t\\(8\\)\r\n\t\r\n\t\r\n\t\\(70\\)\r\n\t\\(3\\)\r\n\t\r\n\t\r\n\t\\(80\\)\r\n\t\\(3\\)\r\n\t\r\n\t\r\n\t\\(90\\)\r\n\t\\(1\\)\r\n\t\r\n\t\r\n\t\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\nGraphical representations of central tendency\r\nWe can visualise measures of central tendency using bell curves and stem-and-leaf plots. These tools make it easier to interpret data and understand its key characteristics.\r\n\r\nBell curves\r\nA bell curve or normal distribution graphically shows the mean in the centre with data symmetrically distributed around it. This shape highlights how values are spread out, with most clustering around the mean. \r\nConsider this example. It is a symmetrical, bell-shaped distribution graph, where the mode, median and mean all have the same value, \\(100\\).\r\n\r\nLearn more about normal distributions on this page.\r\n\r\nStem and leaf plots\r\nStem and leaf plots organise data to display the shape and distribution. They let us quickly see the median and mode, as well as the overall spread of the data. \r\nConsider the following dataset.\r\n\\[1,1,,2,7,11,13,13,14,18,20,22,\\ldots,86,94,96\\]\r\nWe can organise this data in a stem and leaf plot. \r\n\r\n\tThe first row with the stem \\(0\\) gives us the leaf values \\(1\\), \\(1\\), \\(2\\) and \\(7\\). This just tells us that when we have \\(0\\) in the tens column, we have the values \\(1\\), \\(1\\), \\(2\\) and \\(7\\) in the ones column.\r\n\tFor the second row with the stem \\(1\\), we have all the values with a \\(1\\) in the tens column, so numbers ranging from \\(10\\) to \\(19\\). This would be \\(11\\), \\(13\\), \\(13\\), \\(14\\) and \\(18\\). The \\(1\\) in the tens column is written as the stem. And the leaves are the numbers in the ones columns: \\(1\\), \\(3\\), \\(3\\), \\(4\\) and \\(8\\).\r\n\r\n\\[\\begin{array}{ccccccccccc} & 0 & \\mid & & 1 & 1 & 2 & 7\\\\ \r\n& 1 & \\mid & & 1 & 3 & 3 & 4 & 8\\\\ \r\n& 2 & \\mid & & 0 & 2 & 3 & 7 & 9\\\\\r\n& 3 & \\mid & & 2 & 3 & 4 & 5 & 5 & 5 & 8\\\\ \r\n& 4 & \\mid & & 1 & 4 & 7 & 8 & 8 & 9 & 9\\\\\r\n& 5 & \\mid & & 0 & 2 & 4 & 5 & 7 & 8\\\\\r\n& 6 & \\mid & & 1 & 4 & 5 & 6 & 9\\\\\r\n& 7 & \\mid & & 4 & 7 & 8\\\\\r\n& 8 & \\mid & & 0 & 2 & 5 & 6\\\\\r\n& 9 & \\mid & & 4 & 6 \r\n\\end{array}\\]\r\n\r\nCumulative count\r\nSometimes a stem and leaf plot will include an extra column with a cumulative count. To do this:\r\n\r\n\tFind the median row. This will represent the middle of the bell curve.\r\n\tStart at the lowest stem (first row) and count the leaves.\r\n\tGo to the next stem (second row) and count the leaves. Add this to the number of leaves in the first row to get a cumulative total.\r\n\tMove on to the next stem, count the leaves and add it to the cumulative total from the previous row. Keep adding until you reach the median.\r\n\tRepeat this starts from the highest stem (last row).\r\n\r\n\t\r\nFor the previous example, the median row has a stem of \\(4\\). The first stem has \\(4\\) leaves, so we write a \\(4\\) to the left of the stem. The next stem has \\(5\\) leaves. \\(4+5=9\\), so we write a \\(9\\) to the left of the stem. The one after that has \\(5\\) leaves, too. \\(9+5=14\\), so we write \\(14\\) to the left of the stem.\r\nWe continue this until we reach the stem of \\(4\\). The median stem does not have a cumulative total.\r\nFor the last stem, we have \\(2\\) leaves, so we write a \\(2\\) to the left of the stem. The second last stem has \\(4\\) leaves. \\(4+2=6\\) so we write the cumulative total of \\(6\\) to the left of the stem, and so on.\r\n\r\n\\[\\begin{array}{ccccccccccccc} \\mathit{4} & & \\mid & 0 & \\mid & & 1 & 1 & 2 & 7\\\\ \r\n\\mathit{9} & & \\mid & 1 & \\mid & & 1 & 3 & 3 & 4 & 8\\\\ \r\n\\mathit{14} & & \\mid & 2 & \\mid & & 0 & 2 & 3 & 7 & 9\\\\\r\n\\mathit{21} & & \\mid & 3 & \\mid & & 2 & 3 & 4 & 5 & 5 & 5 & 8\\\\ \r\n& & \\mid & 4 & \\mid & & 1 & 4 & 7 & 8 & 8 & 9 & 9\\\\ \r\n\\mathit{20} & & \\mid & 5 & \\mid & & 0 & 2 & 4 & 5 & 7 & 8\\\\ \r\n\\mathit{14} & & \\mid & 6 & \\mid & & 1 & 4 & 5 & 6 & 9\\\\ \r\n\\mathit{9} & & \\mid & 7 & \\mid & & 4 & 7 & 8\\\\ \r\n\\mathit{6} & & \\mid & 8 & \\mid & & 0 & 2 & 5 & 6\\\\ \r\n\\mathit{2} & & \\mid & 9 & \\mid & & 4 & 6 \r\n\\end{array}\\]\r\nInterpreting stem and leaf plots\r\nFrom the stem and leaf plot, we can see that there are \\(21\\) scores less than or equal to \\(38\\) and \\(20\\) scores greater than or equal to \\(50\\). There are also \\(7\\) scores in the forties. This gives us \\(48\\) scores in total.\r\nTo find the median, we calculate \\(\\dfrac{n+1}{2}=\\dfrac{48+1}{2}=24.5\\). This means that the median value is between the \\(24\\)th and \\(25\\)th values. We know the \\(21\\)st score is \\(38\\). The median is therefore \\(47.5\\) (halfway between \\(47\\) and \\(48\\)).\r\nWe can also quickly see that the mode is \\(35\\); it occurs \\(3\\) times.\r\n\r\nExercise \u2013 interpreting graphicval representations of central tendency\r\nFind the mode and median for the data displayed in the stem and leaf plot for which the smallest score is \\(10\\) and the largest \\(69\\).\r\n\\[\\begin{array}{ccccccccccc} 1 & & \\mid & & 0 & 7 & 9\\\\\r\n2 & & \\mid & & 1 & 1 & 3 & 4 & 6 & 7 & 8\\\\ \r\n3 & & \\mid & & 0 & 1 & 3 & 5 & 6 & 7 & 7\\\\ \r\n4 & & \\mid & & 0 & 1 & 1 & 1 & 2\\\\\r\n5 & & \\mid\\\\ 6 & & \\mid & & 9 \r\n\\end{array}\\]\r\n\r\n\r\n\r\n","n":0.028},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":329,"$":[{"v":"Measures of spread","n":0.577},{"v":"\r\nBy looking at how data in a set is distributed, we can see how much the data varies. This helps us understand the data better and make good decisions based on it. Use this resource to learn about the measures of spread.\r\n\r\nThe measure of spread shows how spread out the numbers in a set of data are. It tells us if the data values are close together or far apart. \r\n\r\nHere, we will look at three measures of spread:\r\n\r\n\trange\r\n\tinterquartile range\r\n\tstandard deviation.\r\n\r\n\r\nRange\r\nThe most basic measure of spread is the range. This is the distance from the smallest to the largest value.\r\nLet's consider two datasets:\r\n\r\n\t\\(4,4,5,5,5,6,6\\)\r\n\t\\(1,3,4,5,6,7,9\\)\r\n\r\nBoth sets A and B have mean = median = \\(5\\), but the data sets are quite different. The values in set A are less spread out than those in set B.\r\nFor set A:\r\n\\[\\begin{align*} \\textrm{Range} & = \\textrm{Highest value}-\\textrm{Lowest value}\\\\\r\n& = 6-4\\\\\r\n& = 2\r\n\\end{align*}\\]\r\nFor set B:\r\n\\[\\begin{align*} \\textrm{Range} & = \\textrm{Highest value}-\\textrm{Lowest value}\\\\\r\n& = 9-1\\\\\r\n& = 8\r\n\\end{align*}\\]\r\nWe can see that set B has greater spread than set A. But the problem with the range is that it uses only two of the values in the dataset. One of these may be an odd or unusual value called an outlier.\r\nConsider the two sets of values below:\r\n\r\n\t\\(1,1,2,2,2,2,2,100\\)\r\n\t\\(1,18,23,41,59,63,87,100\\)\r\n\r\nThe range for both is \\(99\\) because set Y has one unusual value, the \\(100\\). Most of the data are ones or twos. The value of \\(100\\) is unusual and is an outlier. This skews the range.\r\n\r\nInterquartile range\r\nThe interquartile range (IQR) is the distance between the first quartile \\(\\textrm{Q}_{1}\\) and the third quartile \\(\\textrm{Q}_{3}\\):\r\n\\[\\textrm{IQR}=\\textrm{Q}_{3}-\\textrm{Q}_{1}\\]\r\nThe first and third quartiles are values that are \\(\\dfrac{1}{4}\\) and \\(\\dfrac{3}{4}\\) of the way through the ordered data.\r\n\\(\\textrm{Q}_{1}\\) is the median of the lower half of the data and \\(\\textrm{Q}_{3}\\) is the median of the upper half of the data.\r\nLet's consider sets Y and Z again. When we order the data, we can identify the median of each set, indicated by the arrow. This will cut the data into a lower half and an upper half.\r\nFor set Y:\r\n\\[1\\quad1\\quad2\\quad2\\downarrow\\quad2\\quad2\\quad2\\quad100\\]\r\nWe can then identify the median of each half, which will be \\(\\textrm{Q}_{1}\\) and \\(\\textrm{m}Q_{3}\\).\r\n\\[1\\quad\\underbrace{1\\quad\\downarrow\\quad2}_{\\textrm{Q}_{1}}\\quad2\\quad\\downarrow\\quad2\\quad\\underbrace{2\\quad\\downarrow\\quad2}_{\\textrm{Q}_{3}}\\quad100\\]\r\nWe can calculate the medians.\r\n\\[\\begin{align*} \\textrm{Q}_{1} & = \\frac{1+2}{2}\\\\\r\n& = 1.5\r\n\\end{align*}\\] \\[\\begin{align*} \\textrm{Q}_{3} & = \\frac{2+2}{2}\\\\ \r\n& = 2\r\n\\end{align*}\\]\r\nThe interquartile range is:\r\n\\[\\begin{align*} \\textrm{IQR} & = \\textrm{Q}_{3}-\\textrm{Q}_{1}\\\\\r\n& = 2-1.5\\\\\r\n& = 0.5\r\n\\end{align*}\\]\r\nLet's find the IQR for set Z.\r\n\\[\\quad1\\quad\\underbrace{18\\quad\\downarrow\\quad23}_{\\textrm{Q}_{1}}\\quad41\\quad\\downarrow\\quad59\\quad\\underbrace{63\\quad\\downarrow\\quad87}_{\\textrm{Q}_{3}}\\quad100\\]\r\nWe have:\r\n\\[\\begin{align*} \\textrm{Q}_{1} & = \\frac{18+23}{2}\\\\\r\n& = 20.5\r\n\\end{align*}\\] \\[\\begin{align*} \\textrm{Q}_{3} & = \\frac{63+87}{2}\\\\\r\n& = 75\r\n\\end{align*}\\]\r\nThe interquartile range is:\r\n\\[\\begin{align*} \\textrm{IQR} & = \\textrm{Q}_{3}-\\textrm{Q}_{1}\\\\\r\n& = 75-20.5\\\\\r\n& = 54.5\r\n\\end{align*}\\]\r\nThe IQR for set Z is much larger than for set Y, which reflects the data more accurately because it ignores extreme values (i.e. it excludes outliers). For these two datasets, the mean and IQR are better summaries of the data than the range.\r\n\r\nStandard deviation\r\nStandard deviation (SD) gives the most comprehensive measure of spread. It measures how data points are spread out from the mean, rather than focusing on the span or central portion of the data like range and IQR do, respectively.\r\nIt takes into account all of the data by giving an indication of the typical or average distance of each score from the mean for the data. \r\nStandard deviation, denoted by \\(\\sigma\\) for population data and \\(s\\) for sample data, can be calculated using the formulas:\r\n\\[s=\\sqrt{\\frac{\\sum(x-\\overline{x})^{2}}{n-1}}\\quad\\textrm{or}\\quad\\sigma=\\sqrt{\\frac{\\sum(X-\\overline{x})^{2}}{N}}\\]\r\nwhere \\(n\\) is the number of values in the sample (or sample number) and \\(N\\) is the number of values in the population.\r\nIt is usually much more convenient to use your calculator or a computer.\r\nSome statistical tests use a measure of spread called variance which is the square of the standard deviation:\r\n\\[\\begin{align*} \\textrm{Variance} & = s^{2}\\\\\r\n& = \\frac{\\sum(x-\\overline{x})^{2}}{n-1}\r\n\\end{align*}\\]\r\nLet's calculate sample standard deviation for sets A and B. For set A: \r\n\\[4\\quad4\\quad5\\quad5\\quad5\\quad6\\quad6\\]\r\n\\[\\begin{align*} \\overline{x} & = \\frac{4+4+5+5+5+6+6}{7}\\\\\r\n& = 5\r\n\\end{align*}\\]\r\nFor set B:\r\n\\[1\\quad3\\quad4\\quad5\\quad6\\quad7\\quad9\\]\r\n\\[\\begin{align*} \\overline{x} & = \\frac{1+3+4+5+6+7+9}{7}\\\\\r\n& = 5\r\n\\end{align*}\\]\r\nThe scores in set A are typically \\(0.82\\) away from the mean, but the scores in set B are typically \\(2.65\\) away from the mean. Although set A and B have the same average value, data points in set B are clearly more dispersed or have greater spread than set A.\r\n\r\nExercise \u2013 calculating measures of spread\r\n\r\n\tFind the standard deviation for the following dataset: \\(12,12,13,14,14,15,15,15,16,\\).\r\n\tA class of \\(22\\) students gained the following scores, out of \\(10\\), on a test: \\(5,7,8,7,6,5,6,4,7,4,8,3,7,9,4,9,7,3,6,8,7,5\\). Find the:\r\n\t\r\n\t\trange\r\n\t\tIQR\r\n\t\tstandard deviation.\r\n\t\r\n\tPistol Pete is the star full-forward for the local football team. Last season he played \\(20\\) games and kicked the following number of goals in each game: \\(5,6,6,5,7,4,3,1,3,8,7,8,6,0,5,2,7,6,5,6\\).\r\n\t\r\n\t\tFind the mean and the standard deviation for the number of goals that Pete kicked per game.\r\n\t\tThis season, the mean number of goals Pete kicks per game is \\(5\\), with a standard deviation of \\(2.7\\). In which season was his performance more consistent?\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.036},[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":330,"$":[{"v":"Logarithms","n":1},{"v":"\r\nThe modeling of growth and decay in areas such as finance, epidemiology and science makes use of equations with logarithms and exponentials. The laws for working with logarithms enable us to solve equations that cannot be solved with other algebraic techniques. Use this resource to learn about them.\r\n\r\nVideo tutorial \u2013 logarithms\r\nWatch this video to learn about logarithms.\r\n\r\n\r\nLogarithmic expressions\r\nThe logarithm (or log) of a number is the power that the base must be raised to, to give that number. For example, the logarithm of \\(16\\) with a base of \\(2\\) is \\(4\\), because \\(2^{4}= 16\\).\r\nA log is an index. Every index or log expression contains a number, a base and an index. For example:\r\n\r\nWe say \"eight is equal to two to the power of three\".\r\nWe can write this in its logarithmic form:\r\n\r\nWe say \u201cthe log to base two of eight is equal to three\u201d (the power of two that gives eight is three).\r\nIn general,\r\n\\[a^{x}=n\\Leftrightarrow\\log_{a}n=x\\]\r\nwhere \\(a>0\\), \\(n>0\\) and \\(a\\neq1\\).\r\n\r\nExercise \u2013 converting to logarithm form\r\nWrite the following in logarithm form.\r\n\r\n\t\\(3^{2}\\)\r\n\t\\(10^{4}\\)\r\n\t\\(10^{-2}\\)\r\n\t\\(e^{a}=b\\)\r\n\r\n\r\n\r\n\r\nProperties of logarithms\r\nThere are three important properties of logarithms.\r\n\r\n\r\n \tFor \\(a>0,\\), \\(a=1\\Leftrightarrow\\log_{a}1=0\\). In other words, for any \\(a>0,\\) the log of \\(1\\) is \\(0\\).\r\n \tFor \\(a>0\\), \\(a=a^{1}\\Leftrightarrow\\log_{a}a=1\\).\r\n \tIf \\(\\log_{a}m=\\log_{a}p\\), then \\(m=p\\).\r\n\r\n\r\nThese will come in handy when you evaluate logarithms, so memorise them!\r\n\r\nEvaluating logarithms with base \\(10\\) or \\(e\\)\r\nThe base of a log can be any positive number except \\(1\\), but the most common are \\(10\\) and \\(e\\). When these bases are used, you can solve or evaluate the logarithm using a calculator.\r\n\r\n\tTo find the power of \\(10\\) that produces \\(50\\), we use the LOG button on the calculator: \\(\\log_{10}50\\approx1.7\\) and \\(10^{1.7}\\approx50\\).\r\n\tTo find the power of \\(e\\) that produces \\(50\\), we use the LN button on the calculator: \\(\\log_{e}50=\\ln50\\approx3.9\\) and \\(e^{3.9}\\approx50\\).\r\n\r\n\r\nYou can also solve logarithms by changing them to index form.\r\n\r\nExample 1 \u2013 solving logarithms with base \\(10\\) or \\(e\\)\r\nEvaluate \\(\\log_{10}1000\\).\r\n\\[\\begin{align*} \\textrm{Let}\\log_{10}1000 & =x\\quad\\textrm{change to index form}\\\\\r\n10^{x} & = 1000\\quad\\textrm{express 1000 as a power of 10}\\\\\r\n10^{x} & = 10^{3}\\quad\\textrm{equate indices}\\\\ \r\nx & = 3\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\nEvaluating logarithms using the logarithm laws\r\nJust like how there are three index laws, there are three laws of logarithms that help you to manipulate logarithms. They can be used to simplify and evaluate logarithmic expressions.\r\nFor these laws to be true, \\(m,n>0\\), \\(a>0\\) and \\(a\\neq1\\).\r\nFirst logarithm law\r\nTo find the log of a product, find the log of each factor separately and add them together.\r\n\\[\\log_{a}\\left(mn\\right)=\\log_{a}m+\\log_{a}n\\]\r\n\r\nSecond logarithm law\r\nTo find the log of a quotient, find the log of the dividend and subtract the log of the divisor.\r\n\\[\\log_{a}\\frac{m}{n}=\\log_{a}m-\\log_{a}n\\]\r\n\r\nThird logarithm law\r\nTo find the log of a number raised to a power, multiply the logarithm by the power.\r\n\\[\\log_{a}m^{p}=p\\log_{a}m\\]\r\n\r\nExample 1 \u2013 using the logarithm laws\r\nEvaluate \\(\\log_{a}15\\).\r\n\\[\\begin{align*} \\log_{a}15 & = \\log_{a}\\left(5\\times3\\right)\\\\\r\n& = \\log_{a}5+\\log_{a}3\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 evaluating logarithms\r\n\r\n\tEvaluate the following without using a calculator.\r\n\t\r\n\t\t\\(\\log_{7}49\\)\r\n\t\t\\(\\log_{10}\\sqrt{10}\\)\r\n\t\t\\(\\log_{5}1\\)\r\n\t\t\\(\\log_{10}100000\\)\r\n\t\t\\(\\log_{5}5\\)\r\n\t\r\n\tSimplify the following.\r\n\t\r\n\t\t\\(\\log_{4}8+\\log_{4}3-\\log_{4}2\\)\r\n\t\t\\(\\dfrac{1}{2}\\log_{10}25-\\log_{10}4+2\\log_{10}3\\)\r\n\t\t\\(\\log_{e}e^{2}+2\\log_{e}3-\\log_{e}18\\)\r\n\t\t\\(\\log_{a}4+2\\log_{a}3-2\\log_{a}6\\)\r\n\t\t\\(\\dfrac{1}{2}\\log_{10}a^{2}+3\\log_{10}b-\\log_{10}3ab^{2}\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on logarithms.\r\n\r\nA logarithm is an index but let\u2019s look at an index first. Here\u2019s the number eight, eight is the same as two to the power of three. Three things here, eight is the number, two is the base and three is the index, two to the power of three is two times two times two which is eight. So here it is again, eight equals two to the power of three. Now this can be written as log of eight to the base two equals three. Notice the three in red, the index three and the logarithm three. Also note the base two, both two to the three and log of eight to the base two. In other words we can write the equivalence here, A to the power of X is equal to N is the same as log of N to the base A is X. Notice the index in red and the logarithm in red.\r\n\r\nNow look at some examples, five to the power of three is 125 which is the same as log of 125 to the base five is three. Notice in each case as I go through these that the index and the logarithm are shown in red. Here\u2019s another one, three to the negative two is equal to one over nine, log of one over nine to the base three is negative two and finally 25 to the power of one half is five is the same as log of five to the base 25 is equal to the half. Study these carefully before moving on to the next slide.\r\n\r\nNow let\u2019s work out log of 64 to the base four. We\u2019ll write the equivalent statement first and beneath that log of 64 to the base four which equals X is the same as four to the power of X is 64. In other words what we\u2019re trying to do here is working out what the X value is, the power of four. So four cubed gives us 64, four times four times four, in other words X is equal to three, therefore log of 64 to the base four is equal to three.\r\n\r\nHere\u2019s another problem but this time we\u2019re solving log of X to the base two equals five. In the previous slide we were working out what the logarithm was, here we\u2019re working out what the number is, which is X in this case. First of all let\u2019s write the equivalent statement and then plugging our log over X to the base two equals five into that equivalent statement gives us two to the power of five is equal to X. In other words we\u2019re working out what two to the power of five is, which is two times two times two times two times two, which is 32, in other words X is equal to 32.\r\n\r\nNow here are two important logarithm rules. The first rule is A to the power of zero is equal to one, which the same as log of one to the base A is equal to zero. In other words for any base the log of one is zero. Second rule is A to the power of one is equal to A or the log of A to the same base A is equal to one, in other words the log of any number with itself as base is equal to one.\r\n\r\nThere are three laws of logarithms that you must know. Here\u2019s the first one. The log of two numbers multiplied together, M N is equal to the log of M plus the log of N. For instance here\u2019s the log of 15 and 15 is five times three, so the log of five times three is equal to the log of five plus the log of three.\r\n\r\nHere\u2019s the second law. The log of a quotient, in other words M divided by N is equal to log of M minus log of N. So for instance if we have the log of 12 over five that is the same as log of 12 subtract the log of five and finally the log of a number to a power, in this case log of M to the power of P is the same as P log of M. Notice how the index here is now placed in front of the log. So for instance the log of nine to the power of two is the same as two log of nine but notice nine is the same as three squared so we can go one step further and write two log three squared, again using the law three squared we can place the two in front so it now becomes two times two log of three which is four log of three.\r\n\r\nNow try some questions for yourselves. The answers to these problems are on the next slide. Thanks for watching this short movie.\r\n","n":0.028},[{"v":"Surds","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Logarithms","i":1,"n":1},{"v":"Indices","i":0,"n":1}]]},{"i":331,"$":[{"v":"Indices","n":1},{"v":"Index notation is a powerful mathematical tool for expressing repeated multiplication concisely. Learn how to work with indices, understand the laws governing them, and convert expressions involving powers and roots. This will help you simplify complex expressions and solve problems across a wide range of mathematical and scientific disciplines.\r\n\r\nVideo tutorial \u2013 indices\r\nWatch this video to learn how the laws of indices work.\r\n\r\n\r\nIndex notation\r\nIndex (or exponential) notation consists of two components:\r\n\r\n\tThe base is the number that is multiplied by itself.\r\n\tThe index or exponent is the number of times that the number is multiplied by itself.\r\n\r\nFor example, in \\(4^{2}\\), the base is \\(4\\) and the index is \\(2\\). This means that \\(4\\) is multiplied by itself \\(2\\) times, i.e. \\(4\\times4\\) which equals \\(16\\).\r\nFor the notation \\(3^{4}\\), the base is \\(3\\) and the index is \\(4\\). This means that \\(3\\) is multiplied by itself \\(4\\) times, i.e. \\(3\\times3\\times3\\times3\\) which equals \\(81\\).\r\nIn general:\r\n\\[ a^{n}=\\underbrace{a\\times a\\times a\\times\\ldots\\times a}_{n\\textrm{ factors}} \\]\r\n\r\n\r\nThe letter \\(n\\) in \\(a^{n}\\) can be referred to in one of three ways:\r\n\r\n\t\\(n\\) is the index in \\(a^{n}\\) with the base \\(a\\).\r\n\t\\(n\\) is the exponent or power to which the base \\(a\\) is raised.\r\n\t\\(n\\) is the logarithm, with \\(a\\) as the base.\r\n\r\nIndex is the singular form, and indices is the plural.\r\n\r\n\r\nIndex laws\r\nThere are laws that tell you how to multiply and divide indices, as well as how to deal with negative indices. These are known as the universal index laws.\r\n\r\nFirst index law\r\nTo multiply index expressions, you add the indices.\r\n\\[ a^{m}\\times a^{n}=a^{m+n} \\]\r\nFor example:\r\n\\[\\begin{align*} 2^{3}\\times2^{2} & = (2\\times2\\times2)\\times(2\\times2)\\\\\r\n& = 2\\times2\\times2\\times2\\times2\\\\\r\n& = 2^{5} \r\n\\end{align*}\\]\r\nTherefore, \\(2^{3}\\times2^{2}=2^{3+2}=2^{5}\\).\r\n\r\nSecond index law\r\nTo divide index expressions, you subtract the indices.\r\n\\[ a^{m}\\div a^{n}=\\frac{a^{m}}{a^{n}}=a^{m-n}\\quad a\\neq0 \\]\r\nFor example:\r\n\\[\\begin{align*} 3^{5}\\div3^{3} & =\\frac{3^{5}}{3^{3}}\\\\ \r\n& =\\frac{3\\times3\\times3\\times3\\times3}{3\\times3\\times3}\\quad\\textrm{cancel \\(3\\) lots of \\(3\\) from top and bottom}\\\\\r\n& =\\frac{3\\times3}{1}\\\\ \r\n& =3^{2} \r\n\\end{align*}\\]\r\nTherefore, \\(3^{5}\\div3^{3}=\\frac{3^{5}}{3^{3}}=3^{5-3}=3^{2}\\).\r\nExpressions in index notation can only be multiplied or divided if they have the same base.\r\n\r\nThird index law\r\nTo raise an expression in index form to a power, you multiply the indices.\r\n\\[ \\left(a^{m}\\right)^{n}=a^{m\\times n} \\]\r\nFor example:\r\n\\[\\begin{align*} (5^{2})^{3} & = 5^{2}\\times5^{2}\\times5^{2}\\\\ \r\n& = 5^{2+2+2}\\quad\\textrm{use the first index law}\\\\ \r\n& = 5^{6} \r\n\\end{align*}\\]\r\nTherefore, \\((5^{2})^{3}=5^{2\\times3}=5^{6}\\).\r\nThis also leads to the expression:\r\n\\[(a^{m}b^{p})^{n} = a^{mn}b^{pn}\\]\r\n\r\nBe careful as this is true for multiplication and division ONLY, not addition or subtraction.\r\n\\[(a+b)^{n}\\neq a^{n}+b^{n}\\]\r\n\r\nExample 1 \u2013 using the index laws\r\nSimplify \\(x^{5}\\times x^{6}\\).\r\nUse the first index law.\r\n\\[\\begin{align*} x^{5}\\times x^{6} & = x^{5+6}\\\\\r\n& = x^{11}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using the index laws\r\nSimplify the following.\r\n\r\n\t\\(c^{5}\\times c^{3}\\times c^{7}\\)\r\n\t\\(3\\times 2^{2}\\times 2^{3}\\)\r\n\t\\(a^{3}\\times a^{2}b^{3}\\times ab^{4}\\)\r\n\t\\(3^{6}\\div 3^{4}\\)\r\n\t\\(a^{8}\\div a^{3}\\)\r\n\t\\(x^{4}y^{6}\\div x^{2}y^{3}\\)\r\n\t\\((x^{3})^{4}\\)\r\n\t\\((x^{m}y^{n})^{5}\\)\r\n\r\n\r\n\r\n\r\n\r\nZero index\r\nSo far, we have only considered expressions with positive, whole number indices. The index laws also apply when the index is zero, negative or a fraction. Here, we will look at the first two.\r\n\r\nConsider \\(2^{3}\\div2^{3}\\).\r\n\r\n\tWhen we work it out by expanding the indices:\r\n\t\\[\\frac{2^{3}}{2^{3}}=\\frac{2\\times2\\times2}{2\\times2\\times2}=\\frac{8}{8}=8\\div8=1\\]\r\n\tWhen we work it out using the second index law: \r\n\t\\[2^{3}\\div2^{3}=2^{3-3}=2^{0}\\]\r\n\r\nTherefore, \\(2^{0}=1\\).\r\nIn general, any expression with a zero index is equal to \\(1\\). Since \\(0^{0}\\) is undefined, we do not allow \\(a=0\\) when dealing with the zero index law.\r\n\\[ a^{0}=1,\\quad a\\neq0 \\]\r\n\r\nNegative indices\r\nWhen you have a negative index, it is the same as taking the reciprocal of the power.\r\nConsider \\(2^{0}\\div2^{4}\\).\r\n\r\n\tWhen we work it out using fractions:\r\n\t\\[2^{0}\\div2^{4} = \\frac{2^{0}}{2^{4}} =\\frac{1}{2^{4}}\\]\r\n\t\r\n\tWhen we work it out using the second index law:\r\n\t\\[2^{0}\\div2^{4}=2^{0-4}=2^{-4}\\]\r\n\t\r\n\r\nTherefore, \\(2^{-4}=\\dfrac{1}{2^{4}}\\).\r\nSince \\(0^{0}\\) is undefined, we also do not allow \\(a=0\\) when dealing with negative indices.\r\n\\[ a^{-n}=\\frac{1}{a^{n}}\\textrm{ which also leads to }\\frac{1}{a^{-n}}=a^{n}\\quad a\\neq0\\]\r\n\r\nExample 1 \u2013 using the negative index law\r\n\\[2^{-3}\\]\r\nTake the reciprocal power:\r\n\\[\\begin{align*} 2^{-3} & = \\frac{1}{2^{3}}\\\\\r\n& = \\frac{1}{8}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using the negative index law\r\nWrite the following with positive indices and evaluate if possible.\r\n\r\n\t\\(x^{-6}\\)\r\n\t\\(250^{0}\\)\r\n\t\\(3ab^{-5}\\)\r\n\t\\((pq)^{-2}\\)\r\n\t\\((5xy)^{-3}\\)\r\n\t\\(\\dfrac{2y}{z^{-5}}\\)\r\n\t\\(2^{-5}\\)\r\n\t\\((-2)^{-3}\\)\r\n\t\\(-(3^{-2})\\)\r\n\t\\(2\\times(-5)^{-2}\\)\r\n\r\n\r\n\r\n\r\nCombining index laws\r\nMake sure that you remember the index laws.\r\n\r\n \t\\(a^{m}\\times a^{n}=a^{m+n}\\)\r\n \t\\(a^{m}\\div a^{n}=a^{m-n}\\,\\quad a\\neq0\\)\r\n \t\\((a^{m})^{n}=a^{mn}\\)\r\n \t\\(a^{0}=1\\,\\quad a\\neq0\\)\r\n \t\\(a^{-n}=\\dfrac{1}{a^{n}}\\,\\quad a\\neq0\\)\r\n\r\nIn your studies, you are likely to need to combine index laws when simplifying complex expressions.\r\n\r\nExample 1 \u2013 combining index laws\r\nSimplify \\((4a^{2}b)^{3}\\div b^{2}\\).\r\n\\[\\begin{align*} & = (4a^{2}b)^{3}\\div b^{2}\\quad\\textrm{use the third index law}\\\\\r\n& = (4^{3}a^{6}b^{3})\\div b^{2}\\quad\\textrm{use the second index law}\\\\\r\n& = 4^{3}a^{6}b^{1}\\\\ \r\n& = 4^{3}a^{6}b\\\\ \r\n& = 64a^{6}b \r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 combining index laws\r\nSimplify the following.\r\n\r\n\t\\(2a^{3}b^{2}\\times a^{-1}\\times b^{3}\\)\r\n\t\\((5x^{-2}y)^{-3}\\)\r\n\t\\((3x^{3}y^{-1})^{5}\\)\r\n\t\\((a^{-4}b^{-5})^{-2}\\)\r\n\t\\(\\dfrac{a^{2}b^{3}c^{-4}}{a^{4}bc^{5}}\\)\r\n\t\\(\\dfrac{a^{7}\\times a^{8}\\times a^{3}}{a^{2}\\times a^{5}}\\)\r\n\t\\(x(x-x^{-1})\\)\r\n\t\\(\\dfrac{(2^{4})^{n}}{2^{3}}\\)\r\n\t\\(\\dfrac{15a^{2}b}{3a^{4}b}\\times \\dfrac{4a^{5}b^{2}}{5a^{3}b^{4}}\\)\r\n\t\\(2^{4}-2^{3}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and learning centre at RMIT University. This is a short movie on indices.\r\n\r\nStart by looking at what we mean by a base and an index. Here we have two to the power of three, in other words the two is the base, the three is the index or the power. Notice also that students tend to confuse this with the same as two times three, which it is not. Two to the power of three means two multiplied by itself three times, in other words two times two times two, which gives us an answer of eight.\r\n\r\nLet\u2019s look at some examples; X to the power of four, well X is the base, four is the index, so it\u2019s X multiplied by itself four times, X times X times X times X. Here we have A squared times B to the power of five. Notice here that there are two different bases, A and B, so multiplying them out, A square is A times A, and B to the power of five is B times B times B times B times B. Finally we have four times M cubed times N cubed, so we leave the four alone and we multiply out the M three times, M times M times M, and the N is multiplied out three times as well, N times N times N, again the bases are different.\r\n\r\nWe\u2019ll now go through some laws of indices. Here\u2019s the first law of indices which says that X to the power of A times X to the power of B is the same as X to the power of A plus B. Notice here that the base must be the same but also that when we multiply terms with indices we actually add the indices.\r\n\r\nLet\u2019s do an example to illustrate this. We have A square times A to the five, so using the law that\u2019s the same as A to the two plus five. Notice here the base is the same so we can add the indices two plus five, so the answer is A to the seven. Similarly with numbers, three square times three cubed, the base are the same which means that we can add the indices, three to the two plus three which is three to the five, three to the five gives us an answer of 243.\r\n\r\nAnd, finally, here\u2019s a problem where we have a number and a fractional index, four times X to the half times X to the half. So we leave the four alone, the base is the same, so we can add the indices, a half plus a half which is one, so we have four X to the one. Notice another important thing here is that X to the power of one is exactly the same as just X so we can leave our answer as four X.\r\n\r\nNow let\u2019s look at the second law of indices; X to the A divided by X to the B is the same as X to the A minus B. A couple of things to notice here, the bases must be the same and when the bases are the same and we\u2019re dividing terms we subtract the indices, so for example A to the five divided by A to the two, the base are the same therefore we can subtract the indices, five take away two is A cubed. Similarly with numbers, two to the seven divided by two to the four, the bases are the same therefore we can subtract the indices, two to the seven minus four is two to the three, which is two times two times two, which gives us an answer of eight.\r\n\r\nAnd finally, here are two numbers, 10 and five, 10 to the X to 11 divided by five X to the six. Here we can divide the numbers, 10 divided by five gives us two, that\u2019s one part of this problem. The other part is dividing the terms with indices, X to the 11 divided by X to the six, the bases are the same therefore we can subtract the indices, so it\u2019s X to the 11 minus six which gives us five. In other words our answer is two X to the five.\r\n\r\nThe third law of indices says that X to the power of zero is equal to one. Now notice that any base with a power of zero will always give us an answer of one. Let\u2019s look at the reason behind this. Let\u2019s look at this problem, A cubed divided by A cubed, if I do this the long way I\u2019ve got the fraction as shown, A times A times A on the top, A times A times A on the bottom, notice they all cancel giving us an answer of one. But there\u2019s a quicker way of doing this using this law; A cubed divided by A cubed, same base therefore we can subtract the indices, three minus three is zero, that gives us A to the power of zero and the third law says that anything with a power of zero to it gives us an answer of one.\r\n\r\nNow let\u2019s look at the fourth law of indices. Notice here that we have brackets with an index inside and an index outside. So when we have this situation we multiply the indices, we don\u2019t add them, we multiply them, so here X to the A all inside brackets with an index of B outside of the brackets gives us an answer of X to the A times B. So for example four cubed in brackets to the power of two would give us four times four times four multiplied by four times four times four, this is the long way of doing this, which gives us an answer of four to the six. But if we use the law we multiply out the indices, in other words four cubed to the power of two is four to the three times two which is four to the six, as you can see a much quicker way of working out this problem.\r\n\r\nNotice also when we have a fraction as a base we do exactly the same as before. In other words the numerator is four, the denominator is five, so they both get indexed, in other words four cubed divided by five cubed and that gives us four times four times four which is 64 divided by five times five times five which is 125, so remember, with a fraction you use both the numerator and the denominator will have the base to it.\r\n\r\nFinally, let\u2019s look at square roots and cubed roots as indices. Notice a couple of things here; that the index inside the root is the numerator of our new index and the index outside the root is the denominator of our new index. Let\u2019s illustrate this with an example, here we have the square root of X which is the same as X to the power of one inside the square roots and we could put a two outside the square root as well and using the rule above that will give us an answer of X to the one over two, or X to the half, so the square root of X is the same as X to the half.\r\n\r\nSimilarly here we have the cubed root of X squared, so the square is on the top of our new index, the cubed root is on the bottom of our new index, so that gives us an answer of X to the two over three or X to the two thirds.\r\n\r\nNow try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.023},[{"v":"Surds","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Logarithms","i":1,"n":1},{"v":"Indices","i":0,"n":1}]]},{"i":332,"$":[{"v":"Solving linear equations","n":0.577},{"v":"Earlier, we looked at how to transpose formulas. Now, let's combine those skills with arithmetic to solve linear equations. By mastering the techniques for solving these equations, you'll be equipped to tackle a wide range of problems and build a strong basis for more advanced mathematical concepts.\r\n\r\nVideo tutorial \u2013 solving linear equations\r\nWatch this video to learn how to solve linear equations.\r\n\r\n\r\n\r\nLinear equations\r\nA linear equation has one unknown variable and may be solved by transposing the equation to make the variable the subject of the equation. For example, the equation:\r\n\\[\\begin{align*} 3(2x-7) & =-15 \r\n\\end{align*}\\]\r\nmay be transposed to get:\r\n\\[\\begin{align*} x & =6\r\n\\end{align*}\\]\r\nTransposition involves changing the equation via operations like addition, subtraction, multiplication and division. Let's look at how to solve simple and more complex linear equations.\r\n\r\nSolving simple equations\r\nRemember that when you transpose formulas, whatever you do to one side, you must do to the other.\r\nIf you add, subtract, multiply or divide one side of the equation by a number, you must do the same to the other side. This keeps the equation balanced, or preserves equality.\r\n\r\nExample 1 \u2013 solving simple equations\r\nSolve \\(m+4=-2\\).\r\nWe need to make the variable, \\(m\\), the subject of the equation. \\(4\\) is being added to \\(m\\), so we need to subtract \\(4\\) from both sides.\r\n\\[\\begin{align*} m+4 & = -2\\quad\\textrm{subtract \\(4\\) from both sides}\\\\\r\nm+4-4 & = -2-4\\\\\r\nm & = -6\r\n\\end{align*}\\]\r\nThe solution is \\(m=-6.\\) You can always check your solution by substituting your answer into the original equation and making sure the left-hand side (LHS) of the equation is equal to the right-hand side (RHS).\r\n\\[\\begin{align*} \\textrm{LHS} & = (-6)+4\\\\\r\n& = -2\\\\\r\n& = \\textrm{RHS}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 solving basic linear equations\r\n\r\n\tSolve the following equations.\r\n\t\r\n\t\t\\(x+3=7\\)\r\n\t\t\\(5-j=-2\\)\r\n\t\t\\(3c=12\\)\r\n\t\t\\(-r=-12\\)\r\n\t\t\\(\\dfrac{m}{2}=-7\\)\r\n\t\t\\(-8u=12\\)\r\n\t\t\\(4g+4=16\\)\r\n\t\t\\(7-2w=1\\)\r\n\t\t\\(\\dfrac{e}{2}-5=-8\\)\r\n\t\t\\(21-3t=12\\)\r\n\t\t\\(\\dfrac{y}{5}-9=-5\\)\r\n\t\t\\(3-\\dfrac{u}{2}=-7\\)\t\t\r\n\t\r\n\tSolve the following equations.\r\n\t\r\n\t\t\\(5i+2=i+10\\)\r\n\t\t\\(10p-11=2p-3\\)\r\n\t\t\\(5a-12=3a+6\\)\r\n\t\t\\(10d+10=0\\)\r\n\t\t\\(f+6=6f-9\\)\r\n\t\t\\(8-g=5g+14\\)\r\n\t\t\\(5h-2=7h-12\\)\r\n\t\t\\(6j+13=4j+13\\)\r\n\t\r\n\r\n\r\n\r\nSolving more complex equations\r\nWith more complex equations, you will need to perform several operations. Remember the following:\r\n\r\n\r\n \tIf the variable appears in terms on both sides of the equation, you need to get these terms to one side of the equation and then solve for the variable.\r\n \tIf an equation involves brackets, the brackets should be expanded first.\r\n \tIf an equation involves fractions, you need to use a common denominator.\r\n\r\n\r\nThe strategy for solving equations is not part of the solution. Here, it is just shown to help you learn. As you become more experienced, you may even perform several operations in one step.\r\n\r\nExample 1 \u2013 solving more complex equations\r\nSolve \\(2w-3=-17\\).\r\nWe need to make the variable, \\(w\\), the subject of the equation. \\(w\\) is multiplied by \\(2\\), then \\(3\\) is subtracted. We need to work backwards to isolate \\(w\\).\r\n\\[\\begin{align*} 2w-3 & = -17\\quad\\textrm{add \\(3\\) to both sides}\\\\\r\n2w-3+3 & = -17+3\\\\\r\n2w & = -14\\quad\\textrm{divide both sides by \\(2\\)}\\\\\r\n\\frac{2w}{2} & = \\frac{-14}{2}\\quad\\textrm{cancel out \\(2\\) on left}\\\\\r\nw & = -7\r\n\\end{align*}\\]\r\nThe solution is \\(w=-7\\).\r\nWith more complex equations, it is a good idea to check your result. However, this is not necessary unless you are asked to check. In this example:\r\n\\[\\begin{align*} \\text{LHS} & =2w-3\\\\\r\n& =2\\times(-7)-3\\\\\r\n& =-14-3\\\\\r\n& =-17\\\\\r\n& =\\text{RHS}\r\n\\end{align*}\\]\r\nOur solution is correct.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 solving more complex linear equations\r\n\r\n\r\n\tSolve the following equations.\r\n\t\r\n\t\t\\(3(2k-4)=18\\)\r\n\t\t\\(5(2z+9)=15\\)\r\n\t\t\\(3(x+4)=6\\)\r\n\t\t\\(3(c+3)+2(c-5)=4\\)\r\n\t\t\\(3(2v-3)+2(v-4)=-25\\)\r\n\t\t\\(3(b+4)=2(4b+1)\\)\r\n\t\r\n\tSolve the following equations.\r\n\t\r\n\t\t\\(\\dfrac{9n}{5}-4=5\\)\r\n\t\t\\(\\dfrac{4m}{3}-11=9\\)\r\n\t\t\\(1-\\dfrac{9q}{2}=-8\\)\r\n\t\t\\(\\dfrac{w-4}{2}=2\\)\r\n\t\t\\(\\dfrac{3-2e}{11}=1\\)\r\n\t\t\\(\\dfrac{3r+9}{5}=-3\\)\r\n\t\t\\(\\dfrac{3t}{8}+4=1\\)\r\n\t\t\\(\\dfrac{y}{3}=\\dfrac{2}{9}\\)\r\n\t\t\\(\\dfrac{5u-4}{4}=\\dfrac{u-5}{5}\\)\r\n\t\t\\(\\dfrac{2i+1}{7}=\\dfrac{3i-5}{4}\\)\r\n\t\t\\(\\dfrac{p+1}{3}+1=4\\)\r\n\t\t\\(2-\\dfrac{5a-4}{4}=4\\)\r\n\t\t\\(\\dfrac{d-3}{3}-4=\\dfrac{d-2}{2}\\)\r\n\t\t\\(\\dfrac{1-m}{5}-m=\\dfrac{2m-1}{2}\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n \r\nSolving equations using subtraction; let\u2019s look at a simple equation, X plus three equals seven. So we\u2019re saying what number instead of X which, when added to three, gives us seven. Well obviously the answer is four because four plus three equals seven. An equation is like a balance, the numbers on the left hand side must balance the numbers on the right hand side. But let\u2019s do this another way, let\u2019s get X by itself, in other words we\u2019re doing it mathematically. To do this we subtract the three from the left hand side to isolate the X, so now the left hand side is X plus three minus three which gives us X. Let\u2019s bring in the right hand side, but remember we have to keep the equation balanced, in other words X plus three minus three on the left hand side is equal to seven minus three on the right hand side. You see how the minus three undoes the plus three, so on the left hand side we\u2019ve got X and the right hand side we\u2019ve got four. Notice that the three has been subtracted from both sides, in other words what we do to one side we must do to the other.\r\n\r\nSo the rule is that the opposite or inverse operation to plus is minus. In other words subtracting undoes adding. Let\u2019s solve an equation using addition. So we\u2019ll start with Y minus six equals 13, so we\u2019re saying what number instead of Y which when added to minus six gives us 13. Well the answer is 19 because 19 take away six equals 13. Again notice the balancing act, 19 minus six on the left hand side is equal to 13 on the right hand side. But let\u2019s do this mathematically. Let\u2019s get Y by itself and to do this we need to add the six from the left hand side to isolate the Y, remember we had to subtract the number with the previous problem, so the left hand side is now Y minus six plus six, which gives us Y, that\u2019s the left hand side. What about the right hand side. Well what we do to one side we do to the other so Y minus six plus six is equal to 13 plus six, notice plus six undoes minus six. In other words Y is equal to 19. Notice again that the six has been added to both sides, so the rule here is that the opposite or inverse operation to minus is plus, in other words adding undoes subtracting.\r\n\r\nLet\u2019s solve an equation using division. Let\u2019s look at the equation four K equals 12. Remember four K means four times K, so we\u2019re saying what number instead of K which when multiplied by four gives us 12, well three because four times three is equal to 12, so we have 12 equals 12, our balancing act again. But let\u2019s do this another more mathematical way, let\u2019s get that K by itself and to do this we divide by four on the left hand side to isolate the K. To balance this we must do the same thing on the right hand side, so here we have the left hand side which is four K divided by four, the right hand side we must divide that by four too, so it\u2019s 12 divided by four, there\u2019s the calculation where we\u2019re dividing by four and so we\u2019ve got 12 divided by four is equal to three. Notice this time I\u2019ve just checked the solution to this equation, in other words if K equals three and we substitute four K into the original equation, four times three is equal to 12. So the rule here is that the opposite or inverse operation to multiplication is division. In other words dividing undoes multiplying.\r\n\r\nLet\u2019s solve an equation using multiplication. Now here we have an equation M divided by four is equal to minus five, in other words what number instead of M which when divided by four gives us minus five. This one isn\u2019t so easy to work out so we have to use the following rule, in other words let\u2019s use a little bit of mathematics. What we\u2019re doing here is multiplying both sides by four, so M over four times four over one, notice how the fours cancel, is equal to minus five times four over one, or four. In other words minus five times four is minus 20. So let\u2019s check the solution. If the answer is minus 20 then minus 20 divided by four is equal to minus five. So the rule here, for multiplication, is that the opposite or inverse operation to division is multiplication, multiplying undoes dividing.\r\n\r\nNow let\u2019s move on to more complicated operations where we have more than one. For instance, let\u2019s look at the equation two P plus five equals seven. The unknown value P is multiplied by two and then five is added to obtain the answer of seven. To find P we undo two P plus five in the reverse order to which it was built up. In other words we subtract five first and then divide by two, which is the opposite to what we did in getting from two P plus five to seven. So let\u2019s subtract five first, two P plus five minus five equals seven minus five, which gives us two. Then let\u2019s divide by two, notice we\u2019re doing it on both sides of the equation, two times P over two equals two over two, the right hand side is now one because we cancelled the two top and bottom, so two P equals two then P equals one.\r\n\r\nSo let\u2019s look at another equation, but this time it\u2019s a division equation for starters, three X minus one divided by two is equal to one. In order to solve the equation the operations is to build up the equation must be undone in the reserve order. In other words we start by dividing by two and then we add the one. Let\u2019s look at this in more detail. Three X minus one divided by two, let\u2019s multiply both sides by two to get rid of the two on the bottom of the fraction on the left hand side so we end up with three X minus one equals two. Now let\u2019s get rid of the minus one on the left hand side, so three X minus one plus one equals two plus one, notice we\u2019re doing the same operation on both left and right hand sides of the equation. That leaves us with three X equals three here, so the final step is to remove the three on the left hand side and because we have three times X we must divide by three so three X over three equals three over three. Notice the three cancels out and we\u2019re left with X equals one.\r\n\r\nSo when we solve equations using more than one operation we first of all look at how the operation was built up, then we solve the equation by undoing it in the reverse order to which it was built up. Now go to the exercise sheet and have a go at the problems for yourself. Thank you.\r\n","n":0.024},[{"v":"Maths equations","i":2,"n":0.707},{"v":"Maths","i":1,"n":1},{"v":"Equations","i":0,"n":1}]]},{"i":333,"$":[{"v":"The quotient rule","n":0.577},{"v":"\r\nThe quotient rule helps us differentiate functions that are one function divided by another. Use this resource to learn how to apply the quotient rule.\r\n\r\nVideo tutorial \u2013 using the quotient rule\r\nWatch this video to learn how to use the quotient rule to find the derivative of a function.\r\n\r\n\r\nThe quotient rule for differentiation\r\nThe quotient rule states that if \\(y=f(x)=\\dfrac{u(x)}{v(x)}\\), then:\r\n\\[\\frac{dy}{dx}=\\frac{v(x)\\cdot\\frac{du}{dx}-u(x)\\cdot\\frac{dv}{dx}}{(v(x))^{2}}\\]\r\nYou may see this written as \\(y'=\\dfrac{vu'-v'u}{v^{2}}\\).\r\n\r\nExample 1 \u2013 differentiating using the quotient rule\r\nIf \\(y=\\dfrac{1+x}{x^{2}-3}\\), find \\(\\dfrac{dy}{dx}\\).\r\nFirst, let \\(u=1+x\\) and \\(v=x^{2}-3\\). The derivatives of \\(u\\) and \\(v\\) are:\r\n\\[u'=1\\] \\[v'=2x\\]\r\nNow, we can use the quotient rule.\r\n\\[\\begin{align*} \\frac{dy}{dx} & = \\frac{vu'-v'u}{v^{2}}\\\\\r\n& = \\frac{(x^{2}-3)\\cdot(1)-(2x)\\cdot(1+x)}{(x^{2}-3)^{2}}\\\\\r\n& = \\frac{x^{2}-3-2x-2x^{2}}{(x^{2}-3)^{2}}\\\\\r\n& = \\frac{-x^{2}-2x-3}{(x^{2}-3)^{2}}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExercise \u2013 differentiating using the quotient rule\r\nFind the derivatives of the following functions with respect to \\(x\\).\r\n\t\r\n\t\t\\(f(x)=\\dfrac{2x+1}{4x-3}\\)\r\n\t\t\\(f(x)=\\dfrac{3}{3x^{2}+1}\\)\r\n\t\t\\(y=\\dfrac{\\sqrt{x}}{1-\\sqrt{x}}\\)\r\n\t\t\\(y=\\dfrac{e^{x}}{\\sin^{2}x}\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on the quotient rule.\r\n\r\nThe quotient rule is used when we want to differentiate the quotient or fraction of two functions. So if f of x is equal to u of x divided by v of x then the derivative f dash x is equal to v of x times u dash x minus v dash x times u of x all divided by v of x all squared.\r\n\r\nNow that formula is rather unwieldy to remember, so its often abbreviated to y dash is equal to v u dash minus u v dash all over v squared, and as with the product rule notice that f dash x and y dash mean the same thing.\r\n\r\nSo lets do an example. Here we want to differentiate y equals 1 plus x divided by x squared minus 3. Note first of all that we have a quotient, ie, a fraction with the numerator 1 plus x and the denominator x squared minus 3. So, we substitute, we let u equals 1 plus x, the derivative u dash is 1 and the denominator v equals x squared minus 3 from which the derivate v dash is equal to 2x. There is our formula for y dash for the quotient rule and now by substituting into that formula we get the following lines of working: so the first line was substituting, the next line we remove the brackets (be careful with the minus sign, the minus 2x squared) and finally we\u2019ve simplified to get the answer in blue at the bottom of the slide (y dash equals minus x squared minus 2x minus 3 divided by bracket x squared minus 3 squared).\r\n\r\nLets do another example we want to differentiate y equals x squared divided by the log of x. Note that the log of x is the same as log of x to the base e. Again as with the previous example we have a quotient. X squared is the numerator, log of x is the denominator, so we use the quotient rule. Substituting u equals x squared, therefore u dash equals 2x and v is the log of x from which the derivative is v dash 1 over x. There is our formula for the quotient rule (y dash equals v u dash minus uv dash divided by v squared) and now substituting we get these two lines of working. Notice that when we get to the final line we need to, as before, simplify, particularly the right hand term in the numerator where one of the x\u2019s cancels top and bottom. So that\u2019s our answer (y dash equals 2x log of x minus x divided by bracket log x squared).\r\n\r\nNow try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.041},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":334,"$":[{"v":"The product rule","n":0.577},{"v":"\r\nThe product rule helps us differentiate functions that are one function multiplied by another. Use this resource to learn how to apply the product rule.\r\n\r\nThe product rule for differentiation\r\nThe product rule states that if \\(y=f(x)=u(x)\\cdot v(x)\\), then:\r\n\\[\\frac{dy}{dx} = u(x)\\cdot\\frac{dv}{dx}+v(x)\\cdot\\frac{du}{dx}\\]\r\nYou may see this written as \\(y'=uv'+u'v\\).\r\nExample 1 \u2013 differentiating using the product rule\r\nFind the derivative of \\(f(x)=(x+3)^{6}(2x-1)\\).\r\n\r\nLet \\(u=(x+3)^{6}\\) and \\(v=2x-1\\). For \\(u\\), we need to use the chain rule, so let's do that first. We can let \\(w=x+3\\) and \\(u=w^{6}\\).\r\n\\[\\begin{align*} u' & = \\frac{du}{dw}\\times\\frac{dw}{dx}\\\\\r\n& = 6w^{5}\\times1\\quad\\textrm{substitute in }w=x+3\\\\\r\n& = 6(x+3)^{5}\r\n\\end{align*}\\]\r\n\r\nNow, we can use the product rule.\r\n\\[\\begin{align*} f'(x) & = u\\cdot v' + u'\\cdot v\\\\\r\n& = ((x+3)^{6}\\cdot 2) + (6(x+3)^{5}\\cdot(2x-1)\\\\\r\n& = (x+3)^{5}\\left(2(x+3)+6(2x-1)\\right)\\\\\r\n& = 14x(x+3)^{5}\r\n\\end{align*}\\]\r\n\r\n\r\nExercise \u2013 differentiating using the product rule\r\n\r\n \tUse the product rule to differentiate the following.\r\n\r\n \t\\(y=(x-2)(6x+7)\\)\r\n \t\\(f(x)=(2x^{2}+4)(x^{5}+4x^{2}-2)\\)\r\n \t\\(y=(\\sqrt{x}-1)(x^{2}+1)\\)\r\n \t\\(y=(x^{3}-4x+\\sqrt{x})(3x^{4}+2)\\)\r\n\r\n\r\n \tFind the derivative of the following.\r\n\r\n \t\\(y=e^{x}\\tan(x)\\)\r\n \t\\(y=x{^2}\\log_{e}(x)\\)\r\n \t\\(y=\\sin(x)\\cos(x)\\)\r\n \t\\(y=\\dfrac{e^{x}}{x}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.085},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":335,"$":[{"v":"The chain rule","n":0.577},{"v":"\r\nThe chain rule helps us differentiate larger functions composed of smaller functions. Use this resource to learn how to apply the chain rule.\r\n\r\nVideo tutorial \u2013 using the chain rule\r\nWatch this video to learn how to use the chain rule to find the derivative of a function.\r\n\r\n\r\nThe chain rule for differentiation\r\nThe chain rule states that if \\(y=g(u)\\) where \\(u=h(x)\\), then:\r\n\\[\\frac{dy}{dx}=\\frac{dy}{du}\\times\\frac{du}{dx}\\]\r\nIn other words, to find the derivative of function \\(y\\), you take the derivative of the outer function \\(g(u)\\) and multiply it by the derivative of the inner function \\(h(x)\\).\r\n\r\nExample 1 \u2013 differentiating using the chain rule\r\nDifferentiate \\(y=(2x-1)^{4}\\).\r\nFirst, we need to define the parts of the composite function. \r\n\r\n\tThe inner function is \\(u\\) or \\(h(x)\\), so \\(u=2x-1\\).\r\n\tThe outer function is \\(y\\) or \\(g(u)\\), so \\(y=u^{4}\\).\r\n\r\nWe can then find the derivative of functions \\(u\\) and \\(y\\).\r\n\\[\\frac{du}{dx}(2x-1)=2\\] \\[\\frac{dy}{du}(u^{4})=4u^{3}\\]\r\nTo put it altogether, we use the chain rule.\r\n\\[\\begin{align*} \\frac{dy}{dx} & = \\frac{du}{dx} \\times \\frac{dy}{du}\\\\\r\n& = 4u^{3} \\times 2\\\\\r\n& = 8u^{3}\r\n\\end{align*}\\]\r\nTo achieve the final answer, we must substitute \\(u=2x-1\\) back into the derivative.\r\n\\[8u^{3} = 8(2x-1)^{3}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 differentiating using the chain rule\r\nFind the derivatives of the following functions.\r\n\t\r\n\t\t\\(y=\\tan(3x)\\)\r\n\t\t\\(f(x)=\\log_{e}\\left(\\dfrac{x}{2}\\right)\\)\r\n\t\t\\(y=\\sin\\left(\\dfrac{\\pi}{4}-2x\\right)\\)\r\n\t\t\\(y=\\cos^{2}x\\)\r\n\t\t\\(f(x)=e^{\\sin(x)}\\)\r\n\t\t\\(y=\\cos^{2}(10x)\\)\r\n\t\r\n\r\n\r\n\r\n\r\n \r\nHi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short calculus movie on the chain rule. The chain rule is used to differentiate a function which is the composition of two simpler functions, in other words if Y equals G U where U equals H X then D Y D X is D Y D U times D U D X.\r\n\r\nLet\u2019s differentiate this function; Y equals two X minus one or to the power of four. So we let U equals two X minus one, then Y equals U to the power of four, differentiating, D U D X equals two, D Y D U equals four U cubed, substituting into the chain rule we have four U cubed times two which gives us eight U cubed and substituting back for U we get an answer of eight brackets two X minus one or to the power of three.\r\n\r\nLet\u2019s look at a trig function; Y equals sign to the five X, so we let U equal five X giving us Y equal sign U, differentiating D U D X is five, D Y D U is cos of U and substituting into the chain rule we get cos U times five which is five cos of five X, again substituting back for U which is, in this case, is five X.\r\n\r\nLet\u2019s look at a more complicated function here. We have Y equals one over the cubed root of 5 T square plus two T plus one, notice first of all that we\u2019ve changed the variable here, instead of D Y D S we\u2019ve got D Y D T, it doesn\u2019t matter we can still use the same process as before. First of all always changed cube roots, square roots and so on into index form to make the differentiation easier, using our substitution, differentiating D U D T, D Y D U, and now using the chain rule which in this case D Y D T equals D Y D U times D U D T and substituting we get that, which gives us, after simplifying that answer.\r\n\r\nHere\u2019s another trig function but note first of all that instead of Y and D Y D X we\u2019ve got F X and F dash X. First of all recall that if Y equals cos cubed X that\u2019s the same as cos of X or to the power of three. Using our substitution Y equals U cubed where U equals cos of X, now we differentiate, D Y D U equals three U squared, D U D X equals minus sign of X. Notice I\u2019ve moved back from F X and F dash X to Y and D Y D X. Now let\u2019s do the chain rule, substituting and substituting back for U we get an answer of minus three sign X cos squared X. Note that the chain rule can be extended to three or more functions. Up to now we\u2019ve been looking at two functions.\r\n\r\nHere\u2019s an example of a three function problem; Y equals log of four X or to the power of three, the three here are Y equals U cubed, U equals log of V and V is equal to four X, differentiating the three, D Y D U is three U squared, D U D V is one over V and D V D X is four, substituting for D Y D U, D U D V and D V D X we get the following, and substituting back we get an answer of three over X brackets log of four X or to the power of two after simplifying.\r\n\r\nNow try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.035},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":336,"$":[{"v":"Differentiation from first principles","n":0.5},{"v":"\r\nLearn how to take a derivative of a function using first principles. Using this method is the best way to understand the concepts around differentiation.\r\n\r\nDerivative of a function\r\nThe derivative of a function \\(f(x)\\) is denoted by \\(f'(x)\\). It is defined as:\r\n\\[f'(x)=\\lim_{h\\rightarrow0}\\left(\\frac{f(x+h)-f(x)}{h}\\right)\\quad h\\neq0\\]\r\nUsing this definition is called differentiating from first principles, but note that it's not always the most convenient method. It should only be used if you are asked to use it.\r\n\r\nAlternative notations\r\nAs well as \\(f'(x)\\), there are other ways to denote derivatives. When \\(y\\) is a function of \\(x\\), i.e. \\(y=f(x)\\), the derivative function may be written as \\(y'\\) or \\(\\dfrac{dy}{dx}\\) or \\(f'(x)\\).\r\nIt is important to understand that \\(\\dfrac{d}{dx}\\) is a symbol that means \"take the derivative of something with respect to \\(x\\)\". \r\nFor example, \\(\\dfrac{d}{dx}\\left(x^{2}\\right)\\) means to take the derivative of \\(x^{2}\\) with respect to \\(x\\).\r\n\r\nVideo tutorial \u2013 differentiating from first principles\r\nWatch this video to learn how to find the derivate of a function using first principles.\r\n\r\n\r\nExample 1 \u2013 differentiating from first principles\r\nIf \\(f(x)=x^{2}\\), find the derivative of \\(f(x)\\) from first principles.\r\n\\[\\begin{align*} f'(x) & = \\lim_{h\\rightarrow0}\\left(\\frac{f(x+h)-f(x)}{h}\\right)\\\\ \r\n& = \\lim_{h\\rightarrow0}\\left(\\frac{(x+h)^{2}-x^{2}}{h}\\right)\\\\ \r\n& = \\lim_{h\\rightarrow0}\\left(\\frac{x^{2}+2xh+h^{2}-x^{2}}{h}\\right)\\\\\r\n& = \\lim_{h\\rightarrow0}\\left(\\frac{2xh+h^{2}}{h}\\right)\\\\\r\n& = \\lim_{h\\rightarrow0}\\left(\\frac{h(2x+h)}{h}\\right)\\\\\r\n& = \\lim_{h\\rightarrow0}\\left(2x+h\\right)\\\\\r\n& = 2x\r\n\\end{align*}\\]\r\nThe derivative of \\(x^{2}=2x\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 differentiating from first principles\r\nFind the derivative of the following functions using differentiation from first principles.\r\n\t\r\n\t\t\\(f(x)=3x\\)\r\n\t\t\\(f(x)=5x^{2}\\)\r\n\t\t\\(f(x)=2x^{2}-\\dfrac{1}{x}\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on differentiation from first principles.\r\n\r\nThe process of finding the derivative f-x is equal to the limit as h approaches zero of f, of x plus h, minus f of x, divided by h, is called differentiation from first principles. Just be aware that f-x is the same as dydx.\r\n\r\nLet\u2019s look at an example. We want to differentiate f of x as equal to s\u00b2 minus x from first principles. Notice in the formula that we have three terms to investigate. We have the f of x plus h term, the f of x term, and the h term. So let\u2019s start with the f of x plus h term first. Knowing f of x is equal to x\u00b2 minus x what we do is we substitute x plus h wherever we see x. In other words, x\u00b2 becomes x plus h all squared, and x becomes x plus h. So we have x plus h all squared, minus x plus h, which on expansion gives u x\u00b2 plus 2xh, plus h\u00b2 minus x minus h.\r\n\r\nNow let\u2019s look at the fx term. Well we already know that f of x is equal to s\u00b2 minus x, so we can simply leave this expression alone. In other words, s\u00b2 minus x is x\u00b2 minus x. So, let\u2019s substitute f of x plus h and f of x into the numerator of the formula at the top of the slide, and we get this. Notice on removing the brackets the negative sign at the end of the line two becomes a plus x in line three. Notice also that the x\u00b2 terms disappear and the h terms cancel out. So we\u2019re left with 2xh plus h\u00b2 minus h. So let\u2019s now return to the formula and substitute. We have in our numerator 2xh plus h\u00b2 minus h.\r\n\r\nNotice also that the numerator factorises, so we can take h out, giving h brackets 2x plus h minus one. Notice also that there\u2019s an h on the bottom of this fraction, so the h\u2019s cancel, leaving us the bottom line f-x is equal to the limit as h approaches zero of 2x plus h minus one.\r\n\r\nNow let\u2019s look at the h approaches zero part of our formula. And what that means is that as h approaches zero, the value of h gets smaller and smaller and smaller. So working your way down the slide notice h gets smaller and smaller and smaller until it approaches zero. So we\u2019re left with f-x is equal to 2x minus one.\r\n\r\nSo, to conclude, the derivative of fx is equal to a\u00b2 minus x using first principles is f-x equals 2x minus one.\r\n\r\nNow try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.038},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":337,"$":[{"v":"Limit of a function","n":0.5},{"v":"\r\nLimits help us explore how functions behave as they approach specific points or infinity, providing the foundation for concepts like continuity and derivatives. You need this skill in many real-world applications, from engineering to physics, where predicting behaviour is crucial. Use this resource to learn how to calculate and interpret limits.\r\n\r\nVideo tutorial \u2013 finding the limit of a function\r\nWatch this video to learn how to find the limit of a function.\r\n\r\n\r\nFinding the limit of a function\r\nThe limit of a function is the value that a curve approaches at a particular point. While we can pinpoint a precise location using a value with many decimal places, limits involve approaching values that can be infinitely close, without necessarily reaching them. \r\nInstead, we can describe the limit of function as the value we approach from both the left and right of the function. That is what we mean by approaching the limit.\r\nAs we approach a particular value \\(x=a\\) and the value of the function \\(a\\) is the same regardless of whether we approach from the left or the right, then the limit exists. Consider the following function, \\(f(x)\\).\r\n\r\nThe limit of \\(f(x)\\) as \\(x\\) approaches \\(2\\) is \\(4\\). We denote this mathematically as:\r\n\\[\\lim_{x\\rightarrow2}f(x)=4\\]\r\nThe function does not need to be continuous or even defined at the point of interest.\r\n\r\nExample 1 \u2013 finding the limit of a function\r\nFind the limit of the function \\(f(x)=x+2\\) as \\(x\\) approaches \\(2\\).\r\nThe behaviour of \\(f(x)\\) as \\(x\\rightarrow2\\) is shown in the table.\r\n\r\nThe table shows that as \\(x\\) approaches \\(2\\), \\(f(x)\\) approaches \\(4\\).\r\nAnother way to find the limit of \\(f(x)\\) is by substituting \\(x=2\\) into the function.\r\n\\[\\begin{align*} f(x) & = x+2\\\\\r\nf(2 ) & = 2+2\\\\\r\n& = 4\r\n\\end{align*}\\]\r\nTherefore, \\(\\lim_{x\\rightarrow2}\\left(x+2\\right)=4\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the limit of a function\r\nDetermine the following limits.\r\n\r\n\t\\(\\lim_{x\\rightarrow0}\\left(4xh+2\\right)\\)\r\n\t\\(\\lim_{x\\rightarrow0}\\left(\\dfrac{9-x^{2}}{4}\\right)\\)\r\n\t\\(\\lim_{h\\rightarrow0}\\left(\\dfrac{xh-2h}{h}\\right)\\)\r\n\t\\(\\lim_{x\\rightarrow0}\\left(\\dfrac{\\tan(x)}{x}\\right)\\)\r\n\r\n\r\n\r\n\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is another movie on differentiation. This time we\u2019re focusing on limits.\r\n\r\nThe limit of a function by definition describes the behaviour of this function in the vicinity of a particular value of the variable, which is usually x. Let\u2019s look at the function x2 plus one, and investigate the limit of this function as it approaches x equals one. Let\u2019s first sketch the function fx equals x2 plus one. As you can see, it\u2019s a parabler, y equals x2, which has been shifted up the y axis one unit. Point of interest here is the point at x equals one and y equals two, and we will refer back to this slide a little later.\r\n\r\nWhat we do first of all is to find values of x as it approaches one from below. In other words, we\u2019re moving from the left-hand side of the graph up to the value of x equals one. Go back to the previous slide. If we do that as x approaches one from the left or below the values of the function will approach two. In other words, as x approaches one from below, x2 plus one approaches two.\r\n\r\nNow let\u2019s investigate what happens to x as it approaches one from above. In other words, we\u2019re moving from the right of the graph to the value of x equals one. Go back to the slide again to see the red arrows indicating this. When we plug in our values of x as they approach one from the right or above, the values of fx approach as you can see, two. So as x approaches one from above, x2 plus one approaches two.\r\n\r\nSo in both cases, below one and above one, as x approaches one the value of x2 plus one approaches two. So we say more formally that the limit of x2 plus one as x approaches one is equal to two.\r\n\r\nNow let\u2019s investigate the limit of a more complicated function. Here the function fx is equal to x2minus one divided by x minus one, again as it approaches x equals one. Let\u2019s investigate this function more closely. Notice that if we let x equals one, the denominator of this function will be zero. In other words, it\u2019s not defined at x equals one. We say it is not continuous at x equals one. If we restrict our function to all values except for x is equal to one, we can now progress and factorise this function, x2 minus x becomes x brackets x minus one. Notice that we can cancel terms top and bottom. In other words, the function fx is equal to x.\r\n\r\nSo sketching the function fx equals x as you would expect, we have a straight line for all values except for x equals one, and this is the point at which we will be investigating the limit. In other words, f of one does not exist, but as x approaches one, the value of f of x approaches one as well. So more formally since f of x if equal to x, then the limit of x2 minus x over x minus one as x approaches one is equal to one.\r\n\r\nNow try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.","n":0.034},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":338,"$":[{"v":"Completing the square","n":0.577},{"v":"\r\nSo far, you will have come across a few situations where a quadratic cannot be further factorised into two linear factors using common factors, perfect squares, or the difference of two squares. In these cases, you can use a different method where you complete the square.\r\n\r\nBefore you begin, make sure you are confident with factorising using common factors and the difference of two squares rule.\r\n\r\nVideo tutorial \u2013 factorising: completing the square\r\nWatch this video to learn about completing the square.\r\n\r\n\r\nCompleting the square when \\(a=1\\)\r\nCompleting the square is a method that involves making the quadratic a perfect square, so that you can then factorise using \\((x+b)^{2}=x^{2}+2bx+b^{2}\\) or \\((x-b)^{2}=x^{2}-2bx+b^{2}\\). It allows any quadratic to be factorised.\r\nYou can complete the square using the following steps:\r\n\r\n\r\n\tTake \\(\\dfrac{1}{2}\\) the coefficient in front of \\(x\\) and square it.\r\n\tAdd the term to the expression. Remember that if you add a term, you must also subtract it to keep the expression balanced.\r\n\tGroup the terms so that you can use the DOTS rule.\r\n\r\n\r\nNote that DOTS can only be used if you are looking at the DIFFERENCE between squares. If the expression is the SUM of two squares, there will be no real linear factors.\r\nSee how this is applied to expressions in the form \\(ax^{2}+bx+c\\) where \\(a=1\\) with the following examples.\r\n\r\nExample 1 \u2013 completing the square when \\(a=1\\)\r\nFactorise \\(x^2+8x-5\\) by completing the square.\r\nFind the term by calculating \\(\\dfrac{1}{2}\\) of the coefficient in front of the \\(x\\), squared:\r\n\\[\\left(\\frac{8}{2}\\right)^{2} = (4)^{2}\\]\r\nAdd and subtract the term:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 competing the square when \\(a=1\\)\r\nFactorise by completing the square.\r\n\r\n\t\\(x^{2}+6x-5\\)\r\n\t\\(x^{2}+4x+2\\)\r\n\t\\(a^{2}-8a-2\\)\r\n\t\\(x^{2}+6x+10\\)\r\n\t\\(x^{2}-5x+5\\)\r\n\t\\(y^{2}+3y+4\\)\r\n\t\\(a^{2}-5a-1\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\nCompleting the square when \\(a\\neq1\\)\r\nWhen \\(a\\neq1\\), we need to first take the \\(a\\) out by dividing each term by \\(a\\) before factorising the quadratic. That is:\r\n\\[ax^{2}+bx+c=a\\left[x^{2}+\\frac{b}{a}x+\\frac{c}{a}\\right]\\]\r\nWe complete the square on the term in the square brackets. Don\u2019t forget to include the \\(a\\) in your final answer!\r\n\r\nExample \u2013 completing the square when \\(a\\neq1\\)\r\nFactorise \\(2x^{2}-10x+2\\).\r\nDivide all terms in the quadratic by \\(a\\). Here, \\(a=2\\).\r\n\\[\\begin{align*} 2x^{2}-10x+2 & = 2\\left[x^{2}-\\frac{10}{2}x+\\frac{2}{2}\\right]\\\\ \r\n& = 2\\left[x^{2}-5x+1\\right]\r\n\\end{align*}\\]\r\nComplete the square. Find the term by calculating \\(\\dfrac{1}{2}\\) of the coefficient in front of the \\(x\\), squared:\r\n\r\n\r\n\r\nExercise \u2013 completing the square when \\(a\\neq1\\)\r\nFactorise by completing the square.\r\n\r\n\t\\(4x^{2}+8x-20\\)\r\n\t\\(x^{2}-6x+2\\)\r\n\t\\(12-4x-2x^{2}\\)\r\n\r\n\r\n \r\n\r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is another factorisation movie. This one is on completing the square.\r\n\r\nLet\u2019s first look at X plus B all squared, which is the same as X squared plus two B X plus B squared, or putting it around the other way, X squared plus two B X plus B squared is the same as X plus B all squared. Notice the term B is half of the co\u2011efficient of X, the middle term. In other words X plus B all squared is called a perfect square. So as an example, X squared minus 24 X plus 144, look at the middle term which is minus 24 X, half of that is minus 12 and thus that is our perfect square. So knowing how to find a perfect square is very important when completing the square.\r\n\r\nLet\u2019s do an example. Factorise X squared plus eight X minus nine. Step one, half the co-efficient of X, in other words half of eight is four, then we square the result, four squared is 16. Step three, add and subtract 16, in other words X square plus eight X plus 16 is our perfect square, then we subtract the 16 minus nine, notice that expression there is the same as the original expression, eight square plus eight X minus nine. Step four, complete the square and simplify. In other words eight square plus eight X plus 16 is X plus four all squared minus 16 minus nine is minus 25 and then finally factorise using the difference of two squares, which is X plus four plus five brackets X plus four minus five, which is the same as X plus four X minus one.\r\n\r\nLet\u2019s do another one. X squared minus six X minus four. Step one, half the co\u2011efficient of X which is half of six is three, square the result, three squared is nine. Step three, add and subtract nine, that gives us X squared minus six X plus nine, there\u2019s our perfect square, then we have to subtract nine to make sure that we have the original X squared minus six X minus four. Step four, complete the square and simplify, in other words, X squared minus six X plus nine is X minus three all squared minus nine minus four is minus 13 and then finally factorise using the difference of two squares, which is X minus three plus root 13, X minus three minus root 13.\r\n\r\nNow let\u2019s look at a slightly harder problem. Notice here we\u2019re factorising two X squared minus 10 X plus two. We can only complete the square if the co-efficient of X squared is one, so the first step we have to do is to remove the factor of two, thus we have two brackets X squared minus five X plus one. Now we go through the same process as before, half the co-efficient of X, but here the co-efficient of X is now five, so that half of that is five over two, square the result, five over two squared is 25 over four, then we add and subtract 25 over four which gives us that expression. Continuing on from the previous slide we have that expression, now we need to take the difference of two squares which gives us two brackets X minus five over two all squared minus 21 over four, but that can be simplified into the final expression two brackets X minus five over two plus root 21 over two brackets X minus five over two minus root 21 over two.\r\n\r\nNow try some questions for yourself. The answers to these problems are on the next slide. Thanks for watching this short movie.\r\n","n":0.032},[{"v":"Maths equations","i":1,"n":0.707},{"v":"Algebra","i":0,"n":1}]]},{"i":339,"$":[{"v":"Quadratic factorisation","n":0.707},{"v":"Quadratic expressions are fundamental in mathematics and appear frequently in various applications, from physics to finance. Use this resource to learn what quadratics are and how to factorise them. The ability to do this is crucial for solving quadratic equations and simplifying complex algebraic problems.\r\n\r\nVideo tutorial \u2013 factorisation: quadratics\r\nWatch this video to learn what quadratics are, review expansion, and understand how to factorise quadratics. \r\n\r\n\r\nQuadratic expressions\r\n\r\nSometimes, expansion results in a second-degree algebraic expression, or quadratic expression. This is when the highest power of an algebraic term is \\(2\\). The term \"quadratic\" comes from the Latin quad which means \"to make square\". Making a number a square is the same as taking it to the power of \\(2\\).\r\nSome examples of quadratic expressions are:\r\n\r\n\t\\(x^{2}+5x+6\\)\r\n\t\\(4x^{2}+8\\)\r\n\t\\(2x^{2}-3x\\)\r\n\t\\(7x^{2}-x+15\\)\r\n\r\nCompare this to a linear expression, where an algebraic term has a term with the highest power of \\(1\\).\r\nLinear expressions have the form \\(ax+by+c\\). Quadratic expressions take the general form:\r\n\\[ax^{2}+bx+c\\]\r\nwhere \\(a\\), \\(b\\) and \\(c\\) are constants and \\(a \\neq 0\\)\r\nTo start with, we will work with expressions that have \\(a=1\\) so the expression becomes \\(x^{2}+bx+c\\).\r\n\r\nExample \u2013 expanding algebraic expressions\r\nExpand \\((x+2)(x+3)\\).\r\nTo expand the expression, multiply each term in the first bracket by each term in the second bracket.\r\n\\[\\begin{align*} (x+2)(x+3) & = x(x+3)+2(x+3)\\\\\r\n& =x^{2}+3x+2x+6\\\\\r\n& =x^{2}+5x+6\r\n\\end{align*}\\]\r\n\r\nFactorising quadratic expressions\r\nJust like factorising simpler algebraic expressions, factorising quadratics involves expressing them as products of their factors. This is the reverse of expansion.\r\nThe basic rule is:\r\n\\[x^{2}+bx+c = (x+m)(x+n)\\]\r\nwhere \\(m\\times n=c\\) and \\(m+n=b\\)\r\n\\(x+m\\) and \\(m+n\\) are the factors. The order in which they are written does not matter. That is:\r\n\\[\\begin{align*} x^{2}+bx+c & = (x+m)(x+n)\\\\\r\n& = (x+n)(x+m)\r\n\\end{align*}\\]\r\nThese factors are called linear factors because they have the form \\(ax+by+c\\).\r\n\r\nExample 1 \u2013 factorising quadratics when \\(a=1\\)\r\nFactorise \\(x^{2}+5x+6\\).\r\nMultiplying the first term in each bracket gives the term \\(x^{2}\\), so the first terms must be \\(x\\).\r\nMultiplying the last term in each bracket gives the constant term \\(c\\). \\(c=+6\\), so:\r\n\\[m\\times n=6\\]\r\n\\(m\\) and \\(n\\) are factors of \\(+6\\) and could therefore be \\(+1\\) and \\(+6\\), or \\(+2\\) and \\(+3\\). Their negative versions could also be factors, such as \\(-1\\) and \\(-6\\), since \\((-1)\\times (-6)=+6\\).\r\nThe coefficient \\(b\\) is the sum of the last term in each bracket. In this case, \\(b=5\\), so:\r\n\\[m+n=5\\]\r\nLooking at the possible values for \\(m\\) and \\(n\\), we can pick out the ones that add up to \\(5\\). This would be \\(+2\\) and \\(+3\\).\r\nTherefore, the factorised form of \\(x^{2}+5x+6\\) is \\((x+2)(x+3)\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 factorising quadratics when \\(a=1\\)\r\nFactorise the following expressions.\r\n\r\n\t\\(x^{2}+10x+21\\)\r\n\t\\(z^{2}+11z+18\\)\r\n\t\\(x^{2}+5x-14\\)\r\n\t\\(m^{2}-m-72\\)\r\n\t\\(x^{2}+6x+9\\)\r\n\t\\(a^{2}-15a+44\\)\r\n\t\\(x^{2}-2x-24\\)\r\n\t\\(y^{2}-10y+16\\)\r\n\t\\(z^{2}+4z-60\\)\r\n\t\\(n^{2}+6n-16\\)\r\n\t\\(a^{2}+5a+10\\)\r\n\t\\(s^{2}+2s-48\\)\r\n\t\\(y^{2}+7y+19\\)\r\n\t\\(x^{2}+16x+39\\)\r\n\t\\(x^{2}-14x+45\\)\r\n\r\n\r\n\r\n\r\n\r\nFactorising when \\(a\\neq1\\)\r\nExpressions of the type \\(ax^{2}+bx+c\\), where \\(a\\neq1\\), can be factorised using a technique similar to what we have used for \\(x^{2}+bx+c\\).\r\nWhen we factorise these types of quadratics, the coefficient of \\(x\\), in at least one bracket, will not equal \\(1\\).\r\nThe examples shown use the trial-and-error method to factorise the quadratics. When there are many factors, the process can get quite tedious. In these cases, we can use other methods. You will learn about them later.\r\n\r\nExample 1 \u2013 factorising quadratics where \\(a\\neq1\\)\r\nFactorise \\(2x^{2}+7x+6\\).\r\nMultiplying the first term in each bracket gives the term \\(2x^{2}\\). The factors of \\(2x^{2}\\) are \\(+1\\) and \\(+2x\\), and its negative versions.\r\nThis gives us:\r\n\\[\\begin{align*} 2x^{2}+7x+6 & = (2x+m)(x+n)\\\\\r\n& = 2x^{2}+2nx+mx+mn\\\\\r\n& = 2x^{2}+(2n+m)x+mn\r\n\\end{align*}\\]\r\n\\(c=+6\\), so:\r\n\\[m\\times n=6\\]\r\n\\(m\\) and \\(n\\) are factors of \\(+6\\) and could therefore be \\(+1\\) and \\(+6\\), \\(+2\\) and \\(+3\\), or their negative versions.\r\nThe coefficient \\(b\\) is the sum of the \\(x\\) terms in the expansion. \\(b=+7\\), so:\r\n\\[2n+m=7\\] \r\nWe can check which of the factors work with this equation. Here, we are checking whether we can get \\(b=+7\\).\r\nFor \\(m=1\\) and \\(n=6\\):\r\n\\[\\begin{align*} 2n+m & = (2\\times6)+1\\\\\r\n& = 13\r\n\\end{align*}\\]\r\nFor \\(m=6\\) and \\(n=1\\):\r\n\\[\\begin{align*} 2n+m & = (2\\times1)+6\\\\\r\n& = 8\r\n\\end{align*}\\]\r\nFor \\(m=-1\\) and \\(n=-6\\):\r\n\\[\\begin{align*} 2n+m & = (2\\times(-6)+(-1)\\\\\r\n& = -13\r\n\\end{align*}\\]\r\nFor \\(m=-6\\) and \\(n=-1\\):\r\n\\[\\begin{align*} 2n+m & = (2\\times(-1)+(-6)\\\\\r\n& = -8\r\n\\end{align*}\\]\r\nFor \\(m=2\\) and \\(n=3\\):\r\n\\[\\begin{align*} 2n+m & = (2\\times3)+2\\\\\r\n& = 8\r\n\\end{align*}\\]\r\nFor \\(m=3\\) and \\(n=2\\):\r\n\\[\\begin{align*} 2n+m & = (2\\times2)+3\\\\\r\n& = 7\r\n\\end{align*}\\]\r\n\\(m=3\\) and \\(n=2\\) satisfies the condition \\(b=+7\\). In other cases, you might have to continue and try the remaining possibilities.\r\nSubstituting \\(m=3\\) and \\(n=2\\) into the expression, the final factorised expression is:\r\n\\[(2x+3)(x+2)\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 factorising quadratics when \\(a\\neq 1\\)\r\nFactorise the following.\r\n\r\n\t\\(5x^{2}+13x+6\\)\r\n\t\\(2x^{2}+x-15\\)\r\n\t\\(3m^{2}-m-2\\)\r\n\t\\(3y^{2}-10y+8\\)\r\n\t\\(2a^{2}+11a+12\\)\r\n\t\\(6x^{2}-11x+5\\)\r\n\r\n\r\n\r\nThe discriminant\r\nAs you saw in one of the earlier examples, sometimes a quadratic does not have real linear factors. But, you don't really want to be completing time-consuming calculations only to find out that you cannot factorise the quadratic.\r\nOne way to avoid this is to first check whether it has any linear factors. We do this using the discriminant.\r\nThe discriminant, denoted \\(\\Delta\\), for the general quadratic \\(ax^{2}+bx+c\\) is:\r\n\\[\\Delta=b^{2}-4ac\\]\r\nIf:\r\n\\[\\begin{align*} \\Delta & = 0\\textrm{, there is one repeated linear factor}\\\\\r\n\\Delta & > 0\\textrm{, there are two distinct linear factors}\\\\\r\n\\Delta & < 0\\textrm{, there are no real linear factors}\r\n\\end{align*}\\]\r\n\r\n\r\nGraphically, a quadratic is a parabola (a U-shaped curve). The discriminant tells us how many times the parabola touches or intersects the \\(x\\)-axis. When \\(\\Delta=0\\), the graph just touches the \\(x\\)-axis at just one point. When \\(\\Delta>0\\), the graph touches the \\(x\\)-axis at two points. When \\(\\Delta<0\\), the graph does not intersect the \\(x\\)-axis.\r\n\r\nExample \u2013 using the discriminant\r\nFactorise \\(x^{2}+6x+12\\).\r\nWe can use the discriminant to determine whether the quadratic has real linear factors.\r\nFor this quadratic, \\(a=+1\\), \\(b=+6\\) and \\(c=+12\\).\r\n\\[\\begin{align*} \\Delta & = b^{2}-4ac\\\\\r\n& = (6)^{2}-(4\\times1\\times12)\\\\\r\n& = 36-48\\\\\r\n& = -12\r\n\\end{align*}\\]\r\n\\(\\Delta<0\\), so \\(x^{2}+6x+12\\) has no real linear factors. \r\nHi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on factorising quadratics. The general form of a quadratic expression is given by A X squared plus B X plus C. Here A, B and C are constants and X is the unknown. For instance here\u2019s an expression, two X squared minus three X minus four, A is plus two, B is minus three and C is minus four. Note carefully the sign for each of the numbers.\r\n\r\nLet\u2019s go back a step first and look at revision of expansion, in other words removing brackets. Here\u2019s an expression, X plus two X plus three and when you expand that you multiply each term in the first bracket by the terms in the second bracket to get an answer of X squared plus five X plus six. We\u2019ve done this in a previous movie.\r\n\r\nNow factorisation is the reverse of expansion, in other words X square plus five X plus six can be expressed to the product of two factors, X plus two and X plus three. In other words as you go from left to right in this equation we\u2019re factorising, as we go from right to left we\u2019re expanding. A few important things to notice about factorisation; first of all multiplying the first terms in each bracket gives the X squared in the expression, in other words X times X is X squared. Multiplying the last terms in each bracket gives the constant term plus six in the expression, in other words plus two times plus three is plus six. Finally the co-efficient of the X terms, in other words the middle term, is the sum of the last term in each bracket, in other words here plus two and plus three is plus five. Now we\u2019ll be using these steps to work out some of these examples that follow.\r\n\r\nSo in general for the expression A squared plus B X plus C where A is equal to one multiply two numbers to get C, we add the same two numbers to get B, in other words we insert one number into each of the brackets and secondly the order is not important, in other words the second brackets could be the first and vice versa, it doesn\u2019t matter which way around you place the brackets.\r\n\r\nSo here\u2019s our first example, X squared plus nine X plus 14. So straight away we can place two brackets, as I\u2019ve done, to the right of the equal signs. Using the rules we need two numbers that when multiplied together gives us the last term, 14, and those same two numbers that, when added, gives us the co-efficient of X, the middle term, which is nine. So we can look at 14 and one, seven and two and so on until we get those two numbers that when multiplied gives us 14 but also added gives us nine and those two numbers are seven and two, seven times two is 14, seven plus two is nine. Notice if I use 14 and one multiplying them is fine but adding them will not give us nine.\r\n\r\nLet\u2019s look at another one; Y squared minus seven Y plus 12. Here we have a negative sign in the middle term which makes it slightly more difficult but we still use the same idea as in the previous example, in other words two numbers that multiply gives us 12, the same two numbers that when added gives us minus seven, so we can look at 12 and one, six and two, four and three. Notice when we look at those different combinations, those factors of 12, there\u2019s only one that will satisfy the two conditions and that is minus three and minus four. Notice minus three times minus four is plus 12, minus three plus minus four is minus seven, so our answer is Y minus three Y minus four.\r\n\r\nHere\u2019s another one. Notice the terms are separated by negative signs, P squared minus five P minus 14. Again, using the rules, multiplying the two you must get minus 14, adding the two numbers we must get minus five. Here it\u2019s not so difficult because we only have 14 and one, that gives us 14, but 14 and one there\u2019s no way in which, when adding those two numbers, whether they be negative or positive, will give us minus five, so the solution to this is P minus seven and P plus two. Be extra careful with the negative signs here, notice that minus seven times plus two gives us minus 14 and minus seven plus two is minus five.\r\n\r\nAnd finally here\u2019s another one; A squared plus six A minus seven. We need two numbers that multiply to give minus seven and two numbers that add to give plus six. So we don\u2019t have much choice here, the only factors we have of seven are seven and one, so they must be seven and one, but the difficult problem here is to work out where the negative signs go and this usually comes just with trial and error, in other words A plus seven A minus one, seven times minus one is minus seven, plus seven minus one is plus six. If you tried minus seven and plus one you would notice you would not get plus six A, so try the different methods to make sure you\u2019re correct.\r\n\r\nNow try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.024},[{"v":"Maths equations","i":1,"n":0.707},{"v":"Algebra","i":0,"n":1}]]},{"i":340,"$":[{"v":"Difference of two squares","n":0.5},{"v":"\r\nLet's learn about the difference of two squares rule. Recognising and applying this pattern allows you to factorise expressions quickly and solve equations more efficiently.\r\n\r\nVideo tutorial \u2013 factorisation: difference of two squares\r\nWatch this video to learn about the difference of two squares rule.\r\n\r\n\r\nDifference of two squares (DOTS) rule\r\nThe difference of two squares (or DOTS) rule states that:\r\n\\[(a+b)(a-b)=a^{2}-b^{2}\\]\r\nIf you have a perfect square subtracted from another perfect square, you can factorise them quickly according to the DOTS rule. The terms in the brackets differ only in the sign of the second term. It will save you a lot of time if you memorise this.\r\nWe can prove this rule by expanding the expression:\r\n\\[\\begin{align*}(a+b)(a-b) & = a^{2}-ab+ab-b^{2}\\\\\r\n& = a^{2}-b^{2}\r\n\\end{align*}\\]\r\n\r\nExample 1 \u2013 factorising using DOTS\r\nFactorise \\(a^{2}-36\\).\r\n\r\n\\((a+6)(a-6)\\) is the same as \\((a-6)(a+b)\\); the order in which they are written does not matter.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 factorising using DOTS\r\nFactorise the following.\r\n\r\n\t\\(x^{2}-4\\)\r\n\t\\(a^{2}-100\\)\r\n\t\\(49-x^{2}\\)\r\n\t\\(64x^{2}-1\\)\r\n\t\\(121x^{2}-49y^{2}\\)\r\n\t\\(a^{2}b^{2}-25\\)\r\n\t\\(5x^{2}-20\\)\r\n\t\\(a^{2}+100\\)\r\n\t\\(x^{2}y^{2}-36y^{2}\\)\r\n\t\\((x+2)^{2}-y^{2}\\)\r\n\t\\((x-5)^{2}-36\\)\r\n\t\\((a+1)^{2}-(b-2)^{2}\\)\r\n\r\n\r\n\r\n\r\n \r\nHi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on factorisation, the difference of two squares or dots.\r\n\r\nLet\u2019s first of all review our knowledge of removing brackets, X plus five X minus five in this example. And if you go down three lines we end up with an answer of X squared minus 25. Notice that is the same as X squared minus five squared, in other words what we have here is terms in brackets differ only in the sign in each bracket, on the left hand side, and the expansion on the right hand side is the difference of two terms both of which are perfect squares, eight squared and five squared. So in general we can state that the dots rule says A plus B times A minus B is the same as A squared minus B squared, so the terms in brackets on the left hand side differ only in the sign, plus or minus, in each bracket and the expansion on the right hand side is the difference of two terms, both of which are perfect squares.\r\n\r\nLet\u2019s look at an example. Here we have X squared minus 36, and the first thing to do is to see if both these terms are perfect squares. Well they are because X squared is the same as X times X, X squared, and 36 is the same as six times six which is 36. So we can now go on and state the rule which says that X squared minus 36 is the same as X plus six X minus six, the only difference is the sign in each brackets.\r\n\r\nLet\u2019s look at another one. Four A squared minus Y squared. The first question are each of the terms perfect squares, well they are, four A squared is the same as two A all squared or two A times two A and Y squared is the same as Y times Y, so the expression is the difference of two squared or dots. So we can use the rule which says that four A squared minus Y squared is the same as two A minus Y two A plus Y, the only difference is the sign in each brackets.\r\n\r\nNow let\u2019s look at a slightly more difficult question. We have three X squared minus 48. So the first question is, is the expression a dots, well it\u2019s not because neither of those terms are perfect squares, however when we factorise this expression by taking three out of the brackets we have three brackets X squared minus 16, but notice this time that X squared minus 16 is a perfect square, X times X and 16 is four times four, in other words three X squared minus 48 is three, open up big sets of brackets in blue, then we have X squared in red minus four squared in red which gives us three brackets X plus four X minus four.\r\n\r\nNow, in this final example we have X plus two all squared minus nine. First question is are these two terms perfect squares. Well the first term must be because it\u2019s X plus two all squared, so that is. Nine is a perfect square because nine is three times three, so we now have the expression X plus two all squared minus three all squared. And the rule, that gives us X plus two plus three, notice the blue plus, minus X plus two minus three, notice the blue minus, which simplifies to X plus five brackets X minus one and that\u2019s our answer.\r\n\r\nNow try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.036},[{"v":"Maths equations","i":1,"n":0.707},{"v":"Algebra","i":0,"n":1}]]},{"i":341,"$":[{"v":"Factorisation","n":1},{"v":"While expansion is the process of removing brackets from algebraic expressions and equations, factorisation is the opposite. Both are important to make solving algebraic problems easier. Use this resource to learn the basics of factorisation.\r\n\r\nVideo tutorial \u2013 factorisation: common factors\r\nWatch this video to learn about factorisation using common factors.\r\n\r\n\r\nFactorising algebraic expressions\r\nFactorisation is the process of writing a number of algebraic expression as a product of numbers or expressions. Something that can be helpful to grasp this concept is to remember that factors are numbers we can multiply together to get another number.\r\nFactorisation involves putting terms into brackets. This is the opposite of expansion, which involves removing brackets.\r\nThe simplest way to factorise an algebraic expression is using common factors. A common factor is a number or pronumeral that is shared by every term in an algebraic expression. Being able to identify common factors is critical for factorisation.\r\n\r\nExample 1 \u2013 identifying common factors\r\nIdentify the factors in \\(42\\) and \\(-2xyz\\).\r\n\\(42\\) is \\(6\\times7\\) and also \\(2\\times3\\times7\\), so the factors are: \\(2\\), \\(3\\), \\(6\\) and \\(7\\).\r\n\\(-2xyz\\) is \\(-1\\times2\\times x\\times y\\times z\\), so the factors are: \\(-1\\), \\(2\\), \\(x\\), \\(y\\) and \\(z\\).\r\nRemember that combinations are factors as well, i.e. \\(-2\\), \\(xy\\), \\(xyz\\), etc.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nFactorising using common factors\r\nTo factorise using common factors, you can use the following steps:\r\n\r\n \tLook at each term in the expression and find all of their factors.\r\n\tIdentify the factors that every expression has in common.\r\n \tIf there is more than one common factor, multiply them to give highest common factor (HCF).\r\n\tDivide each term by the HCF and write them in brackets. Make sure to keep the original operations that connect the terms.\r\n \tPlace the HCF before the bracket.\r\n\r\nLet's look at some examples where these steps are applied.\r\n\r\nExample 1 \u2013 factorising using common factors\r\nFactorise \\(5y+10\\).\r\nThe factors of \\(5y\\) are \\(5\\) and \\(y\\).\r\nThe factors of \\(10\\) are \\(2\\), \\(5\\) and \\(10\\).\r\nThe HCF is \\(5\\).\r\nDividing each term by the HCF:\r\n\\[\\begin{align*} 5y \\div 5 & =y\\\\\r\n10 \\div 5 & =2\r\n\\end{align*}\\]\r\nWhen we keep the original operations and place the HCF before the bracket, we get:\r\n\\[5(y+2)\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 factorising using common factors\r\n\r\n\tFactorise the following expressions.\r\n\t\r\n\t\t\\(3x+3y\\)\r\n\t\t\\(2a-2b\\)\r\n\t\t\\(8a-8b+8c\\)\r\n\t\t\\(xy-5x\\)\r\n\t\t\\(x^{2}-x\\)\r\n\t\t\\(7x+21y\\)\r\n\t\t\\(5m-2n\\)\r\n\t\t\\(c^{2}-2bc-3c\\)\r\n\t\t\\(5mn-10n\\)\r\n\t\t\\(3m^{2}-3mnp\\)\r\n\t\t\\(7x+21x^{2}\\)\r\n\t\t\\(12m^{2}-18mn\\)\r\n\t\t\\(5xy-10xz\\)\r\n\t\t\\(5pq-pq^{2}-3pqr\\)\r\n\t\t\\(2ab^{2}c+6abc^{2}\\)\r\n\t\t\\(rst+5rst^{2}-2rs\\)\r\n\t\t\\(5mn+10m-pqr\\)\r\n\t\t\\(5xyz-x^{2}yz^{2}+10x\\)\r\n\t\r\n\tFactorise the following expressions by removing a negative factor.\r\n\t\r\n\t\t\\(-3x-6y\\)\r\n\t\t\\(-15xy+25xz\\)\r\n\t\t\\(-2xy+4xyz\\)\r\n\t\t\\(14xyz-7xy\\)\r\n\t\t\\(-6xyz-15yz-3xy^{2}z\\)\r\n\t\t\\(7x-21y\\)\r\n\t\r\n\r\n\r\n\r\n \r\nHi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on finding common factors in algebra. Let\u2019s start by recalling that expansion or removing brackets is all about multiplying all the terms inside the brackets by those outside. For instance A brackets B plus C is the same as A times B plus A times C, if A brackets two Y minus four A is equal to 10 A Y minus 20 A X.\r\n\r\nFactorisation is the reverse of expansion or removing brackets. For example, here\u2019s an example, 10 A Y minus 20 A X. At this stage just look at the right hand side which are the factors, five A and brackets two Y minus four X, they are the factors, so moving from right to left, multiplying five A by the brackets will give you 10 A Y minus 20 A X. Similarly the factors for this expression are three and X minus Y and multiplying those terms will give us three X minus three Y, so to factorise an algebraic expression we look for a common factor, if there is more than one common factor we multiply them to get the highest common factor, HCF, the highest common factor is then placed before the bracket and the terms inside the brackets are found by dividing each term by the highest common factor.\r\n\r\nLet\u2019s do some examples to explain these four important steps. Here\u2019s an example, five Y plus 10. First of all let\u2019s look for the common factor and here the common factor of five Y and 10 is five, in other words five will divide into five Y and five will divide into 10. In other words that\u2019s the same as five times Y plus five times two, it\u2019s the same as five brackets Y plus two, so note the four steps, we factorise, the common factor is five, we place the five before the brackets and what goes after that five will be the other factor, Y plus two, which is inside the brackets.\r\n\r\nHere\u2019s another example, but this time we\u2019re using pro-numerals, P squared plus P. What is the common factor of P squared plus P, well it\u2019s P, P will go into P squared and P will go into P, so that\u2019s the same as P times P plus P times one which again is the same as P brackets P plus one. There are two factors P and P plus one, the common factor P, P is placed before the brackets, the P plus one is placed inside the brackets.\r\n\r\nHere is a slightly more complicated one. We\u2019re mixing numbers and pro-numerals here. What\u2019s the common factor of two A B and 12 A, well two goes into both those terms but also A does as well so the common factor is two and A, so we call that the highest common factor here, we put the two together and call them one thing, two A, same as two A times B minus two A times six, which can be written in another way, two A brackets B minus six. So to repeat, factorising, highest common factor is two A, two A before the brackets, B minus six is placed inside the brackets, which is the other factor.\r\n\r\nNow do some questions for yourself. The answers to these problems are on the next slide. Thanks for watching this short movie.\r\n","n":0.033},[{"v":"Maths equations","i":1,"n":0.707},{"v":"Algebra","i":0,"n":1}]]},{"i":342,"$":[{"v":"Algebraic substitution","n":0.707},{"v":"Algebraic substitution is a powerful technique for simplifying algebraic expressions and solving equations by replacing the pronumerals with known values. This skill is essential for tackling complex problems in mathematics, science, and engineering, enabling you to evaluate and manipulate expressions with ease.\r\n\r\nAlgebraic expressions\r\nAn algebraic term consists of a number, pronumeral or both. When algebraic terms are combined using addition and/or subtraction, they become an algebraic expression.\r\nFor example, \\(3x^{3}\\), \\(pq\\) and \\(8ab^{2}c\\) are algebraic terms. \\(\\dfrac{a+b}{2}\\), \\(5x-2\\) and \\(3p+v\\) are algebraic expressions.\r\n\r\n\r\nAlgebraic equations\r\nWhen algebraic expressions are written as a mathematical statement, this is an algebraic equation. One algebraic equation you might have encountered before is the equation for the area of a circle:\r\n\\[A=\\pi r^{2}\\]\r\nwhere \\(A\\) is the area of a circle, \\(\\pi\\) (pi) is a constant equal to approximately \\(3.14\\) and \\(r\\) is the radius of the circle.\r\n\r\n\r\nVideo tutorial \u2013 algebraic substitution\r\nWatch this video to learn about algebraic substitution.\r\n\r\n\r\nSubstitution\r\nSubstitution is when you put a number into an algebraic expression or formula in place of the pronumerals.\r\n\r\nExample 1 \u2013 substituting values into algebraic expressions\r\nCalculate the area of a circle with a radius of \\(5\\textrm{ cm}\\).\r\nWe can substitute the number for the radius of the circle into the algebraic equation \\(A=\\pi r^{2}\\) to calculate the area of the circle. In this question, \\(r=5\\textrm{ cm}\\).\r\n\\[\\begin{align*} A & =\\pi r^{2}\\\\\r\n& =\\pi\\times(5)^{2}\\\\\r\n& =25\\pi\\\\\r\n& \\approx78.54\\textrm{ cm}^{2}\r\n\\end{align*}\\]\r\nRemember that area is represented in square units, thus \\(\\textrm{cm}^{2}\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 substituting values into algebraic expressions\r\n\r\n \tEvaluate the following.\r\n\t\r\n\t\t\\(-4k\\) if \\(k=7\\)\r\n\t\t\\(2mn\\) if \\(m=4\\) and \\(n=-2\\)\r\n\t\t\\(e^{2}-5\\) if \\(e=2\\)\r\n\t\t\\(5-b-b^{2}\\) if \\(b=3\\)\r\n\t\t\\(2k^{2}+4\\) if \\(k=-6\\)\r\n\t\t\\(-3ab^{2}\\) if \\(a=4\\) and \\(b=2\\)\r\n\t\t\\(\\dfrac{n}{4}+2\\) if \\(n=10\\)\r\n\t\t\\(\\dfrac{u}{5v}\\) if \\(u=-20\\) and \\(v=2\\).\r\n\t\r\n\tEvaluate the following if \\(a=-1\\), \\(b=6\\), \\(c=3\\), \\(m=-2\\) and \\(n=2\\).\r\n\t\r\n\t\t\\(3a^{2}-7\\)\r\n\t\t\\(3a-b^{2}\\)\r\n\t\t\\((2m+1)^{2}\\)\r\n\t\t\\(3(a-b^{2})\\)\r\n\t\t\\(\\dfrac{2m}{n}\\)\r\n\t\t\\((m-n)^{2}\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nHi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on substitution in algebra. Replacing a pro-numeral or variable or letter in an expression or formula with a number is called substitution. Instance, here\u2019s an expression, three P, and we want to work this out, in other words evaluate, if P equals minus five. So substituting P equals minus five into the expression, three P is the same as three times P, which is three times minus five which is minus 15.\r\n\r\nHere are two more examples. Evaluate A plus five over B, if A equals minus nine and B equals two, here you notice we have two pro-numerals A and B, so substituting A and B into the formula or, sorry, into the expression, we have minus nine plus five over two which gives us minus four over two which is minus two.\r\n\r\nHere\u2019s another one. We want to evaluate W squared minus two Z, if W equals minus one and Z equals five, so substituting into the expression we have minus one all squared minus two times five which gives us one minus one times minus one is one minus 10 is minus 9.\r\n\r\nLet\u2019s look at a more complicated expression. Here we have the formula C equals five brackets F minus 32 close brackets over nine and that\u2019s the formula that converts a temperature in Fahrenheit into one in Centigrade, so here we want to convert a temperature of 100 degrees in Fahrenheit, so substituting F equals 100 into the formula we get 37.78 degrees Celsius, so again, as with the previous examples, we\u2019re just substituting the unknown, or the pro-numeral, which in this case is F, into the formula.\r\n\r\nNow try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n\r\n","n":0.042},[{"v":"Maths equations","i":1,"n":0.707},{"v":"Algebra","i":0,"n":1}]]},{"i":343,"$":[{"v":"Inverse of a 2\u00d72 matrix","n":0.447},{"v":"\r\nIn matrix algebra, we can add, subtract and multiply matrices as long as the matrix  order is correct. Unlike traditional arithmetic, we cannot divide matrices. Instead, we multiply by the inverse matrix. Inverse matrices have many applications, including computer animation, encryption and digital image transformations.\r\n\r\nInverse matrices\r\nAn inverse matrix is the square matrix of the same order that you multiply by to get an identity matrix.\r\nFor example, if \\(A\\) is a square matrix and \\(B\\) is another square matrix of the same order such that \\(AB=BA=I\\), then we call \\(B\\) the inverse of \\(A\\).\r\nThe inverse of \\(A\\) is denoted by the symbol \\(A^{-1}\\).\r\n\\[AA^{-1}=A^{-1}A=I\\]\r\nCompare this to traditional arithmetic, where an inverse is the reciprocal, or \\(1\\) divided by the number. In matrix algebra, \\(A^{-1}\\neq\\dfrac{1}{A}\\) is not defined.\r\nNot every square matrix has an inverse. If the determinant of a matrix equals \\(0\\), the inverse does not exist and the matrix is called singular. If the determinant is not \\(0\\), the inverse exists and we call the matrix non-singular or invertible.\r\n\r\nInverting \\(2\\times2\\) matrices\r\nIf \\(A\\) is a \\(2\\times2\\) matrix, then \\(A^{-1}\\) is also a \\(2\\times2\\) matrix such that: \r\n\\[AA^{-1} = A^{-1}A = \\left[ \\begin{array}{cc} 1 & 0\\\\\r\n0 & 1\r\n\\end{array} \\right] \\]\r\nThere is a simple formula to find the matrix of a \\(2\\times2\\) matrix.\r\n\r\nFor: \\[A = \\left[ \\begin{array}{cc} a & b\\\\\r\nc & d \r\n\\end{array} \\right] \\] the inverse matrix of \\(A\\), or \\(A^{-1}\\), is given by: \\[A^{-1} = \\frac{1}{ad-bc}\\left[ \\begin{array}{cc} d & -b\\\\ \r\n-c & a \r\n\\end{array} \\right] \\]\r\nYou may notice that \\(ad-bc\\) is the determinant of the matrix \\(A\\). We can rewrite the formula using \\(\\det A\\) or \\(\\left|A\\right|\\).\r\n\\[A^{-1} = \\frac{1}{\\det A}\\left[ \\begin{array}{cc} d & -b\\\\ \r\n-c & a\r\n\\end{array} \\right] \\]\r\nIf \\(\\det A=0\\), we have:\r\n\\[A^{-1} = \\frac{1}{0}\\left[ \\begin{array}{cc} d & -b\\\\ \r\n-c & a \r\n\\end{array} \\right] \\]\r\nBut, \\(\\dfrac{1}{0}\\) is undefined, so the inverse does not exist when \\(\\det A=0\\).\r\n\r\nExample 1 \u2013 finding the inverse of \\(2\\times2\\) matrices\r\nFind the inverse of the matrix \\(A = \\left[ \\begin{array}{cc} 2 & 3\\\\ \r\n2 & 4 \r\n\\end{array} \\right]\\).\r\nFirst, check if the inverse of \\(A\\) exists. To do this, calculate the determinant and determine whether \\(A\\) is singular, i.e. \\(\\det A=0\\).\r\n\\[\\begin{align*} \\det A & = (2\\times4)-(2\\times3)\\\\\r\n& = 8-6\\\\ \r\n& = 2 \r\n\\end{align*}\\]\r\n\\(\\det A\\neq0\\), so \\(A\\) is not singular and the inverse exists. We can use the formula to calculate the inverse. \\[\\begin{align*} A^{-1} & = \\frac{1}{\\det A}\\left[ \\begin{array}{cc} 4 & -3\\\\\r\n-2 & 2 \r\n\\end{array} \\right]\\\\\r\n& = \\frac{1}{2}\\left[ \\begin{array}{cc} 4 & -3\\\\\r\n-2 & 2 \r\n\\end{array} \\right]\\\\ \r\n& = \\left[\\begin{array}{cc} 2 & -\\frac{3}{2}\\\\\r\n-1 & 1 \r\n\\end{array} \\right] \r\n\\end{align*}\\] \r\nWe can check that \\(A^{-1}\\) satisfies \\(AA^{-1}=I\\) and \\(A^{-1}A=I\\).\r\n\\[\\begin{align*} AA^{-1} & = \\left[ \\begin{array}{cc} 2 & 3\\\\\r\n2 & 4 \r\n\\end{array} \\right] \r\n\\left[ \\begin{array}{cc} 2 & -\\frac{3}{2}\\\\ \r\n-1 & 1 \r\n\\end{array} \\right]\\\\\r\n& = \\left[\\begin{array}{cc} (2\\times2)+(3\\times(-1)) & (2\\times(-\\frac{3}{2}))+(3\\times1)\\\\\r\n(2\\times2)+(4\\times(-1)) & (2\\times(-\\frac{3}{2}))+(4\\times1) \r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} 1 & 0\\\\\r\n0 & 1 \r\n\\end{array} \\right]\r\n\\end{align*}\\] \\[\\begin{align*} A^{-1}A & = \\left[ \\begin{array}{cc} 2 & -\\frac{3}{2}\\\\\r\n-1 & 1 \r\n\\end{array} \\right] \r\n\\left[ \\begin{array}{cc} 2 & 3\\\\\r\n2 & 4 \r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} (2\\times2)+((-\\frac{3}{2})\\times2) & (2\\times3)+((-\\frac{3}{2})\\times4)\\\\\r\n(-1\\times2)+(1\\times2) & ((-1)\\times3)+(1\\times4)\r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} 1 & 0\\\\\r\n0 & 1\r\n\\end{array} \\right]\r\n\\end{align*}\\] \r\nTherefore, \\(A^{-1}=\\left[ \\begin{array}{cc} 2 & -\\frac{3}{2}\\\\ \r\n-1 & 1\r\n\\end{array} \\right]\\).\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the inverse of \\(2\\times2\\) matrices\r\n\r\n\tFind, if possible, the inverses of the following matrices:\r\n\t\r\n\t\t\\(\\left[ \\begin{array}{cc} 2 & -1\\\\\r\n\t\t-4 & 3\r\n\t\t\\end{array} \\right] \\)\r\n\t\t\\(\\left[ \\begin{array}{cc} 0 & 4\\\\\r\n\t\t2 & 5\r\n\t\t\\end{array} \\right] \\)\r\n\t\t\\(\\left[ \\begin{array}{cc} -2 & -3\\\\\r\n\t\t6 & 9\r\n\t\t\\end{array} \\right] \\)\r\n\t\t\\(\\left[ \\begin{array}{cc} -3 & 4\\\\\r\n\t\t3 & 1\r\n\t\t\\end{array} \\right] \\)\r\n\t\r\n\tFor \\(A=\\left[ \\begin{array}{cc} -1 & -3\\\\\r\n\t2 & 5\r\n\t\\end{array} \\right]\\) and \\(B=\\left[ \\begin{array}{cc} -2 & 0\\\\\r\n\t4 & 1\r\n\t\\end{array} \\right]\\):\r\n\t\r\n\t\tFind \\(A^{-1}\\) and \\(B^{-1}\\).\r\n\t\tShow that \\((AB)^{-1}=B^{-1}A^{-1}\\).\r\n\t\r\n\r\n\r\n\r\n","n":0.041},[{"v":"Matrices","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":344,"$":[{"v":"Augmented matrices and elementary row operations","n":0.408},{"v":"\r\nMatrices can be used to solve systems of equations using elementary row operations and the augmented matrix. This method of solving systems of equations is handy when you have many variables and equations. You may see it being used in fields like economics, statistics and machine learning.\r\n\r\nSystems of equations\r\nIn algebra, a system of equations is a set of linear equations where the solutions (or values for the variables) satisfy all equations simultaneously. Depending on the type of equations, there may be one solution, no solution, or infinitely many solutions.\r\nTo solve systems of equations using matrices, we must first rewrite them as an augmented matrix.\r\n\r\nAugmented matrices\r\nAn augmented matrix contains a coefficient matrix and a constant matrix next to it, separated by a vertical line.\r\nThe best way to understand this is with an example. Let's consider the following system of equations:\r\n\\[\\begin{align*} x+2y-z & = -3\\\\\r\n2x-3y+2z & = 13\\\\\r\n-x+5y-4z & = -19\r\n\\end{align*}\\]\r\nThis can be written in matrix form:\r\n\\[\\left[ \\begin{array}{ccc} 1 & 2 & -1\\\\\r\n2 & -3 & 2\\\\\r\n-1 & 5 & -4\r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{c} x \\\\\r\ny\\\\\r\nz\r\n\\end{array} \\right]\r\n=\r\n\\left[ \\begin{array}{c} -3\\\\\r\n13\\\\\r\n-19\r\n\\end{array} \\right] \\]\r\nThe corresponding augmented matrix for this system is:\r\n\\[\\begin{align*} \\left[ \\begin{array}{ccc} 1 & 2 & -1\\\\ \r\n2 & -3 & 2\\\\ \r\n-1 & 5 & -4 \r\n\\end{array} \r\n\\left| \r\n\\begin{array}{c} -3\\\\ \r\n13\\\\\r\n-19 \r\n\\end{array}\\right.\\right]\r\n\\end{align*}\\]\r\n\r\n\tThe first row is made up of the three coefficients in front of \\(x\\), \\(y\\) and \\(z\\) (\\(1,2,-1\\)) and the right-hand side \\(\\left(-3\\right)\\) from the first equation.\r\n\tThe second row is made up of the three coefficients in front of \\(x\\), \\(y\\) and \\(z\\) (\\(2,-3,2\\)) and the right-hand side \\(\\left(13\\right)\\) from the second equation.\r\n\tThe third row is made up of the three coefficients in front of \\(x\\), \\(y\\) and \\(z\\) (\\(-1,5,-4\\)) and the right-hand side \\(\\left(-19\\right)\\) from the third equation.\r\n\r\nJust as when you work with algebraic equations, when working with augmented matrices, everything we do to the left-hand side, you do to the right-hand side.\r\nFor augmented matrices, we manipulate the elements using elementary row operations.\r\n\r\nCoefficient matrix\r\nThe coefficient matrix tells us some helpful information about the number of solutions to a system of equations. Consider the previous example:\r\n\\[\\begin{align*} x+2y-z & = -3\\\\\r\n2x-3y+2z & = 13\\\\\r\n-x+5y-4z & = -19\r\n\\end{align*}\\]\r\nHere, the coefficient matrix is:\r\n\\[\\left[ \\begin{array}{ccc} 1 & 2 & -1\\\\\r\n2 & -3 & 2\\\\\r\n-1 & 5 & -4\r\n\\end{array} \\right] \\]\r\nIf the matrix is:\r\n\r\n\tnon-singular, i.e. the determinant \\(\\neq0\\), there will be a unique solution.\r\n\tsingular, i.e. the determinant is \\(0\\), there will be an infinite number of solutions or no solutions.\r\n\r\n\r\nElementary row operations\r\nElementary row operations are operations that can be used to simplify an augmented matrix when solving systems of equations. There are three operations that we are allowed to perform on augmented matrices:\r\n\r\n \tinterchange any two rows\r\n \tmultiply a row by any constant\r\n \tadd or subtract a multiple of one row to/from another row.\r\n\r\n\r\nSolving equations with unique solutions\r\nTo solve equations with unique solutions using augmented matrices and elementary row operations, we follow a method called row reduction:\r\n\r\n \tCreate the augmented matrix.\r\n \tUse row operations to reduce the augmented matrix to the following form (called the reduced augmented matrix), where the elements under the leading diagonal are \\(0\\): \\[\\begin{align*} \\left[ \\begin{array}{ccc} a & b & c\\\\ \r\n\t0 & e & f\\\\ \r\n\t0 & 0 & h \r\n\t\\end{array} \r\n\t\\left| \r\n\t\\begin{array}{c} d\\\\\r\n\tg\\\\ \r\n\tk \r\n\t\\end{array} \\right. \\right]\r\n\t\\end{align*}\\]\r\n \tFrom row \\(3\\), \\(hz=k\\) and so \\(z=\\dfrac{k}{h}\\).\r\n \tFrom row \\(2\\), we have \\(ey+fz=g\\). Since \\(z\\) is known (from Step 3), we can rearrange the equation to find \\(y\\).\r\n \tFrom row \\(1\\), we have \\(ax+by+cz=d\\). Since \\(x\\) and \\(y\\) are known from Steps 3 and 4, we can find \\(x\\).\r\n\r\nSteps 3 to 5 are called back substitution.\r\nTo understand how to use elementary row operations, it is best to look at some examples.\r\n\r\nExample 1 \u2013 solving equations with unique solutions\r\nSolve the following system of equations for \\(x\\), \\(y\\) and \\(z\\):\r\n\\[\\begin{align*} x+2y-z & = -3 & (1)\\\\ \r\n2x-3y+2z & = 13 & (2)\\\\\r\n-x+5y-4z & = -19 & (3)\r\n\\end{align*}\\]\r\n\r\nFirst, write the augmented matrix:\r\n\\[\\begin{align*} \\left[ \\begin{array}{ccc} 1 & 2 & -1\\\\ \r\n2 & -3 & 2\\\\ \r\n-1 & 5 & -4\r\n\\end{array} \r\n\\left|\r\n\\begin{array}{c} -3\\\\\r\n13\\\\\r\n-19 \r\n\\end{array} \\right. \\right]\r\n\\end{align*}\\]\r\nThen, use row operations to get the \\(0\\)'s under the leading diagonal. Remember that we can only do the following:\r\n\r\n\tinterchange any two rows\r\n \tmultiply a row by any constant\r\n \tadd or subtract a multiple of one row to/from another row.\r\n\r\nWe will work from left to right, and get the \\(0\\)'s in row \\(2\\) and row \\(3\\) of column \\(1\\) first.\r\nTo do this for row \\(2\\), we can multiply row \\(1\\) \\(\\left(R_{1}\\right)\\) by \\(2\\) and subtract it from row \\(2\\) \\(\\left(R_{2}\\right)\\) to get the new row \\(2\\) or \\(R_{2}'\\). This operation has been written to the left of the augmented matrix for your reference.\r\n\\[\\begin{align*} \\begin{array}{c} \\\\ \r\nR_{2}' = R_{2}-2R_{1}\\\\\r\n\\\\ \r\n\\end{array}\r\n\\left[ \\begin{array}{ccc} 1 & 2 & -1\\\\\r\n0 & -7 & 4\\\\\r\n-1 & 5 & -4 \r\n\\end{array} \r\n\\left|\r\n\\begin{array}{c} -3\\\\ \r\n19\\\\ \r\n-19 \r\n\\end{array} \\right. \\right]\r\n\\end{align*} \\]\r\nThen, to get a \\(0\\) in row \\(3\\), we can add rows \\(1\\) and \\(3\\) to get \\(R_{3}'\\).\r\n\\[\\begin{align*} \\begin{array}{c} \\\\ \r\n\\\\\r\nR_{3}'=R_{3}+R_{1} \r\n\\end{array}\r\n\\left[ \\begin{array}{ccc} 1 & 2 & -1\\\\ \r\n0 & -7 & 4\\\\ \r\n0 & 7 & -5 \r\n\\end{array}\r\n\\left|\r\n\\begin{array}{c} -3\\\\\r\n19\\\\ \r\n-22 \r\n\\end{array} \\right. \\right]\r\n\\end{align*}\\] \r\nNext, we can get a \\(0\\) in row \\(3\\) column \\(1\\) by adding rows \\(2\\) and \\(3\\):\r\n\\[\\begin{align*} \\begin{array}{c}\\\\\r\n\\\\ \r\nR_{3}'=R_{3}+R_{2} \r\n\\end{array} \r\n\\left[ \\begin{array}{ccc} 1 & 2 & -1\\\\\r\n0 & -7 & 4\\\\\r\n0 & 0 & -1 \r\n\\end{array}\r\n\\left|\r\n\\begin{array}{c} -3\\\\\r\n19\\\\\r\n-3 \r\n\\end{array} \\right. \\right]\r\n\\end{align*}\\]\r\nNow that the augmented matrix is in the correct form, we can start the back substitution.\r\nStarting from row \\(3\\), we have: \\[\\begin{align*} -1z & = -3\\\\\r\nz & = \\frac{-3}{-1}\\\\\r\n& = 3\r\n\\end{align*}\\]\r\nFrom row \\(3\\), we have \\(-7y+4z=19\\) and we can substitute in \\(z=3\\) to find \\(y\\):\r\n\\[\\begin{align*} -7y+4(3) & = 19\\\\\r\n-7y+12 & = 19\\\\\r\n-7y & = 19-12\\\\\r\n-7y & = 7\\\\\r\ny & = \\frac{7}{-7}\\\\\r\n& = -1\r\n\\end{align*}\\]\r\nFrom row \\(1\\), we have \\(x+2y-z=-3\\) and we can substitute in \\(z=3\\) and \\(y=-1\\) to find \\(x\\): \\[\\begin{align*} x+2(-1)-(3) & = -3\\\\\r\nx-2-3 & = -3\\\\\r\nx-5 & = -3\\\\\r\nx & = -3+5\\\\\r\n& = 2\r\n\\end{align*}\\]\r\nThe solution to the system of equations is therefore, \\(x=2\\), \\(y=-1\\) and \\(z=3\\).\r\nYou can check your solution by substituting the values for \\(x\\), \\(y\\) and \\(z\\) into the original equations and making sure it is equal to the right-hand side.\r\nFor equation (\\(1\\)): \\[\\begin{align*} x+2y-z & = 2+2(-1)-3\\\\\r\n& = 2-2-3\\\\\r\n& = -3\\\\\r\n& = \\textrm{RHS}\r\n\\end{align*}\\] For equation (\\(2\\)): \\[\\begin{align*} 2x-3y+2z & = 2(2)-3(-1)+2(3)\\\\\r\n& = 4+3+6\\\\\r\n& = 13\\\\\r\n& = \\textrm{RHS}\r\n\\end{align*}\\] For equation (\\(3\\)): \\[\\begin{align*} -x+5y-4z & = -2+5(-1)-4(3)\\\\\r\n& = -2-5-12\\\\\r\n& = -19\\\\\r\n& = \\textrm{RHS}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 solving systems of equations using augmented matrices and elementary row operations\r\nFind the solutions to the following systems of equations.\r\n\r\n\t\\(x-y-2z = -6\\)\r\n\t\\(2x-3y+3z = 1\\)\r\n\t\\(3x-y+4z = 4\\)\r\n\t\\(2x+y+2z = 4\\)\r\n\t\\(2x+4y-2z = 6\\)\r\n\t\\(-y+3z = 1\\)\r\n\t\\(x_{1}-x_{2}-2x_{3} = 1\\)\r\n\t\\(-3x_{1}+2x_{2}+4x_{3} = -3\\)\r\n\t\\(2x_{1}+x_{2}+x_{3} = 1\\)\r\n\t\\(x+2y-z=5\\)\r\n\t\\(3x-y+2z=1\\)\r\n\t\\(2x-y+3z=0\\)\r\n\r\n\r\n\r\n\r\n\r\nSolving equations with an infinite number of solutions\r\nIn all the examples above, we had a unique solution. That is, \\(x\\), \\(y\\) and \\(z\\) were associated with a single number. However, a set of simultaneous equations can have an infinite number of solutions.\r\nImagine after using row operations, you get an augmented matrix like:\r\n\\[\\begin{align*} \\left[ \\begin{array}{ccc} a & b & c\\\\ \r\n0 & e & f\\\\\r\n0 & 0 & 0 \r\n\\end{array}\r\n\\left|\r\n\\begin{array}{c} d\\\\\r\ng\\\\\r\n0 \r\n\\end{array} \\right. \\right]\r\n\\end{align*}\\]\r\nIn this case, row \\(3\\) tells us nothing. We essentially have 2 equations and 3 unknowns. Therefore, there are infinitely many solutions, and the variables \\(x\\), \\(y\\) and \\(z\\) will be expressed in terms of a parameter.\r\nWhat this means is we define one variable as a set of values. We usually say something like, let \\(x=t\\), where \\(t\\in\\mathbb{R}\\). In other words, the variable \\(x\\) can be any real number, \\(\\mathbb{R}\\). This will give us an infinite number of solutions.\r\nIn general, if the reduced augmented matrix has one or more rows of \\(0\\)'s, there will be an infinite number of solutions to the system of equations.\r\n\r\nExample 1 \u2013 solving equations with infinite solutions\r\nSolve the following set of equations:\r\n\\[\\begin{align*} x+y+z & = 3\\\\\r\n2x-y+z & = 5\\\\\r\n3x+2z & = 8\r\n\\end{align*}\\]\r\nThe augmented matrix is:\r\n\\[\\begin{align*} \\left[ \\begin{array}{ccc} 1 & 1 & 1\\\\\r\n2 & -1 & 1\\\\\r\n3 & 0 & 2\r\n\\end{array}\r\n\\left|\r\n\\begin{array}{c} 3\\\\\r\n5\\\\\r\n8 \r\n\\end{array} \\right. \\right]\r\n\\end{align*}\\] \r\nPerforming row operations, we get:\r\n\\[\\begin{align*} \\begin{array}{c} \\\\\r\nR_{2}'=R_{2}-2R_{1}\\\\\r\n\\\\ \r\n\\end{array} \r\n\\left[ \\begin{array}{ccc} 1 & 1 & 1\\\\ \r\n0 & -3 & -1\\\\\r\n3 & 0 & 2 \r\n\\end{array}\r\n\\left|\r\n\\begin{array}{c} 3\\\\\r\n-1\\\\ \r\n8 \r\n\\end{array} \\right. \\right]\r\n\\end{align*}\\] \\[\\begin{align*} \\begin{array}{c} \\\\\r\n\\\\ \r\nR_{3}'=R_{3}-3R_{1}\r\n\\end{array}\r\n\\left[ \\begin{array}{ccc} 1 & 1 & 1\\\\\r\n0 & -3 & -1\\\\ \r\n0 & -3 & -1 \r\n\\end{array}\r\n\\left|\r\n\\begin{array}{c} 3\\\\ \r\n-1\\\\\r\n-1 \r\n\\end{array} \\right. \\right]\r\n\\end{align*}\\] \\[\\begin{align*} \\begin{array}{c} \\\\\r\n\\\\ \r\nR_{3}'=R_{3}-R_{2}\r\n\\end{array}\r\n\\left[ \\begin{array}{ccc} 1 & 1 & 1\\\\\r\n0 & -3 & -1\\\\\r\n0 & 0 & 0\r\n\\end{array}\r\n\\left|\r\n\\begin{array}{c} 3\\\\\r\n-1\\\\ \r\n0 \r\n\\end{array} \\right. \\right]\r\n\\end{align*}\\] \r\nNow, we use back substitution. Row \\(3\\) tells us nothing.\r\nFrom row \\(2\\), we have:\r\n\\[\\begin{align*} -3y-z & = -1\\\\ \r\n3y+z & = 1\r\n\\end{align*}\\]\r\nThis is an equation with two unknowns. In this case, we let one of the variables, either \\(y\\) or \\(z\\) be free\u2014that is, we let \\(y\\) or \\(z\\) be equal to some parameter \\(t\\in\\mathbb{R}\\). It doesn\u2019t matter which variable you choose to be free, though the answer will look different. We will set \\(z=t\\). Then, we have:\r\n\\[\\begin{align*} 3y+z & = 1\\\\ \r\n3y+t & = 1\\\\ \r\n3y & = 1-t\\\\\r\ny & = \\frac{1-t}{3}\r\n\\end{align*}\\] \r\nFrom row \\(1\\) and substituting \\(z=t\\) and \\(y=\\dfrac{1-t}{3}\\), we have:\r\n\\[\\begin{align*} x+y+z & = 3\\\\\r\nx+\\frac{1-t}{3}+t & = 3\\\\\r\n3x+1-t+3t & = 9\\\\ \r\n3x & = 8-2t\\\\ \r\nx & = \\frac{8-2t}{3}\r\n\\end{align*}\\] \r\nSo, the solution here is \\(x=\\frac{8-2t}{3}\\), \\(y=\\frac{1-t}{3}\\) and \\(z=t\\) where \\(t\\in\\mathbb{R}\\). This gives us an infinite number of solutions because \\(t\\) can be any real number.\r\nTo check the solution, you can pick any value of \\(t\\), determine \\(x\\), \\(y\\) and \\(z\\) and substitute in the original equations. For example, let \\(t=0\\).\r\nSubstituting in equation \\(1\\) gives:\r\n\\[\\begin{align*} x+y+z & = \\frac{8}{3}+\\frac{1}{3}+0\\\\\r\n& = \\frac{9}{3}\\\\ \r\n& = 3\\\\\r\n& = \\textrm{RHS}\r\n\\end{align*}\\]\r\nYou can substitute into the second and third equation to check this solution is correct.\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 solving equations with infinite solutions\r\nFind the solutions to the following system of equations.\r\n\r\n\t\\(x-2y+z-w = 0\\)\r\n\t\t\\(2x+4y-3z = 0\\)\r\n\t\t\\(3x+2y+2z-w = 0\\)\r\n\t\\(2x-y+3z=-3\\)\r\n\t\t\\(x+2y+z=-4\\)\r\n\t\t\\(4x+3y+5z=-11\\)\r\n\t\\(x+2y-z=3\\)\r\n\t\t\\(-x+y-2z=-3\\)\r\n\t\t\\(3x+3z=9\\)\r\n\r\n\r\n\r\n\r\nSolving equations with no solutions\r\nIn some cases, the coefficients in a row are all \\(0\\)'s, but the constant on the other side of the vertical line is a number. This is contradictory because \\(0x+0y+0z\\) cannot give us a non-zero number; it must equal \\(0\\).\r\nGraphically, what this means is that the lines for the equations do not intercept or touch, and so there is no solution for \\(x\\), \\(y\\) and \\(z\\) that satisfies the system of equations. The system of equations is therefore considered inconsistent.\r\n\r\nExample \u2013 solving equations with no solutions\r\nSolve the following system of equations:\r\n\\[\\begin{align*} x+2y-2z & = 3\\\\\r\n2x+y+z & = 4\\\\\r\n3y-5z & = -1\r\n\\end{align*}\\]\r\nThe augmented matrix for this system is:\r\n\\[\\left[ \\begin{array}{ccc} 1 & 2 & -2\\\\\r\n2 & 1 & 1\\\\\r\n0 & 3 & -5\r\n\\end{array}\r\n\\left|\r\n\\begin{array}{c} 3\\\\\r\n4\\\\\r\n-1\r\n\\end{array} \\right. \\right] \\]\r\nUsing elementary row operations, we get:\r\n\\[\\begin{array}{c} \\\\\r\nR_{2}'=R_{2}-2R_{1}\\\\\r\n\\end{array}\r\n\\left[ \\begin{array}{ccc} 1 & 2 & -2\\\\\r\n0 & -3 & 5\\\\\r\n0 & 3 & -5\r\n\\end{array}\r\n\\left|\r\n\\begin{array}{c} 3\\\\\r\n-2\\\\\r\n-1\r\n\\end{array} \\right. \\right] \\] \\[\\begin{array}{c} \\\\\r\n\\\\ \r\nR_{3}'=R_{3}+R_{2}\r\n\\end{array}\r\n\\left[ \\begin{array}{ccc} 1 & 2 & -2\\\\\r\n0 & -3 & 5\\\\\r\n0 & 0 & 0 \r\n\\end{array}\r\n\\left|\r\n\\begin{array}{c}\r\n3\\\\\r\n-1\\\\\r\n-3\r\n\\end{array} \\right. \\right] \\]\r\nIn this case, row \\(3\\) says that \\(0x+0y+0z=-3\\), which is not possible. This means there are no solutions for the system of equations, i.e. it is inconsistent.\r\n\r\nExercise \u2013 solving equations with no solutions\r\nDetermine whether the following system of equations has a solution.\r\n\\[\\begin{align*} x+2y-2z & = 7\\\\\r\n2x-y-z & = -3\\\\\r\n-x-2y+2z & = 1\r\n\\end{align*}\\]\r\n","n":0.023},[{"v":"Matrices","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":345,"$":[{"v":"Determinant of a matrix","n":0.5},{"v":"\r\nA determinant is a number that can be calculated for any square matrix. The determinant is used in calculating vector cross products, eigenvalues, eigenvectors and solving simultaneous equations. Use this resource to learn how to find the determinant of \\(2\\times2\\) and \\(3\\times3\\) matrices.\r\n\r\nDeterminant of a \\(2\\times2\\) matrix\r\nThe determinant of a \\(2\\times2\\) matrix is called a second order determinant.\r\nFor: \\[A = \\left[ \\begin{array}{cc} a & b\\\\\r\nc & d \r\n\\end{array} \\right]\\] the determinant is denoted by \\(\\det A\\) or \\(\\left|A\\right|\\), where: \r\n\\[\\begin{align*} \\left| A\\right| & = \\left| \\begin{array}{cc} a & b\\\\ \r\nc & d \r\n\\end{array} \\right|\\\\ \r\n& = ad-bc\r\n\\end{align*}\\]\r\nA matrix that has a determinant of \\(0\\) is called a singular matrix. Singular matrices do not have an inverse; you will learn about inverses later. If the determinant is not \\(0\\), the matrix is called non-singular.\r\n\r\nExample 1 \u2013 finding the determinant of \\(2\\times2\\) matrices\r\n\\[\\begin{align*} \\left| \\begin{array}{cc} 2 & 3\\\\\r\n1 & 5\r\n\\end{array} \\right| \r\n& = \\left( 2\\times5 \\right)-\\left( 1\\times3 \\right)\\\\\r\n& = 10-3\\\\\r\n& = 7\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the determinant of \\(2\\times2\\) matrices\r\nEvaluate the following determinants.\r\n\r\n\t\\(\\left| \\begin{array}{cc} 2 & 5\\\\\r\n\t1 & 2\r\n\t\\end{array} \\right| \\)\r\n\t\\(\\left| \\begin{array}{cc} -1 & 1\\\\\r\n\t-2 & 5\r\n\t\\end{array} \\right| \\)\r\n\t\\(\\left| \\begin{array}{cc} 2 & -3\\\\\r\n\t-2 & 6\r\n\t\\end{array} \\right| \\)\r\n\t\\(\\left| \\begin{array}{cc} 0 & -3\\\\\r\n\t1 & 0 \r\n\t\\end{array} \\right| \\)\r\n\t\\(\\left| \\begin{array}{cc} 1 & -3\\\\\r\n\t-3 & 9\r\n\t\\end{array} \\right| \\)\r\n\r\n\r\n\r\n\r\nDeterminant of a \\(3\\times3\\) matrix\r\nThe determinant of a \\(3\\times3\\) matrix is called a third order determinant.\r\n\r\nFor: \\[A = \\left[ \\begin{array}{ccc} a & b & c\\\\\r\nd & e & f\\\\\r\ng & h & i \r\n\\end{array} \\right] \\] the determinant is denoted by \\(\\det A\\) or \\(\\left| A \\right|\\), where: \\[\\left| A \\right| = a\\left| \\begin{array}{cc} e & f\\\\\r\nh & i\r\n\\end{array} \\right| \r\n-b\\left| \\begin{array}{cc} d & f\\\\\r\ng & i\r\n\\end{array} \\right|\r\n+c\\left| \\begin{array}{cc} d & e\\\\\r\ng & h\r\n\\end{array} \\right| \\]\r\n\r\n\r\n\tTo get the first term, we multiply \\(a\\) by the determinant of the \\(2\\times2\\) matrix formed when we remove everything in the same column and row as \\(a\\)\u2014that is, \\(b\\), \\(c\\), \\(d\\) and \\(g\\).\r\n\tTo get the second term, we multiply \\(b\\) by the determinant of the \\(2\\times2\\) matrix formed when we remove everything in the same column and row as \\(b\\)\u2014that is, \\(a\\), \\(c\\), \\(e\\) and \\(h\\).\r\n\tTo get the final term, we multiply \\(c\\) by the determinant of the \\(2\\times2\\) matrix formed when we remove everything in the same column and row as \\(c\\)\u2014that is, \\(a\\), \\(b\\), \\(f\\) and \\(i\\).\r\n\r\nThis is just one way of expanding out the matrices. You don't have to choose the first row elements \\(a\\), \\(b\\) and \\(c\\) like we have here. You could choose to start with the second row: \\(d\\), \\(e\\) and \\(f\\), or even the first column: \\(a\\), \\(d\\) and \\(g\\). But to get the maths right, you need to consider place signs.\r\nPlace signs\r\nPlace signs tell you whether you need to add or subtract the expanded values to find the determinant.\r\nEach element in a matrix has a place sign. They alternate in the following pattern.\r\n\\[ \\left|\\begin{array}{ccc} + & - & +\\\\ - & + & -\\\\ + & - & + \\end{array}\\right| \\]\r\nWe apply these signs to the elements. Using the same example as before, but expanding \\(d\\), \\(e\\) and \\(f\\) instead, we get the following:\r\n\\[\\left| A \\right| = \\boldsymbol{-}d\\left| \\begin{array}{cc} b & c\\\\ h & i \\end{array} \\right| \\boldsymbol{+}e\\left| \\begin{array}{cc} a & c\\\\ g & i \\end{array} \\right| \\boldsymbol{-}f\\left| \\begin{array}{cc} a & b\\\\ g & h \\end{array} \\right| \\]\r\nSee how the signs match up with where the element is in the matrix.\r\n\r\nExample 1 \u2013 finding the determinant of \\(3\\times3\\) matrices\r\nCalculate the determinant of the following matrix.\r\n\\[ A = \\left[ \\begin{array}{ccc} 2 & 1 & 3\\\\\r\n1 & -1 & 2\\\\\r\n2 & 1 & 4 \r\n\\end{array} \\right]\\]\r\nLet's look at two ways to calculate the determinant. For the first method, let's expand along row \\(1\\).\r\n\r\nNow, let's try expanding along column \\(2\\).\r\n\r\nAs you can see, no matter which column or row you choose to expand, you can achieve the same correct answer as long as you use the correct place signs.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the determinant of \\(3\\times3\\) matrices\r\nEvaluate the following determinants.\r\n\r\n\t\\(\\left| \\begin{array}{ccc} -1 & 1 & 3\\\\ \r\n\t2 & 3 & 1\\\\\r\n\t-1 & 1 & 5 \r\n\t\\end{array} \\right|\\)\r\n\t\\(\\left| \\begin{array}{ccc} 3 & -1 & -3\\\\\r\n\t1 & 4 & 1\\\\\r\n\t3 & 1 & 1 \r\n\t\\end{array} \\right|\\)\r\n\t\\(\\left| \\begin{array}{ccc} -1 & 5 & 2\\\\\r\n\t3 & -5 & -6\\\\ \r\n\t-2 & 3 & 4 \r\n\t\\end{array} \\right|\\)\r\n\t\\(\\left| \\begin{array}{ccc} -2 & 3 & -3\\\\\r\n\t6 & 0 & 2\\\\ \r\n\t2 & 0 & 4 \r\n\t\\end{array} \\right|\\)\r\n\t\\(\\left| \\begin{array}{ccc} a & 0 & 0\\\\ \r\n\t0 & b & 0\\\\ \r\n\t0 & 0 & c \r\n\t\\end{array} \\right|\\)\r\n\r\n\r\n","n":0.037},[{"v":"Matrices","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":346,"$":[{"v":"Matrix multiplication","n":0.707},{"v":"\r\nA strong understanding of how to multiply matrices is critical in many fields, like computer science, robotics, statistics and economics. Just as there are rules for adding and subtracting matrices, we can multiply matrices with each other when certain requirements are met. Use this resource to learn how to multiply matrices.\r\n\r\nBefore you begin, make sure you understand the basics of matrices, including how to identify the shape or order of a matrix.\r\n\r\n\r\n\r\nTwo matrices can be multiplied if the number of columns of the first matrix is the same as the number of rows of the second matrix.\r\nMatrix multiplication results in a new matrix with the number of rows from the first matrix and the number of columns from the second matrix. This operation is only defined if the number of columns in the first matrix equals the number of rows in the second matrix.\r\nFor instance, if matrix \\(A\\) is \\(m\\times p\\) and matrix \\(B\\) is \\(q\\times n\\), then \\(A\\times B=AB\\) is possible if \\(p=q\\). If \\(p\\neq q\\), the product is undefined or does not exist. For example, a \\(2\\times3\\) matrix can't be multiplied by a \\(1\\times4\\) matrix, but can be multiplied by a \\(3\\times2\\) matrix to produce a \\(2\\times2\\) matrix.\r\nThe process of multiply matrices is a bit confusing. We can summarise it using the figure, but let's look at some examples.\r\n\r\n\r\nExample 1 \u2013 multiplying matrices\r\nConsider matrices \\(A\\) and \\(B\\).\r\n\\[A = \\left[ \\begin{array}{ccc} 2 & 4 & 5\\\\\r\n7 & 8 & 9 \\\r\n\\end{array} \\right]\\quad\r\nB = \\left[ \\begin{array}{cc} 1 & 6\\\\ \r\n3 & 1\\\\ \r\n4 & 2 \r\n\\end{array} \\right] \\] Find the product \\(AB\\), if it exists.\r\nFirst, identify the orders of the matrices to determine whether multiplication is possible. \\(A\\) is a \\(2\\times3\\) matrix and \\(B\\) is a \\(3\\times2\\) matrix. Matrix multiplication is possible since \\(A\\) has \\(3\\) columns and \\(B\\) has \\(3\\) rows. The result will be a matrix with the same number of rows as \\(A\\) and columns as \\(B\\), so \\(2\\times2\\).\r\n\r\nWe'll work through the multiplication one element at a time. First, we find the first element in row \\(1\\) of the final matrix \\(AB\\). To do this, we multiply each element in row \\(1\\) of \\(A\\) by the corresponding element in column \\(1\\) of \\(B\\), then add them altogether.\r\n\\[\\begin{align*} AB & = \\left[ \\begin{array}{ccc} 2 & 4 & 5\\\\ \r\n7 & 8 & 9 \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{cc} 1 & 6\\\\\r\n3 & 1\\\\\r\n4 & 2 \r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} (2\\times1)+(4\\times3)+(5\\times4) & \\textrm{element 2 of row 1}\\\\\r\n\\textrm{element 1 of row 2} & \\textrm{element 2 of row 2}\r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} 2+12+20 & \\textrm{element 2 of row 1}\\\\\r\n\\textrm{element 1 of row 2} & \\textrm{element 2 of row 2}\r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} 34 & \\textrm{element 2 of row 1}\\\\ \r\n\\textrm{element 1 of row 2} & \\textrm{element 2 of row 2}\r\n\\end{array} \\right] \r\n\\end{align*}\\]\r\n\r\nTo find the second element in row \\(1\\) of matrix \\(AB\\), we multiply each element in row \\(1\\) of \\(A\\) by the corresponding element in column \\(2\\) of \\(B\\), then add them altogether.\r\n\\[\\begin{align*} AB & = \\left[ \\begin{array}{ccc} 2 & 4 & 5\\\\ \r\n7 & 8 & 9 \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{cc} 1 & 6\\\\ \r\n3 & 1\\\\ \r\n4 & 2 \r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} 34 & (2\\times6)+(4\\times1)+(5\\times2)\\\\ \r\n\\textrm{element 1 of row 2} & \\textrm{element 2 of row 2}\r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} 34 & 12+4+10\\\\ \r\n\\textrm{element 1 of row 2} & \\textrm{element 2 of row 2}\r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} 34 & 26\\\\ \r\n\\textrm{element 1 of row 2} & \\textrm{element 2 of row 2}\r\n\\end{array} \\right]\r\n\\end{align*}\\]\r\n\r\nTo find the first element in row \\(2\\), we multiply each element in row \\(2\\) of \\(A\\) by the corresponding element in column \\(1\\) of \\(B\\), then add them together.\r\n\\[\\begin{align*} AB & = \\left[ \\begin{array}{ccc} 2 & 4 & 5\\\\ \r\n7 & 8 & 9 \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{cc} 1 & 6\\\\\r\n3 & 1\\\\\r\n4 & 2 \r\n\\end{array}\\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} 34 & 26\\\\\r\n(7\\times1)+(8\\times3)+(9\\times4) & \\textrm{element 2 of row 2} \r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} 34 & 26\\\\ \r\n7+24+36 & \\textrm{element 2 of row 2}\r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} 34 & 26\\\\\r\n67 & \\textrm{element 2 of row 2}\r\n\\end{array} \\right]\r\n\\end{align*}\\]\r\n\r\nFinally, to find the second element in row \\(2\\), we multiply each element in row \\(2\\) of \\(A\\) by the corresponding element in column \\(2\\) of \\(B\\), then add them together.\r\n\\[\\begin{align*} AB & = \\left[ \\begin{array}{ccc} 2 & 4 & 5\\\\\r\n7 & 8 & 9 \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{cc} 1 & 6\\\\\r\n3 & 1\\\\ \r\n4 & 2 \r\n\\end{array}\\right]\\\\\r\n& = \\left[ \\begin{array}{cc} 34 & 26\\\\\r\n67 & (7\\times6)+(8\\times1)+(9\\times2)\r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} 34 & 26\\\\ \r\n7+24+36 & 42+8+18 \r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} 34 & 26\\\\\r\n67 & 68\r\n\\end{array} \\right]\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 multiplying matrices\r\nConsider the following matrices.\r\n\\[\\begin{align*} A & =\\left[ \\begin{array}{cc} 4 & -1\\\\\r\n1 & 3 \r\n\\end{array} \\right]\\quad\r\nB = \\left[ \\begin{array}{cc} 1 & 3\\\\ \r\n-2 & -4\\\\\r\n2 & 0 \r\n\\end{array} \\right]\\quad\r\nC = \\left[ \\begin{array}{ccc} 3 & 1 & -5\\\\\r\n2 & 0 & 4\\\\\r\n1 & 6 & 3 \r\n\\end{array} \\right]\\\\\r\nD & = \\left[ \\begin{array}{c} 3\\\\ \r\n2\\\\ \r\n-2 \r\n\\end{array} \\right]\\quad\r\nE = \\left[ \\begin{array}{ccc} -3 & 2 & 0\\\\ \r\n0 & 5 & -1 \r\n\\end{array} \\right]\r\n\\end{align*}\\] \tFind, if they exist:\r\n\r\n\t\t\\(BA\\)\r\n\t\t\\(DB\\)\r\n\t\t\\(CD\\)\r\n\t\t\\(AA=A^{2}\\)\r\n\t\t\\(BB=B^{2}\\)\r\n\t\t\\(AD\\)\r\n\t\t\\(BE\\)\r\n\t\t\\(EA\\)\r\n\t\t\\(CB\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\nMultiplying by the unit matrix\r\nRemember that a square matrix with \\(1\\)'s along its leading diagonal and zeros elsewhere is called a unit or identity matrix. The order of the unit matrix is sometimes noted with a subscript. For example:\r\n\\[\\begin{align*} I_{1} & = \\left[1\\right]\\quad \r\nI_{2} = \\left[ \\begin{array}{cc} 1 & 0\\\\ \r\n0 & 1 \r\n\\end{array} \\right]\\quad\r\nI_{3} = \\left[ \\begin{array}{ccc} 1 & 0 & 0\\\\\r\n0 & 1 & 0\\\\\r\n0 & 0 & 1 \r\n\\end{array} \\right]\\\\\r\nI_{4} & = \\left[ \\begin{array}{cccc} 1 & 0 & 0 & 0\\\\ \r\n0 & 1 & 0 & 0\\\\ \r\n0 & 0 & 1 & 0\\\\ \r\n0 & 0 & 0 & 1 \r\n\\end{array} \\right]\\quad\r\nI_{n} = \\left[ \\begin{array}{ccccc} 1 & 0 & 0 & \\cdots & 0\\\\ \r\n0 & 1 & 0 & \\cdots & 0\\\\ \r\n0 & 0 & 1 & \\cdots & 0\\\\\r\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ \r\n0 & 0 & 0 & \\cdots & 1 \r\n\\end{array} \\right]\r\n\\end{align*}\\]\r\nThe unit matrix acts like the number \\(1\\) in the real number system. When you multiply by the unit matrix, the original matrix remains unchanged.\r\nWhen you multiply a square matrix \\(n\\times n\\) by a unit matrix \\(I_{n}\\), the order of multiplication does not matter. That is:\r\n\\[AI_{n}=I_{n}A\\]\r\n\r\nExample \u2013 multiplying by the unit matrix\r\nConsider matrix \\(A\\).\r\n\\[A = \\left[ \\begin{array}{cc} 1 & 2\\\\\r\n3 & 4\r\n\\end{array} \\right]\\] Show that \\(AI\\) is the same as \\(IA\\).\r\nSince \\(A\\) has two columns, we need the unit matrix to have two rows and two columns. Therefore, \\(I=\\left[\\begin{array}{cc} 1 & 0\\\\ 0 & 1 \\end{array}\\right]\\).\r\n\\[\\begin{align*} AI & = \\left[ \\begin{array}{cc} 1 & 2\\\\ \r\n3 & 4 \r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{cc} 1 & 0\\\\\r\n0 & 1 \r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} (1\\times1)+(2\\times0) & (1\\times0)+(2\\times1)\\\\ \r\n(3\\times1)+(3\\times0) & (3\\times0)+(4\\times1) \r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} 1 & 2\\\\\r\n3 & 4 \r\n\\end{array} \\right]\\\\\r\n& = A \\end{align*}\\] \\[ \\begin{align*} IA & = \\left[ \\begin{array}{cc} 1 & 0\\\\\r\n0 & 1\r\n\\end{array} \\right]\r\n\\left[ \\begin{array}{cc} 1 & 2\\\\\r\n3 & 4 \r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} (1\\times1)+(0\\times3) & (1\\times2)+(0\\times4)\\\\ \r\n(0\\times1)+(1\\times3) & (0\\times2)+(1\\times4)\r\n\\end{array} \\right]\\\\\r\n& =\\left[ \\begin{array}{cc} 1 & 2\\\\\r\n3 & 4\r\n\\end{array} \\right]\\\\\r\n& = A\r\n\\end{align*}\\] So, \\(AI=IA=A\\).\r\n\r\n","n":0.029},[{"v":"Matrices","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":347,"$":[{"v":"Addition and subtraction of matrices","n":0.447},{"v":"\r\nThe addition and subtraction of matrices is important across many applications, from computer graphics and image processing to statistics. Use this resource to learn the rules for adding and subtracting matrices.\r\n\r\nAddition of matrices\r\nThere is one key rule for adding matrices.\r\nMatrices of the same shape (same number of rows and columns) may be added by adding the corresponding elements.\r\n\r\nLet's consider the three matrices, \\(A\\), \\(B\\) and \\(C\\).\r\n\\[ A = \\left[\\begin{array}{ccc} 3 & 2 & -1\\\\\r\n1 & 5 & -4 \r\n\\end{array}\\right]\\quad\r\nB = \\left[\\begin{array}{ccc} 1 & 0 & 2\\\\\r\n3 & 1 & -2\r\n\\end{array}\\right]\\quad\r\nC = \\left[\\begin{array}{cc} 2 & -3\\\\ \r\n4 & 3 \r\n\\end{array}\\right] \\]\r\nNotice that \\(A\\) and \\(B\\) are the same shape \\(\\left(2\\times3\\right)\\), whereas \\(C\\) is \\(\\left(2\\times2\\right)\\).\r\nWe can add \\(A\\) and \\(B\\) together.\r\n\\[\\begin{align*} A+B & = \\left[\\begin{array}{ccc} 3 & 2 & -1\\\\\r\n1 & 5 & -4 \r\n\\end{array}\\right]\r\n+\\left[\\begin{array}{ccc} 1 & 0 & 2\\\\\r\n3 & 1 & -2 \\end{array}\\right]\\\\\r\n& = \\left[\\begin{array}{ccc} 3+1 & 2+0 & -1+2\\\\ \r\n1+3 & 5+1 & -4+(-2) \r\n\\end{array}\\right]\\\\ \r\n& = \\left[\\begin{array}{ccc} 4 & 2 & 1\\\\\r\n4 & 6 & -6 \r\n\\end{array}\\right] \r\n\\end{align*}\\]\r\n\\(A+B\\) is the same as \\(B+A\\). On the other hand, \\(A+C\\) and \\(B+C\\) are not defined as these pairs of matrices have different shapes.\r\n\r\nSubtraction of matrices\r\nThe rule for subtracting matrices is similar to adding them.\r\nOne matrix may be subtracted from another of the same shape by subtracting the corresponding elements.\r\n\r\nLet's consider the same matrices \\(A\\), \\(B\\) and \\(C\\).\r\n\\[ A = \\left[\\begin{array}{ccc} 3 & 2 & -1\\\\\r\n1 & 5 & -4 \r\n\\end{array}\\right]\\quad\r\nB = \\left[\\begin{array}{ccc} 1 & 0 & 2\\\\\r\n3 & 1 & -2\r\n\\end{array}\\right]\\quad\r\nC = \\left[\\begin{array}{cc} 2 & -3\\\\ \r\n4 & 3 \r\n\\end{array}\\right] \\]\r\nWe can find \\(A-B\\) as they are both \\(\\left(2\\times3\\right)\\) matrices.\r\n\\[\\begin{align*} A-B & = \\left[\\begin{array}{ccc} 3 & 2 & -1\\\\ \r\n1 & 5 & -4 \r\n\\end{array}\\right]\r\n-\\left[\\begin{array}{ccc} 1 & 0 & 2\\\\ \r\n3 & 1 & -2 \r\n\\end{array}\\right]\\\\\r\n& = \\left[\\begin{array}{ccc} 3-1 & 2-0 & -1-2\\\\\r\n1-3 & 5-1 & -4-(-2) \r\n\\end{array}\\right]\\\\ \r\n& = \\left[\\begin{array}{ccc} 2 & 2 & -3\\\\\r\n-2 & 4 & -2\r\n\\end{array}\\right]\r\n\\end{align*}\\]\r\n\\(A-C\\) and \\(B-C\\) are not defined as \\(A\\) and \\(B\\) are \\(\\left(2\\times3\\right)\\) and \\(C\\) is \\(\\left(2\\times2\\right)\\). Remember that you cannot subtract matrices that have different shapes.\r\nWe can also calculate \\(B-A\\).\r\n\\[\\begin{align*} B-A & = \\left[\\begin{array}{ccc} 1 & 0 & 2\\\\ \r\n3 & 1 & -2 \r\n\\end{array}\\right]\r\n-\\left[\\begin{array}{ccc} 3 & 2 & -1\\\\ \r\n1 & 5 & -4 \r\n\\end{array}\\right]\\\\ \r\n& = \\left[\\begin{array}{ccc} 1-3 & 0-2 & 2-(-1)\\\\ \r\n3-1 & 1-5 & -2-(-4)\r\n\\end{array}\\right]\\\\\r\n& = \\left[\\begin{array}{ccc} -2 & -2 & 3\\\\ \r\n2 & -4 & 2 \r\n\\end{array}\\right]\r\n\\end{align*}\\]\r\nThe order in which matrices are subtracted is important just as it is when subtracting real numbers. This means that \\(A-B\\neq B-A\\). What you might notice, however, is that \\(B-A=-\\left(A-B\\right)\\).\r\n\r\nExercise \u2013 adding and subtracting matrices\r\nConsider the matrices.\r\nFind, if possible:\r\n\r\n\t\\(A+B\\)\r\n\t\\(B-C\\)\r\n\t\\(A-D\\)\r\n\t\\(B+E\\)\r\n\t\\(E-D\\)\r\n\r\n\r\n\r\n\r\n\r\nScalar multiplication of matrices\r\nScalar multiplication of matrices is the same as adding the matrix by itself, multiple times.\r\nFor scalar multiplication of matrices, each element in the matrix is multiplied by the scalar (number).\r\n\r\nHere, let's consider matrices \\(D\\) and \\(E\\).\r\n\\[ D = \\left[\\begin{array}{cc} 1 & 2\\\\\r\n5 & 4 \r\n\\end{array} \\right]\\quad\r\nE = \\left[\\begin{array}{cc} 2 & -3\\\\\r\n0 & -1\r\n\\end{array} \\right] \\]\r\nWe can calculate \\(5D\\) which is \\(5\\times D\\).\r\n\\[\\begin{align*} 5D & = 5\\times\\left[ \\begin{array}{cc} 1 & 2\\\\\r\n5 & 4 \r\n\\end{array} \\right]\\\\\r\n& = \\left[\\begin{array}{cc} 5\\times1 & 5\\times2\\\\\r\n5\\times5 & 5\\times4\r\n\\end{array} \\right]\\\\ \r\n& = \\left[\\begin{array}{cc} 5 & 10\\\\\r\n25 & 20 \r\n\\end{array} \\right]\r\n\\end{align*}\\]\r\nAnother example is \\(-2D\\) or \\(-2\\times D\\).\r\n\\[\\begin{align*} -2D & = -2\\times\\left[ \\begin{array}{cc} 1 & 2\\\\ \r\n5 & 4 \r\n\\end{array} \\right]\\\\\r\n& = \\left[\\begin{array}{cc} -2\\times1 & -2\\times2\\\\ \r\n-2\\times5 & -2\\times4\r\n\\end{array} \\right]\\\\ \r\n& = \\left[\\begin{array}{cc} -2 & -4\\\\ \r\n-10 & -8\r\n\\end{array} \\right]\r\n\\end{align*}\\]\r\nYou can also use scalar multiplication with addition or subtraction. For example, \\(5D-2E\\) is \\(\\left(5\\times D\\right)-\\left(2\\times E\\right)\\).\r\n\\[\\begin{align*} 5D-2E & = 5\\times\\left [\\begin{array}{cc} 1 & 2\\\\ \r\n5 & 4 \r\n\\end{array}\\right]\r\n-2\\times\\left[ \\begin{array}{cc} 2 & -3\\\\\r\n0 & -1 \r\n\\end{array}\\right]\\\\\r\n& = \\left [ \\begin{array}{cc} 5\\times1 & 5\\times2\\\\ \r\n5\\times5 & 5\\times4 \r\n\\end{array} \\right]\r\n-\\left[ \\begin{array}{cc} 2\\times2 & 2\\times\\left(-3\\right)\\\\ \r\n2\\times0 & 2\\times\\left(-1\\right) \r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} 5 & 10\\\\ \r\n25 & 20 \r\n\\end{array} \\right]\r\n-\\left[ \\begin{array}{cc} 4 & -6\\\\ \r\n0 & -2 \r\n\\end{array} \\right]\\\\\r\n& = \\left[ \\begin{array}{cc} 5-4 & 10-\\left(-6\\right)\\\\ \r\n25-0 & 20-\\left(-2\\right)\r\n\\end{array} \\right]\\\\ \r\n& = \\left[ \\begin{array}{cc} 1 & 16\\\\ \r\n25 & 22 \r\n\\end{array} \\right]\r\n\\end{align*}\\]\r\n\r\nExercise \u2013 multiplying matrices by scalars\r\nConsider the matrices. \r\n\\[\\begin{align*} A & = \\left[ \\begin{array}{ccc} 2 & 3 & 1\\\\\r\n0 & 1 & 2\r\n\\end{array} \\right]\\quad\r\nB = \\left[ \\begin{array}{c} 4\\\\\r\n-1\\\\\r\n2 \r\n\\end{array} \\right]\\quad\r\nC = \\left[ \\begin{array}{ccc} 1 & -1 & 2\\\\\r\n2 & 0 & 1 \r\n\\end{array} \\right]\\\\\r\nD & = \\left[ \\begin{array}{c} 1\\\\ \r\n0\\\\ \r\n-1 \r\n\\end{array} \\right]\\quad\r\nE = \\left[ \\begin{array}{ccc} 0 & 1 & 3\\\\ \r\n1 & 0 & 2 \r\n\\end{array} \\right]\r\n\\end{align*}\\] Find, if possible.\r\n\r\n\t\t\\(2A+B\\)\r\n\t\t\\(4D-B\\)\r\n\t\t\\(3C-A\\)\r\n\t\t\\(B+2D+E\\)\r\n\t\t\\(3C-A+3E\\)\r\n\r\n\r\n\r\n\r\n","n":0.037},[{"v":"Matrices","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":348,"$":[{"v":"Introduction to matrices","n":0.577},{"v":"\r\nUse this resource to learn about matrices, their order, and the different types of matrices. A sound understanding of matrices is essential for a wide range of real-world applications, like machine learning, natural language processing, robotics, image processing, urban planning and infrastructure development.\r\n\r\nMatrices\r\nMatrices are arrays of numbers or variables arranged in rows and columns. There are a few key features of matrices that you should understand:\r\n\r\n\tA matrix is a rectangular array of elements.\r\n \tMatrices are usually denoted by upper case letters.\r\n \tThe elements are usually written within brackets.\r\n \tThe order or shape of the matrix is determined by the number of rows and columns of the matrix.\r\n \tThe number of rows is always given first, then the number of columns.\r\n\r\nAn example of a matrix is: \r\n\\[A=\\left[\\begin{array}{ccc} 1 & 2 & -9\\\\\r\n2 & 5 & -3\r\n\\end{array}\\right]\\]\r\nMatrix \\(A\\) has \\(2\\) rows and \\(3\\) columns. Is it called a \\(2\\times3\\) (or \\(2\\) by \\(3\\)) matrix. \\(2\\times3\\) is also the order of matrix \\(A\\).\r\n\r\nSquare matrix\r\nA matrix with the same number of rows and columns is called a square matrix. An example is:\r\n\\[B =\\left[\\begin{array}{cc} 2 & 3\\\\\r\n2 & 5 \r\n\\end{array}\\right]\\]\r\nMatrix \\(B\\) is a square matrix and its order is \\(2\\times2\\).\r\n\r\nUnit matrix\r\nA unit (or identity) matrix is a square matrix with diagonal elements equal to \\(1\\), and all other elements equal to \\(0\\). The unit matrix is usually denoted by \\(I\\), with the number of columns/rows indicated by a subscript.\r\nFor example, \\(I_{3}\\) is a \\(3\\times3\\) unit matrix.\r\n\\[I_{3}=\\left[\\begin{array}{ccc} 1 & 0 & 0\\\\\r\n0 & 1 & 0\\\\ \r\n0 & 0 & 1 \r\n\\end{array}\\right]\\]\r\n\\(I_{2}\\) is a \\(2\\times2\\) unit matrix.\r\n\\[I_{2}=\\left[\\begin{array}{cc} 1 & 0\\\\\r\n0 & 1\r\n\\end{array}\\right]\\]\r\n\r\nRow matrix\r\nA matrix with one row is called a row matrix. For example:\r\n\\[D =\\left[\\begin{array}{cccc} 2 & 1 & 0 & 4\\end{array}\\right]\\]\r\nMatrix \\(D\\) is a \\(1\\times4\\) row matrix.\r\n\r\nColumn matrix\r\nA matrix with one column is called a column matrix. For example:\r\n\\[E=\\left[\\begin{array}{c} 2\\\\ \r\n-4\\\\\r\n1 \\end{array}\\right]\\]\r\nMatrix \\(E\\) is a \\(3\\times1\\) column matrix.\r\n\r\nZero matrix\r\nA zero matrix has all elements equal to zero. A zero matrix can simply be written as \\(0\\). For example:\r\n\\[0=\\left[\\begin{array}{cc} 0 & 0\\\\ 0 & 0 \\end{array}\\right]\\]\r\nThis is a \\(2\\times2\\) zero matrix.\r\nA \\(3\\times3\\) zero matrix would also be denoted \\(0\\).\r\n\\[0=\\left[\\begin{array}{ccc} 0 & 0 & 0\\\\\r\n0 & 0 & 0\\\\\r\n0 & 0 & 0\r\n\\end{array}\\right]\\]\r\n\r\nExercise \u2013 determining the order of matrices\r\n\r\n \tDetermine the order of the following matrices.\r\n\t\r\n\t\t\\(\\left[\\begin{array}{ccc} 7 & -5 & 0\\\\ \r\n\t\t6 & 2 & -1 \r\n\t\t\\end{array}\\right]\\)\r\n\t\t\\(\\left[\\begin{array}{cc} 0 & 2\\\\ \r\n\t\t1 & 1 \r\n\t\t\\end{array}\\right]\\)\r\n\t\t\\(\\left[\\begin{array}{c} 2\\\\\r\n\t\t-4\\\\ \r\n\t\t1\\\\ \r\n\t\t1 \\end{array}\\right]\\)\r\n\t\t\\(\\left[\\begin{array}{cc} 1 & 1\\\\\r\n\t\t3 & 0\\\\ \r\n\t\t-2 & 3 \r\n\t\t\\end{array}\\right]\\)\r\n\t\r\n\r\n\r\n\r\nEqual matrices\r\nFor two matrices to be equal, they must be the same shape and the corresponding elements must be equal.\r\nConsider matrices \\(A\\) and \\(B\\).\r\n\\[\\begin{align*} A & =\\left[\\begin{array}{ccc} 2 & 5 & b\\\\\r\n5 & 3 & 1\\\\ \r\n2 & 0 & -2 \r\n\\end{array}\\right] \r\n& B & =\\left[\\begin{array}{ccc} 2 & 5 & 7\\\\ \r\n5 & a & 1\\\\ \r\n2 & 0 & -2 \r\n\\end{array}\\right]\r\n\\end{align*}\\]\r\nIf \\(A\\) and \\(B\\) are equal, then \\(a=3\\) and \\(b=7\\).\r\n\r\nExercise \u2013 identifying equal matrices\r\nIdentify which of the following matrices are equal.\r\n\r\n\t\\(\\left[\\begin{array}{cc} 3 & 0\\\\ \r\n\t1 & -2 \r\n\t\\end{array}\\right] \\)\r\n\t\\(\\left[\\begin{array}{cc} 3 & 1\r\n\t\\end{array}\\right] \\)\r\n\t\\(\\left[\\begin{array}{cc} 3 & 0\r\n\t\\end{array}\\right] \\)\r\n\t\\(\\left[\\begin{array}{cc} 3 & 0\\\\ \r\n\t1 & -2 \r\n\t\\end{array}\\right] \\)\r\n\r\n\r\n\r\n","n":0.044},[{"v":"Matrices","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":349,"$":[{"v":"Atomic structure","n":0.707},{"v":"Atomic structure underpins the behaviour of elements, their interactions, and the formation of molecules. These are fundamental to chemistry, physics, and many applied sciences. This resource explains the structure of atoms, atomic numbers and mass numbers, and isotopes and atomic masses.\r\n\r\nThe atomic model\r\nAll atoms are composed of a small, positively charged nucleus containing protons and neutrons surrounded by a larger negatively charged electron cloud. This is the atomic model, which describes the structure of atoms.\r\n\r\n\r\nProtons, neutrons and electrons are subatomic particles. Subatomic particles present in the nucleus are known as nucleons (e.g.\u00a0protons and neutrons). Properties of the subatomic particles are given in the table. You can see that protons and neutrons are almost identical in mass, whereas electrons are much smaller. Atoms are uncharged species due to the presence of an equal number of postiively charged protons and negatively charged electrons.\r\n\r\nNuclide notation\r\nAtoms are represented using nuclide notation (also called isotope notation). This includes the chemical symbol for the element, its atomic number and mass number.\r\n\r\n\r\n\r\nAtomic number\r\nAn atomic number (Z) is the number of protons present in the nucleus of an atom. Since an atom contains an equal number of protons and electrons, the atomic number also indicates the number of electrons present in an atom.\r\n\r\n \tAtomic number (Z) = Number of protons = Number of electrons\r\n\r\n\r\nMass number\r\nA mass number (A) is the sum of the number of protons and neutrons present in the nucleus of an atom. It is the total number of nucleons.\r\n\r\n \tMass number (A) = Number of protons + Number of neutrons\r\n\r\nThe number of subatomic particles present in an atom can be determined using its atomic number and mass number.\r\n\r\n \tNumber of protons = Atomic number (Z)\r\n \tNumber of electrons = Atomic number (Z)\r\n \tNumber of neutrons = Mass number (A) \u2013 Atomic number (Z)\r\n \tTotal number of subatomic particles = Mass number (A) + Atomic number (Z)\r\n\r\n\r\nIsotopes and atomic masses\r\nIsotopes are atoms of an element with the same number of protons and electrons, but different numbers of neutrons. Examples are:\r\n\r\n\tcarbon which has three isotopes: \\(\\ce{_{6}^{12}C}\\), \\(\\ce{_{6}^{13}C}\\), \\(\\ce{_{6}^{14}C}\\)\r\n\tchlorine which has two isotopes: \\(\\ce{_{17}^{35}Cl}\\), \\(\\ce{_{17}^{37}Cl}\\)\r\n\thydrogen which as three isotopes: \\(\\ce{_{1}^{1}H}\\), \\(\\ce{_{1}^{2}H}\\), \\(\\ce{_{1}^{3}H}\\).\r\n\r\n\r\nAtomic mass is the weighted average mass of an atom of an element. To calculate the atomic mass, the relative percentage abundance and atomic mass of each isotope is considered. The atomic number and atomic mass are typically represented alongside the chemical symbol in the periodic table. You will learn about the periodic table later.\r\n\r\n\r\n\r\n\r\nExample 1 \u2013 determining the number of subatomic particles\r\nCalculate the number of protons, electrons and neutrons present in the atom \\(\\ce{_{18}^{40}Ar}\\).\r\nStep 1: Find the atomic number (Z) and mass number (A) from the nuclide notation. Z is the number written at the bottom. A is the number at the top.\r\n\\[Z=18\\] \\[A=40\\]\r\nStep 2: Determine the number of protons. This is the same as the atomic number (Z).\r\n\\[\\textrm{Number of protons}=Z=18\\]\r\nStep 3: Determine the number of electrons. This is the same as the atomic number (Z).\r\n\\[\\textrm{Number of electrons}=Z=18\\]\r\nStep 4: Determine the number of neutrons. This is the mass number (A) minus the atomic number (Z).\r\n\\[\\begin{align*}\\textrm{Number of neutrons} & =A-Z\\\\\r\n&=40-18\\\\\r\n&=22\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExample 4 \u2013 determining the atomic mass\r\nChlorine naturally exists in two isotopic forms: \\(\\ce{_{17}^{35}Cl}\\) and \\(\\ce{_{17}^{37}Cl}\\). Each isotope's relative percentage abundance and atomic mass are given in the following table. Calculate the atomic mass of chlorine.\r\n\r\nStep 1: For each isotope, multiply the atomic mass by its percentage abundance.\r\n\\[\\begin{align*}\\textrm{Chlorine-35} & = \\textrm{percentage abundance}\\times\\textrm{atomic mass}\\\\\r\n& = 75.53\\textrm{%}\\times34.97\\textrm{ amu}\\\\\r\n& = 26.41\\textrm{ amu}\r\n\\end{align*}\\] \\[\\begin{align*}\\textrm{Chlorine-37} & = \\textrm{percentage abundance}\\times\\textrm{atomic mass}\\\\\r\n& = 24.47\\textrm{%}\\times36.97\\textrm{ amu}\\\\\r\n& = 9.047\\textrm{ amu}\r\n\\end{align*}\\]\r\nStep : Find the sum of the values calculated in Step 1.\r\n\\[\\begin{align*}\\textrm{Atomic mass of chlorine} & = 26.41\\textrm{ amu}+9.047\\textrm{ amu}\\\\\r\n& =35.46\\textrm{ amu}\r\n\\end{align*}\\]\r\n\r\n\r\nYour turn \u2013 atomic structure\r\nTest yourself on your understanding of atomic structure.\r\n\r\n\r\n\r\n const manual = new Freezeframe('.freezeme', { trigger: false }); document .querySelector('.start') .addEventListener('click', () => { manual.start(); }); document .querySelector('.stop') .addEventListener('click', () => { manual.stop(); }); \r\n\r\n \r\nHydrogen-1 (\\(^{1}_{1}\\ce{H}\\)) consists of one electron, with one proton in its nucleus. Carbon-12 (\\(^{12}_{6}\\ce{C}\\)) consists of six electrons, with six protons and six neutrons in its nucleus.\r\n \r\nThe nuclide notation of carbon-12 shows the mass number \\(12\\) at the top, the atomic number \\(6\\) at the bottom, and the chemical symbol \\(\\ce{C}\\).\r\n \r\nThe carbon-12 isotope has an atomic number of \\(6\\), an atomic mass of \\(12.01\\), and the chemical symbol \\(\\ce{C}\\).\r\n","n":0.037},[{"v":"Chemistry","i":1,"n":1},{"v":"Atomic Structures","i":0,"n":0.707}]]},{"i":350,"$":[{"v":"Atoms, molecules and compounds","n":0.5},{"v":"Atoms are the basic building blocks of elements. They are too small to be observed by the naked eye. Let's explore atoms, and molecules and compounds.\r\n\r\nElements are the building blocks of all matter and composed of one type of atom, such as \\(\\ce{He}\\), \\(\\ce{H}\\), \\(\\ce{C}\\), \\(\\ce{Mg}\\), and \\(\\ce{Fe}\\). Some elements exist as molecules, such as \\(\\ce{S}_8\\), \\(\\ce{O}_2\\), and \\(\\ce{Cl}_2\\).\r\n\r\n\r\n\r\nMolecules and compounds\r\nA molecule is a group of two or more atoms that have combined chemically and function as a unit. The simplest molecule that can exist is a diatomic molecule that contains two atoms. Two examples are \\(\\ce{O}_2\\) and \\(\\ce{H}_2\\). A compound contains different types of atoms in fixed proportions.\r\nBoth compounds and molecules are uncharged (neutral) species. Some examples are shown in the table.\r\n\r\n\r\n\r\nMolecules\r\nCompounds\r\n\r\n\r\n\r\n\r\n\\(\\ce{H2}\\)\r\n\\(\\ce{H2O}\\)\r\n\r\n\r\n\\(\\ce{O3}\\)\r\n\\(\\ce{CO2}\\)\r\n\r\n\r\n\\(\\ce{S8}\\)\r\n\\(\\ce{C2H5OH}\\)\r\n\r\n\r\n\\(\\ce{CO2}\\)\r\n\\(\\ce{CH3COOH}\\)\r\n\r\n\r\n\\(\\ce{H2O}\\)\r\n\\(\\ce{FeSO4}\\)\r\n\r\n\r\n\r\n\r\nChemical symbols\r\nEach element has a unique name and chemical symbol. All of the discovered elements are organised in the modern periodic table. Some of these are shown in the table.\r\n\r\n\r\n\r\nElement\r\nChemical symbol\r\n\r\n\r\n\r\n\r\nHydrogen\r\n\\(\\ce{H}\\)\r\n\r\n\r\nCarbon\r\n\\(\\ce{C}\\)\r\n\r\n\r\nCalcium\r\n\\(\\ce{Ca}\\)\r\n\r\n\r\nSilver\r\n\\(\\ce{Ag}\\)\r\n\r\n\r\nCopper\r\n\\(\\ce{Cu}\\)\r\n\r\n\r\n\r\n\r\nChemical formulas\r\nChemical formulas represent the chemical composition of molecules and compounds. They consist of the chemical symbols for the elements present in the compound and numerical subscript which shows the number of atoms of each element involved in the formation of the compound.\r\nConsider the chemical formula for the compound sucrose, \\(\\ce{C12H22O11}\\).\r\n\r\n\tThis formula shows that sucrose contains three different elements: \\(\\ce{C}\\), \\(\\ce{H}\\) and \\(\\ce{O}\\).\r\n\tThe total number of atoms is \\(12+22+11=45\\).\r\n\tIt contains \\(12\\textrm{ C}\\) atoms, \\(22\\textrm{ H}\\) atoms, and \\(11\\textrm{ O}\\) atoms.\r\n\r\n\r\nNow, consider the chemical formula for the compound calcium phosphate: \\(\\ce{Ca3(PO4)2}\\).\r\n\r\n\tThis formula shows that one unit of calcium phosphate is composed of three different elements: \\(\\ce{C}\\), \\(\\ce{P}\\) and \\(\\ce{O}\\).\r\n\tThe total number of atoms is \\(3+2+8=13\\).\r\n\tIt contains \\(3\\textrm{ Ca}\\) atoms, \\(2\\textrm{ P}\\) atoms, and \\(8\\textrm{ O}\\) atoms. For \\(\\ce{O}\\) and \\(\\ce{P}\\), the number of atoms is calculated by multiplying the subscript number by the number written after the brackets.\r\n\r\n\r\nYour turn \u2013 atoms, molecules and compounds\r\nTest yourself on your understanding of atoms, molecules and compounds.\r\n\r\n\r\n\r\n\r\n\r\n","n":0.056},[{"v":"Molecules","i":2,"n":1},{"v":"Chemistry","i":1,"n":1},{"v":"Atomic Structures","i":0,"n":0.707}]]},{"i":351,"$":[{"v":"Right-angled triangle trigonometry","n":0.577},{"v":"\r\nExplore the relationships between the angles and side lengths of a right-angled triangle. You will learn how to use trigonometric ratios to find unknown angles and lengths, enhancing your ability to solve geometric problems. Mastering this concept is vital for fields such as engineering, physics, and architecture, where precise calculations of angles and dimensions are essential.\r\n\r\nTrigonometry\r\nTrigonometry is a branch of math involving the study of triangles. The ability to use and manipulate trigonometric functions is necessary in other branches of mathematics, including calculus, vectors and complex numbers.\r\n\r\nRight-angled triangles\r\n\r\nYou already learned that the longest side of a right-angled triangle is called the hypotenuse. The hypotenuse is always opposite to the right angle.\r\nThe other two sides are named in relation to another known angle (or an unknown angle under consideration).\r\n\r\n\tThe opposite is the side of a right-angled triangle that is opposite to the angle of interest.\r\n\tThe adjacent is the side of a right-angled triangle that is next to the angle of interest.\r\n\r\n\r\n\r\n\r\nTrigonometric ratios\r\nThe angles and lengths of a right-angled triangle are related by three basic trigonometric ratios. In a right-angled triangle, the following ratios are defined for a given angle, \\(\\theta\\):\r\n\r\n\\[\\begin{align*} \\textrm{sine}(\\theta) & = \\frac{\\textrm{opposite length}}{\\textrm{hypotenuse length}}\\\\\r\n\\textrm{cosine}(\\theta) & = \\frac{\\textrm{adjacent length}}{\\textrm{hypotenuse length}}\\\\\r\n\\textrm{tangent}(\\theta) & = \\frac{\\textrm{opposite length}}{\\textrm{adjacent length}}\r\n\\end{align*}\\]\r\n\r\n\r\nThese ratios are abbreviated to \\(\\sin(\\theta)\\), \\(\\cos(\\theta)\\) and \\(\\tan(\\theta)\\), respectively.\r\nA useful way to remember these relationships is using the mnemonic:\r\n\r\n\\[\\textrm{SOH CAH TOA}\\]\r\nwhere \r\n\\[\\sin(\\theta) = \\frac{\\textrm{O}}{\\textrm{H}}\\quad\\cos(\\theta)=\\frac{\\textrm{A}}{\\textrm{H}}\\quad\\tan(\\theta)=\\frac{\\textrm{O}}{\\textrm{A}}\\]\r\n\r\n\r\nExample \u2013 evaluating ratios\r\nIn the right-angled triangle shown, evaluate \\(\\sin(\\theta)\\), \\(\\cos(\\theta)\\), and \\(\\tan(\\theta)\\).\r\n\r\n\r\nThe first thing you should always do is name the sides of the triangle. The hypotenuse is the longest side with length \\(5\\). The opposite is the side opposite to the angle \\(\\theta\\), with length \\(4\\). The adjacent is the side next to the angle \\(\\theta\\), with length \\(3\\).\r\nThen, we can use SOH CAH TOA to find \\(\\sin(\\theta)\\), \\(\\cos(\\theta)\\) and \\(\\tan(\\theta)\\).\r\n\\[\\begin{align*} \\sin(\\theta) & = \\frac{4}{5}\\\\\r\n& = 0.8\\\\\r\n\\cos(\\theta) & = \\frac{3}{5}\\\\\r\n& = 0.6\\\\\r\n\\tan(\\theta) & = \\frac{4}{3}\\\\\r\n& \\approx 1.33\r\n\\end{align*}\\]\r\nExercise \u2013 evaluating ratios\r\nA right-angled triangle is shown.\r\n\r\nFind:\r\n\r\n\t\\(\\sin(\\theta)\\)\r\n\t\\(\\tan(\\theta)\\)\r\n\t\\(\\cos(\\alpha)\\)\r\n\t\\(\\tan(\\alpha)\\)\r\n\r\nHint: Use Pythagoras' theorem to find the hypotenuse.\r\n\r\n\r\n\r\nExample \u2013 finding angles\r\nFind the value of the angle \\(\\theta\\) in the triangle shown.\r\n\r\nName the sides of the triangle. The longest side is the hypotenuse with length \\(19.7\\). The side opposite to the angle \\(\\theta\\) is the opposite with length \\(13.4\\). The side next to the angle is the adjacent with no length provided.\r\nWe have the lengths of the hypothenuse (H) and opposite (O), and the only relationship that connects the two is SOH or \\(\\sin(\\theta)=\\dfrac{\\textrm{O}}{\\textrm{H}}\\).\r\n\\[\\begin{align*} \\sin(\\theta) & = \\frac{\\textrm{O}}{\\textrm{H}}\\\\\r\n& = \\frac{13.4}{19.7}\\\\\r\n& = 0.6802\r\n\\end{align*}\\]\r\nTo undo the \\(\\sin\\) and find \\(\\theta\\), we need to take the inverse \\(\\sin\\), or \\(\\sin^{-1}\\). You can do this on a scientific calculator.\r\n\\[\\begin{align*} \\sin\\theta & = 0.6802\\\\\r\n\\theta & = \\sin^{-1}(0.6802)\\\\\r\n& = 42.9^{\\circ}\r\n\\end{align*}\\]\r\nDon't forget to write the degree symbol!\r\n\r\nExample 1 \u2013 finding side lengths\r\nFind the unknown side length \\(b\\) in the following right-angled triangle.\r\n\r\nHere, we know an angle and the adjacent (A) side. The side length to be determined is the opposite (O) side. The ratio that relates these two sides is the tangent ratio.\r\n\\[\\begin{align*} \\tan(\\theta) & = \\frac{\\textrm{O}}{\\textrm{A}}\\\\ \r\n\\tan(27^{\\circ}) & = \\frac{b}{42}\\\\\r\n(\\tan(27^{\\circ}))\\times42 & = b\\\\ \r\nb & = 21.4\\textrm{ cm}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExercise \u2013 finding angles and side lengths\r\n\r\n\tFind the indicated unknown side length or angle in each of the triangles.\r\n\t\r\n\t\t  \r\n\t\t  \r\n\t\t  \r\n\t\t  \r\n\t\t  \r\n\t\t  \r\n\t\r\n\tIn a right-angled triangle, \\(\\sin(\\phi)=0.55\\) and the hypotenuse is \\(21\\textrm{ mm}\\). Find the length of each of the other two sides.\r\n\r\n\r\n\r\n\r\n\r\n\r\nSpecial angles and exact values\r\nFor right-angled triangles with special angles, the side lengths have exact values rather than decimal approximations. It is a good idea to remember these values. They occur often in maths courses.\r\nFirst is the triangle with \\(45^{\\circ}\\) angles.\r\n\r\n\\[\\begin{align*} \\sin(45^{\\circ}) & = \\frac{1}{\\sqrt{2}}\\\\\r\n\\cos(45^{\\circ}) & = \\frac{1}{\\sqrt{2}}\\\\\r\n\\tan(45^{\\circ}) & = 1\r\n\\end{align*}\\]\r\nSecond is the triangle with angles \\(30^{\\circ}\\) and \\(60^{\\circ}\\).\r\n\r\n\\[\\begin{align*} \\sin(60^{\\circ}) & = \\frac{\\sqrt{3}}{2}\\\\\r\n\\cos(60^{\\circ}) & = \\frac{1}{2}\\\\\r\n\\tan(60^{\\circ}) & = \\sqrt{3}\r\n\\end{align*}\\] \\[\\begin{align*} \\sin(30^{\\circ}) & = \\frac{1}{2}\\\\\r\n\\cos(30^{\\circ}) & = \\frac{\\sqrt{3}}{2}\\\\\r\n\\tan(30^{\\circ}) & = \\frac{1}{\\sqrt{3}}\r\n\\end{align*}\\]\r\n\r\n","n":0.04},[{"v":"Trigonometry","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Functions","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":352,"$":[{"v":"States and properties of matter","n":0.447},{"v":"\r\n\r\nIn the study of chemistry, matter is the essential substance that makes up everything in our universe. Learn the basics of matter using this resource, including its states, properties, classification and how Aboriginal peoples apply their understanding of matter.\r\n\r\nPhysical states of matter\r\nMatter exists in three physical states: solid, liquid and gas.\r\n\r\n\r\n\r\nStart animationStop animation\r\n\r\n\r\n\r\n\r\nProperties of matter\r\nProperties of matter can be classified into two categories depending on whether it is determined by changing the chemical identity or composition of the substance.\r\n\r\nPhysical properties\r\nIf the property can be observed/measured without changing the chemical identity of the substance, it is a physical property. Examples of physical properties include:\r\n\r\n \tShape \u2013 the external form of a substance\r\n \tColour \u2013 the appearance of a substance (its hue, saturation and lightness) due to the reflection of light\r\n \tOdour \u2013 the smell or scent of a substance\r\n \tHardness \u2013 the ability of a substance to resist being scratched\r\n \tMelting point \u2013 the temperature at which the substance changes from a solid to a liquid\r\n \tBoiling point \u2013 the temperature at which the substance changes from a liquid to a gas or vapour\r\n \tPhysical state \u2013 whether the substance is a liquid, solid or gas.\r\n\r\n\r\nChemical properties\r\nIf the property can only be observed or measured by altering the chemical identity of the substance, it is a chemical property. Examples of chemical properties include:\r\n\r\n \tFlammability \u2013 the ability of a substance to ignite or burn\r\n \tCorrosiveness \u2013 the ability of a substance to rust after exposure to moist air for prolonged periods of time\r\n \tAcidity \u2013 the ability of a substance to donate hydrogen ions and decrease the pH of a solution\r\n \tReactivity \u2013 the ability of a substance to undergo a chemical reaction.\r\n\r\n\r\nClassification of matter\r\nMatter can be classified in many ways. You can use the flowchart shown and answer a series of yes/no questions to classify matter.\r\n\r\n\r\nThe states and properties of matter can be applied in a range of ways to obtain or form useful materials.\r\n\r\n\r\n\r\nYour turn \u2013 states and properties of matter\r\nTest yourself on your understanding of the states, properties and classification of matter.\r\n\r\n\r\n\r\n const manual = new Freezeframe('.freeze1', { trigger: false }); document .querySelector('.start') .addEventListener('click', () => { manual.start(); }); document .querySelector('.stop') .addEventListener('click', () => { manual.stop(); }); \r\nconst manual2 = new Freezeframe('.freeze2', { trigger: false }); document .querySelector('.start2') .addEventListener('click', () => { manual2.start(); }); document .querySelector('.stop2') .addEventListener('click', () => { manual2.stop(); });  \r\nIn solids, the particles are tightly packed and arranged in a regular pattern. The particles are locked in place and have minimal movement.\r\nIn liquids, the particles are closely packed but lack a regular pattern. The particles are more free to move.\r\nIn gases, the particles move quickly and are far apart from each other. The particles are very free to move.\r\n Matter: Anything that has mass and occupies space\r\nAre properties and composition constant?\r\nYes\r\nPure substance (a substance that has a uniform chemical composition throughout, e.g. water, sugar, table salt).\r\nCan the pure substance be broken into simpler substances?\r\n\r\n    \r\n        Yes - Chemical compound (Can be chemically broken down into constituent elements, e.g. water, sugar, table salt.)\r\n    \r\n    \r\n        No - Element (Cannot be broken down into simpler substances by chemical or physical means, e.g. hydrogen, oxygen, sulfur.)\r\n    \r\n\r\nNo\r\nMixture (A physical combination of two or more pure substances, e.g. salt\u2013sand mixture, seawater, concrete.)\r\nDoes the mixture have a uniform composition throughout?\r\n\r\n    \r\n        Yes - Homogeneous mixture (One visible phase which has uniform composition throughout, e.g. coffee, air, rainwater.)\r\n    \r\n    \r\n        No - Heterogeneous mixture (Two or more visible phases which have non-uniform composition throughout, e.g. vegetable stew, meat pie, salad.)\r\n    \r\n","n":0.041},[{"v":"Indigenous knowledges","i":3,"n":0.707},{"v":"Matter","i":2,"n":1},{"v":"Chemistry","i":1,"n":1},{"v":"Atoms","i":0,"n":1}]]},{"i":353,"$":[{"v":"NM2: Simpson\u2019s rule","n":0.577},{"v":"\r\nSome integrals cannot be evaluated in terms of the rules of integration or elementary functions. Simpson\u2019s rule is a numerical method that calculates a numerical value for a definite integral.\r\nYou will have evaluated definite integrals such as \\[\\begin{align*} \\int_{1}^{3} & \\left(x^{2}\\right)dx \\end{align*}\\] before. In doing this, you are evaluating the area under the graph of \\(f\\left(x\\right)=x^{2}\\) between \\(x=1\\) and \\(x=3.\\) This is only possible if you can find an antiderivative for \\(x^{2}\\). In this example it is easy, the antiderivative is \\[\\begin{align*} F\\left(x\\right) & =\\frac{1}{3}x^{3}+c,\\text{ where $c$ is a constant.} \\end{align*}\\] and \\[\\begin{align*} \\int_{1}^{3}\\left(x^{2}\\right)dx & =F\\left(3\\right)-F\\left(1\\right)\\\\ & =\\frac{1}{3}3^{3}+c-\\left(\\frac{1}{3}1^{3}+c\\right)\\\\ & =\\frac{26}{3}. \\end{align*}\\]\r\n\r\nSometimes it is not possible to find the antiderivative. In such cases you need to use a numerical method. Two common methods for calculating definite integrals are:\r\n\r\n \tSimpson\u2019s rule, and\r\n \tThe trapezoidal rule.\r\n\r\nThis module considers Simpson\u2019s rule.\r\n\r\nSimpson\u2019s rule\r\nSimpson\u2019s rule estimates the value of the definite integral\r\n\r\n\\[\\begin{align*} \\int_{a}^{b}f\\left(x\\right)dx. \\end{align*}\\] Simpson\u2019s rule estimates the area under the graph of \\(f\\left(x\\right)\\) by approximating the function \\(f\\left(x\\right)\\) by a parabola and calculating the area under the parabola. Each parabolic approximation is done over two strips. The accuracy of the Simpson\u2019s Rule approximation increases as you use more strips. There must be an even number of strips.\r\n\r\nSimpson\u2019s rule has the form: \\[\\begin{align*} \\int_{a}^{b}f\\left(x\\right)dx & \\approx\\frac{1}{3}h\\left[y_{0}+4\\sum_{i=1,3,5,\\ldots}^{n-1}y_{i}+2\\sum_{i=2,4,6,\\ldots}^{n-2}y_{i}+y_{n}\\right]\\\\ & =\\frac{1}{3}h\\left[y_{0}+4y_{1}+2y_{2}+4y_{3}+2y_{4}+\\ldots+4y_{n-1}+y_{n}\\right] & \\left(1\\right) \\end{align*}\\] where\r\n\r\n \t\\(n\\) is the number of strips. Note that \\(n\\) must be even.\r\n \t\\(y_{i}=f\\left(x_{i}\\right)\\) are the values of \\(f\\left(x_{i}\\right)\\) at the points \\(x_{i}\\) where \\(i=0,1,2,3\\ldots,n\\). Note that \\(x_{0}=a,\\) \\(x_{n}=b.\\)\r\n \t\\(h\\) is the width of each strip and \\[\\begin{align*} h & =\\frac{b-a}{n}. \\end{align*}\\]\r\n \t\\(x_{1}=a+h,\\) \\(x_{2}=a+2h,\\) \\(x_{3}=a+3h,\\) \\(\\ldots\\) and so on.\r\n\r\nNote that the value of the function at the odd numbered points are multiplied by four whereas the even numbered points are multiplied by two in equation \\(\\left(1\\right)\\) above.\r\n\r\nThe accuracy of the Simpson\u2019s Rule approximation of an integral increases as you use more strips.\r\n\r\nExample 1\r\nUsing Simpson\u2019s rule, find the value of \\[\\begin{align*} \\int_{1}^{3} & \\left(e^{2x}\\sin\\left(x\\right)\\right)dx \\end{align*}\\] using \\(2\\) strips to an accuracy of \\(3\\) decimal places. This integral may be evaluated and is approximately equal to \\(100.96243981\\).\r\n\r\nSolution:\r\n\r\nConsider the figure below.\r\n\r\n\r\n\r\nSimpson\u2019s rule approximates the real function by a parabola (shown in green/black dotted lines).1 Note that three points are needed to fit a parabola. The area given by Simpson\u2019s rule is the shaded area under the parabola. From the graph we can see that Simpson\u2019s rule will overestimate the area under the real graph from \\(x=1\\) to \\(x=2\\) and underestimate the area from \\(x=2\\) to \\(x=3.\\)\r\n\r\nWe have \\(n=2,\\) \\(a=1,\\) \\(b=3\\) and2 The first strip is from \\(x=1\\) to \\(x=2.\\) The second is from \\(x=2\\) to \\(x=3.\\) \\[\\begin{align*} h & =\\frac{b-a}{2}\\\\ & =\\frac{3-1}{2}\\\\ & =1. \\end{align*}\\]\r\n\r\nCalculating the three points and the value of the function at those points we have:3 Note that the number of points \\(p\\) and the number of strips \\(n\\) are related by \\[\\begin{align*} p & =n+1. \\end{align*}\\] We stop calculating points when \\(x_{n}=b.\\) In this example \\(n=2\\) and there are \\(3=2+1\\) points at which the function must be evaluated. They are \\(x_{0},x_{1}\\) and \\(x_{2}.\\)\r\n\r\n\\(x_{0}=a=1\\) and \\(y_{0}=e^{2\\left(1\\right)}\\sin\\left(1\\right)=6.2176763.\\)\r\n\r\n\\(x_{1}=a+h=1+1=2\\) and \\(y_{1}=e^{2\\left(2\\right)}\\sin\\left(2\\right)=49.6459573.\\)\r\n\r\n\\(x_{2}=a+2h=1+2=3\\) and \\(y_{3}=e^{2\\left(3\\right)}\\sin\\left(3\\right)=56.9318746.\\)\r\n\r\nNow we substitute this in the Simpson\u2019s rule formula \\(\\left(1\\right).\\) We get \\[\\begin{align*} \\int_{1}^{3}\\left(e^{2x}\\sin\\left(x\\right)\\right)dx & =\\frac{1}{3}h\\left[y_{0}+4y_{1}+y_{2}\\right]\\\\ & =\\frac{1}{3}\\left(1\\right)\\left[6.2176763+4\\left(49.6459573\\right)+56.9318746\\right]\\\\ & =\\frac{1}{3}\\left[6.2176763+4\\left(49.6459573\\right)+56.9318746\\right]\\\\ & =87.2444601. \\end{align*}\\] The answer, to three decimal places, is \\[\\begin{align*} \\int_{1}^{3}\\left(e^{2x}\\sin\\left(x\\right)\\right)dx & =87.244. \\end{align*}\\] This is well under the value of \\(100.96243981\\). The percentage error is \\[\\begin{align*} \\text{error} & =\\frac{100.96243981-87.2444601}{100.96243981}\\times100\\\\ & =13.6\\%. \\end{align*}\\]\r\n\r\nExample 2\r\nUsing Simpson\u2019s rule, find the value of \\[\\begin{align*} \\int_{1}^{3} & \\left(e^{2x}\\sin\\left(x\\right)\\right)dx \\end{align*}\\] using \\(4\\) strips to an accuracy of \\(3\\) decimal places. This is exactly the same as Example 1 above except that we are using more strips.\r\nSolution\r\nWe have \\(n=4,\\) \\(a=1,\\) \\(b=3\\) and \\[\\begin{align*} h & =\\frac{b-a}{4}\\\\ & =\\frac{3-1}{4}\\\\ & =0.5\\,. \\end{align*}\\]\r\n\r\nNow calculate the points and the value of the function at those points.\r\n\r\n\\(x_{0}=a=1\\) and \\(y_{0}=e^{2\\left(1\\right)}\\sin\\left(1\\right)=6.2176763.\\)\r\n\r\n\\(x_{1}=a+h=1+0.5=1.5\\) and \\(y_{1}=e^{2\\left(1.5\\right)}\\sin\\left(1.5\\right)=20.0352224\\,.\\)\r\n\r\n\\(x_{2}=a+2h=1+1=2\\) and \\(y_{3}=e^{2\\left(2\\right)}\\sin\\left(2\\right)=49.6459573\\,.\\)\r\n\r\n\\(x_{3}=a+3h=1+1.5=2.5\\) and \\(y_{3}=e^{2\\left(2.5\\right)}\\sin\\left(2.5\\right)=88.8211415\\,.\\)\r\n\r\n\\(x_{4}=a+4h=1+2=3\\) and \\(y_{4}=e^{2\\left(3\\right)}\\sin\\left(3\\right)=56.9318746\\,.\\)\r\n\r\nNow we substitute the \\(y_{n}\\) values in the Simpson\u2019s rule formula \\(\\left(1\\right).\\) We get \\[\\begin{align*} \\int_{1}^{3}\\left(e^{2x}\\sin\\left(x\\right)\\right)dx & =\\frac{1}{3}h\\left[y_{0}+4\\left(y_{1}+y_{3}\\right)+2\\left(y_{2}\\right)+y_{4}\\right]\\\\ & =\\frac{0.5}{3}\\left[6.2176763+4\\left(20.0352224+88.8211415\\right)\\right.\\\\ & \\quad\\quad\\left.+2\\left(49.6459573\\right)+56.9318746\\right]\\\\ & =99.6444869\\,. \\end{align*}\\] So the answer to three decimal places is \\[\\begin{align*} \\int_{1}^{3}\\left(e^{2x}\\sin\\left(x\\right)\\right)dx & =99.645\\,. \\end{align*}\\] This is much closer to the actual value of \\(100.96243981\\). The percentage error is now \\[\\begin{align*} \\text{error} & =\\frac{100.96243981-99.6444869}{100.96243981}\\times100\\\\ & =1.3\\% \\end{align*}\\] which is a huge improvement over the result in Example 1.\r\n\r\nExample 3\r\nUsing Simpson\u2019s rule, find the value of \\[\\begin{align*} \\int_{1}^{3} & \\left(e^{2x}\\sin\\left(x\\right)\\right)dx \\end{align*}\\] using \\(6\\) strips to an accuracy of \\(3\\) decimal places. This is exactly the same as Example 1 and 2 above, except that we are using more strips.\r\nSolution\r\nWe have \\(n=6,\\ \\) \\(a=1,\\) \\(b=3\\) and \\[\\begin{align*} h & =\\frac{b-a}{6}\\\\ & =\\frac{3-1}{6}\\\\ & =\\frac{1}{3}\\,. \\end{align*}\\]\r\n\r\nNow calculate the values at the seven points:\r\n\r\n\\(x_{0}=a=1\\) and \\(y_{0}=e^{2\\left(1\\right)}\\sin\\left(1\\right)=6.2176763.\\)\r\n\r\n\\(x_{1}=a+h=1+1/3=4/3.\\) So \\(y_{1}=e^{2\\left(4/3\\right)}\\sin\\left(4/3\\right)=13.9880487\\,.\\)\r\n\r\n\\(x_{2}=a+2h=1+2/3=5/3.\\) So \\(y_{3}=e^{2\\left(5/3\\right)}\\sin\\left(5/3\\right)=27.9029025\\,.\\)\r\n\r\n\\(x_{3}=a+3h=1+1=2\\,.\\) So \\(y_{3}=e^{2\\left(2\\right)}\\sin\\left(2\\right)=49.6459573\\,.\\)\r\n\r\n\\(x_{4}=a+4h=1+4/3=7/3.\\) So \\(y_{4}=e^{2\\left(7/3\\right)}\\sin\\left(7/3\\right)=76.8948872\\,.\\)\r\n\r\n\\(x_{5}=a+5h=1+5/3=8/3.\\) So \\(y_{5}=e^{2\\left(8/3\\right)}\\sin\\left(8/3\\right)=94.7136211\\,.\\)\r\n\r\n\\(x_{6}=a+6h=1+2=3.\\) So \\(y_{6}=e^{2\\left(3\\right)}\\sin\\left(3\\right)=56.9318746\\,.\\)\r\n\r\nNow we substitute these values into the Simpson\u2019s rule formula \\(\\left(1\\right).\\) We get \\[\\begin{align*} \\int_{1}^{3}\\left(e^{2x}\\sin\\left(x\\right)\\right)dx & =\\frac{1}{3}h\\left[y_{0}+4\\left(y_{1}+y_{3}+y_{5}\\right)+2\\left(y_{2}+y_{4}\\right)+y_{6}\\right]\\\\ & =\\frac{1}{3}\\frac{1}{3}\\left[6.2176763\\right.\\\\ & \\quad\\quad+4\\left(13.9880487+49.6459573+94.7136211\\right)\\\\ & \\quad\\quad+2\\left(27.9029025+76.8948872\\right)+56.9318746\\left.\\!\\right]\\\\ & =100.6817377\\,. \\end{align*}\\] The answer is, to three decimal places, \\[\\begin{align*} \\int_{1}^{3} & \\left(e^{2x}\\sin\\left(x\\right)\\right)dx=100.682\\,. \\end{align*}\\] The percentage error is now \\[\\begin{align*} \\text{error} & =\\frac{100.96243981-100.6817377}{100.96243981}\\times100\\\\ & =0.3\\%. \\end{align*}\\]\r\n\r\nError reduction with increasing number of strips\r\nExamples 1, 2 and 3 above show that the Simpson\u2019s Rule reduces error quickly with increased numbers of steps. This is summarised in the table below.\r\n\r\n\r\n\r\nNumber of strips\r\nValue obtained\r\nError %\r\n\r\n\r\n\r\n\r\n2\r\n87.244\r\n13.5\r\n\r\n\r\n4\r\n99.645\r\n1.3\r\n\r\n\r\n6\r\n100.682\r\n0.3\r\n\r\n\r\n\r\n\r\nExample 4\r\nApproximate \\(\\int_{0}^{2}\\frac{1}{1+x^{3}}dx\\) using Simpson\u2019s rule with \\(4\\) strips to \\(5\\) decimal places.\r\nSolution\r\nWe have \\(a=0\\), \\(b=2\\), \\(n=4\\) and \\[\\begin{align*} h & =\\frac{b-a}{n}\\\\ & =\\frac{2-0}{4}\\\\ & =0.5\\,. \\end{align*}\\] Calculating the points, and the value of the function \\(\\frac{1}{1+x^{3}}\\) at them we have:\r\n\r\n\\(x_{0}=a=0\\) and \\(y_{0}=1\\),\r\n\r\n\\(x_{1}=a+h=0.5\\) and \\(y_{1}=1/\\left(1+0.5^{3}\\right)=0.888889\\),\r\n\r\n\\(x_{2}=a+2h=0+1\\) and \\(y_{2}=1/\\left(1+1^{3}\\right)=0.5\\),\r\n\r\n\\(x_{3}=a+3h=0+1.5=1.5\\) and \\(y_{3}=1/\\left(1+1.5^{3}\\right)=0.228571\\),\r\n\r\n\\(x_{4}=a+4h=0+2=2\\) and \\(y_{4}=1/\\left(1+2^{3}\\right)=0.111111.\\)\r\n\r\nNow we substitute in eqn\\(\\left(1\\right)\\) to get \\[\\begin{align*} \\int_{0}^{2}\\frac{1}{1-x^{3}}dx & =\\frac{1}{3}h\\left[y_{0}+4\\sum_{i=1,3,5,\\ldots}^{n-1}y_{i}+2\\sum_{i=2,4,6,\\ldots}^{n-1}y_{i}+y_{n}\\right]\\\\ & =\\frac{1}{3}h\\left[y_{0}+4\\left(y_{1}+y_{3}\\right)+2\\left(y_{2}\\right)+y_{4}\\right]\\\\ & =\\frac{1}{3}\\left(0.5\\right)\\left[1+4\\left(0.888889+0.228571\\right)\\right.\\\\ & \\quad\\quad\\left.+2\\left(0.5\\right)+0.111111.\\right]\\\\ & =1.0968252\\,. \\end{align*}\\] The answer, to five decimal places is, \\[\\begin{align*} \\int_{0}^{2}\\frac{1}{1-x^{3}}dx & =1.09683. \\end{align*}\\]\r\n\r\nExercises\r\n\r\n \tUse Simpson\u2019s rule to evaluate the integral \\(\\int_{0}^{3}\\left(3x^{2}+4x\\right)dx\\) using 4 strips to two decimal places.\r\n \tUse Simpson\u2019s rule to evaluate the integral \\(\\int_{0}^{1}\\left(3e^{x}\\sin\\left(x\\right)\\right)dx\\) using 2 strips to two decimal places.\r\n\r\n\r\n\u00a0\r\n\r\n","n":0.032},[{"v":"Methodology","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":354,"$":[{"v":"NM1: Newton\u2019s method","n":0.577},{"v":"\r\nYou are used to solving equations using basic algebraic operations and perhaps the quadratic formula. However, some equations cannot be solved using these methods.\r\nFor example the equations: \\[\\begin{align*} e^{-x} & =x & \\left(1\\right) \\end{align*}\\] or \\[\\begin{align*} -\\ln\\left(x\\right) & =x^{3} & \\left(2\\right) \\end{align*}\\] cannot be solved using conventional methods. However they can be solved using numerical methods. The important aspect of equations \\(\\left(1\\right)\\) and \\(\\left(2\\right)\\) is that they can be re-arranged to \\(f\\left(x\\right)=0.\\) That is, the problem can be transformed to finding the root of a function \\(f\\left(x\\right)\\).1 A root of a function \\(f\\left(x\\right)\\) is the value of \\(x\\) for which \\(f\\left(x\\right)=0.\\)\r\n\r\nCommon numerical methods for root finding are:\r\n\r\n \tNewton\u2019s method (also called the Newton-Raphson method)\r\n \tthe Bisection method.\r\n\r\nIn this module we consider Newton\u2019s method.\r\n\r\nThe Newton-Raphson method\r\nNewton\u2019s method (Newton-Raphson) can provide the value of \\(x\\) such that \\(f\\left(x\\right)=0\\).\r\n\r\nIt has the form: \\[\\begin{align*} x_{n+1} & =x_{n}-\\frac{f\\left(x_{n}\\right)}{f'\\left(x_{n}\\right)} & \\left(1\\right) \\end{align*}\\] where \\(n\\) starts at \\(0.\\) The method is iterative, you start at \\(n=0\\) and continue by incrementing \\(n\\) by one until you achieve the desired accuracy.\r\n\r\nNote that:\r\n\r\n \t\\(x_{0}\\) is called the first estimate of the root, or the initial guess,\r\n \t\\(x_{n+1}\\) is the new estimate of the root, and \\(x_{n}\\) is the old estimate of the root,\r\n \t\\(f\\left(x_{n}\\right)\\) is the value of the function at \\(x_{n}\\) and,\r\n \t\\(f'\\left(x_{n}\\right)\\) is the value of the derivative of \\(f\\left(x\\right)\\) at \\(x_{n}.\\)\r\n\r\n\r\nExample 1\r\nFind the solution of \\(e^{-x}=x\\) for \\(0<x<1\\) to three decimal places.\r\nSolution\r\nFirst define the function:2 Note that you can also use \\(f\\left(x\\right)=x-e^{-x}\\) as it is equal to zero at the same value of \\(x\\) as \\(f\\left(x\\right)=e^{-x}-x.\\) \\[\\begin{align*} f\\left(x\\right) & =e^{-x}-x. \\end{align*}\\] We want to find \\(x\\) where \\(f\\left(x\\right)=0\\). To do this we need an initial guess \\(x_{0}\\) of the root. This is easily done by graphing both \\(y=x\\) and \\(y=e^{-x}.\\) The point where the graphs intersect is the point where \\(f\\left(x\\right)=0.\\) The graphs are shown below.\r\n\r\n\r\n\r\nA suitable value for \\(x_{0}\\) would be \\(0.5\\) and\r\n\r\n\\[\\begin{align*} f'\\left(x\\right) & =-e^{-x}-1 \\end{align*}\\]\r\n\r\nUsing eqn \\(\\left(1\\right)\\), we can start the method \\[\\begin{align*} x_{1} & =x_{0}-\\frac{f\\left(x_{0}\\right)}{f'\\left(x_{0}\\right)}\\\\ & =0.5-\\frac{f\\left(0.5\\right)}{f'\\left(0.5\\right)}\\\\ & =0.5-\\frac{e^{-0.5}-0.5}{-e^{-0.5}-1}\\\\ & =0.5663\\\\ x_{2} & =0.5663-\\frac{f\\left(0.5663\\right)}{f'\\left(0.5663\\right)}\\\\ & =0.5663-\\frac{e^{-0.5663}-0.5663}{-e^{-0.5663}-1}\\\\ & =0.5671\\\\ x_{3} & =x_{2}-\\frac{f\\left(x_{2}\\right)}{f'\\left(x_{2}\\right)}\\\\ & =0.5671-\\frac{e^{-0.5671}-0.5671}{-e^{-0.5671}-1}\\\\ & =0.5671. \\end{align*}\\] There has been no change to the first four places so we stop and the answer is \\(x=0.567\\,.\\)\r\n\r\nExample 2\r\nFind the solution of \\(-\\ln\\left(x\\right)=x^{3}\\) correct to five decimal places with an initial guess of \\(x_{0}=1.\\)\r\nSolution\r\nDefine the function \\[\\begin{align*} f\\left(x\\right) & =x^{3}+\\ln\\left(x\\right) \\end{align*}\\] then \\[\\begin{align*} f'\\left(x\\right) & =3x^{2}+\\frac{1}{x}. \\end{align*}\\] In this case we have a value for \\(x_{0}\\) so there is no need to graph the functions. But we do it anyway.3 It is a good idea to graph the functions because it allows us to get an idea of the answer. If Newton\u2019s method gives a result that is different to what we see in the graph, we have to consider that a mistake has been made or Newton\u2019s method has found another root that we did not see on our graph.\r\n\r\n\r\n\r\nFrom the graph we see the answer should be about \\(0.7\\).\r\n\r\nUsing Newton\u2019s method: \\[\\begin{align*} x_{1} & =x_{0}-\\frac{f\\left(x_{0}\\right)}{f'\\left(x_{0}\\right)}\\\\ & =1-\\frac{1^{3}+\\ln\\left(1\\right)}{3\\cdot1^{2}+\\left(\\frac{1}{1}\\right)}\\\\ & =1-\\frac{1+0}{3+1}\\\\ & =0.75\\\\ x_{2} & =0.75-\\frac{\\left(0.75\\right)^{3}+\\ln\\left(0.75\\right)}{3\\cdot\\left(0.75\\right)^{2}+\\left(\\frac{1}{0.75}\\right)}\\\\ & =0.7055775\\\\ x_{3} & =0.7055775-\\frac{\\left(0.7055775\\right)^{3}+\\ln\\left(0.7055775\\right)}{3\\cdot\\left(0.7055775\\right)^{2}+\\left(\\frac{1}{0.7055775}\\right)}\\\\ & =0.7047098\\\\ x_{4} & =0.7047098-\\frac{\\left(0.7047098\\right)^{3}+\\ln\\left(0.7047098\\right)}{3\\cdot\\left(0.7047098\\right)^{2}+\\left(\\frac{1}{0.7047098}\\right)}\\\\ & =0.7047095. \\end{align*}\\] As we have no change in the first five places after the decimal point we stop and take the answer as \\(x=0.70471\\,.\\) Note that we round up the fifth digit.\r\n\r\nConvergence of the Newton-Raphson method\r\nConvergence indicates how quickly you will get an acceptable value of \\(f\\left(x\\right)=0\\). Slow convergence implies many iterations, while fast convergence means few iterations.\r\n\r\nConvergence of the method depends on the function \\(f\\left(x\\right)\\) and the initial guess. If the initial guess is in an area where the gradient of \\(f\\left(x\\right)\\) is nearly zero, convergence can be slow.\r\n\r\nConsider the function \\(f\\left(x\\right)=x^{4}-1.\\) It is graphed below.\r\n\r\n\r\n\r\nObviously the solutions of \\(f\\left(x\\right)=0\\) are \\(x=-1\\) and \\(x=1\\). These can be obtained by basic algebra and there is no need to use Newton\u2019s method. However we will use this function to see what happens with different guesses for \\(x_{0}\\). We assume an accuracy to two decimal places and a calculation tolerance of \\(0.0000001\\).\r\n\r\nLet \\(x_{0}=0.1\\). In this region the gradient of \\(f\\left(x\\right)\\) is close to zero. Newton\u2019s method takes 25 iterations to get the result \\(x=1.00.\\)\r\n\r\nLet \\(x_{0}=0.25\\) then it takes 15 iterations to get the result \\(x=1.00.\\)\r\n\r\nLet \\(x_{0}=0.5\\) then it takes 9 iterations to get the result \\(x=1.00.\\)\r\n\r\nLet \\(x_{0}=0.75\\) then it takes 6 iterations to get the result \\(x=1.00.\\)\r\n\r\nSo we see the number of iterations decrease (higher convergence) when the gradient of \\(f\\left(x\\right)\\) increases.\r\n\r\nConvergence to another root\r\nIf the function \\(f\\left(x\\right)\\) has several roots, Newton\u2019s method may converge to any of these. For example the function \\(f\\left(x\\right)=x^{3}+2x^{2}-5x-6\\) has three roots at \\(x=-3,-1\\) and \\(2\\) as shown below.\r\n\r\n\r\n\r\nIf we set \\(x_{0}=-2\\) we, perhaps, would think the method should find the root \\(x=-3\\text{ or $-1$ }.\\) But Newton\u2019s method finds the root at \\(x=2.\\) This is why it is important to graph the function if possible.\r\n\r\n","n":0.035},[{"v":"Methodology","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":355,"$":[{"v":"NM3: The trapezoidal rule","n":0.5},{"v":"\r\nThe trapezoidal rule is a numerical method that calculates a numerical value for a definite integral. Some integrals cannot be evaluated in terms of the rules of integration or elementary functions.\r\nYou will have evaluated definite integrals such as \\[\\begin{align*} \\int_{1}^{3} & \\left(x^{2}\\right)dx \\end{align*}\\] before. In doing this, you are evaluating the area under the graph of \\(f\\left(x\\right)=x^{2}\\) between \\(x=1\\) and \\(x=3.\\) This is only possible if you can find an antiderivative for \\(x^{2}\\). In this example it is easy, the antiderivative is \\[\\begin{align*} F\\left(x\\right) & =\\frac{1}{3}x^{3}+c,\\text{ where $c$ is a constant.} \\end{align*}\\] and \\[\\begin{align*} \\int_{1}^{3}\\left(x^{2}\\right)dx & =F\\left(3\\right)-F\\left(1\\right)\\\\ & =\\frac{1}{3}3^{3}+c-\\left(\\frac{1}{3}1^{3}+c\\right)\\\\ & =\\frac{26}{3}. \\end{align*}\\]\r\n\r\nSometimes it is not possible to find the antiderivative. In such cases you need to use a numerical method. Two common methods for calculating definite integrals are:\r\n\r\n \tSimpson\u2019s rule, and\r\n \tThe trapezoidal rule.\r\n\r\nThis module considers the trapezoidal rule.\r\n\r\nThe trapezoidal rule\r\nThe trapezoidal rule works by estimating the area under a graph by a series of trapezoidal strips. In the figure below, we see an approximation to \\[\\begin{align*} \\int_{1}^{6} & xe^{-0.5x}dx \\end{align*}\\] using three strips. The approximated area is shown in red.\r\n\r\n\r\n\r\nIn this case, we see the trapezoidal rule will underestimate the first strip, is close in the second strip and will overestimate in the third strip. The trapezoidal rule approximates the area under the curve by adding the areas of the trapezoids. Any number of strips may be used. The accuracy increases as the number of strips increases.\r\n\r\nFor the definite integral \\[\\begin{align*} \\int_{a}^{b}f\\left(x\\right)dx \\end{align*}\\] the trapezoidal rule has the form \\[\\begin{align*} \\int_{a}^{b}f\\left(x\\right)dx & \\approx\\frac{h}{2}\\left[y_{0}+2y_{1}+2y_{2}+\\ldots+2y_{n-1}+y_{n}\\right] & (1) \\end{align*}\\]\r\n\r\nwhere\r\n\r\n \t\\(n\\) is the number of strips and can be any number.\r\n \t\\(y_{n}=f\\left(x_{n}\\right)\\) are the values of \\(f\\left(x_{n}\\right)\\) at the points \\(x_{i}\\) where \\(i=0,1,2,\\ldots,n\\). Note that \\(x_{0}=a,\\) \\(x_{n}=b.\\)\r\n \t\\(h\\) is the width of each strip and \\[\\begin{align*} h & =\\frac{b-a}{n}. \\end{align*}\\]\r\n \t\\(x_{1}=a+h,\\) \\(x_{2}=a+2h,\\) \\(x_{3}=a+3h,\\) \\(\\ldots\\) and so on.\r\n\r\n\r\nExample 1\r\nApproximate \\(\\int_{0}^{6}xe^{-0.5x}dx\\) using the trapezoidal rule with \\(3\\) strips , to \\(3\\) decimal places.\r\n\r\nWe have \\(n=3\\), \\(a=x_{0}=0\\) and \\(b=x_{3}=6\\) so \\[\\begin{align*} h & =\\frac{b-a}{n}\\\\ & =\\frac{6}{3}\\\\ & =2. \\end{align*}\\]\r\n\r\n\\(x_{0}=0\\) and \\(y_{0}=f\\left(x_{0}\\right)=0.\\)\r\n\r\n\\(x_{1}=0+h=2\\) and \\(y_{1}=f\\left(x_{1}\\right)=2e^{-1}=0.73576.\\)\r\n\r\n\\(x_{2}=0+2h=4\\) and \\(y_{2}=f\\left(x_{2}\\right)=4e^{-2}=0.54134.\\)\r\n\r\n\\(x_{3}=0+3h=6\\) and \\(y_{3}=f\\left(x_{3}\\right)=6e^{-3}=0.29872.\\)\r\n\r\nSubstituting into \\(\\left(1\\right)\\) above we get:\r\n\r\n\\[\\begin{align*} \\int_{0}^{6}xe^{-0.5x}dx & \\approx\\frac{2}{2}\\left[y_{0}+2y_{1}+2y_{2}+y_{3}\\right]\\\\ & =0+2\\left(0.73576\\right)+2\\left(0.54134\\right)+0.29872\\\\ & =2.85292. \\end{align*}\\] Hence, to three decimal places, \\(\\int_{0}^{6}xe^{-0.5x}dx\\approx2.853.\\)\r\n\r\nExample 2\r\nApproximate \\(\\int_{0}^{1}\\sqrt{1+x^{3}}dx\\) using the trapezoidal rule with \\(5\\) strips , to 3 decimal places.\r\n\r\nIn this question, \\(n=5\\), \\(a=0\\) and \\(b=1\\) so \\[\\begin{align*} h & =\\frac{b-a}{n}\\\\ & =\\frac{1}{5}\\\\ & =0.2\\,. \\end{align*}\\] So using the formula above we get\r\n\r\n\\(x_{0}=0\\) and \\(y_{0}=f\\left(x_{0}\\right)=1.\\)\r\n\r\n\\(x_{1}=0+h=0.2\\) and \\(y_{1}=f\\left(x_{1}\\right)=\\sqrt{1+\\left(0.2\\right)^{3}}=1.00399.\\)\r\n\r\n\\(x_{2}=0+2h=0.4\\) and \\(y_{2}=f\\left(x_{2}\\right)=\\sqrt{1+\\left(0.4\\right)^{3}}=1.03150.\\)\r\n\r\n\\(x_{3}=0+3h=0.6\\) and \\(y_{3}=f\\left(x_{3}\\right)=\\sqrt{1+\\left(0.6\\right)^{3}}=1.10272.\\)\r\n\r\n\\(x_{4}=0+4h=0.8\\) and \\(y_{4}=f\\left(x_{4}\\right)=\\sqrt{1+\\left(0.8\\right)^{3}}=1.22963.\\)\r\n\r\n\\(x_{5}=0+5h=1.0\\) and \\(y_{5}=f\\left(x_{5}\\right)=\\sqrt{1+\\left(1\\right)^{3}}=1.41421.\\)\r\n\r\nSubstituting into Substituting into \\(\\left(1\\right)\\) above we get:\r\n\r\n\\[\\begin{align*} \\int_{0}^{1}\\sqrt{1+x^{3}}dx & \\approx\\frac{0.2}{2}\\left[y_{0}+2y_{1}+2y_{2}+2y_{3}+2y_{4}+y_{5}\\right]\\\\ & =0.1\\left[1+2\\left(1.00399\\right)+2\\left(1.03150\\right)\\right.\\\\ & \\quad\\quad\\left.+2\\left(1.10272\\right)+2\\left(1.22963\\right)+1.41421\\right]\\\\ & =1.11499. \\end{align*}\\] Hence, to three decimal places, \\(\\int_{0}^{1}\\sqrt{1+x^{3}}dx\\approx1.115.\\)\r\n\r\nExercises\r\n\r\n \tUse the trapezoidal rule to evaluate the integral \\(\\int_{1}^{3}\\left(3x^{2}+4x\\right)dx\\) using 3 strips to two decimal places.\r\n \tUse the trapezoidal rule to evaluate the integral \\(\\int_{0}^{1}\\left(3e^{x}\\sin\\left(x\\right)\\right)dx\\) using 2 strips to two decimal places.\r\n\r\n\r\n\u00a0\r\n\r\n","n":0.046},[{"v":"Methodology","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":356,"$":[{"v":"Finding Australian case law and legislation","n":0.408},{"v":"This interactive video will step you through finding relevant cases and outcomes across Australia. Look for databases and case notes relevant to commonwealth and state cases.\r\n\r\nInteractive video: includes quizzes and links to websites. Quizzes and links appear as boxes in the video window. Select them to test your knowledge or visit the site.\r\n\r\n\r\n\r\n\r\nFinding Australian legislation\r\n\r\nThis interactive video will step you through finding relevant Australian legislation. Find state and commonwealth legislation - both present versions and past iterations. \r\n\r\n\r\n\r\n\r\n\r\nReferencing\r\nAll assessments need to follow the referencing guidelines for that school. Subjects in the law faculty require referencing in AGLC4 format. For other courses, check the assignment instructions.\r\n\r\nUse Easy Cite to check how to correctly reference in a variety of referencing systems, including AGLC4 and Harvard. Simply choose your referencing style, select the item type, and Easy Cite will show you how accurately cite your sources.\r\n\r\nFor more information about when to reference and what to include, visit the Learning Lab's referencing page to learn more. The Library's Law and Justice subject guide has information about referencing and links to law databases.\r\n \r\n\r\nThis video is about searching for Australian legislation.  \r\n\r\nWhen you\u2019re searching for legislation, there are some important things to consider.   \r\n\r\nMost states and territories will have their own legislation covering an area of law, so check that you're searching in the correct jurisdiction. \r\n\r\nEnsure you are accessing the most current or authorised version of the legislation, including all amendments.  \r\n\r\nIf required, you may need to find an older version of the act that was in force at the time the matter occurred. \r\n\r\nEach state, territory and the federal government has a website where you can access their legislation.   \r\n\r\nThese government websites provide the authorised versions of legislation as well as previous or historical versions. \r\n\r\nLawOne is a library database, and you can access it from the Law and Justice Subject Guide, or LibrarySearch. \r\n\r\nLawOne provides full text access to legislation covering all nine Australian Federal, State and Territory jurisdictions, and it\u2019s updated daily. \r\n\r\nYou can find current, repealed and amending Acts and Regulations, as well as draft legislation and extrinsic material. \r\n\r\nYou can Browse or Search LawOne and refine by legislation type and jurisdiction. \r\n\r\nOnce you have found the relevant Act, you can search within the Act or use the + sign on the left-hand side of the screen to expand out the provisions. \r\n\r\nOne of the best features of LawOne is Key Info.  \r\n\r\nFrom here you can access and download the authorised, current and historical versions of the Act as well as commencement details, amendment history and the original Bill. You will also find extrinsic materials that help to explain and interpret the legislation and related instruments, including cases that have considered the legislation. \r\n\r\nTo find out more about searching for legal resources, visit the Law and Justice subject guide. \r\n\r\n","n":0.046},[{"v":"Law","i":2,"n":1},{"v":"Databases","i":1,"n":1},{"v":"Case notes","i":0,"n":0.707}]]},{"i":357,"$":[{"v":"Writing case notes","n":0.577},{"v":"Case notes usually require you to provide a summary of the case which outlines the relevant facts, explains the legal issues and the judge's reasoning (ratio). This is followed by a critique of the judge's decisions and a discussion of the implications of the case. In doing this, you are evaluating the court's decision.\r\n\r\nWriting process\r\nIdentify the purpose of the case notes according to the assignment briefing and the marking rubric. Read the case - skim first to understand the issue and context; reread (as often as necessary) to pull out the key facts and points of law.\r\n\r\nUndertake research using reliable authoritative sources. Begin with relevant course readings; look for existing case notes on the case, peer-reviewed articles or other cases to support your arguments. See the library's Law and Justice subject guide for tips on accessing legal sources.\r\n\r\nBrainstorm your response to the task. Consider the legal principles covered in your lectures and recommended readings for the course.\r\n\r\nPlan the document; organisation can vary. Outline your structure based on the task instructions and decide on a logical order to represent your ideas (e.g. according to judicial reasoning or court rulings).\r\n\r\nWrite your paper using your plan and research. Avoid unnecessary description of case details. Provide an introduction with a brief explanation of the case to lay the foundation for your analysis and critique of the case decision.\r\n\r\nEdit and review. Is it logical, clear and concise? Have you addressed all the elements of the marking rubric? Are all sources cited?\r\n\r\n Proofread. Check for grammatical errors that can reduce clarity. Spelling mistakes, typos and inconsistencies can cost you marks.\r\n\r\nYou can find more information on assignment writing at Getting started with assignments.\r\n\r\nStructure\r\nThere is no one way to structure case notes. The table below outlines requirements and possible elements of case notes. Always refer to assignment instructions.\r\n\r\n\r\nStyle considerations\r\nKey features of the professional, academic style required in writing case notes includes:\r\n\r\n \tformal language; avoid idiom or slang, using I, no contractions\r\n \tan objective tone that avoids emotive language\r\n \theadings, sub-headings and numbering to separate your arguments\r\n \tparagraphs and complete sentences to develop your position, not dot points\r\n \targuments based on evidence from case reports and scholarly sources, which must be cited using the AGLC4 referencing style.\r\n","n":0.052},[{"v":"Law assessments","i":2,"n":0.707},{"v":"Law","i":1,"n":1},{"v":"Case notes","i":0,"n":0.707}]]},{"i":358,"$":[{"v":"Writing a legal memo","n":0.5},{"v":"Legal memorandums that are written as assignments can take three forms: an in-house document, a letter to a client, or a paper on legal policy. All memos provide advice or legal opinion but can have different audiences or intended recipients. They should be clear, concise and informative.\r\n\r\n\r\n\r\n\r\n\r\nThe memo summarises and analyses any relevant laws in relation to the facts of a particular situation. It is the culmination of your legal research. Your memo should be well organised to reflect your analysis and reasoning. There is no one correct way to present your memo. Different lecturers and firms can have their own preferences even though the memo will contain the same elements. At university, check the assignment brief and marking rubric to be clear about the task requirements. If you\u00a0are not sure, ask your lecturer.\r\nWriting process\r\n\r\n \tIdentify the purpose according to the task\r\n \tIdentify the matters and any sub-issues to be discussed\r\n \tUndertake research using reliable authoritative sources\r\n \tPlan the document (content and structure)\r\n \tWrite using your plan and research\r\n \tEdit and review; is it logical and clear?\r\n\r\nVisit the Library's Law and Justice subject guide to access major Australian legal resources and help on searching them.\r\n\r\nAcknowledge sources accurately to avoid plagiarism. Information on referencing can be found on the Learning Lab. See EasyCite for guidance on using AGLC4 referencing.\r\n\r\nAlso, be aware of your reader as this influences the language you use to communicate your reasoning. The style should be formal and professional, and language and tone suitable for the recipient; for example, avoid technical terms and legal jargon when writing for a client.\r\nStructuring the memo\r\nThe following is a suggested model structure. However, this should be adapted according to needs of the memo or the guidelines of your lecturer.\r\n\r\n\r\n\r\n","n":0.059},[{"v":"Professional writing","i":3,"n":0.707},{"v":"Legal","i":2,"n":1},{"v":"Law assessments","i":1,"n":0.707},{"v":"Law","i":0,"n":1}]]},{"i":359,"$":[{"v":"Assessments for law","n":0.577},{"v":"These tutorials will assist you in completing law assessments such as writing a memo or a case note, as well as finding Australian case law and legislation.\r\n\r\n\r\n\r\n\r\n","n":0.192},[{"v":"Law assessments","i":1,"n":0.707},{"v":"Law","i":0,"n":1}]]},{"i":360,"$":[{"v":"The sine rule","n":0.577},{"v":"\r\nThe trigonometric ratios are handy for real-life situations where right-angled triangles are involved, but what about triangles that are not right-angled? We use the sine rule. Use this resource to learn what the sine rule is and how it is applied.\r\n\r\nThe sine rule shows that the ratio of the length of a side, to the sine of its opposite angle, will be the same for all three sides.\r\nConsider the triangle.\r\n\r\nHere:\r\n\r\n\tthe vertices \\(A\\), \\(B\\) and \\(C\\) have the angles \\(A\\), \\(B\\) and \\(C\\), respectively\r\n \t\\(a\\), \\(b\\) and \\(c\\) are the side lengths opposite to the angles \\(A\\), \\(B\\) and \\(C\\), respectively.\r\n\r\nThe sine rule states that:\r\n\r\n\\[\\frac{a}{\\sin(A)} = \\frac{b}{\\sin(B)} = \\frac{c}{\\sin(C)}\\]\r\nor\r\n\\[\\frac{\\sin(A)}{a} = \\frac{\\sin(B)}{b} = \\frac{\\sin(C)}{c}\\]\r\n\r\nThis rule can be used to find angles and sides in any triangle (not just a right-angled triangle) when given:\r\n\r\n \tone side and any two angles; or\r\n \ttwo sides and an angle opposite one of the given sides.\r\n\r\nExample 1 \u2013 using the sine rule\r\nConsider triangle \\(PQR\\).\r\n\r\nFind:\r\n\r\n \tthe side length \\(p\\)\r\n \tthe side length \\(q\\).\r\n\r\nWe have been given two angles and one side length, so we can use the sine rule.\r\n\\[\\begin{align*} \\frac{p}{\\sin(P)} & = \\frac{r}{\\sin(R)}\\\\ \r\n\\frac{p}{\\sin(70^{\\circ})} & = \\frac{15}{\\sin(30^{\\circ})}\\\\ \r\np & = \\frac{15\\times(\\sin70^{\\circ})}{\\sin(30^{\\circ})}\\\\ \r\n& = 28.2\\textrm{ cm}\\\r\n\\end{align*}\\]\r\nThe side length \\(p\\) is \\(28.2\\textrm{ cm}\\).\r\nWe can find the missing angle easily if we remember that the sum of the three interior angles of a triangle is \\(180^{\\circ}\\).\r\n\\[\\begin{align*} \\angle Q & = 180^{\\circ}-\\left(70^{\\circ}+30^{\\circ}\\right)\\\\\r\n& =80^{\\circ}\r\n\\end{align*}\\]\r\nNow that we have angle \\(Q\\), we can use the sine rule to find side length \\(q\\).\r\n\\[\\begin{align*} \\frac{q}{\\sin(Q)} & = \\frac{r}{\\sin(R)}\\\\ \r\n\\frac{q}{\\sin(80^{\\circ})} & = \\frac{15}{\\sin(30^{\\circ})}\\\\\r\nq & = \\frac{15\\times\\sin(80^{\\circ})}{\\sin(30^{\\circ})}\\\\ \r\n& = 29.6\\textrm{ cm}\r\n\\end{align*}\\]\r\nThe side length \\(q\\) is \\(29.6\\textrm{ cm}\\).\r\n\r\n\r\n\r\n\r\n\r\nThe ambiguous case\r\nThe ambiguous case of the sine rule occurs when you're given two side lengths and a non-included angle. This is an angle that is not between the two known side lengths.\r\nThe ambiguous case results in either two valid solutions or no solution. There is an easy way to determine which of these is at play:\r\n\r\n\tIf the original angle and the calculated angle add up to less than \\(180^{\\circ}\\), there are two valid solutions.\r\n\tIf the original angle and the calculate angle add up to more than \\(180^{\\circ}\\), there are no valid solutions.\r\n\r\n\r\nExample 3 \u2013 using the sine rule\r\nConsider triangle \\(ABC\\) with angle \\(A=30^{\\circ}\\), side \\(c=12\\textrm{ cm}\\) and side \\(a=8\\textrm{ cm}\\). Find angle \\(C\\).\r\n\r\n\r\nWhen we use the sine rule:\r\n\\[\\begin{align*} \\frac{\\sin(A)}{a} & = \\frac{\\sin(C)}{c}\\\\ \r\n\\frac{\\sin(30^{\\circ})}{8} & = \\frac{\\sin(C)}{12}\\\\\r\n\\frac{\\sin(30^{\\circ})}{8}\\times12 & = \\sin(C)\\\\ \r\n\\sin(C) & = \\frac{1}{16}\\times12\\\\\r\n& = 0.75\\\\ \r\nC & = \\sin^{-1}(0.75)\\\\\r\n& = 48.6^{\\circ}\r\n\\end{align*}\\]\r\nTo determine whether the ambiguous case applies, we can add up the original angle given and the new calculated angle.\r\n\\[\\begin{align*} A+C & = 30^{\\circ}+48.6^{\\circ}\\\\\r\n& = 78.6^{\\circ}\\\\\r\n& < 180^{\\circ}\r\n\\end{align*}\\]\r\nThe original angle and the calculated angle add up to less than \\(180^{\\circ}\\), so there are two valid solutions.\r\nTo find the other solution, we just need to minus the calculated angle from \\(180^{\\circ}\\).\r\n\\[\\begin{align*} C & =180^{\\circ}-48.6^{\\circ}\\\\\r\n& = 131.4^{\\circ}\r\n\\end{align*}\\]\r\nIf \\(C=48.6^{\\circ}\\), the triangle is \\(ABC_{2}\\). If \\(C=131.4^{\\circ}\\), the triangle is \\(ABC_{1}\\).\r\n\r\nExercise \u2013 using the sine rule\r\n\r\n\tFind the unknown sides for the following triangles.\r\n\t\r\n\t\t  \r\n\t\t  \r\n\t\t  \r\n\t\r\n\tFind the unknown sides and angles for the following triangles.\r\n\t\r\n\t\t  \r\n\t\t  \r\n\t\r\n\tGiven triangle \\(ABC\\), where angle \\(A=35\\), side \\(a=16\\) and side \\(c=21\\), find the magnitude of angles \\(B\\) and \\(C\\) and the length of side \\(b\\).\r\n\r\n\r\n\r\n\r\n\r\n","n":0.043},[{"v":"Trigonometry","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Functions","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":361,"$":[{"v":"Pythagoras\u2019 theorem","n":0.707},{"v":"\r\nPythagoras\u2019 theorem shows the relationship between the sides of a right-angled triangle. Knowing the length of two sides of a right-angled triangle, the length of the third side can be calculated. Understanding Pythagoras' theorem helps you solve problems in fields such as architecture, engineering, and computer graphs, where precise calculations of distances and lengths are crucial but can be difficult to physically measure.\r\n\r\nVideo tutorial \u2013 Pythagoras' theorem\r\nWatch this video to learn how to use Pythagoras' theorem.\r\n\r\n\r\n\r\nRight-angled triangles\r\nTriangles are plane shapes with three straight sides. A right-angled triangle contains an angle of \\(90^{\\circ}\\), that is, a right angle. \r\nA right angle is formed when two lines meet at an angle of \\(90^{\\circ}\\). Usually, this is shown by two perpendicular lines.\r\n\r\nIn a right-angled triangle, the longest side is opposite the right angle and is called the hypotenuse.\r\n\r\nPythagoras\u2019 theorem\r\nPythagoras' theorem states that, in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. This is represented by:\r\n\\[a^{2}+b^{2}=h^{2}\\]\r\nwhere \\(h\\) is the length of the hypotenuse, and \\(a\\) and \\(b\\) are the lengths of the other two sides.\r\n\r\nIn some other resources, you may see that the hypotenuse is given the letter \\(c\\) instead of \\(h\\).\r\nPythagoras\u2019 theorem does not apply to any triangle. It is only true for right-angled triangles.\r\nPythagoras\u2019 theorem may be used to find the length of the third side of a triangle if you know the length of the other two sides.\r\n\r\nExample 1 \u2013 using Pythagoras' theorem\r\nFind the length of the hypotenuse in the triangle shown.\r\n\r\n\\[\\begin{align*} h^{2} & = a^{2}+b^{2}\\\\ \r\n& = 6^{2}+8^{2}\\\\\r\n& = 36+64\\\\ \r\n& = 100\\\\\r\nh & = \\sqrt{100}\\\\\r\n& = 10\r\n\\end{align*}\\]\r\nDon\u2019t forget to include the unit of measurement in your answer! The length of the hypotenuse is \\(10\\textrm{ cm}\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 using Pythagoras' theorem\r\n\r\n\tFind the missing side length in the triangle.\r\n\r\n\tFind the missing side length in the triangle.\r\n\r\n\r\n\r\n\r\nPythagorean triples\r\nIn some right-angled triangles, the lengths of all three sides are integers. When this occurs, the lengths are called a Pythagorean triple.\r\nSome examples are \\(\\left(3,4,5\\right)\\), \\(\\left(5,12,13\\right)\\), \\(\\left(7,24,25\\right)\\) and \\(\\left(8,15,17\\right)\\). Multiples of these numbers are also Pythagorean triples.\r\n\r\n \r\nThis is a short video on Pythagoras\u2019 theorem. Let\u2019s start with the definition of a right angled triangle first. It\u2019s a triangle with one angle that\u2019s 90 degrees, the longer side is opposite the right angle and is called the hypotenuse. Now let\u2019s state Pythagoras\u2019 theorem. It states that for any right angled triangle the square on the hypotenuse, h, is equal to the sum of the squares on the two shorter sides, a and b. In other words, h2 equals a2 plus b2.\r\n\r\nLet\u2019s do an example. Let\u2019s find the length of the hypotenuse of this particular triangle. We\u2019re given the two shorter sides, a and b, so write the formula first at the top, h2 equals a2 plus b2 and substitute in for a and b, four and three. So we have 42 plus 32. Notice 42 is four times four, not four times two. So four time four is 16, 32 is three times three which is nine. So h2 equals 25. So if h2 is 25, then h is the square root of 25, which is five. In other words, the length of the hypotenuse is five.\r\n\r\nNow let\u2019s find the length of one of the shorter sides of the triangle. Here we want to find the length of a. We\u2019re given the hypotenuse. So as before we substitute into the formula, but notice I\u2019ve turned the formula around this time and you\u2019ll see why in a moment. In other words, I\u2019ve got a2 plus b2 equals h2. Substituting b is six, h is 10, so a2 plus 36 equals 100. Now I want to remove 36 from the left hand side to isolate a. And that\u2019s why I\u2019ve turned the formula around this time, to make it easier, \u2018cause I\u2019m leaving a on the left hand side. So I\u2019m subtracting 36, but remember we must subtract from both sides. So 36 minus 36 on the left hand side, 100 minus 36 on the right hand side. Therefore a2 equals 64, from which a is the square root of 64, which is eight. So the length of side a is 8.\r\n\r\nNow let\u2019s look at the last triangle and we want to find the length of b, which is sometimes called the perpendicular side, because it\u2019s at right angles to another side, in this case, a. a will be the horizontal line, b will be the perpendicular line, and the angle between the two will be 90 degrees. So substituting into the formula as before, we have a equals 24, and h is 25. The numbers are a bit larger but it still doesn\u2019t change the method by which we\u2019re doing the problem. So 242 plus b2 equals 252. 576 is 24 times 24, plus b2 equals 625, which is 25 times 25. And as before we remove the number from the left hand side to isolate the b. So we subtract 576 from the right hand side. 625 minus 576 gives us 49. If b2 equals 49 b is now the square root of 49 which is seven. In other words. The length of side b is seven.\r\n","n":0.034},[{"v":"Trigonometry","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Functions","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":362,"$":[{"v":"Linear motion","n":0.707},{"v":"Linear motion is how an object moves in a straight line. To describe these motions, we need to know some technical terms like displacement, distance, velocity, speed and acceleration, and how to represent how they change over time. Use this resource to learn about the terms and their relationships to one another.\r\n\r\nBefore you begin, make sure you understand the basics of scalar and vector quantities.\r\n\r\nDescribing linear motion\r\nIn linear motion, these quantities help us describe not just how far something moves, but how fast it moves and how its motion changes over time. Let's define each quantity and show how they are connected when an object travels in a straight line.\r\n\r\nDisplacement \\(\\left(x\\right)\\)\r\nWhen an object is moved from one point to another, it is said to be displaced. If we start in Melbourne and travel to Dandenong, we can say that we have been displaced \\(32\\,\\text{km}\\) from Melbourne. To then define our position, we need to specify a direction: we are \\(32\\,\\text{km}\\) South East of Melbourne. That is, we give a magnitude \\(\\left(32\\,\\text{km}\\right)\\) and a direction (South East). This makes position, or displacement, a vector quantity. Displacement is often used to show the position of a something relative to a specific point.\r\nIn physics, we assign a positive sign to a specific direction. For example, if East is the positive direction, then travelling \\(2\\,\\text{km}\\) East is \\(+2\\,\\text{km}\\). This would make travelling \\(2\\,\\text{km}\\) West, \\(-2\\,\\text{km}\\).\r\n\r\nDistance \\(\\left(d\\right)\\)\r\nDistance is the magnitude (or size) of the displacement. It is the length of the path taken by something. It has no direction, so it is a scalar quantity. For example, you can say your distance is \\(60\\,\\text{km}\\) away, but your displacement would be \\(60\\,\\text{km}\\) in a direction, such as \\(60\\,\\text{km}\\) West.\r\n\r\nVelocity \\(\\left(v\\right)\\)\r\nVelocity is the rate of change of displacement, \\(x\\). That is, for an object moving at constant velocity:\r\n\\[v=\\frac{x}{t}\\]\r\nwhere \\(t\\) is the time taken.\r\nSince displacement is a vector quantity, so is velocity.\r\nThe units of velocity depend on the units of displacement and time. If displacement is in metres \\(\\left(\\text{m}\\right)\\) and time is in seconds \\(\\left(\\text{s}\\right)\\), the unit of velocity is metres per second \\((\\text{m/s}\\) or \\(\\text{m}\\,\\text{s}^{-1})\\).\r\n\r\nSpeed\r\nThe speed of an object is the magnitude of its velocity. It can also be calculated by:\r\n\\[\\text{speed}=\\frac{d}{t}\\]\r\nwhere \\(t\\) is the time taken.\r\nIt has no direction, making it a scalar quantity.\r\n\r\nAcceleration \\(\\left(a\\right)\\)\r\nAcceleration is the rate of change of velocity:\r\n\\[a=\\frac{v}{t}\\]\r\nIf \\(v\\) is in metres per second, the unit of acceleration is metres per second per second: \\(\\text{m/s}^{2}\\) or \\(\\text{m}\\,\\text{s}^{-2}\\). Since velocity is a vector quantity, so is acceleration. If acceleration is constant, we use the expression 'the acceleration is uniform'.\r\n\r\nRepresenting linear motion\r\nWe can describe linear motion using graphs that show how displacement, velocity and acceleration change over time.\r\n\r\nDistance vs time graphs\r\nLet\u2019s suppose a person, starting from rest, runs a distance \\(d\\) of \\(12\\,\\text{m}\\) in a time \\(t\\) of \\(2\\,\\text{s}\\). Assuming that their speed is constant, the distance\u2013time graph would look like this:\r\n\r\nThe gradient of a distance (a scalar quantity) versus time graph gives the speed (also a scalar quantity). In this case, the speed is given by:\r\n\\[\\begin{align*} \\text{speed} & = \\frac{d}{t} \\\\[6pt]\r\n& = \\frac{12}{2}\\\\[6pt]\r\n& = 6\\text{ m/s}\r\n\\end{align*}\\]\r\n\r\nDisplacement vs time graphs\r\nConsider the displacement\u2013time graph:\r\n\r\nThe gradient of a displacement (a vector quantity) versus time graph gives velocity (also a vector quantity). Expand the accordions to learn what is happening between each point on the graph.\r\n\r\n\r\n\r\n\r\n\r\nVelocity vs time graphs\r\nThe velocity\u2013time graph shows a car accelerating uniformly from rest to \\(60\\,\\text{m/s}\\) in \\(20\\,\\text{s}\\), then travels at a constant velocity of \\(60\\,\\text{m/s}\\) for the next \\(20\\,\\text{s}\\), then decelerates uniformly to rest in \\(20\\,\\text{s}\\). The total journey takes \\(60\\,\\text{s}\\).\r\n\r\nFrom A to B, the object accelerates from \\(0\\,\\text{m/s}\\) to \\(60\\,\\text{m/s}\\). From B to C, the object moves at a constant velocity of \\(60\\,\\text{m/s}\\) and from C to D, the object slows to a stop from \\(60\\,\\text{m/s}\\).\r\nLet's look at some features of the graph in detail: the area under the graph and its gradient.\r\n\r\nArea under velocity\u2013time graph\r\nThe area is the product of velocity and time, so the units are \\(\\text{m}/\\text{s}\\,\\times \\text{s}=\\text{m}\\). That is, the area under the graph gives displacement.\r\n\r\nThe area under a velocity\u2013time graph gives displacement.\r\nThe area under a speed\u2013time graph gives distance.\r\n\r\nConsider the velocity\u2013time graph again. The area has been divided into three regions: \\(1\\), \\(2\\) and \\(3\\). Let the area of each region be \\(A_{1}\\), \\(A_{2}\\) and \\(A_{3}\\), respectively. We have:\r\n\\[\\begin{align*} A_{1} & = \\frac{1}{2}\\times b\\times h \\\\[6pt]\r\n& = \\frac{1}{2}\\times20\\times60 \\\\[6pt]\r\n& = 600\\,\\text{m} \\end{align*}\\] \\[\\begin{align*} A_{2} & = b\\times h \\\\[6pt]\r\n& = 20\\times60\\\\[6pt]\r\n& = 1200\\,\\text{m} \\end{align*}\\] \\[\\begin{align*} A_{3} & = \\frac{1}{2}\\times b\\times h \\\\[6pt]\r\n& = \\frac{1}{2}\\times20\\times60\\\\[6pt]\r\n& = 600\\,\\text{m} \\end{align*}\\]\r\nThe total displacement is therefore: \\(A_{1}+A_{2}+A_{3}=600+1200+600=2400\\,\\text{m}\\).\r\nCheck out the further resources at the end of this page if you need a refresher on calculating the area of triangles and rectangles.\r\n\r\nGradient of a velocity\u2013time graph\r\nThe gradient (or slope) of a velocity\u2013time graph gives the acceleration of the object. It is calculated using:\r\n\\[m=\\frac{\\text{rise}}{\\text{run}}\\]\r\nExpand the accordions to learn what is happening between each point on the graph.\r\n\r\n\r\n\r\n\r\nWhat the gradient and area under each graph represents is summarised in the following table.\r\n\r\nExample 1 \u2013 sketching and describing linear motion graphs\r\nA car accelerates from a stationary position. The rate of acceleration is constant. Sketch graphs of the acceleration versus time, the velocity versus time and the displacement versus time.\r\nSince the acceleration is constant, the acceleration\u2013time graph will simply be a horizontal line.\r\n\r\nAs the car is accelerating, this means that its velocity is increasing over time. The velocity\u2013time graph is a linear graph with a gradient equal to acceleration.\r\n\r\nDisplacement is the total distance travelled, so we get it by looking at how fast the car is going during each time interval. At the start, the car is moving slowly, so it does not cover much distance in the first few seconds. Later, it is moving faster, so it covers more distance in each second than before.\r\nThis means the displacement does not increase at a steady rate \u2013 it increases more and more quickly as time goes on, creating a curve.\r\n\r\nThe gradient (slope) at any point on the curve is the velocity at that time, or the instantaneous velocity. You can find the approximate value of the gradient by drawing a tangent at the point of interest and then calculating the gradient of the tangent.\r\n\r\n\r\n\r\nExercise \u2013 sketching and describing linear motion graphs\r\n\r\n \tThe graph shows the position (displacement) of a dancer moving in a straight line across a stage. The dancer\u2019s movements are designated by sections A to D.\r\n\t\r\n\t\t\r\n\t\t\tDetermine the dancer's starting position.\r\n\t\t\tDetermine in which section (A\u2014D) the dancer is at rest.\r\n\t\t\tDetermine in which section the dancer is moving in a positive direction.\r\n\t\t\tDetermine in which sections the dancer is moving with a negative velocity.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tThe graph represents the straight line motion of a radio-controlled toy car.\r\n\t\r\n\t\t\r\n\t\t\tDescribe, in words, the motion of the car.\r\n\t\t\tDetermine the position of the car at:\r\n\t\t\t\t\r\n\t\t\t\t\t\\(2\\,\\text{s}\\)\r\n\t\t\t\t\t\\(4\\,\\text{s}\\)\r\n\t\t\t\t\t\\(6\\,\\text{s}\\)\r\n\t\t\t\t\t\\(10\\,\\text{s}\\).\r\n\t\t\t\t\r\n\t\t\t\r\n\t\t\tDetermine at what time the car returned to its starting point.\r\n\t\t\tCalculate the velocity of the car:\r\n\t\t\t\t\r\n\t\t\t\t\tduring the first \\(2\\,\\text{s}\\)\r\n\t\t\t\t\tat \\(3\\,\\text{s}\\)\r\n\t\t\t\t\tfrom \\(4\\,\\text{s}\\) to \\(6\\,\\text{s}\\)\r\n\t\t\t\t\tat \\(8\\,\\text{s}\\)\r\n\t\t\t\t\tfrom \\(8\\,\\text{s}\\) to \\(9\\,\\text{s}\\).\r\n\t\t\t\t\r\n\t\t\t\r\n\t\t\tDuring its \\(10\\,\\text{s}\\) motion, calculate the car's:\r\n\t\t\t\t\r\n\t\t\t\t\tdistance travelled\r\n\t\t\t\t\tdisplacement.\r\n\t\t\t\t\r\n\t\t\t\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tThe following position\u2013time graph is for a cyclist travelling along a straight road.\r\n\t\r\n\t\t\r\n\t\t\tDescribe, in words, the motion of the cyclist.\r\n\t\t\tCalculate the velocity of the cyclist during the first \\(30\\,\\text{s}\\).\r\n\t\t\tCalculate the cyclist\u2019s velocity during the final \\(10\\,\\text{s}\\).\r\n\t\t\tEstimate the cyclist\u2019s instantaneous velocity at \\(35\\,\\text{s}\\).\r\n\t\t\tDetermine the average velocity of the cyclist between \\(30\\,\\text{s}\\) and \\(40\\,\\text{s}\\).\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tConsider the velocity\u2013time graphs, A to E. \r\n\t\r\n\tIdentify which velocity-time graph best represents the motion of:\r\n\t\t\r\n\t\t\ta car coming to a stop at a traffic light\r\n\t\t\ta swimmer moving at a constant speed\r\n\t\t\ta cyclist accelerating from rest with constant acceleration\r\n\t\t\ta car accelerating from rest and changing through its gears.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tThe following graph shows the motion of a dog running along a footpath in a northerly direction.\r\n\t\r\n\t\t\r\n\t\t\tDescribe the motion of the dog during each section of the graph (A to F).\r\n\t\t\tCalculate the displacement of the dog after:\r\n\t\t\t\t\r\n\t\t\t\t\t\\(2\\,\\text{s}\\)\r\n\t\t\t\t\t\\(7\\,\\text{s}\\)\r\n\t\t\t\t\t\\(10\\,\\text{s}\\)\r\n\t\t\t\t\r\n\t\t\t\r\n \t\t\tPlot a position\u2013time graph for the dog\u2019s motion.\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tThe straight line motion of a train is shown in the following velocity\u2013time graph.\r\n\t\r\n\t\t\r\n\t\t\tDetermine how long the train takes to reach its cruising speed.\r\n\t\t\tCalculate the acceleration of the train \\(10\\,\\text{s}\\) after starting. (Hint: You can only approximate this. Take a tangent to the graph at \\(10\\,\\text{s}\\) and calculate its gradient.)\r\n\t\t\tCalculate the acceleration of the train \\(40\\,\\text{s}\\) after starting\r\n\t\t\tCalculate the approximate displacement of the train after \\(120\\,\\text{s}\\).\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n\r\n\tThe velocity\u2013time graphs for a bus and a bicycle travelling along the same straight stretch of road are shown in the graph. The bus is initially at rest and starts moving as the bicycle passes it.\r\n\t\r\n\t\t\r\n\t\t\tCalculate the initial acceleration of the bus.\r\n\t\t\tDetermine at what time the bus first starts gaining ground on the bicycle.\r\n\t\t\tDetermine at what time the bus overtakes the bicycle.\r\n\t\t\tDetermine how far the bicycle has traveled before the bus catches up to it.\r\n\t\t\tDetermine the average velocity of the bus during the first \\(8\\,\\text{s}\\).\r\n\t\t\tDraw an acceleration\u2013time graph for the bus.\r\n\t\t\tUse your acceleration\u2013time graph to determine the change in velocity of the bus over the first \\(8\\,\\text{s}\\).\r\n\t\t\r\n\t\r\n\r\n\r\n\r\n","n":0.026},[{"v":"Kinematics","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Force","i":1,"n":1},{"v":"Physics","i":0,"n":1}]]},{"i":363,"$":[{"v":"Static equilibrium","n":0.707},{"v":"When a bridge holds steady under traffic, or a bookshelf stays put under a heavy load, the structure is in static equilibrium. This means every force and every turning effect is perfectly balanced \u2014 nothing accelerates, nothing rotates. Use this resource to explore the two conditions for static equilibrium and learn how to apply them to real situations.\r\n\r\nWhen objects are at rest, they are said to be in static equilibrium: the net force and moment on the object are both zero, meaning that:\r\n\r\n\tthe sum of the forces on the object in each direction is zero, i.e. there is translational equilibrium\r\n\tthe sum of the torques (or moments) in each direction is zero, i.e. there is rotational equilibrium.\r\n\r\n\r\nTranslational equilibrium\r\n\r\nAn object is in translational equilibrium when the net force acting on it is zero:\r\n\\[\\sum F=0\\]\r\nThis means the forces balance in every direction. For a two-dimensional problem, it is usually easiest to resolve the forces into horizontal and vertical components:\r\n\r\nForces to the left = Forces to the right\r\nForces upwards = Forces downwards\r\n\r\n\r\nConsider a set of traffic lights of mass \\(m\\), suspended above an intersection by two wires of negligible mass.\r\n\r\n\r\n\r\nBecause the traffic lights are stationary, the upward and downward forces must balance. The upward components of the wire tensions are \\(T_{1}\\sin30^{\\circ}\\) and \\(T_{2}\\sin60^{\\circ}\\). The downward force is the weight, \\(W=mg\\).\r\n\r\n\\[\\begin{align*} \\text{Forces upwards} & = \\text{Forces downwards} \\\\[6pt]\r\nT_{1}\\sin30^{\\circ}+T_{2}\\sin60^{\\circ} & = W \\\\[6pt]\r\nT_{1}\\sin30^{\\circ}+T_{2}\\sin60^{\\circ} & = mg \\end{align*}\\]\r\n\r\nThe horizontal components must also balance. The left wire pulls to the left and the right wire pulls to the right:\r\n\\[\\begin{align*} \\text{Forces to the left} & = \\text{Forces to the right} \\\\[6pt]\r\nT_{1}\\cos30^{\\circ} & = T_{2}\\cos60^{\\circ} \\end{align*}\\]\r\n\r\nBoth conditions are needed for translational equilibrium. If either condition were not satisfied, the traffic lights would accelerate horizontally or vertically.\r\n\r\nExample \u2013 applying translational equilibrium\r\nConsider the traffic lights from the previous example. One wire is experiencing a tension force \\(\\mathbf{T_{1}}\\) of \\(\\mathbf{1000}\\,\\textbf{N}\\) at \\(\\mathbf{30^{\\circ}}\\) from the horizontal. The second wire is experiencing a tension force \\(\\mathbf{T_{2}}\\) of \\(\\mathbf{2000}\\,\\textbf{N}\\) at \\(\\mathbf{60^{\\circ}}\\) from the horizontal.\r\nUsing vertical forces only and \\(\\mathbf{g=9.8}\\,\\textbf{m s}^{-2}\\), calculate:\r\n\r\n \tthe maximum weight of the traffic lights that can be supported by the cables.\r\n\r\nTo find the maximum weight, we use the vertical components of the two tensions.\r\n\\[\\begin{align*} W & = T_{1}\\sin30^{\\circ}+T_{2}\\sin60^{\\circ} \\\\[6pt]\r\n& = 1000\\sin30^{\\circ}+2000\\sin60^{\\circ} \\\\[6pt]\r\n& = 500+1732.1 \\\\[6pt]\r\n& \\approx 2232\\,\\text{N} \\end{align*} \\]\r\nThe maximum weight of the traffic lights is approximately \\(2232\\,\\text{N}\\).\r\n\r\n\r\n \tthe maximum mass of the traffic lights that can be supported by the cables.\r\n\r\nWe can rearrange \\(W=mg\\) to find the mass:\r\n\\[\\begin{align*} m & = \\frac{W}{g} \\\\[6pt]\r\n& = \\frac{2232}{9.8} \\\\[6pt]\r\n& \\approx 228\\,\\text{kg} \\end{align*} \\]\r\nThe maximum mass of the traffic lights is approximately \\(228\\,\\text{kg}\\).\r\n\r\nRotational equilibrium\r\nAn object is in rotational equilibrium when the net moment about any chosen point is zero:\r\n\\[\\sum M = 0\\]\r\nFor problems with clockwise and anticlockwise moments, this can be written as:\r\n\r\nClockwise moments = Anticlockwise moments\r\n\r\n\r\nConsider a seesaw with two people sitting on opposite sides of its pivot (point X).\r\n\r\n\\(F_{1}\\) and \\(F_{2}\\) are the downward forces exerted by the two people. These forces are usually their weights. The distances \\(x_{1}\\) and \\(x_{2}\\) are the perpendicular distances from the pivot to the lines of action of the forces.\r\nIn this example, the forces are vertical, so there are no horizontal force components to consider. Translational equilibrium is still required for the complete seesaw system, but the balancing condition for the turning effect is:\r\n\\[\\begin{align*} \\text{Clockwise moment} & = \\text{Anticlockwise moment} \\\\[6pt]\r\nF_{2}x_{2} & = F_{1}x_{1} \\end{align*}\\]\r\nThe support force \\(F_{R}\\) is the upward reaction force exerted by the pivot on the seesaw. For vertical translational equilibrium, it balances the total downward force:\r\n\\[F_{R}=F_{1}+F_{2}\\]\r\nBecause \\(F_{R}\\) acts at the pivot, its moment about the pivot is zero. It therefore does not appear in the moment equation.\r\n\r\nExample \u2013 applying rotational equilibrium\r\nTwo children are balanced on a seesaw supported in the middle. One child weighs \\(\\mathbf{200}\\textbf{N}\\) and is seated \\(\\mathbf{1.2}\\,\\textbf{m}\\) from the axis. The other child is seated \\(\\mathbf{1.5}\\,\\textbf{m}\\) from the axis. Calculate how much the second child weighs, ignoring the mass of the seesaw. Use \\(\\mathbf{g=9.8}\\,\\textbf{m s}\\mathbf{^{-2}}\\).\r\nWe will take moments about the pivot. This removes the support force from the calculation because its distance from the pivot is zero.\r\nThe clockwise and anticlockwise moments must be equal, so:\r\n\\[\\begin{align*} \\text{Clockwise moment} & = \\text{Anticlockwise moment} \\\\[6pt]\r\n200\\times1.2 & = W_{2}\\times1.5 \\end{align*} \\]\r\nNext, we rearrange to find the weight of the second child:\r\n\\[\\begin{align*} W_{2} & = \\frac{200\\times1.2}{1.5} \\\\[6pt]\r\n& = 160\\,\\text{N} \\end{align*}\\]\r\nThe second child weighs \\(160\\,\\text{N}\\).\r\n\r\nExercise \u2013 applying static equilibrium\r\nFor the following questions, use \\(g=9.8\\,\\text{m s}^{-2}\\).\r\n\r\n\tA uniform \\(4.0\\,\\text{m}\\) beam has a weight of \\(200\\,\\text{N}\\) acting at its centre. It is supported at both ends. Calculate the upward reaction force at each support.\r\n\r\n\r\n\r\n\r\n \tA picture is hung as shown in the following diagram.\r\n\t\r\n\tIf the hanging wire has a breaking strength of \\(40\\,\\text{N}\\), determine the maximum possible mass of the picture.\r\n\r\n\r\n\r\n\r\n \tTwo children want to make a seesaw from a \\(5.0\\,\\text{m}\\) plank of wood. The children weigh \\(25\\,\\text{kg}\\) and \\(20\\,\\text{kg}\\). They both want to sit right on the ends of the plank. Determine where the plank should be supported for it to balance.\r\n\r\n\r\n\r\n\r\n\tA sign of weight \\(120\\,\\text{N}\\) is held stationary by two cables. The cables make angles of \\(30^{\\circ}\\) and \\(45^{\\circ}\\) with the horizontal. Calculate the tension in each cable.\r\n\r\n\r\n\r\n\r\n \tA \\(100\\,\\text{g}\\) pendant light is supported by two cables \u2013 one at an angle of \\(60^{\\circ}\\) with the ceiling, and one perpendicular to the wall.\r\n\t\r\n\tAssuming the mass of each cable is negligible, calculate the tension in each cable.\r\n\r\n\r\n\r\n\r\n\tA \\(300\\,\\text{N}\\) load is placed \\(0.8\\,\\text{m}\\) from the pivot of a balanced beam. A second load is placed \\(1.2\\,\\text{m}\\) from the pivot on the opposite side. Calculate the size of the second load.\r\n\r\n\r\n\r\n\r\n","n":0.033},[{"v":"Maths","i":2,"n":1},{"v":"Statics","i":1,"n":1},{"v":"Physics","i":0,"n":1}]]},{"i":364,"$":[{"v":"The cosine rule","n":0.577},{"v":"\r\nLike the sine rule, the cosine rule comes in handy for triangles that are not right-angled. Use this resource to learn about the cosine rule, how it is applied to find unknown sides and angles, and its importance in solving complex geometric problems involving any type of triangle.\r\n\r\nThe cosine rule is a generalisation of Pythagoras\u2019 theorem. If you have any two sides of a triangle, as long as you know the angle between them, you can calculate the length of the third side.\r\nConsider the triangle.\r\n\r\nHere:\r\n\r\n\tthe vertices \\(A\\), \\(B\\) and \\(C\\) have the angles \\(A\\), \\(B\\) and \\(C\\), respectively\r\n\t\\(a\\), \\(b\\) and \\(c\\) are the side lengths opposite to the angles \\(A\\), \\(B\\) and \\(C\\), respectively.\r\n\r\nThe cosine rule states that:\r\n\\[\\begin{align*} a^{2} & = b^{2}+c^{2}-2bc\\,\\cos (A)\\\\\r\nb^{2} & = a^{2}+c^{2}-2ac\\,\\cos(B)\\\\\r\nc^{2} & = a^{2}+b^{2}-2ab\\,\\cos(C)\r\n\\end{align*}\\]\r\n\r\nYou might notice that the side of the triangle on the left-hand side of the equation is opposite the angle at the end of the equation:\\[\\boldsymbol{\\underline{a^{2}}} = b^{2}+c^{2}-2bc\\,\\cos(\\boldsymbol{\\underline{A}})\\]\r\nThis rule can be used to find angles and sides in any triangle (not just a right-angled triangle) when given:\r\n\r\n \ttwo sides and the angle between them; or\r\n \tall three sides of the triangle.\r\n\r\n\r\nExample 1 \u2013 using the cosine rule\r\nFind the value of \\(a\\) in the triangle shown.\r\n\r\n\\[\\begin{align*} a^{2} & = b^{2}+c^{2}-2bc\\,\\cos(A)\\\\\r\na^{2} & = 12^{2}+15^{2}-2\\times12\\times15\\times\\cos(83^{\\circ})\\\\ \r\na^{2} & = 144+225-360\\times\\cos(83^{\\circ})\\\\ \r\na^{2} & = 369-43.87\\\\ \r\na^{2} & = 325.13\\\\\r\na & = 18.03 \r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\nExercise \u2013 using the cosine rule\r\n\r\n \tFind the unknown sides for the following triangles.\r\n\t\t\r\n\t\t \r\n\t\t \r\n\t\t \r\n\t\t  \r\n\t\r\n\tFind the unknown angle for the following triangles.\r\n\t\r\n\t\t \r\n\t\t \r\n\t\r\n\r\n\r\n\r\n\r\n\r\n","n":0.063},[{"v":"Trigonometry","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Functions","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":365,"$":[{"v":"The unit circle","n":0.577},{"v":"\r\nAngles are commonly measured in degrees but sometimes, it is useful to define them in terms of the length around the unit circle. This resource introduces radians as a measure of angle. Being able to use radians is crucial for solving problems in engineering, such as analysing the motion of gears and wheels, and in computer graphics, where they help in rendering angles and rotations accurately.\r\n\r\nRadians\r\nAngles can be measured in units known as radians. One radian is the angle created by bending the radius length around the arc of a unit circle. A unit circle has a radius of \\(1\\). You will learn more about the unit circle here.\r\nConsider the circle centred at the origin. The red arc is the same length as the radius of the circle. The angle subtended at (or formed from) the origin \\(\\theta\\) is one radian and is denoted by \\(1^{c}\\).\r\n\r\nIn the examples on this page, we show the unit symbol for radian (\\(^{c}\\)) but note that this is often omitted.\r\nOne radian is approximately \\(57.3^{\\circ}\\).\r\n\r\nConverting between degrees and radians\r\nThe circumference of a unit circle is given by the formula \\(C=2\\pi r\\). Because of this, we know \\(2\\pi\\) radians (or \\(2\\pi^{c}\\)) is a complete rotation and the same as \\(360\\) degrees (or \\(360^{\\circ}\\)). Similarly, half a rotation or \\(180\\) degrees \\(=\\pi\\) radians (\\(180^{\\circ}=\\pi^{c}\\)).\r\n\r\nWe can express angles that represent fractional parts of a circle in terms of \\(\\pi\\).\r\n\r\n\r\n\r\n\r\n\tAngle in degrees\r\n\tAngle in radians\r\n\r\n\r\n\r\n\r\n\t\\(90\\)\r\n\t\\(\\dfrac{\\pi}{2}\\)\r\n\r\n\r\n\t\\(60\\)\r\n\t\\(\\dfrac{\\pi}{3}\\)\r\n\r\n\r\n\t\\(45\\)\r\n\t\\(\\dfrac{\\pi}{4}\\)\r\n\r\n\r\n\t\\(30\\)\r\n\t\\(\\dfrac{\\pi}{6}\\)\r\n\r\n\r\n\t\\(270\\)\r\n\t\\(3\\times90^{\\circ}=3\\times\\dfrac{\\pi}{2}=\\dfrac{3\\pi}{2}\\)\r\n\r\n\r\n\t\\(360\\)\r\n\t\\(2\\pi\\)\r\n\r\n\r\n\r\n\r\nWe can easily convert between degrees and radians using the conversion factors:\r\n\\[1^{c}=\\frac{180^{\\circ}}{\\pi^{c}}\\quad\\textrm{and}\\quad1^{\\circ} =\\frac{\\pi^{c}}{180^{\\circ}}\\]\r\n\r\nExample 1 \u2013 converting between degrees and radians\r\nConvert \\(60^{\\circ}\\) to radians.\r\nMultiply the original quantity by the correct conversion factor. This is the conversion factor with the units we want in the numerator (top of the fraction) and the units we have in the denominator (bottom of the fraction).\r\nWe need to convert from degrees to radians, so the correct conversion factor is \\(\\dfrac{\\pi^{c}}{180^{\\circ}}\\).\r\n\\[\\begin{align*} 60^{\\circ} & = 60^{\\circ}\\times\\frac{\\pi^{c}}{180}\\\\ \r\n& = \\frac{60^{\\circ}\\pi^{c}}{180^{\\circ}}\\quad\\textrm{cancel out degrees on top and bottom}\\\\\r\n& = \\frac{\\pi^{c}}{3}\\\\\r\n& \\approx\\frac{3.142^{c}}{3}\\\\ \r\n& \\approx1.05^{c} \r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 converting between degrees and radians\r\n\r\n \tConvert the following degrees to radians.\r\n\t\r\n \t\t\\(30^{\\circ}\\)\r\n\t\t\\(270^{\\circ}\\)\r\n\t\t\\(20^{\\circ}\\)\r\n\t\t\\(450^{\\circ}\\)\r\n\t\t\\(135^{\\circ}\\)\r\n\t\t\\(57.3^{\\circ}\\)\r\n\t\r\n\tConvert the following radians to degrees.\r\n\t\r\n \t\t\\(\\dfrac{\\pi}{2}\\)\r\n\t\t\\(\\dfrac{5\\pi}{4}\\)\r\n\t\t\\(\\dfrac{11\\pi}{6}\\)\r\n\t\t\\(3.5\\pi\\)\r\n\t\t\\(\\pi\\)\r\n\t\t\\(1\\)\r\n\t\r\n\r\n\r\n\r\n","n":0.054},[{"v":"Trigonometry","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Functions","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":366,"$":[{"v":"Linear functions","n":0.707},{"v":"Linear functions are the most basic type of function. They are used extensively across STEM and other disciplines, especially to model the rate of physical, biological or economic change. Use this resource to learn about linear functions and graphs.\r\n\r\nLinear functions create a straight line when graphed. They have the form:\r\n\\[y=mx+c\\] where \\(m\\) is the slope of the line (or the gradient) and \\(c\\) is where the line crosses the \\(y\\)-axis (or the \\(y\\)-intercept).\r\nWhen a function is linear, there is a constant rate of change. In other words, changing the \\(x\\) value causes a proportional change in the \\(y\\) value.\r\n\r\nGraphs of linear functions\r\nAs you saw in Functions and relations, functions can be represented graphically. When you sketch a graph, you may not need to plot large numbers of points. Often, you just need the basic shape of the graph with some important points clearly labeled.\r\nIn the case of linear graphs, only two points are needed to sketch the graph. The intercepts are also important \u2013 that is, where the graph crosses the \\(x\\)- and \\(y\\)-axes.\r\n\r\n\tThe \\(x\\)-intercept is where the graph cuts the \\(x\\)-axis (\\(y=0\\)).\r\n\tThe \\(y\\)-intercept is where the graph cuts the \\(y\\)-axis (\\(x=0\\)).\r\n\r\nFor any linear graph, it should be clear where these intercepts are.\r\n\r\nExample \u2013 graphing linear functions\r\nSketch the function \\(y=3x+6\\).\r\nWe need to determine two points on the line. The \\(x\\)- and \\(y\\)-intercepts are often the easiest to find.\r\nFor the \\(y\\)-intercept, we substitute \\(x=0\\) into the equation for the function:\r\n\\[\\begin{align*} y & = 3x+6\\\\\r\n& = 3(0)+6\\\\\r\n& = 6\r\n\\end{align*}\\]\r\nThe \\(y\\)-intercept is at \\((0,6)\\).\r\nFor the \\(x\\)-intercept, we substitute \\(y=0\\) into the equation for the function:\r\n\\[\\begin{align*} y & = 3x+6\\\\ \r\n(0) & = 3x+6\\\\ \r\n3x & = -6\\\\ \r\nx & = -2\r\n\\end{align*}\\]\r\nThe \\(x\\)-intercept is at \\((-2,0).\\)\r\nWe have the points \\((0,6)\\) and \\((-2,0)\\) which we can plot on the Cartesian plane and connect with a straight line. The \\(x\\)- and \\(y\\)-intercepts should be marked on the graph. \r\n\r\nIf you need help, go to Plotting points on a graph.\r\n\r\nExercise \u2013 graphing linear functions\r\nSketch the graphs of the following equations.\r\n\r\n\t\\(y=4x+3\\)\r\n\t\\(2y=-x+6\\)\r\n\t\\(3y+2x=9\\)\r\n\t\\(y=2x\\)\r\n\t\\(y=6\\)\r\n\t\\(2y+x=0\\)\r\n\r\n\r\n\r\n\r\nGradient\r\nThe gradient \\(m\\) indicates how steep the line of a graph is. It tells us how much the \\(y\\) value changes for every \\(1\\) unit increase in \\(x\\). \r\nTo calculate \\(m\\), we find the ratio of the vertical change (rise) to the horizontal change (run) between two points \\((x_{1},y_{1})\\) and \\((x_{2},y_{2})\\) on the line.\r\n\\[\\begin{align*} m & = \\frac{\\textrm{Rise}}{\\textrm{Run}}\\\\\r\n& = \\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\r\n\\end{align*}\\]\r\n\r\nThe appearance of a graph can tell you key information about the gradient.\r\n\r\n \tA line that slopes up to the right has a positive gradient.\r\n \tA line that slopes down to the right has a negative gradient.\r\n \tParallel lines have the same gradient.\r\n \tWhen you multiply the gradients of two mutually perpendicular lines, you get \\(-1\\).\r\n \tA line parallel to the \\(x\\)-axis has a gradient of \\(0\\).\r\n \tA line parallel to the \\(y\\)-axis does not have a defined gradient. In fact, the line is not a function; it is a relation.\r\n\r\n\r\nExample 1 \u2013 calculating the gradient of a linear function\r\nFind the gradient of the line \\(y=3x+6\\).\r\nThis is the same function as Example 1 \u2013 graphing linear functions. We found two points, the intercepts: \\((0,6)\\) and \\((-2,0)\\). We can use these to calculate the gradient.\r\nLet \\((x_{1},y_{1})\\) be \\((0,6)\\) and \\((x_{2},y_{2})\\) be \\((-2,0)\\).\r\n\\[\\begin{align*} m & = \\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\\r\n& = \\frac{0-6}{-2-0}\\\\\r\n& = \\frac{-6}{-2}\\\\\r\n& = 3\r\n\\end{align*}\\]\r\nIt doesn't matter which points we assigned \\((x_{1},y_{1})\\) and \\((x_{2},y_{2})\\), as long as they lie on the line.\r\nIf the equation is in the form \\(y=mx+c\\), you can easily tell what the gradient \\(m\\) is. \\(y=3x+6\\) is in this form, so \\(m=3\\).\r\n\r\n\r\n\r\nExercise \u2013 calculating the gradient of a linear function\r\nFind the gradient of the following straight lines.\r\n\r\n\t\\(y=-5x\\)\r\n\t\\(2y=6x+9\\)\r\n\t\\(2x-3y=0\\)\r\n\t\\(3y+2x-2=0\\)\r\n\t\\(5y=8\\)\r\n\t\\(x=-2\\)\r\n\r\n\r\n\r\n\r\nEquation of a straight line\r\nAs well as \\(y=mx+c\\), the equation of a straight line can be written in point\u2014slope form:\r\n\\[y-y_{1}=m(x-x_{1})\\] where \\(m\\) is the gradient and \\((x_{1},y_{1})\\) is a point on the line.\r\nThis shows that as long as you know the slope and one point on the line, you can find the equation of the line.\r\nHere is how the point\u2014slope form is derived.\r\nIf \\((x_{1},y_{1})\\) is a point on the line with a gradient of \\(m\\), then it must satisfy \\(y=mx+c\\). This means by substituting \\(x=x_{1}\\) and \\(y=y_{1}\\) into the equation:\r\n\\[y_{1}=mx_{1}+c\\]\r\nWe can then subtract this from \\(y=mx+c\\) to get:\r\n\\[\\begin{align*} y-y_{1} & = mx+c-(mx_{1}+c)\\\\\r\n& = mx-mx_{1}+c-c\\\\\r\n& = m(x-x_{1})\r\n\\end{align*}\\]\r\n\r\nExample 1 \u2013 finding the equation of a line\r\nFind the gradient and the equation of the straight line shown in the graph.\r\n\r\nThe graph has a positive gradient because it slopes up towards the right. You don\u2019t have to write this down but it is a good check to have in mind. If you calculate a negative gradient, you know something has gone wrong.\r\nWe have been given two points, \\((1,3)\\) and \\((3,7)\\). We can let these be \\((x_{1},y_{1})\\) and \\((x_{2},y_{2})\\), respectively, then calculate the gradient.\r\n\\[\\begin{align*} m & = \\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\\r\n& = \\frac{7-3}{3-1}\\\\\r\n& = \\frac{4}{2}\\\\\r\n& = 2\r\n\\end{align*}\\]\r\nTo find \\(c\\), we can substitute either \\((x_{1},y_{1})\\) or \\((x_{2},y_{2})\\) into the point\u2014slope form, but we should also remember that \\(c\\) is the \\(y\\)-intercept... which we have been given in the graph! The graph intersects the \\(y\\)-axis at \\(1\\), so \\(c=1\\).\r\nTherefore, the equation of the line is \\(y=2x+1\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding the equation of a line\r\n\r\n\tFind the equations of the following straight lines.\r\n\t\r\n\t\tGraph 1\r\n\t\t\r\n\t\tGraph 2\r\n\t\t\r\n\t\r\n\tFind the equation of the line with a gradient of \\(3\\), passing through the point \\((0,-5)\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.034},[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]]},{"i":367,"$":[{"v":"Antidifferentiation","n":1},{"v":"Antidifferentiation is the reverse operation of differentiation. You need it to solve problems involving area under a curve, total accumulated change, and reversing rates of change. By mastering antidifferentiation, you'll be able to reconstruct original functions and solve real-world problems in physics, engineering, and beyond.\r\nIf you have the derivative of a function \\(f'(x)\\), you can find the original function \\(f(x)\\) using integration or antidifferentiation.\r\nHere, the original function is called the antiderivative.\r\n\r\n\r\n\r\nVideo tutorial \u2013 integrating functions\r\nWatch this video to learn about how to integrate functions or find antiderivatives.\r\n\r\n\r\nAntiderivatives\r\nIn general, an antiderivative of \\(f'(x)\\) is given by \\(f(x)+c\\) where \\(c\\) is a constant. For example:\r\n\r\n \tIf \\(f(x)=\\dfrac{x^{3}}{3}\\), then \\(f'(x)=x^{2}\\), so \\(\\dfrac{x^{3}}{3}\\) is an antiderivative of \\(x^{2}\\).\r\n \tIf \\(f(x)=\\dfrac{x^{3}}{3}+1\\), then \\(f'(x)=x^{2}\\), so \\(\\dfrac{x^{3}}{3}+1\\) is an antiderivative of \\(x^{2}\\).\r\n \tIf \\(f(x)=\\dfrac{x^{3}}{3}+2\\), then \\(f'(x)=x^{2}\\), so \\(\\dfrac{x^{3}}{3}+2\\) is an antiderivative of \\(x^{2}\\).\r\n\r\nYou can see that adding a constant to \\(\\dfrac{x^{3}}{3}\\) does not change the fact it is an antiderivative of \\(x^{2}\\). This is because the derivative of a constant is zero. We often write \\(c\\in\\mathbb{R}\\), which means that \\(c\\) is a real number.\r\n\r\nFinding antiderivatives\r\nAntidifferentiation is more complicated than differentiation. But there are some rules to help us. One of the most important is the power rule which says:\r\nIf \\(f'(x)=x^{n}\\), \\(n\\neq-1\\), the antiderivative \\(f(x)=\\dfrac{1}{n+1}x^{n+1}+c\\), where \\(c\\) is a constant.\r\nIf \\(n=-1\\), \\(\\dfrac{1}{n+1}\\) would be \\(\\dfrac{1}{0}\\) which has no meaning. Apart from this restriction, \\(n\\) can be any number.\r\n\r\nIntegrals\r\nWe express antiderivatives using integrals. For example, \\(\\dfrac{1}{3}x^{3}\\) is the antiderivative of \\(y=x^{2}\\). Using intervals, we would use the mathematical notation:\r\n\\[ \\int x^{2}dx=\\frac{1}{3}x^{3}+c,\\,c\\in\\mathbb{R}\\]\r\nWe read this as \"the indefinite integral of \\(x^{2}\\) with respect to \\(x\\) is equal to \\(\\dfrac{1}{3}x^{3}+c\\)\". The \\(\\int\\) sign means indefinite integral and \\(dx\\) means with respect to \\(x\\). The words \"indefinite integral\" are the same as \"antiderivative\".\r\n\r\nExample 1 \u2013 finding antiderivatives\r\nGiven \\(\\dfrac{dy}{dx}=x\\), find the antiderivative.\r\nHere, \\(n=1\\) because \\(x=x^{1}\\). We add \\(1\\) to the power of \\(1\\), divide by the new power, and add a constant \\(c\\).\r\n\\[\\begin{align*} y & = \\frac{x^{1+1}}{1+1}+c,\\,c\\in\\mathbb{R}\\\\\r\n& = \\frac{x^{2}}{2}+c\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding antiderivatives\r\n\r\n\tFind antiderivatives for the following.\r\n\t\r\n\t\t\\(x^{3}\\)\r\n\t\t\\(s^{8}\\)\r\n\t\t\\(\\sqrt[3]{x}\\)\r\n\t\t\\(x^{-5}\\)\r\n\t\t\\(6\\)\r\n\t\t\\(m^{-2}\\)\r\n\t\t\\(p^{-\\frac{1}{2}}\\)\r\n\t\r\n\tIntegrate the following.\r\n\t\r\n\t\t\\(\\int 3dx\\)\r\n\t\t\\(\\int 5xdx\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nWe know how to take derivatives of functions. If I apply the derivative operator to x squared, I get 2x. Now, if I also apply the derivative operator to x squared plus 1, I also get 2x.\r\nIf I apply the derivative operator to x squared plus pi, I also get 2x. The derivative of x squared is 2x. Derivative, with respect to x of pi of a constant, is just 0. Derivative, with respect to x of 1, is just a constant, is just 0. So once again, this is just going to be equal to 2x.\r\nIn general, the derivative, with respect to x of x squared plus any constant, is going to be equal to 2x. The derivative of x squared, with respect to x, is 2x. Derivative of a constant, with respect to x, a constant does not change with respect to x, so it's just equal to 0. So you have\u2013 You apply the derivative operator to any of these expressions and you get 2x.\r\nNow, let's go the other way around. Let's think about the antiderivative. And one way to think about it is we're doing the opposite of the derivative operator. The derivative operator, you get an expression and you find it's derivative.\r\nNow, what we want to do, is given some expression, we want to find what it could be the derivative of. So if someone were to tell\u2013 or give you 2x\u2013 if someone were to say 2x\u2013 let me write this. So if someone were to ask you what is 2x the derivative of? They're essentially asking you for the antiderivative. And so you could say, well, 2x is the derivative of x squared. But you could also say 2x is the derivative of x squared plus 1. You could also say that 2x is the derivative of x squared plus pi, I think you get the general idea.\r\nSo if you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant. So this is what you would consider the antiderivative of 2x.\r\nNow, that's all nice, but this is kind of clumsy to have to write a sentence like this, so let's come up with some notation for the antiderivative. And the convention here is to use kind of a strange looking notation, is to use a big elongated s looking thing like that, and a dx around the function that you're trying to take the antiderivative of.\r\nSo in this case, it would look something like this. This is just saying this is equal to the antiderivative of 2x, and the antiderivative of 2x, we have already seen, is x squared plus c.\r\nNow, you might be saying, why do we use this type of crazy notation. It'll become more obvious when we study the definite integral and areas under curves and taking sums of rectangles in order to approximate the area of the curve.\r\nHere, it really should just be viewed as a notation for antiderivative. And this notation right over here, this whole expression, is called the indefinite integral of 2x, which is another way of just saying the antiderivative of 2x.\r\n","n":0.034},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":368,"$":[{"v":"Rules for differentiation","n":0.577},{"v":"Operational rules for differentiation allow us to handle complex expressions by breaking them down into simpler parts. They are used in real-world applications, such as calculating the speed of a moving car or determining the growth rate of a population. Understanding these rules makes it easier to calculate derivatives in various situations, helping us solve problems involving rates of change and slopes.\r\n\r\nIt is not always convenient to use differentiation from first principles to find a derivative function. Instead, we can use rules that are established from these principles. They make finding derivative functions much more straightforward.\r\n\r\nVideo tutorial \u2013 differentiating using operational rules\r\nWatch this video to learn how to differentiate functions using operational rules.\r\n\r\n\r\nOperational rules\r\nThere are four operational rules that you can use to quickly find the derivative of a function.\r\nRule 1 \r\nIf \\(g(x)=kf(x)\\) where \\(k\\) is a constant, then:\r\n\\[g'(x)=kf'(x)\\textrm{ or }\\dfrac{dg}{dx}=k\\dfrac{df}{dx}\\]\r\nRule 2 \r\nIf \\(f(x)=k\\) where \\(k\\) is a constant, then:\r\n\\[f'(x)=0\\]\r\nRule 3\r\nIf \\(f(x)=g(x)+h(x)\\), then:\r\n\\[f'(x)=g'(x)+h'(x)\\textrm{ or }\\dfrac{df}{dx}=\\dfrac{dg}{dx}+\\dfrac{dh}{dx}\\]\r\nRule 4\r\nIf \\(y=x^{n}\\) where \\(n\\) is a real number, then:\r\n\\[\\frac{dy}{dx}=nx^{n-1}\\]\r\n\r\nBefore we look at some examples, make sure you are comfortable working with Algebra, Indices, surds and logarithms and Trigonometry.\r\n\r\nExample 1 \u2013 differentiating using operational rules\r\nIf \\(y=x^{7}\\), find \\(\\dfrac{dy}{dx}\\).\r\nWe can use operational rule \\(4\\).\r\n\\[\\begin{align*} \\frac{dy}{dx} & = 7x^{7-1}\\\\\r\n& = 7x^{6}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nSpecial rules\r\nThere are special rules we have to follow to find the derivative of some functions. These are represented in the table.\r\n\r\n\r\n\r\n\t\r\n\t\tFunction\r\n\t\tDerivative\r\n\t\r\n\r\n\r\n\t\r\n\t\t\\(f(x)=e^{x}\\)\r\n\t\t\\(f'(x)=e^{x}\\)\r\n\t\r\n\t\r\n\t\t\\(f(x)=\\log_{e}x\\)\r\n\t\t\\(f'(x)=\\dfrac{1}{x}\\)\r\n\t\r\n\t\r\n\t\t\\(f(x)=\\sin(x)\\)\r\n\t\t\\(f'(x)=\\cos(x)\\)\r\n\t\r\n\t\r\n\t\t\\(f(x)=\\cos(x)\\)\r\n\t\t\\(f'(x)=-\\sin(x)\\)\r\n\t\r\n\t\r\n\t\t\\(f(x)=\\tan(x)\\)\r\n\t\t\\(f'(x)=\\sec^{2}x\\)\r\n\t\r\n\r\n\r\n\r\nExample 1 \u2013 differentiating using special rules\r\nIf \\(\\sin(x)+3x^{2}\\), find \\(\\dfrac{dy}{dx}\\).\r\n\\[\\begin{align*} \\frac{dy}{dx} & = \\cos(x)+(3\\times 2x^{2-1})\\\\\r\n& = \\cos(x)+6x\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 differentiating using special rules\r\n\r\n\tDifferentiate the following.\r\n\t\r\n\t\t\\(y=x^{7}\\)\r\n\t\t\\(y=x^{\\frac{1}{5}}\\)\r\n\t\t\\(y=x^{-19}\\)\r\n\t\t\\(y=\\dfrac{1}{x^{4}}\\)\r\n\t\t\\(y=53\\)\r\n\t\t\\(y=\\sqrt[4]{x}\\)\r\n\t\t\\(y=5x^{6}\\)\r\n\t\t\\(y=9x^{-5}\\)\r\n\t\t\\(y=\\dfrac{\\sqrt{5}}{x}\\)\r\n\t\t\\(y=3x^{2}+2x\\)\r\n\t\r\n\tDifferentiate the following.\r\n\t\r\n\t\t\\(y=\\sin(x)-\\cos(x)\\)\r\n\t\t\\(y=10-\\log_{e}(x)\\)\r\n\t\t\\(y=\\tan(x)-\\sqrt{x}\\)\r\n\t\t\\(y=3\\cos(x)-\\dfrac{1}{x^{2}}\\)\r\n\t\t\\(y=\\dfrac{e^{x}}{6}-x^{7}\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nHere's a short video on differentiation. The general form is y equals x to the n. And if y equals x to the n, then the derivative, dydx, is nx to the n minus one. So if we have a function y equals x to the seven, then here n equals seven, hence dydx equals 7x to the 6th, because n minus one is seven minus one.\r\n\r\nLet's look at another example. Here we have two functions, y equals x\u00b3, plus x to the 6th. When we differentiate them we still differentiate them as though they were separate functions, in other words, x\u00b3 becomes three 8ths squared, and x to the 6th becomes 6x to the five, using the rule in the top right hand corner.\r\n\r\nLet's look at a trickier one. Here we have the function y equals one over x\u00b2. Notice the x\u00b2 is on the bottom of the fraction and this is quite difficult to do at this stage, so we now put x\u00b2 up on the top. In other words, we remove the denominator and express it as x to the minus two. Now we have a negative index. And when we differentiate that n is now negative two. So the derivative dydx is negative two x to the negative three, because n minus one is minus two minus one. Notice we finally revert back to the fraction because x to the minus three is minus two over x\u00b3.\r\n\r\nLet's look at this function, y equals the square root of x. We express y equals the square root of x as an index so we write it as x to the half, and we always do this before we differentiate. Now proceeding with the differentiation, dydx will be a half x to the minus half, because here n is a half, so n minus one is a half minus one, which is minus a half. So our final answer is one over two to the square root of x.\r\n\r\nNow let's look at a more complicated fraction. The function here is now x to the four minus one, all over the square root of x. Here I'm breaking it up into two separate fractions, in other words, it's x to the four over x to the half minus one over x to the half. Notice I've replaced the square root sign by an index. And now I write h to the four over h to the half as one term, which is x to the seven over two. One over x to the half is the same as x to the minus a half. In other words, I'm dividing through by x to the half.\r\n\r\nThe derivative is now calculated separably because we have two functions here to differentiate. So x to the seven over two becomes seven over 2x to the five over two, and x to the minus a half becomes a half x to the minus three over two. Notice we have two negative signs, one before the negative x and the index is negative a half, so the two negatives gives us a positive as our answer. So that we have the answer.\r\n\r\nWhat about a derivative of a function which has a constant in front of it? K is the constant. If we have a constant in front of the function then the derivative is the constant multiplied by the rule that we've used before, nx to the n minus one. As a good example, the function here is y equals 7h to the six, so the derivative, dydx is seven, we leave the constant alone, and now we just differentiate h to the six, which becomes 6x to the five. The final answer is given by seven times 6x to the five. Seven sixes are 42, so the answer is now 42x to the five.\r\n\r\nFinally, the derivative of a function which is simply a constant. So here if y equals a constant, then the derivative is always zero. Probably the easiest one of the lot to remember. In other words, if you have any number which is a constant, such as 99, y equals 99, then the derivative is equal to zero.\r\n\r\nNow try some questions for yourself. Here are 10 questions that I want you to differentiate. The answers are on the next slide. Thanks for watching this video.\r\n","n":0.032},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":369,"$":[{"v":"Transposing formulas with brackets and fractions","n":0.408},{"v":"\r\nExpand your transposition skills by exploring how to rearrange formulas involving brackets and fractions. By doing this, you will develop the skills needed to isolate variables in more complex expressions and prepare yourself well for advanced problem-solving in STEM.\r\n\r\nYou will have learned the necessary theory when you were first introduced to transposition of formulas, so this page focuses on examples.\r\n\r\nExample 1 \u2013 transposing formulas with brackets and fractions\r\nThe formula to calculate simple interest is:\r\n\\[A=P(1+rn)\\]\r\nwhere \\(A\\) is the total amount, \\(P\\) is the principal or initial amount, \\(r\\) is the interest rate and \\(n\\) is the time or duration.\r\nExpress the formula in a way that helps calculate the time taken to generate an amount of interest.\r\nWe need to undo the operations to get \\(n\\) on its own on one side of the equal sign. On the right, \\(n\\) is multiplied by \\(r\\), added to \\(1\\), then multiplied by \\(P\\). To undo this, we need to work backwards and divide by \\(P\\) first, then subtract \\(1\\), and finally, divide by \\(r\\).\r\n\\[\\begin{align*} A & = P(1+rn)\\quad\\textrm{divide by \\(P\\) on both sides}\\\\\r\n\\frac{A}{P} & = \\frac{P(1+rn)}{P}\\quad\\textrm{cancel out \\(P\\) on right}\\\\\r\n\\frac{A}{P} & = 1+rn\\quad\\textrm{subtract \\(1\\) on both sides}\\\\\r\n\\frac{A}{P}-1 & = 1+rn-1\\\\\r\n\\frac{A}{P}-1 & = rn\\quad\\textrm{divide by \\(r\\) on both sides}\\\\\r\n\\frac{\\frac{A}{P}-1}{r} & = \\frac{rn}{r}\\quad\\textrm{cancel out \\(r\\) on right}\\\\\r\n\\frac{\\frac{A}{P}-1}{r} & = n\r\n\\end{align*}\\]\r\nThe fraction on the left is a bit messy, so we can rewrite it in a different way. Remember that dividing by something is the same as multiplying by its reciprocal. We can rewrite \\(\\div r\\) as \\(\\times \\dfrac{1}{r}\\). So:\r\n\\[\\left(\\frac{A}{P}-1\\right)\\times\\frac{1}{r} = n\\]\r\nThis can be rewritten with \\(n\\) on the left. We can also remove the \\(\\times\\) sign, since when we use brackets, it is assumed that any term outside the bracket is multiplied by whatever is inside the brackets.\r\n\\[n=\\frac{1}{r}\\left(\\frac{A}{P}-1\\right)\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 transposing formulas with brackets and fractions\r\n\r\n\tFind \\(A\\) if \\(S=C(A+B)\\).\r\n\tFind \\(A\\) if \\(V=\\dfrac{Ah}{3}\\).\r\n\tFind \\(a\\) if \\(A=\\dfrac{h(a+b)}{2}\\).\r\n\tFind \\(B\\) if \\(A=\\dfrac{2B+C}{P}\\).\r\n\tFind \\(C\\) if \\(A=\\dfrac{2P(B-C)}{3}\\).\r\n\tFind \\(r\\) if \\(I=\\dfrac{Mr^{2}}2\\).\r\n\tFind \\(b\\) if \\(H=k(1-bt)\\).\r\n\tFind \\(h\\) if \\(t=2\\pi\\sqrt{\\dfrac{h+k}{g}}\\).\r\n\tFind \\(u\\) if \\(v^{2}=u^{2}+2as\\).\r\n\tFind \\(y\\) if \\(m=\\sqrt{\\dfrac{x+y}{z}}\\).\r\n\r\n\r\n\r\n\r\n\r\n","n":0.055},[{"v":"Maths equations","i":3,"n":0.707},{"v":"Maths","i":2,"n":1},{"v":"Fractions","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":370,"$":[{"v":"Multiplying and dividing algebraic fractions","n":0.447},{"v":"\r\nLet's build on your understanding of basic fraction operations and algebraic expressions by looking at multiplying and dividing algebraic fractions. By mastering these skills, you'll be able to simplify and manipulate complex fractions, which is essential for solving more advanced equations and applying algebra to a variety of contexts.\r\n\r\nVideo tutorial \u2013 algebraic fractions: multiplication and division\r\nWatch this video to learn about multiplying and dividing algebraic fractions.\r\n\r\n\r\nSimplifying algebraic fractions\r\nWe can use the same technique to simplify algebraic fractions as we did with simplifying ordinary fractions.\r\n Simplifying algebraic fractions may involve cancelling out common algebraic terms, as well as numbers. You can only divide the denominator and numerator if ALL the terms have a common factor.\r\nThis is the only real difference from simplifying ordinary fractions.\r\n\r\nExample 1 \u2013 simplifying algebraic fractions\r\nSimplify \\(\\dfrac{4x^{2}y}{6y^{2}}\\).\r\n\\[\\begin{align*} & = \\frac{4x^{2}y}{6y^{2}}\\quad\\textrm{divide top and bottom by \\(2\\)}\\\\\r\n& = \\frac{2x^{2}y}{3y^{2}}\\quad\\textrm{divide top and bottom by \\(y\\)}\\\\\r\n& = \\frac{2x^{2}}{3y}\r\n\\end{align*}\\]\r\nThe HCFs are \\(2\\) and \\(y\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 simplifying algebraic fractions\r\nSimplify the following fractions.\r\n\r\n\t\\(\\dfrac{12ab^{2}}{8bc}\\)\r\n\t\\(\\dfrac{5x-20}{5}\\)\r\n\t\\(\\dfrac{9u-18}{2u-4}\\)\r\n\t\\(\\dfrac{6t-9}{12-8t}\\)\r\n\t\\(\\dfrac{b}{b^{2}+7b}\\)\r\n\t\\(\\dfrac{(j+4)(j-4)}{3j+12}\\)\r\n\t\\(\\dfrac{2(5-v)}{3v-15}\\)\r\n\t\\(\\dfrac{9r^{2}-3r}{16r-48r^{2}}\\)\r\n\r\n\r\n\r\nMultiplying algebraic fractions\r\nYou can also use these ideas with algebraic fractions.\r\nWhen you multiply algebraic fractions, you simply multiply the numerator and denominators of the fractions.\r\n\r\nExample 1 \u2013 multiplying algebraic fractions\r\n\\[\\begin{align*} & = \\frac{5a}{7}\\times\\frac{14}{a}\\quad\\textrm{divide top and bottom by \\(a\\)}\\\\\r\n& = \\frac{5}{7}\\times\\frac{14}{1}\\quad\\textrm{divide top and bottom by \\(7\\)}\\\\\r\n& = \\frac{5}{1}\\times\\frac{2}{1}\\\\\r\n& = \\frac{10}{1}\\\\\r\n& = 10\r\n\\end{align*}\\]\r\nStart by dividing the \\(5a\\) in the top line and the \\(a\\) in the bottom line by a common factor \\(a\\). Then, divide the \\(14\\) in the top line and the \\(7\\) in the bottom line with a common factor \\(7\\).\r\nThe final simplified fractions can be multiplied to give \\(\\dfrac{10}{1}\\) which is the same as \\(10\\).\r\nIn most of these examples, the full working is shown to step you through the thinking process. You do not need to lay your working out like this, and you may take fewer steps when you work through problems yourself.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 multiplying algebraic fractions\r\nSimplify the following.\r\n\r\n\t\\(\\dfrac{4a}{3}\\times\\dfrac{9}{a}\\)\r\n\t\\(\\dfrac{32h^{2}}{9j}\\times\\dfrac{27j}{48h}\\)\r\n\t\\(\\dfrac{3d-2}{3}\\times\\dfrac{4}{3d-2}\\)\r\n\t\\(\\dfrac{2r+4}{3r-9}\\times\\dfrac{5r-15}{7r+14}\\)\r\n\t\\(\\dfrac{10p-5}{3}\\times\\dfrac{3q+3}{2p-1}\\)\r\n\t\\(\\dfrac{4g^{2}-6g}{8}\\times\\dfrac{3}{6g-9}\\)\r\n\t\\(\\dfrac{3-2y}{33y-11}\\times\\dfrac{18y^{2}-6y}{7-2y}\\)\r\n\r\n\r\n\r\n\r\nDividing algebraic fractions\r\nWe use the same technique for dividing algebraic fractions as we use for dividing numerical fractions.\r\nWhen dividing algebraic fractions, we change the divide sign to times and invert the last fraction.\r\n\r\nExample 1 \u2013 dividing algebraic fractions\r\nSimplify \\(\\dfrac{7p}{12}\\div\\dfrac{3}{8}\\). \r\n\\[\\begin{align*} & = \\frac{7p}{12}\\div\\frac{3}{8}\\quad\\textrm{change sign and invert last fraction}\\\\\r\n& = \\frac{7p}{12}\\times\\frac{8}{3}\\quad\\textrm{divide top and bottom by \\(4\\)}\\\\\r\n& = \\frac{7p}{3}\\times\\frac{2}{3}\\\\\r\n& = \\frac{14p}{9}\r\n\\end{align*}\\]\r\nSince dividing by a fraction is the same as multiplying by its reciprocal, we simply invert the \\(\\dfrac{3}{8}\\) and make it a \\(\\dfrac{8}{3}\\). The sign then changes from \\(\\div\\) to \\(\\times\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 dividing algebraic fractions\r\nSimplify the following.\r\n\r\n\t\\(\\dfrac{4m-16}{m}\\div\\dfrac{8m-32}{8m}\\)\r\n\t\\(\\dfrac{6xy-5y^{2}}{4x+10y}\\div\\dfrac{12x^{2}-10xy}{12x+30y}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nHi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on multiplying and dividing algebraic fractions. Let\u2019s start by looking at a normal fraction, 18 over 24. Notice that 18 and 24 are divisible by six, six into 18 goes three, sixes into 24 goes four, in other words the answer is three over four. See that six is a common factor of 18 and 24. Similarly 120 over 70 has a common factor of 10 because 10 will divide into both 120 and 70 leaving us with an answer of 12 over seven.\r\n\r\nHere\u2019s an algebraic fraction. Notice how I\u2019ve ... N minus N squared becomes N minus N minus N and in doing so the factor N minus N will cancel both top and bottom. Four eight squared Y over six Y squared, notice here how I\u2019ve expanded the X squared and the Y squared to look for factors, here there\u2019s a factor of Y top and bottom and there\u2019s a factor of two that divides both into four and six. And finally six X plus nine Y in the denominator will actually factorise into three brackets two X plus Y three Y and in doing so the threes now become factors top and bottom.\r\n\r\nNow let\u2019s multiply algebraic fractions. Here\u2019s a simple fraction where I\u2019ve looked for factors both top and bottom, once that\u2019s done you can cancel the factors in the numerator with factors in the denominator, as done with the previous slide. Similarly with an algebraic fraction the As will cancel because they are common factors, there\u2019s also a factor of seven.\r\n\r\nHere\u2019s a more complicated algebraic fraction, but notice the method is still the same, X minus two is a factor, X is a factor and likewise with the second example, I\u2019ve taken a factor of three out of three M plus 12 giving me three brackets M plus four, the M squared plus four M will factorise into M brackets M plus four, again notice in doing this I can now factorise the top and the bottom factors M plus four.\r\n\r\nDividing algebraic fractions is very much the same but remember the rule, when you divide by a fraction you change the division sign to a multiplication sign and then you invert the fraction that comes afterwards. Here I\u2019ve done this and I\u2019ve factored the eight and 12. Always cancel where possible.\r\n\r\nThe second example I\u2019ve done exactly the same, the factor of four is on the top and 24 on the bottom, fours into four, fours into 24 and finally here\u2019s another example where I\u2019ve taken a factor of two outside of two A plus four allowing me to factorise the top and bottom of the fraction, the factor here being the A plus two.\r\n\r\nNow try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.034},[{"v":"Maths equations","i":3,"n":0.707},{"v":"Functions","i":2,"n":1},{"v":"Arithmetic","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":371,"$":[{"v":"Adding and subtracting algebraic fractions","n":0.447},{"v":"\r\nLet's combine your knowledge of fractions and algebra, and learn how to add and subtract algebraic fractions. By learning to find common denominators and simplify expressions, you'll enhance your ability to work with complex algebraic fractions, a skill crucial for solving equations and tackling advanced mathematical challenges.\r\n\r\nVideo tutorial \u2013 algebraic fractions: addition and subtraction\r\nWatch this video to learn about adding and subtracting algebraic fractions.\r\n\r\n\r\nAdding and subtracting algebraic fractions\r\nAlgebraic fractions are fractions that contain pronumerals. For example, \\(\\dfrac{a}{2}\\), \\(\\dfrac{2b}{3}\\), \\(\\dfrac{1}{2c}\\), \\(\\dfrac{e+1}{2}\\), \\(\\dfrac{3}{a-2}\\), \\(\\dfrac{a}{b}\\) and \\(\\dfrac{a}{b+c}\\) are all algebraic fractions.\r\nAlgebraic fractions can be added and subtracted just like ordinary fractions. Their denominators must be the same before you add or subtract.\r\n\r\nExample 1 \u2013 adding or subtracting algebraic fractions\r\n\\[\\begin{align*} \\frac{h}{6}+\\frac{2h}{9} & = \\left(\\frac{h}{6}\\times\\frac{3}{3}\\right)+\\left(\\frac{2h}{9}\\times\\frac{2}{2}\\right)\\\\ \r\n& = \\frac{3h}{18}+\\frac{4h}{18}\\\\\r\n& = \\frac{3h+4h}{18}\\\\\r\n& = \\frac{7h}{18}\r\n\\end{align*}\\]\r\nHere, the lowest common denominator for \\(6\\) and \\(9\\) is \\(18\\). We multiply the first fraction by \\(\\dfrac{3}{3}\\) and the second fraction by \\(\\dfrac{2}{2}\\). You could also use the common denominator found by multiplying the two denominators, \\(6\\times9=54\\), and then simplify later.\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 adding or subtracting algebraic fractions\r\nSimplify the following.\r\n\r\n\t\\(\\dfrac{x}{3}-\\dfrac{x}{5}\\)\r\n\t\\(\\dfrac{2p}{7}-\\dfrac{p}{4}\\)\r\n\t\\(\\dfrac{2g}{3}+\\dfrac{g+1}{4}\\)\r\n\t\\(\\dfrac{d+3}{2}+\\dfrac{1-d}{4}\\)\r\n\t\\(\\dfrac{5}{9}-\\dfrac{3}{b}\\)\r\n\t\\(\\dfrac{3x+2}{5}-\\dfrac{x-3}{10}\\)\r\n\t\\(\\dfrac{3}{v}+\\dfrac{2}{v+1}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nHi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on adding and subtracting algebraic fractions. Let's start by looking at a simple fraction, 18 over 24. Notice that 18 and 24 are divisible by six, six into 18 goes three, six into 24 goes four; we say that 18 and 24 have a common factor of six, so the answer is three quarters. Similarly 120 over 70 is divisible by 10, in other words 120 and 70 have a common factor of 10, that gives us an answer of 12 over seven.\r\n\r\nDon't forget that only fractions that have a common denominator can be added or subtracted. For instance, in this case, seven tenths minus three sevenths, the common denominator here is 70, in other words 10 will divide into 70 and so will seven, so we have to carry out that process shown on the screen. At the bottom of the slide we are now subtracting these two fractions to give us an answer of 19 over 70.\r\n\r\nNow let's move on to algebraic fractions. Again we're using the same idea as with normal fractions. Here we have H over six plus two H over nine, notice that the six and nine are different so we need a common denominator which, in this case, is 18 and thus we go through the process of adding two algebraic fractions with a common denominator of 18. Once we get to the bottom of our problem we then add the two terms up, three H plus four H which gives us seven H, notice the terms here are alike.\r\n\r\nHere's another example, E plus one over two plus E over five, two and five must divide into a number, the common denominator here would have to be 10 so that equals E plus one over two times five over five plus E over five times two over two. Notice the next line down we have brackets, E plus one, be careful here we're multiplying five by E plus one, so it's five brackets E plus one plus two E over 10. Once we expand the brackets we get five E plus five plus two E, that gives us seven E plus five. Notice I'm adding the like terms together and there are two different terms here, an E term and a number, which is five.\r\n\r\nNow let's look at subtracting algebraic fractions, five over two A minus three over four, so we need a common denominator, a number, that both two A and four will divide into. So four A will be that common denominator so we'd need to multiply two A by two and four by A to get our common denominator of four A. Then we have in the bottom line of our slide 10 minus three A, notice here that the terms are different so we can't do anything about them, we leave them alone, so the answer is 10 minus three A over four A.\r\n\r\nNow try some questions for yourselves. The answers to these problems are on the next slide. Thanks for watching this short movie.\r\n\r\n","n":0.038},[{"v":"Functions","i":2,"n":1},{"v":"Arithmetic","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":372,"$":[{"v":"Algebraic operations","n":0.707},{"v":"Algebraic operations are critical to many fields of study. In science, they are used to calculate forces to accelerate a car or determine the rate of a chemical reaction. In engineering, they can be used to design safe and efficient systems. They also underpin algorithms and data structures, which are fundamental to software development and solving complex computational problems. This resource introduces you to the basic skills for adding (\\(+\\)), subtracting (\\(-\\)), multiplying (\\(\\times\\)) and dividing (\\(÷\\)) algebraic terms.\r\nAlgebraic terms\r\nBefore we dive into algebraic operations, let's recap the components of an algebraic term. An algebraic term consists of a pronumeral and, often, a coefficient.\r\n\r\n\tA pronumeral is a letter that represents a number or range of numbers.\r\n\tA coefficient is the number written before the pronumeral. It represents how many lots of the pronumeral are present.\r\n\r\nFor example, \\(3x\\) is an algebraic term with \\(3\\) as the coefficient and \\(x\\) as the pronumeral. What this really means is \\(3\\times x\\) or \\(3\\) lots of \\(x\\).\r\nAnother example is \\(5ab\\). This is \\(5\\times a \\times b\\). It is the same as \\(5ba\\). The order of the letters and numbers is not important but we usually write the numbers first and use alphabetical order for the letters. \\(5ba\\) would be correct, but it is more common to write \\(5ab\\).\r\nWhen there is no coefficient written before the pronumeral, we assume that the coefficient is \\(1\\).\r\n\r\nVideo tutorial \u2013 algebraic operations\r\nWatch this video to learn about algebraic operations.\r\n\r\n\r\nLike and unlike terms\r\nLike terms contain exactly the same pronumerals, raised to the same power. For example, \\(x\\) and \\(5x\\) are like terms, but \\(x\\) and \\(x^{2}\\) are unlike terms. The following table gives some examples of like and unlike terms.\r\n\r\n\r\n\t\r\n\t\tLike terms\r\n\t\tUnlike terms\r\n\t\r\n\r\n\r\n\t\r\n\t\t\\(3x,\\,5x\\)\r\n\t\t\\(3x,\\,5y\\)\r\n\t\r\n\t\r\n\t\t\\(2a,\\,3a\\)\r\n\t\t\\(3a,\\,3\\)\r\n\t\r\n\t\r\n\t\t\\(m^{2},\\,7m^{2}\\)\r\n\t\t\\(m^{2},\\,7m\\)\r\n\t\r\n\t\r\n\t\t\\(2ab,\\,3ab\\)\r\n\t\t\\(2a^{2}b,\\,3ab^{2}\\)\r\n\t\r\n\t\r\n\t\t\\(3xyz,\\,5xyz\\)\r\n\t\t\\(3xyz,\\,3xy\\)\r\n\t\r\n\t\r\n\t\t\\(3ef,\\,3fe\\)\r\n\t\t\\(3ef,\\,3eg\\)\r\n\t\r\n\t\r\n\t\t\\(6\\alpha\\beta,\\,\\beta\\alpha\\)\r\n\t\t\\(6\\alpha\\beta\\gamma,\\,6\\alpha\\beta\\)\r\n\r\n\r\n\r\n\r\nExercise \u2013 finding like and unlike terms\r\nFor each row in the table, identify the term in the second column that is a like term for the term in the first column.\r\n\r\n\r\n\r\n\t\r\n\t\tTerm\r\n\t\tWhich is the like term?\r\n\t\r\n\r\n\r\n\t\r\n\t\t\\(3x\\)\r\n\t\t\\(3,\\,2x,\\,3x^{2},\\,2xy,\\,4x^{2}a\\)\r\n\t\r\n\t\t\\(2ab\\)\r\n\t\t\\(2a,\\,2b,\\,3x^{2},\\,6abc,\\,12ab\\)\r\n\t\r\n\t\t\\(2x^{2}\\)\r\n\t\t\\(x,\\,2x,\\,5x^{2},\\,4x,\\,4x^{3}\\)\r\n\t\r\n\t\r\n\t\t\\(3xy^{2}\\)\r\n\t\t\\(3,\\,3xy,\\,3x^{2}xy^{2},\\,3y^{2},\\,y^{2}x\\)\r\n\t\r\n\t\r\n\t\t\\(2m^{2}n\\)\r\n\t\t\\(n,\\,mn^{2},\\,8mn,\\,2m^{2},\\,4nm^{2}\\)\r\n\t\r\n\t\r\n\t\t\\(4ab^{2}c\\)\r\n\t\t\\(4,\\,2ab^{2}c,\\,4abc,\\,8b^{2}c,\\,4cba\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\nAddition and subtraction\r\nThe essential rule for adding or subtracting terms is:\r\nOnly like terms may be added or subtracted.\r\n\r\nExample 1 \u2013 adding and subtracting algebraic terms\r\n\\[\\begin{align*} 7e+10e & = (7\\times e)+(10\\times e)\\\\\r\n& = (7+10)e\\\\\r\n& = 17e\r\n\\end{align*}\\]\r\nHere, the terms have been broken up into their coefficients and pronumerals to show you what they mean. In practice, we do not do this and just go straight to the answer \u2013 that is, \\(7e+10e=17e\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 adding and subtracting algebraic terms\r\n\r\n\t\\(13x+4-3x-1\\)\r\n\t\\(10mn+5m+12mn+6m\\)\r\n\t\\(3xy^{2}+2xy+5xy^{2}+3xy\\)\r\n\t\\(5xy+6m-2xy-2m\\)\r\n\t\\(x+y+2x-y\\)\r\n\t\\(3a+5b-a-6b\\)\r\n\t\\(x+4y-x-2y\\)\r\n\t\\(7x-4m-5x-3m\\)\r\n\t\\(4x-5x-3y+5x\\)\r\n\t\\(9mn-3m-n+4m^{2}\\)\r\n\r\n\r\n\r\nMultiplication\r\nBoth like and unlike terms can be multiplied.\r\nWhen multiplying two or more terms:\r\n\r\n\tDetermine the sign of the answer.\r\n\t\r\n\t\t\\((+\\textrm{ve})\\times (+\\textrm{ve})=(+\\textrm{ve})\\)\r\n\t\t\\((+\\textrm{ve})\\times (-\\textrm{ve})=(-\\textrm{ve})\\)\r\n\t\t\\((-\\textrm{ve})\\times (-\\textrm{ve})=(+\\textrm{ve})\\)\r\n\t\r\n\tCalculate the product of the coefficients.\r\n\tUse index laws to find the powers of the pronumerals.\r\n\r\n\r\nExample 1 \u2013 multiplying algebraic terms\r\n\\[(-4) \\times (-3b) = 12b\\]\r\n\\((-\\textrm{ve})\\times (-\\textrm{ve})=(+\\textrm{ve})\\), so the sign of the answer is \\(+\\). We take the product of the coefficients: \\(4\\times 3=12\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 multiplying algebraic terms\r\n\r\n\t\\(5\\times 2k\\)\r\n\t\\(4a \\times 3ab\\)\r\n\t\\(y \\times 3y\\)\r\n\t\\(4m \\times (-3mn)\\)\r\n\t\\(m \\times 3p \\times 5\\)\r\n\t\\(2ab \\times 3bc \\times (-4)\\)\r\n\t\\(2ab^{2} \\times 3ac\\)\r\n\t\\(4m \\times (-5kmp)\\)\r\n\r\n\r\n\r\nDivision\r\nWhen we divide numbers, we can rewrite them as fractions where the first number is the numerator and the second number that it is divided by is the denominator. For example:\r\n\\[2\\div5=\\frac{2}{5}\\]\r\nWe can do the same for algebraic terms:\r\n\\[2x^{2}yz \\div 3xy^{2}= \\frac{2x^{2}yz}{2xy^{2}}\\]\r\nLike and unlike terms can be divided.\r\nWhen dividing algebraic terms:\r\n\r\n\tDetermine the sign of the answer.\r\n\t\r\n\t\t\\((+\\textrm{ve})\\div (+\\textrm{ve})=(+\\textrm{ve})\\)\r\n\t\t\\((+\\textrm{ve})\\div (-\\textrm{ve})=(-\\textrm{ve})\\)\r\n\t\t\\((-\\textrm{ve})\\div (+\\textrm{ve})=(-\\textrm{ve})\\)\r\n\t\t\\((-\\textrm{ve})\\div (-\\textrm{ve})=(+\\textrm{ve})\\)\r\n\t\r\n\tCalculate the quotient of the coefficients or cancel any common factors.\r\n\tUse index laws to find the powers of the pronumerals.\r\n\r\n\r\nExample 1 \u2013 dividing algebraic terms\r\nDivide \\(-12wyz\\) by \\(3yz\\).\r\n\\[\\begin{align*} -12wyz \\div 3yz & = \\frac{-12wyz}{3yz}\\\\\r\n& = - \\frac{12wyz}{3yz}\\\\\r\n& = -4w\r\n\\end{align*}\\]\r\nAfter rewriting the question as a fraction, the sign was determined: \\(-\\textrm{ve} \\div +\\textrm{ve} = -\\textrm{ve}\\).\r\nThen, the quotient of the coefficients was found: \\(12 \\div 3=4\\). This is the same as identifying the common factor of \\(3\\) and dividing both the numerator and denominator by \\(3\\). \\(12 \\div 3=4\\) so the numerator becomes \\(4\\) and \\(3 \\div 3=1\\) so the denominator becomes \\(1\\).\r\nLooking at the pronumerals, the numerator has \\(wyz\\) and the denominator has \\(yz\\). The \\(y\\) and \\(z\\) can be cancelled out to leave \\(w\\) in the numerator.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 dividing algebraic terms\r\n\r\n\t\\(18ef \\div 6f\\)\r\n\t\\(-100uvw \\div 100w\\)\r\n\t\\(24gh^{2} \\div 8gh\\)\r\n\t\\(3m^{2}n \\div 12mn^{2}\\)\r\n\t\\(rs \\times 2st \\div 2s\\)\r\n\t\\(3jk \\times 12km \\div 9jkm\\)\r\n\t\\(10p \\times 3qp \\div 16pq\\)\r\n\t\\(4yz \\times 5w^{2}z \\div 10wy\\)\r\n\r\n\r\n\r\nOrder of operations\r\n\r\nJust like in basic arithmetic, the order of operations in algebraic operations are important.\r\n\r\nExample \u2013 applying BIMDAS\r\n\\[\\begin{align*} 3st-3s \\times 4t & = 3st-12st\\\\\r\n& = -9st\r\n\\end{align*}\\]\r\nAccording to BIMDAS, the multiplication is completed first: \\(3s \\times 4t=12st\\). Then, the subtraction: \\(3st-12st=-9st\\).\r\n\r\nExercise \u2013 applying BIMDAS\r\nSimplify the following expressions:\r\n\r\n\t\\(3m+2\\times 3m\\)\r\n\t\\(6ab-3a\\times 4b\\)\r\n\t\\(16gh-4gh\\times 4\\)\r\n\t\\(9b-3b\\times 2k+2k\\times b\\)\r\n\r\n\r\n\r\n\r\n \r\n\r\nM: Hi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on operations with algebra. First of all make sure you can tell the difference between like terms and unlike terms. Like terms contain exactly the same pronumerals, in other words letters or variables, so look at the like terms down the left hand column and compare them with the unlike terms in the right hand column, in particular look at the last term, seven E F and eight F E, they are in fact like terms. Only like terms can be added or subtracted. For instance, in this example, seven E plus 10 E, the terms are alike, therefore they can be added to give you 17 E. But look at this one, be careful because here we have two terms, the X squared term and the X term so we have to add them separately, so three X squared minus X squared minus four X squared gives us minus two X squared, and two X, which is different to the X squared term must be left alone so the answer is minus two X squared plus two X.Similarly if you look at this example in red, there's a U V term which is the same as V U and that is different to the U term therefore that gives us minus two U V plus three U, we keep them separate, we can't add minus two U V to three U. And finally six R squared S minus two R S squared, they look similar but they are not the same, these are two unlike terms so we can't simplify this expression.Now before we move on to multiplying and dividing algebraic terms make sure you are familiar with the multiplication and dividing of positive and negative numbers on this slide. So study this carefully before moving onto the next slide.\r\n\r\nBut let's multiply some pronumerals. Here's the first example, minus four times three B, multiply the numbers together, minus four and minus three gives you plus 12 and then just tag on the B afterwards so the answer is 12 B. Here's another one, multiply out the numbers first, three times minus 15 gives you minus 15, now multiply the pronumeral, E and E squared is E cubed, so the answer is minus 15 E cubed.\r\n\r\nThe next example in red multiply out, as before, the numbers, minus two times minus four is plus eight, then the pronumerals, there's a U squared term and there are Vs there, V times V is a V squared, so the answer is eight U squared V squared. Notice in these examples that because we're multiplying the actual terms can be different, so make sure you know the difference between adding and subtracting where the pro\u2011numerals must be exactly the same, multiplying and dividing the pronumerals can be different.\r\n\r\nFinally minus three P Q times minus two Q times P, multiply out the numbers, that gives you plus six, P times P is P squared, Q times Q is Q squared, so the answer is six P squared Q squared. Now in this slide we're dividing pronumerals, so the first term minus 12 W Y Z divided by the second term three Y Z, express that as a fraction first and look for terms that will cancel top and bottom, in this case the Ys cancel top and bottom and the Zs cancel top and bottom, notice three will divide into 12 four times, so the answer is minus four W.\r\n\r\nHere's another one slightly more complicated. Now in this slide we're dividing pro\u2011numerals, so the first term minus 12 W Y Z divided by the second term three Y Z, express that as a fraction first and look for terms that will cancel top and bottom, in this case the Ys cancel top and bottom and the Zs cancel top and bottom, notice three will divide into 12 four times so the answer is minus four W.\r\n\r\nHere's another one slightly more complicated. Writing it as a fraction, minus two M squared N over six M N squared, and then in the third term I've expanded the M squared so it's M times N and then I've tacked the N on the end of that and I've done the same thing with the denominator with N squared N times N, so you can clearly see where the Ms and the Ns cancel, giving you an answer of minus M over three N. Notice minus on the top divided by plus on the bottom gives you an answer of minus.\r\n\r\nAnd finally in blue there's a multiplication and a division so you again express that as a fraction, expand the A squared and then cancel top and bottom terms giving you an answer of two A. Now we'll move on to some slightly harder problems with algebra, but notice the rule that we use first of all, in other words if an expression has brackets in it that must be done first, secondly if there are indices you do that next, thirdly you do multiplying and dividing and lastly you do adding and subtraction.\r\n\r\nSo let's do some examples. Three S T minus three S times four T, notice we have a negative sign and a multiplication sign, so the times is worked out first, so three S times four T is 12 S T, notice now we have subtraction, three S T minus 12 S T, notice the terms are alike therefore we can subtract them, three minus 12 is minus nine S T.\r\n\r\nLet's look at this one. We've got two terms in brackets and a division, so the brackets are worked out first, even though there's a plus inside the brackets must be worked out first, which gives us six X Y, then we divide that by two X, express that as a fraction, cancel any terms, Xs cancel here, twos go into six three times so the answer is three Y.\r\n\r\nAnd finally look at this one. We have brackets, multiplication, addition and division. So the brackets is worked out first, that gives as minus 10 A squared B squared in the second line, then I write out the rest, plus two A cubed B cubed divided by A B, what do we do next, well we have division which becomes ... comes before addition, therefore we divide the two A cubed B cubed by A B, writing that as a fraction, simplifying by cancelling like terms and then in the final line we've just got two terms which we can add together. Notice again that the terms are exactly the same, A squared B squared, therefore we can add them, minus 10 plus two gives us minus eight A squared B squared.\r\n\r\nNow try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short video.\r\n\r\n","n":0.023},[{"v":"Maths equations","i":3,"n":0.707},{"v":"Functions","i":2,"n":1},{"v":"Arithmetic","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":373,"$":[{"v":"Expansion","n":1},{"v":"Brackets often appear in algebraic expressions and equations, but what do you do with them? Whatever is around the brackets affects everything inside the brackets, and being able to remove them helps set the stage for solving equations. The process of doing this is called expansion.\r\n\r\nBrackets in algebra\r\nBrackets are commonly used to express mathematical formulae. They are used to group numbers, pronumerals and operations together as a whole. To manipulate expressions containing brackets, you need to remove them first.\r\n\r\nVideo tutorial \u2013 removing brackets\r\nWatch this video to learn about removing brackets.\r\n\r\n\r\nThe distributive law\r\nThe distributive law states that:\r\n\\[a(b+c) = ab+ac\\]\r\nwhere \\(a(b+c)\\) is the same as \\(a \\times (b+c)\\).\r\nIt indicates that when a group of numbers or terms that are added together is multiplied by one number, \\(a\\), it is the same as multiplying the numbers or terms separately. In other words, the \\(a\\) outside the brackets is multiplied by everything inside the brackets.\r\nThe process of removing brackets based on the distributive law is called expansion.\r\n\r\nExample 1 \u2013 expanding algebraic expressions\r\nExpand \\(3(x+2)\\).\r\n\\[\\begin{align*} 3(x+2) & =(3\\times x)+(3\\times2)\\\\\r\n& = 3x+6\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 expanding algebraic expressions\r\n\r\n\tExpand the following.\r\n\t\r\n\t\t\\(2(x-5)\\)\r\n\t\t\\(3(6-b)\\)\r\n\t\t\\((b+5)c\\)\r\n\t\t\\(2x(2a+2b)\\)\r\n\t\t\\(-6(p-2q+4)\\)\r\n\t\t\\(3y(yz-2y+1)\\)\r\n\t\r\n\tExpand and simplify the following.\r\n\t\r\n\t\t\\(3(x+1)+2(x+2)\\)\r\n\t\t\\(2(p-1)-(p-3)\\)\r\n\t\t\\(2(4m-3)-5(9-2m)\\)\r\n\t\t\\(2w(3w+1)+3w(2w-5)\\)\r\n\t\t\\(2a(a-1)-a(a-3)\\)\r\n\t\t\\(5j(3j+2)-2j(4j-1)\\)\r\n\t\t\\(3q(5q+4)+6(8-3q)\\)\r\n\t\t\\(2r(3r-1)-3r(3+2r)\\)\r\n\t\r\n\tExpand and simplify the following by applying the correct order of operations.\r\n\t\r\n\t\t\\(4(2m-3)+8\\)\r\n\t\t\\(9-3(4b+3)\\)\r\n\t\t\\(1-4(x-1)\\)\r\n\t\t\\(4-(5-2x)\\)\r\n\t\t\\(8m-3(1-2m)+6\\)\r\n\t\t\\(9c(4-c)+2(c-7)\\)\r\n\t\r\n\r\n\r\n\r\nBinomial products\r\nA binomial is an algebraic expression made up of the sum or difference of two terms. For example, \\((m+2)\\) and \\((m+3)\\) are binomials. \\((a+3)\\), \\((1+c)\\), \\((a+b)\\) and \\((3x+2y)\\) are also binomials.\r\nA binomial product is therefore the product of two binomial terms. Using the first two examples, \\((m+2)(m+3)\\) would be the binomial product.\r\n\r\nTo expand the two brackets, multiply each term in the first bracket by each term in the second bracket. Then simplify if possible.\r\nFor \\((m+2)(m+3)\\):\r\n\\[\\begin{align*} (m+2)(m+3) & = m(m+3)+2(m+3)\\\\\r\n& = m^{2}+3m+2m+6\\\\\r\n& = m^{2}+5m+6\r\n\\end{align*}\\]\r\n\r\nSymbolically we can write:\r\n\\[(a+b)(c+d)=ac+ad+bc+bd\\]\r\n\r\nSpecial cases\r\nWhen the binomial product consists of two of the same binomials, there are two special cases that occur.\r\nFor \\((a+b)(a+b)=(a+b)^{2}\\), we can expand and simplify as shown:\r\n\\[\\begin{align*} (a+b)^{2} & = (a+b)(a+b)\\\\\r\n& = a^{2}+ab+ba+b^{2}\\\\\r\n& = a^{2}+2ab+b^{2} & (1)\r\n\\end{align*}\\]\r\nThus:\r\n\r\nFor \\((a-b)(a-b)=(a-b)^{2}\\), similarly, we can expand and simplify:\r\n\\[\\begin{align*} (a-b)^{2} & (a-b)(a-b)\\\\\r\n& = a^{2}-ab-ba+b^{2}\\\\\r\n& = a^{2}-2ab+b^{2} & (2)\r\n\\end{align*}\\]\r\nThus:\r\n\r\nThese occur often in mathematics so you should try your best to remember them.\r\n\r\nExample 1 \u2013 expanding binomial products\r\nExpand \\((2f+3)(g+5h)\\).\r\n\r\nIn this example, no further simplification is possible.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 expanding binomial products\r\n\r\n\tExpand the following binomial products.\r\n\t\r\n\t\t\\((x+5)(x+3)\\)\r\n\t\t\\((u-5)(u+3)\\)\r\n\t\t\\((k+6)(j+1)\\)\r\n\t\t\\((y+2)(y-2)\\)\r\n\t\t\\((e-7)(e-8)\\)\r\n\t\t\\((t-3)(5-t)\\)\r\n\t\t\\((3q+2)(q+1)\\)\r\n\t\t\\((8a+5)(a-3)\\)\r\n\t\t\\((2c-d)(3c+2d)\\)\r\n\t\t\\((3a-2b)(3a+2b)\\)\r\n\t\r\n\tExpand the following.\r\n\t\r\n\t\t\\((d+3)^{2}\\)\r\n\t\t\\((x-2)^{2}\\)\r\n\t\t\\((3v+2)^{2}\\)\r\n\t\t\\((y-3z)^{2}\\)\r\n\t\t\\((3-2q)^{2}\\)\r\n\t\t\\((4f-3gh)^{2}\\)\r\n\t\r\n\r\n\r\n\r\nIndices in algebra\r\nThe index laws may be used to simplify algebraic expressions.\r\nExample 1 \u2013 simplifying algebraic expressions with indices\r\nSimplify \\(\\left(x^{3}y^{9}\\right)^{\\frac{2}{3}}\\).\r\n\\[\\begin{align*} \\left(x^{3}y^{9}\\right)^{\\frac{2}{3}} & = x^{3\\times\\frac{2}{3}}y^{9\\times\\frac{2}{3}}\\\\\r\n& = x^{2}y^{6}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\nExercise \u2013 simplifying algebraic expressions with indices\r\nSimplify the following expressions. Give your answers in index notation with positive indices.\r\n\r\n\t\\(\\sqrt{a}\\times\\sqrt[4]{a}\\times\\sqrt[3]{a^{2}}\\)\r\n\t\\(\\left(125a^{6}b\\right)^{\\frac{1}{3}}\\times b^{\\frac{2}{3}}\\)\r\n\t\\(\\dfrac{\\left(2xy^{3}\\right)^{\\frac{1}{2}}}{2}\\times \\left(\\dfrac{x^{\\frac{3}{2}}}{y^{2}}\\right)^{4}\\)\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nHi, I'm Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on removing brackets in algebra.\r\n\r\nThe law that allows us to expand expressions with brackets in maths is called the distributive law. For instance, A brackets B plus C, is the same as A times B plus A times C, you're multiplying the term outside the brackets by the term inside the brackets, an alternate way of writing A times brackets B plus C, same thing, A times B plus A times C.\r\n\r\nBefore we get into some of the examples just recall some of our work we've done in the past, multiplying positive and negative numbers. Remember if the signs are the same, plus plus or minus minus the answer is positive, if the signs are different, plus and minus, then the answer is negative.\r\n\r\nSo let's use this law and multiply out this particular expression just with numbers; two times brackets three plus five, so we're going two times three plus two times five, two times three is six, two times five is ten, answer is 16. Let's look at another example but here we're using algebra; three brackets X plus two, we use the same law again, three times X plus three times two, which is three X plus six, notice we can't add three X to six because the terms are different. Similarly with negative signs, minus five brackets M plus four gives us minus five times M plus minus five times four, which is minus five M minus 20, the terms are different so they must be left alone. And finally, with two negative signs, so be careful here, minus two brackets P minus seven is minus two times P plus minus two times minus seven, which is minus two P plus 14.\r\n\r\nLet's now move on with two sets of brackets, four brackets M minus three plus two brackets M plus one. We use the same ideas as before giving us four M minus 12 plus two M plus two, collect the like terms and we've got six M minus 10. Similarly, two sets of brackets but this time we have to be really careful because there's a negative sign. Minus six X brackets two X plus three minus three X brackets X minus three gives us minus 12 X squared minus 18 X minus 3 X squared and there it is we've got plus 9 X at the end, two negatives give us a positive, collecting up the like terms we've got minus 15 X squared minus 9 X.\r\n\r\nNow let's looked at two sets of brackets that are placed side by side, in other words we're multiplying the first set of brackets by the second set of brackets. And if you look at the example in blue the idea is to multiply the first term in the first set of brackets by the second set of brackets and the second term in the first set of brackets by the second set of brackets, in other words X multiplied by X plus three plus two multiplied by X plus three. If we expand these brackets further we have X squared plus three X plus two X plus six and collecting the like terms gives us an answer of A squared plus five X plus six.\r\n\r\nHere are two more examples that have been written out for you. The one in blue at the top, notice there's a negative sign so be careful here with the negative signs, giving us an answer of M squared minus M minus 20 and the example in red there are two negative signs here, so be doubly careful when multiplying out these sets of brackets, giving us an answer of Q squared minus nine Q plus 18.\r\n\r\nNow try some questions for yourself. The answers to these problems are on the next slide. Thanks for watching this short movie.\r\n","n":0.031},[{"v":"Maths equations","i":3,"n":0.707},{"v":"Functions","i":2,"n":1},{"v":"Arithmetic","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":374,"$":[{"v":"Arguments and evidence","n":0.577},{"v":"\r\nSelecting credible sources for your assignments enhances the quality and reliability of your work, making your arguments more persuasive and impactful. On the other hand, choosing the wrong sources can make your arguments weaker.\r\nFind out the difference between a strong and weak argument, and what information sources are considered reliable.\r\n\r\nStart animationStop animation\r\n\r\n\r\nArgument/evidence gif, by RMIT, licensed under CC BY-NC 4.0\r\n\r\n\r\n \r\n\r\nWeak arguments\r\nWeak arguments rely on insufficient, unreliable, or irrelevant evidence, leading to unconvincing conclusions.\r\nExamples\r\nHere are some, but not all, types of information sources that can weaken your work if relied upon too heavily. Understanding these can help you make more informed choices in your research.\r\n\r\n \tOpinion pieces: These are based on personal views and may lack factual support. For example, a blog post about climate change without scientific evidence. When reading opinion pieces, keep in mind who the author is and if they have any strong beliefs or affiliations which could lead to bias.\r\n \tOutdated sources: Information from a decade ago might not reflect current knowledge. This is particularly the case in medical and technology fields.\r\n \tAnecdotal pieces: These rely on personal stories rather than data. For example, websites that make claims about conspiracies may not list their sources about where their information has come from and should be treated with caution.\r\n \tUnverified online sources: Websites without credibility checks. For example, a random wiki page edited by anyone.\r\n \tNon-peer-reviewed articles: Articles that haven\u2019t been evaluated by experts, for example, a self-published paper on a complex scientific topic, should also be treated with caution. This is because the way that the study was conducted, the results and the interpretation of the results have not been checked and verified by industry experts.\r\n \tAssumptions: These are ideas presented without evidence. For example, if I assume that all students learn the same way without considering individual differences, this can weaken my argument.\r\n\r\nStrong evidence\r\nStrong information sources provide reliable, relevant, and well-supported data that makes your work more convincing. Below are some key types of strong information sources to guide you in selecting the most effective materials for your research.\r\nExamples\r\n\r\n \tPeer-reviewed journal articles: These articles undergo evaluation by experts in the field, ensuring accuracy and validity. For example, a research article published in well-respected journals like\u202fNature\u202for\u202fThe Lancet.\r\n \tGovernment data and official reports: These sources offer authoritative and reliable information. For instance, the Australian Bureau of Statistics provides comprehensive data on demographics and economic indicators, which can be crucial for evidence-based research.\r\n \tBooks by experts: Authored by recognised authorities, these books provide in-depth analysis and insights. An example is a comprehensive text on economics by a renowned economist.\r\n \tAcademic conference papers: These are often peer-reviewed and they usually present current research and developments. For example, recordings from the International Conference on Artificial Intelligence could be a strong source of information for students of information technology subjects.\r\n \tReputable news outlets: Articles from news sources that are known for fact-checking and reliability, such as articles from the ABC, the BBC or The Guardian, often offer well-researched information.\r\n\r\nResponding to authors' claims\r\nWhile you read books and articles or watch videos to find information for your assignments, remember to think about how the authors support their claims. Are they using evidence from reputable sources or are they relying on unchecked sources and assumptions?\r\n\r\nOnce you have found information that you would like to cite in your work, here are some techniques that can be used. In these examples, the writer will use weak points on which to base their counter arguments and strong points to support their argument.\r\nConcession + counter claim\r\nExample: Admittedly, there are benefits to Bako\u2019s (2022) plan for overhauling the education system. However, they do not justify the costs that will be borne by taxpayers.\r\n\r\nRefute with evidence\r\nExample: Cienfuegos (2024) claims that intermittent fasting can reduce inflammation. This is despite a lack of support from recent empirical studies.\r\n\r\nAcknowledge opposing views\r\nExample: A number of academics (Daalman, 2021; Aronov & Weller, 2021; Stojanovski 2019) are of the view that an excess of sugar can cause hyperactivity in children, but a new study by Zapatero (2024) shows that this may not be the case.\r\n\r\nIdentify a lack of evidence\r\nExample: Whittaker (2023) does not support their claim with evidence from a reliable source.\r\n\r\nComparative analysis\r\nExample: Smith (2020) uses anecdotal evidence from a personal experience to conclude that positive education is completely ineffective. On the other hand, Christopoulos (2023) conducted a systematic review that shows that positive education can be an effective classroom tool.\r\n\r\n const manual = new Freezeframe('.freezeme', { trigger: false }); document .querySelector('.start') .addEventListener('click', () => { manual.start(); }); document .querySelector('.stop') .addEventListener('click', () => { manual.stop(); }); ","n":0.036},[{"v":"Media literacy","i":4,"n":0.707},{"v":"Researching","i":3,"n":1},{"v":"Critical thinking","i":2,"n":0.707},{"v":"Critical reading","i":1,"n":0.707},{"v":"Analysis","i":0,"n":1}]]},{"i":375,"$":[{"v":"Persuasive language techniques","n":0.577},{"v":"Understanding how the persuasive language authors use can help us to understand their potential motivations and biases. On this page, we'll examine how an author has used persuasive language in their argument.\r\nFirst of all, read the article The cost of change. Then, read the Persuasive language analysis section to understand the techniques used by the writer.\r\n\r\nThe cost of change\r\nToday\u2019s guest columnist is Senator David Davidson, from the Very Conservative Party (VCP).\r\nToday, people are so obsessed with change that they rarely take the time to reflect on why they want things to change. We are programmed to eagerly anticipate the next technological development, the next philosophical shift, the next cause of political and social outrage\u2026\r\n\r\nWe have lost the ability to be satisfied with what we have right now and to understand how good we truly have it. Spend five minutes on social media and you will be bombarded with a tidal wave of discontent. Everyone, it seems, has something to complain about.\r\n\r\nUnless you have been living under a rock, you will have heard of phenomena such as \u2018cancel culture\u2019 and \u2018Me Too\u2019- if you are more unfortunate, you may even have been \u2018cancelled\u2019 yourself. A recent study from a prominent Australian university discovered that the emotional toll of being \u2018cancelled\u2019 is causing everyday Australians severe anguish. In fact, the study revealed that 78.7% of people are so worried that they will be \u2018cancelled\u2019 that they are now fearful to speak their minds in a public forum.\r\n\r\nThat\u2019s why the VCP will be introducing a bill to Parliament this week that proposes changes to our current Freedom of Speech laws. The changes would remove penalties for harmless comments about people\u2019s physical attributes, their personal beliefs and their cultural background.\r\n\r\nMany experts in the area of anthropology have identified that countries where people have to watch their words carefully have higher rates of depression. Our proposed laws will bring this country back to being a happy country where people can speak without consequences.\r\n\r\nThe time has come to act. Our country is being overrun by wannabe socialists with left-leaning political agendas. We urge you to petition your local member of Parliament to support our bill before decent, conservative-minded Australians like you and I are no longer able to live our lives the way we always have.\r\n\r\nTogether we can stop the nonsense. Join us and we can get this country back on track.\r\nPersuasive language analysis\r\nThe author of The cost of change uses a number of persuasive techniques to influence the reader, such as:\r\n\r\n \tEmotional appeal: The author taps into feelings of nostalgia and fear, suggesting that people are losing satisfaction with the direction of society by mentioning \"severe anguish\" and \"fearful to speak\".\r\n \tBandwagon: Phrases like \"unless you have been living under a rock\" suggest that the issues are universally acknowledged, encouraging readers to align with the majority.\r\n \tStatistics: The use of a specific percentage (78.7%) from a \"recent study\" attempts to add credibility. However, the specific source is not mentioned, and the author tells us the information came from a \"prominent Australian university.\"\r\n \tAppeal to authority: The article references \"many experts in the area of anthropology\" to support the argument but does not mention who the experts are.\r\n \tFear of consequences: The text warns of a \"left-leaning political agenda\" and suggests that conservative values are under threat, urging immediate action.\r\n \tCall to action: The author strongly encourages readers to \"petition your local member of Parliament,\" creating a sense of urgency.\r\n \tRidicule/attack: The term \"wannabe socialists\" is used to ridicule and discredit opponents, framing them as a threat to traditional values. The phrase \u201ceveryone has something to complain about\u201d suggests that people who do not hold the same view as the author somehow lack resilience.\r\n\r\n","n":0.04},[{"v":"Media literacy","i":4,"n":0.707},{"v":"Critical thinking","i":3,"n":0.707},{"v":"Critical reading","i":2,"n":0.707},{"v":"Analysis","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":376,"$":[{"v":"Academic integrity and preparing for exams","n":0.408},{"v":"Each exam has a set of rules and regulations that students need to follow. If they do not, this can lead to academic misconduct. On this page, you'll learn how academic integrity relates to exams and how to find out what is and isn't allowed in exams.\r\n\r\nExam rules\r\nNot all exams have the same rules. For example, in some exams, you may be allowed to use notes, but in others this will not be allowed. If you have an exam coming up, check your Canvas modules and email tutors/lecturers to see if there are any special conditions for your exam.\r\n\r\nKnowing the conditions as early as possible will help you to prepare for your exam more effectively.\r\n\r\nSome things to check are:\r\n\r\n \twhat technology, if any, can be brought into the exam (e.g. smart phones or watch, calculators).\r\n \tif notes may be brought into the exam, and if so, are there any conditions in terms of quantity?\r\n \twhat you are required to do with personal items such as phones. For example, you may not be permitted to keep them on your desk even when switched off.\r\n\r\nFor online exams, you should consider academic integrity by checking the instructions on what technologies are permitted. Also, you may be required to use a lockdown browser to prevent you from navigating away from the exam screen.\r\nExample exam instructions\r\nThese instructions explain the specific rules that apply to this exam.\r\n\r\n\r\nExam details\r\n\r\n\r\nCourse code: ISYS2056\r\n\r\nCourse description: Business Information Systems\r\n\r\nDate of exam: 1/11/2025 Start time: 9.15 Duration of exam: 2hr 15min\r\n\r\nTotal number of pages (incl. this cover sheet): 22\r\n\r\nAllowable materials and instructions to candidates\r\n\r\n\r\n \tWrite your full name and student number on each exam booklet together with the number of exam books used.\r\n \tYou must not write, mark in any way any exam materials, read any other text other than the exam paper or do any calculations during reading time.\r\n \tAll mobile phones must be switched off and placed under your desk. You are in breach of exam conditions if it is on your person (i.e. in your pocket).\r\n \tYou can be reported for misconduct if you breach exam regulations.\r\n \tThis is a CLOSED BOOK Exam.\r\n \tCalculators are not allowed.\r\n \tBi-lingual dictionaries are not allowed.\r\n \tStudents must write all answers within the Exam Paper in the spaces provided.\r\n \tThis Exam Paper contains Sections A, B & C. Refer to instructions in each section.\r\n \tThis Exam Paper is worth 100 marks & comprises 50% of the total marks allocated in this course.\r\n\r\n\r\n\r\n\r\nAfter your exam\r\nIn many cases, students are not allowed to take exam papers out of the exam room to protect the integrity of the exam. Removing or sharing exam materials can be considered academic misconduct and lead to serious consequences.\r\n\r\nWant to learn more? Visit the academic integrity section of the Learning Lab.\r\n","n":0.047},[{"v":"Exam preparation","i":1,"n":0.707},{"v":"Academic integrity","i":0,"n":0.707}]]},{"i":377,"$":[{"v":"Argument analysis","n":0.707},{"v":"Argument analysis is the process of breaking down an argument to determine if it is logical and well-supported. This helps you think critically, which is one of the key skills required in university study.\r\nIn this section, we will start by learning about the key steps and elements of argument analysis before providing you with an example to test your new knowledge.\r\n\r\nThe steps of argument analysis\r\n\r\nHere are the key steps to help you break down and assess an argument effectively.\r\nStep one: What is the argument?\r\nAn argument is the position or claim that the writer is putting forward. \u00a0An argument can be recognised via:\r\n\r\nlanguage\r\nevidence \r\ntext comparisons.\r\nAn argument will assume an audience's awareness of the topic and also consider their assumptions, prior knowledge and interest.\r\n\r\nSee also Reading skills.\r\n\r\n\r\n\r\nStep two: Identifying bias\r\nBias can be recognised via language, sentence structure, evidence used for supporting claims (including quality of evidence), lack of evidence for refuting claims, repetition of themes and language.\r\n\r\nIt is also important to examine your own biases. When all you can think about is your own perspective on an argument, you blind yourself to the actual argument before you. Therefore, it can be useful to identify your own feelings on the topics and be aware of them. Also, remember to keep an open mind while you read.\r\nStep three: Who is the audience? Who is the writer?\r\nThe audience is the assumed readership that the author wants to persuade or influence. Think about who the writer is writing for, and why they are writing to that audience. Is their intention to persuade, inform or entertain them? \r\n\r\nSome texts provide useful information about the author, such as their level of expertise, profession, or affiliations. If this information is not provided then a search for their name can often turn up the results you need. Knowing the writer's profession or affiliations can help us understand their motives, and knowing their level of expertise can inform us about whether or not we should take their view seriously.\r\nStep four: How does the author influence the audience?\r\nSelected vocabulary (which could include colloquial or slang words), language techniques and sentence construction are used to appeal to a particular audience. Arguments are carefully considered to ensure they are convincing to the target audience.\r\nStep five: Critically analyse by identifying claims, evidence and assumptions\r\nCritical analysis starts with these questions:\r\n\r\n \tWhat is the overall claim of the author?\r\n \tIs the claim backed up by credible evidence?\r\n \tIs the claim based on assumptions?\r\n\r\nStep six: What is the evidence?\r\nEvidence includes information or data from elsewhere to support the claims the writer makes, or refute the claims made by others. Evidence can be judged for quality by the credibility of its origin. Evidence needs to come from a source that has authority and is up-to-date and objective.\r\nIs the evidence credible?\r\nCredible evidence comes from a source that the reader can believe is factual and valid. The key things to look for are:\r\n\r\n \tAuthor: Who does the author work for or write for? What other publications have they been involved in? What is their area and level of expertise in the topic\r\n \tDate: Is the information current given the topic?\r\n \tSource: Is the publication source well known or respected in the industry? Peer reviewed? Relevant to the topic?\r\n\r\n\r\n\r\n","n":0.043},[{"v":"Media literacy","i":4,"n":0.707},{"v":"Academic writing","i":3,"n":0.707},{"v":"Research sources","i":2,"n":0.707},{"v":"Analysis","i":1,"n":1},{"v":"Critical thinking","i":0,"n":0.707}]]},{"i":378,"$":[{"v":"Referencing an oral presentation","n":0.5},{"v":"Referencing is important for any type of assignment you do, and presentations are no different. On this page, we'll cover the ways that you can cite and reference the ideas, quotes and images you use in your presentations.\r\n\r\nWhich referencing style?\r\nThe first step for referencing is to find out which style you are required to use for your assignment. This information is usually included in your assignment instructions. If you have any doubt, ask your lecturer or tutor. If you use the incorrect referencing style or make errors in formatting, you could lose marks.\r\n\r\nCitation\r\nFor oral presentations, there are two ways that you can cite information that has been taken from other sources. They are:\r\n\r\nOn slides: the same way that you would cite in a written assignment. The way you format your citations will depend on the style you have been asked to use. In APA 7th edition, for example, your citation will need to include the author(s)' surname(s) and the year of publication for paraphrases, and for direct quotes, you will also need to include the page number where you found the original text.\r\n\"Ecosystems are the silent architects of our world, weaving the web of life that sustains us all.\" (Kumar, 2024, p. 299)\r\nOrally: by mentioning the author(s) of the original material out loud. \r\n\r\nFor more information, visit Understanding citation.\r\nReferences list\r\nFor your presentation, you should include a slide with a reference list. This list should be formatted in the same way that you would format a reference list for a written assignment.\r\n\r\nReferencing images\r\nReferencing images can be confusing at times. Luckily, RMIT has developed a great referencing resource called Easy Cite that guides students through the referencing process. It has rules and examples for referencing and citing in all of the major referencing styles used at RMIT, and the sections for each of these styles have information for referencing images and audiovisual material. \r\n\r\nTips for managing references\r\n\r\nAs you research, note down the publication details of books, articles, videos, etc. This will save you from having to search for it later, which can be very difficult.\r\nCite as you write- put citations into your writing from the beginning. This will help you avoid confusion about which information belongs to which author.\r\nIf you have doubts about how to reference a resource, contact Study Support. They have advisors who specialise in academic writing and referencing.\r\n\r\nTest your knowledge\r\n\r\n","n":0.051},[{"v":"Referencing","i":1,"n":1},{"v":"Presentations","i":0,"n":1}]]},{"i":379,"$":[{"v":"Transposing formulas","n":0.707},{"v":"The transposition of formulas involves rearranging equations to solve for a specific pronumeral or variable. You need to be able to first isolate your desired variable in order to solve problems various fields. Mastering transposition will enhance your ability to manipulate and understand mathematical relationships effectively.\r\n\r\nVideo tutorial \u2013 transposition of formulas\r\nWatch this video to learn about the transposition of formulas.\r\n\r\n\r\nFormulas\r\n\r\nFormulas set out the mathematical relationships between variables. You may have learned that you can easily substitute numbers or values into formulas to find an unknown.\r\nExamples include:\r\n\r\n\t\\(A=\\pi r^{2}\\)\r\n\t\\(s=ut+\\dfrac{1}{2}at^{2}\\)\r\n\t\\(S=P(1+i)^{n}\\)\r\n\r\nIn these examples, \\(A\\), \\(s\\) and \\(S\\) are the subjects of the formulas. A subject is shown on one side of the equal sign of the formula, followed by an equal sign.\r\n\r\nTransposing formulas\r\nSometimes, a formula is given in a particular form and you need to rearrange it to make a different variable the subject.\r\nFor example, you might know the area of a circle (\\(A\\)) and want to find the radius (\\(r\\)). It is perfectly reasonable to substitute the area into the formula \\(A=\\pi r^{2}\\), but making \\(r\\) the subject will make problem-solving much more efficient.\r\nTo transpose formulas, we need to \"undo\" the operations to make a different variable the subject of the formula. You do this by doing the opposite operation. The following table tells you what you need to do to undo certain operations.\r\n\r\n\r\n\t\r\n\t\tOperation\r\n\t\tOpposite operation\r\n\t\r\n\r\n\r\n\t\r\n\t\tsubtraction (\\(-\\))\r\n\t\taddition (\\(+\\))\r\n\t\r\n\t\r\n\t\tmultiplication (\\(\\times\\))\r\n\t\tdivision (\\(\\div\\))\r\n\t\r\n\t\r\n\t\tsquare root (\\(\\sqrt{x}\\))\r\n\t\tsquare (\\(x^{2}\\))\r\n\t\r\n\r\n\r\nNote that when taking the square root of a term\u2014let's say, \\(x^{2}\\)\u2014the answer will be \\(\\pm x\\). This is because \\(x\\times x=x^{2}\\) but also \\((-x)\\times(-x)=x^{2}\\). Since we may not know whether the \\(x\\) is positive or negative, we must indicate that \\(\\sqrt{x^{2}}=\\pm x\\).\r\n\r\nThere is a basic rule you can follow to transpose formulas:\r\nWhen rearranging equations and formulas, whatever you do on one side of the equal sign, you must do on the other.\r\nWhen you encounter questions asking you to transpose a formula, you may see the terms \"rearrange\", \"transform\", \"make \\(x\\) the subject\", or \"express in terms of \\(x\\)\". These are all instructions transpose formulas.\r\n\r\nExample 1 \u2013 transposing formulas\r\nMake \\(A\\) the subject in the formula \\(A+B=C\\).\r\nWe need to undo the operations to get \\(A\\) on its own on one side of the equal sign. On the left, \\(B\\) is added to \\(A\\), so we can remove it by subtracting \\(B\\) on both sides.\r\n\\[\\begin{align*} A+B & = C\\quad\\textrm{subtract \\(B\\) from both sides}\\\\\r\nA+B-B & = C-B\\\\\r\nA & = C-B\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 transposing formulas\r\n\r\n\tFind \\(n\\) if \\(m=n-2\\).\r\n\tFind \\(C\\) if \\(A=2B+C\\).\r\n\tFind \\(B\\) if \\(A=2B+C\\).\r\n\tFind \\(k\\) if \\(P=\\dfrac{k}{V}\\).\r\n\tFind \\(V\\) if \\(PV=k\\).\r\n\tFind \\(a\\) if \\(v=u+at\\).\r\n\tFind \\(t\\) if \\(v=u+at\\).\r\n\tFind \\(A\\) if \\(r=\\sqrt{\\dfrac{A}{\\pi}}\\).\r\n\tFind \\(x\\) if \\(A=x^{2}\\).\r\n\tFind \\(r\\) if \\(A=\\pi r^{2}\\).\r\n\r\n\r\n\r\n\r\n\r\n \r\nA formula is a general rule giving the relationship between variables. The variables are represented by letters which provide us with a short hand way of writing the rule. Sometimes a formula is given in an inconvenient form and it's necessary to change the subject of the formula. When we do this we need to use certain operations, so remember the following rules; minus undoes plus, plus undoes minus, division undoes multiplication, multiplication undoes division, square root undoes square and square undoes square root.\r\n\r\nLet's look at some examples to explain this. Let's look at the formula A plus B equals C. Now let's make A the subject, because we want to obtain A on its own we start by subtracting B so let's use the rule minus undoes plus, so we have A plus B minus B equals C minus B, so from this we obtained A equals C minus B. Note we have to use minus B on both sides of the equation, remember the saying what we do to one side of the equation we must do to the other side of the equation so we have transposed the formula to find an expression for A.\r\n\r\nLet's look at another example. We have X minus Y equals Z and we want to make Y the subject. Notice we have a minus Y so let's use the rule plus undoes minus, we write X minus Y plus Y equals Z plus Y. Note we have used plus Y on both sides of the equation so this leaves us with X equals Z plus Y, but we want Y on its own so we use the rule minus undoes plus again to get rid of Z so we have X minus Z equals Z minus Z plus Y which leaves us with X minus Z equals Y or rewriting that with Y on the left hand side we have Y equals X minus Z.\r\n\r\nLet's look at the formula for the area of a rectangle. The area of a rectangle is A equals L times W where A is the area, L is the length and W is the width of the rectangle. We say that A is the subject of the formula, let's say we want to find the width, W, so we use the rule division undoes multiplication by dividing both sides by L. So let's do that. We divide the left hand side by L, which is A over L and we divide the right hand side by L, so it's L times W over L. So A over L is equal to W, notice that the Ls have cancelled out on the right hand side. If we swap the left hand side and the right hand side we have W equals A divided by L.\r\n\r\nLet's look at the formula M equals Q over P. M is the magnification of an object, Q is the size of the image and P is the size of the object. We want to find Q, the image size, so let's use the rule multiplication undoes division, so if we multiply both sides by P we get M times P equals Q over P multiplied by P and this leaves us with M times P equals Q and so if we swap both left hand side and right hand side we get Q equals M times P.\r\n\r\nThat completes the film clip of transposition of formulas. Now go to the worksheet and try some exercises for yourself. Thank you.\r\n","n":0.031},[{"v":"Maths equations","i":3,"n":0.707},{"v":"Functions","i":2,"n":1},{"v":"Arithmetic","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":380,"$":[{"v":"Differential equations","n":0.707},{"v":"Differential equations help us capture the relationships between changing variables. They provide a framework for analysing how different factors interact over time and let us to predict future behavior. Use this resource to learn how to solve differential equations.\r\n\r\n\r\n\tIntroduction to differential equations\r\n\tFirst order separable\r\n\tFirst order linear\r\n\tSecond order homogeneous\r\n\tSecond order non-homogeneous\r\n\r\n","n":0.141},[{"v":"Maths","i":1,"n":1},{"v":"Differential equations","i":0,"n":0.707}]]},{"i":381,"$":[{"v":"Report checklist","n":0.707},{"v":"This checklist is a great way to make sure you have included the necessary features in your report.\r\n\r\nCheck that your report has the following features:\r\n\r\n\r\n\tAre all parts of your assignment task covered? In relevant sections?\r\n\tHave you used conventional report format and structure?\r\n\tAre there well-structured sections where the information is organised logically under content headings? Do headings clearly indicate the ideas that follow?\r\n\tAre there clear headings and sub-headings for each section? Have you used heading styles to format the headings?\r\n\tAre there clear, structured paragraphs?\r\n\tIs there a title page that includes title of the report, your name and student number, name of the course and date?\r\n\tDo you have a table of contents with page numbers?\r\n\tIs there an executive summary or abstract that summarises the whole report?\r\n\tDoes the introduction outline the context, background and purpose of the report?\r\n\tIs there a method section where the research techniques used are clearly explained?\r\n\tDo you have a results/findings section that presents and summarises your findings?\r\n\tIs there a discussion section explaining and interpreting the results?\r\n\tDoes the conclusion sum up what was covered and contain the \u2018answer\u2019? No new information?\r\n\tDo you have recommendations that suggest possible future actions or solutions? (may use bullets here)\r\n\tHave you included a reference list that includes all references used in your report, in the appropriate style?\r\n\tHave you included extra information in an appendix/appendices?\r\n\r\n","n":0.068},[{"v":"Assessment","i":3,"n":1},{"v":"Postgraduate","i":2,"n":1},{"v":"Reports","i":1,"n":1},{"v":"Report structure","i":0,"n":0.707}]]},{"i":382,"$":[{"v":"Understanding your assignment","n":0.577},{"v":"\r\nThe first step to assignment success is having a clear understanding of what you are required to do. This involves identifying the type of assignment you have been given, analysing the assignment instructions and taking formatting information into account. \r\n\r\nCommon assignment types\r\nHere's a list of some of the assignments that you may be required to complete in tertiary education. Follow the links to find more in-depth information about these assignment types:\r\n\r\n\r\nannotated bibliography\r\nacademic reflection\r\nreport\r\nliterature review\r\ncase study\r\npresentation\r\nessay.\r\n\r\n\r\nKeep in mind that different names are sometimes used to describe the same assignment type. Sometimes, there will be variations in what is required for the assignment. If you ever have any doubts, read your assignment instructions again and consult your teacher. For example:\r\n\r\n\tessay: argumentative essay, composition\r\n\tliterature review: scoping review\r\n\tacademic reflection: personal reflection, reflective journal\r\n\tcase study: case analysis, case report\r\n\treport: research report, analytical report\r\n\tpresentation: talk, demonstration, workshop\r\n\r\n\r\nUnpacking assignment instructions\r\nHaving a clear understanding of your assignment topic will help you to maintain focus on and meet the assignment's objectives. It will help you to stay on topic and manage your time more efficiently.\r\n\r\nLet's explore some techniques for analysing assignment topics.\r\n\r\nContent words and phrases:\r\nThese are the key terms in an assignment that indicate the main topics or concepts that need to be addressed. They help you focus on the specific subject matter you should explore and discuss in your response. They are typically nouns or phrases.\r\n\r\n\r\n\r\nInstruction words and phrases:\r\nThese words guide you on how to approach the assignment by specifying the type of response required, such as \"analyse,\" \"compare,\" \"discuss,\" or \"evaluate.\" Understanding these words is crucial as they dictate the method and depth of your analysis. They are typically verbs.\r\n\r\n\r\nLimiting words and phrases:\r\nThese words narrow the scope of your assignment by setting boundaries on what you should include, such as specific time periods, locations, or populations. They help you focus your research and writing on particular aspects of the topic, ensuring your response remains relevant and concise.\r\n\r\n\r\nOther key information\r\nYour assignment instructions should tell you the date that your assignment is due, the referencing style you are required to use and the word count.\r\nOn top of this, they usually list requirements about how your work should be presented, and failing to follow these instructions can lead to the loss of marks. Elements covered include:\r\n\r\nfont size and style\r\nspacing\r\nmode of submission\r\nfile type (e.g. Word doc, PDF, PPT, etc.)\r\nwhether or not a cover page is required, and what should be included on it.\r\n\r\n\r\nSample assignment instructions\r\nLet's unpack the following sample assignment brief from a psychology subject. The assignment type and the content, instruction and limiting words and phrases have been highlighted. Other key considerations are also highlighted.\r\n\r\n\r\n\r\nMaking a checklist\r\nA great way to ensure that your assignment covers all of the requirements is to create a checklist. To do this, take the information from the assignment instructions that you have identified and put it into short, concise dot points. \r\n\r\nThe checklist can be used while you are writing your assignment, and also when you are reviewing it to make sure that no important tasks have been missed. \r\n\r\nHere's an example of a checklist that has been made from the sample assignment instruction from the previous section: \r\n\r\n\r\nAssignment 3 checklist\r\n\r\n  synthesise 10+ articles\r\n        \r\n            peer reviewed\r\n            published after 2023\r\n  \t\t\r\n\t\tcompare and contrast authors' ideas\r\n  provide critical analysis\r\n  demonstrate engagement with literature\r\n  clear structure\r\n  academic writing standards\r\n  referencing: APA 7th edition\r\n  1500 words\r\n  Word doc\r\n       \r\n            Times New Roman\r\n            12 pt\r\n\r\nTest your knowledge\r\n","n":0.042},[{"v":"Researching","i":2,"n":1},{"v":"Research sources","i":1,"n":0.707},{"v":"Assignments","i":0,"n":1}]]},{"i":383,"$":[{"v":"Do I have enough information?","n":0.447},{"v":"When researching for assignments, you may gather a lot of information from websites, books, e-journals, etc. How do you know when you have enough information for your assignment? On this page we give some practical tips.\r\n\r\n\r\nKnowing if you have enough information can be tricky but the following tips should help you.\r\n\r\n \tFind out the requirements of your assignment. Consult your course guide, information about the assignment in Canvas, or your lecturer.\r\n \tGo beyond what is required. If your assignment specifies the number of sources to use, for example 'Use three scholarly journal articles', try to find more than the minimum number and then evaluate your sources to find the best ones to use.\r\n \tRead and gather enough information to provide an in-depth response to your assessment task.\r\n \tIf no new ideas are emerging from your sources, you might need to look for sources that cover a different aspect of the topic.\r\n\r\nOnce you have started the first draft of your assignment, you may discover that you need more information. At this stage you can do some additional reading.\r\n\r\n\r\n\r\n\r\n\r\nAm I research ready?\r\nThis self-assessment checklist provides a guide to help you get research ready.\r\nDo you know how to:\r\n\r\n\r\n\r\n Analyse an assignment topic? \r\n Develop a search strategy? \r\n Find the right information source? \r\n Search smarter on the web? \r\n Search the Library resources? \r\n Find and search databases? \r\n Evaluate all your information sources?\r\n\r\n\r\n","n":0.066},[{"v":"Researching","i":2,"n":1},{"v":"Information sources","i":1,"n":0.707},{"v":"Assignments","i":0,"n":1}]]},{"i":384,"$":[{"v":"Evaluate information sources","n":0.577},{"v":"When you choose information for your assignments, there should be a continual evaluation process occurring. Evaluate information for its relevance and usefulness to your work, and its quality.\r\n\r\nNot all information that is published is credible and reliable, especially if you find that information on the web. As a smart researcher, you need to be able to judge the value of any information you find, printed or online, before you use it.\r\n\r\nEvaluating information sources, whether they are from a book, scholarly database or the web, is an important step in your research. Anyone can publish online so you need to choose your information carefully.\r\nCRAAP test\r\nNot all information is created equal. Learn how to apply the CRAAP (Currency, Relevance, Authority, Accuracy, Purpose) test to your sources.\r\n\r\nCurrency: the timeliness of the information\r\n\r\n \tWhen was the information published or posted?\r\n \tHas the information been revised or updated?\r\n \tDoes your topic require current information, or will older sources work as well?\r\n \tAre the links functional?\r\n\r\nRelevance: the importance of the information for your needs\r\n\r\n \tDoes the information relate to your topic or answer your question?\r\n \tWho is the intended audience?\r\n \tIs the information at an appropriate level (i.e. not too elementary or advanced for your needs)?\r\n \tHave you looked at a variety of sources before determining this is one you will use?\r\n \tWould you be comfortable citing this source in your research paper?\r\n\r\nAuthority: the source of the information\r\n\r\n \tWho is the author/publisher/source/sponsor?\r\n \tWhat are the author's credentials or organizational affiliations?\r\n \tIs the author qualified to write on the topic?\r\n \tIs there contact information such as a publisher or email address?\r\n \tDoes the URL reveal anything about the author or source? e.g. .com .edu .gov .org .net\r\n\r\nAccuracy: the reliability, truthfulness and correctness of the content\r\n\r\n \tWhere does the information come from?\r\n \tIs the information supported by evidence?\r\n \tHas the information been reviewed or refereed?\r\n \tCan you verify any of the information in another source or from personal knowledge?\r\n \tDoes the language or tone seem unbiased and free of emotion?\r\n \tAre there spelling, grammar or typographical errors?\r\n\r\nPurpose: the reason the information exists\r\n\r\n \tWhat is the purpose of the information? Is it to inform, teach, sell, entertain or persuade?\r\n \tDo the authors/sponsors make their intentions or purpose clear?\r\n \tIs the information fact, opinion or propaganda?\r\n \tDoes the point of view appear objective and impartial?\r\n \tAre there political, ideological, cultural, religious, institutional or personal biases?\r\n\r\n\r\nHow to evaluate resources\r\nFor a quick overview of the CRAAP test, watch this video from McMaster University.\r\n\r\n\r\nThe SIFT method\r\nAnother method of evaluating information is the SIFT model. This is a technique that allows you to explore content online related to a claim or source of information. It is particularly useful when analysing websites and social media.\r\n\r\nFor more information on SIFT, check the Learning Lab page on Engaging Critically with Social Media.\r\n","n":0.046},[{"v":"Media literacy","i":3,"n":0.707},{"v":"Research sources","i":2,"n":0.707},{"v":"Essays","i":1,"n":1},{"v":"Critical reading","i":0,"n":0.707}]]},{"i":385,"$":[{"v":"Choose valid sources","n":0.577},{"v":"In this tutorial you will learn why it is important to use high quality information. Hint: you'll get better grades! You will learn the difference between scholarly and non-scholarly information as well as strategies to help you critically evaluate information.\r\n\r\nWhat is scholarly information?\r\nMany lecturers require students to use scholarly or peer-reviewed journal articles for assignments. As a smart researcher, you need to be able to tell whether an article is scholarly.\r\n\r\nScholarly articles are in-depth, accurate, well-researched, written by academics and published in academic journals. They are good and credible sources of information because the articles have been through a peer-review process. This means the article has been formally evaluated by equally qualified academics and experts.\r\n\r\nPlay this video to find out 'What's a scholarly resource?'\r\n\r\n\r\nNon-scholarly information\r\nSometimes your assignment requires you to use other types of information other than scholarly resources. Non-scholarly sources can also be useful for your research. They can provide useful information about:\r\n\r\n \tcurrent events\r\n \tpopular and community information\r\n \tindustry, trade and the professions\r\n \tgovernment policy and legal information\r\n\r\nThey are usually written by professionals who have specialist knowledge in a field, or by people with a special interest in a topic who have developed reputation for their opinions.\r\n\r\nSome common examples of non-scholarly sources are:\r\n\r\n \tnews sources, newspapers, and materials that are time-based and get updated frequently\r\n \tprimary sources\r\n \tdata and statistical publications and compilations\r\n \ttrade and professional sources\r\n \taudio-visual resources and videos\r\n\r\nCheck with a librarian and/or your lecturer if you are unsure of whether your non-scholarly source is appropriate information for your assignment. You can also use the Ask the Library service.\r\nGoogle and Wikipedia\r\nBoth Google and Wikipedia can be used for a basic overview and to find background or general information about your topic. However, they do have serious limitations for academic research.\r\n\r\nWatch the video to find out more about using Google and Wikipedia for your research.\r\n\r\n\r\n\r\nTo see a list of the best databases your topic area, choose your Subject Guide.\r\n\r\nUse RMIT Google Scholar, that way your search results will include resources from some of the RMIT Library online collections.\r\n\r\nYou might like to bookmark these!\r\n\r\n\r\nActivity: Which information source is best?\r\nTry this quick activity to test your knowledge about appropriate information sources.\r\n\r\n\r\n \r\n\r\nNarrator: Unidentified female.\r\nVisual: RMIT University Logo and website www.rmit.edu.au.\r\nVisual: What\u2019s a Scholarly Source?\r\nVisual: Colour cartoon of a blue cat standing on two paws in the centre of the screen. Lots of pages of blank paper are flying around the screen and are landing in five piles around the cat. Gradually, the piles grow to be taller than the cat. The cat\u2019s eyes are rolling around looking at all the paper.\r\nNarrator: There's a lot of information out there but it's not all created equal. Sometimes you will be asked to ...\r\nVisual: Same image of the blue cat standing between the five piles of paper. The word Scholarly appears to the left of screen.\r\nNarrator: ... use a scholarly ...\r\nVisual: Same image as before but without the word Scholarly; now the word Academic appears to the right of the screen.\r\nNarrator: ... academic ...\r\nVisual: Same image as before but without the word Academic; now the word Peer reviewed appears to the left of the screen.\r\nNarrator: ... or peer-reviewed source in your assignments.\r\nVisual: Same image as before of blue cat standing between the five piles of paper. A red question mark appears above the cat\u2019s head. At the top of the screen appears the sentence: What is a scholarly source?\r\nNarrator: But what does that mean? What is a scholarly source?\r\nVisual: Colour cartoon of the blue cat sitting at a desk typing on a laptop. The cat is wearing a mortarboard and reading glasses. There is a pile of papers on each side of the desk and behind the cat is a drawing of a bookshelf. A sentence appears to the right of the screen beside the cat at the desk as Narrator speaks: Written by academics or researchers.\r\nNarrator: Scholarly information is written by academics or researchers.\r\nVisual: Same image as before ie colour cartoon of the blue cat sitting at a desk typing on a laptop. The cat is wearing a mortarboard and reading glasses. There is a pile of papers on each side of the desk and behind the cat is a drawing of a bookshelf. A sentence appears to the right of the screen beside the cat at the desk as the Narrator speaks: in depth, well researched.\r\nNarrator: It is in-depth, well-researched and usually ...\r\nVisual: Computer drawings of three journals, then beside those a book with its pages being opened, then beside those a folder marked conference with three pages being placed inside the folder.\r\nNarrator: ... found in journals, books or conference papers.\r\nVisual: Cartoon drawing of three cats sitting behind a desk. The blue cat is in the centre. They are all wearing mortar boards and holding a paper and each has a red buzzer on the desk in front of them. There are also three unlit lights on the wall behind each cat. The yellow cat has glasses and is sitting to the right of screen and the grey cat is sitting to the left of the screen. The words: peer-reviewed, refereed appear to the right of screen as the Narrator speaks.\r\nNarrator: Often, this kind of research is peer-reviewed or refereed.\r\nVisual: Same carton drawing as before of the three cats sitting at the desk. In sequence of right to left, the cats touch their buzzer and a green light turns on behind their heads.\r\nNarrator: This means that a group of experts in the field have verified the quality of the research.\r\nVisual: Cartoon drawing of the blue cat wearing a lab coat, holding a pointer and standing in front of a whiteboard with a cute meow formula written on the board.\r\nNarrator: Scholarly sources report the results ...\r\nVisual: The whiteboard morphs into a circle of the earth, Australia is in the centre, Asia is near the top of the globe and there are black mortarboards located all over the land on the map. The cat spins around and lands in Australia.\r\nNarrator: ... of research to the rest of the academic world.\r\nVisual: Cartoon of the blue cat (small) standing beside a huge pile of papers. The cat looks up at the top and does a little jump.\r\nNarrator: This is how the body of knowledge in a subject grows over time.\r\nVisual: Cartoon of a yellow cat standing on two paws, wearing a lab coat and glasses. The cat is holding a book titled Birds and the cat has a purple feather in his mouth which he blows out and it falls to the ground.\r\nNarrator: When you use scholarly sources in your assignments, you know that the information is considered ...\r\nVisual: Same image as before of yellow cat. He winks and gives the thumbs up sign.\r\nNarrator: ... reliable by experts in the field.\r\nVisual: Cartoon of blue cat siting on a stool at a desk wearing a baseball cap backwards. He is typing on a laptop with a book to one side of him and a pile of papers on the other side.\r\nNarrator: This adds weight to your arguments ...\r\nVisual: Same image as before. A paper with HD written on the bottom appears beside the cat. He stops typing and bounces up and down, smiling with his paws in the air.\r\nNarrator: ... and is likely to get you a better mark.\r\nVisual: Cartoon drawing of a scholarly paper with a red question mark beside it.\r\nNarrator: So what does a scholarly source look like? It may be scholarly if ...\r\nVisual: Same image as before. To the right of the screen, beside the scholarly paper, appears the sentence written in red and in sync with the narrator\u2019s words: it's based on a study conducted by the author. A heading in bold on the scholarly paper is circled in red.\r\nNarrator: ... it's based on a study conducted by the author ...\r\nVisual: Same image as before. To the right of the screen beside the scholarly paper appears words written in red: it's been peer-reviewed.\r\nNarrator: ... it's been peer-reviewed ...\r\nVisual: Same image as before. To the right of the screen beside the scholarly paper appears words written in red: it has a summary or abstract and extensive references. The summary on the scholarly paper is circled in red.\r\nNarrator: ... it has a summary or abstract and extensive references ...\r\nVisual: The first page of the scholarly paper is removed from screen, underneath it and now fully visible is a page with two graphs on it which are circled in red.\r\nNarrator: ... or it contains tables, graphs or statistics. VISIUAL: Cartoon of the blue cat looking at a paper it is holding, squinting at it and rubbing its head.\r\nNarrator: Sometimes, it's hard to tell whether or not a source you've found is scholarly. Other reliable sources of information are available ...\r\nVisual: Cartoon drawing of blue cat wearing a hard hat and workman\u2019s belt. The words: trade journals appears to the right of the screen.\r\nNarrator: ... including trade journals ...\r\nVisual: Cartoon drawing of blue cat wearing judge\u2019s wig, robe and gavel. The words: industry information appears to the right of screen.\r\nNarrator: ... industry information ...\r\nVisual: Cartoon drawing of a yellow cat sitting behind a desk typing at a laptop and wearing a fedora hat. The words: popular writing appears to the right of screen.\r\nNarrator: ... or popular writing, however, these are not ...\r\nVisual: Same image as before, yellow cat typing at desk wearing fedora. The words: these are not scholarly appears to the right of screen.\r\nNarrator: ... scholarly. So how can you find scholarly ...\r\nVisual: Same image as before ie yellow cat typing at desk wearing fedora. A big red questions mark appears to the right of screen.\r\nNarrator: ... sources?\r\nVisual: Written in red in centre of screen: LibrarySearch. Underneath that is a small square grey box and in smaller print the words: Limit your search results to peer-reviewed articles. The grey box gets a big black tick put in it.\r\nNarrator: LibrarySearch; you can limit your search results to peer reviewed articles which are usually scholarly.\r\nVisual: A computer image of LibrarySearch\u2019s login page with a search box, a Browse section for subject and collection guides and also four bullet point options under the heading Launch: Databases A-Z; Course reserve; Google scholar; and Research repository. As the Narrator speaks, the following words are increased to double the size: Databases A-Z; then Google Scholar.\r\nNarrator: Databases; many library databases also have the option to limit by scholarly, academic or peer-reviewed articles. Google Scholar is a good source of scholarly material. Many, but not all, results will be scholarly.\r\nVisual: Cartoon of grey cat sitting at desk looking at three laptop screens. The website is written across the bottom of the screen in red: www.rmit.edu.au/library/askalibrarian.\r\nNarrator: Librarians are trained to find scholarly information so if you're not sure, ask a librarian.\r\nVisual: RMIT University Logo and website www.rmit.edu.au.\r\nEnd of video\r\n\r\n \r\n\r\nMeet Katrina. Katrina is a health science student at RMIT University. Recently, Katrina received her marks back for an assignment and she was shocked to find she hadn\u2019t done as well as she\u2019d hoped. Katrina had done what most students do: She had just Googled her topic.\r\n\r\nFeeling down, Katrina asks for feedback from her lecturer, Professor Felinopolous. He explains that while Google is a good starting point to get an overview of a topic, you can\u2019t just use Google and expect to get good marks for your assignment.\r\n\r\nSo, why is that?\r\n\r\nFirstly, the internet is not organised. When you search a term in Google, you\u2019ve probably noticed that you get huge amounts of information returned to you.\r\n\r\nThese results are based on popularity. In other words, Google is a popularity contest and search results are based on Google\u2019s own ranking, rather than credibility or reliability. Google also knows a bit about you. Its \u201cpersonalisation\u201d feature means that your search results will be affected by your internet history, location, and IP address, too.\r\n\r\nSecondly, there is little quality control on Google. Anyone can write anything on the internet. The information may be biased, inaccurate, or have a hidden agenda. Information could also be sponsored by sources with vested interests. And lots of information is not authoritative or credible. It\u2019s not written by experts or peer-reviewed.\r\n\r\nOn Google, it can be hard to identify sources. Who wrote it? How do you verify the sources? Trying to find this information can waste valuable research time.\r\n\r\n","n":0.022},[{"v":"Study skills","i":3,"n":0.707},{"v":"Research sources","i":2,"n":0.707},{"v":"Critical thinking","i":1,"n":0.707},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":386,"$":[{"v":"Electronic resources and databases","n":0.5},{"v":"The Library has an extensive collection of electronic resources which includes e-books, e-journals, streaming videos, and images. These are available on and off campus using your RMIT login credentials.\r\nThe Library's electronic resources are available through the Library collections webpage and also via LibrarySearch.\r\n\r\nThe Library also has large collections of online videos, audio recordings, DVDs and CDs. The audiovisual collections can also be searched through LibrarySearch.\r\n\r\nRMIT Vietnam students have access to the same e-resources, which can be accessed from the Vietnam Library home page using the RMIT login.\r\nMore resources\r\nPlay this video about e-books in academic libraries.\r\n\r\n\r\n\r\nDatabases\r\nThe Library subscribes to many databases of online resources. These databases can include journal articles, news articles, ebooks, streaming videos, and specialised information such as standards, theses, case studies and images.\r\n\r\nSo what is a library database?\r\n\r\n\r\nChoosing a database\r\nDatabases are a rich source of information in the subject areas taught at RMIT University, but there are many different databases. How do you know which one to choose?\r\n\r\nThe Subject Guides have been written by librarians to help you select the most relevant databases for your topic.\r\n\r\nIf you know the name of the database you want, you can find it using the Database A-Z list from LibrarySearch.\r\nSearching databases\r\nMost databases offer both a basic and an advanced search option. If you need help searching a particular database, look for the Help menu in the database to find tips and 'How to' videos. You may also find database tutorials on YouTube, and from some Library Subject Guides.\r\n\r\nFor an advanced database search, you may need to use Boolean operators. Boolean operators connect your search terms together in specific ways (AND, OR, NOT), and also help you to narrow or broaden your search. You can consult the \u2018Develop your search strategy\u2019 page to understand how to create a search using Boolean operators and other techniques. \r\n\r\n\r\n\r\nVisual description\r\nNarration\r\n\r\n\r\n\r\n\r\ne-Books in Academic Libraries.\r\n\r\nWhat are they? How do they work? Why are they different? RMIT University logo\r\n\r\n\r\n\r\nComputer image colour drawing of a female sitting against a tree trunk in its shade and reading a book. There are other trees beside her.\r\n\r\nCamera pans in on this image to focus on the female sitting, reading her book.\r\nImagine a book as a container.\r\n\r\n\r\nComputer image colour drawing of a large red book, bound with ring around it of gold brass that is fixed with a padlock. There is a row of books in the background.\r\n\r\nLock is broken and the book lid is opened up (like a chest), streams of binary code ie 0 and 1 are coming out of the book like a fountain.\r\nWith a traditional print book, the information or content and the container are locked together.\r\n\r\nWith an e-book, this physical connection is broken and the information is released from the container.\r\n\r\n\r\nComputer image colour drawing of girl sitting against trunk of tree reading a book (same as seen earlier in video) and two other students, one male and one female who are both sitting leaning against their own tree trunk.\r\n\r\nHe has an open book beside him on the grass.\r\n\r\nThree streams of binary code fall from the top of the screen onto each person's device starting with the female sitting in the centre of screen looking at her laptop, then the female sitting reading her book and lastly, the male looking at his smartphone.\r\n\r\nThe stream dissipates until it's finished in the same order as it fell, i.e. finishing last with the male student.\r\nThis is still dependable quality information but you can now access it in a variety of formats.\r\n\r\nThis makes it ...\r\n\r\n\r\nComputer image colour drawing of a hand moving across lines representing sentences on a tablet and highlighting the lines in yellow.\r\n... more convenient to work with.\r\n\r\n\r\nComputer image colour drawing of a female sitting at a desk in a library looking at a laptop.\r\nIn university libraries e-books work differently from the ones you buy for personal use.\r\n\r\n\r\nComputer image colour drawing of six volumes of red covered books. They are standing on a shelf.\r\n\r\nThe shortest books is to the left of screen, with each book gradually increases in height to the tallest at the right of screen. The book spines are labelled Volume 1 (being the shortest on the left) to Volume 6 (being the tallest on the right).\r\n\r\nA xylophone stick hits each book in order starting at the shortest as if it's playing an xylophone scale.\r\nAs with print books ...\r\n\r\n\r\nComputer image colour drawing of a computer and scanner in the library. The computer screen shows a picture of how to scan.\r\n\r\nA red book is passed over the scanner.\r\n\r\nThe red book is to the left of the scanner and the screen now shows a green tick for successful borrowing.\r\n... you borrow e-books from the library. This means that even if you've ...\r\n\r\n\r\nComputer image colour drawing of the downloading icon i.e. an arrow pointing down in a circle.\r\n... downloaded a book it will disappear from your ...\r\n\r\n\r\nComputer image colour graphic of a tablet leaning on its stand, with a pile of books in the background.\r\n\r\nA digital timer with red numbers is counting down very quickly from 2:00 to zero.\r\n... computer or device at the end of the ...\r\n\r\n\r\nAs per previous image but timer has been removed and a yellow stream of binary code is evaporating from the tablet.\r\n... borrowing period. And not all e-books are the same.\r\n\r\n\r\nComputer graphics of the word e-books written in five different texts and colours is gradually overlaid over each other on the screen.\r\nIndividual publishers decide how their e-books can be used ...\r\n\r\n\r\nComputer image graphic of five clock faces in different colours, each has different times, the minute and second hours are spinning around quickly.\r\n... which leads to differences in how long you can borrow for ...\r\n\r\n\r\nComputer graphic of a tablet with horizontal lines representing the sentences.\r\n\r\nFirst, the left page and then the right page on the tablet is highlighted in yellow.\r\n... how much of the book you can copy, print ...\r\n\r\n\r\nComputer graphic of a printer on the left of the screen and computer graphic of the download arrow icon on the right of screen.\r\n... and download and also ...\r\n\r\n\r\nComputer graphic of a Reader software eye icon. A hand pointer moves across the screen and clicks on the eye icon then disappears.\r\n... whether you need to get extra software ...\r\n\r\n\r\nSame as previous computer graphic image, magnifying glass zooms in on the Reader software icon.\r\n... to view the content.\r\n\r\n\r\nComputer graphic drawing of a male wearing sunglasses and an Aloha shirt sitting on a deckchair at the beach, drinking a green cocktail with a straw and raising his eyebrows as he slurps up his drink.\r\n\r\nHe has a laptop open on his lap.\r\nPlus you don't have to go into the library ...\r\n\r\n\r\nSame image as previous, male gives thumbs up sign and a yellow plane enters the sky to the right of the screen behind him.\r\n... to borrow or return e-books ...\r\n\r\n\r\nYellow plane flies past screen pulling a banner which reads No fines.\r\n... and there are no fines. E-books are a great ...\r\n\r\n\r\nComputer graphic of an addition/plus symbol, following by the word Library and then the word e-books.\r\n... addition to what libraries offer but when it comes to ...\r\n\r\n\r\nComputer graphic of library shelves, seven shelves high, filled with books.\r\nPrint books are still really important in university libraries.\r\n\r\n\r\nComputer graphic image of a dark red book titled World of Frogs.\r\nBooks for some subject areas ...\r\n\r\n\r\nAs per previous image plus a brighter red book titled Text is placed beside the World of Frogs book.\r\n... and most text books are still only published in print format.\r\n\r\n\r\nVideo computer graphic of pile of books, laptop on desk.\r\n\r\nCamera pans back to show female sitting and working at the laptop in a library.\r\nSo make sure you research widely for all your assignments.\r\n\r\n\r\nVISUAL: RMIT University Logo and website www.rmit.edu.au.\r\n\r\n\r\n\r\n\r\n \r\n\r\nI got my first uni assignment and my lecturer told me to look in the library databases for journal articles.\r\n\r\nI was like, THE WHAT?\r\n\r\nBut when I got to the University Library, a librarian explained it to me.\r\n\r\nShe said, \"First you have to get your head around academic journals\".\r\n\r\nThese are not your 'dear diary' journals, right?\r\n\r\nAcademic journals are collections of articles written by experts in an academic discipline.\r\n\r\nThey've usually been reviewed by other academics in the field, who have said that this research is original, and important, and at a high standard.\r\n\r\nThis means that, if you use them, you'll get better marks than if you use Wikipedia, or a random website you've found on Google.\r\n\r\nIn the past, journals were always printed; articles were written and collected in journals which were published at regular intervals.\r\n\r\nLibraries collected these and filled lots of shelves with them.\r\n\r\nThen the internet came along and changed the way journals were produced.\r\n\r\nNow, they are more often found online as e-journals.\r\n\r\nOnline e-journals are organised in library databases.\r\n\r\nA library database is like a bucket: each holds a collection of e-journals. Some databases hold heaps of journals in them while others don't have that many.\r\n\r\nThe University Library subscribes to databases that focus on relevant subjects for RMIT students and researchers.\r\n\r\nNo single database will have all the information you need, so you need to find out which ones are best for what you're researching.\r\n\r\nThe Library has subject guides that show you which databases are best for your area.\r\n\r\nIt's so much easier to find information when you know where to look and why!\r\n\r\n","n":0.026},[{"v":"Researching","i":2,"n":1},{"v":"Information sources","i":1,"n":0.707},{"v":"Digital media","i":0,"n":0.707}]]},{"i":387,"$":[{"v":"LibrarySearch and subject guides","n":0.5},{"v":"RMIT's LibrarySearch is a search engine that can help you find print and electronic resources held by RMIT University Library. It provides access to most of the Library's books, e-books, journal and e-journal articles, news articles, case studies, streaming videos and more.\r\n\r\nLibrarySearch is available from the search box on the  Library home page. RMIT Vietnam students also have access to LibrarySearch through the Online Resources section of the Vietnam Library website.\r\n\r\nFor more help visit the Tips for using LibrarySearch page on the RMIT University Library website.\r\n\r\nPlay this short video on RMIT's LibrarySearch to learn more.\r\n\r\n\r\n\r\nPrint books\r\nRMIT students in Australia can use LibrarySearch to find print books relevant to their research. Print books can be accessed at every campus library, and you can also request books from other campuses to your own using the \u2018Place a Hold\u2019 function in LibrarySearch. \r\n\r\nPlay the video below to find out how to find a book in the Library.\r\n\r\n\r\n\r\nLibrary Subject Guides\r\nLibrary Subject Guides can help you find subject dictionaries and reference resources for background reading that is relevant to what you're studying. Library subject guides will also direct you to the most relevant and valuable databases for your area of study and research, as well as providing links the most appropriate ebooks, websites and other sources including videos and images. \r\n\r\nPlay the following video to find out how Library Subject Guides can help you.\r\n\r\n\r\n \r\nLibrarySearch is a discovery tool. It allows you to search resources held by the Library, including print books, e-books, journal articles - and lots more! You can also use LibrarySearch to place holds, and save searches.\r\n\r\nLibrarySearch is a great way to start finding information on a topic. You\u2019ll find lots of quality academic sources in LibrarySearch, and it gives a good indication of what\u2019s available on your topic.\r\n\r\nHowever, LibrarySearch does not search everything in the Library. It shouldn\u2019t be the only tool you use to find information.\r\n\r\nLibrary Subject Guides point you towards the most relevant resources for your particular subject area -- including those that aren\u2019t covered by LibrarySearch.\r\n\r\nUsing both LibrarySearch and your Subject Guide is a winning combination for getting the most out of the Library.\r\n\r\nIf you are having any trouble finding information, we are only a click away. Ask a Librarian by live chat, or email.\r\n \r\n\r\n\r\n\r\n\r\nVisual description\r\nNarration\r\n\r\n\r\nHow to find a book in the Library.\r\n\r\nAn easy guide to understanding call numbers and finding books on the shelf.\r\n\r\nRMIT University logo.\r\nTo find the book you want ...\r\n\r\n\r\nImage of search page for RMIT University LibrarySearch with a search engine box.\r\n\r\nThe following words are typed gradually into the box: EU-Asean Global finance.\r\n\r\nUnder the box are the words: or your Library catalogue.\r\n... start with Library Search or your library catalogue.\r\n\r\n\r\nMultiple video clips of call number 337.405 E86 which have been written on the palm of a hand, the back of a male\u2019s hand and on a male\u2019s bald head.\r\nWrite down the call number.\r\n\r\n\r\nVideo of male student walking in library.\r\n\r\nThe following title is overlaid on top of the video of a male student walking in a library: 000 Computing, General.\r\nBooks in the library ...\r\n\r\n\r\nThe following titles are overlaid on top of the video of a male student walking in a library: 100 Philosophy, Psychology.\r\n... are organised by ...\r\n\r\n\r\nVideo of same male student standing with his back to camera looking at an information sign in the library for book locations and room locations.\r\n\r\nOverlay of writing: 200 Religion.\r\n... call number ...\r\n\r\n\r\nVideo of same male student walking past book shelves in library.\r\n\r\nWriting overlays: 300 Social Sciences; and then 400 Language.\r\n... and are grouped together by subject area.\r\n\r\n\r\nVideo of male\u2019s feet walking past library shelves.\r\nTo find the right area in the library ...\r\n\r\n\r\nVideo showing ends of library shelves with signs indicating range of call numbers per shelf.\r\n\r\nVideo stops on sign on end of book shelf \u2013 334 to 338.064.\r\n... look at the first three numbers in the call number.\r\n\r\nOur call number begins with 337.\r\n\r\n\r\nMale walks to appropriate shelf for 334 to 338.064 and looks at books.\r\nRead the numbers ...\r\n\r\n\r\nVideo of five bookshelves joined together, seven selves high.The books can be seen under a sheer white computer graphic covering.\r\n\r\nThe books on each shelf are gradually being revealed in sections until all books are fully revealed i.e. uncovered.\r\n... on each shelf from left to right like a page from a book.\r\n\r\nThe numbers increase as you read across the shelf.\r\n\r\nNow that you have located the books marked ...\r\n\r\n\r\nSame image as before of the bookshelf but now has 337 overlaid on top of image of bookshelves.\r\n... 337, you\u2019ll need to look at the ...\r\n\r\n\r\nVideo of books on shelf with 337 plus more of the call numbers on spine stickers.\r\n... numbers after the dot or decimal point ...\r\n\r\n\r\nVideo stops on the image of several spines of books on a shelf and the label on the spine of a book is enlarged and it reads: 337.405 E86.\r\n\r\nThe size of 405 is increased as this word is spoken.\r\n... in our case, 405.\r\n\r\nThe numbers after the decimal point in a call number are read individually, not as a whole number.\r\n\r\nThat means you read the number as ...\r\n\r\n\r\nAs narrator says each number, the number on the enlarged label with is highlighted in green.\r\n... four, zero, five ...\r\n\r\n\r\nSame image of the enlarged book spine label: 337.405 E86.\r\n... not 405.\r\n\r\nSo to find 337 point 405, look between ...\r\n\r\n\r\nPhoto of books in background with overlay of graphic spine label 337.405 E86 in centre (as per previous clip).\r\n\r\nThen appearing to the left of screen is graphic spine label 337.4.\r\n... 337 point four ...\r\n\r\n\r\nPhoto of books in background with overlay of graphic spine label 337.405 E86 in centre and graphic label 337.4 to left (as per previous clip) then 337.41 on a graphic label appears to the right of screen.\r\n... and 337 point four one.\r\n\r\n\r\nSame video as previous clip but 337.405 is highlighted in green.\r\nNow that you\u2019ve got the right number ...\r\n\r\n\r\nSame video as previous clip but now E is highlighted in green.\r\n... look for the letter after the number. These are in alphabetical order.\r\n\r\n\r\nLeft and right spine labels are removed and only centre graphic spine label remains ie 337.405 E86.\r\n\r\nHighlighting the numeral eight, followed by the numeral six.\r\nYou then read the following numbers individually.\r\n\r\n\r\nUnidentified person\u2019s hands takes out the book from the shelf which has been the subject of the video\u2019s search and shows the book\u2019s cover to the camera: EU-ASEAN.\r\n\r\nThe person then opens the book.\r\nGot it.\r\n\r\n\r\nRMIT University Logo and website www.rmit.edu.au.\r\n\r\n\r\n\r\n\r\nEnd of video\r\n \r\n\r\nAs a student do you ever feel like you're drowning in a sea of information?\r\n\r\nI know I did.\r\n\r\nThere's so much out there but I never knew how to tell whether it was any good.\r\n\r\nThen the library threw me a lifeline.\r\n\r\nI found my subject guide!\r\n\r\nSubject guides are mini websites for each subject area.\r\n\r\nThey show you all the best places to look for information on what you're studying.\r\n\r\nThis includes the best journals, websites, conference papers, e-books, books and more, whatever is relevant to what you're studying.\r\n\r\nSubject guides and made by librarians, experts in finding information in your area, so you know what you're searching will be good quality.\r\n\r\nFor my next assignment I skipped Google and started with my subject guide.\r\n\r\nI saved heaps of time by heading straight for the quality resources.\r\n\r\nNow assignment time isn't overwhelming because I know I can get to the right information when I need it.\r\n\r\nIt makes all the difference when you know where to start.\r\n","n":0.029},[{"v":"Study skills","i":2,"n":0.707},{"v":"Researching","i":1,"n":1},{"v":"Academic English","i":0,"n":0.707}]]},{"i":388,"$":[{"v":"Find the information","n":0.577},{"v":"Finding the right information for your assignment topic may seem daunting at first. This section will cover the 'where' and the 'how' of finding information to get you off to a strong start.\r\n\r\nBackground reading\r\nDoing some background reading will give you a better understanding of your topic and provide direction for your research. It helps you define terms, gather facts and get an overview of the topic.\r\nWikipedia is a good starting place for a topic overview, but don't rely on it as a key reference for your assignments. Wikipedia articles can be written by anyone regardless of their knowledge or qualifications.\r\nHere are some ideas on how to find background information:\r\n\r\n \tDictionaries can help you define terms used in your subject area - use general and subject-based dictionaries.\r\n \tUse books and textbooks for broad overviews and detailed information on the main aspects of a topic.\r\n \tJournal review articles written by experts on the topic you are researching can help you identify the main issues in that topic.\r\n\r\n\r\n\r\nUsing the web for searching\r\nGoogle and other search engines\r\nThe web is a rich source of information. Students usually use search engines like Google to find information about an assignment topic, but finding useful resources among many pages of search results can be challenging.\r\n\r\nGoogle and other search engines are easy to search, but they produce a large number of results and many that are not relevant to the topic.\r\n\r\nFortunately, you can conduct a much more efficient search and save yourself lots of time and effort by using a few simple search commands. Try  Lifehack's 20 Tips to use Google Search efficiently.\r\n\r\nGoogle is a good place to find background or general information, but you need to be aware of its limitations for academic research.\r\nWhy can't I just Google it?\r\nThis question is covered in more detail in the next tutorial.\r\nUsing Wikipedia for academic research\r\nWhen you do a Google search about a topic you will most likely find Wikipedia listings at the top of your search results. However, Wikipedia is not considered a reliable source of information. This is because we do not know who has added information, what their expertise on the subject is or if the information they have added is accurate.\r\n\r\nWikipedia can be a great place to start your search as it can give you a general understanding of a topic and you can use the sources listed at the bottom of the page as part of your search. However, it is not recommended that you cite Wikipedia articles in your assignments.\r\nGoogle Scholar\r\nAnother way to search the web more effectively is to use Google Scholar as your search engine. Google Scholar is a simple way to search for scholarly literature in articles, theses, books, abstracts and other formats.\r\n\r\nIf you access Google Scholar from the RMIT Library website using your RMIT login, your search results will link you back to the full text resources held by the Library. Go to the Library Google Scholar page for more information.\r\nFor more help on using Google Scholar for research go to Google Scholar search tips or find a video on YouTube.\r\n\r\nWe will explore more resources that can help you find better, more reliable information than you would find on the web, by looking at LibrarySearch and databases.","n":0.043},[{"v":"Researching","i":1,"n":1},{"v":"Information sources","i":0,"n":0.707}]]},{"i":389,"$":[{"v":"Develop your search strategy","n":0.5},{"v":"Although conducting research online can be straightforward, your university assignments will require more academic and reliable sources. This is when LibrarySearch and library databases become valuable tools.\r\n\r\nHowever, these database search systems work differently than Google and other internet search engines, so creating an effective search strategy will help you find higher-quality, more relevant results for your academic projects.\r\n\r\nTo search effectively, you need to develop a clear search strategy that will find all the relevant information on your topic.\r\nTo do this, you need to:\r\n\r\n \tIdentify the keywords that describe the main ideas in your topic or assignment.\r\n \tExpand the list of keywords by finding alternative words. This includes synonyms, related words, abbreviations, acronyms and words that are specific to your subject area or profession.\r\n\r\nExample\r\nLet's use the following topic as an example:\r\nCompare team models in two contemporary organisations. Discuss the effectiveness of these models with reference to current management theories.\r\nNote the key words and possible alternative words that could be used for your search strategy:\r\n\r\n\r\n\r\nKeyword\r\nAlternative words\r\n\r\n\r\n\r\n\r\nteam models\r\ngroups, teamwork\r\n\r\n\r\neffectiveness\r\neffective, functional, useful, operational, performance\r\n\r\n\r\nmanagement theories*\r\n*Refine your search by using specific management theories\r\n\r\n\r\n\r\nTip: Use a dictionary or a thesaurus to help you find alternative words.\r\nActivity: Identifying keywords\r\n\r\n\r\nCreate a search query\r\nMost databases offer both a basic and an advanced search option. If you need help searching a particular database, look for the Help menu in the database to find tips and 'How to' videos. You may also find database tutorials on YouTube, and from some Library Subject Guides.\r\n\r\nFor an advanced database search, you may need to use Boolean operators. Boolean operators connect your search terms together in specific ways (AND, OR, NOT), and also help you to narrow or broaden your search.\r\n\r\nBoolean operators\r\nSome databases and catalogues allow you to combine keywords using Boolean operators (advanced operators).\r\nFor example:\r\n\r\n\r\n\r\n\r\n\r\nTruncation and wildcards\r\nSome databases and catalogues also allow you to use truncation and wildcard symbols.\r\nFor example:\r\nbehaviour* = behavioural, behaviours, behaviour.\r\nYou can use truncation to broaden your search by substituting a word ending with a symbol. In this example, we use an asterisk (*) as the truncation symbol. However, different databases use different symbols, so you need to check the search tips in a database to find out what symbol it uses.\r\nWildcards\r\nYou can use wildcards to broaden your search by substituting letters in words. In this example, we have used a question mark (?) as the wildcard symbol. However, different databases use different symbols, so you will need to check the search tips in a database to find out what symbol it uses.\r\n\r\nFor example:\r\norgani?ational behavio?r = organisational, organizational, behaviour, behavior.\r\n\r\nSearching for a phrase\r\nSome databases and catalogues allow you to search for phrases. Quotation marks placed at the beginning and end of a phrase narrows a search. Only results that include the exact phrase will appear.\r\n\r\n\r\n \t\"organisational behaviour\" = results with the exact words, \"organisational behaviour\"\r\n\r\n","n":0.046},[{"v":"Understanding the question","i":1,"n":0.577},{"v":"Researching","i":0,"n":1}]]},{"i":390,"$":[{"v":"Academic reflection activities","n":0.577},{"v":"Use the quizzes on this page to test your knowledge of reflective writing.\r\n\r\nAcademic reflection: Quick quiz\r\n\r\n\r\nFeatures of DIEP\r\n\r\n\r\n","n":0.236},[{"v":"Academic writing","i":2,"n":0.707},{"v":"Assessment","i":1,"n":1},{"v":"Reflective writing","i":0,"n":0.707}]]},{"i":391,"$":[{"v":"Sample reflection","n":0.707},{"v":"This page features an academic reflection sample showing the DIEP process.\r\n\r\n\r\nA reflective blog: an academic reflection in four paragraphs (from Master of Project Management)\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nLinking experiences to evidence\r\nLinking the experiences you have had to ideas from the literature produces higher quality reflective writing. Read the sample reflection to see how this student has integrated theories from literature into the reflection.\r\n\r\n\r\n\r\nAuthentic reflections\r\nReflections sound authentic and relevant when specific events are described. Read the following reflection and note the highlighted example from the learning situation that prompted the writer's reflection and learning development.\r\n\r\n","n":0.105},[{"v":"Sample writing","i":1,"n":0.707},{"v":"Reflective writing","i":0,"n":0.707}]]},{"i":392,"$":[{"v":"How to use DIEP (Describe Interpret Evaluate Plan)","n":0.354},{"v":"DIEP is a framework that can be used to guide your academic reflections. This page takes you through the different sections of the DIEP framework.\r\n\r\n\r\n \r\n \r\n\r\n","n":0.192},[{"v":"Reflective writing","i":1,"n":0.707},{"v":"DIEP","i":0,"n":1}]]},{"i":393,"$":[{"v":"Essays: Review","n":0.707},{"v":"It's important to revise your draft to ensure it addresses the task and is written clearly. Use the checklists on this page to get you started.\r\n\r\nEssay editing checklist\r\nUse the checklist below to guide your revision.\r\n\r\nStructure\r\n\r\n      Does the introduction clearly state the topic and what will be covered? Is there a thesis statement? \r\n      Do the paragraphs contain topic sentences that refer to the thesis stated in the introduction? \r\n      If you read only the topic sentence of each paragraph, does the structure of the whole piece become clear? \r\n      Are the links between paragraphs and ideas clear? \r\n      Does your conclusion answer the question and sum up the argument? \r\n\r\n\r\nTopic\r\n\r\n      Have you answered the question? \r\n      Check your plan against the original question. \r\n      Compare the finished assignment with the plan. Does it follow the plan? \r\n      Have you provided enough evidence and detail to support your ideas? \r\n      Is your content accurate, detailed and concise? \r\n\r\n\r\nArgument\r\n\r\n      Is your argument clearly stated in the introduction? \r\n      Is your argument thorough and logical? \r\n      Is your argument supported by appropriate evidence? \r\n\r\n\r\nUnity and coherence\r\n\r\n      Is there a logical flow within each paragraph? \r\n      Does every sentence in the paragraph relate to the topic sentence? \r\n      Do you use appropriate linking words? (e.g. firstly, however, therefore) \r\n      Are key terms clearly defined, consistent and clear? \r\n      Are relationships between ideas clear? \r\n\r\n\r\nReferencing\r\n\r\n      Are your sources referenced in the correct style? \r\n      Are all the sources cited in the text included in your reference list/bibliography? \r\n      Are your references in the correct order? (alphabetical - Harvard & APA; order of appearance \u2013 footnote) \r\n      Are all the bibliographic details provided? (author, date, titles, publisher, place) \r\n      Do you have enough references? \r\n      Are your sources reliable, current and appropriate? \r\n\t  Are paraphrases worded differently enough from the original source? \r\n\r\n\r\nLanguage and grammar\r\n\r\n\t  Have you checked for errors in spelling, punctuation and sentence structure? \r\n","n":0.056},[{"v":"Essays","i":2,"n":1},{"v":"Editing","i":1,"n":1},{"v":"Checklists","i":0,"n":1}]]},{"i":394,"$":[{"v":"Structure of an essay","n":0.5},{"v":"An essay has three parts: the introduction, a body and a conclusion. Write using your detailed plan. Expect to write several drafts.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.213},[{"v":"Essays","i":2,"n":1},{"v":"Essay structure","i":1,"n":0.707},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":395,"$":[{"v":"Plan and research","n":0.577},{"v":"A good place to start planning your essay is by researching your topic. From there, you can move on to develop a mind map that helps you formulate a linear essay plan.\r\n\r\nStart early and plan how you will manage your time. Tools like the Assignment planner can be useful for this. You will need to move back and forth between steps as you refine your essay plan.\r\n\r\nResearch the topic\r\n\r\n \tRe-read your course notes then do some initial reading on the topic from the suggested reading list.\r\n \tUse LibrarySearch and subject guides to find additional, relevant material.\r\n \tTake notes.\r\n \tKeep a record of all relevant sources to avoid plagiarism.\r\n\r\nMindmapping\r\n\r\nOrganise your research by creating a mind map.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nLinear essay plan\r\n\r\nTurn your mindmap into a detailed linear plan that includes references. Several options for the structure may be available at this stage. Your choice might need modification or review later on. Look at the following examples.\r\nExample 1\r\n\r\n \tIntroduction\r\n \tDefinitions and background\r\n\r\n \tTypes of organisations\r\n\r\n \tdaily operations and goals\r\n\r\n\r\n\r\n\r\n \tOrganisational strategic planning\r\n \tPurpose of teams (habitual/traditional or purpose formed)\r\n\r\n \tspecific team purpose and how they operate (in business context)\r\n \ttheories; Idealistic model (normative)\r\n\r\n \tBerlin, Carlstrom & Sandberg 2012\r\n\r\n\r\n \tproblem solving teams (describe structure and purpose in Co. A)\r\n \tcross functional teams (describe structure and purpose in Co. B)\r\n\r\n\r\n \tTeam effectiveness (cohesiveness)\r\n\r\n \tMathieu et al 2014\r\n\r\n\r\n \tThreats to effectiveness\r\n \tConclusion\r\n\r\nExample 2\r\n\r\n \tIntroduction\r\n \tdefinitions and background\r\n\r\n \tCo. A and Co. B operations and goals\r\n \tCo. A and Co. B basic description of team models\r\n\r\n\r\n \tEffective teams\r\n\r\n \tHow do we measure the effectiveness of a team? (...according to theorists)\r\n \tWhat makes a team effective (according to theorists - adaptability? incusivity? positive dynamics? collaboration?...and so on)\r\n\r\n\r\n \tComparison of team models in Co. A and Co. B\r\n\r\n \tHow do the two different models in each organisation compare? (this is the major section. There should be lots of paragraphs starting with topic sentences such as:)\r\n\r\n \tThe team model in Co. A is more effective than that in Co. B because...\r\n \tAccording to...the team model of Co. B has the advantage of...\r\n\r\n\r\n\r\n\r\n \tConclusion\r\n\r\n\r\n\r\n\r\n \r\n\r\nIdeas:\r\n\r\n \tPurpose of team\r\n \tEffectiveness\r\n \tModels\r\n \tRelevance and adaptability\r\n\r\nKeywords:\r\n\r\n \tprojects daily operations\r\n \tstaff moral\r\n \tprofessional development\r\n \tteam leaders\r\n \tmember skills\r\n \tmanagement\r\n \tteam members\r\n \tdifferent departments\r\n \tcommunication\r\n \tresources\r\n \tauthoritarian\r\n \tparticipatory\r\n \tfunction\r\n \ttypes of teams\r\n \tlife span\r\n \tgoals\r\n \tunderstandings\r\n \tpermanent\r\n \ttemporary\r\n \tgroup dynamics\r\n \tfixing ineffective teams\r\n \twho measures goal success?\r\n \tindividual gains\r\n \twhy companies use teams\r\n\r\n \tteam reporting\r\n \tdealing with unproductive teams\r\n \tteam disputes\r\n \tforming a team\r\n \tdismantling a team\r\n \tno team structure\r\n\r\n\r\n\r\n\t \r\n\r\nUse and effectiveness (mindmap central heading):\r\n\r\n \tEffectiveness (subheading)\r\n\r\n \tprofessional development\r\n \tstaff moral\r\n \tcommunication\r\n \tfixing ineffective teams\r\n \tresources\r\n \tgroup dynamics\r\n \twho measures goal success?\r\n \tunderstandings\r\n\r\n\r\n \tPurpose of team (subheading)\r\n\r\n \tfunction\r\n \tgoals\r\n \tprojects daily operations\r\n \tindividual gains\r\n\r\n\r\n \tModels (subheading)\r\n\r\n \tmanagement\r\n \tteam leaders\r\n \tmember skills\r\n \tpermanent\r\n \ttemporary\r\n \ttypes of teams\r\n \tparticipatory\r\n \tauthoritarian\r\n\r\n\r\n \tRelevance and adaptability (subheading)\r\n\r\n \tdifferent departments\r\n \tlife span\r\n \twhy companies use teams\r\n \tteam members\r\n\r\n\r\n\r\n\t","n":0.046},[{"v":"Researching","i":2,"n":1},{"v":"Essays","i":1,"n":1},{"v":"Essay structure","i":0,"n":0.707}]]},{"i":396,"$":[{"v":"Case studies: Editing and proofing","n":0.447},{"v":"Before submitting any assessment, it is important that you spend some time editing and proofreading your work. Use this checklist to guide you through this process.\r\nMake sure you have followed the task instructions, addressed the marking criteria and presented your work as well as possible. Use the following checklist to make sany revisions before submitting your assignment:\r\n\r\n\r\n    Have you identified the key issues and problems?\r\n    Have you used theory, concepts and/or models to analyse these issues and problems?\r\n    Have you included specific examples from the case to support your analysis?\r\n    Do the solutions/recommendations you propose address the key issues (problems) you identified?\r\n    Are there references to indicate where you've integrated theories/concepts from your research?\r\n    Have you provided a reference list using the appropriate referencing style?\r\n    Have you checked for spelling and grammar errors?\r\n    Is your writing clear and concise?\r\n\r\n\r\n\r\n","n":0.085},[{"v":"Proofreading","i":3,"n":1},{"v":"Editing","i":2,"n":1},{"v":"Checklists","i":1,"n":1},{"v":"Case studies","i":0,"n":0.707}]]},{"i":397,"$":[{"v":"Make recommendations","n":0.707},{"v":"The recommendations that you make should be based upon your solutions for each of the identified problems. Here are some examples for how you can do this.\r\nThe following example shows how headings and subheadings make the structure clear and how theory is used to support the recommendations. The shows how a student has used the solutions to make specific recommendations.\r\nNote that recommendations are not always required as part of your assessment task. Always check your assessment criteria.\r\n\r\n\r\n","n":0.114},[{"v":"Sample writing","i":2,"n":0.707},{"v":"Case studies","i":1,"n":0.707},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":398,"$":[{"v":"Case studies: Analyse the task","n":0.447},{"v":"The first step of any assignment should be to become familiar with the assignment instructions. This will help you to know exactly what is required by the task and help you to focus your energy on the right tasks. The following example explains how to analyse a task in order to address the assignment requirements.\r\n\r\nCase study assessment task\r\n\r\nAssignment instructions\r\nThe HealthCo Board has asked you to write a report recommending what management approaches it should adopt in order to meet the challenges that the company currently faces.\r\n\r\nYou will need to critically analyse the current situation, develop and justify a range of logical solutions and recommend a strategy to address the priority issue you have determined that HealthCo is facing.\r\n\r\nUse management theory, approaches, concepts and/or models to support your analysis and expected outcomes.\r\n\r\nThe report will be 1500 words.\r\n\r\n\r\nBreaking down assignment instructions\r\nAll assignment instructions contain instruction words. These are words that tell us what we need to do and how we should do it. One way we can find these words is to search for the verbs in the instructions. Let\u2019s look at the same instructions, this time with the instruction words highlighted. \r\n\r\nAssignment instructions\r\n\r\nCreating a checklist\r\nNext, you can use the highlighted words to create a checklist of items that need to be included in your assignment. When you do this, try to use as few words as possible so that it will be easy to read later. \r\n\r\nThis list can be used as you write to remind you what needs to be included, and also when you have finished writing your draft to make sure we haven\u2019t missed any of the requirements. \r\nHere's an example of a checklist: \r\n\r\nChecklist for my report:\r\n\t\r\n\t\tRecommend management approaches to meet challenges\r\n\t\tCritically analyse the current situation\r\n\t\tDevelop solutions\r\n\t\tJustify solutions\r\n\t\tRecommend a strategy to address the issue\r\n\t\tUse management theory to support:\r\n \t\t\t\r\n\t\t\t\tThe analysis\r\n \t\t\t\tThe expected outcomes\r\n\t\t\t\r\n\t\r\n\r\n\r\n","n":0.057},[{"v":"Analysis","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":399,"$":[{"v":"Case studies: Developing solutions","n":0.5},{"v":"Once you have analysed the problems, you need to develop possible solutions using research, theory and industry practice.\r\nYou then need to evaluate these solutions to make recommendations for action. The exampless on this page will help you get started with this process.\r\n\r\nPotential solutions should be developed for each of the key problems you have identified. Often there is more than one solution, so it is useful to evaluate each solution in terms of its advantages and disadvantages.\r\n\r\nYour evaluation of a solution may need to consider constraints such as:\r\n\r\n \teffectiveness\r\n \tcosts\r\n \ttime\r\n \tresources.\r\n\r\nThe solutions section should be clear and concise. Use headings and subheadings where appropriate to set out the advantages and disadvantages of a course of action.\r\n\r\n\r\n   \r\n\r\n\r\n\r\n\r\n","n":0.092},[{"v":"Sample writing","i":2,"n":0.707},{"v":"Report structure","i":1,"n":0.707},{"v":"Case studies","i":0,"n":0.707}]]},{"i":400,"$":[{"v":"Case studies: Research and link to theory","n":0.378},{"v":"Begin your research with information from your course, for example: lectures, course notes and readings.\r\nResearch the specific issues you identified to find additional academic and authoritative sources to support your analysis of the case. As your understanding of the topic develops, your research becomes more focused and effective.\r\nFor more information about using the library search tools and study guides, go to Library orientation.\r\n\r\nThe next step is to link the identified problems to theory. In fact, integrating theory into your case study answer is vital. It allows you to:\r\n\r\n \testablish how theory relates to the actual issues/problems\r\n \tdemonstrate your understanding of your course content.\r\n\r\nHow to map case evidence to relevant theory\r\nUse a simple strategy such as a table to organise evidence from the case and link it to the broader problem or issue the evidence describes.\r\n\r\nThis will help you target your research to relevant theory, models or practice which are directly linked to your case evidence.\r\n\r\nYour research will also help you select appropriate solutions to the problems that you first identified.\r\n\r\nThe following example uses information from a management case study.\r\n\r\nExample case study\r\n\r\nThe example from the following management case study shows how the student has identified an issue based on actual examples from the case. They have then used theory to explain how this behaviour affects the organisation. Note how the student has used only one issue in their paragraph, which is a good model to follow.\r\nCase study\r\n\r\n","n":0.065},[{"v":"Theory","i":2,"n":1},{"v":"Researching","i":1,"n":1},{"v":"Case studies","i":0,"n":0.707}]]},{"i":401,"$":[{"v":"Identify key issues","n":0.577},{"v":"Identifying the major issues and their causes is vital to proposing appropriate solutions later. Let's explore how we can do this.\r\nFrom your annotation and note-taking you should have an understanding of what the problems are.\r\nSummarise or list the issues in your own words\r\nMake sure you:\r\n\r\n \tsort the major problems from the minor problems\r\n \tidentify evidence from the case which relates to each of the problems\r\n \tidentify underlying causes of the problems.\r\n\r\nA useful strategy to start with is to represent the problems and their relationships as a mind map.\r\n\r\n \r\n \r\nThe image shows handwritten text. At the top there are two groups labelled \"Staff\" and \"Communication\". They both have arrows pointing to a large piece of text, \"Problems/issues\", in a circle in the centre.\r\nThe group labelled \"Staff\" has the following sub-topics:\r\n\r\n    treated as numbers\r\n    only there for the money\r\n    no initiative allowed\r\n    unsupported\r\n    underutilised\r\n\r\nThe group labelled \"Communication\" has the following sub-topics:\r\n\r\n    top down only\r\n    'memos'\r\n    no feedback\r\n    no updates on company policy changes\r\n\r\nFrom the \"Problems/issues\" circle there are arrows to two more groups, labelled \"Management culture\" and \"Leadership\". In addition, there are arrows between each of these groups.\r\nThe group labelled \"Management culture\" has the following sub-topics:\r\n\r\n    top down: hierarchical\r\n    distrust staff\r\n    no respect\r\n    controlling / no initiative allowed\r\n    gender: no senior women\r\n    no understanding of what frontline staff do\r\n    staff are 'low-skilled clerks'\r\n\r\nThe group labelled \"Leadership\" has the following sub-topics:\r\n\r\n    CEO/ Owner\r\n    autocratic style\r\n    results focussed\r\n    controls\r\n    no interaction with staff\r\n\r\nFinally, each of these groups has arrows which lead to the last group, labelled \"Control\". It has the following sub-topics:\r\n\r\n    'super snoops' are spies\r\n    monitor computer hits to measure productivity\r\n    no initiative to make improvements\r\n    'us and them' mentality\r\n\r\n \r\n\r\nThe image shows handwritten text. At the top there are two groups labelled \"Drug therapies\" and \"Diet therapies\". There's also an additional label \"what is UC (define)?\". All three arrows pointing to circled text, \"Ulcerative Colitis\", which then points to large circled text, \"Case study, Mr Rogers\".\r\nThe group labelled \"Drug therapies\" has the following sub-topics:\r\n\r\n    Corticosteroids\r\n        \r\n            Vedolizumab (side effects)\r\n            Mesalazine\r\n            Budesonide (oral)\r\n            Prednisolone & aminosalicylates\r\n        \r\n    \r\n    Antimicrobials\r\n    Consider:\r\n        \r\n            side effects\r\n            cost\r\n            practicality\r\n        \r\n    \r\n\r\nThe group labelled \"Diet therapies\" has the following sub-topics:\r\n\r\n    Use with drug therapies\r\n    Education & lifestyle changes\r\n    Exclude caffeine, high fat spicy food, ALCOHOL\r\n    Probiotics (?)\r\n    Healthy food habits\r\n\r\nFrom the \"Case study, Mr Rogers\" circle there are arrows to two more groups, labelled \"GORD\" and \"Colonoscopy\". In addition, there are arrows between each of these groups.\r\nThe group labelled \"Management culture\" has the following sub-topics:\r\n\r\n    top down: hierarchical\r\n    distrust staff\r\n    no respect\r\n    controlling / no initiative allowed\r\n    gender: no senior women\r\n    no understanding of what frontline staff do\r\n    staff are 'low-skilled clerks'\r\n\r\nThe group labelled \"GORD\" has the following sub-topics:\r\nWhat is GORD (define)?\r\nSymptom control\r\n\r\n    Impact on quality of life (difficult to work, travel and exercise)\r\n    Require diet and drug therapies\r\n        \r\n            (PPI) Proton Pump Inhibitor\r\n            (H2RA) Histamine 2-receptor antagonists\r\n            Antacids\r\n            Surgery\r\n        \r\n    \r\n\r\n\r\nColonoscopy\r\n\r\n    Surgical procedure\r\n        \r\n            analgesia\r\n            allergies (amoxycillin)\r\n        \r\n    \r\n    Hypotension\r\n        \r\n            laxatives\r\n        \r\n    \r\n\r\n","n":0.045},[{"v":"Sample writing","i":3,"n":0.707},{"v":"Note-taking","i":2,"n":1},{"v":"Critical reading","i":1,"n":0.707},{"v":"Case studies","i":0,"n":0.707}]]},{"i":402,"$":[{"v":"Read the case","n":0.577},{"v":"The next step is to read the case to identify what the issues are and try to understand how, or why, they might have occurred. Let's find out how we can do this.\r\n\r\n\r\nA proper analysis requires thorough reading more than once.\r\n\r\nThe three reading strategies to use are:\r\n\r\n \tSkimming: To gain a general idea of the content and structure of the case\r\n \tScanning: To pick up important key words, linking words, names and numbers (including dates)\r\n \tDeep reading: Slow and careful reading with a critical (questioning) approach\r\n\r\nStep 1: Skimming and scanning\r\nDo not stop to make notes. The purpose is to understand the larger concepts, overall structure and scan for keywords, names and dates. When you have finished, ask yourself the following questions:\r\n\r\n \tWhat is the case about?\r\n \tWhat issues can I see now?\r\n \tDoes anything immediately connect to my course notes?\r\n\r\nStep 2: Detailed reading\r\nRe-read the case again, more slowly and carefully, for a deeper understanding. Some questions you could ask are:\r\n\r\n \tWhat critical events are important? When and why did they happen?\r\n \tWho are the people involved and who is responsible?\r\n \tWhat are the key problems? What evidence from the case shows this?\r\n \tWhat are the constraints or obstacles to potential solutions?\r\n \tWhat do I need to research?\r\n\r\n\r\nStep 3: Annotation/note-taking\r\nNow, begin to annotate the case and continue to ask yourself questions as you read.\r\n\r\nHighlight, underline and circle text, and add names, labels and comments to:\r\n\r\n \tidentify important points and supporting points (examples, reasons, statistics)\r\n \tdraw connections between ideas\r\n \tmark important definitions and terminology\r\n \tidentify any points that need clarification\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.062},[{"v":"Sample writing","i":3,"n":0.707},{"v":"Note-taking","i":2,"n":1},{"v":"Critical reading","i":1,"n":0.707},{"v":"Case studies","i":0,"n":0.707}]]},{"i":403,"$":{"0":{"v":"Chemistry","n":1},"2":[{"v":"Chemistry","i":0,"n":1}]}},{"i":404,"$":[{"v":"Sample report structures","n":0.577},{"v":"Reports from different areas and industries vary in their structure. This page features some examples of common report structures.\r\n\r\nThe highlighted headings and subheadings show some of the differences between reports in different fields of practice or for different purposes.\r\n\r\n\r\n","n":0.16},[{"v":"Academic writing","i":3,"n":0.707},{"v":"Assessment","i":2,"n":1},{"v":"Reports","i":1,"n":1},{"v":"Report structure","i":0,"n":0.707}]]},{"i":405,"$":[{"v":"Overall structure of a report","n":0.447},{"v":"There are as many structures for a report as there are reasons for writing one. The differences between reports occur chiefly in the body. This page provides you with information about the different sections that can be found in a report and what information should be included in them.\r\nReports follow a principle of moving from a problem or issue through to solutions. The structure (headings, sub-headings) needs to be tailored to fit the context (rather than vice versa).\r\n\r\nYour lecturer may provide a structure by giving headings or questions to answer. Follow the instructions of your teacher first.\r\n\r\nYou need to determine the most logical and efficient way to:\r\n\r\n \tchunk the information\r\n \torder the information\r\n \tlabel the information using headings or subheadings.\r\n\r\nReports vary in style and purpose but the following sections are common to most reports. The sections described in the accordions below should be used as a guide and are not prescriptive.\r\nBeginning\r\nYour report should start by setting the context for the body of the report. It is the section where the reader is informed of what they need to know to understand the rest of the report.\r\n\r\n\r\n\r\n\r\n\r\n\r\nMiddle (body)\r\nOrganise the sections in a logical sequence based on:\r\n\r\n \thow you investigated the issue or problem (method, methodology, materials, participants, length of study time)\r\n \twhat you found through the investigation, (results)\r\n \twhat judgements/ interpretations you have made about the process or the findings (discussion, analysis)\r\n\r\n\r\n\r\n\r\n\r\n\r\nEnd\r\nReports are often about solving problems or exploring issues. The end of the report should give the reader a sense of outcome and future direction after the body.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.062},[{"v":"Academic writing","i":3,"n":0.707},{"v":"Assessment","i":2,"n":1},{"v":"Reports","i":1,"n":1},{"v":"Report structure","i":0,"n":0.707}]]},{"i":406,"$":[{"v":"Writing professional emails","n":0.577},{"v":"Emails are an important form of written communication in the workplace. This resource covers how to structure and write clear and concise professional emails.\r\nAs emails are short pieces of communication, there is a set structure you should use. One feature of this structure is putting important information first.\r\n\r\nWhy use this approach?\r\n\r\n \tPeople are busy and do not have enough time to read everything.\r\n \tThe reader will look at the beginning and decide whether to read on.\r\n \tIt helps keep the email short and to the point.\r\n\r\nRemember to:\r\n\r\n \tput the most important information first\r\n \tremove any unnecessary information.\r\n\r\nEmails refer to immediate situations and need to quickly get to the point.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nOriginal email:\r\n\r\nDear all,\r\nRMIT is a large university. All staff are on the email system. There have been occasions when an email virus has been sent to out employees. Many employees were not sure how to respond to this problem. Some employees forwarded these emails to others and the virus was able to spread. The university now has a policy in place to prevent the spread of such problems in future. All staff should notify a ITS Service and Support Officer immediately if they suspect they have received a virus.\r\nITS Service and Support\r\n\r\nThe main point is:\r\n\r\nAll staff should notify an ITS Service and Support Officer immediately if they suspect they have received a virus.\r\n\r\nRewritten email:\r\n\r\nDear all,\r\nAll staff are advised to contact a ITS Service and Support Officer immediately on suspicion of an email virus. Early awareness and detection should prevent future problems.\r\nITS Service and Support\r\n\r\n \r\nPoor readability\r\nHello Saul,\r\n\r\nI hope this finds you well. I'm writing about the self-access HR materials for casual staff. Currently, these materials are only available in hard copy and this presents two problems: the materials often go missing and they become redundant as soon as policies are updated. The result is that people often do not have access to the policies and have to devote time and energy calling HR. It is also expensive in terms of printing costs as we constantly have to re-print the materials. Would it be possible to post these materials on the HR webpage? If your team could look into this, I'd be most grateful.\r\n\r\nBest wishes\r\nNadia\r\nImproved readability\r\nHello Saul,\r\n\r\nI hope this finds you well.\r\n\r\nI'm writing about the self-access HR materials for casual staff.\r\n\r\nCurrently, these materials are only available in hard copy and this presents two problems:\r\n\r\n \tthe materials often go missing and\r\n \tthey become redundant as soon as policies are updated.\r\n\r\nThe result is that people often do not have access to the policies and have to devote time and energy calling HR. It is also expensive in terms of printing costs as we constantly have to re-print the materials.\r\n\r\nWould it be possible to post these materials on the HR webpage? If your team could look into this, I'd be most grateful.\r\n\r\nBest wishes\r\nNadia\r\n\r\n \r\n\r\nNadia and Saul work for the same company. They have a good professional relationship. Saul works in Human Resources (HR). Nadia manages a group of casual staff.\r\n\r\nThe brackets show the features of a good email.\r\n\r\nTo: saul.bihari@modernflair.com\r\n\r\nFrom: nadia.rajapakse@modernflair.com\r\n\r\nSubject: Self-access HR materials (clear, accurate and concise subject line)\r\n\r\nHello Saul (use of informal greeting is appropriate as they know one another)\r\n\r\nI hope this finds you well.\r\n\r\nI'm writing about the self-access HR materials for casual staff. (purpose for writing the email is stated clearly at the beginning)\r\n\r\nCurrently, these materials are only available in hard copy and this presents two problems: (layout of email is easy to read with use of bullet points)\r\n\r\n \tthe materials often go missing and\r\n \tthey become redundant as soon as policies are updated.\r\n\r\nThe result is that people often do not have access to the policies and have to devote time and energy calling HR. It is also expensive in terms of printing costs as we constantly have to re-print the materials.\r\n\r\nWould it be possible to post these materials on the HR webpage? If your team could look into this, I\u2019d be most grateful. (Nadia politely suggests a solution in a professional neutral tone)\r\n\r\nBest wishes (appropriate informal ending as they know each other)\r\nNadia\r\n\r\nNadia Rajapakse (Ms.) (Most organisations require employees to use email templates including ways of formatting your signature)\r\nOperational Manager\r\nModern Flair Designs\r\nUnit 3, Melbourne\r\nVIC 3000\r\ntel: 03 - 852 - 1773\r\n\r\n","n":0.038},[{"v":"Professional writing","i":1,"n":0.707},{"v":"Email communication","i":0,"n":0.707}]]},{"i":407,"$":[{"v":"Writing for the workplace","n":0.5},{"v":"When writing for the workplace, using plain English will make your texts easy to read. The following tips guide you through how to write in plain English.\r\n\r\nShort and simple sentences\r\nOne of the ways to make your writing easier to read and understand is to use short and simple sentences. Keep your sentences short (15\u201320 words) by removing unnecessary words.\r\n\r\nWhy?\r\n\r\n \tThey are easier to read and understand.\r\n \tLonger sentences can lead to misunderstandings and confusion if read quickly.\r\n \tLong sentences make ideas less clear.\r\n\r\nCompare:\r\n\r\n\r\n\r\n\r\n\r\n\r\nThe second sentence is easier to read. Sometimes long sentences may be necessary, however too many make it difficult for the reader to comprehend.\r\nUse clear and concise language\r\nUse simple, clear and direct language. Use strong verbs (doing/action words) rather than noun phrases.\r\nCompare:\r\n\r\n\r\n\r\nBoth sentences have the same meaning. Using the verb 'compared' rather than the noun phrase 'undertook a comparison' makes the second sentence clear and direct.\r\n\r\n\r\nMain idea at the beginning\r\nWhere possible, put the main idea or key information at the start of the sentence to make it easier to comprehend.\r\nCompare:\r\n\r\n\r\n\r\nIn sentence 1b, the key information has been placed at the beginning of the sentence for easier comprehension.\r\n\r\n\r\n\r\nIn the example above, the tone in sentence 2a is accusatory. Sentence 2b would be more appropriate as it does not attribute blame.\r\nUse verbs instead of abstract nouns\r\nIf you are looking for a way to make your writing easier to read, one method is using verbs instead of abstract nouns. This pages explains how this can be done.\r\nSome commonly used abstract nouns and their verb equivalents are listed below.\r\n\r\n\r\n\r\nAbstract noun\r\nVerb\r\n\r\n\r\n\r\n\r\nimplementation\r\nimplement\r\n\r\n\r\ninvestigation\r\ninvestigate\r\n\r\n\r\nagreement\r\nagree\r\n\r\n\r\ndiscussion\r\ndiscuss\r\n\r\n\r\narrangement\r\narrange\r\n\r\n\r\n\r\nIn these pairs of sentences, the B sentences use verbs instead of abstract nouns. As a result, they are easier to read.\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nReplace these \u2018wordy\u2019 phrases with a more concise alternative.\r\n\r\n1. At this point in time = _\r\n\r\n2. Due to the fact that = _\r\n\r\n3. As a matter of fact = _\r\n\r\n4. At all times = _\r\n\r\n5. At the present time = _\r\n\r\n6. By means of = _\r\n\r\n7. For the purpose of = _\r\n\r\n8. In spite of the fact that = _\r\n\r\n9. In the event that = _\r\n\r\n10. In the final analysis = _\r\n\r\n11. A sufficient number = _\r\n\r\n12. Have the ability to = _\r\n\r\n13. Until such time as = _\r\n\r\n14. In the nature of = _\r\nAnswers\r\n1. At this point in time = now\r\n\r\n2. Due to the fact that = because\r\n\r\n3. As a matter of fact = in fact\r\n\r\n4. At all times = always\r\n\r\n5. At the present time = currently\r\n\r\n6. By means of = by\r\n\r\n7. For the purpose of = for\r\n\r\n8. In spite of the fact that = although\r\n\r\n9. In the event that = if\r\n\r\n10. In the final analysis = finally\r\n\r\n11. A sufficient number = enough\r\n\r\n12. Have the ability to = able to\r\n\r\n13. Until such time as = until\r\n\r\n14. In the nature of = like\r\n\r\n \r\n\r\nRewrite the following sentences using clear, concise language. Eliminate any unnecessary words, and change nouns to verbs where possible.\r\n\r\nSentence 1: \r\n\r\nStaff are hoping that management will make a decision in regard to the air conditioning soon.\r\n\r\nSuggested answer 1:\r\n\r\nStaff hope management will decide about the air conditioning soon.\r\n\r\nSentence 2: \r\n\r\nThe manager had a discussion with the technicians yesterday for the purpose of solving the computer problems.\r\n\r\nSuggested answer 2:\r\n\r\nThe manager discussed solutions to the computer problems with the technicians yesterday.\r\n\r\nSentence 3: \r\n\r\nAt this point in time, the computer network is unstable and technical staff are carrying out an investigation.\r\n\r\nSuggested answer 3:\r\n\r\nThe computer network is currently unstable and technical staff are investigating.\r\n\r\nSentence 4: \r\n\r\nIn spite of the fact that the new email system has been installed, staff are advised to give consideration to the possibility that some problems may occur.\r\n\r\nSuggested answer 4:\r\n\r\nAlthough the new email system has been installed, some problems may still occur.\r\n\r\nSentence 5:\r\n\r\nWe are hoping to bring this market research report to a conclusion in the shortest possible time.\r\n\r\nSuggested answer 5:\r\n\r\nWe hope to conclude the market research report shortly.\r\n\r\nSentence 6:\r\n\r\nIt should be pointed out that all staff have worked hard to give assistance to management in this very busy time.\r\n\r\nSuggested answer 6:\r\n\r\nWe acknowledge that staff have worked had to assist management.\r\n\r\nSentence 7:\r\n\r\nAt Thursday's meeting, management had a discussion with regards to staff punctuality and developed guidelines to provide a solution and solve this problem.\r\n\r\nSuggested answer 7:\r\n\r\nAt Thursday's meeting, management discussed staff punctuality and developed guidelines to solve the problem.\r\n\r\nSentence 8:\r\n\r\nIn the final analysis, all offshore staff have the ability and the possibility to make a contribution regarding this matter.\r\n\r\nSuggested answer 8:\r\n\r\nAll offshore staff can contribute to this matter.\r\n\r\n","n":0.037},[{"v":"Professional writing","i":0,"n":0.707}]]},{"i":408,"$":[{"v":"Company law: Mind maps","n":0.5},{"v":"Mind maps are a useful visual tool for organising, revising and remembering Company Law topics.\r\nThey are useful for:\r\n\r\n \tgaining an understanding of a topic, its subtopics and legal issues each week\r\n \tpreparing for exams.\r\n\r\nMaking a mind map in company law\r\nFor each topic, follow these steps to make a mind map:\r\n\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n\r\nFurther use of mind maps\r\nOngoing revision\r\nAfter each topic make a mind map of key topics, subpoints, legal issues and relevant statute and case law. Complete these mind maps after each lecture topic when the content is still fresh in your mind. This will help build your knowledge of the law.\r\nExam preparation\r\nTo revise your knowledge of Company Law redo your mind maps after each topic. Your mind maps can be used as a guide when answering tutorial questions and as preparation for exams.\r\nTake notes while reading your textbook. Highlight the important points and make notes in the lecture PowerPoint handout.\r\n\r\n The image features a central handwritten label, \"Types of companies,\" with branches leading to various classifications and descriptions of companies.\r\n\r\n    Holding or Subsidiary:\r\n        \r\n            Ultimate Holding\r\n            Test for control/subsidiary\r\n        \r\n    \r\n    Public Company:\r\n        \r\n            Public PTY\r\n                \r\n                    150\r\n                    Issues Capital/S497(1)\r\n                \r\n            \r\n            Public Limited:\r\n                \r\n                    150\r\n                    Distribution List: small and large firms\r\n                    S240(1)\r\n                    S660A(4)\r\n                    S660B(3)\r\n                    Note: involvement\r\n                \r\n            \r\n        \r\n    \r\n    No Liability Company:\r\n        \r\n            150\r\n            Feature\r\n            S501BA 2015\r\n            S501BA and S744 results\r\n        \r\n    \r\n    Company Limited by Guarantee:\r\n        \r\n            150\r\n            S922\r\n            S924\r\n            S930\r\n            Liability only when wound up (e.g., charities)\r\n        \r\n    \r\n    Others:\r\n        \r\n            Foreign entities\r\n            Exemptable bodies\r\n            Trustee companies\r\n        \r\n    \r\n\r\nBranches lead to explanations of concepts such as \"Concept of control,\" \"Liability,\" and their respective tests, statutory numbers, and legislation references are noted throughout.","n":0.06},[{"v":"Study skills","i":2,"n":0.707},{"v":"Mind mapping","i":1,"n":0.707},{"v":"Company Law","i":0,"n":0.707}]]},{"i":409,"$":[{"v":"Company law: Managing the reading","n":0.447},{"v":"Manage your reading when studying company law. Start from the big picture \u2013 gain an overview of the topic.\r\n\r\nStep 1: Outline of the topic\r\nThe lecture PowerPoints provide a guide to what is important for each company law topic \u2013 read, understand and make your own notes before the lecture and before reading the textbook.\r\nStep 2: Read the chapter outlines\r\nLook at the table of contents in each chapter \u2013 this gives you an overview of the areas and sub-areas in the chapter.\r\n\r\n \r\n\r\n\r\n\r\n\r\nStep 3: Surveying\r\nGo through headings in the chapter to understand the structure of the topic as well as any charts, diagrams, flowcharts and tables which explain the text.\r\nStep 4: Read the relevant sections of the chapter\r\nRead relevant sections of the chapter highlighting key points, legislation and cases.\r\nStep 5: Make a mind map and/or summary\r\nMake a mind map and/or summary of the key headings and subheadings in this topic and their relationship with each other.\r\nStep 6: Test your knowledge\r\nTest your knowledge by redoing quick mind maps and summaries. Each time you redo a summary or mind map, you remember more. You also find out where the gaps in your knowledge are.\r\nDon't just summarise the whole chapter. Instead use the lecture PowerPoints as a guide for your reading and notes. This way you can focus on the topics and subtopics that are important for the lecture. Not everything is relevant to remember in each chapter of your textbook.\r\n\r\n \r\nA sample title page of a book chapter. Certain features of the page have been highlighted: the chapter title, the table of contents, and the areas and sub areas within the table of contents. \r\n\r\nOn another page, the 'key points' section has been highlighted with the text in the margin reading: \"Read the key points and overview at the beginning of the chapter\". A heading on the same page has been highlighted with text in the margin reading: \"Go through the headings\".\r\n","n":0.056},[{"v":"Study skills","i":2,"n":0.707},{"v":"Law","i":1,"n":1},{"v":"Active reading","i":0,"n":0.707}]]},{"i":410,"$":[{"v":"Company law: Learning a legal language","n":0.408},{"v":"There are two types of vocabulary that you need to learn in company law: commonly used vocabulary and topic-specific vocabulary.\r\n\r\nCommonly used vocabulary\r\nThe following words are often used when explaining the law in case studies. Your lecturers and tutors may use these words to explain the law in lectures and tutorials. These words may also be useful when writing your legal case study answers.\r\n\r\nTopic-specific vocabulary\r\nEach topic in company law has its own legal language. It is important to understand and remember these terms in order to build your understanding of company law.\r\nStrategies for remembering law vocabulary\r\nCreate a list of key words for each topic, organised in meaningful groups.\r\nExample of a topic sheet\r\n\r\nMake sure you understand and revise all of the legal terms in your topic sheets.\r\nQuiz\r\nRead the following statements and identify if they are true or false. For any question that you answer incorrectly, revise the vocabulary list that your learned at the beginning of this page.\r\n\r\n \r\n\r\nThe company is entitled to enforce the statutory contract.\r\nUnder common law the veil of incorporation could be lifted if a company is set up to evade an existing legal obligation.\r\nS 246 D is concerned with the right to challenge a variation of class rights.\r\nThe company has contravened a provision of the Corporations Act.\r\nFailure to comply with S 136(5) is an offence of strict liability.\r\n\r\n","n":0.067},[{"v":"Vocabulary","i":2,"n":1},{"v":"Legal","i":1,"n":1},{"v":"Company Law","i":0,"n":0.707}]]},{"i":411,"$":[{"v":"Company law: Writing a legal scenario","n":0.408},{"v":"This resource will help you to develop strategies for learning and remembering the course content and demonstrate a four-step technique for writing case study answers.\r\nCompany law is a challenging course for students because it involves:\r\n\r\n \tdifficult legal terminology\r\n \tcomplex and detailed content\r\n \ta particular writing style.\r\n\r\nIRAC structure\r\nWhen answering a company law case study problem, it is useful to apply the IRAC structure so that you address all areas required. There are four steps:\r\n\r\n\r\n\r\n    1. Identify the issue: Define the legal relevant issue.\r\n \t2. Relevant law: Apply legal principles from section(s) of the Corporations Act, cases and statutes. Explain the law, don't just state it.\r\n \t3. Application to the facts: Develop legal arguments by applying the law to the facts of the issue.\r\n \t4. Conclusion: Arrive at a considered conclusion.\r\n\r\n\r\nNote: You are more likely to gain the most marks by explaining the relevant law and then applying it to the facts.\r\n\r\n\r\nExample\r\n\r\n\r\n\r\n Advise the directors. (This a five-mark question)\r\n\r\nAnswer\r\n\r\n\r\n\r\nIssue\r\nThe issue in this case is whether the corporate veil can be lifted.\r\n\r\n\r\nRelevant law\r\nThe legal principle to apply, established in Saloman's case, is that a company is a separate legal identity from its directors and members. However, there are exceptions when the corporate veil can be lifted. In the case of Guildford Motor Co Ltd v Horne it was decided that the veil of incorporation can be lifted if a wholly owned subsidiary was set up by a company to avoid a legal obligation under contract.\r\n\r\n\r\nApplication to facts\r\nIn this case, it appears that Bingo Ltd set up the subsidiary company to avoid its contractual obligations to supply its goods exclusively to Melvin Ltd.\r\n\r\n\r\nConclusion\r\nTherefore, the directors can ask the court to lift the veil of the wholly owned subsidiary of Bingo and sue Bingo for breach of contract.\r\n\r\n\r\n\r\nPlease note: This is a simple low mark answer (5 marks) to illustrate the use of IRAC only. Many Company Law problems will involve multiple issues. In these scenarios each issue would need to be addressed. For example: The first issue is... The second issue is whether...\r\nActivity 1\r\n\r\n\r\n\r\n\r\nNow read the following text and try to identify the IRAC structure. Check your understanding by opening the Activity 1 answer accordion folder.\r\n\r\nThe issue here is whether Gerald has rights to enforce. There are three sources of company law rights: the statute, the common law and the internal rules. As a matter of internal rights and obligations the parties need to turn to the internal rules. Without a constitution to provide internal rules, the corporate parties Rules. 202A (2) provides \"the company may also pay the director's travelling and other expenses that they properly incur: ... (c) in connection with the company's business\". This section provides discretion to the company regarding the payment of expenses. This means that the issue is a matter of private contract which is not addressed in the scenario. Accordingly, Gerald has no statutory remedy and no remedy arising from the internal rules.\r\n\r\nActivity 2\r\nRead the following question, then identify which part of the IRAC structure best fits each text excerpt by selecting from the drop down menus.\r\n\r\n\r\n\r\n\r\n\r\n","n":0.044},[{"v":"Law assessments","i":2,"n":0.707},{"v":"IRAC","i":1,"n":1},{"v":"Company Law","i":0,"n":0.707}]]},{"i":412,"$":[{"v":"Writing a legal argument: The case study","n":0.378},{"v":"In this section, we are going to use a negligence model plan to answer a case study. After reading the details of the Bert V Groovy Clothing Store case, use the model plan to identify the legal issues and relevant laws.\r\n\r\nIntroduction to Bert V Groovy Clothing Store\r\n\r\n\r\n\r\n\r\nThe details\r\nYou will need the details below to complete the following exercises. Use the case information as detailed below to guide you through this tutorial and to fill out the the Negligence Model Plan.\r\n\r\n\r\n\r\nBert may have a cause for action in the tort of negligence against Groovy Clothing Store (GCS). To prove that Groovy Clothing Store was negligent in this case, Bert must establish that:\r\n\r\n1) GCS owed him a duty of care\r\n\r\n2) GCS breached the applicable standard of that duty; and\r\n\r\n3) GCS's breach of such duty of care resulted in Bert's suffering loss or damage.\r\n\r\nLord Atkin's neighbour test suggested that person A owes a duty of care if B is sufficiently proximate to A. In other words, the test will hold if A's actions or omissions may affect B in a reasonably foreseeable manner and consequently cause damage or suffering to B. In this case, Groovy Clothing Store owes Bert a duty of care because Bert was in their premises and such duty of care is non delegable (and hence could not be discharged) to Groovy Clothing Store's contractor who was hired to renovate the premises. According to the neighbour test, occupiers of land owe a duty of care to their entrants (neighbours) in respect of premises because of their control over the premise Australian Safeway Stores v Zaluzna [1987] HCA 7. Since Bert's presence in the store was organised and controlled by the store, and it is reasonably foreseeable that Groovy Clothing Store's actions and omissions could cause damage to Bert and other shoppers, it is hence sufficient for Groovy Clothing Store to owe Bert a duty of care.\r\n\r\nThe standard of care owed by Groovy Clothing Store is that of an ordinary, reasonable and prudent occupier who takes reasonable care to avoid foreseeable risks of harm to entrants (not just shoppers) of the premises Bolton v Stone [1951] 1 All ER 1078. In this case, no information was given as to the standards usually required of storeowners or whether Groovy Clothing Store had complied with the retail industry's general standards of practice. However it is reasonably foreseeable that harm could occur: allowing too many people to climb the stairs at the same time could very likely cause serious injuries, especially when people started pushing against each other. In addition, Groovy Clothing Store should have foreseen (as it had intended to do) that an opening sale would attract a huge crowd to the store. However we have no information to support whether or not the stairs would collapse when too many people were stepping on it. Clearly such an accident could have been avoided if Groovy Clothing Store had simply made an announcement using the public address system or by putting up signs reminding shoppers not to rush up the stairs. Also, Groovy Clothing Store could have hired extra staff to control the customer flow at the stairs. Given the likelihood of serious harm to the stair users, it would have been reasonable for ordinary and prudent storeowners to perform the above precautions. Groovy Clothing Store omitted these precautions, failed to warn Bert and other shoppers and as a result caused Bert extensive injuries. It appears that Groovy Clothing Store breached its duty of care to Bert.\r\n\r\nAlthough a shopper entering a shop like Groovy Clothing Store would not assume voluntarily (volenti non fit injuria) the risk of falling down a collapsed staircase, in defence of its breaching duty of care, Groovy Clothing Store could probably claim that Bert was partly liable for his injury due to his failure to take reasonable care of himself (contributory negligence) on the grounds that an ordinary, reasonable and prudent person would not force him or herself up the stairs at the same time along with so many other people because of the foreseeability of an accident Wyong Shire Council v Shirt [1980] HCA 12. If this were the case, then not all the losses and damages Bert suffered would be recoverable.\r\n\r\nIf Groovy Clothing Store failed to prove that it was partly Bert's negligence that caused himself such injuries, it would then be completely liable to Bert for any foreseeable loss or damage (Wagon Mound Cases) which he suffered as a result of the breach. But for the breach, Bert would not have suffered extensive injuries and be hospitalised for six months Cork v Kirby MacLean LTD [1952] 2 AII ER 402; Chappel v Hart [1998] HCA 55. Hence the medical expenses for Bert's hospitalisation and any foreseeable and reasonable physical or other economic damages (e.g. loss of earning capacity), though not mentioned in the case, would be recoverable. It appears that Bert would be successful if he sued Groovy Clothing Store for negligence. However, if Bert was held partly liable for his injury due to contributory negligence, he might be able to recover only part of these damages.\r\nNegligence model plan\r\nUse the model plan to identify the legal issues and relevant laws, and apply the laws to the facts.\r\n\r\n\r\n\r\nIssues to be identified\r\nDiscussion of appropriate law\r\nAppropriate application of law\r\n\r\n\r\nWas a duty of care owed?\r\n\r\n\r\n\r\n\r\nWhat was the standard \r\nof that duty of care?\r\n\r\n\r\n\r\n\r\nDamages\r\n\r\n\r\n\r\n\r\nDefences\r\n\r\n\r\n\r\n\r\n\r\n","n":0.033},[{"v":"Law","i":1,"n":1},{"v":"Case studies","i":0,"n":0.707}]]},{"i":413,"$":[{"v":"Contract law: Model case study","n":0.447},{"v":"\r\nThis section will take you through the process of constructing a legal argument in contract law. It provides a model for you to follow. This section is based on a case study used in a contract law exam. You can complete the case study as an exam if you wish.\r\n\r\nModel a legal answer\r\nIRAC method\r\n\t\r\n\t\tIdentify: The introduction. Establishes the relevant area of law.\r\n\t\tRules: Define and examine the rules of the relevant issues.\r\n\t\tApply: The rules to the facts for each relevant issue. Always use relevant case(s) to support your application of the facts for each issue.\r\n\t\tConclusion: In some instances, you may need to identify more than one issue and be prepared to define and apply all aspects of the relevant law, always using a case reference where possible to support your answer (conclusion).\r\n\t\r\n\r\n\r\nOnce you have analysed the model outline, apply it to the case study.\r\nMerryn case study\r\nThis is a case study exercise. Read the case and answer the task below.\r\nRemember to apply the structure model to your answer to make sure that you cover all the relevant areas. If you wish to complete this exercise as a test, you can apply the following conditions:\r\n\r\n \ttime allocated: 1.5 hours\r\n \topen book exam.\r\n\r\n\r\n\r\n \r\n\r\n\r\nApply the model\r\nHaving read the case study, analyse the case in respect to the following categories. Formulate your answer using these categories.\r\n\r\nCase outline\r\n\r\nIdentify - introduction - identifies relevant area of the law and elements of a simple contract\r\n\r\nRules - define and examine the laws and principles of relevant issues\r\n\r\n \tIntention\r\n \tOffer/Acceptance\r\n \tConsideration\r\n\r\nApply - apply the laws and principles to the facts\r\n\r\n \tRules relating to the offer (include relevant cases)\r\n \tRules of acceptance (include relevant cases)\r\n \tIssue consideration (include relevant cases)\r\n\r\nConclusion\r\n\r\nNote: Once you have finished your response using this outline, check your answer against the answer guide to see how you went and adjust your answer as needed.\r\n\r\n","n":0.057},[{"v":"Contract law","i":2,"n":0.707},{"v":"Case studies","i":1,"n":0.707},{"v":"Business Law","i":0,"n":0.707}]]},{"i":414,"$":[{"v":"Contract law: Concept map","n":0.5},{"v":"A concept map is a way of sequencing, ordering or breaking down information into its constituent parts. It helps you to work out how a topic is put together and what areas you need to think about in order to analyse the topic.\r\n\r\n1. Formal contracts\r\nFormal contracts are particular types of contracts which must be completed in a particular way to be valid, e.g., they may require to be formed in writing.\r\n\r\n2. Simple contracts\r\nSimple contracts are very common types of contracts which can be completed in a variety of ways, e.g., can be formed verbally or in writing.\r\n\r\n3. Apparent simple contract\r\nThe first three elements of a simple contract must be present at all times for the contract to be valid.\r\n\r\n    Intention: There are two presumptions: firstly, that business agreements have an intention to create legal relations and secondly that social, domestic, or family agreements do not. Both these presumptions can be rebutted (argued against).\r\n    Agreement: Offer & acceptance\r\n    Consideration: Something of value (legal) promised/exchanged by both parties.\r\n\r\n\r\n4. Elements of simple contract which may affect validity\r\nThe last three elements of a simple contract may be relevant when they become an issue which could invalidate the agreement.\r\n\r\n    Capacity: Capacity relates to restrictions which may be placed on people's ability to enter into contracts, e.g., young people, mentally disabled, etc.\r\n    Certainty: Relates to the terms of a contract which must be clear.\r\n    Legality: Legality relates to the objects of the contract which must be legal.\r\n\r\n\r\n5. Simple contracts - forms\r\n'Some' simple contracts are required to be in writing, e.g., land contracts. It is important to check whether there are any 'form' requirements to your particular simple contract before you enter into it.\r\n\r\n\r\n","n":0.06},[{"v":"Concept Maps","i":1,"n":0.707},{"v":"Business Law","i":0,"n":0.707}]]},{"i":415,"$":[{"v":"Contract law: Language focus","n":0.5},{"v":"The use of legal terms is very important when developing answers to legal problems. These words are common contract law expressions.\r\n\r\nGlossary of contractual terms\r\nThis list will take you through contractual terms. It is necessary to understand these terms if you are to successfully understand contract law.\r\n\r\n\r\nQuiz 1\r\nRead the following questions and choose the right answer. Check your answers when you have answered all of the questions.\r\n\r\n\r\n\r\n\r\n\r\nQuiz 2\r\nRead the following statements and identify if they are true or false. For any question that you answer incorrectly, revise the vocabulary list that your learned at the beginning of this page.\r\n\r\n\r\n\r\n\r\n\r\nMore terms\r\nHere are some more legal terms for you to become familiar with. Complete the table with the missing information. In some cases, you will need to enter the verb form of the word, in other, the noun form of the word.\r\n\r\n","n":0.085},[{"v":"Vocabulary","i":2,"n":1},{"v":"Contract law","i":1,"n":0.707},{"v":"Business Law","i":0,"n":0.707}]]},{"i":416,"$":[{"v":"Negligence law: Model case study","n":0.447},{"v":"This section will take you through the process of constructing a legal argument around the tort of negligence.\r\nIt provides a case study using a series of steps called \"IRAC\". You can complete the case study as an exam if you wish.\r\nModel a legal answer\r\nIdentify the relevant area of law in your introduction.\r\nRules - define and examine the rules of the relevant issues.\r\nApply the rules to the facts for each relevant issue. Always use relevant case(s) to support your application of the facts for each issue.\r\nConclusionIn some instances, you may need to identify more than one issue and be prepared to define and apply all of the relevant law, always using a case reference where possible to support your answer (conclusion).\r\n\r\nOnce you have analysed the model outline, apply it to the negligence case study.\r\nCase study: Negligence sample exercise\r\nThis is a case study exercise. Read the case and complete the task below.\r\n\r\nRemember to apply the IRAC structure model to your answer to make sure that you cover all the relevant areas. If you wish to complete this exercise as a test, you can apply the following conditions:\r\n\r\n \ttime allocated: 1.5 hours\r\n \topen-book exam\r\n\r\n\r\n \r\n\r\nApply the model\r\nHaving read the case study, analyse the case in respect to the following categories. Formulate your answer using these categories.\r\n\r\nCase outline\r\n.\r\n\r\n\r\n\r\n\r\nDoes the relationship fall within an established category such as doctor/patient or lawyer/client? If not, must establish foreseeability and consider salient (other relevant) features of the case.\r\nApply the law to the facts.\r\n\r\n\r\n\r\nNegligence calculus\r\n\r\n \tDegree of risk against the likelihood of injury\r\n \tGravity of injury\r\n \tEffort required to remove the risk\r\n\r\nApply the law to the facts.\r\n\r\n\r\n\r\nApply the law to the facts.\r\n\r\n\r\nOnce you have finished your response using this outline, check your answer against the answer guide to see how you went and adjust your answer as needed.\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.058},[{"v":"Negligence","i":2,"n":1},{"v":"Case studies","i":1,"n":0.707},{"v":"Business Law","i":0,"n":0.707}]]},{"i":417,"$":[{"v":"Business law – Negligence","n":0.5},{"v":"This resource introduces you to the language and vocabulary of negligence law. It provides practice activities as well as a case study exercise.\r\n\r\n\r\n\r\nGlossary of negligence terms\r\n\r\nThe following glossary of negligence terms will help you fully understand the tort of negligence. After you have studied the terms and their definitions, you can complete the practice activity.\r\n\r\n tr td:first-child {font-weight: bold;} \r\n\r\n\r\nNegligence concept map\r\nThis negligence concept map will help you to explore sequence, order or break down information into its constituent parts. \r\n\r\n1. Tort\r\nTort means civil wrongs, that is, where one person is harmed by the actions of another (not arising from contract).\r\n\r\n2. Definition\r\nNegligence means the failure of a person to take care which leads to injury (or pure economic loss) to another.\r\n\r\n3. Elements\r\nTo successfully lodge a claim of negligence against another party, three elements must be proven:\r\n\r\n\tDuty of care (two-stage test):Does the relationship between the parties fall within an established category?\r\n\t\t\r\n\t\t\tReasonable foreseeability\r\n\t\t\tSalient features\r\n\t\t\r\n\t\r\n\tBreach of duty:Reasonable person test (special skills)Negligence calculus\r\n\t\t\r\n\t\t\tLikelihood of injury\r\n\t\t\tGravity of injury\r\n\t\t\tEffort required to remove the risk\r\n\t\t\r\n\t\r\n\tDamage:\r\n\t\t\r\n\t\t\tCausation - 'But for test'\r\n\t\t\tScope of liability\r\n\t\t\r\n\t\r\n\r\n4. Defences\r\nSomeone who has to defend a negligence claim can argue that the plaintiff contributed to their own injury (contributory negligence) or that they were fully aware of all the risks involved in their actions (voluntary assumption of risk).\r\n\r\n5. Vicarious liability\r\nAn employee is held responsible for their employee's negligence.\r\n\r\n\r\n \r\n\r\nRead the following definitions and type the correct term into the text box. Correct spelling is required. Click on the \u2018Check Answer\u2019 button to check your answer.\r\n\r\nDefinition: Individuals involved in a legal case\r\n\r\nAnswer: party\r\n\r\nDefinition: The person who has been found responsible for a legal wrong\r\n\r\nAnswer: liable\r\n\r\nDefinition: Arguments/reasons put forward in a court by the defendant in a negligence case to reduce or avoid a payout to the plaintiff\r\n\r\nAnswer: defences\r\n\r\nDefinition: The particular damage suffered by a plaintiff must be closely related to the defendant\u2019s action\r\n\r\nAnswer: remoteness\r\n\r\nDefinition: The plaintiff must show that the damage suffered was a direct result of the defendant\u2019s actions\r\n\r\nAnswer: causation\r\n\r\nDefinition: The serious nature of an injury due to the defendant\u2019s actions\r\n\r\nAnswer: gravity\r\n\r\n","n":0.055},[{"v":"Negligence","i":2,"n":1},{"v":"Case studies","i":1,"n":0.707},{"v":"Business Law","i":0,"n":0.707}]]},{"i":418,"$":[{"v":"Business law: Concept maps","n":0.5},{"v":"A concept map is a way of sequencing, ordering or breaking down information into its constituent parts. It helps you to work out how a topic is put together and what areas you need to think about in order to analyse the topic.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.152},[{"v":"Concept Maps","i":1,"n":0.707},{"v":"Business Law","i":0,"n":0.707}]]},{"i":419,"$":[{"v":"Writing a legal argument: Language focus","n":0.408},{"v":"Modal and conditional language is important in legal argument as it allows arguments to be constructed in terms of possibility, probability and conditions. It is the difference between saying 'it is because' versus 'it could be because'.\r\n\r\nModal verbs\r\nThe conditional nature of statements in legal case studies means that you may need to use words that express degrees of probability, possibility and certainty. The following chart on modal verbs may be useful when writing case studies.\r\n\r\n\r\nConditional language\r\nOften the advice in legal answers outlines a possible outcome in a future court case. Conditional language is very useful in case studies, especially when giving advice on possible or probable solutions to legal problems. A conditional is a two part sentence, e.g. if this happens, then that may/might happen.\r\n\r\nA common conditional term is 'if'. If clauses mean 'on the condition that'. It is a two part sentence and if is usually followed by a 'then'.\r\n\r\nIf the damage was reasonably foreseeable by the defendant, then liability will flow.\r\n\r\nIf this is the case, then not all the losses and damages Bert suffered would be recoverable.\r\n\r\nExample\r\nConditional language is often used in conjunction with modals, e.g. would, could, may etc. Look at how the modal and conditional work together in the following examples taken from the negligence case study.\r\n\r\nActivity\r\n\r\n","n":0.069},[{"v":"Writing structures","i":3,"n":0.707},{"v":"Legal","i":2,"n":1},{"v":"Law","i":1,"n":1},{"v":"English language skills","i":0,"n":0.577}]]},{"i":420,"$":[{"v":"Writing a legal argument","n":0.5},{"v":"The example paragraphs below demonstrate the things you need to include to write a successful legal argument.\r\n\r\nTo successfully write a legal case study you need to:\r\n\r\n \tidentify relevant legal issues\r\n \tapply the law to the facts\r\n \tstructure your answer clearly and logically (use the model plan)\r\n \tuse appropriate language for a legal argument.\r\n\r\nIdentify relevant legal issues and apply the law to the facts\r\nThese model paragraphs show how a student has successfully identified the legal issues and applied those issues to the facts of the law.\r\n\r\nExample 1\r\n\r\n\r\nExample 2\r\nHere the application of the law to the facts and the identification of legal issues have been interwoven together in one sentence.\r\n\r\n\r\n\r\n","n":0.096},[{"v":"Writing structures","i":3,"n":0.707},{"v":"Law assessments","i":2,"n":0.707},{"v":"IRAC","i":1,"n":1},{"v":"Academic English","i":0,"n":0.707}]]},{"i":421,"$":[{"v":"Writing a legal argument: Planning your answer","n":0.378},{"v":"Legal argument is a process of identifying legal issues, connecting it to appropriate laws then applying those laws to the facts. It is a process that weaves a legal answer together. You must find and cite supporting cases and laws to argue your position.\r\n\r\nExample model plan\r\n1 Legal issue2 Cases cited to support legal issue\r\n","n":0.136},[{"v":"Planning","i":3,"n":1},{"v":"Legal","i":2,"n":1},{"v":"Law","i":1,"n":1},{"v":"Citation","i":0,"n":1}]]},{"i":422,"$":[{"v":"Writing a legal argument: Writing your answer","n":0.378},{"v":"Once you have written your plan, you need to turn this into your answer. There should be one main legal issue per paragraph. Each main issue needs to demonstrate appropriate application of the law to the facts of the case.\r\nHere is the paragraph outline for Bert v Groovy Clothing Store, an example from negligence.\r\n1. Establishes negligence as the area of law relevant to the facts\r\nNegligence:\r\n\r\n \tInvolves establishment of duty of care (DOC)\r\n \tBreach of DOC\r\n \tBreach DOC\u2192damages to Bert\r\n\r\n2. Duty of care \u2014 establishes whether a duty of care was owed Bert from Groovy Clothing Store\r\nLord Atkin's neighbour test\r\nApply law to the facts:\r\n\r\n \tForeseeability \u2014 Reasonably foreseeable that Groovy Clothing Store's actions / omissions could cause damage to Bert\u2192Groovy Clothing Store owes Bert a duty of care Australian Safeway Stores v Zaluzna [1987] HCA 7\r\n \tSalient features of the care are consistent with the existence of a duty of care: Zaluzna's case\r\n\r\n3. Breach in standard of duty of care\r\n\r\n \tEstablish standard owed to entrants of a store of a reasonable, ordinary, prudent occupier\r\n \tForeseeability of harm \u2014 apply to the facts\r\n\r\n \tOpening crowd\r\n \tToo many people on stairs\r\n \tNo warning to shoppers\r\n \t\u2192Breach of duty of care to Bert Bolton v Stone, Mercer's case [1951] 1 All ER 1078. \r\n\r\n\r\n\r\n4. Defences\r\nContributory negligence of Bert\r\nApply law to the facts:\r\n\r\n \tFailure of Bert to take care of himself on the stairs\r\n \tReasonable foreseeability of an accident Wyong Shire Council v Shirt  [1980] HCA 12\r\n\r\n5. Damages\r\n\r\n \tDamages reasonably foreseeable \u2014 apply law to the facts\r\n\r\n \tGroovy Clothing Store liable for damages Wagon Mound Cases \r\n\r\n\r\n \tCausation 'but for' test \u2014 apply law to the facts\r\n\r\n \t'But for' the breach, the damages i.e. Bert's hospitalisation and injuries would not have occurred Cork v Kirby MacLean LTD [1952] 2 AII ER 402 [1998] HCA 55\r\n\r\n\r\n","n":0.058},[{"v":"Sample writing","i":2,"n":0.707},{"v":"Legal","i":1,"n":1},{"v":"Law","i":0,"n":1}]]},{"i":423,"$":[{"v":"Business law","n":0.707},{"v":"\r\nThis tutorial is useful for students who have limited prior knowledge of Australian law. The tutorial introduces you to the fundamentals, principles and language of law.\r\n\r\nIt includes a glossary of common legal terms, language exercises and legal concept maps. The activities will assist you in becoming familiar with the language of the law.\r\n\r\nGlossary of common terms\r\n\r\nStudy the glossary below then complete the activities to test your knowledge of common terms.\r\n\r\n tr td:first-child {font-weight: bold} \r\n\r\n\r\n\r\n\r\nTerms\r\nDefinitions\r\n\r\n\r\n\r\n\r\nLegislation/Acts or statutes\r\nLaws made by both commonwealth and state parliaments\r\n\r\n\r\nCommon or case law\r\nLaws made by judges in courts\r\n\r\n\r\nPlaintiff\r\nThe person(s) taking legal action against another person(s) in civil cases\r\n\r\n\r\nProsecution\r\nThe lawyers presenting a case for the crown against a defendant in a criminal action\r\n\r\n\r\nDefendant\r\nThe person(s) defending legal action against another party\r\n\r\n\r\nEquity\r\nSeparate body of law which supplements the rules of common law based on fairness and justice\r\n\r\n\r\nDoctrine of precedent\r\nDecisions of a higher court are binding on lower courts in the same court hierarchy when judging similar cases\r\n\r\n\r\nRatio decidendi\r\nThe reason for the judge's decision\r\n\r\n\r\nBinding precedent\r\nA previous decision that must be followed\r\n\r\n\r\nPersuasive precedent\r\nA previous decision that influences future decisions but need not be followed outside the court hierarchy or from lower courts within the same hierarchy\r\n\r\n\r\n\r\n\r\nActivity 1\r\nTest your knowledge of the words explained on this page. For any questions that you answer incorrectly, revise the terms and their definitions.\r\n\r\n\r\nActivity 2\r\nRead the following definitions and select the correct terms that match them from the drop down menus.\r\n\r\n\r\n\r\n \r\n\r\nWhat does the term ratio decidendi mean?\r\n\r\n \tA) Rational decisions\r\n \tB) Reasons behind judge's decisions\r\n \tC) Previous reasons for decisions\r\n \tD) A rational defendant\r\n\r\nAnswer: B) Reasons behind judge's decisions\r\n\r\nWhat does the term plaintiff mean?\r\n\r\n \tA) The judge\r\n \tB) The lawyer presenting the case against a defendant in a criminal court\r\n \tC) The person taking legal action against another person(s) in civil actions\r\n \tD) The plain action of a court\r\n\r\nAnswer: C) The person taking legal action against another person(s) in civil actions\r\n\r\nWhat does the term binding precedent mean?\r\n\r\n \tA) A previous decision that must be followed\r\n \tB) A decision that has been bound\r\n \tC) A tight predicament\r\n \tD) A previous decision that influences future decisions but need not be followed\r\n\r\nAnswer: A) A previous decision that must be followed\r\n\r\nWhat does the term prosecution mean?\r\n\r\n \tA) Acts or laws of state or commonwealth parliaments\r\n \tB) To persecute someone through the law\r\n \tC) The lawyers presenting a case against a defendant in a criminal court\r\n \tD) The reason behind the court case\r\n\r\nAnswer: C) The lawyers presenting a case against a defendant in a criminal court\r\n\r\nWhat does the term equity mean?\r\n\r\n \tA) Separate body of law which supplements the rules of common law based on fairness and justice\r\n \tB) Equal payment of court costs\r\n \tC) Fairness in the judge's eyes\r\n \tD) Equal chance to sit on a jury\r\n\r\nAnswer: A) Separate body of law which supplements the rules of common law based on fairness and justice\r\n\r\n \r\n\r\nRead the following definitions, and provide the right term. Answers will be given at the end of each definition.\r\n\r\nDefinition: An Act or law made by both Commonwealth and state parliaments.\r\n\r\nTerm: legislation\r\n\r\nDefinition: Decisions of a higher court are binding on lower courts in the same court hierarchy when judging similar cases.\r\n\r\nTerm: doctrine of precedent\r\n\r\nDefinition: The person defending legal action against another party.\r\n\r\nTerm: defendant\r\n\r\nDefinition: Laws made by judges in court.\r\n\r\nTerm: common law\r\n\r\nDefinition: A previous decision that influences future decisions but need not be followed.\r\n\r\nTerm: persuasive precedent\r\n\r\nDefinition: Any proceeding in a court.\r\n\r\nTerm: legal action\r\n","n":0.042},[{"v":"Business Law","i":0,"n":0.707}]]},{"i":424,"$":[{"v":"Future application","n":0.707},{"v":"The final section of a studio knowledge object speculates about how you can transfer this new knowledge to future projects and professional practice. The aim is to develop your design expertise by applying lessons learned from this studio and from your structured reflection.\r\n\r\nPlanning\r\nOn the Reflecting on your learning page, we learnt about the DIEP framework for reflective writing. As part of DIEP, we need to write a Plan for each of our reflections. Now, in the future applications section of our studio knowledge object, we bring all of our Plan sections together and think of the future application of what we have learnt from our insight as a whole .\r\n\r\nTo write this section, it is helpful to ask yourself the following questions:\r\n\r\n\tWhat aspect(s) of the learning in this studio will you draw on for future design practice?\r\n\tWhat techniques will you transfer to your future practice?\r\n\tWhat meaning and value does this have for your practice?\r\n\tHow might you apply this understanding to your future practice?\r\n\r\nNow, let's explore how students have used their answers to these questions to write about the future application of their insights.\r\n\r\n\r\n\r\n\r\n\r\n","n":0.074},[{"v":"Sample writing","i":2,"n":0.707},{"v":"Reflective writing","i":1,"n":0.707},{"v":"Design","i":0,"n":1}]]},{"i":425,"$":[{"v":"Reflecting on your learning","n":0.5},{"v":"The main section of the knowledge object contains a series of reflections on your studio learning. Reflections identify key learning and insights gained through critical incidents in the design process and activities in the studio.\r\n\r\nArticulation of your learning through reflection supports learning-by-doing and transfer of learning from one studio to another. The aim is to identify general principles from specific learning experiences, linking practice with design theory and methods. It is common for a studio knowledge object assignment to contain three or more reflection paragraphs.\r\nStep-by-step guide for each reflection\r\nThere are a number of ways that reflective writing can be structured. One is using the DIEP framework. Here's how it works:\r\n\r\n Describe the insight: \r\nFirst, identify what the insight is. An insight is a new understanding that emerged from the studio activities or your design process. It may be a practical, theoretical or critical understanding such as a new awareness of certain factors in the design process, a new method that improves your efficiency, a deeper understanding of a particular issue, a connection between a particular theory and practice, a model that supports your interrogation of a problem, or a framework for evaluating design outcomes.\r\n\r\nThen, briefly describe the incident that led to this insight. What was the catalyst?\r\nInterpret the insight: \r\nExplain the connection between this insight and the methods, principles and theory of the design field.\r\nEvaluate the insight:\r\nMake an evaluation of the significance of this new understanding for you. This may relate to the further development of the project or future design projects, or to your design process or the way you think about design.\r\nPlan:\r\nMake a plan for how you might transfer this new understanding to future projects and/or practice. Speculate about how this new insight might change the way you approach a design task in the future.\r\n\r\nLanguage features\r\n Verb tense\r\nVerb tense is used in reflective writing to indicate what you have learnt from the experience. For example, when you speak about your level of knowledge or experience before the insight, often the past tense is used. \r\nOn the other hand, a mixture of the simple present, present continuous, present perfect and future tenses are used to describe your current level of knowledge or experience.\r\nI, me, my\r\nIn most cases, pronouns like I, me and my are avoided in academic writing. However, in reflective writing you are writing about your experiences. This requires a more personal style of writing where I, me and my should be used.\r\n\r\nExample paragraphs\r\nThe following examples of student writing provide a guide to the structure and language of a reflection paragraph. They highlight the identify, describe, interpret, evaluate and plan sections of the paragraphs, as well as pointing out verb tense and the use of I, me and my.\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\nStructure\r\nQuestions to ask:\r\n\r\n \tWhat insight(s) did I gain through the studio? (Insights)\r\n \tWhich significant events, theories, methods, thinkers and practitioners contributed to your insight? (Methods/thinkers)\r\n \tWhat do you now understand and how does it affect your practice? (Meaning/value)\r\n\r\n[Insights] My specific insight from this studio is that I can use strategy to form an enduring connection with an audience. This is important in this studio because we want to affect social change. [End insights] [Meaning/value] My initial strategy was complicated and unrealistic. I was trying to solve too much. I realised my strategy required a great consideration of my chosen archetype. [End meaning/value] [Methods/thinkers] By building skills in mapping I made very clear correlations between climate change and food security, their current effects on third world communities and the very real threat they pose to our future. Mapping my own connection to the victims of these effects formed a strong sense of empathy and responsibility to take action. Strengthening my skills in interviewing and researching helped me to develop a persona archetype. As I sought to understand everything about them to design a more direct mission and informed touch points, I began to see what could be realistically addressed. [End methods/thinkers] [Meaning/value] I think this studio has challenged and strengthened a range of my personal and professional skills as a designer: how my strategy can appeal to the target audience, how they might interact with the strategy and how I can develop a long-term connection between my strategy and the audience. [End meaning/value]\r\n\r\n\r\nLanguage features\r\nFeatures\r\n\r\n \tUse \u2018I' to demonstrate your personal and subjective response to the field of practice\r\n \tA reflective paragraph identifies a change in learning over time from the past to the present/future\r\n\r\n[Present/future] My specific insight from this studio is that I can use strategy to form an enduring connection with an audience. This is important in this studio because we want to affect social change. [End present/future] [Past] My initial strategy was complicated and unrealistic. I was trying to solve too much. I realised my strategy required a great consideration of my chosen archetype. By building skills in mapping I made very clear correlations between climate change and food security, their current effects on third world communities and the very real threat they pose to our future. Mapping my own connection to the victims of these effects formed a strong sense of empathy and responsibility to take action. Strengthening my skills in interviewing and researching helped me to develop a persona archetype. As I sought to understand everything about them to design a more direct mission and informed touch points, I began to see what could be realistically addressed. [End past] [Present/future] I think this studio has challenged and strengthened a range of my personal and professional skills as a designer: how my strategy can appeal to the target audience, how they might interact with the strategy and how I can develop a long-term connection between my strategy and the audience. [End present/future]\r\n\r\n\r\n \r\nStructure\r\nQuestions to ask:\r\n\r\n \tWhat insight(s) did I gain through the studio? (Insights)\r\n \tWhich significant events, theories, methods, thinkers and practitioners contributed to your insight? (Methods/thinkers)\r\n \tWhat do you now understand and how does it affect your practice? (Meaning/value)\r\n\r\n[Insights] The most useful and interesting insight I have gained from this studio is the importance of research methods for creating end-user profiles. [End insights] [Meaning/value] Before this journey, design research (for me) was about making mood boards and identifying basic audience demographics. Without downgrading the importance of these methods, I now recognise that there can be a lot more to research than that. [End meaning/value] [Methods/thinkers] In the Design for Social Change studio I have gained an understanding of specific research methods such as case studies, interviews, customer journey mapping and building comprehensive persona profiles. These methods help designers understand the complexity of the problems and then develop strategies for social change that actually have some validity and meaning. [End methods/thinkers] [Meaning/value] These ways of understanding the user or customer were very helpful in designing my Oxfam Student Month project. I am confident that this new understanding of research methods for identifying and targeting specific users will be beneficial for my future design practice. [End meaning/value]\r\nLanguage features\r\nFeatures\r\n\r\n \tUse \u2018I' to demonstrate your personal and subjective response to the field of practice\r\n \tA reflective paragraph identifies a change in learning over time from the past to the present/future\r\n\r\n[Present/future] The most useful and interesting insight I have gained from this studio is the importance of research methods for creating end-user profiles. [End present/future] [Past] Before this journey, design research (for me) was about making mood boards and identifying basic audience demographics. [End past] [Present/future] Without downgrading the importance of these methods, I now recognise that there can be a lot more to research than that. In the Design for Social Change studio I have gained an understanding of specific research methods such as case studies, interviews, customer journey mapping and building comprehensive persona profiles. These methods help designers understand the complexity of the problems and then develop strategies for social change that actually have some validity and meaning. [End present/future] [Past] These ways of understanding the user or customer were very helpful in designing my Oxfam Student Month project. [End past] [Present/future] I am confident that this new understanding of research methods for identifying and targeting specific users will be beneficial for my future design practice. [End present/future]\r\n\r\n\r\n \r\nStructure\r\nQuestions to ask:\r\n\r\n \tWhat insight(s) did I gain through the studio? (Insights)\r\n \tWhich significant events, theories, methods, thinkers and practitioners contributed to your insight? (Methods/thinkers)\r\n \tWhat do you now understand and how does it affect your practice? (Meaning/value)\r\n\r\n[Insights] From this studio I have a new understanding of how psychology underpins exhibition design for learning and play environments such as zoos, museums and the like. [End insights] [Methods/thinkers] Throughout the course I was exposed to immersive and interactive design principles; that combine sound, visual, text and spatial elements to create positive learning environments. At the Melbourne Museum and the Melbourne Zoo the exhibitions are created with attention to elements such as: goals or challenges to overcome, stimulus of the senses, feelings of control and minimal distractions. Lidwell, Holden and Butler (2003) suggest these elements are necessary to create a successful immersive experience.Stevenson (1994, p.32) highlights the effectiveness of interactivity to 'inspire and provoke exploration ... and to tempt people to look more thoughtfully' at the displays. Through designing my own miniature suitcase exhibition in this studio, I have gained an understanding of how to undertake integrated design for learning and play. Falk and Dierking's contextual model of learning (2012) suggests that a visitor's personal, sociocultural and physical context all work together to shape their experiences and what they decide to absorb and learn from an exhibition. [End methods/thinkers] [Meaning/value] The idea that the visitor contributes to the learning process, which is a feedback loop between museum and visitor, allows for more flexible and positive, interactive play and learning. Appreciating how these elements combine with the use of different mediums and techniques will assist me in my future practice to create memorable, impactful and effective learning and play environments. [End meaning/value]\r\nLanguage features\r\nFeatures\r\n\r\n \tUse \u2018I' to demonstrate your personal and subjective response to the field of practice\r\n \tA reflective paragraph identifies a change in learning over time from the past to the present/future\r\n\r\n[Present/future] From this studio I have a new understanding of how psychology underpins exhibition design for learning and play environments such as zoos, museums and the like. [End present/future] [Past] Throughout the course I was exposed to immersive and interactive design principles; that combined sound, visual, text and spatial elements to create positive learning environments. [End past] At the Melbourne Museum and the Melbourne Zoo the exhibitions are created with attention to elements such as: goals or challenges to overcome, stimulus of the senses, feelings of control and minimal distractions. Lidwell, Holden and Butler (2003) suggest these elements are necessary to create a successful immersive experience. Stevenson (1994, p. 32) highlights the effectiveness of interactivity to \u2018inspire and provoke exploration \u2026 and to tempt people to look more thoughtfully' at the displays. [Present/future] Through designing my own miniature suitcase exhibition in this studio, I have gained an understanding of how to undertake integrated design for learning and play. [End present/future] Falk and Dierking's contextual model of learning (2012) suggests that a visitor's personal, sociocultural and physical context all work together to shape their experiences and what they decide to absorb and learn from an exhibition. The idea that the visitor contributes to the learning process, which is a feedback loop between museum and visitor, allows for more flexible and positive, interactive play and learning. [Present/future] Appreciating how these elements combine with the\r\n","n":0.023},[{"v":"Reflective writing","i":0,"n":0.707}]]},{"i":426,"$":[{"v":"Contextualising a field of enquiry","n":0.447},{"v":"This page explains the process of contextualising a field of enquiry — an important step to creating a studio knowledge object.\r\n\r\nContextualising the field of enquiry\r\nThe first section of a studio knowledge object is where the field of enquiry is contextualised, and this section serves as an introduction. It outlines the particular field of design practice or inquiry investigated in the studio. The design field needs to be described with enough detail to give a non-expert reader a good understanding of the key issues and approaches that your insights relate to.\r\n\r\nThe context outlines:\r\n\r\n \tthe context/focus of the design field or studio\r\n \tthe aims and intent of this field of practice\r\n \tthe key thinkers, practitioners, methods and approaches.\r\n\tthe implications of these aspects on your design practice.\r\n\r\nExamples\r\nRead the following example paragraphs to observe the structure and language that can be used in a context paragraph.\r\n\r\n\r\n\r\n\r\n \r\nExample 1\r\nStructure\r\nQuestions\r\n\r\n \tWhat field of practice does the studio investigate? (Context)\r\n \tWhat is the intent of this field of practice? (Aim/intent)\r\n \tWhat are the key approaches/methods and who are the key thinkers in this field of practice? (Approaches/thinkers)\r\n \tWhat are the implications of these for design practice (yours/others)? (Implications)\r\n\r\n[Context] The studio, Shaping Nature Experiences, is an introduction to the field of interpretive environment design. [End context] [Aim/intent] This is a rapidly changing field; the complexity and range of techniques and subject areas involved in the process of designing an exhibition makes this a dynamic and intriguing discipline. [End aim/intent] [Approaches/thinkers] Exhibition design, or interpretive environment design, is a multi-disciplinary field that involves the interweaving of many different elements of media and communications [End approaches/thinkers] [Aim/intent] to provide transformative experiences for visitors [End aim/intent] (Bayer 1961). [Approaches/thinkers] The field of interpretive environment design is distinguished by three main themes - designing for learning and play, immersion and interactivity. [End approaches/thinkers] [Implications] As a designer, and potentially as a future art director, I am conscious that memorable, effective and impactful interpretive environments can offer experiences that engage and enrich visitors' lives, and are a means of changing their perception of the world. [End implications]\r\nLanguage features\r\n\r\n \tUse \u2018I\u2019 to demonstrate your personal and subjective response to the field of practice\r\n \tGenerally context paragraphs describe the field of practice in the present\r\n\r\nThe studio, Shaping Nature Experiences, is an introduction to the field of interpretive environment design. This is a rapidly changing field; the complexity and range of techniques and subject areas involved in the process of designing an exhibition [Present] makes this a dynamic and intriguing discipline. Exhibition design, or interpretive environment design, [Present] is a multi-disciplinary field that involves the interweaving of many different elements of media and communications to provide transformative experiences for visitors (Bayer 1961).The field of interpretive environment design [Present] is distinguished by three main themes - designing for learning and play, immersion and interactivity. As a designer, and potentially as a future art director, [Present] I am conscious that memorable, effective and impactful interpretive environments can offer experiences that engage and enrich visitors' lives, and are a means of changing their perception of the world.\r\n\r\n\r\nExample 2\r\nStructure\r\nQuestions\r\n\r\n \tWhat field of practice does the studio investigate? (Context)\r\n \tWhat is the intent of this field of practice? (Aim/intent)\r\n \tWhat are the key approaches/methods and who are the key thinkers in this field of practice? (Approaches/thinkers)\r\n \tWhat are the implications of these for design practice (yours/others)? (Implications)\r\n\r\n[Context] This studio, Design for social change: climate change and food security, looks at the social role and ethical practices of communication design. [End context] [Aim/intent] Social change is a major influence on professional communication design as design can help to connect people to global issues which appear incredibly complex. This design field is human-centred; it requires in-depth research on people and issues that affect them, as well the people who can help. [End aim/intent] [Approaches/thinkers] Mapping, creating archetype categories, observing case studies and the six spheres tool help designers to form strategies that are realistic and empathetic. Finally, the presentation from Studio Thick encompassed everything we\u2019ve learnt in this studio and how these skills are implemented in the industry today. [End approaches/thinkers] [Implications] All of this helped me develop a deeper, personal understanding of the strategy which I felt was conveyed in my final presentation. My new understanding of socially responsible and sustainable service design will be used in my future practice. [End implications]\r\nLanguage features\r\n\r\n \tUse \u2018I\u2019 to demonstrate your personal and subjective response to the field of practice\r\n \tGenerally context paragraphs describe the field of practice in the present\r\n \tContext paragraphs may also describe the future use of your learning from this field of practice\r\n\r\nThis studio, Design for social change: climate change and food security, looks at the social role and ethical practices of communication design. [Present] Social change is a major influence on professional communication design as design can help to connect people to global issues which appear incredibly complex. [Present] This design field is human-centred; it requires in-depth research on people and issues that affect them, as well the people who can help. Mapping, creating archetype categories, observing case studies and the six spheres tool help designers to form strategies that are realistic and empathetic. Finally, the presentation from Studio Thick encompassed everything we\u2019ve learnt in this studio and [Present] how these skills are implemented in the industry today. All of this helped me develop a deeper, personal understanding of the strategy which I felt was conveyed in my final presentation. [Future use] My new understanding of socially responsible and sustainable service design will be used in my future practice.\r\n\r\n\r\nExample 3\r\nStructure\r\nQuestions\r\n\r\n \tWhat field of practice does the studio investigate? (Context)\r\n \tWhat is the intent of this field of practice? (Aim/intent)\r\n \tWhat are the key approaches/methods and who are the key thinkers in this field of practice? (Approaches/thinkers)\r\n \tWhat are the implications of these for design practice (yours/others)? (Implications)\r\n\r\n[Context] The Design for Social Change studio is to design for social innovation, reframing complex problems into practical outcomes that can be integrated into people's lives Working with Oxfam we have had the opportunity to explore complex systemic issues of climate change and food security through design and research. [End context] [Aim/intent] The field of practice explores how design can help make people understand how big problems are relevant to their daily lives. [End aim/intent] [Approaches/thinkers] Approaches and methods used in this emerging field demonstrate ways to promote and support human-centred research and design using ethnography-research skills, and holistic and systems thinking. The studio includes case studies and presentations of social-oriented programs, businesses and NGOs. An 'impact design agency' like Studio Thick works with companies such as Telstra and the Commonwealth Bank that require sustainability and service design to meet their social and community responsibilities. [End approaches/thinkers] [Implications] This field of practice is a way to spread information, ideas and messages to contribute to shaping mentalities and making the world a better place to live. I hope the field continues to develop and become recognised as an actual pathway for designers in the future. [Implications]\r\nLanguage features\r\n\r\n \tUse \u2018I\u2019 to demonstrate your personal and subjective response to the field of practice\r\n \tGenerally context paragraphs describe the field of practice in the present\r\n \tContext paragraphs may also describe the future use of your learning from this field of practice\r\n\r\n[Present] The Design for Social Change studio is to design for social innovation, reframing complex problems into practical outcomes that can be integrated into people's lives Working with Oxfam we have had the opportunity to explore complex systemic issues of climate change and food security through design and research. [Present] The field of practice explores how design can help make people understand how big problems are relevant to their daily lives. Approaches and methods used in this emerging field [Present] demonstrate ways to promote and support human-centred research and design using ethnography-research skills, and holistic and systems thinking. The studio includes case studies and presentations of social-oriented programs, businesses and NGOs. An 'impact design agency' like Studio Thick works with companies such as Telstra and the Commonwealth Bank that require sustainability and service design to meet their social and community responsibilities. [Present] This field of practice is a way to spread information, ideas and messages to contribute to shaping mentalities and making the world a better place to live. [Present] I hope the field continues to develop and become recognised as an actual pathway for designers in the future.\r\n\r\n\r\nExample 4\r\nStructure\r\nQuestions\r\n\r\n \tWhat field of practice does the studio investigate? (Context)\r\n \tWhat is the intent of this field of practice? (Aim/intent)\r\n \tWhat are the key approaches/methods and who are the key thinkers in this field of practice? (Approaches/thinkers)\r\n \tWhat are the implications of these for design practice (yours/others)? (Implications)\r\n\r\n[Context] This design studio, Shaping Nature Experiences, provides an advanced level of engagement in communication design for learning institutions like museums and zoos. [End context] [Aim/intent] This design field seeks to engage visitors through immersive and interactive environments and experiences, [End aim/intent] [Approaches/thinkers] based on a constructivist theory of free-choice learning. The guiding principles of interpretation design are to engage, educate and inspire through carefully designed exhibitions. The field of interpretation design is guided by these principles and employs a myriad of overlapping strategies and frameworks. These target exhibition narrative, communication aims and their effectiveness, audience needs, degrees of interaction, and approaches to learning. The frameworks of Falk, Dierking and Adams (2007) aim at allowing the audience to contribute and to engage on multiple levels for an immersive and engaging experience and provide a basis for my design. [End approaches/thinkers] [Implications] As a designer I seek to apply the principles of free-choice learning to exhibition design in order to generate a partnership with the viewer and create a journey through the content that is open to broad as well as individual responses. [End implications]\r\nLanguage features\r\n\r\n \tUse \u2018I\u2019 to demonstrate your personal and subjective response to the field of practice\r\n \tGenerally context paragraphs describe the field of practice in the present\r\n\r\n[Present] This design studio, Shaping Nature Experiences, [Present] provides an advanced level of engagement in communication design for learning institutions like museums and zoos. [Present] This design field seeks to engage visitors through immersive and interactive environments and experiences, based on a constructivist theory of free-choice learning. [Present] The guiding principles of interpretation design are to engage, educate and inspire through carefully designed exhibitions. [Present] The field of interpretation design is guided by these principles and employs a myriad of overlapping strategies and frameworks. These target exhibition narrative, communication aims and their effectiveness, audience needs, degrees of interaction, and approaches to learning. The frameworks of Falk, Dierking and Adams (2007) aim at allowing the audience to contribute and to engage on multiple levels for an immersive and engaging experience and provide a basis for my design. [Present] As a designer I seek to apply the principles of free-choice learning to exhibition design in order to generate a partnership with the viewer and create a journey through the content that is open to broad as well as individual responses.\r\n\r\n","n":0.023},[{"v":"Design","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":427,"$":[{"v":"Critical writing: Structure","n":0.577},{"v":"This section outlines the overall structure of a critical essay comprising an introduction, body and conclusion.\r\nThe structure will help you plan and organise your response to the task - an analysis of a particular work of communication design from the twentieth century.\r\n\r\nOverview\r\nUse the questions provided in the task to create an outline of the body of the essay. Along with the introduction and conclusion, your essay needs to:\r\n\r\n \tDescribe the work to the reader.  What is the work? Who made it? When was it made?\r\n \tContextualise the work. Consider what was happening in the world at that time? Think about what was happening in society and in politics for example, both locally and more broadly.\r\n \tAnalyse the work's relevance for contemporary communication design. \r\n\r\n\r\n\r\n\r\nIntroduction\r\nThe introduction prepares the reader for your analysis.\r\n\r\nThere are three main parts:\r\n\r\n \tGeneral statements to provide context and background information, and to show the importance of the work and the reason for your evaluation.\r\n \tA thesis statement (argument) that identifies your specific topic and your position.\r\n \tAn outline of the scope and organisation of the essay so the reader knows what is covered. This will be determined by your argument, which you support in the body of your essay.\r\n\r\nThe introduction should also include key words from the themes and concepts of the framework. Key words should be used throughout your essay to reinforce cohesion and help with the overall 'flow' of your writing.\r\nUse the examples below to explore the different features of an introduction.\r\n\r\n\r\n\r\n\r\n\r\n\r\nBody paragraphs\r\n\r\nThe body of a critical essay should have a logical sequence. This helps the reader to follow the development of the argument, as stated in the introduction. Your argument determines what evidence you select from your research, how you develop your reasoning, and what citations you include to support your position.\r\n\r\nConsider others' analyses of the work and identify the strengths or weaknesses of your reasoning. Use evidence to support your views and acknowledge any arguments against them.\r\n\r\nParagraphs are used to group and categorise your evidence. In the example below, note how the linking sentence of the first paragraph connects to the topic sentence of the following paragraph.\r\n\r\nEach paragraph should:\r\n\r\n \trelate back to the thesis (argument)\r\n \tdescribe one main idea supported by information and evidence from your research\r\n \tuse keywords to express the concepts and themes from the framework\r\n \tuse a structure such as TEEL to clearly express your ideas to the reader.\r\n\r\n\r\n\r\n\r\n\r\nConclusion\r\nThe conclusion is your final opportunity to persuade the reader of your position based on the information you have presented. An effective conclusion establishes a sense of completeness.\r\n\r\nYour conclusion should:\r\n\r\n \trestate your position\r\n \tsummarise how the most important evidence supports this\r\n \tshow how your position is related to the broader body of knowledge of this field/topic\r\n\r\nExamples\r\n\r\n\r\n \r\n \r\n\r\n \r\nIntroduction\r\nWeingart's philosophy of design practice calls for a commitment to the act of design, to focus on design as an iterative process of experimentation, prototyping and continuous refinement.\r\nParagraph 2: Describe\r\nWhat is the work? Who made it? When was it made?\r\n\r\n \tWolfgang Weingart -1972/1973, 15 covers for Typografische Monatsblatter magazine\r\n \tdescribe the distinctive features and style of the work.\r\n\r\nParagraph 3: Contextualise\r\nWhy was it important at the time?\r\n\r\n \tPost World War II - increasing globalisation\r\n \tNeed for communicative clarity, multilingual formats,\r\n\r\nParagraph 4: Contextualise\r\nWhy was it important at the time?\r\n\r\n \t1968 reaction against conservatism & demand for social change\r\n\r\nParagraph 5: Analyse\r\nWhat relevance does it have for us today?\r\n\r\n \tExplore all potential ideas via experimentation.\r\n\r\nParagraph 6: Analyse\r\nWhat relevance does it have for us today?\r\n\r\n \tmore aware and expressive attitudes in design resulted\r\n\r\nConclusion\r\nWolfgang Weingart's critical questioning of assumed principles of design remains an important element of a designer's philosophy of practice today.\r\n\r\n","n":0.041},[{"v":"TEEL","i":2,"n":1},{"v":"Essays","i":1,"n":1},{"v":"Critical writing","i":0,"n":0.707}]]},{"i":428,"$":[{"v":"Contract law","n":0.707},{"v":"Answering any contract law case question requires you to do the following: identify and discuss legal issues, apply the law to the facts, and\u00a0cite relevant cases.\r\n\r\nReading the question\r\nWhen reading any legal case problem, ask yourself the following questions:\r\n\r\n \tWhat is the general issue here?\r\n \tWhat are the particular legal issues?\r\n \tHow do these issues relate to the facts of the problem?\r\n \tWhat case law is relevant to the facts presented?\r\n\r\n\r\nAnalysing the question\r\nWhen analysing any legal question you are required to:\r\n\r\n \tidentify the legal issues \u2013 general and particular\r\n \tapply the facts to the law\r\n \tcite relevant cases to support your answer.\r\n\r\nThe importance of analysing the general legal issue(s) is to determine what type of law(s) is/are applicable, e.g. contract, negligence, corporation etc. Once the general area has been determined, the particular legal issues relating to that area can be analysed.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nCase citations\r\nTo answer any legal question, you must apply the facts to the law and back up your arguments through citing relevant legal cases.\r\n\r\n \r\n\r\n[Intention\u00a0to\u00a0create\u00a0legal\u00a0relations]James saw a classified [Invitation\u00a0to\u00a0treat]advertisement[end invitation] in a magazine offering to sell Kate\u2019s piano for $4,000. [Offer]James wrote to Kate stating that he would \"take the instrument at your price\". [end offer][end intention]\r\n\r\nOn 1st May, [Counter\u00a0offer]Kate wrote to James stating that she now wanted $5,000[end counter] and would [Consideration]keep her offer open for a week.[end consideration]\r\n\r\n[Revocation\u00a0of\u00a0offer]James received the letter on 4th May and wrote back to Kate stating he agreed to pay the $5,000 \"at your convenience\"[end revocation]. [Postal\u00a0acceptance\u00a0rule]The letter unfortunately didn't arrive until 11th May.[end rule]\r\n\r\nThe [Breach\u00a0of\u00a0contract]next day (5th May)[end breach], [Domestic/family\u00a0arrangement]Kate had lunch with her sister Julie. They discussed the piano and [Breach\u00a0of\u00a0contract]agreed that Julie could have the piano for $3,500[end breach][end arrangement] if she [Vagueness\u00a0of\u00a0terms]minded Kate\u2019s children on Saturday nights[end vagueness].\r\n\r\nOn 6th May, the piano was delivered to Julie's house.\r\n \r\n\r\n[fact: Kate's advertisement] [legal issue: is an invitation to treat. An ad. can be an offer:] [citations: Carlill v Carbolic Smoke Ball], [legal issue: but is more likely to be an invitation to treat: [citations: Partridge v Crittenden]. [fact: Although not exactly business,] [legal issue: there is no reason to doubt intention to create legal relations.]\r\n\r\n[fact: James' first letter is an offer,] [legal issue: a definite promise to] [fact: take the piano at her price.] [citations: (Cp Harvey v Facey:] [legal issue: mere response to a request for information as to the price).]\r\n\r\n[fact: Kate's reply] [legal issues: is a counter offer,] [fact: rejecting James' offer] [citation: (Hyde v Wrench).] [fact: There is no requirement for her to hold it open for a week] [legal issue: as there is no consideration] [fact: provided by James,] [legal issue: so it is not an option] [citation: (Goldsborough Mort v Quinn).] [legal issues: However, if she wishes to revoke the offer, the revocation must be communicated] [fact: before James has accepted] [citation: (Dickinson v Dodds).]\r\n\r\n[fact: Because the offer was made by post, and no method of acceptance prescribed,] [legal issues: postal acceptance is appropriate:] [citation: Adams v Lindsell.] [legal issue: Acceptance is complete upon posting.] [fact: Assuming James posted the letter on the 4th of May,] [legal issue: a contract is formed on that date.]\r\n\r\n[fact: Kate, therefore,] [legal issue: breaches that contract] [fact: by selling the piano to Julie on 5th May. Her agreement with Julie may not be legally binding] [legal issue: as it is a family arrangement, though that presumption can be overcome] [citation: (Wakeling v Ripley).] [legal issue: It may also be too vague to be enforceable:] [fact: for how many Saturday nights must Julie mind Kate's children?]\r\n\r\n","n":0.041},[{"v":"Law assessments","i":2,"n":0.707},{"v":"Contract law","i":1,"n":0.707},{"v":"Case studies","i":0,"n":0.707}]]},{"i":429,"$":[{"v":"Building a coherent argument","n":0.5},{"v":"Writing coherently means creating an easily readable piece of writing that flows smoothly, is logically organised and has ideas that are connected effectively. This means creating paragraphs that develop your main argument and link together to build coherence. Let\u2019s explore how you can make your academic writing more coherent.\r\nStructure your work\r\nThere are many types of academic writing, but one thing they usually have in common is that they have an introduction, a body and a conclusion:\r\n\r\n\r\n \tYour introduction should contain a section which is called a thesis statement. This is where you tell the reader what the main argument is and the topics you will cover.\r\n \tIn the body paragraphs, you develop and support your main argument.\r\n \tThe conclusion reminds your reader what your thesis or main argument is.\r\n\r\nIn the Assessments section, you can find out more about structuring assignments such as essays, reports and literature reviews.\r\n\r\nParagraphs\r\nIn academic writing, each paragraph should focus on a distinct topic, and you should start a new paragraph when introducing a new topic. \r\n\r\nOrganising body paragraphs logically is important; group paragraphs exploring the same theme together before moving on to different themes. The first sentence of your paragraphs (the topic sentence), should clearly state what will be discussed in the paragraph.\r\n\r\nCreating an essay plan can help ensure each paragraph has a clear focus and follows a logical sequence. To do this:\r\n\r\nlist the topics of each paragraph\r\ncheck for logical grouping of topics\r\ncheck that all paragraph topics are relevant to your assignment question and support your argument effectively.\r\n\r\nHere's an example of an essay plan that a student has made to ensure their work flows logically.\r\n\r\n\r\nThesis statement\r\n\tUrbanisation offers employment opportunities but also poses challenges like environmental degradation, social inequality, and infrastructural strain, requiring sustainable planning.\r\n\r\nBody paragraph 1 topic\r\n\r\n\t Urbanisation drives economic growth through job creation and innovation.\r\n\r\nBody paragraph 2 topic\r\n\r\n\tIt often leads to environmental degradation, necessitating green policies.\r\n\r\nBody paragraph 3 topic\r\n\r\n\tUrbanisation can increase social inequality by limiting access to essential services for marginalised communities.\r\n\r\nBody paragraph 4 topic\r\n\r\n\tInfrastructural strain from urbanisation underscores the need for comprehensive urban planning.\r\n\r\nConclusion\r\n\r\n\tUrbanisation's benefits and challenges must be balanced through sustainable and inclusive measures.\r\n\r\n\r\nTEEL paragraph structure\r\nWe have already seen that it is important to have a clear topic sentence for your paragraph. Now, let's explore how the rest of your paragraph can support your topic sentence. \r\n\r\nOne way we can do this is by using the TEEL structure, which is commonly used for body paragraphs in academic writing. It stands for:\r\n\r\n \tTopic sentence\r\n \tExplanation\r\n \tEvidence and/or examples\r\n \tLink (the paragraph to the next paragraph)\r\n\r\nThe Writing paragraphs page has a detailed explanation of TEEL paragraphs.\r\nTEEL paragraph example\r\nIn this example, one of the body paragraphs from the example essay has been divided into the different sections of the TEEL structure.\r\n\r\n\r\n\r\n\r\n\r\n \r\nPost World War II, the increase in international trade, [linking]coupled with[end linking] the growing speed and pace of communication, meant that the need for [repetition]communicative clarity[end repetition], multilingual formats, and universally comprehensible pictographs and glyphs became important (Meggs & Alston 2011). Modernist graphic design had gained strong momentum in Switzerland and by the 1950s a distinctively Swiss graphic language of [repetition]simplicity[end repetition] and [repetition]formal discipline[end repetition] developed (Hollis 2016). Known eventually as '[substitution]Swiss Typography[end substitution]', 'International Typographic Style' or simply, 'International Style', [substitution]it[end substitution] was recognisable particularly for [substitution]its[end substitution] use of sans serif typography, photography, vibrant colour, asymmetry and negative space within a grid structure, which allowed the clear presentation of complex information (Paradis et al. 2013). [substitution]This new clarity and minimalist design[end substitution] was effective in helping satisfy the communication needs of a world in an era of immense change (Meggs & Alston 2011).\r\n","n":0.041},[{"v":"Essay structure","i":2,"n":0.707},{"v":"Design","i":1,"n":1},{"v":"Critical writing","i":0,"n":0.707}]]},{"i":430,"$":[{"v":"Critical incident report for nursing","n":0.447},{"v":"Are you wondering what reflective writing is about? Do you need help deciding what sort of incident you should select? Are you clear on what a variable is? This series of videos covers how to structure and write a critical incident reflection.\r\n\r\nPart 1: The preliminary guide\r\nThis video gives an overview of the assignment, some of the pitfalls to avoid and an outline of a sample critical incident.\r\n\r\n\r\nPart 2: Incorporating the literature\r\nThis assignment involves reflecting on YOUR particular incident, but at the same time you need to integrate relevant information from the literature. This video will provide some guidance of how this can be done.\r\n\r\n\r\nPart 3: Structures for reflective writing\r\nAvoid a rambling stream of consciousness that recounts what happened when. There is a structure to the writing process as well as the reflective process. View this video to see what you need to think about and say in regard to your incident.\r\n\r\n\r\nPart 4: Sample paragraphs\r\nIf you are still unsure about what to do, watch how a sample incident can be analysed.\r\n\r\n\r\n\r\n","n":0.077},[{"v":"Reflective writing","i":2,"n":0.707},{"v":"Nursing","i":1,"n":1},{"v":"Critical incident","i":0,"n":0.707}]]},{"i":431,"$":{"0":{"v":"Law","n":1},"2":[]}},{"i":432,"$":[{"v":"Converting flow rates between mL/hr and dpm","n":0.378},{"v":"How do you convert IV flow rates? In the previous video, we looked at converting between millilitres (ml) and drops. This time we are converting not only the volumes, (drops and mls) but the units for time as well.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nFlow rate keeps a measure of how much liquid is flowing by for each unit of time. We need to know an amount of liquid, be it mls or drops, and the time unit it takes to flow, be it hours or minutes. mls per hour and drops per minute are the units that you will commonly work with. They can express the same amount and you can and should be able to convert between them.\r\n\r\nThe strength of expressing your flow rate in mls per hour is that you get a clearer idea of how much liquid volume is being added to the patient's system. The beauty of having flow rate in drops per minute is that you can more easily calibrate how much medication is going in and how quickly. And also you can judge that visually by counting and timing the drops.\r\n\r\nIt is important to be able to make very fine calibrations like this, especially when dealing with children. In addition, your instrumentation will determine what unit that you must use and so that you need to be able to convert between the units and set your instruments according to their needs.\r\n\r\nIf you want to convert flow rate in mls per hour to flow rate in drops per minute you will need a couple of conversion factors to do this. Firstly, we want to turn mls into drops, and we know that there are 20 drops per ml.\r\n\r\nWe want to get rid of the mls so we should put that at the bottom and the drops value can go on the top. We also want to turn the hours into minutes. One hour equals 60 minutes. So we will put the hour on the top, so we can cancel it and the minutes that we want can go on the denominator of our conversion factor. So, we notice that after the cancelling we have multiplied the numerator by 20 and the denominator by 60. Let's merge those together and 20 over 60 simplifies down to one-third.\r\n\r\nSo if you have a flow rate in mls per hour and you multiply it by third you will get a flow rate in drops per minute, and that's the case for whatever the original flow rate may be, but remember that is only true when 20 drops is equivalent to 1 ml, and that this is when it's for an IV for an adult size patient.\r\n\r\nSo from this we know that flow rate in mls per hour, can be converted to drops per minute by multiplying the value by one-third. If you want to go from drops per minute back to mls per hour, you do the opposite, which is to divide by a third. But that is effectively the same as multiplying by 3\u2026 But remember, this is only true for macro drops, which are the drops used by adult infusions.\r\n\r\nLet's recall our understanding of macro drops and micro drops. Macro drops for adult infusions are of other size, such as there are 20 drops per mil. For paediatric infusions on the other hand - drops are much smaller, there are 60 drops making up that one mil, and the drop factor is 60 drops per ml\r\n\r\nSo, if we are dealing with micro drops and we want to turn flow rate into mls per hour into microdrops per minute, we can follow the same sort of calculation we did before. But we do not use 20 drops-per-mil we use 60-drops-per-mil. The difference is though, this time we are multiplying the numerator by 60 drops, and the denominator by 60 minutes\u2026 Once again we can merge them together, and 60 on 60 is one.\r\n\r\n","n":0.039},[{"v":"Nursing calculations","i":3,"n":0.707},{"v":"Nursing","i":2,"n":1},{"v":"Formulas","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":433,"$":[{"v":"Drop factor: converting between millilitres and drops","n":0.378},{"v":"Drops are just another unit for measuring the amount of fluid flowing into a patient\u2019s system. Discover what \u2018drop factor\u2019 means, and how to convert between millilitres and drops.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nIn getting medication, we are constantly talking about millilitres. It helps to understand what a ml actually is, a ml measures a volume of liquid. If you can imagine a cube of water that measures one centimetre\u2026 by one centimetre... by one centimetre... then this is 1 ml. One cubic centimetre of liquid is one millilitre.\r\n\r\nNow, imagine that same cubic centimetre divided into 20 separate drops, drops that are large enough to pack tightly into that same cubic centimetre of space. Here we have 20 drops in that space, we have 20 drops for that 1 ml of liquid. In other words, we have 20 drop per ml. This is your drop factor for macro drops, take note of the units here because the drop factor has units, the units of drop factor are drops per ml.\r\n\r\nThe unit of drops on its own is another way to express a volume of liquid, just as you can express a price in dollars or you could express it in Japanese yen or euros, both describe the same value but they are in different units, mls or drops. You can convert one to the other if you have the conversion rate - which is a little bit like the drop factor here. But you need to clearly state what you have - weather it's drops or mls or euros or dollars.\r\n\r\nSo, if you have 1 ml of liquid you the equivalent of 20 drops. If you have 2 ml of liquid you have 2 times 20, that is 40 drops. If you have 5 mls of liquid, you have 5 lots of 20, that is 100 drops.\r\n\r\nSee if you can make these conversions: 7 ml is equivalent to 7 times 20, which is 140 drops. 10 ml will give you 10 times 20, which is 200 drops. 50 ml, each containing 20 drops gives you 5 times 2 is 10, and 2 extra zeros, that's a 1000 drops. Same thing with the 80 ml, 8 times 2 is 16 and then 2 extra zeros, becomes 1600 drops. Finally, 120 mls multiplied by 20, gives us 2400 drops.[Stop]\r\n\r\nIn short, if you want to convert from mls to drops and this is when dealing with an adult patient and you are using macro drops, you multiply the number of mls by 20. If on the other you have drops and you want to convert back to mls. You do the opposite to multiply, which is to divide by 20.\r\n\r\nFor example, 600 drops converted back to mls, requires you to divide 600 by 20, after cancelling zeros we get an answer of 30. So 600 drops is equivalent to 30 mls. If you have 430 drops and you want to convert them to mls, you again divide the 430 by 20, it doesn't go evenly this time, it is equal to 21 and a half, which should be expressed as 21.5. 430 drops equals 21.5 mls.\r\n\r\nFinally, please remember that we have been dealing with macro drops, which are the larger size drops for adult infusions. If you are dealing with a paediatric patient, the infusion may be in micro drops, these drops are smaller and they are so small that 60 of them fit into the 1 ml of liquid (where before we had 20). So that means we have 60 drops per ml when dealing with micro drops, and 60 drops per ml is your drop factor for paediatric infusions. This means that converting between mls and drops for micro drops would require multiplying and dividing, not by 20 but rather by 60.\r\n\r\n","n":0.04},[{"v":"Nursing calculations","i":3,"n":0.707},{"v":"Nursing","i":2,"n":1},{"v":"Formulas","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":434,"$":[{"v":"Managing fractions and decimals","n":0.5},{"v":"What do you do if the flow rate calculation involves fractions such as \u00bc of an hour or 0.5 litres? Find out how this is managed mathematically.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nOur first problem is that we have an infusion running over 15 minutes and 50 mls of the infusion is being administered to the patient. We want to calculate the flow rate, and this answer needs to be in mls per hour.\r\n\r\nOur formula for flow rate is volume divided by time, and we know the volume going into the patient, which is 50 mls of infusion, and we know that it is 15 minutes for the time. But if we keep going here, our answer will clearly be in mls per minutes and not mls per hour. We need to change the 15 minutes to hours and we do know that 15 minutes is quarter of an hour. So let's put that in, even though it looks a little strange to have an fraction within a fraction.\r\n\r\nWhat this is actually saying is how many quarters are in 50? Or what is 50 divided by one quarter? Now, you cannot cancel across a division sign and we cannot divide any numbers here. The next step is to turn the division sign into a multiplication sign, and flip the second fraction upside-down. So, as you see here, the problem turns into 50 times 4 because the second fraction - the quarter - gets turned upside-down. Now there is nothing we can cancel, but we can go ahead and just multiply.\r\n\r\nFifty times four is 200. That's no surprise when you think about it, because there are four quarters in everyone and we have 50 of them. So 50, how many quarters, would be 50 times 4. And the mls unit is still on the top and the hours the denominator.\r\n\r\nThis time we have a similar problem, but the infusion this time is administered over 45 minutes. Once again, we need the formula flow rate equals volume over time. Once again the volume is 50 mls, and in this problem the time is 45 minutes. Once again, we need to change the minutes to hours. Similar to last time, we can change the 45 minutes to three quarters of an hour. This looks a little bit more complicated but we treat it the same way. It is really saying 50 mls divided by three-quarters of an hour. Like last time, we changed the division sign to times and flipped the second fraction. Three-quarters upside down is four-thirds, we can try to cancel if we can, but 3 does not divide nicely into 50, so we just multiply through.\r\n\r\n50 multiplied by 4 makes 200, and then divide this by 3. It comes down to a decimal of 66.6 recurring, which can be rounded to 66.7 mls per hour, and if you want to round it to the nearest whole ml, than it will be 67 mls per hour. We have the same problem as we had in the first slide, where we have a 50 mil infusion running over 15 minutes again. Once again, we need to turn the 15 minutes into hours, but this time we will not use fractions, we'll use a decimal. And 15 minutes is a quarter of an hour, and a quarter as a decimal is 0.25. Now, decimals within fractions do not work very well, but what you can do is turn that fraction into an equivalent fraction. That means a fraction that reduces down to the same thing, but the number inside it are proportionally bigger. We just need to use the numbers that are not complicated with a decimal points.\r\n\r\nHere's how we do it: if we jump that annoying decimal point to the right in the 0.25, it is okay as long as we increase the numerator in the same way, that is, we jump the decimal points in both numbers. That will give us 500 over 2.5, which is still difficult, so let's move it another place to the right. And, what you do to the denominator you must do the same to the numerator. And that will give us 5000 over 25. This is equivalent, the numbers are bigger but the fraction overall comes down to the same thing that we started with. And most importantly, we are now dealing with only whole numbers within our fraction. So 5000 mls divided by 25 hours... This is straightforward long division, 5000 divided by 25 equals 200. And as we expected, the answer is 200 mls per hour.\r\n\r\n","n":0.036},[{"v":"Nursing calculations","i":3,"n":0.707},{"v":"Nursing","i":2,"n":1},{"v":"Formulas","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":435,"$":[{"v":"The flow rate formula","n":0.5},{"v":"What does \"flow rate = volume/time\" mean? Flow rate is determined by the volume of liquid that passes by (into a patient) within a certain time period. This is the fundamental formula for all IV problems.\r\n\r\nThis short video uses practical examples to explain the formula for calculating the flow rate of a medication infusion.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nIn the simplest form, this is the formula for finding the flow rate of an infusion. You need to know the volume of the liquid flowing into the patient and divide this by the time it takes. The volume will be measured in units of some variations of litres, millilitres, or microlitres. And time will be measured in hours or minutes, or could theoretically be calculated in seconds... But this would not be a time unit that would be very practical in nursing. What is important to remember, however, is the units put into the one formula are compatible. That means that if the flow rate is measured in mls per hour, the volume must be measured in terms of mls and the time must be in terms of hours.\r\n\r\nIf the flow rate is in microlitres per second, the time will be in seconds, and the volume will be put into the formula in microlitres. If the flow rate is in terms of drops per minute, the volume would have to be in drops. Mls and drops are both ways to measure volume, but they are still different units. To have flow rates in drops per minute, you would have to convert the mls value to a value of drops before you put it in. And likewise, the time needs to be in minutes, so if you have time in hours, you would have to convert it before you put it into the formula.\r\n\r\nThe formula, as it is, gives you a way to calculate the flow rate if you already have the volume and the time. However, you can calculate any of the three factors as long as you have the other two. The formula can be rearranged so that you calculate either the volume or the time, if these are the missing quantities. The triangle on the left hand side of the screen may help you derive any other helpful formulae. We can see how flow rate can be achieved using volume on time.\r\n\r\nIf we want to calculate volume and we are given the flow rate and the time, we would multiply these together\u2026 And if we need to find the time after we are given the volume and flow rate, then time equals volume divided by flow rate.\r\n\r\nLet's put this formula to use. Here we have a problem whereby we know the volume of the liquid to be infused into the patient and we know the flow rate. The missing factor here is the time, the volume is 100 mls, and the flow rate is 25 mls per hour, and the formula for time is volume divided by flow rate. Plugging those values in, 100 mls divided by 25 mls per hour, the mls will cancel.\r\n\r\nNotice here we have a fraction within the fraction and in that the 25 ml per hour is in the denominator. So the denominator of the denominator, that is the hours here, gets flipped up and multiplies out the numerator. 100 divided by 25 is 4 and the unit now at the top is hours.In our next problem, we need to find the time again, but this time we are given the volume in mls and the flow rate in drops per minute. What is more, we for the time in terms of hours. Unfortunately, we cannot have hours and minutes in the same formula, so we will have to get our answer for time in minutes first, and then at the very end we can turn the minutes into hours, if that's what's needed. Similarly, the mls and the drops cannot exist together in the same formula, so one of them will have to change. We could turn 100 mls into 2000 drops, or we could change the 10 drops into half a ml. Either of these will bring us to the same answer.\r\n\r\nI will turn the 10 drops into mls, so we have mls in the formula. This allows for the mls to be cancelled. The minutes at the bottom of the denominator get flipped to the top again, and minutes is clearly the unit that our answer will be in. 100 divided by a half is 200, that is 200 minutes, and 200 minutes is three-and-a-third of a hour, or three hours and 20 minutes.\r\n\r\n","n":0.036},[{"v":"Nursing calculations","i":3,"n":0.707},{"v":"Nursing","i":2,"n":1},{"v":"Formulas","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":436,"$":[{"v":"Medication dosage by body weight","n":0.447},{"v":"How do you deal with a unit like mg/kg? Milligrams per kilogram means that you must give a certain amount (mgs) of a drug for each kilogram of the patient\u2019s body mass. Bigger patients need a bigger dose!\r\nLearn what this unit means and when and how to use it.\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nFrom time-to-time, a medication dosage is expressed in terms of a number of milligrams or millilitres, per kilogram of body weight. This means that a larger person needs a larger dose of the drug. It means that the amount you give is not a standard dosage but, rather, depends on the body mass of the patient.\r\n\r\nIn this example, the patient will need 12 milligrams of the drug for each kilogram of their body, and don't forget that this is what they need per day. Do n't neglect this last unit! Other examples may specify that this has to be given per dose or per hour, so please keep this one in mind.\r\n\r\nLet's just focus on this information first. We know that the patient is 60 kilograms, and as soon as you have a patient with a specific body weight, you need to determine how much of the drug we will give to this particular patient. We know from the dosage label that for every kilo of their body mass, we need to give 12 milligrams of the medication. Clearly, that will be 60 lots of 12 milligrams. 60 times 12 is 720, and remember that is per day.\r\n\r\nYou can visualise this in the following way: Think of the 60 kilogram patient, and then, visualise them consisting of 60 separate kilograms of body mass. To each kilogram, we need to deliver 12 milligrams of the medication. 60 times 12 gives us a total of 720 milligrams. And, as I said before, that is per day.\r\n\r\nYou can show your working in the following way: 60 kilograms multiplied by 12 milligrams per kilogram. The kilograms cancel, so that the unit you are left with is milligrams. That is, 720 milligrams.\r\n\r\nOnce you have achieved this single amount, it is unnecessary to think of the dosage in terms of milligrams per kilogram anymore\u2026 Until you are faced with another patient, of course. Your dosage for this patient is simply 720 milligrams and, as we have said previously, that will be 720 milligrams per day.\r\n\r\n","n":0.051},[{"v":"Nursing calculations","i":3,"n":0.707},{"v":"Nursing","i":2,"n":1},{"v":"Formulas","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":437,"$":[{"v":"Converting between large and small units","n":0.408},{"v":"How do you convert between units? Nurses use units that begin with milli and micro a lot. If you express your dosage in a smaller unit, the number must get bigger, and vice versa.\r\nFind out how to convert dosages between larger and smaller units.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nIf one day, you decide to keep a pet elephant, chances are you will only have room for one. If, on the other hand, you decided to have pet mice, the same space would accommodate probably thousands of them.\r\n\r\nTherefore, when converting from a large unit to a small unit, you can have more of them. This means that the number increases, which means the decimal point will move right. In the example, one microgram which is the larger unit is being converted to mcg which is smaller, from milli to micro you will need to\u00a0 move the decimal point three places, and that is three places to the right. On the other hand, one mouse only takes up a small part of an elephant.So if you're converting from a small unit to a large unit you will not have many... You'll only have a small fraction.\r\n\r\nIn this example, we are changing three milligrams (which is now the smaller of the two units) to grams (which is the larger of the\u00a0 two units). From milligrams to grams there is a factor of 1000, this mean moving the decimal point three places. The number gets smaller by a fraction of 1000 so it is now three places to the left.\r\n\r\nSo when going to a larger unit to a smaller unit\u2014such as grams to milligrams\u2014the number that you have will get larger and this means that the decimal point shifts to the right.\r\n\r\nIn contrast, If you are going to a smaller unit to a larger unit\u2014such as microlitres to litres\u2014the number that you have will get smaller and this means shifting the decimal point to the left. The end, thank you!\r\n\r\n","n":0.056},[{"v":"Nursing calculations","i":3,"n":0.707},{"v":"Nursing","i":2,"n":1},{"v":"Formulas","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":438,"$":[{"v":"Using proportions with liquid solutions","n":0.447},{"v":"If you need double the dose of medication, you need to give double the volume of the liquid that it is dissolved in and so on. This is the simple principle behind the volume required formula.\r\nLearn how you could solve these problems before applying the formula.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nLet's return to the GOFUEL mixture from a previous video. Let's recall that this is a sachet of GOFUEL, it is in powder form, and this sachet contains 40 grams of powder. This is a bottle of water and this bottle contains one litre of water. We're going to empty the sachet into the water so it dissolves evenly. And in this session, we are going to understand what the value of the concentration means.\r\n\r\nHere is our bottle with the sachet mixed evenly through... It's a nice blackcurrant flavour now. So we know that now that there is 40 grams of GOFUEL per litre of water. The concentration is just telling you about the proportion of solution that there is in the liquid or\u2014in the case of liquid medication\u2014what drug strength is in the liquid, if it is syrup-like, cough medicine, or saline for an IV.\r\n\r\nAnd, logically enough, the more liquid you administer, the more of the solute or the drug you receive in a proportional way. We can express the concentration in this way, as 40 grams per liter. One litre is equivalent to 1000 millilitres so we can express it this way\u2026 And that allows us to cancel some zeros from the numerator and the denominator, so that it is reduced to four grams per 100 milliliters.\r\n\r\nBecause four grams is also equivalent to 4000 micrograms, we can use the 4000 micrograms so we can express the concentration in this way. Remember, you always have to state your units, or else the number becomes meaningless. Once again, this allows us to cancel some zeros. And the concentration of our mixtures can be stated most simply this way. However, all of the above concentrations are equal and correct, it all depends on which units you want to use.\r\n\r\nSo the question is just how much go fuel powder does a person get if they drink a certain amount of liquid. It makes sense that if you drink the whole 1 L you'll take in 40 g of go fuel. What if you drink half of a litre, than you'll get half of the 40 g which is 20 mg; or maybe, 250 ml which is a quarter of a litre; then you will get a quarter of the total go fuel, which is 10 g. And if you had 200 mL which is a fifth of a bottle, you will get one-fifth of the 40 grams which is eight grams. Point 75 litres represents three-quarters of the litre bottle, and 75% or three-quarters of the GOFUEL bottle is 30 grams.\r\n\r\nWe could of solve these question in another way (keep in mind the other concentrations we calculated). We know for our sports drink solution that every mL contains 40 mg of go fuel; so if we have 500 millilitres of water, 500 lots of 40 mg is 20000 mg\u2014as we found before\u2014and notice that the mL cancel there leaving only the unit of mg.\r\n\r\nIf we drink 250 ml of the water, we will be receiving 250 lots of the 40 mg per mL, which is 10,000 mg; and the working out for that is similar. Now, if we drank 1250 ml of the liquid and that is more than we have in the bottle, but that's no problem because we can still focus how much go fuel we get if we had more of our blackcurrant mixture. 1250 x 40 is 50,000 mg -and the working out is the same again; and you could express that as 50,000 mg or 50 g if you prefer. But you must state the unit because it makes all the difference.\r\n\r\nSo, in all the concentration gives a measure of the medication for each unit of liquid. Using this you can figure out how much medication or in this case go fuel powder is in these certain amount of liquid. If you have one-third of the liquid for example, you'll have one-third of the medication and this relationship is proportional. What if a fraction you drink or receive out of the total, this is the same as the fraction of drug strength that you take out of the total available.\r\n\r\n","n":0.037},[{"v":"Nursing calculations","i":3,"n":0.707},{"v":"Nursing","i":2,"n":1},{"v":"Formulas","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":439,"$":[{"v":"Need over have","n":0.577},{"v":"This short video is the third of three videos in Nursing calculations: Finding the volume required. It explains the formula for calculating the quantity of a drug in solution for correct dosage by determining the stock strength and volume (have), then calculating a fraction of that stock volume for prescribed dosage (need).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nThis is a sachet of GOFUEL drink hydrate, it is in powder form, and the sachet contains 40 grams of powder... This is a bottle of water and this bottle contains one litre of water... We are going to empty the sachet into the water so that it dissolves evenly.\r\n\r\n\"If I need to ingest 15 grams of GOFUEL\u2026 How much solution should I drink?\"... Most pure drug calculations will be of this form. Our sports coach has recommended that we take 15 grams of the GOFUEL powder. How much liquid do I need to have out of the ready made 1 litre solution? Let's clarify our variables.\r\n\r\nWhat we need is 15 grams of GOFUEL. What we have in the bottle is 40 grams of GOFUEL. The total volume of liquid in the bottle is a 1000 mls\u2026 And the amount of liquid we are required to drink is the quantity we are looking for.\r\n\r\nSo the volume of the liquid that we have is given this equation: the amount of GOFUEL we need over the amount of GOFUEL we have, as a fraction of the bottle's volume. So plugging those values into the formula gives us fifteen-fortieths of the 1000 millilitres.\r\n\r\n15 and 40 is divisible by 5, so they can cancel down to three-eighths... The volume we're required to drink is three-eighths of the 1000 millilitres. Three-eighths converted to a decimal form is 0.375, and multiplying that by 1000 gives us 375 mL. We need to drink 375 mL of the liquid to give us 15 millilitres of GOFUEL.\r\n\r\nNow, returning to our formula of what we need over what we have as a fraction of the bottled volume. This is normally expressed in this form. What we need is strength required and what we have available is the stock strength. The volume of the bottle is normally the stock volume.\r\n\r\nSo the volume of liquid we need to drink or receive is given by... The amount of drug powder that we want, divided by the amount of drug powder we have available in the bottle, as a fraction of the bottle volume that it comes in\u2026 And this is the formula in abbreviated form.\r\n\r\nVR stands for Volume Required, this is usually the amount being sought in the problem. SR represents the Strength Required as in how much go fuel powder you want to ingest, or how much medication you have been prescribed. Stock Strength is given by SS, which is how many grams or milligrams is in the bottle to begin off with. And SV is Stock Volume, which is how much liquid is in the bottle or, in other cases, is perhaps what is in vial or an IV bag.\r\n\r\n","n":0.045},[{"v":"Nursing calculations","i":3,"n":0.707},{"v":"Nursing","i":2,"n":1},{"v":"Formulas","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":440,"$":[{"v":"Calculating from the labels","n":0.5},{"v":"This short video is the second of three videos in Nursing calculations: Finding the volume required. It explains how to calculate medication dosage from labels using the method of mental calculation and proportionality to get the right dosage for drugs in solution.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nIn the hospital environment you will need to calculate the required volume of liquid solution by mental calculation and by using simple proportionality.\r\n\r\nBy proportionality I mean if one thing is doubled so is the other.\r\n\r\nIf one quantity is reduced by a quarter then the other is reduced by a quarter as well.\r\n\r\nYour drug in solution form is a quantity of liquid so if you want more or less of the drug you need proportionally more or less of the liquid.\r\n\r\nLet's look at some of the labels that may appear on a vial of drug solution.\r\n\r\nHere, we have a label for Oxytocin and we have in this vial 10 units and in this vial it is in a solution of 1 ml of liquid. The doctor has ordered 6 units out of the 10, so we need a fraction of 6 over 10.\r\n\r\nWe only need 0.6 of the vial or 60% of the vial, or 6/10ths of the vial if you like. So 6/10ths of one ml is 0.6 ml.\r\n\r\nHere, we have another drug and in this vial we have 300 mg of drug and it is in a solution of 15 mls. The quantity that you see below is actually a simplified version the same fraction, but I suspect that the vial that the label is on here actually has 15 ml in it. 300 over 15 is proportionally the same as 20 over 1, the fraction has just been simplified for you.\r\n\r\nSo, this time, the doctor has ordered 200 mg of the drug. We will not need the whole vial, or all of the amount, and this time 200/300 is a fraction of two-thirds, so we need two-thirds of the bottle. Two-thirds of the 15 ml is equal to 10 ml. We will need 10 ml of solution.\r\n\r\nHere, we have another drug, and this time we have 100 mg of the drug and it is dissolved in 5ml of liquid. Once again, we have another reduction that 20 mg per ml is a reduced form of that fraction, but as we can see from this label, near the bottom, this is actually a 5ml bottle.\r\n\r\nSo, the doctor this time the doctor has order 50 mg out of the 100 mg. So 50/100 mg is a fraction of 1/2. We will only need half of the bottle, and half of the bottle this time means half of the 5 mls which is 2.5 mls.\r\n\r\nLast but not least, we have another drug label. This time, we have a Caffeine Citrate Injection and we have 60 mgs of this medication, which is dissolved in 3 ml of the liquid. Again, we have the 20 mg/1 ml as a simplification of that fraction. So the doctor this time has ordered 80 milligrams. That means we will be needing more than the one vial because the vial has only 3 ml in it and we will need 80 over 60. So 80 over 60 is more than 1, that is equivalent 4 over 3 and that is equal to one and one-third. We will need one and one-third of this medication. 4/3 and 3 ml is 4 milligrams. If we cancel the threes there, we will need 4 ml of the liquid.\r\n\r\n","n":0.042},[{"v":"Nursing calculations","i":3,"n":0.707},{"v":"Nursing","i":2,"n":1},{"v":"Formulas","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":441,"$":[{"v":"Finding the volume required","n":0.5},{"v":"What is volume required formula? This formula tells us how much liquid-form medication we need to give a patient, considering the strength they need and the source that it comes from.\r\nDefining the formula\r\nThis short video is the first of three videos in Nursing calculations: Finding the volume required. It explains how to define the formula for calculating medication dosage, how to understand the formula structure and how to use the formula with practical examples provided.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nThe volume required formula combines two key sorts of values. Firstly, there are the values for drug strength. Think of that as the mass of the drug\u2026 The more of the drug you have\u2026 the stronger the dose is. The amount of medication\u2014which is often in powder form in its original state\u2014gives an indication of the dosage. Mass is measured in units such as kilograms or grams, but, of course, in medication, this is more likely to be in milligrams or micrograms.\r\n\r\nAnother way to measure the amount of mass of a medication is units. That is, there is an alternative to milligrams or micrograms and it's called units.The other value that the formula uses is volume\u2026 Volume measures the amount of liquid. For example, a bottle contains one litre of water or 600 mls of juice. Litres, millilitres, and microlitres are units of volume.\r\n\r\nA certain amount of drug, 100 milligrams for example, can be diluted to any number of volumes. You might put 100 milligrams of drug into 1 ml of liquid for an intravenous injection. Or you might put 100 milligrams of drug into a 500 ml bag of saline, which will be infused into a patient over a period of hours.\r\n\r\nThe milligrams indicates the amount of the drug that the patient receives. The mLs, then, indicates the volume of the liquid. The milligrams per ml indicates the concentration of the solution.Milligrams per ml is the unit for the concentration of the solution. A few milligrams put into a small volume of mls creates a high concentration. A few milligrams put into a large volume creates a low concentration.\r\n\r\nSo, with this formula, you are seeking the volume required... You want to know how much of the medication solution you should give to the patient\u2026 This depends on the strength required, which is how many milligrams or micrograms of the drug the patient must be given.\r\n\r\nThe stock strength is how many milligrams or micrograms of the drug there are in the original bottle, or ampoule, or capsule. And, finally, the stock volume tells us how much liquid is in the original bottle, or ampoule, or capsule.\r\n\r\nAll in all, what we are finding is the proportion of the drug we need out of what we have available to begin with. And that is the proportion\u2014or the fraction\u2014that we want out of the original stock bottle of liquid... Put simply, if we only need half of the drug that is available in the bottle, then we only need half of the liquid solution that it comes in!\r\n\r\nWhen preparing a medication of a prescribed strength, you need to know how much of the liquid solution you need to draw up. The volume of the solution required can be be calculated using this formula. Using the strength of the drug required, the strength available, and finding the volume that is available in the stock bottle, vial, or ampoule\u2026 And it can be written in the abbreviated form like this...\r\n\r\nIt is useful, though, to understand why the formula has this shape. The formula is helpful, but, what is more important is the logic behind the formula, and if you understand that for simple calculations, you may not need to rely on the formula.\r\n\r\nHere's a sample problem to solve: The doctor has ordered 25 milligrams of Drugafloxin. This drug is available in solution form\u2026 And there is 100 milligrams of the drug in 120 mls of liquid solution in the vial. Now, thinking about this first\u2026 We only need 25 out of the available 100 milligrams. That is only one quarter of the strength. We need one quarter\u2014or one fourth\u2014of the available solution. So, consequently, we only need one quarter of the liquid that it comes in. One quarter of 120mls is\u2026 30 mls. Let's demonstrate this.\r\n\r\nWe only need 25 milligrams out of the total 100 milligrams of the Drugafloxin, so we only need the same proportion of the liquid. A quarter of the original 120 mls will be 30 mls\u2026 And hopefully that is the answer that we calculate. Now let's write this down using the formula. So the fraction of the drug needed out of what is available is the same as the fraction of the volume needed, out of what is available. And now for some clever algebra: if we want to get the volume required on one side of the equal side on it's own, then we can multiply that side with the stock volume. But we must do this to both sides if want to keep the equation balanced. And why did we do this? This gives us the opportunity to cancel the stock volume on the left hand side of the equation, and there we have the equation in the form that we normally see it.\r\n\r\nThis is the form that is most useful for finding the value for the volume required, but the important message behind it... Is that whatever fraction you need out of the available drug strength... That is the same fraction you need out of the available volume of the medicated solution. The volume required is what we are looking for. The strength required is 25 milligrams\u2026 And the stock strength is 100 milligrams. Let's finish substituting into the formula and the stock volume is 120 mls.\r\n\r\nNow mg and mg cancel. 25 and 100 can cancel. 25 goes into 25 once, and into 100 four times. There is your quarter. Now we want a quarter of the 120 mls. We can cancel the four with the 120. Four goes into 120... 30 times. The remaining unit is mls. The volume required is 30 mls. We only need 25 out of the available 100 milligrams, that is only one quarter of the strength, so it makes sense that the amount of solution we need is also a quarter of the available solution.\r\n\r\nWe can understand that the proportion of the drug we need is equal to the proportion of the liquid we need in comparison to the drug amount or the liquid amount we have. We know we need 25 out of the 100 milligrams. This is equivalent to needing the 30 out of the 120 mls. Both of these proportions reduce to a quarter.\r\n\r\nLet's rearrange this equation to get the volume required on its own. We can do this if we multiply by stock volume or SV. But whatever we do to one side of the equation we must do to the other so that it keeps the equation balanced. On the left side, the SV will cancel.\u00a0\u00a0 One the right side it will remain. And there we have it, the usual form of the volume required formula. This more common form of the formula\u2013as well as the equal fractions shown above\u2013are both true. They are actually expressions of the same formula\u2026 They have just undergone some rearrangement. The message that is hiding inside these formulas is that the fraction of the drug you need out of what you have is the same fraction in terms of the liquid solution.\r\n\r\n","n":0.028},[{"v":"Nursing calculations","i":3,"n":0.707},{"v":"Nursing","i":2,"n":1},{"v":"Formulas","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":442,"$":[{"v":"Preparing for exams","n":0.577},{"v":"If you're feeling overwhelmed about how to revise for an exam, this tutorial will give you some useful techniques to improve your performance. We'll cover how to revise for your exams and also provide strategies you can use while you are taking the exam.\r\n\r\n\r\nRevision\r\nRevision begins in the first week of the semester. This page explains how topics are presented logically and sequentially so you need to build on your knowledge and understanding throughout the semester.\r\nExamples of regular revision are:\r\n\r\n \treading of required texts\r\n \trevising after each lecture\r\n \tsummarising your notes\r\n \tdeveloping a weekly study timetable\r\n \testablishing regular study habits.\r\n\r\nRevision strategies\r\nJust reading is too passive. Your understanding and ability to recall information will improve if you study more actively. If you actively use what you have read or heard, you will find it much easier to remember and understand in the future. If you don't do anything with the information you have learned in a lecture:\r\n\r\n \tyou will forget 50% within one hour\r\n \tyou will forget 75% within 24 hours.\r\n\r\nSo, don't just read\u2014write notes, talk, listen and make sure what you have seen or heard is learned.\r\nHow to make revision more 'active'\r\n\r\n \tWrite summaries and use mind maps\r\n\t\t\r\n \t\t\tFile your lectures, essays, and research under topics.\r\n \t\t\tWrite bullet point summaries of each topic\u2014use notes from lectures, tutorials, reading and assignment materials.\r\n \t\t\tFurther reduce your notes using highlighting.\r\n \t\t\tRewrite your notes as mind maps\u2014note key ideas and supporting examples or evidence.\r\n\t\t\r\n\t\r\n \tRevise with others\r\n \tPractise answering past exam papers\r\n \tAnalyse practice exam questions by identifying instruction words, key words and limiting statements.\r\n\r\n\t\t\r\nLet's look at the following exam question as an example:\r\nAnalyse the reasons for the fluctuations in the Australian dollar over the past two years. Look at the different components that make up this question.\r\nIn this case:\r\n\r\nanalyse is an instruction word.\r\nthe reasons for, the fluctuation in, and the Australian dollar are key words.\r\nAustralian and over the past two years are limiting statements.\r\n\r\n\r\n\r\n","n":0.056},[{"v":"Time management","i":1,"n":0.707},{"v":"Study skills","i":0,"n":0.707}]]},{"i":443,"$":[{"v":"Writing concisely","n":0.707},{"v":"Good academic writing gets straight to the point. It does not waste words.\r\nWrite concisely by:\r\n\r\n1. Making the actor and the action clear\r\n\r\nAttempts at writing in an academic style often result in the meaning being buried under a pile of nominalisations* and the prepositional phrases** needed to support them. While some nominalisations are useful and appropriate, using too many makes sentences unwieldy and hides the main action.\r\n\r\n\r\n\r\n2. Eliminating redundancies\r\n\r\nWordiness adds to your word count without adding to the meaning. Be brutal: delete meaningless, doubled and implied terms, and replace unnecessarily long phrases with shorter ones.\r\nMaking the actor and action clear\r\nLook at the following example. Who is doing what here? What is the main action? Who or what is doing it?\r\nThere is general agreement among climate scientists as to the elevation of global temperatures as a result of carbon dioxide in the atmosphere.\r\nThe first step is to find the actions by turning the nomalisations into verbs. Remember nominalisations usually end in -tion, -ence, -ity, -ness, -ment, etc.\r\n\r\n\r\n\r\nNominalisations\r\n\u2192\r\nVerbs\r\n\r\n\r\n\r\n\r\nThere is general agreement among climate scientists as to the elevation of global temperatures as a result of carbon dioxide in the atmosphere.\r\n\r\n\r\nagreement \u2192 agree\r\nelevation \u2192 elevate\r\n\r\n\r\n\r\n\r\n\r\n\r\nNow consider the following:\r\n(a) Who is agreeing?\r\n(b) What is elevating what?\r\n\r\nScreen reader users, this text uses visual highlights to indicate different parts of the text.\r\n2 Actors3 Actions4 Target\r\n\r\n\r\n\r\n\r\nOriginal\r\nRevised\r\n\r\n\r\n\r\n\r\n(a) There is general  among  as to the elevation of global temperatures as a result of carbon dioxide in the atmosphere.\r\n  that ...\r\n\r\n\r\n(b) There is general agreement among climate scientists as to the  of  as a result of  in the atmosphere.\r\n is  \r\n\r\n\r\n\r\nRewrite the text, moving the main actors and actions as close to the beginning of the sentence as you can. Note how the extra prepositional phrases will naturally disappear, leaving only those that are necessary.\r\n\r\n\r\n\r\nRevised\r\nFinal\r\n\r\n\r\n\r\n\r\n  that\r\n+\r\n is \r\nMost   that  in the atmosphere is  .\r\n\r\n\r\n\r\nThe resulting sentence is not only more concise, it is also much easier to understand.\r\n\r\n\r\n\r\nOriginal\r\nFinal\r\n\r\n\r\n\r\n\r\nThere is general  among  as to the  of  as a result of  in the atmosphere.\r\nMost   that  in the atmosphere is  .\r\n\r\n\r\n\r\n(Adapted from Williams, J.M. and Bizup, J. (2014). Style: Lessons in clarity and grace.[/hl] (11th ed.) Upper Saddle River, NJ: Pearson.)\r\n\r\n\r\nEliminating redundancies\r\nAvoid padding your writing with unnecessary words.\r\nDelete meaningless terms\r\n\r\n \teach and every\r\n \ttrue and accurate\r\n \tcompletely finish\r\n \tfuture plans\r\n \tIt is important to note that\r\n\r\nReplace unnecessarily long phrases\r\n\r\n \tdue to the fact that \u2192 because, since\r\n \tnot similar \u2192 different\r\n\r\nExample\r\n\r\nTip: be careful when attempting to write concisely. Don\u2019t leave out important details or qualifications. Cut the waffle, not the content.\r\n(Adapted from Williams, J.M. and Bizup, J. (2014). Style: Lessons in clarity and grace. (11th ed.) Upper Saddle River, NJ: Pearson.)\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nRead each of the sentences. Revise them by selecting the most appropriate verb to describe the actions within the square brackets.\r\nSentence 1\r\nAttempts by researchers at identification of the AIDS virus have met with success; however, attempts at development of a vaccine for the immunisation of those at risk have failed.\r\nRevise\r\nAIDS researchers have [eradicated, identified, isolated] the AIDS virus but have failed to [develop, isolate, identify] a vaccine that will [cure, identify, immmunise] those at risk.\r\nAnswer\r\nAIDS researchers have [identified] the AIDS virus but have failed to [developed] a vaccine that will [immunise] those at risk.\r\nSentence 2\r\nAny departures by the members from established procedures may cause termination of membership by the Board.\r\nRevise\r\nIf members [terminate, proceed, depart] from established procedures, their membership may be [departed, terminated, established] by the Board.\r\nAnswer\r\nIf members [depart] from established procedures, their membership may be [terminated] by the Board.\r\n(Adapted from Williams, J.M. and Bizup, J. (2014). Style: Lessons in clarity and grace. (11th ed.) Upper Saddle River, NJ: Pearson.)\r\n \r\nRead the sentence and think about which three words or phrases can be deleted. Answers will be provided: One version with the deleted words in [ ]. The second version with the full answer.\r\nSentence 1\r\nEmployability actually depends on certain factors that basically involve generic skills, such as ability to communicate, as much as any particular technical disciplinary knowledge.\r\nAnswer with deleted words/phrases:\r\nEmployability [actually] depends on [certain factors that basically involve] generic skills, such as ability to communicate, as much as [any particular technical] disciplinary knowledge.\r\nAnswer:\r\nEmployability depends on generic skills, such as ability to communicate, as much as disciplinary knowledge.\r\nSentence 2\r\nGoing forward, it is recommended that such kinds of issues be considered in the future planning process.\r\nAnswer with deleted words/phrases:\r\n[Going forward], it is recommended that such [kinds of] issues be considered in the future planning process.\r\nAnswer:\r\nIt is recommended that such issues be considered in the future planning process.\r\n\r\n","n":0.037},[{"v":"Sentences","i":3,"n":1},{"v":"Paragraphs","i":2,"n":1},{"v":"Style","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":444,"$":[{"v":"Nursing","n":1},{"v":"\r\n\r\n\r\n\r\n\r\n\r\n\r\nFurther resources\r\nAlso, the Learning Lab has sections that will help nursing students with important academic concepts:\r\n\r\n\r\n\r\n\r\n\r\n","n":0.25},[{"v":"Nursing calculations","i":2,"n":0.707},{"v":"Nursing","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":445,"$":[{"v":"Linking words","n":0.707},{"v":"Linking words are an important way of developing a logical flow of ideas. They act as signposts for the reader so they can follow the way the writer has organised their ideas.\r\n\r\nDevelop cohesion\r\nLinking words and phrases show the relationship between different ideas. They can show contrast, comparison, a sequence, add additional information, provide emphasis, and show cause and effect.\r\nExample\r\nIn this example, notice how the writer uses linking words to develop the flow of their argument.\r\nThe Australian preferential voting system ensures the chosen candidate reflects the choice of most voters. In this system, voters rank candidates in order of preference on their ballots, instead of picking just one (Liu, 2024). , if no candidate gets more than half of the first-choice votes, the candidate with the least votes is removed. , their votes are given to the next preferred candidate on those ballots. , this process helps to ensure that the winning candidate has wider support among voters.\r\nShow connections between the ideas\r\nLinking words enable the writer to show the connections between ideas.\r\nExample\r\nIn this example, notice how the use of linking words emphasises the writer\u2019s analysis of their research.\r\nOne factor within the team that seems to be important is the notion of team cohesiveness. , team cohesiveness enables a diverse group of individuals to work towards common goals. Maxwell (2011, p.107) argues that highly cohesive teams \u2018have clear goals and successful outcomes\u2019. , cohesive teams are more likely to work well together, have enthusiasm for their work (Brown 2009) and so achieve a high rate of success (Zheng & Hui 2008). , team cohesiveness has been shown to enhance team effectiveness.\r\nCommon linking words\r\nAll sentences in a paragraph need to relate to the main idea in the topic sentence. The reader should be able to see how each sentence flows from the previous one and how each is connected to the topic sentence. Linking words and phrases weave sentences together to create a cohesive paragraph.\r\n\r\n","n":0.056},[{"v":"Writing structures","i":2,"n":0.707},{"v":"Linking words","i":1,"n":0.707},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":446,"$":[{"v":"Electricity","n":1},{"v":"You already interact with electricity constantly but what's actually happening inside a circuit? These resources build the foundation for understanding and measuring it, introducing the key quantities: charge, current, voltage and power.\r\n\r\n\r\n\r\n","n":0.177},[{"v":"Physics","i":0,"n":1}]]},{"i":447,"$":[{"v":"Kinematics","n":1},{"v":"Before you can explain why something moves, you need to describe how it moves. Kinematics gives you the tools to do that. Use these resources to learn about the language and tools of motion, from linear and angular motion to constant and non\u2011constant acceleration.\r\n\r\n\r\n\r\n","n":0.151},[{"v":"Motion","i":2,"n":1},{"v":"Physics","i":1,"n":1},{"v":"Dynamics","i":0,"n":1}]]},{"i":448,"$":[{"v":"Advanced dynamics and statics","n":0.5},{"v":"What stops a bridge from collapsing, or keeps a spinning shaft from flying apart? Advanced dynamics and statics takes the fundamentals further, into rotation, torque and the forces acting on real structures. You'll work with moments, moment of inertia and equilibrium: the tools engineers use to analyse whether something will hold.\r\n\r\n\r\n\r\n","n":0.14},[{"v":"Physics","i":0,"n":1}]]},{"i":449,"$":[{"v":"Measurement","n":1},{"v":"Measurement forms the basis of many practical applications in STEM fields. Use this resource to develop a solid understanding of fundamental measurement concepts. Working through these pages will equip you with the skills you need for further exploration in science, technology, engineering, and mathematics.\r\n\r\n\r\n\tUnits and conversions\r\n\tDimensions\r\n\tNotation\r\n\r\n\t","n":0.147},[{"v":"Maths","i":1,"n":1},{"v":"Arithmetic","i":0,"n":1}]]},{"i":450,"$":[{"v":"Percentages","n":1},{"v":"Percentages are a powerful tool for expressing proportions, comparisons and changes. Learn how to convert fractions and decimals into percentages using this resource. This will enhance your ability to interpret data and make informed decisions when faced with quantities.\r\n\r\nA percentage is another way to express a part of a whole. It is always a number expressed as a fraction out of \\(100\\). For example, \\(\\dfrac{18}{100}\\) is the same as \\(18\\%\\).\r\n\r\nVideo tutorial \u2013 percentages \r\nWatch this video to learn how to convert between fractions, decimals and percentages.\r\n\r\n\r\nExample 1 \u2013 converting between fractions, decimals and percentages\r\nConvert \\(75\\%\\) into a fraction.\r\n\\[\\begin{align*} 75\\% & = \\frac{75}{100}\\quad\\textrm{divide top and bottom by \\(25\\)}\\\\\r\n& = \\frac{3}{4}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 converting between fractions, decimals and percentages\r\n\r\n\tConvert the following percentages into fractions and simplify.\r\n\t\r\n\t\t\\(35\\%\\)\r\n\t\t\\(6\\%\\)\r\n\t\t\\(180\\%\\)\r\n\t\t\\(62.5\\%\\)\r\n\t\t\\(138\\%\\)\r\n\t\t\\(0.55\\%\\)\r\n\t\r\n\tConvert the following decimals into percentages.\r\n\t\r\n\t\t\\(0.23\\)\r\n\t\t\\(1.69\\)\r\n\t\t\\(0.068\\)\r\n\t\t\\(0.0007\\)\r\n\t\r\n\tConvert the following fractions and decimals into percentages.\r\n\t\r\n\t\t\\(\\dfrac{1}{4}\\)\r\n\t\t\\(\\dfrac{3}{8}\\)\r\n\t\t\\(4\\dfrac{1}{5}\\)\r\n\t\t\\(0.37\\)\r\n\t\t\\(0.06\\)\r\n\t\t\\(1.705\\)\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on percentages.\r\n\r\nA percentage means out of 100. For example, 25 percent means 25 out of 100 or 25 divided by a 100. Seven percent means seven divided by a 100 or seven out of 100. And 98 percent means 98 out of 100 or 98 divided by 100. Let\u2019s start by changing percentages into fractions. So the first one, 75 percent is 75 divided by 100. What we need to do is to break that down into smaller numbers, in other words we\u2019re simplifying the fraction, so let\u2019s divide by 25, top and bottom, and we get three divided by four, so 75 percent, as a fraction is three-quarters, similarly 40 percent is 40 divided by 100. Notice here that there are zeroes top and bottom, cancel one on the top with one on the bottom, that leaves us with four over 10 which cancels even further into two over five, so 40 percent is two-fifths.\r\n\r\nLet\u2019s look at a large percentage and change this to a fraction, 175 percent is the same as 175 divided by 100, but now we turn this into a mixed fraction, which is one and 75 over 100. Notice again that 75 over 100 can be cancelled down, so we divided both top and bottom by 25 leaving us with an answer of one and three-quarters. So 175 percent is one and three-quarters.\r\n\r\nNow let\u2019s change a decimal percentage into a fraction. For instance we have the percentage 17.5 percent, which is 17.5 over 100. What we need to do here is to turn that 17.5 into a whole number so that we can then cancel the terms down. So we\u2019re multiplying 17.5 by 10 and doing that to the top we must do the same thing to the bottom. In other words 17.5 by 10 is 175, 100 times 10 is 1,000, now we look to see if anything cancels and of course it does, we can divide both top and bottom by 25 leaving us with an answer of seven over 40, so 17.5 percent is seven over 40 as a fraction.\r\n\r\nNow let\u2019s look at a small decimal percentage. We have here 1.25 percent, which is the same as 1.25 divided by 100, again we need to turn the 1.25 into a whole number so this time we multiply it by 100 because 1.25 has got two decimal places and similarly we must multiply the denominator by 100 as well, so the top line becomes 125, the bottom line becomes 10,000, which cancels down into one over 80, so 1.25 percent is one over 80 as a fraction.\r\n\r\nNow let\u2019s change percentages into decimals. An example, 15 percent is the same as 15 over 100, change a percentage into a decimal we always move the decimal point two places to the left, so if you move the decimal point two places to the left we end up with 0.15. Similarly with a large percentage, 220 percent is the same as 220 divided by 100, so the quick way to do this is to move the decimal point two places to the left, so we end up with an answer of 2.2. So 220 percent is 2.2 as a decimal.\r\n\r\nFinally 7.9 percent is 7.9 divided by 100 and dividing by 100 you simply move the decimal point two places to the left, so moving it two places to the left leaves us with an answer of 0.079, so 7.9 percent is 0.079.\r\n\r\nNow let\u2019s change fractions into percentages. So here we\u2019re working in reverse to what we were doing in previous slides. Here\u2019s an example, seven over 10. To change a fraction into a percentage we always multiply it by 100 and because the fraction is written as seven over 10 we can write the 100 as a fraction as well, 100 over one. Notice that one zero on the top will cancel with one zero on the bottom, in other words I\u2019m dividing by 10, which leaves us with 70 over one, which gives us an answer of 70 percent. In other words seven-tenths is 70 percent as a percentage.\r\n\r\nHow about changing decimals into percentages, well we use the same idea as before. Here\u2019s an example, 0.55, to change that decimal into a percentage we multiply it by 100 and when you multiply by 100 what you\u2019re actually doing is you\u2019re moving the decimal point two places to the right, so we start at the decimal point, move two places and we finish after the last digit, the five in this case. So we have an answer of 55, which is 55 percent, so 0.55 is 55 percent as a percentage.\r\n\r\nWhat about small decimals, changing those into percentages. We use exactly the same rule as before. Here\u2019s an example, 0.09, to change that into a percentage we multiply by 100, so again, as before, we\u2019re moving the decimal point two places to the right and here the decimal point ends up on the far right hand side after the last digit, the nine, so our answer is nine percent, so 0.09 is nine percent as a percentage.\r\n\r\nNow try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.031},[{"v":"Maths","i":3,"n":1},{"v":"Fractions","i":2,"n":1},{"v":"Decimals","i":1,"n":1},{"v":"Arithmetic","i":0,"n":1}]]},{"i":451,"$":[{"v":"Numbers","n":1},{"v":"Learn the basics for working with integers, including negative numbers. Numerical reasoning skills you will learn here are very fundamental to maths; you will apply them to everything from everyday situations to more complex mathematical challenges.\r\n\r\nIntegers\r\nCounting or whole numbers, also known as integers, can be used in many different ways. Here, you will learn how to multiply and divide large numbers, and complete operations in the correct order so that you get a correct answer.\r\n\r\nPlace values\r\nBefore we dive into this, it is important to remember place values. Numbers consist of digits in different \"places\". A place value is therefore the value of a digit in a number based on its position (or place).\r\nFrom right to left, the places are the \"ones\" column, the \"tens\" column, \"hundreds\", \"thousands\", \"tens of thousands\", and so on.\r\nLet's apply this to the number \\(3429\\). Working from right to left:\r\n\r\n\tThe \\(9\\) is in the ones column. It has a place value of \\(9\\times1=9\\).\r\n\tThe \\(2\\) is in the tens column. It has a place value of \\(2\\times10=20\\).\r\n\tThe \\(4\\) is in the hundreds column. It has a place value of \\(4\\times100=400\\).\r\n\tThe \\(3\\) is in the thousands column. It has a place value of \\(3\\times1000=3000\\).\r\n\r\n\r\n\r\nMultiplying numbers\r\nWhen you multiply or times two numbers together, you get a product. See the examples to learn how to multiply integers.\r\nAnother handy skill to have in your toolkit is multiplying by \\(10\\), \\(100\\), \\(1000\\), or numbers with \\(1\\) followed by any number of \\(0\\)s. All you need to do is count the number of zeroes and add them to the number you are multiplying by.\r\n\r\nVideo tutorial \u2013 multiplying numbers\r\nWatch this video to learn how to multiply numbers.\r\n\r\n\r\nExample 1 \u2013 multiplying numbers\r\n\\[\\begin{array}{r} 52\\\\\r\n\\underline{\\times 4}\\\\\r\n208\r\n\\end{array}\\]\r\nHere, the steps are:\r\n\r\n\t\\(2\\times4=8\\)\r\n\tWrite the \\(8\\) in the ones column.\r\n\t\\(5\\times4=20\\)\r\n\tWrite the \\(0\\) in the tens column. Write the \\(2\\) in the hundreds column.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 multiplying numbers\r\n\r\n\t\\(53\\times6\\)\r\n\t\\(78\\times9\\)\r\n\t\\(96\\times8\\)\r\n\t\\(27\\times11\\)\r\n\t\\(39\\times45\\)\r\n\t\\(89\\times77\\)\r\n\t\\(306\\times21\\)\r\n\t\\(295\\times15\\)\r\n\t\\(783\\times123\\)\r\n\t\\(89\\times10\\)\r\n\t\\(98\\times100\\)\r\n\t\\(73\\times1000\\)\r\n\t\\(664\\times10000\\)\r\n\t\\(3253\\times1000\\)\r\n\r\n\r\n\r\n\r\nDividing numbers\r\nWhen you divide one number (a dividend) by another (the divisor), you get a quotient. \r\nSee the examples to learn how to divide integers.\r\nThe basic approach to long division is divide, multiply, subtract, bring down.\r\n\r\nThis is repeated until you get a remainder of \\(0\\).\r\n\r\nVideo tutorial \u2013 dividing numbers\r\nWatch this video to learn how to divide numbers.\r\n\r\n\r\nExample 1 \u2013 dividing numbers\r\n\\(74400\\div600\\)\r\nYou can make life easier for yourself by crossing out any common \\(0\\)s. So, this becomes \\(744\\div6\\).\r\n\\[\\begin{array}{r} 124\\phantom{)}\\\\\r\n6{\\overline{\\smash{\\big)}\\,744\\phantom{)}}} \\\\\r\n\\underline{-~\\phantom{(}6\\phantom{~~~~)}} \\\\ \r\n14\\phantom{)~~} \\\\ \r\n\\underline{-~\\phantom{(}12\\phantom{~~)}} \\\\\r\n24\\phantom{)}\\\\ \r\n\\underline{-~\\phantom{(}24\\phantom{)}} \\\\\r\n0\\phantom{)}\\\\ \r\n\\end{array}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 dividing numbers\r\n\r\n\t\\(84\\div3\\)\r\n\t\\(376\\div8\\)\r\n\t\\(605\\div11\\)\r\n\t\\(9000\\div60\\)\r\n\t\\(144000\\div600\\)\r\n\t\\(35000\\div10\\)\r\n\t\\(1001\\div13\\)\r\n\t\\(2128\\div19\\)\r\n\r\n\r\n\tA school hall can sit \\(950\\) students. If \\(25\\) student can be seated in each row, how many rows are there?\r\n\tA fire alarm sounds a sharp beep three times a second. If it sounded a total of \\(900\\) beeps, how long did the alarm sound for (in minutes)?\r\n\r\n\r\n\r\nOrder of operations\r\nThe basic operations are addition (\\(+\\)), subtraction (\\(-\\)), multiplication (\\(\\times\\)) and division (\\(\\div\\)).\r\nWhen more than one operation is used in a single maths expression, we use a set of rules called the order of operations to complete the calculation in the right sequence.\r\nLet's consider the following arithmetic problem:\r\n\r\nAt first sight, there are several possible answers.\r\n\r\n\tWorking from left to right:\r\n\t\\[\\begin{align*} 4+3 \\times 2+5 & = 7 \\times 2+5\\\\\r\n\t& = 14+5\\\\\r\n\t& = 19\r\n\t\\end{align*}\\]\r\n\t\r\n\tWorking from right to left:\r\n\t\\[\\begin{align*} 4+3 \\times 2+5 & = 4+3 \\times 7\\\\\r\n\t& = 4+21\\\\\r\n\t& = 25\r\n\t\\end{align*}\\]\r\n\t\r\n\tIf we do the multiplication first:\r\n\t\\[\\begin{align*} 4+3 \\times 2+5 & = 4+6+5\\\\\r\n\t& = 15\r\n\t\\end{align*}\\]\r\n\t\r\n\tIf we do the additions first:\r\n\t\\[\\begin{align*} 4+3 \\times 2+5 &amp = 7 \\times 7\\\\\r\n\t& = 49\r\n\t\\end{align*}\\]\r\n\t\r\n\r\nMathematicians don\u2019t like imprecision. To allow only one answer to the problem, we perform mathematical operations in the following order:\r\n\r\n \tBrackets\r\n \tIndices\r\n \tMultiplication and Division from left to right\r\n \tAddition and Subtraction\r\n\r\nUsing the bolded first letters, we get the abbreviation BIMDAS.\r\nWhen we apply BIMDAS to the previous problem, we get:\r\n\\[\\begin{align*} 4+3 \\times 2+5 & = 4+6+5\\\\\r\n& =15\r\n\\end{align*}\\]\r\nYou may have also heard this being called BOMDAS, where the O stands for \"orders\". You will learn about indices later.\r\n\r\nVideo tutorial \u2013 order of operations\r\nWatch this video to see how to apply these rules.\r\n\r\n\r\nExample 1 \u2013 applying BIMDAS\r\n\\[\\begin{align*}3 \\times 2+4 & = 6+4\\\\\r\n& = 10\r\n\\end{align*}\\]\r\nThe multiplication is completed first, so \\(3 \\times2=6\\). Then, the addition \\(6+4=10\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 applying BIMDAS\r\nSimplify the following expressions:\r\n\r\n\t\\((18+(3\\times (-5))\\)\r\n\t\\(3 \\times (-4) + (8\\times 2)\\)\r\n\t\\(10-5^{2}+3\\)\r\n\t\\((10-5)^{2}+3\\)\r\n\t\\(10^{2}-5^{2}\\)\r\n\t\\((10-5)^{2}\\)\r\n\r\n\r\n\r\n\r\nAdding and subtracting negative numbers\r\nIn many areas of STEM, you may encounter arithmetic involving negative numbers. Profit and loss calculations in economics and finance can involve negative numbers. Physics uses negative numbers when dealing with vector quantities and engineering uses negative numbers for potential differences and current directions.\r\nIt is important to have a solid grasp of how to add and subtract negative numbers so that you are successful in STEM.\r\n\r\nVideo tutorial \u2013 adding and subtracting negative numbers\r\nWatch this video to learn how to add and subtract negative numbers.\r\n\r\n\r\n\r\nExample 1\u00a0\u2013 adding and subtracting negative numbers\r\n\\[+2-4\\]\r\nFor subtraction, we move left along the number line. From \\(+2\\), \\(4\\) numbers to the left is \\(-2\\).\r\n\r\nWe can also think of it as \\(2-2=0\\), then subtracting a further \\(2\\) to give you \\(0-2=-2\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 adding and subtracting negative numbers\r\n\r\n\t\\(4-6\\)\r\n\t\\(7-12\\)\r\n\t\\(-8+3\\)\r\n\t\\(-12+17\\)\r\n\t\\(-11-5\\)\r\n\t\\(-14-7\\)\r\n\t\\(-2-(-8)\\)\r\n\t\\(-9-(-3)\\)\r\n\t\\(-12-(-21)\\)\r\n\t\\(-4-(-3)+2\\)\r\n\r\n\r\n\r\nMultiplying and dividing negative numbers\r\nThere are rules when multiplying and dividing negative numbers.\r\n\r\n\r\n\tThe multiplication and division are done as usual.\r\n\tWhen the signs are the same, the answer is positive.\r\n\tWhen the signs are different, the answer is negative.\r\n\r\n\r\nThis is summarised in the table.\r\n\r\n\t\r\n\t\t\r\n\t\tPositive \\(+\\)\r\n\t\tNegative \\(-\\) \r\n\t\r\n\t\r\n\t\tPositive \\(+\\)\r\n\t\t\\(+\\) \r\n\t\t\\(-\\)\r\n\t\r\n\t\r\n\t\tNegative \\(-\\)\r\n\t\t\\(-\\) \r\n        \\(+\\)\r\n\t\r\n\r\n\r\nVideo tutorial \u2013 multiplying and dividing negative numbers\r\n\r\n\r\nExample 1 \u2013 multiplying and dividing negative numbers\r\n\\[+36\\div+9\\]\r\nPositive divided by positive gives positive. So, \\(36\\div9=+4\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 multiplying and dividing by negative numbers.\r\n\r\n\t\\(11\\times9\\)\r\n\t\\(-7\\times(-8)\\)\r\n\t\\(-5\\times12\\)\r\n\t\\(13\\times(-5)\\)\r\n\t\\(42\\div7\\)\r\n\t\\(-72\\div(-9)\\)\r\n\t\\(-99\\div11\\)\r\n\t\\(108\\div(-3)\\)\r\n\t\\(\\frac{1}{3}\\times(-72)\\)\r\n\t\\(-4\\times(-3)\\times7\\)\r\n\r\n\r\n\r\n\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on multiplying numbers.\r\n\r\nLet\u2019s start by multiplying two numbers by one number, or two digits by one digit. Look at the left hand side, 52 times four. We start with the four and say four times two is eight, place the eight under the four. Then we say four times five is 20 and we place the 20 next to the eight. So 52 times four is 208. Not let\u2019s look on the right hand side example, 49 times 8. We start as before with the eight and we say eight times nine is 72, place the two under the eight and we carry the seven, shown in red. We then go from eight to four, eight times four is 32, and we add the seven to give us 39. So our answer is 392.\r\n\r\nNow let\u2019s multiply two numbers by two numbers or two digits by two digits. 34 times 87. Start on the left hand side and say seven times four is 28, carry the two, shown in red at the top. Then we go from seven to three, seven times three is 21. Now we add the two shown in red to the 21, that gives us 23. So our first row is 238. Now we move to the next row, but before that we place a zero on the right hand side, shown in blue. Start our multiplication by multiplying eight by four, eight fours are 32, two carry three, shown in red at the top. Then we say eight times three is 24, plus three is 27. So our next row is 2720. We now add the two rows up, eight and zero is eight, three plus two is five, two plus seven is nine, and two. So our multiplication is equal to 2958.\r\n\r\nNow let\u2019s multiply number by 10, 100 and 1000. Here\u2019s a number 52 times 10. And 52 times 10 gives us 520. The thing to notice about this multiplication is that because 10 has got one zero we simply add the one zero to the 52 to give us our answer, 520. So we\u2019re adding the zero. What about 52 times 100? The same idea applies, 100 has got two zeros, so we add the two zeros to the 52 to give us 5200. Finally, let\u2019s multiply 52 by 1000. 1000 has three zeros, so we add the three zeros to the right hand side, to give us 52000.\r\n\r\nNow try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this movie.\r\n \r\nHi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on dividing numbers.\r\n\r\nLet\u2019s start by dividing 74,400 by 600. You set it out in the usual way, in other words we say 600 into 74,400. Notice with this question that 600 and 74,400 both have two zeros, so we can in fact cross off the two zeros in both numbers. This simplifies matters a lot, because now what we\u2019re doing is we\u2019re dividing six into 744. So let\u2019s carry out the division. We start by saying six into seven goes once, shown in red, and that gives us one left over. So we place the one next to the four. So the new number becomes 14. So we now say six into 14 goes two, remainder two. The two goes next to the four, giving us a new number of 24. So we now say six into 24 goes four times. So the answer is 124.\r\n\r\nNow let\u2019s divide 9,682 by 47. First of all we set it out in the usual way. We say 47 into 9,682. We now notice that there are no zeros in either number so we can\u2019t cross off zeros, so we go ahead with the division. So we say 47 into nine won\u2019t go, 47 into 96 goes two times, shown in red. So 47 times two is 94. We place the 94 under the 96 and subtract. That gives us two left over. We then bring the eight down as shown with the arrow alongside the two, giving us a number 28. So we now say 47 into 28 won\u2019t go, so we put the zero in red at the top, then we bring down the two shown with the red arrow. So we now have 282. So we divide 47 into 282, and that goes exactly six times. Six times 47 is 282. Do the subtraction, as long as you get zero left over, that\u2019s the end of the division. So the answer is 206 times.\r\n\r\nLet\u2019s look at our final example, 12,065 divided by 19. Let\u2019s rewrite that as 19 into 12,065. So we now carry out the long division. Notice there are no zeros at the end so we have to carry on as normal. 19 into one won\u2019t go, 19 into 12 won\u2019t go, 19 into 120 will go six times. So we say six times 19 is 114. Do the subtraction, 120 take away 114 gives us six. Bring the six down shown with the red arrow, so the new number is now 66. So we now say 19 into 66 goes three times, shown in red. 19 times three is 57, write the 57 under the 66, subtract, that gives us nine left over. Now we bring down the five, shown with the red arrows, to give us our new number of 95. Carry out the long division, 19 into 95 goes five times, five 19s are 95. Carry out the subtraction, that gives us zero. In other words, the answer is 635.\r\n\r\nNow carry out some division questions for yourself. The answers to these questions are on the next slide. Thanks for watching this movie.\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on order of operations.\r\n\r\nWhen we work through a maths problem we have to be aware of following a set of rules. These rules are called order of operations. The first thing to look out for are brackets and if a problem\u2019s got brackets in it, then what\u2019s inside those brackets must be done first. Secondly, if the problem has got multiplication and division, then they must be done next and we carry out the operation from left to right as we work through the problem. And finally, if there are adding and subtracting parts to the problem they must be done at the end, and again they must be done from left to right.\r\n\r\nLet\u2019s do an example. Here\u2019s the question two plus seven minus six divided by two times three. Notice there are no brackets so we now move down to the second level, which is multiplication and division. Here we have six divided by two times three. So this part of the question must be done first, six divided by two is three, times three. Notice we\u2019re moving from left to right, and finally we multiply three by three which gives us nine. So we end up with two plus seven minus nine. Now we have just adding and subtracting to do, which is the last operations. Moving from left to right, two plus seven is nine, then we take away nine and finally we have our answer, zero. So the answer to this question is zero.\r\n\r\nLet\u2019s look at this problem. 16 divided by eight plus four times brackets seven minus three close brackets. Notice here we have brackets, so that\u2019s the first operation that we carry out. So seven take away three is worked out to give us four. Now we have 16 divided by eight plus four times four. The division and multiplication must be carried out next, and we move from left to right, 16 divided by eight is two plus four times four is 16. So we have an answer of 18.\r\n\r\nNow do some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on adding and subtracting negative numbers.\r\n\r\nLet\u2019s start by adding negative numbers. Here\u2019s the example. Plus two minus four, I place the number line at the bottom of the screen, zero at the centre, numbers to the right are positive; numbers to the left are negative. So let\u2019s start with a plus two, plus two means moving two units to the right on the number line from zero, so now we\u2019re at positive two. Now we have to take away four, so starting at the two we move back four places because it\u2019s negative. Notice the answer here will be from where we started to where we finished, in other words from zero to negative two. In other words plus two, take away four, is equal to minus two.\r\n\r\nLet\u2019s do another one. Minus three plus six, notice I\u2019m starting with a negative here, so by going negative I\u2019m moving to the left from the number line, from zero, so I\u2019m now at negative three. Now I have to plus six, being a plus I have to move to the right six units, so starting at negative three, moving six places to the right we end up at positive three. So again the answer is from where we started, which is a zero, to where we finished, which in this place is positive three, so our answer is positive three. So negative three plus six is positive three.\r\n\r\nNow let\u2019s add two negative numbers. Negative one, negative two, so because they\u2019re all negative I\u2019m moving to the left on the number line, negative one and then negative two and the answer is from where we start to where we finish, which will be negative three, so negative one negative two is equal to negative three.\r\n\r\nLet\u2019s look at another one. Here we\u2019re starting with plus one minus negative four. Notice again that the two negatives are next to each other so when they\u2019re next to each other we add, so this problem becomes positive one plus four, which is quite a simple operation to do on the number line, plus one is moving one place to the right, adding four will be adding four more, giving us an answer of five. In other words, positive one minus negative four is positive five.\r\n\r\nNow try some questions for yourselves. The answers to these questions are on the next slide. Thank you for watching this short movie.\r\n \r\n\r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on multiplying and dividing positive and negative numbers.\r\n\r\nLet\u2019s multiply positive and negative numbers first. At the table red pluses and minuses are what we start with, the blue pluses and minuses are the answers. So for instance if I multiply a plus by a plus I trace down from the plus in the top row and I trace across from the plus in the left hand column. Where they intersect is my answer, in other words, a plus times a plus is a plus shown in blue. Similarly do the same thing with the negative signs, the negative in the top row, trace it down, the negative in the left hand column trace it across, where they intersect is the answer and as you can see the answer is a plus. In other words if the signs are the same the answer is positive.\r\n\r\nNow let\u2019s look at the other signs. Let\u2019s multiply a plus by a minus. So we trace down from the plus in the top and we trace across from the minus in the left hand column, where they intersect is our answer in blue and as you can see they intersect at the negative sign. Similarly if I multiply a minus by a plus a minus in the top row by a plus in the left hand column, trace the minus down from the top and the plus sign from the left across until they intersect, where they intersect is the answer and as you notice it\u2019ll be a negative sign again. In other words when the signs are different, in other words plus and minus, the answer is negative.\r\n\r\nSo let\u2019s look at this in more detail. Let\u2019s go through each one step by step. The first one is positive by positive is equal to a positive. Plus times plus equals plus. Here\u2019s an example; plus four times plus four, plus times plus gives us a positive answer, four times seven is 28, in other words the answer is plus 28. So let\u2019s look at this in more detail. Let\u2019s go through each one step by step, the first one is positive by positive is equal to a positive, plus times plus. Now let\u2019s look at negative times negative, minus times a minus is a plus. [Unclear] minus five times minus nine, two negatives gives us a positive answer, notice if the signs are the same the answer is always positive, so minus times a minus is a plus, five times nine is 45, so our answer is plus 45.\r\n\r\nNow we have negative times a positive is a negative, minus times a plus is a minus, notice the signs are different the answer is always negative, so with an example, minus six times plus 11, the signs are different so the answer is negative, six times 11 is 66 so the answer is minus 66.\r\n\r\nFinally positive times a negative is a negative. Plus times a minus is a minus. The signs are different therefore the answer is always negative. Example, plus 12 times minus seven, the signs are different so the answer is negative, 12 sevens are 84 so the answer is negative 84.\r\n\r\nNow let\u2019s divide positive and negative numbers. Look at the table, and look again at the similarity between this table and the table we had for multiplication. For instance if the signs are the same the answer is positive, if the signs are different the answer is negative. In other words the rules that we use for division are exactly the same as the rules that we use for multiplication.\r\n\r\nSo let\u2019s look at them one by one. Plus divided by a plus is a plus, 36 divided by nine is the same as positive 36 divided by positive nine which would give us an answer of positive four. Negative divided by a negative is a positive, minus divided by a minus is a plus, notice if the signs are the same the answer is always positive. So for instance minus 56 divided by minus eight, the signs are the same, the answer must be positive, 56 divided by eight is seven so our answer is plus seven.\r\n\r\nNow let\u2019s look at negative divided by positive, the answer is negative because if the signs are different the answer is always negative. For instance minus 78 divided by plus three, minus divided by a plus is a negative, 78 divided by three is 26, so the answer is negative 26. Finally positive divided by a negative is a negative, plus divided by minus is a minus. Example, plus 96 divided by minus six, the signs are different, therefore the answer must be negative, 96 divided by negative six is 16, so the answer is negative 16.\r\n\r\nNow do some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.017},[{"v":"Maths","i":2,"n":1},{"v":"Decimals","i":1,"n":1},{"v":"Arithmetic","i":0,"n":1}]]},{"i":452,"$":[{"v":"Fractions","n":1},{"v":"Mastering fractions is crucial for tackling a wide range of mathematical problems. The skill is applicable to a huge range of real-world contexts, from cooking and construction to science and finance. Use this resource to learn how to simplify, add, subtract, multiply and divide fractions.\r\n\r\nA fraction is the easiest way to express a part of a whole. It consists of a numerator (the number at the top) and a denominator (the number at the bottom) which are separated by a horizontal bar or line called a vinculum.\r\nFractions help to communicate information. For example, if you got 18 marks correct out of the 20 marks available on your quiz, you will have received \\(\\dfrac{18}{20}\\) for the quiz.\r\nLet's see how to add, subtract, multiply and divide fractions, and reduce fractions to their simplest terms (or smallest numbers).\r\n\r\nSimplifying fractions\r\nSimplifying fractions involves reducing them to their simplest form. Fractions are simplified by dividing the numerator and denominator by the same number until they are their lowest possible whole-number values. In other words:\r\nSimplifying fractions involves dividing the numerator and denominator by the highest common factor (HCF).\r\nWe do this by cancelling out the HCF.\r\nWhen the numerator is larger than the denominator, the fraction is called an improper fraction. We can convert them to a mixed number, which consists of a whole number and a fraction.\r\n\r\nVideo tutorial \u2013 simplifying fractions\r\nWatch this video to learn how to simplify fractions.\r\n\r\n\r\nExample 1 \u2013 simplifying ordinary fractions\r\nSimplify \\(\\dfrac{18}{24}\\).\r\n\\[\\begin{align*} & = \\frac{18}{24}\\quad\\textrm{divide top and bottom by \\(2\\)}\\\\\r\n& = \\frac{9}{12}\\quad\\textrm{divide top and bottom by \\(3\\)}\\\\\r\n& = \\frac{3}{4}\r\n\\end{align*}\\]\r\nThis simplification also can be done in one step if you realise that \\(6\\) divides into \\(18\\) and \\(24\\)\u2014that is, \\(6\\) is the HCF.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 simplifying fractions\r\n\r\n\tSimplify the following.\r\n\t\r\n\t\t\\(\\dfrac{16}{24}\\)\r\n\t\t\\(\\dfrac{30}{55}\\)\r\n\t\t\\(\\dfrac{48}{100}\\)\r\n\t\t\\(\\dfrac{21}{77}\\)\r\n\t\r\n\tWrite the following as mixed numbers and simplify where possible.\r\n\t\r\n\t\t\\(\\dfrac{13}{9}\\)\r\n\t\t\\(\\dfrac{33}{6}\\)\r\n\t\t\\(\\dfrac{35}{3}\\)\r\n\t\t\\(\\dfrac{22}{8}\\)\r\n\t\r\n\tWrite the following as improper fractions.\r\n\t\r\n\t\t\\(4\\dfrac{2}{5}\\)\r\n\t\t\\(9\\dfrac{3}{8}\\)\r\n\t\t\\(6\\dfrac{7}{9}\\)\r\n\t\t\\(2\\dfrac{11}{12}\\)\r\n\t\r\n\r\n\r\n\r\nAdding and subtracting fractions\r\nThe denominator is the bottom of the fraction and the numerator is the top of the fraction. For the fraction \\(\\dfrac{2}{5}\\), the numerator is \\(2\\) and the denominator is \\(5\\).\r\nThe basic concept is that only fractions with a common denominator may be added or subtracted.\r\n\r\nVideo tutorial \u2013 adding fractions\r\nWatch this video to learn how to add fractions.\r\n\r\n\r\nVideo tutorial \u2013 subtracting fractions\r\nWatch this video to learn how to subtract fractions.\r\n\r\n\r\nExample 1 \u2013 adding and subtracting fractions\r\n\\[\\begin{align*} \\frac{1}{4}+\\frac{2}{4} & = \\frac{1+2}{4}\\\\\r\n& = \\frac{3}{4}\r\n\\end{align*}\\]\r\nThe fractions have the same denominator, so the lowest common denominator is \\(4\\).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 adding and subtracting fractions\r\n\r\n\tSimplify the following.\r\n\t\r\n\t\t\\(\\dfrac{4}{5}+\\dfrac{3}{4}\\)\r\n\t\t\\(\\dfrac{3}{4}+\\dfrac{1}{4}\\)\r\n\t\t\\(\\dfrac{1}{6}+\\dfrac{1}{3}\\)\r\n\t\t\\(2\\dfrac{3}{5}+3\\dfrac{7}{10}\\)\r\n\t\t\\(3\\dfrac{4}{7}+\\dfrac{9}{14}\\)\r\n\t\t\\(\\dfrac{5}{9}+\\dfrac{11}{36}\\)\r\n\t\t\\(6\\dfrac{5}{8}+3\\dfrac{3}{4}\\)\r\n\t\t\\(1\\dfrac{7}{9}+5\\dfrac{5}{18}\\)\r\n\t\t\\(7\\dfrac{5}{7}+2\\dfrac{13}{14}\\)\r\n\t\t\\(\\dfrac{7}{8}+1\\dfrac{1}{4}+2\\dfrac{2}{3}\\)\r\n\t\t\\(\\dfrac{3}{4}+1\\dfrac{2}{3}+3\\dfrac{5}{6}\\)\r\n\t\r\n\tSimplify the following.\r\n\t\r\n\t\t\\(\\dfrac{3}{4}-\\dfrac{1}{4}\\)\r\n\t\t\\(\\dfrac{1}{3}-\\dfrac{1}{6}\\)\r\n\t\t\\(3\\dfrac{7}{10}-2\\dfrac{3}{5}\\)\r\n\t\t\\(3\\dfrac{4}{7}-\\dfrac{7}{14}\\)\r\n\t\t\\(\\dfrac{5}{9}-\\dfrac{11}{36}\\)\r\n\t\t\\(6\\dfrac{7}{8}-3\\dfrac{3}{4}\\)\r\n\t\t\\(5\\dfrac{5}{18}-1\\dfrac{2}{9}\\)\r\n\t\t\\(7\\dfrac{5}{7}-2\\dfrac{5}{14}\\)\r\n\t\t\\(2\\dfrac{2}{3}-1\\dfrac{1}{4}\\)\r\n\t\t\\(3\\dfrac{5}{6}-1\\dfrac{2}{3}\\)\r\n\t\r\n\r\n\r\n\r\nMultiplying fractions\r\nWhen we multiply two fractions, we multiply the numerators and the denominators of each fraction.\r\nIt is best to simplify each fraction, if possible, before you do the multiplication. This saves you having to simplify afterwards.\r\n\r\nVideo tutorial \u2013 multiplying fractions\r\nWatch this video to learn how to multiply fractions.\r\n\r\n\r\nExample 1 \u2013 multiplying fractions\r\n\\[\\frac{3}{4}\\times\\frac{7}{5} = \\frac{21}{20}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 multiplying fractions\r\nSolve the following fractions and simplify where possible.\r\n\r\n\t\\(\\dfrac{1}{3}\\times\\dfrac{6}{7}\\)\r\n\t\\(\\dfrac{1}{5}\\times\\dfrac{1}{6}\\)\r\n\t\\(\\dfrac{8}{9}\\times6\\)\r\n\t\\(2\\dfrac{3}{4}\\times3\\dfrac{3}{7}\\)\r\n\t\\(\\dfrac{9}{14}\\times\\dfrac{7}{9}\\)\r\n\t\\(\\dfrac{7}{11}\\times\\dfrac{11}{42}\\)\r\n\t\\(\\dfrac{1}{8}\\times2\\dfrac{2}{9}\\)\r\n\t\\(3\\dfrac{5}{6}\\times1\\dfrac{1}{5}\\)\r\n\t\\(5\\dfrac{2}{3}\\times4\\dfrac{2}{7}\\)\r\n\t\\(1\\dfrac{1}{8}\\times5\\dfrac{1}{3}\\times\\dfrac{1}{9}\\)\r\n\r\n\r\n\r\n\r\n\r\nDividing fractions\r\nDividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of a fraction is just the fraction with its numerator and denominator swapped. This is also called inverting the fraction.\r\nFor example, the reciprocal of \\(\\dfrac{3}{4}\\) is \\(\\dfrac{4}{3}\\).\r\n\r\nVideo tutorial \u2013 dividing fractions\r\nWatch this video to learn how to divide fractions.\r\n\r\n\r\nExample 1 \u2013 dividing ordinary fractions\r\n\\[\\begin{align*} & = \\frac{5}{4}\\div\\frac{19}{8}\\quad\\textrm{change sign to times and invert last fraction}\\\\\r\n& = \\frac{5}{4}\\times\\frac{8}{19}\\quad\\textrm{divide top and bottom by \\(4\\)}\\\\\r\n& = \\frac{5}{1}\\times\\frac{2}{19}\\\\\r\n& = \\frac{10}{19}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 dividing fractions\r\nSolve the following fractions and simplify where possible.\r\n\r\n\t\\(\\dfrac{3}{4}\\div\\dfrac{1}{4}\\)\r\n\t\\(\\dfrac{4}{9}\\div\\dfrac{7}{9}\\)\r\n\t\\(\\dfrac{7}{8}\\div7\\)\r\n\t\\(\\dfrac{3}{4}\\div\\dfrac{4}{5}\\)\r\n\t\\(\\dfrac{8}{6}\\div\\dfrac{2}{3}\\)\r\n\t\\(3\\dfrac{1}{4}\\div\\dfrac{7}{12}\\)\r\n\t\\(3\\dfrac{4}{5}\\div\\dfrac{4}{5}\\)\r\n\t\\(5\\dfrac{1}{3}\\div1\\dfrac{1}{3}\\)\r\n\t\\(4\\dfrac{1}{2}\\div1\\dfrac{1}{4}\\)\r\n\t\\(\\dfrac{7}{8}\\div2\\dfrac{1}{3}\\times\\dfrac{8}{9}\\)\r\n\r\n\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on simplifying fractions.\r\n\r\nLet\u2019s start by simplifying 27 divided by 36. The thing to notice about this fraction is that you can divide both top and bottom by nine, in other words, nine into 27 goes three and 36 divided by nine is 4. So we\u2019ve reduced a more complicated fraction to a much simpler one, three quarters.\r\n\r\nLet\u2019s write this another way. 27 over 36 is divisible both top and bottom by nine. Nine into 27 goes three. Notice here we\u2019ve cancelled the 27, in other words, we\u2019re dividing by nine. And what we do to the top we do to the bottom, so we cancel the 36 and divide by nine, giving us four. Thus the answer is three quarters.\r\n\r\nHere\u2019s another example. We have to write 16 over five as a mixed number, or sometimes called a mixed fraction. 16 over five as you can see is a top heavy fraction, the number on the top is bigger than the number on the bottom. First of all what we need to do is divide the five into the 16, which gives us three and that gives us a remainder of one. And the remainder one is divided by the five, because the five is on the bottom of the fraction 16 over five. In other words, 16 over five is equal to three and one 5th.\r\n\r\nNow let\u2019s work in the reverse to what we were doing in the previous slide. Here we start with a mixed fraction or a mixed number and we need to turn it into an improper fraction. So we want to write two and four 7ths as an improper fraction. There\u2019s our two and four 7ths and what we do is this. We multiply the seven by the two to give us 14. Once we\u2019ve got the 14 we then add it to the four on the top of the fraction part of this number. So 14 plus four is 18. There\u2019s the arithmetic on the right hand side. Two times seven in brackets, which we do first, then we add the four. That gives us fourteen plus four over seven, which gives us 18 over seven. In other words, the mixed fraction two and four 7ths is 18 over seven as an improper fraction.\r\n\r\nNow try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on adding fractions.\r\n\r\nLet\u2019s start by adding two basic fractions, a quarter plus two quarters. Notice in this case that the number on the bottom called the denominator is the same. We call that the common denominator, in other words, the number they both have in common is four. Once the denominator is the same then we can just go ahead and start adding the top numbers, called the numerators, in other words, one plus two gives us three over four. So we leave the denominator alone and we add the numerators. So the answer is three quarters.\r\n\r\nBut what happens if the denominators are not the same? Well, we need to now find the common denominator of five and four, meaning what number does five and four divide into. That number is 20, so the common denominator is 20. And this is how we go about doing it. Five goes into 20 four times, so we multiple five by four in the left hand fraction. If we multiple the bottom by four, we must multiply the top of the fraction by four, so two times four is on the top. With our three quarters fraction, we\u2019re multiplying four by five to give us 20, therefore we multiply the top by five, three times five. In other words, we now have adding the eight and the 15 to give us the answer for our numerator, so eight plus 15 is 23, divided by 20 will give us one and three 20ths.\r\n\r\nLet\u2019s add whole number fractions or mixed fraction. One and one 3rd, plus two and three 5ths. The first job is to separate out the whole numbers from the fractions. So we now have one plus two, and the fractions are one 3rd plus three 5ths. So the whole numbers must be added up and the fractions must be added up. So we now have three, and now we go ahead and find the common denominator of three and five, which is 15. So as with the previous slide, we say three times five is 15, one times five is five. The second fraction is five times three is 15, and therefore we multiply the top of the fraction by three, three times three. This gives us three plus five over 15, plus nine over 15. And five over 15 plus nine over 15 gives us 14 over 15. So our answer is three and 14 15ths.\r\n\r\nNow try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n \r\nHi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on subtracting fractions.\r\n\r\nLet\u2019s start with an easy fraction subtraction; two quarters minus one quarter. The thing to notice about this particular fraction subtraction is that the denominator for both fractions is in fact the same, which is four. If the denominator is the same then we can simply carry out the subtraction of the numerators for both fractions to take away one giving us an answer of one quarter, but just to reiterate in this particular case the fraction denominators are the same, so the subtraction for the numerators can be carried out straight away. However what if the two fractions have different denominators, in this case four and five are different denominators, so what we do first of all is look for a common denominator, that is the lowest number in which the four and the five will divide into, that number is 20, so we then carry out a little operation here. What I\u2019ve done is to multiply the four by the five and in the second fraction I\u2019ve multiplied the five by the four, so what I do to the bottom I must do to the top, so I\u2019ve multiplied the three by the five for the first fraction and I\u2019ve multiplied the two by the four for the second fraction. Now I have the same denominator on both, for both fractions, which is 20, because the denominator is the same I can then simply subtract the numerators, 15 minus eight.\r\n\r\nWhat about whole number fraction subtractions, in this case two and three-fifths take away one and one-third. What we do first of all is to subtract the whole numbers, in this case it\u2019s two take away one, then we go ahead and subtract the fractions, three\u2011fifths minus one-third. Here the common denominator is 15 and so with the first fraction, three over five, I\u2019m multiplying that by three to make me 15, therefore I multiply the top by three. The second fraction, one-third, I multiply the bottom by five to give us 15, therefore I multiply the top by five, this gives me one plus nine\u2011fifteenths minus five-fifteenths, which is an answer of one and four-fifteenths.\r\n\r\nNow try some questions for yourself. The answers to the questions are on the next slide. Thanks for watching this short movie.\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on multiplying fractions.\r\n\r\nLet\u2019s start by looking at two simple fractions, two 3rds times four 5ths. Notice with this fraction problem that there are no terms top and bottom that cancel. In other words, what we do is to multiply out the top line and multiply out the bottom line, in other words, the numerator which is the top line is multiplied out, two times four, and the denominator, the bottom line, is multiplied out, three times five. Again stress the point here that there are no terms that cancel. So this gives us four times two is eight, three times five is 15. So our answer to this problem is eight 15ths.\r\n\r\nBut what happens if we have terms that do cancel? Here we have the problem five over eight times two over five. And you should notice that there are terms that cancel here. The first one which is the obvious one are the fives top and bottom. So we go ahead with our cancelling. Five into five goes one, five into five goes one. So we now have one times two over eight times one. But notice again we have cancelling of the two and the eight. Two into two goes once, and two into eights goes four. So we\u2019re cancelling more terms. That\u2019s about as far as we can go, so we multiply out the top line, one times one is one, the bottom line is four times one is four. So the answer to this problem is a quarter.\r\n\r\nHow about multiplying out mixed numbers? Here we have mixed numbers one and three quarters multiplied by two and two 3rds. When we multiply mixed numbers we first of all turn them into improper fractions. So one and three quarters is seven over four, two and two 3rds is eight over three. So now we look to see whether any of these terms cancel. And as you can see four goes into four once, four goes into eight twice. Nothing else cancels, so we then multiply out the top line, seven times two is 14, the bottom line, one times three is three. But notice we give our answer as a mixed number, not an improper fraction, so 14 over three is four and two 3rds.\r\n\r\nNow try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on dividing fractions.\r\n\r\nLet\u2019s start by dividing two simple fractions, four 3rds divided by two 5ths. We start by changing the division sign to a multiplication sign and then we take the reciprocal of the two 5ths, the last term. In other words, we flip the two over five upside, which becomes five over two. We then carry out the operation multiplying four over three by five over two, and as you can see, the twos cancel, two into two goes one, two into four goes two. Nothing else cancels, so we multiply out top and bottom, 10 over three and leave our answer as a mixed fraction. So the answer to this question is three and one 3rd.\r\n\r\nLet\u2019s divide two mixed fractions. Two and one quarter divided by one and one 5th. Before we actually carry out the division we must express our mixed fractions as improper fractions, so two and one quarter is nine over four divided by one and one 5th is six over five. Now we carry out our division by first of all changing the division sign to a multiplication sign and then taking the reciprocal of six over five, which becomes five over six. From there we look to see if anything cancels, and as you can see the nine and the six will divided by three, three into six goes two, three into nine goes three. Nothing else cancels, so we multiply out the top line, three times five is 15, the bottom line, four times two is eight. Thus an improper fraction which must be converted into a mixed fraction, which is one and seven 8ths. So the answer to this mixed fraction question is one and seven 8ths.\r\n\r\nOccasionally you may see questions like this where there is a fraction divided by a whole number, and before you carry out the division problem you must express the whole number as a fraction. In other words, three over four divided by six, which is six over one. So you express the whole number as six over one. Then we carry out the operation by changing the division sign to a multiplication sign and taking the reciprocal of six over one, which is one over six. Look again for terms that cancel, here three goes into three once, three goes into six twice. Nothing else cancels, multiply out the numerator, one, the denominator, four times two is eight. So our answer to this question is one over eight.\r\n\r\nNow carry out some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.019},[{"v":"Maths","i":2,"n":1},{"v":"Fractions","i":1,"n":1},{"v":"Arithmetic","i":0,"n":1}]]},{"i":453,"$":[{"v":"Decimals","n":1},{"v":"Decimals are used to express fractions in a more precise and easily understandable form. They are essential for handling money, measuring quantities, and applying mathematical concepts in everyday life and various professional fields. Use this resource to learn how to add, subtract, multiply, divide and convert decimals.\r\n\r\nA decimal is a way of representing numbers that are not whole, using a decimal point to separate the whole number part from the fractional part.\r\n\r\nPlace values are important. Remember that to the left of a decimal point, we use places: ones, tens, hundreds, thousands, etc. To the right of a decimal point, we use: tenths, hundredths, thousandths, etc.\r\n\r\nFor example, in the number \\(32.157\\):\r\n\r\n\t\\(3\\) is in the tens column and has a place value of \\(30\\).\r\n\t\\(2\\) is in the ones column and has a place value of \\(2\\).\r\n\t\\(1\\) is in the tenths column and has a place value of \\(0.1\\).\r\n\t\\(5\\) is in the hundredths column and has a place value of \\(0.05\\).\r\n\t\\(7\\) is in the thousandths column and has a place value of \\(0.007\\).\r\n\r\n\r\n\r\nWhole numbers simply have a \\(0\\) in the tenths, hundredths, etc, place. So, \\(17\\) would be the same as \\(17.000\\).\r\n\r\nAdding decimals\r\nThere is a simple rule for adding decimals.\r\nWhen you add decimals, the decimal points must line up.\r\n\r\nVideo tutorial \u2013 adding decimals\r\nWatch this video to learn how to add decimals.\r\n\r\n\r\nExample \u2013 adding decimals\r\n\\[3.04+17+21.479\\]\r\n\\(21.479\\) has the most decimal places, so we write the other numbers to the same number of decimal places. \\(3.04\\) becomes \\(3.040\\) and \\(17\\) becomes \\(17.000\\).\r\n\\[\\begin{array}{r} 3.040\\\\\r\n17.000\\\\\r\n\\underline{+21.479}\\\\\r\n41.519\r\n\\end{array}\\]\r\n\r\nExercise \u2013 adding decimals\r\nCalculate the following.\r\n\r\n\t\\(1.5+4.7\\)\r\n\t\\(6.54+2.3\\)\r\n\t\\(347.3+37.74\\)\r\n\t\\(27.58+34\\)\r\n\t\\(8.186+2.05\\)\r\n\t\\(13.09+9.876+4.87\\)\r\n\t\\(46.48+9.94+23.6\\)\r\n\t\\(29.54+0.0076+35.65+2.3\\)\r\n\t\\(333.54+7.22+29.5+20.899\\)\r\n\t\\(436.4+2.002+23.5+10.98\\)\r\n\r\n\r\n\r\nSubtracting decimals\r\nThe rule for subtracting decimals is the same as when you add them.\r\nWhen you subtract decimals, the decimal points must line up.\r\n\r\nVideo tutorial \u2013 subtracting decimals\r\nWatch this video to learn how to subtract decimals.\r\n\r\n\r\nExample \u2013 subtracting decimals\r\n\\[825.709-272.456\\]\r\nBoth have the same number of decimal places, so we don't need to rewrite either of them.\r\n\\[\\begin{array}{r} 825.709\\\\\r\n\\underline{-272.456}\\\\\r\n553.253\r\n\\end{array}\\]\r\n\r\nExercise \u2013 subtracting decimals\r\nCalculate the following.\r\n\r\n\t\\(18.5-4.3\\)\r\n\t\\(6.54-2.63\\)\r\n\t\\(354.3-31.7\\)\r\n\t\\(27.58-19.48\\)\r\n\t\\(8.127-2.09\\)\r\n\t\\(13.09-9.87\\)\r\n\t\\(46.48-9.94\\)\r\n\t\\(29.54-15\\)\r\n\t\\(333-7.22\\)\r\n\t\\(436.4-2.999\\)\r\n\r\n\r\n\r\n\r\nMultiplying decimals\r\nThere are some rules you need to follow to correctly place the decimal point in your answer after you multiply decimals.\r\n\r\nIgnore the decimal points when you multiply the numbers.\r\nCount the number of decimal places in the original numbers to find the total.\r\nPlace the decimal point so that the final product has the same number of total decimal places.\r\n\r\n\r\n\r\nVideo tutorial \u2013 multiplying decimals\r\nWatch this video to learn how to multiply decimals.\r\n\r\n\r\nExample 1 \u2013 multiplying decimals\r\n\\[5.2\\times10\\]\r\nWhen multiplying by multiples of \\(10\\), i.e. \\(10\\), \\(100\\), \\(1000\\), etc, all you need to do is move the decimal point to the right depending on how many zeroes the multiple has.\r\nFor example, if you multiply by \\(10\\), the number \\(10\\) has one \\(0\\), so we move the decimal point one place to the right. When you multiply by \\(1000\\), you move the decimal point three places to the right, corresponding to the three \\(0\\)'s in \\(1000\\). \r\nIn this case, we multiply by \\(10\\), so we move the decimal point one place to the right.\r\n\\[5.2\\times10=52\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 multiplying decimals\r\nCalculate the following.\r\n\r\n\t\\(8.3\\times10\\)\r\n\t\\(7.85\\times10\\)\r\n\t\\(9.625\\times100\\)\r\n\t\\(27.44\\times100\\)\r\n\t\\(395.6\\times100\\)\r\n\t\\(89.76\\times1000\\)\r\n\t\\(0.306\\times1000\\)\r\n\t\\(2.9\\times5\\)\r\n\t\\(7.8\\times8\\)\r\n\t\\(1.6\\times3.4\\)\r\n\t\\(3.5\\times4.5\\)\r\n\t\\(7.38\\times1.6\\)\r\n\t\\(6.64\\times1.33\\)\r\n\t\\(3.253\\times6.66\\)\r\n\r\n\r\n\r\n\r\nDividing decimals\r\nThere are also rules you need to follow to correctly place the decimal point in your answer after you divide decimals.\r\n\r\nTurn the divisor (the number that you are dividing by) into a whole number by moving the decimal point.\r\nMove the decimal point in the dividend (the number that you are dividing into) the same number of decimal places as you have done for the divisor.\r\nDivide the numbers.\r\nPlace the decimal point directly above its position in the dividend.\r\n\r\n\r\n\r\nWhen you move the decimal point one spot to the right to make the divisor a whole number, this is the same as multiplying by \\(10\\). Two spots to the right is the same as multiplying by \\(100\\), and so on.\r\nWhen you move the decimal point for the dividend, you multiply by the same power of ten.\r\n\r\nVideo tutorial \u2013 dividing decimals\r\nWatch this video to learn how to divide decimals.\r\n\r\n\r\nExample 1 \u2013 dividing decimals\r\n\\[53.7\\div10\\]\r\nWhen dividing by multiples of \\(10\\), i.e. \\(10\\), \\(100\\), \\(1000\\), etc, all you need to do is move the decimal point to the left depending on how many zeroes the multiple has.\r\nFor example, if you divide by \\(10\\), the number \\(10\\) has one \\(0\\), so we move the decimal point one place to the left. When you divide by \\(1000\\), you move the decimal point three places to the left, corresponding to the three \\(0\\)'s in \\(1000\\). \r\nIn this case, we divide by \\(10\\), so we move the decimal point one place to the left.\r\n\\[53.7\\div10=5.37\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nConverting between fractions and decimals\r\nFractions and decimals are both effective ways to express numbers that are not whole. In some cases, it might be easier to work with fractions. In others, decimals might be more useful. It is helpful to be able to convert between them.\r\n\r\n\r\n\tTo convert a fraction to a decimal, you divide the numerator by the denominator.\r\n\tTo convert a decimal to a fraction, you express the decimal as a fraction with a power of ten as the denominator and simplify if necessary.\r\n\r\n\r\n\r\nVideo tutorial \u2013 converting between decimals and fractions\r\nWatch this video to learn how to convert between decimals and fractions.\r\n\r\n\r\nExample 1 \u2013 converting between fractions and decimals\r\nConvert \\(0.455\\) into a fraction and simplify.\r\nLook at the number of decimal places to determine which power of ten is placed at the bottom of the fraction. \\(0.455\\) has three decimal places, so the denominator needs to be \\(10^{3}\\) or \\(1000\\).\r\nWe can write the fraction with \\(455\\) as the numerator and \\(1000\\) as the denominator, and simplify.\r\n\\[\\begin{align*} 0.455 & = \\frac{455}{1000}\\quad\\textrm{divide top and bottom by \\(5\\)}\\\\\r\n& = \\frac{91}{200}\r\n\\end{align*}\\]\r\n\r\n\r\n\r\n\r\n\r\n\r\nExercise \u2013 converting between fractions and decimals\r\n\r\n\tConvert the following decimals into fractions and simplify.\r\n\t\r\n\t\t\\(0.8\\)\r\n\t\t\\(3.75\\)\r\n\t\t\\(1.125\\)\r\n\t\t\\(0.07\\)\r\n\t\t\\(7.10\\)\r\n\t\t\\(0.385\\)\r\n\t\r\n\tConvert the following fractions into decimals and simplify. Give your answer to no more than three decimal places.\r\n\t\r\n\t\t\\(2\\dfrac{3}{4}\\)\r\n\t\t\\(3\\dfrac{1}{12}\\)\r\n\t\t\\(\\dfrac{4}{9}\\)\r\n\t\t\\(4\\dfrac{5}{11}\\)\r\n\t\t\\(6\\dfrac{1}{6}\\)\r\n\t\t\\(5\\dfrac{1}{15}\\)\r\n\t\r\n\r\n\r\n\r\n \r\nHello. I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on adding decimals.\r\n\r\nLet\u2019s start with this example, 3.04 plus 17 plus 21.479. First of all set out the decimals as shown. A couple of things to notice here. The first thing is that the 17 has been turned into a decimal, in other words, it\u2019s now 17.000. Why? Because the 21.479 decimal has got three numbers after the decimal point. They\u2019re called decimal places. So the 17 has three decimal places as well. Notice also the 3.04 has got an additional decimal place to make it three decimal places as well. So notice how the columns line up, each of the decimals has now got three decimal places.\r\n\r\nThe other thing to notice are the decimal points have been lined up in one column, and that\u2019s most important. So let\u2019s go ahead and do the addition. Starting with the right-hand column, zero plus zero plus nine is nine. The next column, four plus seven is 11, which is one and we carry the one shown in red and put it next to the four. The next column, zero plus zero plus four plus one is five. Now moving onto the other side of the decimal points, three plus seven plus one is 11, which is one and we carry the one as we did with the other one. In other words, we now have to add one plus two plus one which gives us four. So the answer to this decimal addition is 41.519.\r\n\r\nNow do some questions for yourself. The answers to these questions are on the next slide. Thank you for watching this short movie.\r\n \r\n\r\nI\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on subtracting decimals.\r\n\r\nLet\u2019s start with an example. 825.709 take away 272.456. Make sure at the outset that you line up the decimal points. Now let\u2019s carry out the subtraction. Nine take away six is three. The next column isn\u2019t so easy, we\u2019re saying zero take away five. In this case we have to add a one which effectively is a 10 to the top number. So we\u2019re now saying 10 take away five is five. Now we carry the one at the bottom number, in other words the bottom number is four plus one which is five. So the subtraction now is seven take away five which is two. Now moving over to the left hand side of the decimal point, five take away two is three. The next column is two take away seven, which as before with a zero we can\u2019t do, so what we do, we add the one or effectively the 10. So the 10 plus the two on the top becomes the number 12. So we\u2019re saying 12 take away seven which is five. As before make sure you carry the one on the bottom, so the number on the bottom now becomes two plus one which is three. So we say eight take away three is five. So the subtraction to this problem is 553.253.\r\n\r\nNow do some questions for yourself. The answers to these problems are on the next slide. Thanks for watching this short movie.\r\n\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on multiplying decimals.\r\n\r\nLet\u2019s start by multiplying decimals by 10, 100 and 1000. So let\u2019s start with the decimal 5.2 times 10. When we multiple by 10 notice that we have one zero, and when you multiply by 10 you simply move the decimal point one place to the right. In other words, 5.2 times 10 becomes 52.0, or just 52. In other words, the rule is move the decimal point one place to the right for each zero. Another one, 62.7 times 100. Notice two zeros, so we simply move the decimal point two places to the right, so this answer becomes 6270. Finally, let\u2019s move 3.14 times 1000. Here we move the decimal point three places to the right because we have three zeros, 1000 has got three zeros. So the answer is 3140.\r\n\r\nNow let\u2019s look at a multiplication which does not involve multiplying by 10, 100 or 1000. We have 8.5 times four. It\u2019s set out as shown with the four underneath the five on the right hand side. Notice also that there is no decimal point in this multiplication. We\u2019ll come back to that in a moment. So we start by multiplying the right hand columns, four times five is 20, zero under the four, carry the two, which I placed in red above the eight. Now I say four times eight is 32 and now I add the two to the 32 to give us 34. So we now have an answer with 340. Now we have to decide where to place the decimal point, and what you do now is to go up to the question again and count up how many numbers there are after the decimal point. Here we have one number, the five, which is after the decimal point. There are no numbers after the four. In other words we have one decimal place. So what we do is we move back from the right hand side one decimal place also. So the decimal point is now between the four and the zero. So our answer is 34.0.\r\n\r\nNow let\u2019s look at a slight more complicated multiplication, 4.9 times 2.8. First of all set it out as shown. Note again there are no decimal points, 49 times 28. First of all we start on the right hand side by multiplying eight times nine which is 72, carry the seven, shown in red at the top. Multiply eight times four which is 32, add the seven and we get 39. So our first row is 392. Move onto the next row. Here we\u2019re multiplying by 2, so first of all we place a zero in the right hand column. Now we multiply two by nine which is 18, carry the one shown at the top. Two times four is eight plus one is nine. Now we have our second row 980. All we need to do now is add up the two rows. Two and zero is two, nine and eight is 17, carry one, nine and three is 12 plus one is 13. Finally we decide where to place the decimal point. Remember the rule, how many decimal places are there in the question. There are two, there\u2019s a nine and there\u2019s an eight, so there are two decimal places, which mean that we must move the decimal point back two places from the right hand side. In other words, the answer to this question is 13.72.\r\n\r\nNow try some questions for yourself. The answers to these questions are on the next slide. Thank you for watching this short movie.\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on dividing decimals.\r\n\r\nLet\u2019s start by dividing decimals by 10, 100 and 1000. Here are some examples. Let\u2019s look at the first example, 53.7 divided by 10. Notice here that 10 has got one zero. So what we do when we divide by 10 is to move the decimal point one place to the left. In other words, the decimal point is now between the five and the three, giving us an answer of 5.37. Second example, 462.9 divided by 100. Here 100 has got two zeros, so we move the decimal point two places to the left. So the decimal point is now between the four and the six, giving us an answer of 4.629. Finally, 3782 divided by 1000. Notice here that 1000 has got three zeros, so we move the decimal point three places to the left, giving us a decimal point between three and seven, so the answer now is 3.782.\r\n\r\nNow let\u2019s look at some decimal divisions where we\u2019re not dividing by 10, 100 or 1000. Here\u2019s a fairly straight forward one, 8.4 divided by four. Setting it out as shown, four into 8.4, notice I\u2019ve lined up the decimal points to begin with. Four into eight goes two, and four into four goes one, giving us an answer of 2.1.\r\n\r\nWhat happens if we\u2019re not dividing by a whole number? Here\u2019s an example, 8.48 divided by 0.4. Notice 0.4 is not a whole number. What we do is to turn that 0.4 into a whole number and to do that we have to move the decimal point one place to the right. In other words, instead of the decimal point between zero and four, it\u2019s now after the four, or just four. But what we do to one decimal we have to do the other decimal, the 8.48 has to be changed as well. So moving the decimal point one place to the right give the new decimal 84.8. So we now have 84.8 divided by 4. So now we go ahead and carry out the division, four into 84.8. Four into eight goes two, four into four goes one. Notice the decimal points are lined up, and finally four into eight goes two, giving us an answer of 21.2.\r\n\r\nNow let\u2019s extend this decimal division to a slightly harder problem, 145.25 divided by 0.25. Again notice the problem. The problem is 0.25 is not a whole number. So we first of all have to turn that into a whole number. So what we do is move the decimal point two places to the right, giving us 25. Similarly, because we\u2019ve done that to the 0.25, we must move the decimal point two places to the right for our 145.25, giving us an answer of 14525. So what we\u2019re carrying out now is a division 14525 divided by 25. So let\u2019s go ahead and carry out that division 25 into 14525. 25 into one won\u2019t go, 25 into four won\u2019t go, 25 into 145 goes five times and that gives us a remainder of 20. So we place the 20 next to the two. So we\u2019re now dividing 25 into 202, which goes eight times and that gives us a remainder of two, which we place next to the five. So we\u2019re now saying 25 into 25 goes one exactly. So the answer to this question 145.25 divided by 0.25 is 581.\r\n\r\nNow do some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n \r\nHi, I\u2019m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on converting between fractions and decimals.\r\n\r\nLet\u2019s start by converting a decimal into a fraction. Here\u2019s the decimal 0.455. The first thing to notice about this decimal is that there are three digits after the decimal point. The first digit is the 10th, the second digit the 100th, and the third digit is the 1000th. It\u2019s the third digit you look at because the 1000th is the number that you place on the bottom of the fraction. In other word, what goes on the bottom of this fraction is 1000. What goes on the top is the decimal without the decimal point, in other words it\u2019s 455. So 0.455 as a fraction is 455 over 1000. Now you look to see if anything cancels top and bottom. And as you can see here, you can divide both top and bottom by five. Five into the top goes 91. Five into 1000 goes 200. So 0.455 as a fraction is 91 over 200.\r\n\r\nNow let\u2019s look at a whole number decimal. Here\u2019s the decimal, 3.75. As before look at the digits, the first digit before the decimal point is a one. The digit after the decimal point is a 10th. The second digit is the 100th. Concentrate on that last digit, which is a 100th, in other words, the denominator of your fraction will be 100. The top of the fraction will be the decimal without the decimal point, in other words, 375. So 3.75 is 375 over 100. Now look to see if anything cancels and as you can see fives will divide into both top and bottom, 75 and 20. But also notice that 75 over 20 will cancel again, again five will divide into both top and bottom, giving us 15 over four. Turn that improper fraction into a mixed fraction, in other words, 15 over four is three and three quarters. So 3.75 as a decimal is three and three quarters as a fraction.\r\n\r\nNow let\u2019s convert a fraction into a decimal. Here\u2019s the fraction, two and three 8ths. The first thing we do is to convert this mixed fraction into an improper fraction, so two and three 8ths is 19 over eight. 19 over eight means 19 divided by eight, so when we turn it into a decimal we have to use the process as shown. Notice that the eight goes outside of the division and 19 goes inside. So now we carry out the division, eights into 19 goes two times. Notice here we have two times with a remainder of three. Now if there\u2019s a remainder then we have to place a decimal point and zeros after that decimal point. So carrying on, because we remainder three, we place the three next to the zeros, the number becomes 30. So we divide eight into 30, goes three times, three eights are 24, remainder six. Place the six next to the zero. The number is now 60. Eight into 60 goes seven, eight sevens are 56, remainder four. So the new number is 40. Eight into 40 goes five times exactly. Notice we keep going until we do not have a remainder. So our fraction two and three 8ths is 2.375 as a decimal.\r\n\r\nNow do some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.\r\n","n":0.017},[{"v":"Measurement","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Decimals","i":1,"n":1},{"v":"Arithmetic","i":0,"n":1}]]},{"i":454,"$":[{"v":"Indices, surds and logarithms","n":0.5},{"v":"Indices, surds and logarithms extend our understanding of numbers and their relationships. Use these resources to enhance your comprehension of these powerful mathematical tools.\r\n\r\n\r\n\tIndices\r\n\tSurds\r\n \tFractional indices\r\n \tLogarithms\r\n\r\n\r\n\r\n\t\r\n ","n":0.192},[{"v":"Surds","i":3,"n":1},{"v":"Maths","i":2,"n":1},{"v":"Logarithms","i":1,"n":1},{"v":"Indices","i":0,"n":1}]]},{"i":455,"$":[{"v":"Equation solving","n":0.707},{"v":"Solving algebraic equations involves finding the values of variables that satisfy mathematical statements. Build on your foundational understanding of algebra by tackling how to solve basic equations.\r\n\r\n\r\n\tSolving linear equations\r\n\tSolving simultaneous equations\r\n\r\n","n":0.18},[{"v":"Maths equations","i":2,"n":0.707},{"v":"Maths","i":1,"n":1},{"v":"Equations","i":0,"n":1}]]},{"i":456,"$":[{"v":"Algebra","n":1},{"v":"Algebra focuses on understanding and manipulating mathematical symbols and expressions. Here, you'll explore key techniques such as simplification, factorisation, and rearranging formulas, which are essential for solving equations and understanding mathematical relationships.\r\n\r\n\r\n\tAlgebraic operations\r\n\t\tAlgebraic substitution\r\n\t\tExpansion\r\n\t\tFactorisation\r\n\t\tAdding and subtracting algebraic fractions\r\n\t\tMultiplying and dividing algebraic fractions\r\n\t\tTransposing formulas\r\n\t\tTransposing formulas with brackets and fractions\r\n\r\n","n":0.144},[{"v":"Maths equations","i":3,"n":0.707},{"v":"Functions","i":2,"n":1},{"v":"Arithmetic","i":1,"n":1},{"v":"Algebra","i":0,"n":1}]]},{"i":457,"$":[{"v":"Writing cohesively","n":0.707},{"v":"Good academic writing has a clear flow. Each phrase or sentence is linked to those that come before and after. Let's explore what makes academic writing cohesive.\r\nWrite cohesively by using:\r\n\r\n \tOrganic structure: used as the basis for paragraph organisation.\r\n \tCohesive devices: repetition, substitution, and linking words are then used to further enhance the flow.\r\n \tAvoid ambiguity: good sentence structure prevents confusion.\r\n\r\nThese strategies help to build a 'lexical chain', which creates a natural flow.\r\n\r\nOrganic structure\r\nAn organic structure is used as the basis for paragraph organisation. Each new point relates either directly to the last part of the previous point, or back to the main idea.\r\n\r\nRead the following paragraph, then open the Organic structure example folder to see how each sentence complements other sentences in the paragraph.\r\n\r\nComplete paragraph\r\n\r\n\r\nCohesive devices\r\nHere are some of the cohesive devices that can be used to create greater cohesion in your writing. \r\nKeyword and phrases\r\nKeywords and concepts should connect the topic and linking sentences in each paragraph to the next. This can be done through the repetition of keywords or terms, and by using synonyms (words and phrases that have a similar meaning).\r\nLinking words and phrases\r\nUsing linking words and phrases explains to your reader the connection between your ideas. For example, some linking words and phrases show contrast (e.g. however, in spite of), while others show addition (e.g. additionally, moreover) and others introduce examples (e.g. to illustrate, for example).\r\n\r\nFor more information, visit the Linking words page.\r\nDemonstrative adjectives\r\nThis, that, these and those are used to refer to something that has previously been mentioned. For example:\r\n\r\nPemberton (2023) describes a system that could be implemented into the education of high school students. This system has many benefits.\r\n\r\nRelative pronouns\r\nWords such as which, that, who and where are relative pronouns. They are used to link parts of a sentence together, creating cohesion in the sentence. In the following example, the first part of the sentence mentions micromanagement. Later, which also refers to micromanagement:\r\n\r\nMicromanagement is a management style which can lead to inefficiencies and high turnover rates in an organisation.\r\n\r\nExample of language features and cohesion\r\nIn the following paragraph, the language features that give the paragraph cohesion are highlighted.\r\n\r\nAvoid ambiguity\r\nBy using clear sentence structures with well-defined subjects and objects, and providing specific information, you can remove ambiguity from your writing, which improves the flow and meaning of your text.\r\nExample 1: Clarify pronouns\r\nPronouns are often used to replace nouns or noun phrases. However, sometimes who or what the pronoun refers to can be confusing, particularly if more than one noun or noun phrase has been used in the sentence. Reread your sentences to check for clarity.\r\n\r\nIn the original sentence, it is not clear who he refers to; is it the manager or the staff member? The revised sentence uses a more specific term instead of he to avoid ambiguity.\r\nExample 2: Who did what?\r\nThe passive voice is frequently used in academic writing to avoid mentioning who or what performed an action. However, in some cases it is necessary to mention the agent to avoid ambiguity or confusion.\r\n\r\nExample 3: Be specific\r\nAvoid vague terms like many, a lot, few, etc. Give your reader more specific information so that they won't misinterpret your meaning.\r\n\r\n\r\n\r\n\r\n \r\n\r\nIt is useful to distinguish two types of [repetition] learning strategy [end repetition] : language [repetition] learning strategies [end repetition] and skill [repetition] learning strategies [end repetition]. [substitution] The former [end substitution], as defined by Tarone (1980), are concerned with the learners' attempts to master new linguistic and sociolinguistic information about the target language. [substitution] The latter [end substitution] are concerned with the learners' attempts to become skilled listeners, speakers, readers, or writers. Writers, [linking words] for instance [end linking words], employ a variety of [repetition] strategies [end repetition] for exploiting teachers' feedback on [substitution] their [end substitution] written compositions (see Cohen 1987; 1991). [linking words] However [end linking words], the distinction is not an easy one and the literature on [repetition] learning strategies [end repetition] does not always distinguish clearly between [substitution] these [end substitution] two types. Cohen (1990:15), [linking words] for instance [end linking words], refers to [repetition] learning strategies [end repetition] directed at the 'language skill' of 'vocabulary learning', [linking words] although [end linking words] [substitution] this [end substitution] is clearly an aspect of linguistic knowledge.\r\n(Source: Adapted from Ellis, R. (1994). The study of second language acquisition. Oxford: O.U.P. pp.530-1)\r\n","n":0.037},[{"v":"Sentences","i":3,"n":1},{"v":"Style","i":2,"n":1},{"v":"Paragraphs","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":458,"$":[{"v":"Becoming a critical thinker","n":0.5},{"v":"Critical thinking doesn't just happen. It takes time to develop and practice the necessary combination of knowledge, skills and attitude. Let's take a look at how we can develop our critical thinking skills.\r\n\r\n\r\n\r\nAs a critical thinker, you will:\r\n\r\n \task questions\r\n \tanalyse and evaluate information/theories/practice\r\n \tbe aware of your own biases, beliefs and values.\r\n\r\nIn developing your skills as a critical and independent thinker at university, the following traits are important.\r\n\r\n \tDevelop a critical attitude\r\n \tDevelop a critical approach to texts.\r\n\r\nDeveloping a critical attitude\r\nIt takes a conscious effort to develop a critical attitude. Before you can think critically, you need to be aware of what's stopping you. Attitudes that prevent us from thinking critically include the following:\r\n\r\n \tI want you to tell me what I need to know, not figure it out myself.\r\n \tI 'know' what is right so I don't have to think about it.\r\n \tI don't usually try to think about why things happen the way they do.\r\n \tI'm not an expert but this other person is, so I can't say anything.\r\n \tI don't like to be criticised.\r\n \tI don't want to get into an argument.\r\n \tI don't want to be disrespectful.\r\n\r\nReflecting on your learning is a useful activity that can help you develop a more critical attitude. It allows you to think deeply about your experiences of learning and the attitudes you bring to it. Critical or academic reflection is a common assessment task at university.\r\n\r\n\r\n","n":0.065},[{"v":"Study skills","i":2,"n":0.707},{"v":"Critical thinking","i":1,"n":0.707},{"v":"Critical reading","i":0,"n":0.707}]]},{"i":459,"$":[{"v":"Propositional writing","n":0.707},{"v":"Propositional writing sets out a plan of action or design intent. It presents clearly what the project or idea is, outlines why certain actions or design elements are significant (the reasoning behind the project) and how they can be achieved.\r\n\r\nWhat is an abstract?\r\nAn abstract is a short piece of writing which relates to a larger body of work or writing. It can be written as a reflective summary after a larger work, or as propositional abstract to articulate the intent and vision of a current or future work.\r\n\r\nThe abstract should outline what you are intending to address, how you will do this and why you are doing this (why it is important).\r\n\r\nKeywords should be used to express and build the concepts or themes that run through a given framework. These themes give structure to your work and describe the ways you have used them.\r\n\r\nStructure\r\n\r\nThis example demonstrates how the writer has analysed a specific site, using a Modernist framework. The key words develop the concepts and themes that are outlined in the abstract and the drawing.\r\n\r\nThe example below uses the RMIT Library Harvard referencing guide.\r\n\r\n\r\nReference\r\nTreib M (1993) \u2018Axioms for a modern landscape architecture\u2019, in M Treib (ed) Modern landscape architecture: a critical review, MIT Press, Cambridge.\r\n\r\n \r\nAmid growing populations and expanding urban sprawls, it is essential for logical, functional design to supersede a merely aesthetic, classical approach to landscape architecture. [What] By examining the intersection site using modernist principles outlined in Marc Treib\u2019s (1993), Axioms for a modern landscape architecture, latent, modernist, spatial arrangements embedded within the site are revealed and drawn out.[End what]\r\nThe most significant element to consider is the circulation of people as they move through the site and their reasons for doing so.\r\n[How] Using a rational approach, the lines, spaces and forms of the site are selected to emphasise them.\u2018Form [is] derived from the accommodation of the program\u2019 (Treib 1993:62) to include the specific requirements of the client, the site and its uses. The layout of the plan is developed and the spaces between are formed. The hierarchy of circulation is represented through tonal variation; the darkest tone relating to the dominance of the tramline, and the lightest pertaining to the pedestrian traffic. Identification and representation of this rational hierarchy of separated programs indicates an inherent order within the site, as the modernist landscape\r\n[Why] \u2018ultimately concerns making outdoor places for human use\u2019 (Treib 1993:55).\r\n\r\n[How] The absence of a central axis in the site allows for an omnidirectional use of the space, accommodating both formal and informal occupation. All elements of the site follow modernist principles. Even the trees are positioned rationally; they appear in isolation rather than in naturalistic clusters, to heighten their sculptural qualities. This is a clear\r\n[Why] break from a formal, classical style as [Why] Tunnard (cited in Treib 1993:56) explains, [Why] \u2018selection, not massing for picturesque effect is the requirement\u2019 of an ordered, modernist approach.\r\n\r\nBy adopting tools and techniques from modernist principles, the intersection site is reordered and redrawn in plan,\r\n\r\nReference list\r\nTreib M (1993) \u2018Axioms for a modern landscape architecture\u2019, in M Treib (ed) Modern landscape architecture: a critical review, MIT Press, Cambridge.\r\n\r\n","n":0.044},[{"v":"Essays","i":3,"n":1},{"v":"Design","i":2,"n":1},{"v":"Art","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":460,"$":[{"v":"Critical writing","n":0.707},{"v":"Critical writing involves analysis and review, supported by evidence. In design, it may take the form of an essay, visual analysis or journal article.\r\n\r\nThe process of writing an essay incorporates a number of elements: analysing the task, research, developing an organised structure through mind mapping, writing a first draft, revising and rewriting.\r\n\r\nPlanning\r\nPlanning is critical to the success of your essay. It helps guide your writing; allows you to focus on your topic and shapes the direction of your research.\r\n\r\nThe process generally follows the steps in the diagram below but there is overlap between stages and at times you will return to a previous stage as you refine and focus your argument.\r\n\r\n\r\n\r\n\r\n\r\nYour first written draft will generally shape your essay. Look back at the essay topic; have you responded to all the elements you identified as important? Does the essay \u2018flow\u2019? Are your points relevant and do they contribute to your overall argument? You may ask for feedback from someone else as well. Redraft and refine your essay through critical editing.\r\n\r\nFinally, proofread for punctuation and grammar errors. Confirm that your referencing and citations are correct before submitting your essay.\r\nFor more information, visit the Essay planning tutorial.\r\nMind mapping\r\nMind mapping is a visual representation of the focus and structure of a task. The advantage of creating and developing a mind map is that it gives you a holistic view of the task and helps you identify relationships between ideas.\r\n\r\nThere are three stages in the process.\r\n\r\n\r\n\r\n\r\n\t\r\n\t\r\n\t\r\n\r\n\r\n\r\n\r\n\r\n The image features \"Wolfgang Weingart\" at the centre, circled in red. Surrounding it are various phrases and concepts related to Weingart's influence on typography and design. These include:\r\n\t\r\n\t\tUrbanism (post WWII)\r\n\t\tText as art\r\n\t\tInternational typography styles\r\n\t\tIterative design\r\n\t\tAdvances in technology\r\n\t\tPhototypesetting\r\n\t\tVietnam War\r\n\t \r\n\t\r\n\tLike the first image, this features \"Wolfgang Weingart\" at the centre, circled in red. The various phrases and concepts surrounding it are colour coded to group them together. A couple of topics are crossed out. \r\n\t\r\n\t \r\n\t\r\n\tAt the centre of the image is \"Wolfgang Weingart,\" with categories branching out, each representing a different aspect of his influence:\r\n\t\r\n    History (contextualise)\r\n        \r\n            Vietnam War\r\n            social environment\r\n            urbanism (post WWII)\r\n            opposing corporatisation\r\n            International typography styles\r\n            political change\r\n            Swiss typography in 50's & 60's\r\n            Swiss type\r\n        \r\n    \r\n    Example (describe)\r\n        \r\n            new media combinations\r\n            photolithography\r\n            woodtype\r\n            phototypesetting\r\n            colour\r\n            TM covers (1972 & 1973 key years, 15 covers)\r\n        \r\n    \r\n    Philosophy (analyse)\r\n        \r\n            move from anonymity to individualism in design\r\n            text as art\r\n            basic typographic / design principles\r\n            use relevant technology\r\n            experimentation\r\n            iterative design\r\n            rapid change\r\n        \r\n    \r\n    Education (contextualise)\r\n        \r\n            other teachers\r\n            Basel school\r\n            students of Weingart\r\n            schedule for gestaltung\r\n            encouraged experimentation (in his students)\r\n        \r\n    \r\n    Concepts (analyse)\r\n        \r\n            advances in technology\r\n            type aesthetics\r\n            magazines\r\n            aesthetic archetypes\r\n            graphic design and advertising\r\n        \r\n    \r\n\r\n\t\r\n\t","n":0.047},[{"v":"Essays","i":2,"n":1},{"v":"Design","i":1,"n":1},{"v":"Critical writing","i":0,"n":0.707}]]},{"i":461,"$":{"0":{"v":"Statistics and probability","n":0.577},"2":[{"v":"Statistics","i":1,"n":1},{"v":"Maths","i":0,"n":1}]}},{"i":462,"$":[{"v":"Artist statement","n":0.707},{"v":"Writing an artist statement is an essential professional skill for any creative arts or design practice. It can help to identify what is important in your creative work.\r\nHere you will find information and learning activities that will help you to:\r\n\r\nUnderstand what an artist statement is and why it's useful.\r\nIdentify different types of artist statements and when to use them.\r\nAnalyse examples of real professional artist statements from our RMIT creative communities and art collections.\r\nPractice how to write an artist statement in your own words.\r\nCommunicate your creative practice to others, develop your writing skills and find your 'voice'.\r\n\r\nWhat is an artist statement?\r\nAn artist statement is a short text that explains your creative work in your own words. Your artist statement can help others to understand your work and can be a way to engage and connect with your audience.\r\n\r\nImagine your artist statement as something that could represent you if you weren't present to explain your work.\r\n\r\nYour artist statement will change over time as you evolve your creative practice \u2013 you will need to write it more than once! You will also need different types of statements for different purposes.\r\nWho needs one and why?\r\nWho needs an artist statement? Anyone who works in creative fields such as:\r\n\r\n \tvisual & creative arts\r\n \tgraphic & product design\r\n \tfashion & textile design\r\n \twriting & literary arts\r\n \tmusic & performing arts\r\n \tscreen & digital media\r\n \tinterior & architectural design\r\n\r\nAn artist statement is useful for a portfolio of work, as part of a creative project, or as part of an application process for a job, project, exhibition, performance, grant, or other professional opportunity.\r\n\r\nMany artists and designers now have an online presence with websites and social media platforms showcasing their work. This is a useful way to promote yourself and form communities of practice. A relevant and current artist statement can be part of your brand and might be one of the first things people encounter when they view your work online, so it's important to know how to write about your practice in an accurate and engaging way.","n":0.054},[{"v":"Professional writing","i":3,"n":0.707},{"v":"Design","i":2,"n":1},{"v":"Artist statement","i":1,"n":0.707},{"v":"Art","i":0,"n":1}]]},{"i":463,"$":[{"v":"Vectors","n":1},{"v":"Vectors are quantities that have both magnitude (size) and direction. This branch of maths is fundamental to physics and engineering to represent physical quantities that have a direction. Use this resource to get the basics down.\r\n\r\n\t\tIntroduction to vectors\r\n\t\tResolution of vectors\r\n\t\tScalar or dot product\r\n\t\tProjection of vectors\r\n\r\n\r\n","n":0.149},[{"v":"Vectors","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":464,"$":[{"v":"Numerical methods","n":0.707},{"v":"Numerical methods are a collection of techniques used to solve mathematical problems that are difficult or impossible to solve with exact solutions. These worksheets are filled with information that will improve your skills in numerical methods.\r\n\r\n\r\n \tNM1 Newton\u2019s method\r\nSome equations cannot be solved using algebra or other mathematical techniques. In this case we need to use numerical methods. Newton\u2019s method is one such method and allows us to calculate the solution of f(x)=0.\r\n \tNM2 Simpson\u2019s rule\r\nSome integrals cannot be evaluated in terms of the rules of integration or elementary functions. Simpson\u2019s rule is a numerical method that calculates a numerical value for a definite integral.\r\n \tNM3 The trapezoidal rule\r\nThe trapezoidal rule is a numerical method that calculates a numerical value for a definite integral. Some integrals cannot be evaluated in terms of the rules of integration or elementary functions.\r\n\r\n","n":0.085},[{"v":"Methodology","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":465,"$":[{"v":"Matrices","n":1},{"v":"Matrices are used in a wide range of fields and applications to represent and manipulate data in a structured way. Begin your journey with these resources.\r\n\r\n\t\tIntroduction to matrices \r\n \t\tAddition and subtraction of matrices\r\n \t\tMatrix multiplication\r\n\t\tDeterminant of a matrix\r\n\r\n\r\n","n":0.16},[{"v":"Matrices","i":1,"n":1},{"v":"Maths","i":0,"n":1}]]},{"i":466,"$":[{"v":"Laplace transforms","n":0.707},{"v":"Transforms are another means of solving some differential equations that may prove too difficult to solve using other methods.\r\n\r\nDownload the laplace transforms worksheets to improve your skills in these areas.\r\n\r\n \tLT1 Basic definition of laplace transforms (PDF)\r\n \tLT2 Table of transforms (PDF)\r\n \tLT3 Solving differential equations (PDF)\r\n \tLT4 Convolutional theorem (PDF)\r\n \tLT5 Electrical circuits (PDF)\r\n","n":0.135},[{"v":"Maths","i":3,"n":1},{"v":"Laplace transforms","i":2,"n":0.707},{"v":"Electrical circuits","i":1,"n":0.707},{"v":"Differential equations","i":0,"n":0.707}]]},{"i":467,"$":[{"v":"Complex numbers","n":0.707},{"v":"Complex numbers are a group of numbers that help us to get mathematical solutions where real numbers (which include positive and negative, counting numbers, fractions and decimals) just can\u2019t work. Use this resource to get started with complex numbers.\r\n\r\n\tIntroduction to complex numbers\r\n\r\n","n":0.154},[{"v":"Maths equations","i":2,"n":0.707},{"v":"Complex numbers","i":1,"n":0.707},{"v":"Maths","i":0,"n":1}]]},{"i":468,"$":[{"v":"Academic word list 1 & 2","n":0.408},{"v":"Read or download the sublists 1 & 2 from the academic word list. If you are unsure of the meaning of any of the words, look them up in a dictionary and find examples of how they can be used in academic writing.\r\n\r\nDifferent activities focus on:\r\n\r\n    word meaning\r\n    grammatical word form\r\n    collocation (words that typically pair with the target word)\r\n    spelling\r\n\r\n\r\n \r\n\r\nActivities\r\nNow try the activities which will help you become familiar with the words and expand your vocabulary.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.113},[{"v":"Vocabulary","i":1,"n":1},{"v":"Academic word lists","i":0,"n":0.577}]]},{"i":469,"$":[{"v":"Group work: Wrapping up","n":0.5},{"v":"This stage is often the most complex as the team brings together all its information and work into the finished product. Here are some steps you can take to wrap up your assignment.\r\n\r\nAssign specific production tasks to members\r\nHave one or two group members take responsibility for each task:\r\n\r\n \tensuring the content meets the task requirements\r\n \tcreating a cohesive 'voice' throughout the work\r\n \tmaintaining a consistent visual style (e.g. the same formatting, heading styles, colour scheme, and layout throughout).\r\n\r\nReview the final product\r\nAllocate enough time to review and edit the final submission.Even if one or two group members are completing the final draft, the whole team should review the work, including proof-reading, before submitting.\r\nDon't forget to reference\r\nEnsure all ideas and information are referenced correctly. For more information, visit Group work and academic integrity.","n":0.087},[{"v":"Wrapping up","i":1,"n":0.707},{"v":"Group work","i":0,"n":0.707}]]},{"i":470,"$":[{"v":"Group work: Doing the work","n":0.447},{"v":"Once you have built a strong foundation for group work by using the strategies previously outlined, it's time to get down to work.\r\nTo meet the task requirements, group members will be busy with different responsibilities at the same time. Focus on communicating and collaborating to successfully manage yourselves and the task.\r\n\r\nLet's explore some strategies you can use.\r\n\r\nHold regular meetings\r\n\r\n \tUse an agenda to keep meetings focussed on the task.\r\n \tActively contribute to discussions using inclusive and respectful language.\r\n \tBe open to different ideas and different perspectives.\r\n \tEnsure everyone has a say. Listen when others are speaking and encourage the quiet members to contribute.\r\n \tCheck and measure progress against the timeline.\r\n \tMeetings are an opportunity for the group to reconnect, review progress, provide feedback and make any necessary changes.\r\n\r\nGive each other constructive feedback\r\n\t\r\n\tGive positive feedback to motivate and encourage team members. Always start with genuine, positive feedback: say what you appreciate about their work before offering critical feedback.\r\n\tBe sensitive to how others may respond to criticism. Be nice.\r\n\tUse the S.M.A.R.T. principles to focus your feedback on the task:\r\n\t\t\r\n\t\t\tSpecific: Focus on actual examples of the work or performance. General or vague statements are unhelpful.\r\n\t\t\tMotivating: Give reasons to continue or change behaviour. Why is it important or useful?\r\n\t\t\tAction-oriented: Align actions with the task requirements. If a change is required, what needs to happen?\r\n\t\t\tRealistic: Is the change something that the person can do? Do they need help? Who is going to help them?\r\n\t\t\tTimely: Provide feedback when it is requested or required. Waiting until the end of the task may affect everyone in the group and could allow a small problem to escalate.\r\n\t\t\r\n\t\r\nDeal with challenging behaviours and conflict\r\n\r\n \tIdentify and acknowledge any conflict or dissatisfaction early. Disagreements arise when members lack clear directions or feel their knowledge and skills are ignored or criticised.\r\n \tCommunicate any problems to the whole group and try to resolve the problem quickly, as a group. You will need to negotiate and compromise.\r\n\r\nThe diagram shows how issues can often build upon one another.\r\n\r\n\r\nManage group dynamics\r\nGroups are made up of people with different expectations. For example, some students are happy just to pass while others aim for high marks.\r\n\r\nSometimes these differences can cause tension. This is normal \u2013\u00a0all groups experience highs and lows when working on complex projects over a period of time.\r\n\r\nDo you recognise any of these people and their behaviours and attitudes in your group?\r\n\r\n \tTeam player: Collaborative, communicative, committed.\r\n \tHitchhiker: Along for the ride, does minimal work but shares the results.\r\n \tSuperhero: Comes to the rescue by taking over every task. A group of one!\r\n \tZombie: Unengaged and unmotivated. May disappear for weeks or completely.\r\n\r\nYou are always going to encounter diversity within groups. The best way to ensure an effective group work experience is to develop a strong process with your group members.\r\n\r\nSeek help from your lecturer or other academic support services if your group is unable to agree on a solution.\r\nManaging the task\r\nUse agendas in meetings\r\n\r\n \tUse the agenda as a checklist to ensure that all the task criteria have been met.\r\n \tRecord decisions for actions, roles and responsibilities to provide clear direction to group members.\r\n \tMeasure agenda actions against the timeline to help the team stay on track. Be prepared to adjust the timeline as the project progresses.\r\n\r\nManage tasks and resources\r\n\r\n \tOrganise information so that it is shareable and accessible to everyone. There are many online collaborative tools and apps to choose from. Remember to keep it simple!\r\n \tCreate the final product. Your final submission may have specific requirements. Delegate responsibilities among group members for completing the work. Follow the strategies for Wrapping up.\r\n\r\n \r\nA pyramid with three levels. They are:\r\n\r\n\tFrustration: \"Why won't they listen to me?\"\r\n\tUnclear responsibilities: \"Am I supposed to do this? I don't know what I have to do.\"\r\n\tPoor planning: \"What do I do next? Are we working together? Separately?\"\r\n\r\n","n":0.04},[{"v":"Time management","i":2,"n":0.707},{"v":"Study skills","i":1,"n":0.707},{"v":"Group work","i":0,"n":0.707}]]},{"i":471,"$":[{"v":"Group work: Communicating and sharing","n":0.447},{"v":"How well your group communicates with each other affects how you work together and how task information is stored, managed and shared. Good communication is an advantage for any group. Let's learn how to foster effective communication.\r\n\r\nAgree on how to work together\r\n\r\n \tSuccessful groups make decisions as a group and expect all members to contribute equally to the task.\r\n \tShare your contact details and agree on the best way of keeping in touch.\r\n \tMeet face to face where possible to make decisions and work together effectively.\r\n\tWrite a group charter to outline how you will work together, what expectations members have and how you will approach the task. Group members need to agree on the charter and sign it.\r\n\r\nHere's an example of a group charter:\r\n\r\n\r\n\r\n\r\nSharing ideas\r\nHopefully, in your first meeting you will have decided how you will communicate as a group. Here are some tips for what can be covered in your next meetings.\r\n\r\nYou should:\r\n\r\n\treport on your work: what you have done, what you have learned, what you still need to do\r\n\taddress problems and talk about possible solutions (product and process problems) as they emerge\r\n\tensure everyone has a chance to speak without interruption\r\n\tmake decisions about individual responsibilities for actions; e.g. the development of the group's points for the report / presentation\r\n\taddress the issue of members not contributing ideas, if relevant, and stress that all opinions are needed and valued.\r\n\r\n\r\n\r\nSet an agenda\r\nBefore the meeting, the note taker can send out a request to group members to suggest any topics that need to be discussed. They can then create an agenda that that has a description of each item that needs to be discussed by the group, a field for where the name of the person/people who is responsible for the task and space the actions that need to be completed.\r\n\r\nHere's an example of a meeting agenda:\r\n\r\n\r\n\t\r\n\t  \r\n\t\t\r\n\t\t  Item\r\n\t\t  Responsibility\r\n\t\t  Action\r\n\t\t\r\n\t  \r\n\t  \r\n\t   The task\r\n\t\t  All members\r\n\t\t \r\n\t\t Brainstorm the task: each member articulates their understanding of the taskSummarise group understanding of the taskDistribute summary with other notes from meeting\r\n\t\t\r\n\t\t  Assignment part 1 and 2: information search\r\n\t\t  All members\r\n\t\t  \r\n\t\t  Assign two members to each sectionDiscuss search strategy databasesSet deadline\r\n\t\t\r\n\t\t\r\n\t\t\r\n\t\t  Attending meetings\r\n\t\t  Josefa (leader)\r\n\t\t  \r\n\t\t  Discussion of importance of attending meetingsDiscussion of how absence affects the groupReminder of the group charter\r\n\t\t\r\n\t\t\r\n\t\t  Discuss schedule for upcoming meetings\r\n\t\t  Lin (note-taker)\r\n\t\t  \r\n\t\t  Discuss group member avialability for the next weekSend meeting invites\r\n\t\t\r\n\t  \r\n\r\n\r\n\r\n","n":0.05},[{"v":"Plagiarism","i":1,"n":1},{"v":"Group work","i":0,"n":0.707}]]},{"i":472,"$":[{"v":"Group work: Getting started","n":0.5},{"v":"Getting off to a strong start is one of the keys to successful group work. This page will guide you through the initial stages of group assignments.\r\nHere\u2019s a guide to help you through the initial stages:\r\nThe first steps\r\nInitial meeting goals\r\nIntroduce yourselves to each other. Discuss your strengths and hopes for the group.\r\nBrainstorm the task\r\nShare your understanding of the assessment goal. Aim to reach a consensus on the requirements.\r\nPlanning and organisation\r\nDevelop an agenda\r\nList topics, actions required, responsible members, and deadlines.\r\nDefine research tasks\r\nDecide on tasks and allocate them among members.\r\nCreate a timeline\r\nSet deadlines for specific tasks.\r\nCommunication and meetings\r\nDecide on communication tools\r\n\r\n \tOptions include Google Docs, a wiki, email, instant messaging, etc.\r\n \tExchange contact details.\r\n\r\nSchedule meetings\r\n\r\n \tArrange dates, times, and locations that suit everyone.\r\n \tRecord the meeting details.\r\n\r\nEstablish a group charter\r\nAgree on principles and expectations for group behaviour.\r\n\r\nAssign roles\r\nGroups usually have a number of set roles to help them fuction well and it is best to assign these roles at your first team meeting. They include:\r\n\r\n\r\n","n":0.078},[{"v":"Group work","i":1,"n":0.707},{"v":"Assessment","i":0,"n":1}]]},{"i":473,"$":[{"v":"Group work: Planning","n":0.577},{"v":"Good planning lays the groundwork for successfully completing group task. Let's find out how to make a strong start.\r\nStarting a group project is the same as starting an individual project. You will need to:\r\n\r\n \tanalyse the task\r\n \tdetermine what needs to be done\r\n \tcreate a plan for completing the task.\r\n\r\nIt is important to establish good working relationships at the beginning, with protocols and agreements for communication, behaviour and responsibilities. Good communication is especially important for online groups. Here are some strategies for starting strong.\r\nAnalyse the task as a group\r\n\r\n \tAgree on what the task is asking you to do. Review the task instructions, marking criteria/rubric and any other information you have.\r\n \tIdentify if the task has to be completed in a particular order (e.g. parts 3 or 4 rely on work done for parts 1 and 2). This determines how you might plan your timeline and delegate responsibilities for completing the assignment. If your group has any questions about the task, contact your lecturer as soon as possible.\r\n\r\nCreate a timeline\r\n\r\n \tAssess the time required for a group assignment. Groups often underestimate how long an assignment may take. A timeline identifies key milestones and helps you manage the workload to meet deadlines. Be realistic and be prepared to adapt your timeline as you progress.\r\n \tMembers of online groups may be working in different time zones. Factor this into your timeline.\r\n \tInclude enough time for research. Your research will be a large part of the task. You may also need to allow time for additional research during the drafting/production stage of the task.\r\n \tAllow time for drafting and producing the final product. Editing and proofreading is an important final stage of any task. Allow time for the group to review the overall product for accuracy and consistencys.\r\n\r\nThe example below shows a group's schedule over four weeks. Some group work tasks may last a whole semester.\r\n\r\n","n":0.057},[{"v":"Understanding the question","i":2,"n":0.577},{"v":"Presentations","i":1,"n":1},{"v":"Assignment planning","i":0,"n":0.707}]]},{"i":474,"$":[{"v":"Annotated bibliographies","n":0.707},{"v":"An annotated bibliography is a list of the information sources you have selected for your research topic (e.g. journal articles or book chapters) accompanied by comments (annotations) on how each item contributes to your research. \r\n\r\nThis page explains the purpose and structure of an annotated bibliography, what to include in an annotation, and provides a sample example.\r\nPurpose of an annotated bibliography\r\nRegardless of your unit of study, the purpose of an annotated bibliography is to:\r\n\r\n \tdevelop critical thinking skills by reviewing the issues and noting the main arguments in a particular research area.\r\n \tdevelop deeper research skills by engaging with individual sources.\r\n \tfind and reflect on the literature that has been published about your research topic.\r\n\r\nStructure of an annotated bibliography\r\nAn annotated bibliography includes two main sections:\r\n\r\n \ta reference (bibliographic information or citation).\r\n \tan annotation (description and comments on the source).\r\n        The annotation section contains:\r\n\t\t\r\n\t\t\ta summary or description \r\n\t\t\tan evaluation of each source\r\n\t\t\tan indication of how that source contributes to your research topic.\r\n\t\t\r\n\t\r\n\r\nWriting an annotation\r\nTo write an annotated bibliography first check the assignment guide instructed by your lecturer. Keep in mind that each course may have specific requirements for what annotated bibliographies contain and how they should be structured. Depending on the requirements, an annotated bibliography may include one or all the components below:\r\n\r\n \ta summary or description of each of the information sources\r\n \ta critical analysis of each of the information sources\r\n \ta reflection of how each source contributes to the overall research topic.\r\n\r\n\r\n ","n":0.064},[{"v":"Referencing","i":1,"n":1},{"v":"Annotated bibliography","i":0,"n":0.707}]]},{"i":475,"$":[{"v":"Topic sentences","n":0.707},{"v":"A well-written paragraph contains a topic sentence. A topic sentence controls the main idea discussed in a paragraph. It provides the reader with a summary of the key idea and the point of view the writer is expressing. It is often the first sentence as it acts as an introduction to what follows, making it easier for the reader to follow. \r\nFor the writer, topic sentences help you to ensure that each paragraph relates directly to the assignment topic. They help you to organise your ideas logically and to check that you have included appropriate and relevant support material. \r\nPoor example\r\n  \r\n\tRMIT programs enable students to develop successful and appropriate pathways. Programs at RMIT include degree and postgraduate programs in Business, Art & Design, Engineering and Education as well as many other areas. In addition RMIT offers many technical and vocational programs at Certificate and Diploma level. RMIT therefore has a lot to offer its students.\r\n\r\nWhat is this paragraph about?\r\nThe topic is that RMIT offers a range of programs. A possible topic sentence could be: \"RMIT offers a wide range of programs from Certificate to Post Graduate levels\".\r\nImproved example\r\n  \r\n\tRMIT offers a wide range of programs from Certificate to Postgraduate level. RMIT programs enable students to develop successful and appropriate pathways. Programs at RMIT include degree and postgraduate programs in Business, Art & Design, Engineering and Education as well as many other areas. In addition RMIT offers many technical and vocational programs at Certificate and Diploma level. RMIT therefore has a lot to offer its students\r\n\r\n","n":0.063},[{"v":"Topic sentences","i":3,"n":0.707},{"v":"Paragraphs","i":2,"n":1},{"v":"Essays","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":476,"$":[{"v":"Sentence fragments","n":0.707},{"v":"A sentence fragment is a sentence that is missing some of its parts. This page explains the three main reasons why a sentence may be incomplete.\r\n\r\n1. Missing subject\r\nOriginal: Becoming extinct because of rising sea temperatures.\r\n\r\nRevised: Phytoplankton could become extinct because of rising sea temperatures.\r\nSentences beginning with words like so, as, because, who, which, that, are often incomplete.\r\n2. Missing verb\r\nOriginal: Significantly, one particular form of Western Australian finch.\r\n\r\nRevised: Significantly, one particular form of Western Australian finch has decreased in numbers.\r\nVerbs are 'doing' words. They describe an action (e.g. run, walk, study) or a state of being(is, has been, will be). Each complete sentence should have at least one verb.\r\n3. Incomplete thought\r\nOriginal: In a recent article about loss of habitat due to climate change.\r\n\r\nRevised: In a recent article about loss of habitat due to climate change, Australian animals were shown to be particularly vulnerable.\r\nRead your sentences to see if you have expressed your point clearly. If the point is missing, it is likely that the sentence is incomplete.\r\nPoor example\r\n\r\n\r\n\r\nImproved example\r\n\r\n\r\n\r\n","n":0.077},[{"v":"Sentences","i":3,"n":1},{"v":"Sample writing","i":2,"n":0.707},{"v":"Grammar","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":477,"$":[{"v":"Art and design","n":0.577},{"v":"Resources to help you with critical and reflective writing, artist statements, essays and more.\r\n\r\n\r\n\r\n","n":0.267},[{"v":"Design","i":1,"n":1},{"v":"Art","i":0,"n":1}]]},{"i":478,"$":[{"v":"Academic reflection","n":0.707},{"v":"Reflective writing requires you to think deeply and write about an experience, event or course content.\r\nThis involves writing about:\r\n\r\n    What happened (positive or negative) and what you personally learned from the experience.\r\n    What it means, and how it changes the way you think or understand something.\r\n    What you can change or how you can apply the new learning in the future.\r\n\r\nReflective writing records the development of your insights and ideas. It focuses on a specific new understanding and its application to future practice.\r\nThe following diagram illustrates the reflective process of DIEP.\r\n\r\n\r\n Insight to transfer: top to bottom.\r\nDescribe: What happened? What did you learn?\r\nInterpret: What does the experience mean? Why?\r\nEvaluate: How valuable was the learning experience?\r\nPlan: How will you apply your learning?","n":0.091},[{"v":"Reports","i":2,"n":1},{"v":"Reflective writing","i":1,"n":0.707},{"v":"Critical incident","i":0,"n":0.707}]]},{"i":479,"$":[{"v":"Case studies","n":0.707},{"v":"This section will help you navigate your way through the process of reading, analysing and responding to a case study.\r\nCase studies describe real-life situations and events over a period of time. They may centre on a person, group, project or organisation. A case study assessment task requires you to analyse actual or potential problems and identify solutions. For this, you need to consider:\r\n\r\n \twhat is happening\r\n \twhy it's happening\r\n \thow to solve the problem.\r\n\r\nYou will need to use theories to develop solutions to practical situations.\r\n\r\nThere are three broad stages involved in a case study assignment:\r\n\r\n \tidentify key issues and problems from the case\r\n \tresearch then link the problems to relevant theory\r\n \tdevelop solutions that lead to actionable recommendations.\r\n\r\nStructure\r\nCase study analyses are generally presented in a report-style format. They are usually written for a professional audience (e.g. a client or manager) as well as your lecturer.\r\n\r\nA case study should have:\r\n\r\n \ta logical structure\r\n \tsections with headings and subheadings.\r\n\r\nCheck your assignment instructions carefully to see what structure you need to use. If you have any doubts, contact your course teaching staff.\r\nA business case study is usually highly structured, as shown in the following example.\r\n\r\n\r\nSee the Reports page for more information.\r\nStages in the process\r\nGenerally speaking, there are three stages to writing a case study. They are:\r\nAnalyse the task\r\n\r\n \tfocus on task instructions/questions and marking criteria/rubrics\r\n \tread the case in detail\r\n \tidentify key issues and problems.\r\n\r\nGo to Stage 1: Analyse the task\r\n\u00a0Research and link to theory\r\n\r\n \tresearch the topic and surrounding areas\r\n \tlink to theory - link problems to relevant theory, models or practice.\r\n\r\nGo to Stage 2: Research and link to theory\r\nDeveloping solutions\r\n\r\n \tevaluate and prioritise for most effective solutions, consider constraints/obstacles\r\n \tmake recommendations\u00a0- All recommendations need to be practical and actionable\r\n \tedit/proofread\r\n \tsubmit.\r\n\r\nGo to Stage 3: Developing solutions\r\n","n":0.058},[{"v":"Case studies","i":1,"n":0.707},{"v":"Assignments","i":0,"n":1}]]},{"i":480,"$":[{"v":"Reading skills","n":0.707},{"v":"Throughout your time at university, you will be required to read and engage with many articles, books and other information sources. On this page, you will find tips to help you manage your reading load by reading both efficiently and effectively.\r\n\r\nReading efficiently: surveying, skimming and scanning\r\nStudents can be easily overwhelmed by the amount of reading they are required to do at university. This short video gives you an overview of what makes an\u00a0efficient reader and shows you how to become one yourself.\r\n\r\n\r\n\r\nSurveying\r\nSurveying is a pre-reading technique that allows you to get a general idea of the contents of a book, article or other types of text.\r\n\r\nOnce you have surveyed, you can often decide whether to then read the text in detail, or if it doesn't have the information you need, move on to the next text.\r\n\r\nFor books and articles, you start by looking at:\r\n\r\n \tblurbs, abstracts and summaries\r\n \tintroduction and conclusion paragraphs\r\n \theadings and subheadings\r\n \tany images, graphs or charts and their captions\r\n \tchapter outlines or key points, if they have been provided\r\n\r\nWith your surveying done, you can then move on to skimming and scanning. In some of your courses, you may find that you are faced with long lists of required readings. The truth is, you are not required to read every book, every journal article, or every research paper word-for-word. Instead, you need to learn how to survey, skim, and scan texts. This will allow you to get the most out of texts, whilst saving you a lot of time in the long run.\r\n\r\nSo, how do I survey a text? Surveying texts is straightforward; in fact, you're probably already using this reading technique without realising. To survey a text, you read the title, the cover blurb, the contents page, and the index. Then, you quickly look over the chapter headings.\r\n\r\nHow do I skim? When you're skimming a text, you're looking for just the main ideas.\r\n\r\nHow do I scan? When scanning, you're looking for the important keywords in a text, you are quickly looking for the detail.\r\n\r\nLet's look at the example.\r\n\r\n\r\nSo, what next? So you've surveyed, skimmed, and scanned everything like you were supposed to. Now what? Often this is enough to get the main idea of a text, but sometimes you will need to read the text in detail. However, since you now know what the key ideas and keywords are, and you have a general idea of where to find this information, you'll find it easier to read only the stuff that's relevant.\r\n\r\nSo, let's recap:\r\n\r\nWhen we survey, we read the title, cover blurb, contents page, index, and chapter headings of a text.\r\nWhen we skim, we just look for the main ideas of a text.\r\nAnd, when we scan, we look for the keywords of a text.\r\n\r\n\r\n","n":0.047},[{"v":"Reading skills","i":3,"n":0.707},{"v":"Critical thinking","i":2,"n":0.707},{"v":"Critical reading","i":1,"n":0.707},{"v":"Active reading","i":0,"n":0.707}]]},{"i":481,"$":[{"v":"Plotting points on a graph","n":0.447},{"v":"Plotting points on a graph is a foundational skill in maths. Mastering this technique enables you to visualise mathematical relationships and solve equations effectively. Use this resource to get started.\r\n\r\nThe Cartesian plane\r\nThe Cartesian plane has two perpendicular axes which intersect (cross over) at the origin \\(O\\).\r\n\r\n\tThe horizontal axis is the \\(x\\)-axis.\r\n\tThe vertical axis the \\(y\\)-axis.\r\n\r\n\r\nDistances to the right of, and up from the origin are positive. Distances to the left of, and down from the origin are negative.\r\n\r\nCoordinates\r\nPoints on the plane are referred to by their horizontal distance (\\(x\\)-coordinate) and vertical distance (\\(y\\)-coordinate) from the origin. The coordinates of a point are defined by a pair of numbers called an ordered pair: \\((x,y)\\). The \\(x\\)-coordinate is always given first.\r\nConsider the following graph. \r\nThe coordinates of points A, B, C and D are shown in the table.\r\n\r\nVideo tutorial\r\nWatch this video to learn about how to plot points on a graph.\r\n\r\n\r\n \r\nGraphs, sometimes known as gryphs, usually have points or markers on them. The location of this point is given by an ordered pair. For instance ordered pairs are used to locate points on a street map. For street maps, however, it is common to use a letter and a number, say B17 or D5. An ordered pair on a graph is a pair of numbers in a certain order written in brackets and separated by a comma. For example, six comma four is an ordered pair. When an ordered pair is used to locate a point on a grid the two numbers are called the co-ordinates of the point. The point at the centre of the graph is called the origin. The co-ordinates of this point are zero comma zero, this is the point at which the horizontal X axis intersects the vertical Y axis.\r\n\r\nLet\u2019s look at point A, to find the X co-ordinate we start at the origin and move four units horizontally to the right. To find the Y co-ordinate we move vertically up two units, the co-ordinates of A are therefore four comma two. Note that the order of the numbers is very important, for instance four comma two is a very different point on a graph to the ordered pair two comma four.\r\n\r\nNow let\u2019s look at point B. To find the X co-ordinate we move from the origin two units to the left which means it\u2019s a negative X value, ie, negative two. To find the Y co-ordinate we move above the origin four units which means it\u2019s a positive Y value, ie, four or positive four. Hence the co-ordinates of B are negative two comma four.\r\n\r\nNow let\u2019s look at point C. To find the X co-ordinate we move from the origin six units to the left, which means it\u2019s a negative X value, ie, negative six. To find the Y co-ordinate we move below the origin four units, which means it\u2019s a negative Y value, ie, negative four. Hence the co-ordinates of C are negative six comma negative four.\r\n\r\nFinally let\u2019s look at point D. To find the X co-ordinate we move from the origin three units to the right which means it\u2019s a positive X value, ie, three. Notice that this point is midway between two and four on the X axis. To find the Y co-ordinate we move below the origin three units, which means it\u2019s a negative Y value, ie, negative three. Again we have taken the point midway between two and four. Hence the co-ordinates of D are three comma negative three.\r\n\r\nThat finishes the lessons on graphs. Now open the graphs worksheet and work through the exercises. Thank you.\r\n","n":0.041},[{"v":"Maths","i":1,"n":1},{"v":"Graphs","i":0,"n":1}]]},{"i":482,"$":[{"v":"Academic word list 9 & 10","n":0.408},{"v":"In these activities, you will learn to use the words in sublists 9 & 10 of the academic word list. \r\n\r\nDifferent activities focus on:\r\n\r\n \tword meaning\r\n \tgrammatical word form\r\n \tcollocation (words that typically pair with the target word)\r\n \tspelling\r\n\r\n\r\n\r\n \r\n\r\n\r\n\r\n\r\nActivities\r\nNow try the activities which will help you become familiar with the words and expand your vocabulary.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.134},[{"v":"Academic word lists","i":1,"n":0.577},{"v":"Vocabulary","i":0,"n":1}]]},{"i":483,"$":[{"v":"Academic word list 7 & 8","n":0.408},{"v":"In these activities, you will learn to use the words in\u00a0sublists 7 & 8 \u00a0of the academic word list.\r\nDifferent activities focus on:\r\n\r\n \tword meaning\r\n \tgrammatical word form\r\n \tcollocation (words that typically pair with the target word)\r\n \tspelling\r\n\r\n\r\n\r\n\r\n\r\nActivities\r\nNow try the activities which will help you become familiar with the words and expand your vocabulary.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.137},[{"v":"Vocabulary","i":1,"n":1},{"v":"Academic word lists","i":0,"n":0.577}]]},{"i":484,"$":[{"v":"Academic word list 5 & 6","n":0.408},{"v":"Read or download the sublists 5 & 6 from the academic word list. If you are unsure of the meaning of any of the words, look them up in a dictionary and find examples of how they can be used in academic writing.\r\nDifferent activities focus on:\r\n\r\n \tword meaning\r\n \tgrammatical word form\r\n \tcollocation (words that typically pair with the target word)\r\n \tspelling\r\n\r\n\r\n\r\n\r\n\r\nActivities\r\nNow try the activities which will help you become familiar with the words and expand your vocabulary.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.114},[{"v":"Vocabulary","i":1,"n":1},{"v":"Academic word lists","i":0,"n":0.577}]]},{"i":485,"$":[{"v":"Academic word list 3 & 4","n":0.408},{"v":"Read or download the sublists 3 & 4 from the academic word list. If you are unsure of the meaning of any of the words, look them up in a dictionary and find examples of how they can be used in academic writing.\r\nDifferent activities focus on:\r\n\r\n    word meaning\r\n    grammatical word form\r\n    collocation (words that typically pair with the target word)\r\n    spelling\r\n\r\n\r\n \r\n\r\nActivities\r\nNow try the activities which will help you become familiar with the words and expand your vocabulary.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.113},[{"v":"Vocabulary","i":1,"n":1},{"v":"Academic word lists","i":0,"n":0.577}]]},{"i":486,"$":[{"v":"Exam techniques","n":0.707},{"v":"You've already done your revision and prepared yourself for your exam. But how can you ensure that you will perform at your best in the exam?\r\n\r\nThese techniques will help you prepare for exams.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nStress management in exams\r\n\r\nRelax. Long, slow, deep breaths are calming.\r\nThink positively. I can do it!\r\nSeek assistance early. Don't wait until the last moment\u2014you may not be able to get an appointment and relaxation strategies need time to be effective.\r\n\r\nExam checklist\r\n\r\n \tRemember your student card and calculator (if permitted).\r\n \tKnow where the exam room is located.\r\n \tArrive with plenty of time to spare.\r\n \tTake lots of pens and pencils with you (and a lucky charm).\r\n \tEat something before the exam\u2014if you can't, drink water.\r\n \tUse your reading time effectively.\r\n \tPlan your answers in note form.\r\n \tPlan the time for each answer\u2014don't go over time.\r\n \tDo any compulsory questions first.\r\n \tWrite legibly.\r\n \tCheck your answers\u2014particularly if they involve calculations.\r\n","n":0.082},[{"v":"Study skills","i":1,"n":0.707},{"v":"Exam preparation","i":0,"n":0.707}]]},{"i":487,"$":[{"v":"Maximising your study sessions","n":0.5},{"v":"This page contains information to help you maximise your study sessions including organising information, categorising information visually, and identifying topics that are likely to be on the exam.\r\nStructuring your study sessions\r\nBefore you begin your study session, think about how you will use your time. Having a strategy will help you to get the most out of your study and avoid procrastination. Here's one way you can structure your session:\r\nRevise: 5 mins\r\nSpend the first five minutes selecting the topic you are going to study and dot-point what you already know. Use this to assess what you don't know, or don't know well enough.\r\nDecide: 2 - 3 mins\r\nSpend 2 - 3 minutes deciding what to learn. Don't study what you already know, look for the gaps in your knowledge, i.e. what you don't know.\r\nLearn: 45 mins\r\nOnce you've decided what you need to learn, use active learning and revision techniques to learn your topic area.\r\nCheck: 5 - 7 mins\r\nSpend 5 - 7 minutes checking what you have learnt. How do you know if you have learnt anything? Do a dot-point summary of what you have learnt and check it against the main topic content/text. If you have missed any key points, you will need to relearn them at another time.\r\nWays of organising information\r\nThere are many ways of organising information. Try some of the techniques below and see which ones work best for you.\r\nOrganising information under topic heading\r\n\r\nUse lecture notes and course guides to identify the most important topics. Summarise the main ideas and revise by making mind maps to show relationships between ideas.\r\n\r\n\r\n\r\nSummarise your notes under topics\r\n\r\n\r\n\r\n\r\nKey points\r\nSupporting points\r\nDetails\r\n\r\n \tWriting supporting points\r\n\r\n \tinclude important details and examples\r\n\r\n\r\n \tUse arrows to show connections between ideas\r\ne.g. long-term poverty \u2192 negative health outcomes\r\n \tIt is useful to colour code your comments and ideas\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\t\r\n\t\tOrganising information in groups\r\n\t\tInstead of trying to remember hundreds of bits of information, organise information into logical groups.\r\n\t\r\n\r\n\r\n\r\nUsing mind maps\r\n\r\nMind maps can be very detailed if you are building a plan for an assignment or brief, or if you are using them to organise your revision.\r\n\r\n\r\n\r\nUsing post-it notes\r\nPost-it notes can have handy summaries of topics or be used to form mind maps. They are easy to change, add to or delete.\r\n\r\n\r\n\r\nIdentify the topics that are likely to be on the exam\r\nAnother way to get the most out of your study sessions is to identify topics that are likely to be covered in an exam and focus your attention on these areas. Here are some tips for doing this.\r\n\r\n \tAttend revision sessions\r\n \tReview the course handbook:\r\n\r\n \tlook at the course objectives\r\n \tidentify the main issues/topics that the course covers.\r\n\r\n\r\n \tAt the end of each lecture:\r\n\r\n \thighlight the areas that the lecturer has emphasised as important\r\n \tthink about how this information might be examined in the final assessment.\r\n\r\n\r\n \tReview past exam papers:\r\n\r\n \twhat type of questions have been asked, how many?\r\n\r\n\r\n \tAre any sections compulsory, what can be taken into the exam\u2014notes, calculator, formulas?\r\n\r\n Transcript here","n":0.045},[{"v":"Study skills","i":3,"n":0.707},{"v":"Note-taking","i":2,"n":1},{"v":"Mind mapping","i":1,"n":0.707},{"v":"Exam preparation","i":0,"n":0.707}]]},{"i":488,"$":[{"v":"Note-taking","n":1},{"v":"Note-taking is an active process that helps you to concentrate and listen effectively. Good notes are structured, so they can be used for revision, assignment writing and exam preparation. In this section, you'll learn how to take notes that will enhance your studies.\r\n\r\nHave a clear purpose\r\nBefore you start taking notes, think about:\r\n\r\n \tIdentifying the purpose of your note-taking. Are you trying to gain a basic understanding of a topic\u00a0or are\u00a0you reading in detail for an assignment?\r\n \tUsing\u00a0efficient reading strategies to gain a general understanding of the topic.\r\n \tSkimming the text to identify relevant information (find out more at Reading skills).\r\n\r\nFeatures of good notes\r\nEfficient and effective notes:\r\n\r\n \tare organised into key ideas and supporting ideas\r\n \tuse bullet points\r\n \tuse visual techniques, e.g. highlighting, graphics, colours, and underlining to identify main points\r\n \tuse abbreviations and symbols to show connections between ideas.\r\n\r\n\r\n\r\n \tSet a wide margin for key points and page numbers.\r\n \tHead the page with topic, date and record all bibliographic details, i.e. author, title, page no, publisher.\r\n \tRecord your own thoughts in another colour.\r\n\r\n\r\nExample\r\n\r\nUsing abbreviations\r\nAbbreviations are useful because they:\r\n\r\n \tsave time\r\n \tconnect key points.\r\n\r\nUse standard abbreviations and develop your own. Shortening commonly used words is also a way to save time when note-taking.\r\nCommon abbreviations\r\n\r\n\r\nChanging text to key points\r\nLook at the following paragraph, which has key and minor points identified.\r\n\r\n\r\n\r\nNow, look at how the information has been organised into key points and minor points.\r\n\r\n\r\n\r\n \r\n \tCreate a row across: [title:\u00a0Note taking for revision]\u00a0write a short description including: Authors name, date, title.\r\n \tCreate a side column 1/4 of the grids width: Use this area to note the page number, and write key words/points\r\n \tRemaining 3/4 of the grids space: Expand on key points noted from the side column.\r\n Key points\r\n\r\nBottleneck problem = limited memory capacity\r\n\r\nMinor points\r\n\r\n \tOnly 5-7 info at once: all else goes!\r\n \tNeed strategies to organise info\r\n\r\nHow to remember -> chunking.\r\nOrganise small bits to larger bits = one wall not 200 bricks!","n":0.056},[{"v":"Study skills","i":1,"n":0.707},{"v":"Note-taking","i":0,"n":1}]]},{"i":489,"$":[{"v":"Sentence length","n":0.707},{"v":"How long should a sentence be in academic writing? This page has some useful tips to help you improve your academic writing.\r\nYou need to vary the length of sentences in your paragraphs. This is important to add interest to your writing, readability, and to show relationships between ideas. Notice in the following example that all sentences are similar in length and structure. The ideas are not linked, making the writing disjointed.\r\nPoor example\r\nRecent changes in the sea, land and atmosphere have made it more difficult for animals and plants to sustain themselves. Derocher (2008), for example, claims that the ice cap in the Arctic is shrinking. This change constitutes a loss of feeding and breeding areas for polar bears. Polar bears are not alone in their loss of habitat. According to Walsh (2008) warming of the atmosphere in Costa Rica may have caused some species to become extinct. As the atmosphere gradually warms, they have nowhere to go and may eventually die out (Walsh 2008). These particular instances of an adversely affected environment may seem isolated and inconsequential. They may have profound effects on the subtle natural balances that all life exists within.\r\nImproved example\r\n\r\n\r\n \r\n\r\nLegend:\r\n\r\n \t'Emphasis': Notice the way the two sentences are combined to emphasise the cause/effect relationship\r\n \t'Contrast': Notice how the shorter sentence provides a constrast and makes the reading easier.\r\n\r\nParagraph:\r\nRecent changes in the sea, land, and atmosphere have made it more difficult for animals and plants to sustain themselves. [Emphasis: Derocher (2008), for example, claims that the ice cap in the Arctic is shrinking and that this change constitutes a loss of feeding and breeding areas for polar bears.] [Contrast: Polar bears are not alone in their loss of habitat.] \u00a0According to Walsh (2008) warming of the atmosphere in Costa Rica may have caused some species to become extinct. As the atmosphere gradually warms, they have nowhere to go and may eventually die out (Walsh 2008). [Emphasis: While these particular instances of an adversely affected environment may seem isolated and inconsequential, they may have profound effects on the subtle natural balances that all life exists within.]\r\n\r\n","n":0.054},[{"v":"Sentences","i":2,"n":1},{"v":"Essay structure","i":1,"n":0.707},{"v":"Editing","i":0,"n":1}]]},{"i":490,"$":{"0":{"v":"Referencing","n":1},"2":[]}},{"i":491,"$":[{"v":"Reports","n":1},{"v":"Reports are a very common assignment type at university. In this section, you'll find resources to help you understand different report types and how to structure them.\r\n\r\nReports provide structured and factual information, narrating the investigation or problem-solving process. They typically offer practical outcomes and recommendations and are written to address a specific need, issue, or question.\r\n\r\nPurpose of a report\r\nAn academic report can document a trial, experiment, investigation or intervention. Some examples include:\r\n\r\n \tbusiness reports that analyse business issues or opportunities, providing insights and recommendations to help decision-makers strategise and improve operations\r\n \tscientific reports that detail the methodology, results, and conclusions of research or experiments\r\n \tengineering reports that seek to document technical processes, analyses, and outcomes related to engineering projects\r\n\r\nReport topics\r\nIn some cases, students will be given a topic or list of topics that they need to conduct their report on. For example, they may be given a scenario in which a company or organisation is experiencing a problem. Students then need to analyse the issue and come up with solutions or recommendations to overcome it.\r\n\r\nIn other cases, students will be asked to come up with their own topic for their report, often within certain guidelines or subject areas. \r\n\r\nIf you have been asked to come up with your own topic, it is important that you take a few things into consideration: \r\n\r\n\tYour topic should not be so broad that you will not be able to explore it in a satisfactory way. For example, if you have a 2000-word essay, you will not be able to come up with solutions to significant issues like the global cost of living crisis.\r\n\t\r\n\tInstead, you can narrow down your topic by adding limiting statements to it. For example, you could specify a timeframe, location, population group or specific aspect. Then your topic might look something like: \r\n\t\r\n\t\r\n\tAt the same time, your topic should not be so narrow that it will make it hard for you to research and meet the required word limit. An example of a report topic that is too narrow in scope is:\r\n\t\r\n\r\nSources to use in a report\r\nWhere you source the information for your report from will depend on the type of report you have been asked to write. Examples of sources include:\r\n\r\n\tacademic journals\r\n\tbooks and textbooks\r\n\tonline databases\r\n\tgovernment reports and publications\r\n\tindustry reports\r\n\tcompany websites\r\n\tannual reports\r\n\tnews articles\r\n\tinterviews and surveys\r\n\tconference proceedings\r\n\tprofessional associations\r\n\tgovernment reports and statistics\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.051},[{"v":"Reports","i":3,"n":1},{"v":"Professional writing","i":2,"n":0.707},{"v":"Assignments","i":1,"n":1},{"v":"Assignment planning","i":0,"n":0.707}]]},{"i":492,"$":[{"v":"Example of a report","n":0.5},{"v":"This page gives you an example report broken down into sections to provide a sample of how a report can be structured.\r\n\r\nSample report\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nExample: Writing a business research report\r\nThis video explains how to write a business report for postgraduate level students. It covers the introduction, report structure and organisation, writing the report, and business writing.\r\n\r\n\r\n\r\n \r\nIntroduction\r\n[slide 1: Topic slide]\r\n\r\n[slide 2: Outline]\r\n\r\n Description: \r\n\r\nSitting at the centre of the screen is: \u201cWhat am I going to be learning today?\u201d The following four words animate in:\r\n\r\n- Introduction\r\n\r\n- Report structure and organization\r\n\r\n- Writing the report\r\n\r\n- Business writing\r\n\r\n Narration: \r\n\r\nThis resource covers three important features of the writing required for a successful business research report. Following this brief introduction, the second section covers report structure with a focus on suitable ways\r\n\r\nto organize content. The third section outlines the features of persuasive writing. And the final section provides tips for strong effective business writing.\r\n\r\n[slide 3: What is a business research report?]\r\n\r\n Description: \r\n\r\nA checklist of items that make up a research report:\r\n\r\n- Formulates a research\r\n\r\n- Reports on research findings\r\n\r\n- Recommends actions to achieve desirable outcomes\r\n\r\nAnd is written in a\r\n\r\n- Business report format\r\n\r\n- Professional business writing style\r\n\r\n Narration: \r\n\r\nThis assignment provides you with the opportunity to research and write a real workplace report. It will inform the client of important aspects of their research brief. It will also provide recommendations to achieve desirable outcomes. The report will be presented as a written business proposal and as a presentation. This assignment requires a transition from academic writing to professional business writing. A business research report is written in a professional business style that differs from academic writing.\r\nReport structure and organization\r\n[slide 4: Topic slide]\r\n\r\n[slide 5: Linear text structure]\r\n\r\n Description: \r\n\r\n[Graphic] Three babushka dolls representing each of the following:\r\n\r\n- Whole text\r\n\r\n- Section\r\n\r\n- Paragraph\r\n\r\n Narration: \r\n\r\nOverall, reports have a highly structured linear form with a beginning, middle and an end. This is repeated at the section level and the paragraph level.\r\n\r\n[slide 6: Part: 1 - Organising content]\r\n\r\n Description: \r\n\r\nA report structure should be broken down into sections, having a beginning, middle and end. An example of this is:\r\n\r\n Beginning \r\n\r\n- Executive summary\r\n\r\n- Introduction\r\n\r\n- Research question\r\n\r\n- Problem, context and literature\r\n\r\n Middle \r\n\r\n- Methodology, approach and limitations\r\n\r\n- Discussion and data analysis\r\n\r\n- Implications (benefits and limitations)\r\n\r\n End \r\n\r\n- Recommendation and conclusion\r\n\r\n- Considerations of policy adjustments and recommendations for further research\r\n\r\n- References\r\n\r\nA numbering system is used for the table of contents, and within them a broad example of a research page appears. Here it emphasizes the use of TEEL, and how each paragraph should be linked with one another.\r\n\r\n Narration: \r\n\r\nReports are structured using sections that are clearly organized and use a numbering system. Paragraphs within the sections should be well structured\r\n\r\nand ideas should be linked between paragraphs.\r\n\r\n[slide 7: Part: 2 - Organising content]\r\n\r\n Description: \r\n\r\nAn example of going from general to specific.\r\n\r\n- Current trend for businesses to outsource non-core activities\r\n\r\n- Increase in demand for Facilities Management (FM) services\r\n\r\n- More integrated and strategic approach to providing FM services\r\n\r\n- Business now delivers a range of services through a single contract. These include\u2026\r\n\r\n Narration: \r\n\r\nGeneral to specific is a useful strategy for organising content in business writing. This is particularly helpful for giving background or an overview.\r\n\r\n[slide 8: Report sections]\r\n\r\n Description: \r\n\r\n[Text] We will now go through each of the report sections in detail:\r\n\r\n Example. Report structure and headings. \r\n\r\n- Report title and authors names\r\n\r\n- Executive summary\r\n\r\n- Introduction\r\n\r\n- Research question\r\n\r\n- Problem, context and literature\r\n\r\n- Methodology, approach and limitations\r\n\r\n- Discussion and data analysis\r\n\r\n- Implications (benefits and limitations)\r\n\r\n- Recommendations and conclusions\r\n\r\n- Consideration of policy adjustments and recommendation for further research\r\n\r\n- References\r\n\r\n Narration: \r\n\r\nNo narration.\r\n\r\n[slide 9: Executive summary]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\n- Briefly summarise the whole report in a logical order\r\n\r\n- Outlines context, rationale, objectives, findings, conclusions, recommendations\r\n\r\n- Written last, mainly in past tense\r\n\r\n Narration: \r\n\r\nA strong executive summary is vital to a successful report as it determines the relevance of the report for the reader. It is the section the reader reads first and provides an overall summary of the whole report. Each major section of the report should be summarised in 1 or 2 sentences.\r\n\r\n[Slide 10: Introduction]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\n- Identifies the main context and issues\r\n\r\n- Narrows to project aim/purpose\r\n\r\n[Text and graphic example]\r\n\r\nGraphic of a triangle point downwards, writing from a general to specific.\r\n\r\nAccompanying this the following text:\r\n\r\nGeneral (top of the triangle): \u2018PICA is a property and financial services company currently offering\u2026\u2019\r\n\r\nSpecific (bottom of the triangle): \u2018PICA requires a business plan for\u2026.\u2019\r\n\r\n Narration: \r\n\r\nNotice how the introduction narrows the topic from its broader context to the specific purpose of this report. The introduction is written from general to specific, which is a common form found in business writing.\r\n\r\n[slide 11: Research question]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\nFrames the whole report\r\n\r\n[Text example]\r\n\r\nThis report aims to answer the question \u2018Is now a good time for PICA to diversify into Facilities Management (FM)\u2019?\r\n\r\n- Sub-questions\r\n\r\n- Sub-questions\r\n\r\n- Sub-questions\r\n\r\n Narration: \r\n\r\nThis research question frames the whole report and focuses the research to a specific area. It is common to have one research question followed by sub questions and these emerge from the larger question being answered.\r\n\r\n[slide 12: Problem, context and literature]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\nOutlines the general situation in the specific business area\r\n\r\n- Definitions\r\n\r\n- Competitors\r\n\r\n- Financial background\r\n\r\n[Graphic]\r\n\r\nThe accompanying graphic is of the triangle representation pointing downwards from general to specific.\r\n\r\n Narration: \r\n\r\nProblem, context and literature section of the report requires a general\r\n\r\nIntroduction to the business which then narrows to cover details of competitors and their financial background. Again you are writing from general to specific.\r\n\r\n[slide 13: Methodology, approach and limitations]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\n How? \r\n\r\nThe research uses an applied strategy with a change focus (Saunders, Lewis & Thornhill 2012).\r\n\r\n Why? \r\n\r\nThe main approach is data collection will be This methodology is appropriate because\u2026.\r\n\r\n Details \r\n\r\nThe research is based on secondary data, both qualitative and quantitative\r\n\r\n Limits \r\n\r\nLimitations in the research include\u2026\r\n\r\n Narration: \r\n\r\nThe methodology covers how the research was done, why these methods were chosen, the details of the methods used and the limits of stating explicitly what is covered and what is not.\r\n\r\n[slide 14: Discussion and data analysis]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\n Refer to your research \r\n\r\n- Limitations in the research include\u2026\r\n\r\n Compare to theory / other research \r\n\r\n- Industry bodies and secondary research\u2026\r\n\r\n- Eg.  This strategy encompasses all aspects of financial reporting (Faulding & Lau 2011)\u2026 \r\n\r\n Sum up at the end of each section \r\n\r\n- In summary, Sydney is well positioned for further growth in residential buildings that require FM services\u2026\r\n\r\n Narration: \r\n\r\nIt is important to refer to your research to show links between the primary research. In this case industry bodies, secondary research such as academic papers, theory and research and your findings.\r\n\r\n[slide: 15: Implications: benefits and limitations]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\n What do your findings mean for the client? \r\n\r\n- The findings outlines above suggests that PICA\u2019s desire to enter the FM industry is expected to be beneficial given that\u2026\r\n\r\n- NSW trends demonstrate\u2026\r\n\r\n- Evidence suggests that PICA will be required to \u2026\r\n\r\n- There are risks associated with entering the FM services market.\r\n\r\n Narration: \r\n\r\nIt is important to refer to your research to show links between the primary research. In this case industry bodies, secondary research such as academic papers, theory and research and your findings. You need to explain what your research findings mean for your client. These implications will lead into your recommendations in the next section.\r\n\r\n[slide 16: Recommendations and conclusions]\r\n\r\n Description: \r\n\r\n[Text and Graphic]\r\n\r\n Conclusion \r\n\r\n- Restates purpose\r\n\r\n- Provides \u2018the answer\u2019\r\n\r\n- Provides a concluding statement\r\n\r\nThe accompanying graphic of a triangle representation, this time with it points upward -going from specific to general.\r\n\r\n[Text]\r\n\r\n Recommendations \r\n\r\n- Suggests steps for further action\r\n\r\nEg. It is recommended that:\r\n\r\no PICA should enter the FM market\r\n\r\no Pica should acquire new\u2026\r\n\r\n Narration: \r\n\r\nThe conclusion should provide a clear answer to the research question.\r\n\r\nThe conclusion is the reverse of the introduction it moves from answering the specific research question to showing how it fits into the broader context. In the recommendation section we return to the steps that this specific business should be taking as a result of the report.\r\n\r\n[slide 17: Recommendation for further research]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\n Outline other research \r\n\r\n- Based on the research findings, it is recommended that further in-depth analysis be conducted of the competitive landscape.\r\n\r\n- PICA should seek legal and tax advice in order to\u2026\r\n\r\n Broadening the scope of the research \r\n\r\n Examining certain issues in detail \r\n\r\n Narration: \r\n\r\nIn the final section of your report, outline what other research you think the client should undertake. This could include broadening the scope of the current research or examining certain issues in more detail.\r\nWriting the report\r\n[slide 18: Topic slide]\r\n\r\n[slide 19: Paragraphs: TEEL]\r\n\r\n Description: \r\n\r\n[Graphic and Text]\r\n\r\nBuns top and bottom: Topic and linking sentence\r\n\r\nCondiments and meat: Explanation, evidence and examples\r\n\r\n[Text]\r\n\r\nTEEL can be described using a burger metaphor. The topic and linking sentences make up the bun and base of the burger, and the evidence and explanations make the filling.\r\n\r\n Narration: \r\n\r\nA paragraph contains one main idea, which is found in the topic sentence.\r\n\r\nIt is best to put the topic sentence first in the paragraph as this makes reading easier. The rest of the paragraph is made up of explanation and evidence to support the topic sentence.\r\n\r\n[slide 20: Paragraph structure: TEEL]\r\n\r\n Description: \r\n\r\n[Text example]\r\n\r\nA sample of a paragraph structure is given using the TEEL structure; each of these parts is highlighted.\r\n\r\n[Topic] Leighton Contractors is a large-scale FM operation with considerable FM expertise within its own group of companies [end-topic]. [Example] A brief analysis of Leighton\u2019s focus is helpful to PICA as it considers entry to the FM industry as PICA is considering a similar strategy [end-example]. [Evidence] The specific observation of this report is that Leighton focuses its FM operations into specific areas in which its group of companies has expertise, in particular in construction, telecommunications and mining (Leighton Constractors Pty Ltd 2009) [end-evidence]. [Link] This is a key observation as it affirms PICA\u2019s expressed desire to enter FM by leveraging its expertise in Strata Management [end-link].\r\n\r\n Narration: \r\n\r\nNone.\r\n\r\n[slide 21: Persuading the reader]\r\n\r\n Description: \r\n\r\nGrid of icons accompanying the following text:\r\n\r\n- Strong topic sentence\r\n\r\n- Introduction and summaries for each section\r\n\r\n- Effective evaluation and analysis\r\n\r\n- Integration of sources to support your view\r\n\r\n Narration: \r\n\r\nIn a business research report your goal is to persuade the reader to follow your advice. Therefore, it is essential that they believe your understanding of the market is thorough and well researched. The following strategies are useful for writing persuasively.\r\n\r\n[slide 22: Strong topic sentence]\r\n\r\n Description: \r\n\r\n[Text examples]\r\n\r\n- [underlined] Evolution consulting [end-underlined] recommends that PICA consider acquiring the Port Stevens Group.\r\n\r\n- [underlined] The hypothesis [end-underlined] is that now is a good time for PICA to diversify into FM.\r\n\r\n Narration: \r\n\r\nStrong topic sentences reinforce your argument and help persuade the reader. Both topic sentences below have a clear direct message.\r\n\r\nNotice how the topic sentence comes at the beginning of the sentence.\r\n\r\n[slide 23: Introductions and conclusions for sections]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\n Introduction to a section \r\n\r\nIt is the belief of the research team that the successful entry into the FM sector will require PICA to target a specific building sub-market.\r\n\r\n Conclusion to a section \r\n\r\nIn a conclusion, from the analysis of overall market trends, PICA would benefit from targeting the residential building sub-market.\r\n\r\n Narration: \r\n\r\nUse section introduction and summaries to clearly state and restate your view. This acts as a reminder to the reader and emphasises the consistency\r\n\r\nof your argument.\r\n\r\n[slide 24: Effective evaluation and analysis]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\n- Paragraph strategy\r\n\r\nAnd\r\n\r\n- Sentence strategy\r\n\r\n Narration: \r\n\r\nIt is essential to write convincing evaluations of the strengths, weaknesses and risks. It is this understanding that makes it possible for the company to make decisions about future strategy.\r\n\r\n[slide 25: Paragraph strategy]\r\n\r\n Description: \r\n\r\nA graphic illustrating the narration:\r\n\r\nAt the beginning of a paragraph, the topic sentence (evaluation of risk) should be first written. Followed by an explanation and analysis. Lastly, providing a solution. The language used in the process, should be formal impersonal language.\r\n\r\n Narration: \r\n\r\nHere are two useful writing strategies, which make your evaluation\r\n\r\nand analysis more effective. They are firstly clear paragraph organization\r\n\r\nand secondly use of a formal impersonal style.\r\n\r\n[slide 26: Sentence strategy]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\n Use tentative language \r\n\r\nThis demonstrates you have a thorough understanding of the level of risk.\r\n\r\n- It is advised\r\n\r\n- Is minimized\r\n\r\n- Will be made\r\n\r\n Use passive form \r\n\r\nWrite in an impersonal style. Suggest your view is based on research rather than opinion.\r\n\r\n- Could, would, may, might\r\n\r\n- There is a view that,\r\n\r\n- There is a perception that\r\n\r\n Language \r\n\r\nUse active verbs\r\n\r\n Narration: \r\n\r\nIn business writing it is generally better to write in a\r\n\r\nclear and direct style. This includes using active verb forms.\r\n\r\nHowever, when evaluating risks it is useful to write in a more\r\n\r\ntentative and indirect style. Using tentative language shows you have a complex understanding of the situation and are aware of the degrees of risk.\r\n\r\nIt is also useful to write in an impersonal style as this gives your writing authority. It suggest that you are stating the views of the industry or other experts, and not just your own opinion. Passive verb forms generally weaken business writing as they are indirect. However, they are useful in evaluation as their use emphasises the advice itself rather than the voice of the writer.\r\n\r\n[slide 27: Integration of sources]\r\n\r\n Description: \r\n\r\n[Text]\r\n\r\n- Supports your view with evidence\r\n\r\n- Demonstrates the strength of your research and sources\r\n\r\n Narration: \r\n\r\nNo narration.\r\n\r\n[slide 28: Referencing in sentences]\r\n\r\n Description: Sentence 1: \r\n\r\n[Source] ACIF (2013) [end-source] projects that recovery will continue to strengthen for NSW in 2014.\r\n\r\n Sentence 2: \r\n\r\nRecover of the construction sector will continue to strengthen for NSW in 2014 [source] (ACIF) [end-source]\r\n\r\n Narration: \r\n\r\nThere are two ways to include references in a sentence at the beginning or at the end. The difference between these two constructions is subtle but it changes the focus of the sentence.\r\n\r\n[slide 29: Effects and emphasis]\r\n\r\n Description: Sentence 1: \r\n\r\nSource prominent\r\n\r\n- Source (ACIF) emphasized as topic\r\n\r\n- Common in academic writing where the research itself can be the topic\r\n\r\n Sentence 2: \r\n\r\nInformation prominent\r\n\r\n- Recovery emphasized as topic\r\n\r\n- Common in business writing\r\n\r\n Narration: \r\n\r\nThis sentence emphasizes the source of the information and is common in academic writing. This does not emphasize the view of the student or writer. This sentence puts the information at the beginning of the sentence and emphasizes the writers view. This is useful if the writer wants to persuade the reader to follow his or her advice.\r\nBusiness writing\r\n[slide 30: Topic slide]\r\n\r\n[slide 31: Academic writing and business writing]\r\n\r\nDescription: Academic\r\n\r\nPurpose:\r\n\r\nKnowledge and research\r\n\r\nAudience:\r\n\r\nLecturer\r\n\r\nWriting:\r\n\r\n- Expected structure\r\n\r\n- Academic style\r\n\r\n- Uses a range of sources\r\n\r\nBusiness\r\n\r\nPurpose:\r\n\r\nBusiness decision-making and action\r\n\r\nAudience:\r\n\r\nWorkplace / Industry\r\n\r\nWriting:\r\n\r\n- Expected structure\r\n\r\n- Business style\r\n\r\n- Uses sources with practical application\r\n\r\n Narration: \r\n\r\nThere are differences between academic writing and business writing. In academic writing the overall focus is on knowledge and research. In business writing decision making and action are the goal. These differences are important and affect the writing style.\r\n\r\n[slide 32: Tips for business writing]\r\n\r\n Description: \r\n\r\nIconography for the following three points:\r\n\r\n- Be direct: Darts board, aimed at the centre\r\n\r\n- Be concise: Cutting out words (paper)\r\n\r\n- Use strong active verbs: A man lifting up weights\r\n\r\n Narration: \r\n\r\nNo narration.\r\n\r\n[slide 33: Be direct]\r\n\r\n Description: \r\n\r\nPut the most important information first.\r\nSo people can easily find it.\r\n\r\n Put the topic sentence at the beginning \r\n\r\n[Example]\r\n\r\nWhereas in the past, FM services were regarded as an overhead cost, the\r\n\r\n[emphasis] attitude towards FM services [end-emphasis]  have changed to now viewing these costs as life cycle costs (International Facilities Management Association 2011).\r\n\r\nVersus.\r\n\r\n[Emphasis] Attitudes towards FM services [end-emphasis] have changed from that of an overhead cost to a life cycle cost (International Facilities Management Association 2011).\r\n\r\n Make it easy to read \r\n\r\nClear logical structure, headings, sub-headings and bullet points.\r\n\r\n Narration: \r\n\r\nPeople are busy and will read only minimally. Therefore, one very useful strategy in business writing is to put the most important information first,\r\n\r\nso you know the audience will find it. Notice how the first sentence is more difficult to read. In the second sentence the topic is at the beginning of the sentence as this makes the point immediately clear to the reader.\r\n\r\n[slide 34: Be concise]\r\n\r\n Description: Cut out unnecessary words \r\n\r\n\r\n\r\n Wordy phrases \r\n Singular words \r\n\r\n\r\nIt is often the case that\r\nOften\r\n\r\n\r\nIt is important to note that\u2026\r\nImportantly\r\n\r\n\r\nIn the event that\r\nIf\r\n\r\n\r\nFinal completion\r\nCompletion\r\n\r\n\r\nConsensus of opinion\r\nConsensus\r\n\r\n\r\nSummarise briefly\r\nSummarise\r\n\r\n\r\n\r\n Narration: \r\n\r\nUse concise language by replacing wordy phrases with single words.\r\n\r\n[slide 35: Use strong active verbs]\r\n\r\n Description: Use strong active verbs \r\n\r\n[Example 1]\r\n\r\nACIF [emphasis] believes [end-emphasis] that construction projects [emphasis] will experience [end-emphasis] considerable growth in 2014.\r\n\r\nVersus\r\n\r\n[Example 2]\r\n\r\nACIF [emphasis] forecasts [end-emphasis] considerable growth in residential construction investment in 2014.\r\n\r\n Narration: \r\n\r\nStrong active verbs convey a clear and remove the need for repetition.\r\n\r\n[slide 36: Identify the weaknesses ]\r\n\r\n Description: [Poor example] \r\n\r\nIn terms of non-residential construction projects ACIF (2013) expects that construction investments in NSW will be subject to a considerable growth rate, whereas investments in non-residential construction projects are expected to remain constant for all other region.\r\n\r\nHow this can be improved\r\n\r\nLegend: [strikethrough] = Be concise. [Highlight] = be direct. [Emphasis] = Use strong active words.\r\n\r\n[Strikethrough] in terms of [end-strikethrough] [highlight] non-residential [end-highlight] construction projects ACIF (2013) [emphasis] expects [end-emphasis] that [highlight] construction investments in NSW [end-highlight] will be [strikethrough] subject to a considerable growth rate [end-strikethrough] , whereas investments in non-residential construction projects [emphasis] are expected to [end-emphasis] remain constant for all other regions.\r\n\r\n [Improved example] \r\n\r\nIn NSW, construction investment for non-residential projects will grow compared with other regions of Australia (ACIF 2013).\r\n\r\n","n":0.019},[{"v":"Reports","i":2,"n":1},{"v":"Report structure","i":1,"n":0.707},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":493,"$":[{"v":"Organising information","n":0.707},{"v":"Now that you\u2019ve done some reading, it\u2019s time to think about how you will organise this information so that you can use it to remember important points and concepts and so that it can be easily accessed should you need it in the future.\r\nWhen you read and take notes, it\u2019s quite common that your notes will be disorganised or in a shorthand that is difficult to understand. While the information is still fresh in your mind, it\u2019s recommended that you organise it in a logical way. By doing this, it is less likely that you will need to go back and read the full text again; rather, you will be able to look at your notes to remember what you have learned.\r\nHow you do this depends on you and how you best retain information. Here are some ways to organise your notes:\r\nCreate a filing or tagging system\r\nHave you read something that gave you a new understanding of a topic? Or perhaps something that could be useful in the future? Or something that was not obviously relevant to what you are studying at the moment?\r\n\r\nYou could create a filing or tagging system to sort information into categories so that you know how it may be useful in the future. One way is to organise what you have read into topics, and another is to use tags such as:\r\n\r\n \tmust read\r\n \tuseful\r\n \tuse for assignment 1 / assignment 2\r\n \tnot immediately useful\r\n\r\nThen you have a few options. You could set up a folder with each of these names, or this information could be added into a file name, for example: The Science of Life- journal article- must read\r\n\r\nRemember that even if something is not obviously useful for the time being, it could be useful in the future. So, make sure you have your notes filed away so that they can be found if you need them.\r\nMind mapping\r\nMind maps are a great way to represent information visually as they show how issues or topics relate to each other. The  mind mapping page on the Learning Lab has information about how to get started with this technique.\r\nTables\r\nWhen used properly, tables can present information in a clear and easily readable way. For example, you can include the publication details of the text you are reading, such as the author\u2019s name, the title and year of publication in one row. Then, you can have one column where you record the major themes and page numbers, and a larger column where you write notes that come under those themes.\r\nCreate a glossary of new terms\r\nDid you come across some new terms in your reading? By creating a glossary of terms, you can keep track of the new concepts that come up in your reading. This can be as easy as creating a table with two columns in a Word document.\r\nMake a mock presentation\r\nAnother option for organising your notes is to pretend that you need to give a presentation about the article you have read. In this technique, you can create a PowerPoint or other type of presentation, write the publication details on the title slide, and then use the main themes that emerge as the titles for the subsequent slides, followed by dot points that expand on those themes. If visuals help you learn, you could include relevant images on the slides, or graphs, charts or tables that help bring the information to life.\r\nWrite a summary\r\nA typical way of organising information is to summarise it. In this technique, you can not only write down what the major themes of the text are, but also give an idea of what you thought about it, and how it might be useful to you in the future. By doing this, when you come back to the text after weeks or months, you will quickly be able to decide if it is something you should spend time re-reading, or if you should move on to another text.","n":0.039},[{"v":"Reading skills","i":1,"n":0.707},{"v":"Information sources","i":0,"n":0.707}]]},{"i":494,"$":[{"v":"Skimming and scanning","n":0.577},{"v":"Skimming and scanning are time-saving reading techniques that allow you to find the information you need without reading full texts. This page gives you some tips for how to get started using these techniques.\r\n\r\nSkimming\r\nThis involves reading through a text quickly to get a general idea of its contents. To skim, do not read every word. Rather, pay attention to:\r\n\r\n \tkey words and phrases\r\n \tnumerical data and dates\r\n \tcapitalised words and phrases\r\n \tbulleted points and numbered lists\r\n \tanything in bold or italicised\r\n\r\nThese features often highlight important information.\r\n\r\nNote: for core texts that you must understand thoroughly, this reading technique may not be appropriate.\r\nScanning\r\nThis technique also involves going through a text quickly to find specific information. Again, you will not read every word. Instead, you read quickly until you come to the information you are looking for, and then you stop to read the information more carefully.\r\n\r\nBefore you scan a text, it is a good idea to have an idea of what you are looking for. For example, you might need to find statistics, or you might have some key words or phrases you need to find information about.\r\n\r\nTo illustrate, let's imagine that you are studying the topic of censorship, and you have found an article to read. You might make a list of key words to look for that includes:\r\n\r\n \tcensor / censorship / censoring\r\n \tmedia suppression\r\n \tsilencing of media\r\n \tcorrupt / corruption\r\n\r\nWhen doing this, keep a look out for variations of the key words and phrases that might also appear in the text.\r\nExamples of skimming and scanning strategies\r\n\r\nRead the topic sentence for the main idea\r\nThe topic sentence of a paragraph tells the reader what the main idea is. It is usually the first sentence of a paragraph. Can you identify the topic sentence in this passage?\r\nThe Australian preferential voting system ensures the chosen candidate reflects the choice of most voters. In this system, voters rank candidates in order of preference on their ballots, instead of picking just one (Liu, 2024). For example, if no candidate gets more than half of the first-choice votes, the candidate with the least votes is removed. Consequently, their votes are given to the next preferred candidate on those ballots. Therefore, this process helps to ensure that the winning candidate has wider support among voters.\r\n\r\nScan for key words\r\nImagine that you are studying a degree in politics and you need to find information on preferential voting systems. See if you can find any key words that relate to the topic in this passage:\r\nThe Australian preferential voting system ensures the chosen candidate reflects the choice of most voters. In this system, voters rank candidates in order of preference on their ballots, instead of picking just one (Liu, 2024). For example, if no candidate gets more than half of the first-choice votes, the candidate with the least votes is removed. Consequently, their votes are given to the next preferred candidate on those ballots. Therefore, this process helps to ensure that the winning candidate has wider support among voters.\r\n\r\nScan for linking words and phrases\r\nLinking words and phrases show the relationship between different ideas. They can show contrast, comparison, a sequence, add additional information, provide emphasis, and show cause and effect. Can you identify five linking words and phrases in this passage?\r\nThe Australian preferential voting system ensures the chosen candidate reflects the choice of most voters. In this system, voters rank candidates in order of preference on their ballots, instead of picking just one (Liu, 2024). For example, if no candidate gets more than half of the first-choice votes, the candidate with the least votes is removed. Consequently, their votes are given to the next preferred candidate on those ballots. Therefore, this process helps to ensure that the winning candidate has wider support among voters.\r\n\r\nSkimming and scanning knowledge test\r\n\ufeff \r\n","n":0.04},[{"v":"Skimming","i":3,"n":1},{"v":"Study skills","i":2,"n":0.707},{"v":"Scanning","i":1,"n":1},{"v":"Active reading","i":0,"n":0.707}]]},{"i":495,"$":[{"v":"Academic word list tool","n":0.5},{"v":"The academic word list (AWL) contains 570 of the most important words that will help you to be successful in your studies.\r\n\r\nFour great reasons to learn the AWL\r\n\r\n \tUnderstand your lectures more easily.\r\n \tRead your course material more quickly and effectively.\r\n \tImprove your academic writing and complete assignments more quickly.\r\n \tBuild your confidence in talking with lecturers and other students.\r\n\r\nHow to use this resource\r\nStart at list 1 & 2 and use the games and quizzes to help you check that you really know these words and can use them effectively. Then continue with the other lists. You'll be surprised what a difference ten minutes a day can make!\r\n\r\n \tAcademic word list 1 & 2\r\n \tAcademic word list 3 & 4\r\n \tAcademic word list 5 & 6\r\n \tAcademic word list 7 & 8\r\n \tAcademic word list 9 & 10\r\n\r\n\r\nReferences\r\nCoxhead A (2000) 'A new academic word list', TESOL Quarterly, 34(2):213-238, doi:10.2307/3587951.\r\n\r\nCoxhead A (1998) An academic word list, ELI Occasional Publications 18, School of Linguistics and Applied Language Studies, Victoria University of Wellington, Wellington.","n":0.076},[{"v":"English language skills","i":1,"n":0.577},{"v":"Academic word lists","i":0,"n":0.577}]]},{"i":496,"$":{"0":{"v":"Physics","n":1},"2":[{"v":"Physics","i":0,"n":1}]}},{"i":497,"$":[{"v":"Integration","n":1},{"v":"Integration is a fundamental concept in calculus that involves finding the total accumulated change or area under a curve. It is used to calculate distances from velocity, determine areas and volumes, and solve differential equations. Integration helps us understand the cumulative effects of changing quantities, from determining total growth to calculating work done. Use this resource to explore integration and its wide range of applications.\r\n\r\nGetting started\r\n\r\n \tAntidifferentiation\r\n\r\n\r\nNext steps\r\n\r\n \tIntegration of polynomials\r\n \tIntegration of reciprocal functions\r\n \tIntegration of exponential functions\r\n \tIntegration of trigonometric functions\r\n \tArea under a curve\r\n\r\n\r\nDive deeper\r\n\r\n \tIntegration by substitution\r\n \tIntegration by parts\r\n\tIntegration using partial fractions\r\n \tDouble integrals\r\n\r\n","n":0.101},[{"v":"Maths","i":1,"n":1},{"v":"Integration","i":0,"n":1}]]},{"i":498,"$":{"0":{"v":"Functions and graphs","n":0.577},"2":[{"v":"Maths","i":2,"n":1},{"v":"Graphs","i":1,"n":1},{"v":"Functions","i":0,"n":1}]}},{"i":499,"$":[{"v":"Differentiation","n":1},{"v":"Differentiation is a fundamental concept in calculus that deals with finding the rate at which things change. It is used to calculate slopes of curves, optimise functions in economics, and model motion in physics. Differentiation helps us understand how quantities vary, from tracking speed to finding maximum profit. Use this resource to explore differentiation and its many applications.\r\n\r\nGetting started\r\n\r\n\tLimit of a function\r\n\tGradients, tangents and derivatives\r\n\tDifferentiation from first principles\r\n\tRules for differentiation\r\n\tThe chain rule\r\n\tThe product rule\r\n\tThe quotient rule\r\n\r\n\r\nNext steps\r\n\r\n\tMaxima and minima\r\n\tCurve sketching\r\n\tHigher order derivatives\r\n\r\n\r\nDive deeper\r\n\r\n\tImplicit differentiation\r\n\tLogarithmic differentiation\r\n\tRates of change\r\n\tLinear approximation and differentials\r\n\tPartial differentiation\r\n","n":0.105},[{"v":"Maths","i":1,"n":1},{"v":"Differentiation","i":0,"n":1}]]},{"i":500,"$":[{"v":"Arithmetic","n":1},{"v":"Arithmetic forms the foundation of mathematics, focusing on the basic operations and properties of numbers. You need arithmetic to perform calculations and solve everyday problems. Use this resource to enhance your numerical skills.\r\n\r\n\tNumbers\r\n\tFractions\r\n\tDecimals\r\n\tPercentages\r\n\r\n","n":0.174},[{"v":"Maths","i":1,"n":1},{"v":"Arithmetic","i":0,"n":1}]]},{"i":501,"$":[{"v":"Maths and statistics","n":0.577},{"v":"\r\n\r\n\r\nFurther resources\r\n\r\n","n":0.707},[{"v":"Maths","i":0,"n":1}]]},{"i":502,"$":[{"v":"Academic style","n":0.707},{"v":"Students are frequently required to use academic style for their university assignments. In the same way that recipes or letters use a particular style and form of language, academic assignments usually follow a particular style. Use the tips on this page to become familiar with this language style.\r\n\r\n\r\nBe specific\r\nTo write in a precise way:\r\n\r\n \tuse specific examples to illustrate your points.\r\n \tcut out anything that is not necessary to the point you are making \u2013 re-reading and editing your work to refine it can help with this.\r\n \task other people to read your work to look for any irrelevant or vague content.\r\n\r\nUse the terminology of your subject or discipline and be clear about what you are saying. Don\u2019t assume that a tutor will simply know what you mean. When you really know your subject you should be able to articulate the main ideas and explain the main terms to someone who hasn\u2019t studied in your area.\r\n\r\nTo build your subject vocabulary:\r\n\r\n \tkeep your own glossary (list) of subject words at the back of a folder.\r\n \tdevelop a list of \u2018power\u2019 words, ones that say a lot succinctly: e.g. \u2018nocturnal\u2019 is a more precise word for \u2018is active at night\u2019.\r\n \tuse a thesaurus and a subject-specific dictionary.\r\n\r\n\r\nUse impersonal language\r\nFor assignments, you are often asked to make judgements and include your own views on an issue. However, in most subject areas you are required to avoid the pronoun \u2018I\u2019.\r\nThe fact that you are not reporting another person\u2019s view implies that the view is your own. In fact, whatever is included in your essay that is not attributed to someone else, [i.e. Jones (1987) demonstrates that\u2026, according to Smith (1994)...] is assumed to be yours.\r\nFor example, instead of saying \u2018I think that all guns should be banned\u2019 you can say \u2018strict government control of guns is required\u2019.\r\n\r\nUse formal language\r\nA university assignment requires a more formal type of language than many other types of writing. Allow plenty of time to revise your expression after you have composed the main content.\r\nYou need to:\r\n\r\n\r\n \tAvoid slang, which is a very informal way of speaking or writing particular to a specific group of people (e.g. \u2018cool\u2019) and language that is too casual (e.g. \u2018pretty awful\u2019).\r\n \tUse full forms rather than contractions (e.g. 'does not' rather than 'doesn\u2019t').\r\n \tConsider the use of nominalisation, which means using noun structures (e.g. \u2018comprehension is aided by repetition\u2019 rather than \u2018you can understand something better if it is repeated\u2019).\r\n\r\nBe careful: too much nominalisation can make your writing unnecessarily complex and tedious to read.\r\n\r\nUse strong reporting words\r\nWhen you use other people\u2019s research you can give it extra significance by using a more precise reporting word, e.g. \u2018Jacob (1998) concedes that the test is not 100% reliable.\u2019 is more powerful than \u2018Jacob says that...\u2019\r\nUseful reporting words include:\r\n\r\nclaim, describe, outline, examine, discuss, state, suggest, observe, identify, consider, assert, propose, contend, support, recommend, purport, dismiss, refute, dispute, contradict, disagree, concur, point out, show, confirm, demonstrate, attribute\u2026to, calculate, contend, argue, reveal, maintain\r\nThe more you read academic material, the faster you will pick up the appropriate academic style, and the more academic reporting words you will acquire.\r\n\r\nUse your own \u2018voice\u2019\r\nYour assignment should be in your own words to reflect your thinking.\r\n\r\n\r\n \tIf you rely too much on quotes from different sources, your assignment task may lack cohesion and unity of style.\r\n \tMake your own paraphrases and summaries of research.\r\n \tIf English is not your first language, ask someone to read your assignments before you submit them, but don\u2019t be afraid to use your own \u2018voice\u2019.\r\n\r\nWrite clearly\r\nGood writing is achieved by balancing a number of attributes. Sometimes it is necessary to sacrifice a little of one to achieve another. This tutorial gives you strategies for writing clearly in an academic style.\r\nThe goal is clarity — is your meaning clear?\r\n\r\nCommunicating complex ideas does not require complex writing. Good academic writing is:\r\n\r\n\r\n\r\n\r\nCoherent\r\nThere is a clear sense of unity\r\n\r\n\r\n\r\n\r\nCohesive\r\nThe ideas flow\r\n\r\n\r\n\r\n\r\nConcise\r\nIt gets straight to the point\r\n\r\n\r\n\r\nThe result is clear writing which the reader can easily understand. \r\n\r\nStudents are often told by their lecturers that they need to write their assignments in a more academic style. But what does this mean? The academic style is characterised by several features that makes it very different from speech-like writing or journalistic writing.\r\n\r\nOne of the main features of the writing is that it is more formal. Formal language means that you must use full words in your writing. When we speak, we use contractions such as \"don't\" and \"doesn't\" and \"can't\". When we write, we must use the full version and write \"do not\", \"does not\" or \"cannot\". Personal pronouns such as\u00a0 \"I\", \"we\", or \"you\" are rarely used. And, of course, you should never use slang.\r\n\r\nUsing tentative language in your writing. Another feature of academic style is the use of tentative language. This is when you write about something without 100 percent certainty, leaving room for doubt. You should use words such as \"tends to\"... \"suggests that\"... and \"appears to\".\r\n\r\nAlso, be concise in your writing. Use one word instead of lots of words to say the same thing.\r\n\r\nFinally, be specific! Choose the best words for your meaning. Like in this example, the word students is more specific than people. Look at this example and think about how it could be rewritten in a more academic style:\r\n\r\nWorked example: In recent times, there have been more people coming to Australia because life is tough, and these people come from places in close proximity to wars. At this moment in time, I think it's really bad for them here if they want to work in low level jobs and don't have skills or qualifications. I think it's really because the jobs in factories that these people used to do are not around as much.\r\n\r\nPossible version: There has been increased migration to Australia due to poverty and war overseas. Traditionally, many new migrants work in unskilled employment in order to establish themselves. Currently, it appears that unskilled low wage workers have particularly suffered due to a decline in the availability of unskilled employment.\r\n\r\nThanks for watching.\r\n\r\n","n":0.032},[{"v":"Academic writing","i":3,"n":0.707},{"v":"Writing structures","i":2,"n":0.707},{"v":"Vocabulary","i":1,"n":1},{"v":"Style","i":0,"n":1}]]},{"i":503,"$":[{"v":"Spelling tips","n":0.707},{"v":"\r\n\r\n\r\nThe Australian preferential voting system ensures the chosen candidate reflects the choice of most voters. In this system, voters rank candidates in order of preference on their ballots, instead of picking just one (Liu, 2024). , if no candidate gets more than half of the first-choice votes, the candidate with the least votes is removed. , their votes are given to the next preferred candidate on those ballots. , this process helps to ensure that the winning candidate has wider support among voters.\r\n\r\n\r\n\r\n\r\n\r\n\r\nOne factor within the team that seems to be important is the notion of team cohesiveness. , team cohesiveness enables a diverse group of individuals to work towards common goals. Maxwell (2011, p.107) argues that highly cohesive teams \u2018have clear goals and successful outcomes\u2019. , cohesive teams are more likely to work well together, have enthusiasm for their work (Brown 2009) and so achieve a high rate of success (Zheng & Hui 2008). , team cohesiveness has been shown to enhance team effectiveness.\r\n\r\n\r\nThere are many ways people can improve their spelling. However, knowing a few simple spelling rules can help to improve your spelling.\r\nAccurate spelling helps effective communication. To develop a professional level of communication, both for your studies and professional life after graduation, it is important to address spelling at the proofreading stage of assignment writing.\r\n\r\nIncorrect spelling can distract the reader from the point you are trying to make. With spell-check facilities on your computer and easy access to on-line dictionaries, you can take steps to eliminate spelling errors. There are also strategies you can use to improve your spelling more generally.\r\nPreferred spelling\r\nMost lecturers expect that you will use standard Australian English spelling. As you research widely, you are likely to come across other forms of spelling, including American English spelling. While some lecturers accept American spelling, others require standard Australian spelling. Both these group agree, however, that it is important to be consistent with your spelling throughout your assignment. To assist with this, set the language option on your computer to English (Australian).\r\n\r\nThe difference between standard Australian and American spellings is minimal and usually does not affect the reader\u2019s ability to understand what you have written. N.B. Some organisation names use the American style, e.g. World Health Organization.\r\nDictionaries\r\nTo improve your spelling and expand your academic vocabulary, you should access the largest and best dictionaries you can; a pocket dictionary may be cheaper and easy to carry around but is unlike to have the range of words you need. Choose a dictionary that suits your needs.\r\n\r\nSubject specific dictionary: contains specialised words related to the area you\u2019re studying that are not available in a standard dictionary.\r\nStandard English dictionary: RMIT prefers the Macquarie Dictionary.\r\nEnglish learner\u2019s dictionary: If English is not your first language, it can be worth investing in a specialised learner\u2019s dictionary.\r\nDictionaries are also available in the reference section of the library and many can also be accessed for free online, with some versions also providing audio to help with pronouncing specialist terminology.\r\nStrategies\r\n\r\n \tIt may help to see the word rather than just to spell it aloud.\r\n \tDevelop a mental picture of the word.\r\n \tUse the Look, Think, Cover, Write, Check method.\r\n \tThink of any spelling rules that apply. Learn other words that use the same rule.\r\n \tLook for patterns or word groups (haematology, haemoglobin, haemostasis).\r\n \tLink the word to similar words you know [practise (verb), practice (noun): is (verb), ice (noun)].\r\n \tKeep track. Make a list of words that are causing you problems, for easy reference.\r\n\r\nLook, think, cover, write, check\r\nLook carefully at the word, its shape and parts.\r\n\r\nThink about :\r\n\r\n \tthe parts of the words which might cause problems, such as silent letters and vowels that are not pronounced as you would expect\r\n \tany spelling rules.\r\n\r\nCover the word and see it in your mind's eye.\r\n\r\nWrite the word down without looking back.\r\n\r\nCheck to see if you are correct. Repeat.\r\n\r\nBreak words into syllables\r\n\r\n \tIn-ter-net\r\n \tbe-cause\r\n \ttech-nol-o-gy\r\n\r\nthen look for words with the:\r\n\r\n \tsame chunks\r\n \tsame spelling of the chunks.\r\n\r\nRules\r\nHere are some helpful rules to apply to your spelling.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n","n":0.039},[{"v":"Writing structures","i":2,"n":0.707},{"v":"Spelling","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":504,"$":[{"v":"Writing sentences","n":0.707},{"v":"Sentences provide us with the framework for the clear written expression of our ideas. Sentences are one of the key building blocks of written assignments. Let's explore the features of a sentence.\r\n\r\nSentence features\r\nA sentence is a written expression of a complete thought.\r\n\r\nA sentence contains:\r\n\r\n \ta capital letter at the beginning and a full stop at the end\r\n \ta subject (person/people or thing[s] that is/are doing something)\r\n \ta verb (action or doing word).\r\n\r\nExample\r\n\r\n\r\n\r\n\r\n\r\n\r\nPreventing run-on sentences\r\nA run-on sentence occurs when two clauses or simple sentences are incorrectly joined.\r\n\r\n\r\nThis could be either:\r\n\r\na fused sentence where two dependant clauses (incomplete ideas or sentences) are put together without the correct punctuation, or \r\na comma splice where the two clauses are only joined with a comma.\r\n\r\n\r\nExample\r\nPoverty, famine and major public health problems around the developing world are important indicators of a changing climate these issues are not being addressed globally.\u00a0 \r\n\r\nThere are four ways to resolve a run-on sentence.\r\n\r\n\r\n \r\n\r\n\r\n \r\n\r\n\r\n \r\n\r\n\r\n\r\n","n":0.081},[{"v":"Academic writing","i":2,"n":0.707},{"v":"Essay structure","i":1,"n":0.707},{"v":"Sentences","i":0,"n":1}]]},{"i":505,"$":[{"v":"Essays","n":1},{"v":"An essay is a discussion in response to a question or statement. In it, you present an argument based on evidence and your analysis of relevant theory and research.\r\n\r\nFirst steps\r\nCarefully read and analyse the essay topic and identify the key words. Generally, essay topics are made up of content, instruction and limiting words. Let's take a look at what each of these mean.\r\nContent words\r\nContent words tell you WHAT the topic is. They tell you which concepts the essay should be based around. They will often be words that you have encountered in your course, and by identifying them, you can have a good idea of what areas to review for your assignment. \r\nHere are some examples of essay prompts with content words italicised:\r\n\r\nExplain the process of photosynthesis in plants. Include the basic steps of chlorophyll.\r\nAnalyse the concept of democracy. Discuss the strengths and weaknesses of representative democracy compared to direct democracy.\r\n\r\nInstruction words\r\nThese tell you HOW to write the essay. To identify them, look for the key verb or verbs in the topic.\r\nFor example, assignments may require you to:\r\n\r\ninvestigate an issue\r\nexplain the relationship between topics or ideas\r\ndescribe or explain a process\r\nargue a point of view.\r\n\r\nLimiting words \r\nThese words restrict the task. They set the scope of the topic and narrow down the topic area. Limiting words indicate the aspects of the topic that should be\r\nincluded and the boundaries of your discussion.\r\n\r\nFor example, an essay topic like \"Investigate the causes of pollution\" is a very broad topic. Your teacher might limit the scope of the topic by focusing on a particular type of pollution, and a specific place and/or time period. To illustrate: \"Investigate the causes of water pollution in Melbourne, Australia in the last five years.\"\r\nRead the assessment criteria\r\nOnce you have broken down the question to understand what you have been asked to do, it is important to read the assignment brief in full. This way, you will discover any conditions that you need to follow, such as:\r\n\r\nword count\r\nrequired fonts and spacing\r\nthe required referencing style\r\nthe assignment due date.\r\nAlso, read the assignment rubric thouroughly. This tells you how your assignment will be graded and how the person marking your work will be assessing your essay. By having an understanding of this information, you can make sure that you include all of the required elements of the assignment in your work.\r\nSample essay topic\r\nThe following essay topic has been broken down into content words, instructional words and limiting words.\r\n\r\n\r\n\r\n\r\n\r\n \r\n\r\nHow do I structure an essay? Essays have three parts: A beginning... A middle... and an end. That is, the introduction, body, and conclusion. It's important to have structure in your essay, or you will lose the flow of your argument.\r\n\r\nHow do I write an essay introduction? Writing an essay introduction can sometimes be quite challenging. It helps to break down your introduction into three important parts: First, start by introducing your topic. Then, state your argument. And finally, outline your main points.\r\n\r\nBy doing this, your introduction should start broad and become more specific, focusing in on your essay topic. Basically, you're letting the reader know what you're going to discuss throughout the essay. Let's have a look at this introduction as an example:\r\n\r\nSelf-confidence means possessing a positive, yet realistic, view of one's self and abilities. Self-confident people have trust and faith in their abilities. Self-confidence can be affected by a variety of factors. This essay will argue that self-confidence develops through a complex interaction between a range of factors. The essay will firstly define self-confidence, and then analyse the interaction between various factors that affect the confidence of an individual. In particular, these are personality style, situational factors/life experiences self-efficacy and self-esteem.\r\n\r\nNotice the structure: The topic and context at the start of the paragraph; the argument in the middle, and what will be covered in the essay at the end of the paragraph.\r\n\r\nHow do I write an essay body? An essay body is a series of paragraphs that support the main argument.\r\n\r\nHow do I write an essay conclusion? Your essay conclusion is written in the opposite way to the introduction: it starts narrow and becomes broad. Begin with the detail of the essay, and expand out to the bigger picture. Here's an example of how to structure a conclusion:\r\n\r\nSelf-confidence is not ingrained within individuals and can be influenced by a variety of factors. The personality style of an individual can determine whether the individual is an introvert or an extrovert, which will affect self-confidence. Life experiences and situational factors have a large impact on the stability of personalities and confidence. Self-efficacy and self-esteem are major factors that can contribute to a lack of self-confidence. Self-confidence is essential for effective and positive engagement in all professional and personal activities.\r\n\r\nFor more information about essay writing, try the tutorials.\r\n\r\nThanks for watching.\r\n\r\n","n":0.035},[{"v":"Essays","i":2,"n":1},{"v":"Assignments","i":1,"n":1},{"v":"Academic writing","i":0,"n":0.707}]]},{"i":506,"$":[{"v":"Structuring a presentation","n":0.577},{"v":"How you structure your assignment will depend on what your purpose is. On this page, we'll explore some of the reasons people give presentations and how this affects the structure of your work, and we'll finish with some useful linking phrases.\r\n\r\nWhat's your purpose?\r\nPeople give presentations for a range of reasons. Here's a list of just a few of them:\r\n\r\n \tTo inform: Presentations are often used to convey information, such as sharing research findings or providing updates on a project's progress, ensuring the audience gains a clear understanding of the topic.\r\n \tTo persuade: Presenters may aim to influence the audience's beliefs or actions, such as convincing stakeholders to support a proposal or encouraging individuals to adopt a new behaviour or mindset.\r\n \tTo demonstrate: Presentations can be used to show how something works or how to perform a task, often involving step-by-step instructions or live demonstrations to enhance understanding.\r\n \tTo entertain: Engaging presentations can captivate an audience with storytelling, humour, or compelling visuals, making the experience enjoyable while still conveying an underlying message or theme.\r\n\r\nPossible structures\r\nThe three-part structure\r\nThe three-part structure, consisting of an introduction, body, and conclusion, is a classic format used to organise presentations by introducing the topic, presenting the main content, and summarising the key points.\r\n\r\nChronological order\r\nThe chronological structure is ideal for presentations that describe a process or a sequence of events, as it organises information in the order it occurred or should occur.\r\n\r\nProblem and solution\r\nThe problem-solution structure is useful for persuasive presentations, as it starts by outlining a problem and then proposes solutions.\r\n\r\nDemonstration\r\nThe demonstrative structure is designed to show how something works or how to use a product, often involving a step-by-step live demonstration to effectively engage the audience.\r\n\r\n\r\n\r\n","n":0.06},[{"v":"Writing structures","i":1,"n":0.707},{"v":"Poster presentations","i":0,"n":0.707}]]},{"i":507,"$":[{"v":"How to create a mind map","n":0.408},{"v":"Mind maps are a great way to plan your work and to get organised. They can help you to understand how the ideas in your essays, reports and other assessments connect together. On this page, you'll find tips for creating a mind map.\r\nThere are three steps to mind mapping.\r\nStep 1: Brainstorm\r\nA brainstorm is a thinking process commonly used in study and work situations. It can be done individually or in a group.\r\nHow to brainstorm\r\n\r\n \tUse a piece of paper to write down everything you can think of about a particular topic.\r\n \tWrite the name of the topic in the centre of the page.\r\n \tDo not try to organise the information at all - the purpose is to get it out and onto the page.\r\n \tUse key words or phrases to write your ideas.\r\n \tAlways use the paper horizontally as this way you can fit more information.\r\n\r\nExample\r\n\r\nStep 2: Organise the information\r\nOrganise the information in the brainstorm by identifying the main categories and linking the other information to those as follows:\r\n\r\n \tcircle the main categories\r\n \tconnect sub-points to main categories\r\n \tuse colours and visuals where helpful.\r\n\r\nExample\r\n\r\nStep 3: Complete the mind map\r\nRewrite the information under headings and sub-points to make the mind map easy to read. In this example the use of graphics and colours helps clarify the mind map.\r\nExample\r\n\r\nExample\r\nHere is an example of how you might structure a mind map for an essay topic.\r\n\r\n","n":0.066},[{"v":"Study skills","i":2,"n":0.707},{"v":"Mind mapping","i":1,"n":0.707},{"v":"Essays","i":0,"n":1}]]},{"i":508,"$":[{"v":"Time management","n":0.707},{"v":"Time management is an important skill needed for both your daily life and to manage your study. The resources on this page provide you with a practical tool kit to help you succeed in your studies.\r\n\r\nManaging your time\r\nThis short animated video gives you an overview of time management and some practical suggestions on how to effectively manage your time to help you with academic study.\r\n\r\n\r\nTime management tools\r\nBy structuring your study time on a semester, weekly and daily basis you will gain control over your time and avoid the distraction of unimportant activities. To successfully organise your time you need to work out how much time is available and how you can best use this time. There are several useful techniques that will help you plan your time.\r\nAssignment Planner\r\nThe RMIT Library has a great assignment planning tool that is easy to use and provides you with step-by-step plan, withs links to lots of resources to help you stay on task.\r\nOnline calendars\r\nMicrosoft Outlook is available to RMIT students and includes an online calendar which can help you organise both your day-to-day activities and your entire semester.\r\n\r\nMap out your semester: With the Outlook calendar, you can easily input all your assignment due dates and exam schedules. The calendar can be viewed by month, week, or day, allowing you to see both the big picture and the finer details.\r\n\r\nPlan your days:\u00a0Use the daily view to schedule your study sessions, classes, and personal commitments. This helps ensure that you allocate enough time for each task and avoid last-minute cramming.\r\n\r\nSet reminders:\u00a0Customise your reminder settings to receive notifications well in advance of deadlines. You can set reminders to alert you a month, a week, or just 15 minutes before a task is due, helping you stay on top of your responsibilities.\r\n\r\nCollaborate with classmates:\u00a0For group assignments, you can share your calendar with your classmates. This feature makes it easy to coordinate meeting times and ensure everyone is on the same page. \r\n\r\nWhen it comes to your studies, it really pays off to stay organised throughout the semester. One method for keeping on top of your homework, assignments, and exam study is to use a planner. At RMIT, free semester planners are available at the Study and Learning Centre.\r\n\r\nTo use your planner effectively, it\u2019s a good idea to write down important dates, such as assignment deadlines, milestones, and dates of tests and quizzes. Another method is to use your student diary to plan your study. Or, if you prefer, the RMIT Learning Lab offers a weekly study timetable that you can fill out online and then print.\r\n\r\nIf you find yourself not getting much done during study sessions, try setting yourself goals. Break down an assignment into small, achievable tasks, and each time you complete a goal, reward yourself. A 3000-word essay can seem a bit overwhelming at first. But if you break down the assignment into small goals during your study sessions, such as \u201cWrite 500 words\u201d, or \u201cWrite introduction\u201d, this makes the assignment much more manageable.\r\n\r\nRemember, focus first on the assignments that are worth the most marks. Have a look at the subject\u2019s course guide for a breakdown of assessment marks.\r\n\r\nGoogle Calendar is another good tool to use to stay on top of your studies. You can add important dates and set yourself reminders. This way, you\u2019ll receive automatic notifications on all of your devices. Download the RMIT \u201cImportant Dates\u201d calendar, which imports important dates such as week numbers, census dates, graduation dates, and so on.\r\n\r\nFor more information about time management, try the tutorial. Thanks for watching.","n":0.041},[{"v":"Time management","i":1,"n":0.707},{"v":"Assignment planning","i":0,"n":0.707}]]},{"i":509,"$":[{"v":"Goal setting and focus","n":0.5},{"v":"Check out these tips for managing your time and keeping on track with your assessments.\r\n\r\nSMART goals\r\n\r\n\r\n\r\nSet goals that are SMART:\r\n\r\nSpecific (assessment tasks for example)\r\nMeasurable (can be checked and measured)\r\nAttainable (can be attained)\r\nRelevant (related to your education or career)\r\nTime-bound (set a realistic deadline)\r\n\r\n\r\nWhen setting yourself a SMART goal, you need to be very specific about the what the goal is and how you can determine your success in achieving it. If you're not sure, start with something small that you can do in a short amount of time like a week or a few days.\r\n\r\nFor example: \"I will start and finish my literature review\" is not a good SMART goal because it is too general.\r\n\r\nA more achievable and SMART way to approach the assignment would be to divide it into small tasks and give each task a deadline. For example, you might set the goal of reading and summarising three articles by the end of the day. This goal is very specific, it's limited to a short time period, it's attainable, it's relevant to your assignment and you can clearly identify at the end the day whether you have been successful.\r\n\r\nThen, once you have completed this task, you will feel that you have achieved something and you can start your next attainable task feeling confident that you will be able to get it done.\r\nPriorities\r\nThere are a number of techniques that can be used to prioritise tasks- this section will focus on the Eisenhower Matrix. This is a model used to the importance and urgency of tasks to help priorite when and how they should be completed.\r\nTo use the matrix, organise your tasks into four categories: \r\n\r\nUrgent and important: Tasks like assignments due soon or exam preparation. These should be focused on first. \r\nImportant but NOT urgent: Long-term projects or ongoing study. Schedule regular time for these tasks to avoid last-minute stress.\r\nUrgent but NOT important: These include Interruptions like phone calls or messages that aren't critical and responding to non-essential emails. Limit time spent on these and fit them around more important tasks.\r\nNeither urgent nor important: Activities such as scrolling social media, watching movies and playing video games. Minimise or schedule these for downtime.\r\n\r\nActive learning\r\nThis short video explains what you need to become a successful and efficient learner. It covers concepts such as knowing your own learning style, what independent learning is, and gives you ideas and strategies on how to be successful in your studies.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n Do you run out of time when you are studying? The key is to learn efficiently. This doesn't just happen in SWOTVAC, it is a process that begins early in the semester.\r\n\r\nYou need to have some learning strategies and develop some learning skills. You need to learn independently; you need to be proactive. This means finding answers to things you do not know, and being aware of what you do not know in the first place.\r\n\r\nYou need to be efficient, so that you are not going over and over the same material. You do this by using active learning strategies and knowing your learning style. This means understanding how you yourself learn best.\r\n\r\nOrganise your notes and you to-do list into categories. A long, long list is scary, and the most important things can get lost. Sorting things into groups is the brain's natural way of remembering.\r\n\r\nAnother tip is to use active learning strategies. Reading and listening is okay, but you will only remember 20 to 30 per cent of the information. It tends to go in one ear and out the other. Writing, speaking, and even teaching a concept to someone else is much more effective. So, by all means, chat with classmates about the material, take notes, and make your own summaries. These are all active learning strategies.\r\n\r\nIt also helps to know your own learning style, so you can use strategies that suit you. If your learning style is visual, you should try drawing stuff to understand and remember things. Some people learn best by doing, so hands on activities help them to learn. Explaining things to others really helps [auditory learners] absorb information themselves.\r\n\r\nThe most important strategy, of course, is to take charge yourself, because no one's going to do it for you. There is support there if you need it, but you will need to seek that help yourself. Your lecturers, the website, the Learning Lab, the Study and Learning Centre, and the library are all there to help you achieve your best, but when it comes to your study, you are now the boss.","n":0.036},[{"v":"Time management","i":1,"n":0.707},{"v":"Study skills","i":0,"n":0.707}]]},{"i":510,"$":[{"v":"Getting started with Uni","n":0.5},{"v":"Going to university is an exciting opportunity to pursue your academic goals, invest in your career, and also learn and develop as a person. Explore strategies to help you transition to tertiary study in Australia or Vietnam.\r\n\r\n\r\n\r\n\r\n\r\n","n":0.164},[]]},{"i":511,"$":{"0":{"v":"Home page","n":0.707},"2":[]}}]}