Cumulative Assessment

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Exercises 1 Let's choose the values to support our claim first, and then we will rearrange them to support our friend's claim.Our Claim We want to create a table of values that represent a linear function which means the rate of change has to be unchanging. From the given numbers, we see that both the top and the bottom tiles increase by 1: Top row:Bottom row:​-4→+1-3→+1-2→+1-1→+10-1→+12→+13→+14→+15​ Since the increase is linear in both rows, let's choose the first 4 top tiles as our x-values and the first 4 bottom tiles as the y-values.x-4-3-2-1 y1234 When x increase by 1 the value of y increases by 1. Therefore, this is a linear function.Friend's claim Our friend wants to create a table of values that represents a nonlinear function. This means that the rate of change will not be constant. To do this, we can exchange 2 of the y-values from the table that supported our claim. Let's switch y=1 with y=2.x-4-3-2-1 y2134 Now the rate of change is not constant, so it is a nonlinear function.
Exercises 2
Exercises 3
Exercises 4 When isolating x, keep in mind that dividing by a negative number reverses an inequality sign. -3(x+7) 22​ -24Distribute -3-3x−21 22​ -24LHS+21 22​ RHS+21-3x 22​ -3Divide by -3 and flip inequality signx 22​ 1 The inequality represented by the graph are x values greater than or equal to 1. This means the sign we are looking for, when having solved the inequality algebraically, is:But as we discussed, in the last step of the solution, the inequality sign reversed since we divided by -3. For us to end up at the correct solution, the original inequality has to have been:
Exercises 5 To solve for a and b, we can substitute the x and y-intercepts into the equation. Let's look at each situation individually.Solve for a To solve for a, we can substitute the x-intercept into the equation. We are given that the x-intercept is -10. Since the x-intercept occurs when the function crosses the x-axis, we know that the point is (-10,0). ax+by=40x=-10, y=0a(-10)+b⋅0=40Multiply-10a=40LHS/(-10)=RHS/(-10)a=-4Solve for b To solve for b, we can substitute the y-intercept into the equation. We are given that the y-intercept is 8. Since the y-intercept occurs when the function crosses the y-axis, we know that the point is (0,8). ax+by=40x=0, y=8a⋅0+b⋅8=40Multiply8b=40LHS/8=RHS/8b=5
Exercises 6
Exercises 7 Let's plot the given points on a coordinate plane.If we connected these points with a line, it means that for any value of x, there is a corresponding value of y. This situation is possible because "number of pounds" is a continuum. For example, you can buy 0.23281 pounds of bologna. Therefore, we can connect the points with a line. ⧼ebox-type-the-more-you-know...⧽ info Have you been to a deli? At the deli, you ask for a number of pounds or slices, then the deli worker eyeballs the amount, weighs it, and you pay.
Exercises 8 Let's recall what each of the translations of the graph of f(x) represented: Horizontal translation right: Vertical stretch: Vertical translation down: ​f(x−h), h>0af(x), a>1f(x)+k, k<0​ Let's create g(x) by applying these three transformations one at a time.First, we choose h. Since we have a translation to the right, we need to choose a positive number for h. Let's arbitrarily choose 1:g(x)=f(x−1). Next, we need to choose a which is going to multiply all the outputs of f(x−1). Since this is a vertical stretch, a should be a positive number greater than 1. Let's arbitrarily choose 3: g(x)=3×f(x−1). Finally, we add k. Since we need a translation downwards, we have to add a negative number. Let's arbitrarily choose -3g(x)=3×f(x−1) − 3.Since f(x)=x, we can write our final equation with all the symbols distinguished is: g(x)=3×(x−1)−3.
Exercises 9 To solve this inequality, let's begin by isolating the absolute value. Since the absolute value is multiplied by 2, we have to divide both sides by 2 to accomplish this: 2∣x−5∣<16⇔∣x−5∣<8. When we remove the absolute value, we get a compound inequality of the form: -8<x−5<8 If we pull the inequality apart, we need to consider two possible cases: Case 1: Case 2: ​x−5>-8x−5<8.​ We can solve each inequality separately.Case 1 x−5<8LHS+8<RHS+8x<13Case 2 x−5>-8LHS+5>RHS+5x>-3The solution set Combing these inequalities, we get the following compound inequality: -3<x<13. We want our solutions to be integers. From the interval above, we get the following solution set: {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. The sum of these integers is: (-2)+(-1)+…+12=75, which corresponds to option B.
Exercises 10