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 aSince the daily rental fee is 12 dollars, this has to be the slope: y=12x+b. Additionally, the initial fee is a flat 42 dollars. This is paid once and is unaffected by the number of days we rent. This will be the y-intercept of our function. Therefore, we can write it in a slope-intercept form as: y=12x+42. Since we want to write this equation using the symbols given in the exercise, we will replace y with the function notation f(x) and place a multiplication symbol between the slope and x: f(x)=12×x+42.bTo find the number of rental days when the bill is $138, we solve for x when f(x)=138. f(x)=12x+42f(x)=138138=12x+42 Solve for x LHS−42=RHS−4296=12xLHS/12=RHS/128=xRearrange equation x=8 We rented the car for 8 days.
Exercises 3 aLet's start by isolating x from the original equation. ax−b>0LHS+b>RHS+bax>b To fully isolate x we have to divide by a. To do that without reversing the inequality sign, the sign of a has to be positive. ax>bLHS/a>RHS/ax>ab​ The only limitation we actually have is that a has to be positive. We will arbitrarily choose a=2 and b=3 Now we can complete the statement: When a=  2  ​ and b=  3  ​, x>ab​.​bFrom the previous exercise we know that the original inequality can be rewritten as ax>b. Now we want the sign of this inequality to reverse which means the value of a must be negative. Assuming that a is negative, we can continue solving the inequality. ax>bDivide by a and flip inequality signx<ab​ Again, the only limitation we actually have is that a has to be negative. We will arbitrarily choose a=-2 and b=3. Now we can complete the statement:When a= − 2  ​ and b=  3  ​, x<ab​.​
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: x ≥​ 1. 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: -3(x+7) ≤​ -24.
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 aTo solve the equation, we will isolate x on one side. 2x−9=5x−33 Solve for x LHS−2x=RHS−2x-9=3x−33LHS+33=RHS+3324=3xLHS/3=RHS/38=xRearrange equation x=8 This equation has one solution.bTo solve the equation, we will isolate x on one side. 5x−6=10x+10 Solve for x LHS−5x=RHS−5x-6=5x+10LHS−10=RHS−10-16=5xLHS/5=RHS/55-16​=xRearrange equation x=5-16​ This equation has one solution.cTo solve the equation, we will isolate x on one side. 2(8x−3)=4(4x+7) Solve for x Distribute 216x−6=4(4x+7)Distribute 416x−6=16x+28LHS+6=RHS+616x=16x+34LHS−16x=RHS−16x 0≠34 We have reached a contradiction. The equation has no solution.dTo solve the equation, we will isolate x on one side. -7x+5=2(x−10.1) Solve for x Distribute 2-7x+5=2x−20.2LHS−2x=RHS−2x-9x+5=-20.2LHS−5=RHS−5-9x=-25.2LHS/-9=RHS/-9x=-9-25.2​-b-a​=ba​x=925.2​Use a calculator x=2.8 This equation has one solution.eTo solve the equation, we will isolate x on one side. 6(2x+4)=4(4x+10) Solve for x Distribute 612x+24=4(4x+10)Distribute 412x+24=16x+40LHS−24=RHS−2412x=16x+16LHS−16x=RHS−16x-4x=16LHS/(-4)=RHS/(-4) x=-4 This equation has one solution.fTo solve the equation, we will isolate x on one side. 8(3x+4)=2(12x+16)Distribute 824x+32=2(12x+16)Distribute 224x+32=24x+32 This statement is always true as the left-hand side and the right-hand side are the same. The equation is an identity, which has infinitely many solutions.
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...⧽ 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 aLet B be the amount of money in our account, which currently is $3000, B=3000. We use this only for paying rent, $625 per month. Therefore, after m months, our balance is B=3000−625m. When our balance drops below $700, we get a text alert. When the balance is greater than or equal to this amount, we pay our rent without receiving the text. We can model this with the inequality: 3000−625m≥700.bTo find the maximum number of months that we can pay our rent without receiving a text alert, we solve the inequality from part A. 3000−625m≥700LHS−3000≥RHS−3000-625m≥-2300Divide by -625 and flip inequality signm≤3.68 The maximum number of months we can pay our rent without receiving a text alert is 3. From the solution set m≤3.68, we know that after 3 months we still have enough money in the account not to trigger the text message. It's not until after the 4th month's payment that the balance dips below $700. First month: 3000−625Second month: 2375−625Third month: 1750−625Fourth month: 1125−625​=2375=1750=1125=500​cFrom the previous part, we know that the maximum number of months we can pay our rent is 3, so the first month is June. We can pay the rent for 2 additional months, which are: July and August.