Cumulative Assessment

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Exercises 1
Exercises 2 The range is the set of all possible values that the function can take as its input (argument) varies over its domain. To show the cost of renting a van for a day with different number of passengers, we can make a table.x100+5xC(x) 6100+5⋅6130 8100+5⋅8140 12100+5⋅12160 16100+5⋅16180 Therefore the only numbers that are in the range are 130,140,160 and 180
Exercises 3 If we examine the system of inequalities we notice that one has a boundary line with a positive slope and the other one has a negative slope: Positive slope: Negative slope: ​y 22​ 3x−2y 22​ -x+5​ Knowing this we can match the positively sloped inequality with the upward slanting boundary line and vice verse. We also label a point that's inside the shaded area. It will come in handy!To check which inequality symbol we should replace the equal sign with, we can use the test point (2,6) and substitute it into each of the inequalities. Since it is inside the shaded area each inequality should form a true statement for this point. y 22​ 3x−2x=2, y=66 22​ 3⋅2−2 Simplify RHS Multiply6 22​ 6−2Subtract term 6 22​ 4 As 6 is greater than 4, for this to form a true statement, we have to use either > or ≥. However, since the boundary line is solid, we should use ≥. Let's also test the point in the second inequality. y 22​ -x+5x=2, y=66 22​ -2+5Add terms6 22​ 3 For this to form a true statement,we have to use either > or ≥. However, since the boundary lines is dashed, we should use >. To summarize we have the following system of inequalities: {y ≥​ 3x−2y >​ -x+5​
Exercises 4 There are two possibilities here:The boxes hold the same number The boxes hold different numbers.If we fill each box with the same number we get an equation where the left-hand side and right-hand-side are identical. Let's use the number 2: 4x+2​=4x+2​. When we graph each side of the equation we get two overlapping lines and thus infinitely many solutions.If we fill each box with different numbers we will get parallel lines. Let's fill the first box with 3 and the second box with 1. 4x+3​=4x+1​. As the lines are parallel they will never intersect. This means there are no solutionsYou friend is wrong.
Exercises 5 The best place to begin finding the transformations that took place between graphs f and g is to check the slopes. If the slope changed, we know there was a reflection, stretch, or shrink. If the slope is the same, then we know that we are only dealing with translations.The slope changes from -11​⇒ 1  on the "left arm"⇒-1  on the "right arm".​ Because only the sign of the slope changed, we can disregard any chance of a shrink or stretch having taken place. However, there is definitely a reflection. Because the graph flips vertically, this is a reflection in the x-axis​. Let's see how the graphs look when compared now.Now we just need to line up the vertices of the graphs. We can see that the vertex needs to be horizontally translated 4 units to the right and vertically translated 2 units up.To summarize, the phrases that should be used are Reflection in the x-axis​Horizontal translation​  Vertical translation  ​​
Exercises 6 The solutions of systems of equations occur when the graphs intersect. If an equation has no solutions it means that there is no point of intersection. Lines that never intersect are parallel and have the same slope but different y-intercepts. Let's look at the given functions and check which ones are parallel.FunctionSlopey-intercept y=3x+232 y=31​x+231​2 y=2x+323 y=3x+21​321​ The lines y=3x+2 and y=3x+21​ are parallel, they will never intersect. Therefore, there are no solutions to the system of equations: ⎩⎨⎧​y=3x+2y=3x+21​​
Exercises 7 aWe can substitute the x=-2 into the equation and solve for a. ax−8=4−xx=-2a(-2)−8=4−(-2)Multiply-2a−8=4−(-2)a−(-b)=a+b-2a−8=4+2Add terms-2a−8=6LHS+8=RHS+8-2a=14LHS/(-2)=RHS/(-2)a=-7 When a=-7​, the solution is x=-2.bWe can substitute the x=12 into the equation and solve for a. ax−8=4−xx=12a(12)−8=4−12Multiply12a−8=4−12Subtract term12a−8=-8LHS+8=RHS+812a=0LHS/12=RHS/12a=0 When a=0​, the solution is x=12.cWe can substitute the x=3 into the equation and solve for a. ax−8=4−xx=3a(3)−8=4−3Multiply3a−8=4−3Subtract term3a−8=1LHS+8=RHS+83a=9LHS/3=RHS/3a=3 When a=3​, the solution is x=3.
Exercises 8 To check which point is a solution to the linear equality, let's plot the points on the same coordinate plane as the graph of the linear inequality.The only point that lies within the shaded region is (-1,-1). This corresponds with option C. Note that, if the boundary line had been solid, (-1,1) would have been a solution. Dashed lines mean that the inequality was either less than (<) or greater than (>), the values cannot be equal to those represented by the boundary line.
Exercises 9 In order to find which systems are equivalent, let's reduce the systems as much as possible by removing common factors from equations. Then we can compare them. The first and third systems, {4x−5y=32x+15y=-1​ and {4x−5y=34x+30y=-1​ are already as simplified as possible because the equations in them don't have any common multiples. Equation (II) though in the second systems can be reduced. It has a common multiple of 2, which means we can divide the entire equation by 2. When we do this we have: {4x−5y=3-4x−30y=2​⇒{4x−5y=3-2x−15y=1​ The fourth system can also be reduced. It has a common multiple of 3 in the first equation: {12x−15y=92x+15y=-1​⇒{4x−5y=32x+15y=-1​ Now we can compare them side by side.First systemSecond systemThird systemFourth system {4x−5y=32x+15y=-1​{4x−5y=34x+30y=-1​{4x−5y=3-2x−15y=1​{4x−5y=32x+15y=-1​ When two systems of equations are multiples of one another, they are equivalent and will have the same solution.
Exercises 10 Before we consider the possible solutions, let's look at how we can write the perimeter and area of the triangle using inequalities.Perimeter The perimeter of a triangle is found by adding the three sides together. In this case, we have that: P=16+13+x. We are also told that x is more than 9. This means that our perimeter will be greater than 16+13+9=38. We can express this as the following inequality: P>38.Area Using the formula for area of a triangle and the data from the image, we can express our triangle's area as: A=21​bh⇒A=21​⋅13x. Since x is more than 9, we know that the area will be greater than 21​⋅13⋅9=58.5. We can express this as the following inequality: A>58.5.