## Engage NY Eureka Math 8th Grade Module 4 End of Module Assessment Answer Key

### Eureka Math Grade 8 Module 4 End of Module Assessment Task Answer Key

Question 1.

Use the graph below to answer parts (a)â€“(c).

a. Use any pair of points to calculate the slope of the line.

Answer:

m = \(\frac{6-3}{0-2}=\frac{3}{-2}\) = –\(\frac{3}{2}\)

b. Use a different pair of points to calculate the slope of the line.

Answer:

m = \(\frac{6-0}{0-4}=\frac{6}{-4}\) = –\(\frac{3}{2}\)

c. Explain why the slopes you calculated in parts (a) and (b) are equal.

Answer:

The slopes are equal because the slope triangle are similar, âˆ†ABC ~ âˆ†AB’C’. Each triangle has 90Â° angle at âˆ ABC & âˆ AB’C’, respectively. They are 90Â° because they are at intersection of the grid lines. Both triangles share âˆ BAC. By the AA criterion âˆ†ABC ~ âˆ†AB’C’, which means their corresponding sides are equal in ratio.

\(\frac{\left|B^{\prime} C^{\prime}\right|}{|B C|}=\frac{\left|A B^{\prime}\right|}{|A B|}\) which is the same as \(\frac{\left|B^{\prime} C^{\prime}\right|}{\left|A B^{\prime}\right|}=\frac{|B C|}{|A B|}\) where –\(\frac{\left|B^{\prime} C^{\prime}\right|}{\left|A B^{\prime}\right|}\) is the slope in (b) and –\(\frac{|B C|}{|A B|}\) is the slope in (a).

Question 2.

Jeremy rides his bike at a rate of 12 miles per hour. Below is a table that represents the number of hours and miles Kevin rides. Assume both bikers ride at a constant rate.

a. Which biker rides at a greater speed? Explain your reasoning.

Answer:

Let Y be the distance triangled and X be the number of hours,

Then for jeremy, \(\frac{y}{x}\) = \(\frac{12}{1}\) â‡’ 12x

For kevin, \(\frac{46-23}{4-2}\) = \(\frac{23}{2}\) = 11.5, then y = 11.5x

When you compare their rates, 12 > 11.5, therefore jeremy rides at a greater speed.

Graphically:

b. Write an equation for a third biker, Lauren, who rides twice as fast as Kevin. Use y to represent the number of miles Lauren travels in x hours. Explain your reasoning.

Answer:

“Twice as Fast” means lauren goes twice the distance in the same time. Then in 2 hours she rides 46 miles and in 4 hours, 92 miles. If y is the total distance in x hours, y = \(\frac{46}{2}\) x

y = 23x

c. Create a graph of the equation in part (b).

Answer:

d. Calculate the slope of the line in part (c), and interpret its meaning in this situation.

Answer:

m = \(\frac{46-23}{2-1}\) = \(\frac{23}{1}\)

The slope is the rate that lauren rides, 23 miles per hour.

Question 3.

The cost of five protractors is $14.95 at Store A. The graph below compares the cost of protractors at Store A with the cost at Store B.

Estimate the cost of one protractor at Store B. Use evidence from the graph to justify your answer.

Answer:

The cost of protractors at store B is probably about $2.99 per protractor and it looks like the slope for store B is about half of the slope for store A.

Question 4.

Given the equation 3x+9y=-8, write a second linear equation to create a system that:

a. Has exactly one solution. Explain your reasoning.

Answer:

4x + 9y = -10

This equation has a slope different from 3x + 9y = -8. So the graphs of the equations will intersect.

b. Has no solution. Explain your reasoning.

Answer:

x + 3y = 10

This equation has the same slopes as 3x + 9y = -8, And no common points (solutions) Therefore the graphs of the equation are parallel lines.

c. Has infinitely many solutions. Explain your reasoning.

Answer:

6x + 18y = -16

This equation defines the same line as 3x + 9y = -8 and the graphs of the equations will coincide.

d. Interpret the meaning of the solution, if it exists, in the context of the graph of the following system of equations.

-5x+2y=10

10x-4y=-20

Answer:

-5x+2y=10 m = \(\frac{5}{2}\) (0, 5)

10x-4y=-20 m = \(\frac{5}{2}\) (0, 5)

This system will have infinitely many solutions because the graphs of these linear equations are the same line. Each equation has a slope of m = \(\frac{5}{2}\) and a y-intercept at (0, 5). There exists only one line through a point and a given slope. Therefore this system graphs as the same line and has infinitely many solutions.

Question 5.

Students sold 275 tickets for a fundraiser at school. Some tickets are for children and cost $3, while the rest are adult tickets that cost $5. If the total value of all tickets sold was $1,025, how many of each type of ticket was sold?

Answer:

Let X be the # of kids tickets

Let Y be the # of adults tickets

x + y = 275

3x + 5y = 1025

y = 100

x + 100 = 275

x = 175

(175, 100)

175 children’s tickets and 100 adults tickets were sold.

Question 6.

a. Determine the equation of the line connecting the points (0,-1) and (2,3).

Answer:

m = \(\frac{3-(-1)}{2 – 0}\) = \(\frac{4}{2}\) = 2

y = 2x – 1

b. Will the line described by the equation in part (a) intersect the line passing through the points (-2,4) and (-3,3)? Explain why or why not.

Answer:

m = \(\frac{4-3}{-2-(3)}\) = \(\frac{1}{1}\)

Yes, The lines will intersect because they have different slopes they will eventually intersect.

Question 7.

Line l_{1} and line l_{2} are shown on the graph below. Use the graph to answer parts (a)â€“(f).

a. What is the y-intercept of l_{1}?

Answer:

(0, 4)

b. What is the y-intercept of l_{2}?

Answer:

(0, 2)

c. Write a system of linear equations representing lines l_{1} and l_{2}.

Answer:

l_{1} : y = \(\frac{1}{2}\)x + 4

l_{2} : y = x + 2

d. Use the graph to estimate the solution to the system.

Answer:

(1.2, 3.3)

e. Solve the system of linear equations algebraically.

Answer:

y = \(\frac{1}{2}\)x + 4

y = x + 2

–\(\frac{1}{2}\) x + 4 = x + 2

4 = \(\frac{3}{2}\) x + 2

2 = \(\frac{3}{2}\) x

\(\frac{4}{3}\) = x

y = \(\frac{4}{3}\) + 2

= \(\frac{10}{3}\)

(\(\frac{4}{3}\), \(\frac{10}{3}\))

f. Show that your solution from part (e) satisfies both equations.

Answer:

\(\frac{10}{3}\) = –\(\frac{1}{2}\)(\(\frac{4}{3}\)) + 4

\(\frac{10}{3}\) = –\(\frac{2}{3}\) + 4

\(\frac{10}{3}\) = \(\frac{10}{3}\)

\(\frac{10}{3}\) = \(\frac{4}{3}\) + 2

\(\frac{10}{3}\) = \(\frac{10}{3}\)