Eureka Math

Engage NY Eureka Math 8th Grade Module 5 Lesson 10 Answer Key

Eureka Math Grade 8 Module 5 Lesson 10 Exercise Answer Key

Opening Exercise
a.
i. Write an equation to determine the volume of the rectangular prism shown below.

Answer:
V = 8(6)(h)
= 48h
The volume is 48h mm³.

ii. Write an equation to determine the volume of the rectangular prism shown below.

Answer:
V = 10(8)(h)
= 80h
The volume is 80h in³.

iii. Write an equation to determine the volume of the rectangular prism shown below.

Answer:
V = 6(4)(h)
= 24h
The volume is 24h cm³.

iv. Write an equation for volume, V, in terms of the area of the base, B.
V = Bh

b. Using what you learned in part (a), write an equation to determine the volume of the cylinder shown below.

Answer:
V = Bh
= 4² πh
= 16πh
The volume is 16πh cm³.

Exercises 1–3

Exercise 1.
Use the diagram to the right to answer the questions.

a. What is the area of the base?
Answer:
The area of the base is (4.5)(8.2) in² or 36.9 in².

b. What is the height?
Answer:
The height of the rectangular prism is 11.7 in.

c. What is the volume of the rectangular prism?
Answer:
The volume of the rectangular prism is 431.73 in³.

Exercise 2.
Use the diagram to the right to answer the questions.

a. What is the area of the base?
Answer:
A = π2²
A = 4π
The area of the base is 4π cm².

b. What is the height?
Answer:
The height of the right circular cylinder is 5.3 cm.

c. What is the volume of the right circular cylinder?
Answer:
V = (πr²)h
V = (4π)5.3
V = 21.2π
The volume of the right circular cylinder is 21.2π cm³.

Exercise 3.
Use the diagram to the right to answer the questions.

a. What is the area of the base?
Answer:
A = π6²
A = 36π
The area of the base is 36π in².

b. What is the height?
Answer:
The height of the right circular cylinder is 25 in.

c. What is the volume of the right circular cylinder?
Answer:
V = (36π)25
V = 900π
The volume of the right circular cylinder is 900π in³.

Exercises 4–6

Exercise 4.
Use the diagram to find the volume of the right circular cone.

Answer:
V = \(\frac{1}{3}\) (πr²)h
V = \(\frac{1}{3}\) (π4²)9
V = 48π
The volume of the right circular cone is 48π mm³.

Exercise 5.
Use the diagram to find the volume of the right circular cone.

Answer:
V = \(\frac{1}{3}\) (πr²)h
V = \(\frac{1}{3}\) (π2.3²)15
V = 26.45π
The volume of the right circular cone is 26.45π m³.

Exercise 6.
Challenge: A container in the shape of a right circular cone has height h, and base of radius r, as shown. It is filled with water (in its upright position) to half the height. Assume that the surface of the water is parallel to the base of the inverted cone. Use the diagram to answer the following questions:

a. What do we know about the lengths of AB and AO?
Answer:
Then we know that |AB| = r, and |AO| = h.

b. What do we know about the measure of ∠OAB and ∠OCD?
Answer:
∠OAB and ∠OCD are both right angles.

c. What can you say about △OAB and △OCD?
Answer:
We have two similar triangles, △OAB and △OCD by AA criterion.

d. What is the ratio of the volume of water to the volume of the container itself?
Answer:
Since \(\frac{|A B|}{|C D|}\) = \(\frac{|A O|}{|C O|}\), and |AO| = 2|OC|, we have \(\frac{|A B|}{|C D|}\) = 2\(\frac{2|O C|}{|C O|}\).
Then |AB| = 2|CD|.
Using the volume formula to determine the volume of the container, we have V = \(\frac{1}{3}\) π|AB|² |AO|.
By substituting |AB| with 2|CD| and |AO| with 2|OC| we get:
V = \(\frac{1}{3}\) π(2|CD|)² (2|OC|)
V = 8(\(\frac{1}{3}\) π|CD|² |OC|), where \(\frac{1}{3}\) π|CD|² |OC| gives the volume of the portion of the container that is filled with water.
Therefore, the volume of the water to the volume of the container is 1:8.

Eureka Math Grade 8 Module 5 Lesson 10 Problem Set Answer Key

Question 1.
Use the diagram to help you find the volume of the right circular cylinder.

Answer:
V = πr² h
V = π(1)² (1)
V = π
The volume of the right circular cylinder is π ft³.

Question 2.
Use the diagram to help you find the volume of the right circular cone.

Answer:
V = \(\frac{1}{3}\) πr² h
V = \(\frac{1}{3}\) π(2.8)² (4.3)
V = 11.237333…π
The volume of the right circular cone is about 11.2π cm³.

Question 3.
Use the diagram to help you find the volume of the right circular cylinder.

Answer:
If the diameter is 12 mm, then the radius is 6 mm.
V = πr² h
V = π(6)² (17)
V = 612π
The volume of the right circular cylinder is 612π mm³.

Question 4.
Use the diagram to help you find the volume of the right circular cone.

Answer:
If the diameter is 14 in., then the radius is 7 in.
V = \(\frac{1}{3}\) πr² h
V = \(\frac{1}{3}\) π(7)² (18.2)
V = 297.26666…π
V ≈ 297.3π
The volume of the right cone is about 297.3π in³.

Question 5.
Oscar wants to fill with water a bucket that is the shape of a right circular cylinder. It has a 6-inch radius and 12-inch height. He uses a shovel that has the shape of a right circular cone with a 3-inch radius and 4-inch height. How many shovelfuls will it take Oscar to fill the bucket up level with the top?
Answer:
V = πr² h
V = π(6)² (12)
V = 432π
The volume of the bucket is 432π in³.
V = \(\frac{1}{3}\) πr² h
V = \(\frac{1}{3}\) π(3)² (4)
V = 12π
The volume of shovel is 12π in³.
\(\frac{432 \pi}{12 \pi}\) = 36
It would take 36 shovelfuls of water to fill up the bucket.

Question 6.
A cylindrical tank (with dimensions shown below) contains water that is 1-foot deep. If water is poured into the tank at a constant rate of 20 \(\frac{\mathrm{ft}^{3}}{\mathrm{~min}}\) for 20 min., will the tank overflow? Use 3.14 to estimate π.

Answer:
V = πr² h
V = π(3)² (12)
V = 108π
The volume of the tank is about 339.12 ft³.
V = πr² h
V = π(3)² (1)
V = 9π
There is about 28.26 ft³ of water already in the tank. There is about 310.86 ft³ of space left in the tank. If the water is poured at a constant rate for 20 min., 400 ft³ will be poured into the tank, and the tank will overflow.

Eureka Math Grade 8 Module 5 Lesson 10 Exit Ticket Answer Key

Question 1.
Use the diagram to find the total volume of the three cones shown below.

Answer:
Since all three cones have the same base and height, the volume of the three cones will be the same as finding the volume of a cylinder with the same base radius and same height.
V = πr² h
V = π(2)²3
V = 12π
The volume of all three cones is 12π ft³.

Question 2.
Use the diagram below to determine which has the greater volume, the cone or the cylinder.

Answer:
V = πr² h
V = π4² (6)
V = 96π
The volume of the cylinder is 96π cm³.
V = \(\frac{1}{3}\) πr² h
V = \(\frac{1}{3}\) π6² (8)
V = 96π
The volume of the cone is 96π cm³.
The volume of the cylinder and the volume of the cone are the same, 96π cm³.

Engage NY Eureka Math 8th Grade Module 5 Lesson 9 Answer Key

Eureka Math Grade 8 Module 5 Lesson 9 Exploratory Challenge/Exercise Answer Key

Exploratory Challenge 1/Exercises 1–4
As you complete Exercises 1–4, record the information in the table below.

Answer:

Exercise 1.
Use the figure below to answer parts (a)–(f).

a. What is the length of one side of the smaller, inner square?
Answer:
The length of one side of the smaller square is 6 in.

b. What is the area of the smaller, inner square?
Answer:
6² = 36
The area of the smaller square is 36 in².

c. What is the length of one side of the larger, outer square?
Answer:
The length of one side of the larger square is 8 in.

d. What is the area of the larger, outer square?
Answer:
8² = 64
The area of the larger square is 64 in².

e. Use your answers in parts (b) and (d) to determine the area of the 1 – inch white border of the figure.
Answer:
64 – 36 = 28
The area of the 1 – inch white border is 28 in².

f. Explain your strategy for finding the area of the white border.
Answer:
First, I had to determine the length of one side of the larger, outer square. Since the inner square is 6 in. and the border is 1 in. on all sides, then the length of one side of the larger square is (6 + 2) in = 8 in. Then, the area of the larger square is 64 in². Next, I found the area of the smaller, inner square. Since one side length is 6 in., the area is 36 in². To find the area of the white border, I needed to subtract the area of the inner square from the area of the outer square.

Exercise 2.
Use the figure below to answer parts (a)–(f).

a. What is the length of one side of the smaller, inner square?
Answer:
The length of one side of the smaller square is 9 in.

b. What is the area of the smaller, inner square?
Answer:
9² = 81
The area of the smaller square is 81 in².

c. What is the length of one side of the larger, outer square?
Answer:
The length of one side of the larger square is 11 in.

d. What is the area of the larger, outer square?
Answer:
11² = 121
The area of the larger square is 121 in².

e. Use your answers in parts (b) and (d) to determine the area of the 1 – inch white border of the figure.
Answer:
121 – 81 = 40
The area of the 1 – inch white border is 40 in².

f. Explain your strategy for finding the area of the white border.
Answer:
First, I had to determine the length of one side of the larger, outer square. Since the inner square is 9 in. and the border is 1 in. on all sides, the length of one side of the larger square is (9 + 2) in = 11 in. Therefore, the area of the larger square is 121 in². Then, I found the area of the smaller, inner square. Since one side length is 9 in., the area is 81 in². To find the area of the white border, I needed to subtract the area of the inner square from the area of the outer square.

Exercise 3.
Use the figure below to answer parts (a)–(f).

a. What is the length of one side of the smaller, inner square?
Answer:
The length of one side of the smaller square is 13 in.

b. What is the area of the smaller, inner square?
Answer:
13² = 169
The area of the smaller square is 169 in².

c. What is the length of one side of the larger, outer square?
Answer:
The length of one side of the larger square is 15 in.

d. What is the area of the larger, outer square?
Answer:
15² = 225
The area of the larger square is 225 in².

e. Use your answers in parts (b) and (d) to determine the area of the 1 – inch white border of the figure.
Answer:
225 – 169 = 56
The area of the 1 – inch white border is 56 in².

f. Explain your strategy for finding the area of the white border.
Answer:
First, I had to determine the length of one side of the larger, outer square. Since the inner square is 13 in. and the border is 1 in. on all sides, the length of one side of the larger square is (13 + 2) in = 15 in. Therefore, the area of the larger square is 225 in². Then, I found the area of the smaller, inner square. Since one side length is 13 in., the area is 169 in². To find the area of the white border, I needed to subtract the area of the inner square from the area of the outer square.

Exercise 4.
Write a function that would allow you to calculate the area of a 1 – inch white border for any sized square picture measured in inches.

a. Write an expression that represents the side length of the smaller, inner square.
Answer:
Symbols used will vary. Expect students to use s or x to represent one side of the smaller, inner square. Answers that follow will use s as the symbol to represent one side of the smaller, inner square.

b. Write an expression that represents the area of the smaller, inner square.
Answer:
s²

c. Write an expression that represents the side lengths of the larger, outer square.
Answer:
s + 2

d. Write an expression that represents the area of the larger, outer square.
Answer:
(s + 2)²

e. Use your expressions in parts (b) and (d) to write a function for the area A of the 1 – inch white border for any sized square picture measured in inches.
Answer:
A = (s + 2)² – s²

Exercises 5–6

Exercise 5.
The volume of the prism shown below is 61.6 in³. What is the height of the prism?

Answer:
Let x represent the height of the prism.
61.6 = 8(2.2)x
61.6 = 17.6x
3.5 = x
The height of the prism is 3.5 in.

Exercise 6.
Find the value of the ratio that compares the volume of the larger prism to the smaller prism.

Answer:
Volume of larger prism:
V = 7(9)(5)
= 315
The volume of the larger prism is 315 cm³.
Volume of smaller prism:
V = 2(4.5)(3)
= 27
The volume of the smaller prism is 27 cm³.
The ratio that compares the volume of the larger prism to the smaller prism is 315:27. The value of the ratio is \(\frac{315}{27}\) = \(\frac{35}{3}\).

Exploratory Challenge 2/Exercises 7–10
As you complete Exercises 7–10, record the information in the table below. Note that base refers to the bottom of the prism.

Answer:

Exercise 7.
Use the figure to the right to answer parts (a)–(c).

a. What is the area of the base?
Answer:
The area of the base is 36 cm².

b. What is the height of the figure?
Answer:
The height is 3 cm.

c. What is the volume of the figure?
Answer:
The volume of the rectangular prism is 108 cm³.

Exercise 8.
Use the figure to the right to answer parts (a)–(c).

a. What is the area of the base?
Answer:
The area of the base is 36 cm².

b. What is the height of the figure?
Answer:
The height is 8 cm.

c. What is the volume of the figure?
Answer:
The volume of the rectangular prism is 288 cm³.

Exercise 9.
Use the figure to the right to answer parts (a)–(c).

a. What is the area of the base?
Answer:
The area of the base is 36 cm².

b. What is the height of the figure?
Answer:
The height is 15 cm.

c. What is the volume of the figure?
Answer:
The volume of the rectangular prism is 540 cm³.

Exercise 10.
Use the figure to the right to answer parts (a)–(c).

a. What is the area of the base?
Answer:
The area of the base is 36 cm².

b. What is the height of the figure?
Answer:
The height is x cm.

c. Write and describe a function that will allow you to determine the volume of any rectangular prism that has a base area of
Answer:
36 cm².
The rule that describes the function is V = 36x, where V is the volume and x is the height of the rectangular prism. The volume of a rectangular prism with a base area of 36 cm² is a function of its height.

Eureka Math Grade 8 Module 5 Lesson 9 Problem Set Answer Key

Question 1.
Calculate the area of the 3 – inch white border of the square figure below.

Answer:
17² = 289
11² = 121
The area of the 3 – inch white border is 168 in².

Question 2.
Write a function that would allow you to calculate the area, A, of a 3 – inch white border for any sized square picture measured in inches.

Answer:
Let s represent the side length of the inner square in inches. Then, the area of the inner square is s² square inches. The side length of the outer square, in inches, is s + 6, which means that the area of the outer square, in square inches, is (s + 6)². The function that describes the area, A, of the 3 – inch border is in square inches
A = (s + 6)² – s².

Question 3.
Dartboards typically have an outer ring of numbers that represent the number of points a player can score for getting a dart in that section. A simplified dartboard is shown below. The center of the circle is point A. Calculate the area of the outer ring. Write an exact answer that uses π (do not approximate your answer by using 3.14 for π).

Answer:
Inner ring area: πr² = π(6² ) = 36 π
Outer ring: πr² = π(6 + 2)² = π(8² ) = 64 π
Difference in areas: 64 π – 36 π = (64 – 34)π = 28 π
The inner ring has an area of 36π in². The area of the inner ring including the border is 64π in². The difference is the area of the border, 28π in².

Question 4.
Write a function that would allow you to calculate the area, A, of the outer ring for any sized dartboard with radius r. Write an exact answer that uses π (do not approximate your answer by using 3.14 for π).

Answer:
Inner ring area: πr²
Outer ring: πr² = π(r + 2)²
Difference in areas: Inner ring area: π(r + 2)² – πr²
The inner ring has an area of πr² in². The area of the inner ring including the border is π(r + 2)² in². Let A be the area of the outer ring. Then, the function that would describe that area in square inches is
A = π(r + 2)² – πr².

Question 5.
The shell of the solid shown was filled with water and then poured into the standard rectangular prism, as shown. The height that the volume reaches is 14.2 in. What is the volume of the shell of the solid?

Answer:
V = Bh
= 1(14.2)
= 14.2
The volume of the shell of the solid is 14.2 in³.

Question 6.
Determine the volume of the rectangular prism shown below.

Answer:
6.4 × 5.1 × 10.2 = 332.928
The volume of the prism is 332.928 in³.

Question 7.
The volume of the prism shown below is 972 cm³. What is its length?

Answer:
Let x represent the length of the prism.
972 = 8.1(5)x
972 = 40.5x
24 = x
The length of the prism is 24 cm.

Question 8.
The volume of the prism shown below is 32.7375 ft³. What is its width?

Answer:
Let x represent the width.
32.7375 = (0.75)(4.5)x
32.7375 = 3.375x
9.7 = x
The width of the prism is 9.7 ft.

Question 9.
Determine the volume of the three – dimensional figure below. Explain how you got your answer.

Answer:
2 × 2.5 × 1.5 = 7.5
2 × 1 × 1 = 2
The volume of the top rectangular prism is 7.5 units³. The volume of the bottom rectangular prism is 2 units³. The figure is made of two rectangular prisms, and since the rectangular prisms only touch at their boundaries, we can add their volumes together to obtain the volume of the figure. The total volume of the three – dimensional figure is 9.5 units³.

Eureka Math Grade 8 Module 5 Lesson 9 Exit Ticket Answer Key

Question 1.
Write a function that would allow you to calculate the area in square inches, A, of a 2 – inch white border for any sized square figure with sides of length s measured in inches.

Answer:
Let s represent the side length of the inner square in inches. Then, the area of the inner square is s² square inches. The side length of the larger square, in inches, is s + 4, and the area in square inches is (s + 4)². If A is the area of the 2 – inch border, then the function that describes A in square inches is
A = (s + 4)² – s².

Question 2.
The volume of the rectangular prism is 295.68 in³. What is its width?

Answer:
Let x represent the width of the prism.
295.68 = 11(6.4)x
295.68 = 70.4x
4.2 = x
The width of the prism is 4.2 in.

Engage NY Eureka Math 8th Grade Module 5 Lesson 7 Answer Key

Eureka Math Grade 8 Module 5 Lesson 7 Exploratory Challenge/Exercise Answer Key

Exploratory Challenge/Exercises 1–4
Each of Exercises 1–4 provides information about two functions. Use that information given to help you compare the two functions and answer the questions about them.

Exercise 1.
Alan and Margot each drive from City A to City B, a distance of 147 miles. They take the same route and drive at constant speeds. Alan begins driving at 1:40 p.m. and arrives at City B at 4:15 p.m. Margot’s trip from City A to City B can be described with the equation y = 64x, where y is the distance traveled in miles and x is the time in minutes spent traveling. Who gets from City A to City B faster?
Answer:
Student solutions will vary. Sample solution is provided.
It takes Alan 155 minutes to travel the 147 miles. Therefore, his constant rate is \(\frac{147}{155}\) miles per minute.
Margot drives 64 miles per hour (60 minutes). Therefore, her constant rate is \(\frac{64}{60}\) miles per minute.
To determine who gets from City A to City B faster, we just need to compare their rates in miles per minute.
\(\frac{147}{155}\) < \(\frac{64}{60}\)
Since Margot’s rate is faster, she will get to City B faster than Alan.

Exercise 2.
You have recently begun researching phone billing plans. Phone Company A charges a flat rate of $75 a month. A flat rate means that your bill will be $75 each month with no additional costs. The billing plan for Phone Company B is a linear function of the number of texts that you send that month. That is, the total cost of the bill changes each month depending on how many texts you send. The table below represents some inputs and the corresponding outputs that the function assigns.

At what number of texts would the bill from each phone plan be the same? At what number of texts is Phone Company A the better choice? At what number of texts is Phone Company B the better choice?
Answer:
Student solutions will vary. Sample solution is provided.
The equation that represents the function for Phone Company A is y = 75.
To determine the equation that represents the function for Phone Company B, we need the rate of change. (We are told it is constant.)
\(\frac{60 – 50}{150 – 50}\) = \(\frac{10}{100}\)
= 0.1
The equation for Phone Company B is shown below.
Using the assignment of 50 to 50,
50 = 0.1(50) + b
50 = 5 + b
45 = b.
The equation that represents the function for Phone Company B is y = 0.1x + 45.
We can determine at what point the phone companies charge the same amount by solving the system:
y = 75
y = 0.1x + 45

75 = 0.1x + 45
30 = 0.1x
300 = x
After 300 texts are sent, both companies would charge the same amount, $75. More than 300 texts means that the bill from Phone Company B will be higher than Phone Company A. Less than 300 texts means the bill from Phone Company A will be higher.

Exercise 3.
The function that gives the volume of water, y, that flows from Faucet A in gallons during x minutes is a linear function with the graph shown. Faucet B’s water flow can be described by the equation y = \(\frac{5}{6}\) x, where y is the volume of water in gallons that flows from the faucet during x minutes. Assume the flow of water from each faucet is constant. Which faucet has a faster rate of flow of water? Each faucet is being used to fill a tub with a volume of 50 gallons. How long will it take each faucet to fill its tub? How do you know?

Suppose the tub being filled by Faucet A already had 15 gallons of water in it, and the tub being filled by Faucet B started empty. If now both faucets are turned on at the same time, which faucet will fill its tub fastest?
Answer:
Student solutions will vary. Sample solution is provided.
The slope of the graph of the line is \(\frac{4}{7}\) because (7, 4) is a point on the line that represents 4 gallons of water that flows in 7 minutes. Therefore, the rate of water flow for Faucet A is \(\frac{4}{7}\). To determine which faucet has a faster flow of water, we can compare their rates.
\(\frac{4}{7}\) < \(\frac{5}{6}\)
Therefore, Faucet B has a faster rate of water flow.

Exercise 4.
Two people, Adam and Bianca, are competing to see who can save the most money in one month. Use the table and the graph below to determine who will save the most money at the end of the month. State how much money each person had at the start of the competition. (Assume each is following a linear function in his or her saving habit.)

Answer:
The slope of the line that represents Adam’s savings is 3; therefore, the rate at which Adam is saving money is $3 per day. According to the table of values for Bianca, she is also saving money at a rate of $3 per day:
\(\frac{26 – 17}{8 – 5}\) = \(\frac{9}{3}\) = 3
\(\frac{38 – 26}{12 – 8}\) = \(\frac{12}{4}\) = 3
\(\frac{62 – 26}{20 – 8}\) = \(\frac{36}{12}\) = 3
Therefore, at the end of the month, Adam and Bianca will both have saved the same amount of money.
According to the graph for Adam, the equation y = 3x + 3 represents the function of money saved each day. On day zero, he had $3.
The equation that represents the function of money saved each day for Bianca is y = 3x + 2 because, using the assignment of 17 to 5
17 = 3(5) + b
17 = 15 + b
2 = b.
The amount of money Bianca had on day zero was $2.

Eureka Math Grade 8 Module 5 Lesson 7 Problem Set Answer Key

Question 1.
The graph below represents the distance in miles, y, Car A travels in x minutes. The table represents the distance in miles, y, Car B travels in x minutes. It is moving at a constant rate. Which car is traveling at a greater speed? How do you know?
Car A:

Answer:
Based on the graph, Car A is traveling at a rate of 2 miles every 3 minutes, m = 2/3. From the table, the constant rate that Car B is traveling is
\(\frac{25 – 12.5}{30 – 15}\) = \(\frac{12.5}{15}\) = \(\frac{25}{30}\) = \(\frac{5}{6}\).

Since \(\frac{5}{6}\)>\(\frac{2}{3}\), Car B is traveling at a greater speed.

Question 2.
The local park needs to replace an existing fence that is 6 feet high. Fence Company A charges $7,000 for building materials and $200 per foot for the length of the fence. Fence Company B charges are based solely on the length of the fence. That is, the total cost of the 6 – foot high fence will depend on how long the fence is. The table below represents some inputs and their corresponding outputs that the cost function for Fence Company B assigns. It is a linear function.

a. Which company charges a higher rate per foot of fencing? How do you know?
Answer:
Let x represent the length of the fence and y represent the total cost.
The equation that represents the function for Fence Company A is y = 200x + 7,000. So, the rate is 200 dollars per foot of fence.
The rate of change for Fence Company B is given by:
\(\frac{26,000 – 31,200}{100 – 120}\) = \(\frac{ – 5,200}{ – 20}\)
= 260
Fence Company B charges $260 per foot of fence, which is a higher rate per foot of fence length than Fence Company A.

b. At what number of the length of the fence would the cost from each fence company be the same? What will the cost be when the companies charge the same amount? If the fence you need were 190 feet in length, which company would be a better choice?
Answer:
Student solutions will vary. Sample solution is provided.
The equation for Fence Company B is
y = 260x.
We can find out at what point the fence companies charge the same amount by solving the system
y = 200x + 7000
y = 260x

200x + 7,000 = 260x
7,000 = 60x
116.6666…… = x
116.7 ≈ x
At 116.7 feet of fencing, both companies would charge the same amount (about $30,340). Less than 116.7 feet of fencing means that the cost from Fence Company A will be more than Fence Company B. More than 116.7 feet of fencing means that the cost from Fence Company B will be more than Fence Company A. So, for 190 feet of fencing, Fence Company A is the better choice.

Question 3.
The equation y = 123x describes the function for the number of toys, y, produced at Toys Plus in x minutes of production time. Another company, #1 Toys, has a similar function, also linear, that assigns the values shown in the table below. Which company produces toys at a slower rate? Explain.

Answer:
We are told that #1 Toys produces toys at a constant rate. That rate is:
\(\frac{1,320 – 600}{11 – 5}\) = \(\frac{720}{6}\)
= 120
The rate of production for #1 Toys is 120 toys per minute. The rate of production for Toys Plus is 123 toys per minute. Since 120 is less than 123, #1 Toys produces toys at a slower rate.

Question 4.
A train is traveling from City A to City B, a distance of 320 miles. The graph below shows the number of miles, y, the train travels as a function of the number of hours, x, that have passed on its journey. The train travels at a constant speed for the first four hours of its journey and then slows down to a constant speed of 48 miles per hour for the remainder of its journey.

a. How long will it take the train to reach its destination?
Answer:
Student solutions will vary. Sample solution is provided.
We see from the graph that the train travels 220 miles during its first four hours of travel. It has 100 miles remaining to travel, which it shall do at a constant speed of 48 miles per hour. We see that it will take about 2 hours more to finish the trip:
100 = 48x
2.08333… = x
2.1 ≈ x.
This means it will take about 6.1 hours (4 + 2.1 = 6.1) for the train to reach its destination.

b. If the train had not slowed down after 4 hours, how long would it have taken to reach its destination?
Answer:
320 = 55x
5.8181818…. = x
5.8 ≈ x
The train would have reached its destination in about 5.8 hours had it not slowed down.

c. Suppose after 4 hours, the train increased its constant speed. How fast would the train have to travel to complete the destination in 1.5 hours?
Answer:
Let m represent the new constant speed of the train.
100 = m(1.5)
66.6666…. = x
66.7 ≈ x
The train would have to increase its speed to about 66.7 miles per hour to arrive at its destination 1.5 hours later.

Question 5.
a. A hose is used to fill up a 1,200 gallon water truck. Water flows from the hose at a constant rate. After 10 minutes, there are 65 gallons of water in the truck. After 15 minutes, there are 82 gallons of water in the truck. How long will it take to fill up the water truck? Was the tank initially empty?
Answer:
Student solutions will vary. Sample solution is provided.
Let x represent the time in minutes it takes to pump y gallons of water. Then, the rate can be found as follows:

\(\frac{65 – 82}{10 – 15}\) = \(\frac{ – 17}{ – 5}\)
= \(\frac{17}{5}\)
Since the water is pumping at a constant rate, we can assume the equation is linear. Therefore, the equation for the volume of water pumped from the hose is found by
65 = \(\frac{17}{5}\) (10) + b
65 = 34 + b
31 = b
The equation is y = \(\frac{17}{5}\) x + 31, and we see that the tank initially had 31 gallons of water in it. The time to fill the tank is given by
1200 = \(\frac{17}{5}\) x + 31
1169 = \(\frac{17}{5}\) x
343.8235… = x
343.8 ≈ x
It would take about 344 minutes or about 5.7 hours to fill up the truck.

b. The driver of the truck realizes that something is wrong with the hose he is using. After 30 minutes, he shuts off the hose and tries a different hose. The second hose flows at a constant rate of 18 gallons per minute. How long now does it take to fill up the truck?
Since the first hose has been pumping for 30 minutes, there are 133 gallons of water already in the truck. That means the new hose only has to fill up 1,067 gallons. Since the second hose fills up the truck at a constant rate of 18 gallons per minute, the equation for the second hose is y = 18x.
Answer:
1067 = 18x
59.27 = x
59.3 ≈ x
It will take the second hose about 59.3 minutes (or a little less than an hour) to finish the job.

Eureka Math Grade 8 Module 5 Lesson 7 Exit Ticket Answer Key

Question 1.
Brothers Paul and Pete walk 2 miles to school from home. Paul can walk to school in 24 minutes. Pete has slept in again and needs to run to school. Paul walks at a constant rate, and Pete runs at a constant rate. The graph of the function that represents Pete’s run is shown below.

a. Which brother is moving at a greater rate? Explain how you know.
Answer:
Paul takes 24 minutes to walk 2 miles; therefore, his rate is \(\frac{1}{12}\) miles per minute.
Pete can run 8 miles in 60 minutes; therefore, his rate is \(\frac{8}{60}\), or \(\frac{2}{15}\) miles per minute.
Since \(\frac{2}{15}\)>\(\frac{1}{12}\), Pete is moving at a greater rate.

b. If Pete leaves 5 minutes after Paul, will he catch up to Paul before they get to school?
Answer:
Student solution methods will vary. Sample answer is shown.
Since Pete slept in, we need to account for that fact. So, Pete’s time would be decreased. The equation that would represent the number of miles Pete runs, y, in x minutes, would be
y = \(\frac{2}{15}\)(x – 5).
The equation that would represent the number of miles Paul walks, y, in x minutes, would be y = \(\frac{1}{12}\) x.
To find out when they meet, solve the system of equations:
y = \(\frac{2}{15}\) x – \(\frac{2}{3}\)
y = \(\frac{1}{12}\) x

\(\frac{2}{15}\) x – \(\frac{2}{3}\) = \(\frac{1}{12}\) x
\(\frac{2}{15}\) x – \(\frac{2}{3}\) – \(\frac{1}{12}\) x + \(\frac{2}{3}\) = \(\frac{1}{12}\) x – \(\frac{1}{12}\) x + \(\frac{2}{3}\)
\(\frac{1}{20}\) x = \(\frac{2}{3}\)
(\(\frac{20}{1}\)) \(\frac{1}{20}\) x = \(\frac{2}{3}\) (\(\frac{20}{1}\))
x = \(\frac{40}{3}\)
y = \(\frac{1}{12}\) (\(\frac{40}{3}\)) = \(\frac{10}{9}\) or y = \(\frac{2}{15}\) (\(\frac{40}{3}\)) – \(\frac{2}{3}\)
Pete would catch up to Paul in \(\frac{40}{3}\) minutes, which occurs \(\frac{10}{9}\) miles from their home. Yes, he will catch Paul before they get to school because it is less than the total distance, two miles, to school.

Eureka Math Grade 8 Module 5 Lesson 7 Multi – Step Equations II Answer Key

Question 1.
2(x + 5) = 3(x + 6)
Answer:
x = – 8

Question 2.
3(x + 5) = 4(x + 6)
Answer:
x = – 9

Question 3.
4(x + 5) = 5(x + 6)
Answer:
x = – 10

Question 4.
– (4x + 1) = 3(2x – 1)
Answer:
x = \(\frac{1}{5}\)

Question 5.
3(4x + 1) = – (2x – 1)
Answer:
x = – \(\frac{1}{7}\)

Question 6.
– 3(4x + 1) = 2x – 1
Answer:
x = – \(\frac{1}{7}\)

Question 7.
15x – 12 = 9x – 6
Answer:
x = 1

Question 8.
\(\frac{1}{3}\) (15x – 12) = 9x – 6
x = \(\frac{1}{2}\)

Question 9.
\(\frac{2}{3}\) (15x – 12) = 9x – 6
Answer:
x = 2

Engage NY Eureka Math 8th Grade Module 5 Lesson 5 Answer Key

Eureka Math Grade 8 Module 5 Lesson 5 Exploratory Challenge/Exercise Answer Key

Exploratory Challenge/Exercises 1–3
Exercise 1.
The distance that Giselle can run is a function of the amount of time she spends running. Giselle runs 3 miles in 21 minutes. Assume she runs at a constant rate.
a. Write an equation in two variables that represents her distance run, y, as a function of the time, x, she spends running.
\(\frac{3}{21}\) = \(\frac{y}{x}\)
y = \(\frac{1}{7}\) x

b. Use the equation you wrote in part (a) to determine how many miles Giselle can run in 14 minutes.
Answer:
y = \(\frac{1}{7}\) (14)
y = 2
Giselle can run 2 miles in 14 minutes.

c. Use the equation you wrote in part (a) to determine how many miles Giselle can run in 28 minutes.
Answer:
y = \(\frac{1}{7}\) (28)
y = 4
Giselle can run 4 miles in 28 minutes.

d. Use the equation you wrote in part (a) to determine how many miles Giselle can run in 7 minutes.
Answer:
y = \(\frac{1}{7}\) (7)
y = 1
Giselle can run 1 mile in 7 minutes.

e. For a given input x of the function, a time, the matching output of the function, y, is the distance Giselle ran in that time. Write the inputs and outputs from parts (b)–(d) as ordered pairs, and plot them as points on a coordinate plane.

Answer:
(14, 2), (28, 4), (7, 1)

f. What do you notice about the points you plotted?
Answer:
The points appear to be in a line.

g. Is the function discrete?
Answer:
The function is not discrete because we can find the distance Giselle runs for any given amount of time she spends running.

h. Use the equation you wrote in part (a) to determine how many miles Giselle can run in 36 minutes. Write your answer as an ordered pair, as you did in part (e), and include the point on the graph. Is the point in a place where you expected it to be? Explain.
Answer:
y = \(\frac{1}{7}\) (36)
y = \(\frac{36}{7}\)
y = 5 \(\frac{1}{7}\)
(36, 5 \(\frac{1}{7}\)) The point is where I expected it to be because it is in line with the other points.

i. Assume you used the rule that describes the function to determine how many miles Giselle can run for any given time and wrote each answer as an ordered pair. Where do you think these points would appear on the graph?
Answer:
I think all of the points would fall on a line.

j. What do you think the graph of all the possible input/output pairs would look like? Explain.
Answer:
I know the graph will be a line as we can find all of the points that represent fractional intervals of time too. We also know that Giselle runs at a constant rate, so we would expect that as the time she spends running increases, the distance she can run will increase at the same rate.

k. Connect the points you have graphed to make a line. Select a point on the graph that has integer coordinates. Verify that this point has an output that the function would assign to the input.
Answer:
Answers will vary. Sample student work:
The point (42, 6) is a point on the graph.
y = \(\frac{1}{7}\) x
6 = \(\frac{1}{7}\) (42)
6 = 6
The function assigns the output of 6 to the input of 42.

l. Sketch the graph of the equation y = \(\frac{1}{7}\) x using the same coordinate plane in part (e). What do you notice about the graph of all the input/output pairs that describes Giselle’s constant rate of running and the graph of the equation y = \(\frac{1}{7}\) x?
Answer:
The graphs of the equation and the function coincide completely.

Exercise 2.
Sketch the graph of the equation y = x² for positive values of x. Organize your work using the table below, and then answer the questions that follow.

Answer:

a. Plot the ordered pairs on the coordinate plane.

Answer:

b. What shape does the graph of the points appear to take?
Answer:
It appears to take the shape of a curve.

c. Is this equation a linear equation? Explain.
Answer:
No, the equation y = x² is not a linear equation because the exponent of x is greater than 1.

d. Consider the function that assigns to each square of side length s units its area A square units. Write an equation that describes this function.
Answer:
A = s²

e. What do you think the graph of all the input/output pairs (s, A) of this function will look like? Explain.
Answer:
I think the graph of input/output pairs will look like the graph of the equation y = x². The inputs and outputs would match the solutions to the equation exactly. For the equation, the y value is the square of the x value. For the function, the output is the square of the input.

f. Use the function you wrote in part (d) to determine the area of a square with side length 2.5 units. Write the input and output as an ordered pair. Does this point appear to belong to the graph of y = x²?
Answer:
A = (2.5)²
A = 6.25
The area of the square is 6.25 units squared. (2.5, 6.25) The point looks like it would belong to the graph of y = x²; it looks like it would be on the curve that the shape of the graph is taking.

Exercise 3.
The number of devices a particular manufacturing company can produce is a function of the number of hours spent making the devices. On average, 4 devices are produced each hour. Assume that devices are produced at a constant rate.
a. Write an equation in two variables that describes the number of devices, y, as a function of the time the company spends making the devices, x.
Answer:
\(\frac{4}{1}\) = \(\frac{y}{x}\)
y = 4x

b. Use the equation you wrote in part (a) to determine how many devices are produced in 8 hours.
Answer:
y = 4(8)
y = 32
The company produces 32 devices in 8 hours.

c. Use the equation you wrote in part (a) to determine how many devices are produced in 6 hours.
Answer:
y = 4(6)
y = 24
The company produces 24 devices in 6 hours.

d. Use the equation you wrote in part (a) to determine how many devices are produced in 4 hours.
Answer:
y = 4(4)
y = 16
The company produces 16 devices in 4 hours.

e. The input of the function, x, is time, and the output of the function, y, is the number of devices produced. Write the inputs and outputs from parts (b)–(d) as ordered pairs, and plot them as points on a coordinate plane.

Answer:

(8, 32), (6, 24), (4, 16)

f. What shape does the graph of the points appear to take?
Answer:
The points appear to be in a line.

g. Is the function discrete?
Answer:
The function is not discrete because we can find the number of devices produced for any given time, including fractions of an hour.

h. Use the equation you wrote in part (a) to determine how many devices are produced in 1.5 hours. Write your answer as an ordered pair, as you did in part (e), and include the point on the graph. Is the point in a place where you expected it to be? Explain.
Answer:
y = 4(1.5)
y = 6
(1.5, 6) The point is where I expected it to be because it is in line with the other points.

i. Assume you used the equation that describes the function to determine how many devices are produced for any given time and wrote each answer as an ordered pair. Where do you think these points would appear on the graph?
Answer:
I think all of the points would fall on a line.

j. What do you think the graph of all possible input/output pairs will look like? Explain.
Answer:
I think the graph of this function will be a line. Since the rate is continuous, we can find all of the points that represent fractional intervals of time. We also know that devices are produced at a constant rate, so we would expect that as the time spent producing devices increases, the number of devices produced would increase at the same rate.

k. Connect the points you have graphed to make a line. Select a point on the graph that has integer coordinates. Verify that this point has an output that the function would assign to the input.
Answer:
Answers will vary. Sample student work:
The point (5, 20) is a point on the graph.
y = 4x
20 = 4(5)
20 = 20
The function assigns the output of 20 to the input of 5.

l. Sketch the graph of the equation y = 4x using the same coordinate plane in part (e). What do you notice about the graph of input/output pairs that describes the company’s constant rate of producing devices and the graph of the equation y = 4x?
Answer:
The graphs of the equation and the function coincide completely.

Exploratory Challenge/Exercise 4.
Examine the three graphs below. Which, if any, could represent the graph of a function? Explain why or why not for each graph.
Graph 1:

Answer:
This is the graph of a function. Each input is a real number x, and we see from the graph that there is an output y to associate with each such input. For example, the ordered pair (-2, 4) on the line associates the output 4 to the input -2.

Graph 2:

Answer:
This is not the graph of a function. The ordered pairs (6, 4) and (6, 6) show that for the input of 6 there are two different outputs, both 4 and 6. We do not have a function.

Graph 3:

Answer:
This is the graph of a function. The ordered pairs (-3, -9), (-2, -4), (-1, -1), (0, 0), (1, -1), (2, -4), and (3, -9) represent inputs and their unique outputs.

Eureka Math Grade 8 Module 5 Lesson 5 Problem Set Answer Key

Question 1.
The distance that Scott walks is a function of the time he spends walking. Scott can walk \(\frac{1}{2}\) mile every 8 minutes. Assume he walks at a constant rate.
a. Predict the shape of the graph of the function. Explain.
Answer:
The graph of the function will likely be a line because a linear equation can describe Scott’s motion, and I know that the graph of the function will be the same as the graph of the equation.

b. Write an equation to represent the distance that Scott can walk in miles, y, in x minutes.
Answer:
\(\frac{0.5}{8}\) = \(\frac{y}{x}\)
y = \(\frac{0.5}{8}\) x
y = \(\frac{1}{16}\) x

c. Use the equation you wrote in part (b) to determine how many miles Scott can walk in 24 minutes.
Answer:
y = \(\frac{1}{16}\) (24)
y = 1.5
Scott can walk 1.5 miles in 24 minutes.

d. Use the equation you wrote in part (b) to determine how many miles Scott can walk in 12 minutes.
Answer:
y = \(\frac{1}{16}\) (12)
y = \(\frac{3}{4}\)
Scott can walk 0.75 miles in 12 minutes.

e. Use the equation you wrote in part (b) to determine how many miles Scott can walk in 16 minutes.
Answer:
y = \(\frac{1}{16}\) (16)
y = 1
Scott can walk 1 mile in 16 minutes.

f. Write your inputs and corresponding outputs as ordered pairs, and then plot them on a coordinate plane.

Answer:
(24, 1.5), (12, 0.75), (16, 1)

g. What shape does the graph of the points appear to take? Does it match your prediction?
Answer:
The points appear to be in a line. Yes, as I predicted, the graph of the function is a line.

h. Connect the points to make a line. What is the equation of the line?
Answer:
It is the equation that described the function: y = \(\frac{1}{16}\) x.

Question 2.
Graph the equation y = x³ for positive values of x. Organize your work using the table below, and then answer the questions that follow.

Answer:

a. Plot the ordered pairs on the coordinate plane.

Answer:

b. What shape does the graph of the points appear to take?
Answer:
It appears to take the shape of a curve.

c. Is this the graph of a linear function? Explain.
Answer:
No, this is not the graph of a linear function. The equation y = x³ is not a linear equation.

d. Consider the function that assigns to each positive real number s the volume V of a cube with side length s units. An equation that describes this function is V = s³. What do you think the graph of this function will look like? Explain.
Answer:
I think the graph of this function will look like the graph of the equation y = x³. The inputs and outputs would match the solutions to the equation exactly. For
the equation, the y-value is the cube of the x-value.
For the function, the output is the cube of the input.

e. Use the function in part (d) to determine the volume of a cube with side length of 3 units. Write the input and output as an ordered pair. Does this point appear to belong to the graph of y = x³?
Answer:
V = (3)³
V = 27
(3, 27) The point looks like it would belong to the graph of y = x³; it looks like it would be on the curve that the shape of the graph is taking.

Question 3.
Sketch the graph of the equation y = 180(x – 2) for whole numbers. Organize your work using the table below, and then answer the questions that follow.

Answer:

a. Plot the ordered pairs on the coordinate plane.

Answer:

b. What shape does the graph of the points appear to take?
Answer:
It appears to take the shape of a line.

c. Is this graph a graph of a function? How do you know?
Answer:
It appears to be a function because each input has exactly one output.

d. Is this a linear equation? Explain.
Answer:
Yes, y = 180(x – 2) is a linear equation. It can be rewritten as y = 180x-360.

e. The sum S of interior angles, in degrees, of a polygon with n sides is given by S = 180(n-2). If we take this equation as defining S as a function of n, how do you think the graph of this S will appear? Explain.
Answer:
I think the graph of this function will look like the graph of the equation y = 180(x-2). The inputs and outputs would match the solutions to the equation exactly.

f. Is this function discrete? Explain.
Answer:
The function S = 180(n – 2) is discrete. The inputs are the number of sides, which are integers. The input, n, must be greater than 2 since three sides is the smallest number of sides for a polygon.

Question 4.
Examine the graph below. Could the graph represent the graph of a function? Explain why or why not.

Answer:
This is not the graph of a function. The ordered pairs (1, 0) and (1, -1) show that for the input of 1 there are two different outputs, both 0 and -1. For that reason, this cannot be the graph of a function because it does not fit the definition of a function.

Question 5.
Examine the graph below. Could the graph represent the graph of a function? Explain why or why not.

Answer:
This is not the graph of a function. The ordered pairs (2, -1) and (2, -3) show that for the input of 2 there are two different outputs, both -1 and -3. Further, the ordered pairs (5, -3) and (5, -4) show that for the input of 5 there are two different outputs, both -3 and -4. For these reasons, this cannot be the graph of a function because it does not fit the definition of a function.

Question 6.
Examine the graph below. Could the graph represent the graph of a function? Explain why or why not.

Answer:
This is the graph of a function. The ordered pairs (-2, -4), (-1, -3), (0, -2), (1, -1), (2, 0), and (3, 1) represent inputs and their unique outputs. By definition, this is a function.

Eureka Math Grade 8 Module 5 Lesson 5 Exit Ticket Answer Key

Question 1.
Water flows from a hose at a constant rate of 11 gallons every 4 minutes. The total amount of water that flows from the hose is a function of the number of minutes you are observing the hose.
a. Write an equation in two variables that describes the amount of water, y, in gallons, that flows from the hose as a function of the number of minutes, x, you observe it.
Answer:
\(\frac{11}{4}\) = \(\frac{y}{x}\)
y = \(\frac{11}{4}\) x

b. Use the equation you wrote in part (a) to determine the amount of water that flows from the hose during an 8-minute period, a 4-minute period, and a 2-minute period.
Answer:
y = \(\frac{11}{4}\) (8)
y = 22
In 8 minutes, 22 gallons of water flow out of the hose.
y = \(\frac{11}{4}\) (4)
y = 11
In 4 minutes, 11 gallons of water flow out of the hose.
y = \(\frac{11}{4}\) (2)
y = 5.5
In 2 minutes, 5.5 gallons of water flow out of the hose.

c. An input of the function, x, is time in minutes, and the output of the function, y, is the amount of water that flows out of the hose in gallons. Write the inputs and outputs from part (b) as ordered pairs, and plot them as points on the coordinate plane.

Answer:
(8, 22), (4, 11), (2, 5.5)

Engage NY Eureka Math 8th Grade Module 5 Lesson 4 Answer Key

Eureka Math Grade 8 Module 5 Lesson 4 Example Answer Key

Example 1.
Classify each of the functions described below as either discrete or not discrete.
a. The function that assigns to each whole number the cost of buying that many cans of beans in a particular grocery store.
b. The function that assigns to each time of day one Wednesday the temperature of Sammy’s fever at that time.
c. The function that assigns to each real number its first digit.
d. The function that assigns to each day in the year 2015 my height at noon that day.
e. The function that assigns to each moment in the year 2015 my height at that moment.
f. The function that assigns to each color the first letter of the name of that color.
g. The function that assigns the number 23 to each and every real number between 20 and 30.6.
h. The function that assigns the word YES to every yes/no question.
i. The function that assigns to each height directly above the North Pole the temperature of the air at that height right at this very moment.
Answer:
a) Discrete
b) Not discrete
c) Not discrete
d) Discrete
e) Not discrete
f) Discrete
g) Not discrete
h) Discrete
i) Not discrete

Example 2.
Water flows from a faucet into a bathtub at a constant rate of 7 gallons of water every 2 minutes Regard the volume of water accumulated in the tub as a function of the number of minutes the faucet has be on. Is this function discrete or not discrete?
Answer:
→ Assuming the tub is initially empty, we determined last lesson that the volume of water in the tub is given by y = 3.5x, where y is the volume of water in gallons, and x is the number of minutes the faucet has been on.

→ What limitations are there on x and y?
Both x and y should be positive numbers because they represent time and volume.

→ Would this function be considered discrete or not discrete? Explain.
This function is not discrete because we can assign any positive number to x, not just positive integers.

Example 3.
You have just been served freshly made soup that is so hot that it cannot be eaten. You measure the temperature of the soup, and it is 210°F. Since 212°F is boiling, there is no way it can safely be eaten yet. One minute after receiving the soup, the temperature has dropped to 203°F. If you assume that the rate at which the soup cools is constant, write an equation that would describe the temperature of the soup over time.
Answer:
The temperature of the soup dropped 7°F in one minute. Assuming the cooling continues at the same rate, then if y is the temperature of the soup after x minutes, then, y = 210 – 7x.
→ We want to know how long it will be before the temperature of the soup is at a more tolerable temperature of 147°F. The difference in temperature from 210°F to 147°F is 63°F. For what number x will our function have the value 147?
147 = 210 – 7x; then 7x = 63, and so x = 9.
→ Curious whether or not you are correct in assuming the cooling rate of the soup is constant, you decide to measure the temperature of the soup each minute after its arrival to you. Here’s the data you obtain:

Our function led us to believe that after 9 minutes the soup would be safe to eat. The data in the table shows that it is still too hot.

→ What do you notice about the change in temperature from one minute to the next?
For the first few minutes, minute 2 to minute 5, the temperature decreased 6°F each minute. From minute 5 to minute 9, the temperature decreased just 5°F each minute.
→ Since the rate of cooling at each minute is not constant, this function is said to be a nonlinear function.

→ Sir Isaac Newton not only studied the motion of objects under gravity but also studied the rates of cooling of heated objects. He found that they do not cool at constant rates and that the functions that describe their temperature over time are indeed far from linear. (In fact, Newton’s theory establishes that the temperature of soup at time x minutes would actually be given by the formula y = 70 + 140(\(\frac{133}{140}\))^x.)

Example 4.
Consider the function that assigns to each of nine baseball players, numbered 1 through 9, his height. The data for this function is given below. Call the function G.

Answer:
→ What output does G assign to the input 2?
The function G assigns the height 5′ 4” to the player 2.

→ Could the function G simultaneously assign a second, different output to player 2? Explain.
No. The function assigns height to a particular player. There is no way that a player can have two different heights.
→ It is not clear if there is a formula for this function. (And even if there were, it is not clear that it would be meaningful since who is labeled player 1, player 2, and so on is probably arbitrary.) In general, we can hope to have formulas for functions, but in reality we cannot expect to find them. (People would love to have a formula that explains and predicts the stock market, for example.)

→ Can we classify this function as discrete or not discrete? Explain.
This function would be described as discrete because the inputs are particular players.

Eureka Math Grade 8 Module 5 Lesson 4 Exercise Answer Key

Exercises 1–3

Exercise 1.
At a certain school, each bus in its fleet of buses can transport 35 students. Let B be the function that assigns to each count of students the number of buses needed to transport that many students on a field trip.

When Jinpyo thought about matters, he drew the following table of values and wrote the formula B = x/35. Here x is the count of students, and B is the number of buses needed to transport that many students. He concluded that B is a linear function.

Alicia looked at Jinpyo’s work and saw no errors with his arithmetic. But she said that the function is not actually linear.
a. Alicia is right. Explain why B is not a linear function.
Answer:
For 36 students, say, we’ll need two buses—an extra bus for the extra student. In fact, for 36,37, …, up to 70 students, the function B assigns the same value 2. For 71,72, …, up to 105, it assigns the value 3. There is not a constant rate of increase of the buses needed, and so the function is not linear.

b. Is B a discrete function?
Answer:
It is a discrete function.

Exercise 2.
A linear function has the table of values below. It gives the costs of purchasing certain numbers of movie tickets.

a. Write the linear function that represents the total cost, y, for x tickets purchased.
Answer:
y = \(\frac{27.75}{2}\) x
y = 9.25x

b. Is the function discrete? Explain.
Answer:
The function is discrete. You cannot have half of a movie ticket; therefore, it must be a whole number of tickets, which means it is discrete.

c. What number does the function assign to 4? What do the question and your answer mean?
Answer:
It is asking us to determine the cost of buying 4 tickets. The function assigns 37 to 4. The answer means that 4 tickets will cost $37.00.

Exercise 3.
A function produces the following table of values.

a. Make a guess as to the rule this function follows. Each input is a word from the English language.
Answer:
This function assigns to each word its first letter.

b. Is this function discrete?
Answer:
It is discrete.

Eureka Math Grade 8 Module 5 Lesson 4 Problem Set Answer Key

Question 1.
The costs of purchasing certain volumes of gasoline are shown below. We can assume that there is a linear relationship between x, the number of gallons purchased, and y, the cost of purchasing that many gallons.

a. Write an equation that describes y as a linear function of x.
Answer:
y = 3.65x

b. Are there any restrictions on the values x and y can adopt?
Answer:
Both x and y must be positive rational numbers.

c. Is the function discrete?
Answer:
The function is not discrete.

d. What number does the linear function assign to 20? Explain what your answer means.
Answer:
y = 3.65(20)
y = 73
The function assigns 73 to 20. It means that if 20 gallons of gas are purchased, it will cost $73.00.

Question 2.
A function has the table of values below. Examine the information in the table to answer the questions below.

a. Describe the function.
Answer:
The function assigns those particular numbers to those particular seven words. We don’t know if the function accepts any more inputs and what it might assign to those additional inputs. (Though it does seem compelling to say that this function assigns to each positive whole number the count of letters in the name of that whole number.)

b. What number would the function assign to the word eleven?
Answer:
We do not have enough information to tell. We are not even sure if eleven is considered a valid input for this function.

Question 3.
The table shows the distances covered over certain counts of hours traveled by a driver driving a car at a constant speed.

a. Write an equation that describes y, the number of miles covered, as a linear function of x, number of hours driven.
Answer:
y = \(\frac{141}{3}\) x
y = 47x

b. Are there any restrictions on the value x and y can adopt?
Answer:
Both x and y must be positive rational numbers.

c. Is the function discrete?
Answer:
The function is not discrete.

d. What number does the function assign to 8? Explain what your answer means.
Answer:
y = 47(8)
y = 376
The function assigns 376 to 8. The answer means that 376 miles are driven in 8 hours.

e. Use the function to determine how much time it would take to drive 500 miles.
Answer:
500 = 47x
\(\frac{500}{47}\) = x
10.63829… = x
10.6 ≈ x
It would take about 10.6 hours to drive 500 miles.

Question 4.
Consider the function that assigns to each time of a particular day the air temperature at a specific location in Ithaca, NY. The following table shows the values of this function at some specific times.

a. Let y represent the air temperature at time x hours past noon. Verify that the data in the table satisfies the linear equation y = 92 – 1.5x.
Answer:
At 12:00, 0 hours have passed since 12:00; then, y = 92 – 1.5(0) = 92.
At 1:00, 1 hour has passed since 12:00; then, y = 92 – 1.5(1) = 90.5.
At 2:00, 2 hours have passed since 12:00; then, y = 92 – 1.5(2) = 89.
At 4:00, 4 hours have passed since 12:00; then, y = 92 – 1.5(4) = 86.
At 8:00, 8 hours have passed since 12:00; then, y = 92 – 1.5(8) = 80.

b. Are there any restrictions on the types of values x and y can adopt?
Answer:
The input is a particular time of the day, and y is the temperature. The input cannot be negative but could be intervals that are fractions of an hour. The output could potentially be negative because it can get that cold.

c. Is the function discrete?
Answer:
The function is not discrete.

d. According to the linear function of part (a), what will the air temperature be at 5:30 p.m.?
Answer:
At 5:30, 5.5 hours have passed since 12:00; then y = 92 – 1.5(5.5) = 83.75.
The temperature at 5:30 will be 83.75°F.

e. Is it reasonable to assume that this linear function could be used to predict the temperature for 10:00 a.m. the following day or a temperature at any time on a day next week? Give specific examples in your explanation.
Answer:
No. There is no reason to expect this function to be linear. Temperature typically fluctuates and will, for certain, rise at some point.
We can show that our model for temperature is definitely wrong by looking at the predicted temperature one week (168 hours) later:
y = 92 – 1.5(168)
y = – 160.
This is an absurd prediction.

Eureka Math Grade 8 Module 5 Lesson 4 Exit Ticket Answer Key

Question 1.
The table below shows the costs of purchasing certain numbers of tablets. We can assume that the total cost is a linear function of the number of tablets purchased.

a. Write an equation that describes the total cost, y, as a linear function of the number, x, of tablets purchased.
Answer:
y = \(\frac{10,183}{17}\) x
y = 599x

b. Is the function discrete? Explain.
Answer:
The function is discrete. You cannot have half of a tablet; therefore, it must be a whole number of tablets, which means it is discrete.

c. What number does the function assign to 7? Explain.
Answer:
The function assigns 4,193 to 7, which means that the cost of 7 tablets would be $4,193.00.

Question 2.
A function C assigns to each word in the English language the number of letters in that word. For example, C assigns the number 6 to the word action.
a. Give an example of an input to which C would assign the value 3.
Answer:
Any three – letter word will do.

b. Is C a discrete function? Explain.
Answer:
The function is discrete. The input is a word in the English language, therefore it must be an entire word, not part of one, which means it is discrete.

Engage NY Eureka Math 8th Grade Module 5 Lesson 3 Answer Key

Eureka Math Grade 8 Module 5 Lesson 3 Example Answer Key

Example 1.
In the last lesson, we looked at several tables of values showing the inputs and outputs of functions. For instance, one table showed the costs of purchasing different numbers of bags of candy:

Answer:
→ What do you think a linear function is?
A linear function is likely a function with a rule described by a linear equation. Specifically, the rate of change in the situation being described is constant, and the graph of the equation is a line.

→ Do you think this is a linear function? Justify your answer.
Yes, this is a linear function because there is a proportional relationship:
\(\frac{10.00}{8}\) = 1.25; $1.25 per each bag of candy
\(\frac{5.00}{4}\) = 1.25; $1.25 per each bag of candy
\(\frac{2.50}{2}\) = 1.25; $1.25 per each bag of candy
The total cost is increasing at a rate of $1.25 with each bag of candy. Further justification comes from the graph of the data shown below.

→ A linear function is a function with a rule that can be described by a linear equation. That is, if we use x to denote an input of the function and y its matching output, then the function is linear if the rule for the function can be described by the equation y = mx + b for some numbers m and b.

→ What rule or equation describes our cost function for bags of candy?
The rule that represents the function is then y = 1.25x.

→ Notice that the constant m is 1.25, which is the cost of one bag of candy, and the constant b is 0. Also notice that the constant m was found by calculating the unit rate for a bag of candy.
No matter the value of x chosen, as long as x is a nonnegative integer, the rule y = 1.25x gives the cost of purchasing that many bags of candy. The total cost of candy is a function of the number of bags purchased.

→ Why must we set x as a nonnegative integer for this function?
Since x represents the number of bags of candy, it does not make sense that there would be a negative number of bags. It is also unlikely that we might be allowed to buy fractional bags of candy, and so we require x to be a whole number.

→ Would you say that the table represents all possible inputs and outputs? Explain.
No, it does not represent all possible inputs and outputs. Someone can purchase more than 8 bags of candy, and inputs greater than 8 are not represented by this table (unless the store has a limit on the number of bags one may purchase, perhaps).

Example 2.
Walter walks at a constant speed of 8 miles every 2 hours. Describe a linear function for the number of miles he walks in x hours. What is a reasonable range of x-values for this function?
Answer:
→ Consider the following rate problem: Walter walks at a constant speed of 8 miles every 2 hours. Describe a linear function for the number of miles he walks in x hours. What is a reasonable range of x-values for this function?
Walter’s average speed of walking 8 miles is \(\frac{8}{2}\) = 4, or 4 miles per hour.

→ We have y = 4x, where y is the distance walked in x hours. It seems reasonable to say that x is any real number between 0 and 20, perhaps? (Might there be a cap on the number of hours he walks? Perhaps we are counting up the number of miles he walks over a lifetime?)

→ In the last example, the total cost of candy was a function of the number of bags purchased. Describe the function in this walking example.
The distance that Walter travels is a function of the number of hours he spends walking.

→ What limitations did we put on x?
We must insist that x ≥ 0. Since x represents the time Walter walks, then it makes sense that he would walk for a positive amount of time or no time at all.
→ Since x is positive, then we know that the distance y will also be positive.

Example 3.
Veronica runs at a constant speed. The distance she runs is a function of the time she spends running. The function has the table of values shown below.

Answer:
→ Since Veronica runs at a constant speed, we know that her average speed over any time interval will be the same. Therefore, Veronica’s distance function is a linear function. Write the equation that describes her distance function.
The function that represents Veronica’s distance is described by the equation y = \(\frac{1}{8}\) x, where y is the distance in miles Veronica runs in x minutes and x,y≥0.

→ Describe the function in terms of distance and time.
The distance that Veronica runs is a function of the number of minutes she spends running.

Example 4.
Water flows from a faucet into a bathtub at the constant rate of 7 gallons of water pouring out every 2 minutes. The bathtub is initially empty, and its plug is in. Determine the rule that describes the volume of water in the tub as a function of time. If the tub can hold 50 gallons of water, how long will it take to fill the tub?
Answer:
The rate of water flow is \(\frac{7}{2}\), or 3.5 gallons per minute. Then the rule that describes the volume of water in the tub as a function of time is y = 3.5x, where y is the volume of water, and x is the number of minutes the faucet has been on.
To find the time when y = 50, we need to look at the equation 50 = 3.5x. This gives x = \(\frac{50}{3.5}\) = 14.2857… ≈ 14 . It will take about 14 minutes to fill the tub.

Assume that the faucet is filling a bathtub that can hold 50 gallons of water. How long will it take the faucet to fill the tub?
Since we want the total volume to be 50 gallons, then
50 = 3.5x
\(\frac{50}{3.5}\) = x
14.2857… = x
14 ≈ x
It will take about 14 minutes to fill a tub that has a volume of 50 gallons.

Now assume that you are filling the same 50-gallon bathtub with water flowing in at the constant rate of 3.5 gallons per minute, but there were initially 8 gallons of water in the tub. Will it still take about 14 minutes to fill the tub?
Answer:
No. It will take less time because there is already some water in the tub.

→ What now is the appropriate equation describing the volume of water in the tub as a function of time?
If y is the volume of water that flows from the faucet, and x is the number of minutes the faucet has been on, then y = 3.5x + 8.

→ How long will it take to fill the tub according to this equation?
Since we still want the total volume of the tub to be 50 gallons, then:
50 = 3.5x + 8
42 = 3.5x
12 = x
It will take 12 minutes for the faucet to fill a 50-gallon tub when 8 gallons are already in it.
(Be aware that some students may observe that we can use the previous function rule y = 3.5x to answer this question by noting that we need to add only 42 more gallons to the tub. This will lead directly to the equation 42 = 3.5x.)
→ Generate a table of values for this function:

Example 5.
Water flows from a faucet at a constant rate. Assume that 6 gallons of water are already in a tub by the time we notice the faucet is on. This information is recorded in the first column of the table below. The other columns show how many gallons of water are in the tub at different numbers of minutes since we noticed the running faucet.

Answer:
→ After 3 minutes pass, there are 9.6 gallons in the tub. How much water flowed from the faucet in those 3 minutes? Explain.
Since there were already 6 gallons in the tub, after 3 minutes an additional 3.6 gallons filled the tub.

→ Use this information to determine the rate of water flow.
In 3 minutes, 3.6 gallons were added to the tub, then \(\frac{3.6}{3}\) = 1.2, and the faucet fills the tub at a rate of 1.2 gallons per minute.

→ Verify that the rate of water flow is correct using the other values in the table.
Sample student work:
5(1.2) = 6, and since 6 gallons were already in the tub, the total volume in the tub is 12 gallons.
9(1.2) = 10.8, and since 6 gallons were already in the tub, the total volume in the tub is 16.8 gallons.

→ Write an equation that describes the volume of water in the tub as a function of time.
The volume function that represents the rate of water flow from the faucet is y = 1.2x + 6, where y is the volume of water in the tub, and x is the number of minutes that have passed since we first noticed the faucet being on.

→ For how many minutes was the faucet on before we noticed it? Explain.
Since 6 gallons were in the tub by the time we noticed the faucet was on, we need to determine how many minutes it takes for 6 gallons to flow into the tub. The columns for x = 0 and x = 5 in the table show that six gallons of water pour in the tub over a five-minute period. The faucet was on for 5 minutes before we noticed it.

Eureka Math Grade 8 Module 5 Lesson 3 Exercise Answer Key

Exercises 1–3

Exercise 1.
Hana claims she mows lawns at a constant rate. The table below shows the area of lawn she can mow over different time periods.

a. Is the data presented consistent with the claim that the area mowed is a linear function of time?
Answer:
Sample responses:
Linear functions have a constant rate of change. When we compare the rates at each interval of time, they will be equal to the same constant.
When the data is graphed on the coordinate plane, it appears to make a line.

b. Describe in words the function in terms of area mowed and time.
Answer:
The total area mowed is a function of the number of minutes spent mowing.

c. At what rate does Hana mow lawns over a 5-minute period?
Answer:
\(\frac{36}{5}\) = 7.2
The rate is 7.2 square feet per minute.

d. At what rate does Hana mow lawns over a 20-minute period?
Answer:
\(\frac{144}{20}\) = 7.2
The rate is 7.2 square feet per minute.

e. At what rate does Hana mow lawns over a 30-minute period?
Answer:
\(\frac{216}{30}\) = 7.2
The rate is 7.2 square feet per minute.

f. At what rate does Hana mow lawns over a 50-minute period?
Answer:
\(\frac{360}{50}\) = 7.2
The rate is 7.2 square feet per minute.

g. Write the equation that describes the area mowed, y, in square feet, as a linear function of time, x, in minutes.
Answer:
y = 7.2x

h. Describe any limitations on the possible values of x and y.
Answer:
Both x and y must be positive numbers. The symbol x represents time spent mowing, which means it should be positive. Similarly, y represents the area mowed, which should also be positive.

i. What number does the function assign to x = 24? That is, what area of lawn can be mowed in 24 minutes?
Answer:
y = 7.2(24)
y = 172.8
In 24 minutes, an area of 172.8 square feet can be mowed.

j. According to this work, how many minutes would it take to mow an area of 400 square feet?
Answer:
400 = 7.2x
\(\frac{400}{7.2}\) = x
55.555… = x
56 ≈ x
It would take about 56 minutes to mow an area of 400 square feet.

Exercise 2.
A linear function has the table of values below. The information in the table shows the total volume of water, in gallons, that flows from a hose as a function of time, the number of minutes the hose has been running.

a. Describe the function in terms of volume and time.
Answer:
The total volume of water that flows from a hose is a function of the number of minutes the hose is left on.

b. Write the rule for the volume of water in gallons, y, as a linear function of time, x, given in minutes.
Answer:
y = \(\frac{44}{10}\) x
y = 4.4x

c. What number does the function assign to 250? That is, how many gallons of water flow from the hose during a period of 250 minutes?
Answer:
y = 4.4(250)
y = 1100
In 250 minutes, 1,100 gallons of water flow from the hose.

d. The average swimming pool holds about 17,300 gallons of water. Suppose such a pool has already been filled one quarter of its volume. Write an equation that describes the volume of water in the pool if, at time 0 minutes, we use the hose described above to start filling the pool.
Answer:
\(\frac{1}{4}\) (17300) = 4325
y = 4.4x + 4325

e. Approximately how many hours will it take to finish filling the pool?
Answer:
17300 = 4.4x + 4325
12975 = 4.4x
\(\frac{12975}{4.4}\) = x
2948.8636… = x
2949 ≈ x
\(\frac{2949}{60}\) = 49.15
It will take about 49 hours to fill the pool with the hose.

Exercise 3.
Recall that a linear function can be described by a rule in the form of y = mx + b, where m and b are constants. A particular linear function has the table of values below.

Answer:
a. What is the equation that describes the function?
Answer:
y = 5x + 4

b. Complete the table using the rule.
Answer:

Eureka Math Grade 8 Module 5 Lesson 3 Problem Set Answer Key

Question 1.
A food bank distributes cans of vegetables every Saturday. The following table shows the total number of cans they have distributed since the beginning of the year. Assume that this total is a linear function of the number of weeks that have passed.

a. Describe the function being considered in words.
Answer:
The total number of cans handed out is a function of the number of weeks that pass.

b. Write the linear equation that describes the total number of cans handed out, y, in terms of the number of weeks, x, that have passed.
Answer:
y = \(\frac{180}{1}\) x
y = 180x

c. Assume that the food bank wants to distribute 20,000 cans of vegetables. How long will it take them to meet that goal?
Answer:
20 000 = 180x
\(\frac{20000}{180}\) = x
111.1111… = x
111 ≈ x
It will take about 111 weeks to distribute 20,000 cans of vegetables, or about 2 years.

d. The manager had forgotten to record that they had distributed 35,000 cans on January 1. Write an adjusted linear equation to reflect this forgotten information.
Answer:
y = 180x + 35 000

e. Using your function in part (d), determine how long in years it will take the food bank to hand out 80,000 cans of vegetables.
Answer:
80 000 = 180x + 35 000
45 000 = 180x
\(\frac{45000}{180}\) = x
250 = x
To determine the number of years:
\(\frac{250}{52}\) = 4.8076… ≈ 4.8
It will take about 4.8 years to distribute 80,000 cans of vegetables.

Question 2.
A linear function has the table of values below. It gives the number of miles a plane travels over a given number of hours while flying at a constant speed.

a. Describe in words the function given in this problem.
Answer:
The total distance traveled is a function of the number of hours spent flying.

b. Write the equation that gives the distance traveled, y, in miles, as a linear function of the number of hours, x, spent flying.
Answer:
y = \(\frac{1062.5}{2.5}\) x
y = 425x

c. Assume that the airplane is making a trip from New York to Los Angeles, which is a journey of approximately 2,475 miles. How long will it take the airplane to get to Los Angeles?
Answer:
2475 = 425x
\(\frac{2475}{425}\) = x
5.82352… = x
5.8 ≈ x
It will take about 5.8 hours for the airplane to fly 2,475 miles.

d. If the airplane flies for 8 hours, how many miles will it cover?
Answer:
y = 425(8)
y = 3400
The airplane would travel 3,400 miles in 8 hours.

Question 3.
A linear function has the table of values below. It gives the number of miles a car travels over a given number of hours.

a. Describe in words the function given.
Answer:
The total distance traveled is a function of the number of hours spent traveling.

b. Write the equation that gives the distance traveled, in miles, as a linear function of the number of hours spent driving.
Answer:
y = \(\frac{203}{3.5}\) x
y = 58x

c. Assume that the person driving the car is going on a road trip to reach a location 500 miles from her starting point. How long will it take the person to get to the destination?
Answer:
500 = 58x
\(\frac{500}{58}\) = x
8.6206… = x
8.6 ≈ x
It will take about 8.6 hours to travel 500 miles.

Question 4.
A particular linear function has the table of values below.

a. What is the equation that describes the function?
Answer:
y = 3x + 1

b. Complete the table using the rule.
Answer:

Question 5.
A particular linear function has the table of values below.

a. What is the rule that describes the function?
Answer:
y = x + 6

b. Complete the table using the rule.
Answer:

Eureka Math Grade 8 Module 5 Lesson 3 Exit Ticket Answer Key

The information in the table shows the number of pages a student can read in a certain book as a function of time in minutes spent reading. Assume a constant rate of reading.

a. Write the equation that describes the total number of pages read, y, as a linear function of the number of minutes, x, spent reading.
Answer:
y = \(\frac{7}{2}\) x
y = 3.5x

b. How many pages can be read in 45 minutes?
Answer:
y = 3.5(45)
y = 157.5
In 45 minutes, the student can read 157.5 pages.

c. A certain book has 396 pages. The student has already read \(\frac{3}{8}\) of the pages and now picks up the book again at time x = 0 minutes. Write the equation that describes the total number of pages of the book read as a function of the number of minutes of further reading.
Answer:
\(\frac{3}{8}\) (396) = 148.5
y = 3.5x + 148.5

d. Approximately how much time, in minutes, will it take to finish reading the book?
Answer:
396 = 3.5x + 148.5
247.5 = 3.5x
\(\frac{247.5}{3.5}\) = x
70.71428571… = x
71 ≈ x
It will take about 71 minutes to finish reading the book.

Engage NY Eureka Math 8th Grade Module 5 Lesson 2 Answer Key

Eureka Math Grade 8 Module 5 Lesson 2 Example Answer Key

Exercises 1–5

Exercise 1.
Let D be the distance traveled in time t. Use the equation D = 16t² to calculate the distance the stone dropped for the given time t.

Answer:

a. Are the distances you calculated equal to the table from Lesson 1?
Answer:
Yes

b. Does the function D = 16t² accurately represent the distance the stone fell after a given time t? In other words, does the function described by this rule assign to t the correct distance? Explain.
Answer:
Yes, the function accurately represents the distance the stone fell after the given time interval. Each computation using the function resulted in the correct distance. Therefore, the function assigns to t the correct distance.

Exercise 2.
Can the table shown below represent values of a function? Explain.

Answer:
No, the table cannot represent a function because the input of 5 has two different outputs. Functions assign only one output to each input.

Exercise 3.
Can the table shown below represent values of a function? Explain.

Answer:
No, the table cannot represent a function because the input of 7 has two different outputs. Functions assign only one output to each input.

Exercise 4.
Can the table shown below represent values of a function? Explain.

Answer:
Yes, the table can represent a function. Even though there are two outputs that are the same, each input has only one output.

Exercise 5.
It takes Josephine 34 minutes to complete her homework assignment of 10 problems. If we assume that she works at a constant rate, we can describe the situation using a function.
a. Predict how many problems Josephine can complete in 25 minutes.
Answer:
Answers will vary.

b. Write the two-variable linear equation that represents Josephine’s constant rate of work.
Answer:
Let y be the number of problems she can complete in x minutes.
\(\frac{10}{34}\) = \(\frac{y}{x}\)
y = \(\frac{10}{34}\) x
y = \(\frac{5}{17}\) x

c. Use the equation you wrote in part (b) as the formula for the function to complete the table below. Round your answers to the hundredths place.

After 5 minutes, Josephine was able to complete 1.47 problems, which means that she was able to complete 1 problem, then get about halfway through the next problem.
Answer:

d. Compare your prediction from part (a) to the number you found in the table above.
Answer:
Answers will vary.

e. Use the formula from part (b) to compute the number of problems completed when x = -7. Does your answer make sense? Explain.
Answer:
y = \(\frac{5}{17}\) (-7)
= -2.06
No, the answer does not make sense in terms of the situation. The answer means that Josephine can complete -2.06 problems in -7 minutes. This obviously does not make sense.

f. For this problem, we assumed that Josephine worked at a constant rate. Do you think that is a reasonable assumption for this situation? Explain.
Answer:
It does not seem reasonable to assume constant rate for this situation. Just because Josephine was able to complete 10 problems in 34 minutes does not necessarily mean she spent the exact same amount of time on each problem. For example, it may have taken her 20 minutes to do 1 problem and then 14 minutes total to finish the remaining 9 problems.

Eureka Math Grade 8 Module 5 Lesson 2 Problem Set Answer Key

Question 1.
The table below represents the number of minutes Francisco spends at the gym each day for a week. Does the data shown below represent values of a function? Explain.

Answer:
Yes, the table can represent a function because each input has a unique output. For example, on day 1, Francisco was at the gym for 35 minutes.

Question 2.
Can the table shown below represent values of a function? Explain.

Answer:
No, the table cannot represent a function because the input of 9 has two different outputs, and so does the input of 8. Functions assign only one output to each input.

Question 3.
Olivia examined the table of values shown below and stated that a possible rule to describe this function could be y = -2x + 9. Is she correct? Explain.

Answer:
Yes, Olivia is correct. When the rule is used with each input, the value of the output is exactly what is shown in the table. Therefore, the rule for this function could well be y = -2x + 9.

Question 4.
Peter said that the set of data in part (a) describes a function, but the set of data in part (b) does not. Do you agree? Explain why or why not.

Answer:
Peter is correct. The table in part (a) fits the definition of a function. That is, there is exactly one output for each input. The table in part (b) cannot be a function. The input -3 has two outputs, 14 and 2. This contradicts the definition of a function; therefore, it is not a function.

Question 5.
A function can be described by the rule y = x² + 4. Determine the corresponding output for each given input.

Answer:

Question 6.
Examine the data in the table below. The inputs and outputs represent a situation where constant rate can be assumed. Determine the rule that describes the function.

Answer:
The rule that describes this function is y = 5x + 8.

Question 7.
Examine the data in the table below. The inputs represent the number of bags of candy purchased, and the outputs represent the cost. Determine the cost of one bag of candy, assuming the price per bag is the same no matter how much candy is purchased. Then, complete the table.

Answer:

a. Write the rule that describes the function.
Answer:
y = 1.25x

b. Can you determine the value of the output for an input of x = -4? If so, what is it?
Answer:
When x = -4, the output is -5.

c. Does an input of -4 make sense in this situation? Explain.
Answer:
No, an input of -4 does not make sense for the situation. It would mean -4 bags of candy. You cannot purchase -4 bags of candy.

Question 8.
Each and every day a local grocery store sells 2 pounds of bananas for $1.00. Can the cost of 2 pounds of bananas be represented as a function of the day of the week? Explain.
Answer:
Yes, this situation can be represented by a function. Assign to each day of the week the value $1.00.

Question 9.
Write a brief explanation to a classmate who was absent today about why the table in part (a) is a function and the table in part (b) is not.

Answer:
The table in part (a) is a function because each input has exactly one output. This is different from the information in the table in part (b). Notice that the input of 1 has been assigned two different values. The input of 1 is assigned 2 and 19. Because the input of 1 has more than one output, this table cannot represent a function.

Eureka Math Grade 8 Module 5 Lesson 2 Exit Ticket Answer Key

Question 1.
Can the table shown below represent values of a function? Explain.

Answer:
Yes, the table can represent a function. Each input has exactly one output.

Question 2.
Kelly can tune 4 cars in 3 hours. If we assume he works at a constant rate, we can describe the situation using a function.
a. Write the function that represents Kelly’s constant rate of work.
Answer:
Let y represent the number of cars Kelly can tune up in x hours; then
\(\frac{y}{x}\) = \(\frac{4}{3}\)
y = \(\frac{4}{3}\) x

b. Use the function you wrote in part (a) as the formula for the function to complete the table below. Round your answers to the hundredths place.

Answer:

c. Kelly works 8 hours per day. According to this work, how many cars will he finish tuning at the end of a shift?
Answer:
Using the function, Kelly will tune up 10.67 cars at the end of his shift. That means he will finish tuning up 10 cars and begin tuning up the 11th car.

d. For this problem, we assumed that Kelly worked at a constant rate. Do you think that is a reasonable assumption for this situation? Explain.
Answer:
No, it does not seem reasonable to assume a constant rate for this situation. Just because Kelly tuned up 4 cars in 3 hours does not mean he spent the exact same amount of time on each car. One car could have taken 1 hour, while the other three could have taken 2 hours total.

Engage NY Eureka Math 8th Grade Module 5 Lesson 1 Answer Key

Eureka Math Grade 8 Module 5 Lesson 1 Example Answer Key

Example 1.
Suppose a moving object travels 256 feet in 4 seconds. Assume that the object travels at a constant speed, that is, the motion of the object can be described by a linear equation. Write a linear equation in two variables to represent the situation, and use the equation to predict how far the object has moved at the four times shown.

Answer:

→ Suppose a moving object travels 256 feet in 4 seconds. Assume that the object travels at a constant speed, that is, the motion of the object can be described by a linear equation. Write a linear equation in two variables to represent the situation, and use the equation to predict how far the object has moved at the four times shown.

→ Let x represent the time it takes to travel y feet.
\(\frac{256}{4}\) = \(\frac{y}{x}\)
y = \(\frac{256}{4}\) x
y = 64x

→ What are some of the predictions that this equation allows us to make?
After one second, or when x = 1, the distance traveled is 64 feet.
Accept any reasonable predictions that students make.

→ Use your equation to complete the table.
→ What is the average speed of the moving object from 0 to 3 seconds?
The average speed is 64 feet per second. We know that the object has a constant rate of change; therefore, we expect the average speed to be the same over any time interval.

Example 2.
The object, a stone, is dropped from a height of 256 feet. It takes exactly 4 seconds for the stone to hit the ground. How far does the stone drop in the first 3 seconds? What about the last 3 seconds? Can we assume constant speed in this situation? That is, can this situation be expressed using a linear equation?

Answer:

Provide students time to discuss this in pairs. Lead a discussion in which students share their thoughts with the class. It is likely they will say the motion of a falling object is linear and that the work conducted in the previous example is appropriate.

→ If this is a linear situation, then we predict that the stone drops 192 feet in the first 3 seconds.
Now consider viewing the 10-second “ball drop” video at the following link: http://www.youtube.com/watch?v = KrX_zLuwOvc. Consider showing it more than once.
→ If we were to slow the video down and record the distance the ball dropped after each second, here is the data we would obtain:

Have students record the data in the table of Example 2.
→ Was the linear equation developed in Example 1 appropriate after all?
Students who thought the stone was traveling at constant speed should realize that the predictions were not accurate for this situation. Guide their thinking using the discussion points below.

→ According to the data, how many feet did the stone drop in 3 seconds?
The stone dropped 144 feet.

→ How can that be? It must be that our initial assumption of constant rate was incorrect.
What predictions can we make now?
After one second, x = 1; the stone dropped 16 feet, etc.
→ Let’s make a prediction based on a value of x that is not listed in the table. How far did the stone drop in the first 3.5 seconds? What have we done in the past to figure something like this out?

Eureka Math Grade 8 Module 5 Lesson 1 Exercise Answer Key

Exercises 1–6
Use the table to answer Exercises 1–5.

Exercise 1.
Name two predictions you can make from this table.
Answer:
Sample student responses:
After 2 seconds, the object traveled 64 feet. After 3.5 seconds, the object traveled 196 feet.

Exercise 2.
Name a prediction that would require more information.
Answer:
Sample student response:
We would need more information to predict the distance traveled after 3.75 seconds.

Exercise 3.
What is the average speed of the object between 0 and 3 seconds? How does this compare to the average speed calculated over the same interval in Example 1?
\(\text { Average Speed } = \frac{\text { distance traveled over a given time interval }}{\text { time interval }}\)
Answer:
The average speed is 48 feet per second: \(\frac{144}{3}\) = 48. This is different from the average speed calculated in Example 1. In Example 1, the average speed over an interval of 3 seconds was 64 feet per second.

Exercise 4.
Take a closer look at the data for the falling stone by answering the questions below.
a. How many feet did the stone drop between 0 and 1 second?
Answer:
The stone dropped 16 feet between 0 and 1 second.

b. How many feet did the stone drop between 1 and 2 seconds?
Answer:
The stone dropped 48 feet between 1 and 2 seconds.

c. How many feet did the stone drop between 2 and 3 seconds?
Answer:
The stone dropped 80 feet between 2 and 3 seconds.

d. How many feet did the stone drop between 3 and 4 seconds?
Answer:
The stone dropped 112 feet between 3 and 4 seconds.

e. Compare the distances the stone dropped from one time interval to the next. What do you notice?
Answer:
Over each interval, the difference in the distance was 32 feet. For example, 16+32 = 48, 48+32 = 80, and 80+32 = 112.

Exercise 5.
What is the average speed of the stone in each interval 0.5 second? For example, the average speed over the interval from 3.5 seconds to 4 seconds is
\(\frac{\text { distance traveled over a given time interval }}{\text { time interval }}\) = \(\frac{256-196}{4-3.5}\) = \(\frac{60}{0.5}\) = 120;120 feet per second
Repeat this process for every half-second interval. Then, answer the question that follows.
a. Interval between 0 and 0.5 second:
Answer:
\(\frac{4}{0.5}\) = 8;8 feet per second

b. Interval between 0.5 and 1 second:
Answer:
\(\frac{12}{0.5}\) = 24;24 feet per second

c. Interval between 1 and 1.5 seconds:
Answer:
\(\frac{20}{0.5}\) = 40;40 feet per second

d. Interval between 1.5 and 2 seconds:
Answer:
\(\frac{28}{0.5}\) = 56;56 feet per second

e. Interval between 2 and 2.5 seconds:
Answer:
\(\frac{36}{0.5}\) = 72;72 feet per second

f. Interval between 2.5 and 3 seconds:
Answer:
\(\frac{44}{0.5}\) = 88;88 feet per second

g. Interval between 3 and 3.5 seconds:
Answer:
\(\frac{52}{0.5}\) = 104;104 feet per second

h. Compare the average speed between each time interval. What do you notice?
Answer:
Over each interval, there is an increase in the average speed of 16 feet per second. For example, 8 + 16 = 24, 24 + 16 = 40, 40 + 16 = 56, and so on.

Exercise 6.
Is there any pattern to the data of the falling stone? Record your thoughts below.

Answer:
Accept any reasonable patterns that students notice as long as they can justify their claim. In the next lesson, students learn that y = 16t².
Each distance has 16 as a factor. For example, 16 = 1(16), 64 = 4(16), 144 = 9(16), and 256 = 16(16).

Eureka Math Grade 8 Module 5 Lesson 1 Problem Set Answer Key

A ball is thrown across the field from point A to point B. It hits the ground at point B. The path of the ball is shown in the diagram below. The x-axis shows the horizontal distance the ball travels in feet, and the y-axis shows the height of the ball in feet. Use the diagram to complete parts (a)–(g).

Answer:

a. Suppose point A is approximately 6 feet above ground and that at time t = 0 the ball is at point A. Suppose the length of OB is approximately 88 feet. Include this information on the diagram.
Answer:
Information is noted on the diagram in red.

b. Suppose that after 1 second, the ball is at its highest point of 22 feet (above point C) and has traveled a horizontal distance of 44 feet. What are the approximate coordinates of the ball at the following values of t: 0.25, 0.5, 0.75, 1, 1.25, 1.5, 1.75, and 2.
Answer:
Most answers will vary because students are approximating the coordinates. The coordinates that must be correct because enough information was provided are denoted by a *.
At t = 0.25, the coordinates are approximately (11, 10).
At t = 0.5, the coordinates are approximately (22, 18).
At t = 0.75, the coordinates are approximately (33, 20).
*At t = 1, the coordinates are approximately (44, 22).
At t = 1.25, the coordinates are approximately (55, 19).
At t = 1.5, the coordinates are approximately (66, 14).
At t = 1.75, the coordinates are approximately (77, 8).
*At t = 2, the coordinates are approximately (88, 0).

c. Use your answer from part (b) to write two predictions.
Answer:
Sample predictions:
At a distance of 44 feet from where the ball was thrown, it is 22 feet in the air. At a distance of 66 feet from where the ball was thrown, it is 14 feet in the air.

d. What is happening to the ball when it has coordinates (88,0)?
Answer:
At point (88,0), the ball has traveled for 2 seconds and has hit the ground at a distance of 88 feet from where the ball began.

e. Why do you think the ball is at point (0, 6) when t = 0? In other words, why isn’t the height of the ball 0?
Answer:
The ball is thrown from point A to point B. The fact that the ball is at a height of 6 feet means that the person throwing it must have released the ball from a height of 6 feet.

f. Does the graph allow us to make predictions about the height of the ball at all points?
Answer:
While we cannot predict exactly, the graph allows us to make approximate predictions of the height for any value of horizontal distance we choose.

Eureka Math Grade 8 Module 5 Lesson 1 Exit Ticket Answer Key

Question 1.
A ball is bouncing across the school yard. It hits the ground at (0,0) and bounces up and lands at (1,0) and bounces again. The graph shows only one bounce.

a. Identify the height of the ball at the following values of t: 0, 0.25, 0.5, 0.75, 1.
Answer:
When t = 0, the height of the ball is 0 feet above the ground. It has just hit the ground.
When t = 0.25, the height of the ball is 3 feet above the ground.
When t = 0.5, the height of the ball is 4 feet above the ground.
When t = 0.75, the height of the ball is 3 feet above the ground.
When t = 1, the height of the ball is 0 feet above the ground. It has hit the ground again.

b. What is the average speed of the ball over the first 0.25 seconds? What is the average speed of the ball over the next 0.25 seconds (from 0.25 to 0.5 seconds)?
Answer:
\(\frac{\text { distance traveled over a given time interval }}{\text { time interval }}\) = \(\frac{3-0}{0.25-0}\) = \(\frac{3}{0.25}\) = 12;12 feet per second
\(\frac{\text { distance traveled over a given time interval }}{\text { time interval }}\) = \(\frac{4-3}{0.5-0.25}\) = \(\frac{1}{0.25}\) = 4;4 feet per second

c. Is the height of the ball changing at a constant rate?
Answer:
No, it is not. If the ball were traveling at a constant rate, the average speed would be the same over any time interval.

Engage NY Eureka Math 7th Grade Module 3 Mid Module Assessment Answer Key

Eureka Math Grade 7 Module 3 Mid Module Assessment Task Answer Key

Question 1.
Use the expression below to answer parts (a) and (b).
4x-3(x-2y)+\(\frac{1}{2}\)(6x-8y)
a. Write an equivalent expression in standard form, and collect like terms.
Answer:
4x – 3(x – 2y) + \(\frac{1}{2}\)(6x – 8y)
4x – 3x + 6y + 3x – 4y
4x – 3x + 3x + 6y – 4y
4x + 2y

b. Express the answer from part (a) as an equivalent expression in factored form.
Answer:
4x + 2y
2(2x + y)

Question 2.
Use the information to solve the problems below.
a. The longest side of a triangle is six more units than the shortest side. The third side is twice the length of the shortest side. If the perimeter of the triangle is 25 units, write and solve an equation to find the lengths of all three sides of the triangle.
Answer:

2x + x + x + 6 = 25
4x + 6 = 25
4x + 6 – 6 = 25 – 6
4x + 0 = 19
\(\frac{1}{4}\)(4x) = \(\frac{1}{4}\)(9)
x = \(\frac{19}{4}\)
x = 4\(\frac{3}{4}\)

Smallest side: x = 4\(\frac{3}{4}\)
Largest side: x + 6 = 10\(\frac{3}{4}\)
Third side: 2x = 9\(\frac{1}{2}\)
3 sides are: 4\(\frac{3}{4}\) units, 10\(\frac{3}{4}\) units, 9\(\frac{1}{2}\) units

b. The length of a rectangle is (x+3) inches long, and the width is 3 \(\frac{2}{5}\) inches. If the area is 15 \(\frac{3}{10}\) square inches, write and solve an equation to find the length of the rectangle.
Answer:

Length: x + 3: = 1\(\frac{1}{2}\) + 3 = 4\(\frac{1}{2}\) inches
Width: 3\(\frac{2}{5}\) inches
3\(\frac{2}{5}\)(x + 3) = 15\(\frac{3}{10}\)
3\(\frac{2}{5}\)x + 3(3\(\frac{2}{5}\)) = 15\(\frac{3}{10}\)
\(\frac{17}{5}\)x + 3(\(\frac{17}{5}\)) = 15\(\frac{3}{10}\)
\(\frac{17}{5}\)x + \(\frac{51}{5}\) = 15\(\frac{3}{10}\)
\(\frac{17}{5}\)x + 10\(\frac{1}{5}\) = 15\(\frac{3}{10}\)
\(\frac{17}{5}\)x + 10\(\frac{1}{5}\) – 10\(\frac{1}{5}\) = 15\(\frac{3}{10}\) – 10\(\frac{1}{5}\)
(\(\frac{17}{5}\)x) + 0 = 5\(\frac{1}{10}\)
\(\frac{5}{17}\)(\(\frac{17}{5}\)x) = (\(\frac{51}{10}\))(\(\frac{5}{17}\))
x = \(\frac{3}{2}\)
x = 1\(\frac{1}{2}\)

Question 3.
A picture 10 \(\frac{1}{4}\) feet long is to be centered on a wall that is 14 \(\frac{1}{2}\) feet long. How much space is there from the edge of the wall to the picture?
a. Solve the problem arithmetically.
Answer:

(14\(\frac{1}{2}\) – 10\(\frac{1}{4}\)) ÷ 2
(14\(\frac{2}{4}\) – 10\(\frac{1}{4}\)) ÷ 2
4\(\frac{1}{4}\) ÷ 2
\(\frac{17}{4}\) ÷ 2
\(\frac{17}{4}\) ∙ \(\frac{1}{2}\)
\(\frac{17}{8}\)
2\(\frac{1}{8}\)
The picture is 2\(\frac{1}{8}\) inches from the wall.

b. Solve the problem algebraically.
Answer:

Let x: distance from one side to the picture
x + 10\(\frac{1}{4}\) + x = 14\(\frac{1}{2}\)
2x + 10\(\frac{1}{4}\) = 14\(\frac{1}{2}\)
2x + 10\(\frac{1}{4}\) – 10\(\frac{1}{4}\) = 14\(\frac{1}{2}\) – 10\(\frac{1}{4}\)
2x + 0 = 4\(\frac{1}{4}\)
(\(\frac{1}{2}\)) (2x) = (4\(\frac{1}{4}\))(\(\frac{1}{2}\))
x = (\(\frac{17}{4}\)) (\(\frac{1}{2}\))
x = \(\frac{17}{8}\) = 2\(\frac{1}{8}\)
The picture is 2\(\frac{1}{8}\) inches from the wall.

c. Compare the approaches used in parts (a) and (b). Explain how they are similar.
Answer:
The solutions are the same. The actual operations performed in the equation are the same operations done arithmetically.

Question 4.
In August, Cory begins school shopping for his triplet daughters.
a. One day, he bought 10 pairs of socks for $2.50 each and 3 pairs of shoes for d dollars each. He spent a total of $135.97. Write and solve an equation to find the cost of one pair of shoes.
Answer:
d: cost of shoes
10(2.50) + 3d = 135.97
25 + 3d = 135.97
3d + 25 = 135.97
3d+ 25 – 25 = 135.97 – 25
3d + 0 = 110.97
(\(\frac{1}{3}\))(3d)= (110.97)(\(\frac{1}{3}\))
d = 36.99
The cost of one pair of shoes is 36.99

b. The following day Cory returned to the store to purchase some more socks. He had $40 to spend. When he arrived at the store, the shoes were on sale for \(\frac{1}{3}\) off. What is the greatest amount of pairs of socks Cory can purchase if he purchases another pair of shoes in addition to the socks?
Answer:
Shoes: \(\frac{1}{3}\)(36.99)
12.33 off
New price
36.99 – 12.33 = 24.66
Socks: d
2.50d + 24.66 ≤ 40
2.50d + 24.66 – 24.66 ≤ 40 – 24.66
2.50d + 0 ≤ 15.34
(\(\frac{1}{2.50}\))(2.50d) ≤ (15.34)(\(\frac{1}{2.50}\))
d ≤ 6.136
The greatest amount of socks he can buy is 6 pairs.

Question 5.
Ben wants to have his birthday at the bowling alley with a few of his friends, but he can spend no more than $80. The bowling alley charges a flat fee of $45 for a private party and $5.50 per person for shoe rentals and unlimited bowling.
a. Write an inequality that represents the total cost of Ben’s birthday for p people given his budget.
Answer:
45 + 5.50p ≤ 80

b. How many people can Ben pay for (including himself) while staying within the limitations of his budget?
Answer:
P: number of people invited
45 + 5.50p ≤ 80
5.50p + 45 ≤ 80
5.50p + 45 – 45 ≤ 80 – 45
(\(\frac{1}{5.50}\))(5.50p) ≤ (35)(\(\frac{1}{5.50}\))
P ≤ \(\frac{350}{55}\)
P ≤ \(\frac{70}{11}\)
P ≤ 6\(\frac{4}{11}\)
6 people can attend the party
P ≤ 6

c. Graph the solution of the inequality from part (a).
Answer:

6. Jenny invited Gianna to go watch a movie with her family. The movie theater charges one rate for 3D admission and a different rate for regular admission. Jenny and Gianna decided to watch the newest movie in 3D. Jenny’s mother, father, and grandfather accompanied Jenny’s little brother to the regular admission movie.
a. Write an expression for the total cost of the tickets. Define the variables.
Answer:
d: cost in dollars of 3D admission
r: cost in dollars of regular admission
Jenny Gianna Mother Father Grandfather Brother
d + d + r + r + r + r
2d + 4r

b. The cost of the 3D ticket was double the cost of the regular admission ticket. Write an equation to represent the relationship between the two types of tickets.
Answer:
d = 2r

c. The family purchased refreshments and spent a total of $18.50. If the total amount of money spent on tickets and refreshments was $94.50, use an equation to find the cost of one regular admission ticket.
Answer:
2d + 4r + 18.50 = 94.50
2(2r) + 4r + 18.50 = 94.50
4r + 4r + 18.50 = 94.50
8r + 18.50 = 94.50
8r + 18.50 – 18.50 = 94.50 – 18.50
8r + 0 = 76
(\(\frac{1}{8}\))(8r) = (76)(\(\frac{1}{8}\))
r = 9.5
The cost of one regular admission ticket is $9.50

Question 7.
The three lines shown in the diagram below intersect at the same point. The measures of some of the angles in degrees are given as 3(x – 2)°, \(\frac{3}{5}\) y)°, 12°, 42°.

a. Write and solve an equation that can be used to find the value of x.
Answer:
3(x – 2) = 42
3x – 6 = 42
3x – 6 + 6 = 42 + 6
3x + 0 = 48
(\(\frac{1}{3}\))(3x) = (48)(\(\frac{1}{3}\))
x = 16
OR
\(\frac{1}{3}\)(3(x – 2)) = (42)(\(\frac{1}{3}\))
x – 2 = 14
x – 2 + 2 = 14 + 2
x + 0 = 16
x = 16

b. Write and solve an equation that can be used to find the value of y.
Answer:
\(\frac{3}{5}\) y + 12 + 42 = 180
\(\frac{3}{5}\) y + 54 = 180
\(\frac{3}{5}\) y + 54 – 54 = 180 – 54
\(\frac{3}{5}\) y + 0 = 126
(\(\frac{5}{3}\))(\(\frac{3}{5}\) y) = (126)(\(\frac{5}{3}\))
y = (42)(5)
y = 210

Engage NY Eureka Math 7th Grade Module 3 Lesson 11 Answer Key

Eureka Math Grade 7 Module 3 Lesson 11 Example Answer Key

Example 1.
The following figure shows three lines intersecting at a point. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Confirm your answers by measuring the angle with a protractor.

Answer:
The angles 86°, 68°, and the angle between them, which is vertically opposite and equal in measure to x, are angles on a line and have a sum of 180°.
86 + x + 68 = 180
x + 154 = 180
x + 154 – 154 = 180 – 154
x = 26

Example 2.
In a complete sentence, describe the angle relationships in the diagram. You may label the diagram to help describe the angle relationships. Write an equation for the angle relationship shown in the figure and solve for x. Confirm your answers by measuring the angle with a protractor.

Answer:
The angle formed by adjacent angles a° and b° is vertically opposite to the 77° angle. The angles x°, a°, and b° are adjacent angles that have a sum of 90° (since the adjacent angle is a right angle and together the angles are on a line).
x + 77 = 90
x + 77 – 77 = 90 – 77
x = 13

Example 3.
In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Find the measures of ∠JAH and ∠GAF. Confirm your answers by measuring the angle with a protractor.

Answer:
The sum of the degree measurements of ∠JAH, ∠GAH, ∠GAF, and the arc that subtends ∠JAF is 360°.
225 + 2x + 90 + 3x = 360
315 + 5x = 360
315 – 315 + 5x = 360 – 315
5x = 45
(\(\frac{1}{5}\))5x = (\(\frac{1}{5}\))45
x = 9
m∠JAH = 2(9°) = 18° m∠GAF = 3(9°) = 27°

Example 4.
In the accompanying diagram, the measure of ∠DBE is four times the measure of ∠FBG.

a. Label ∠DBE as y° and ∠FBG as x°. Write an equation that describes the relationship between ∠DBE and ∠FBG.
Answer:
y = 4x

b. Find the value of x.
Answer:
50 + x + 4x = 180
50 + 5x = 180
5x + 50 – 50 = 180 – 50
5x = 130
(\(\frac{1}{5}\))(5x) = (\(\frac{1}{5}\))(130)
x = 26

c. Find the measures of ∠FBG, ∠CBD, ∠ABF, ∠GBE, and ∠DBE.
Answer:
m∠FBG = 26°
m∠CBD = 26°
m∠ABF = 4(26°) = 104°
m∠GBE = 50°
m∠DBE = 104°

d. What is the measure of ∠ABG? Identify the angle relationship used to get your answer.
Answer:
∠ABG = ∠ABF + ∠FBG
∠ABG = 104 + 26
∠ABG = 130
m∠ABG = 130°
To determine the measure of ∠ABG, you need to add the measures of adjacent angles ∠ABF and ∠FBG.

Eureka Math Grade 7 Module 3 Lesson 11 Exercise Answer Key

Opening Exercise
a. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Confirm your answers by measuring the angle with a protractor.

Answer:
The angles marked by x°, 90°, and 14° are angles on a line and have a sum of 180°.
x + 90 + 14 = 180
x + 104 = 180
x + 104 – 104 = 180 – 104
x = 76

b. \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are intersecting lines. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for y. Confirm your answers by measuring the angle with a protractor.

Answer:
The adjacent angles marked by y° and 51° together form the angle that is vertically opposite and equal to the angle measuring 147°.
y + 51 = 147
y + 51 – 51 = 147 – 51
y = 96

c. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for b. Confirm your answers by measuring the angle with a protractor.

Answer:
The adjacent angles marked by 59°, 41°, b°, 65°, and 90° are angles at a point and together have a sum of 360°.
59 + 41 + b + 65 + 90 = 360
b + 255 = 360 – 255
b = 105

d. The following figure shows three lines intersecting at a point. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for z. Confirm your answers by measuring the angle with a protractor.

Answer:
The angles marked by z°, 158°, and z° are angles on a line and have a sum of 180°.
z + 158 + z = 180
2z + 158 = 180
2z + 158 – 158 = 180 – 158
2z = 22
z = 11

e. Write an equation for the angle relationship shown in the figure and solve for x. In a complete sentence, describe the angle relationship in the diagram. Find the measurements of ∠EPB and ∠CPA. Confirm your answers by measuring the angle with a protractor.

Answer:
∠CPA, ∠CPE, and ∠EPB are angles on a line and their measures have a sum of 180°.
5x + 90 + x = 180
6x + 90 = 180
6x + 90 – 90 = 180 – 90
6x = 90
(\(\frac{1}{6}\))6x = (\(\frac{1}{6}\))90
x = 15
∠EPB = 15°
∠CPA = 5(15°) = 75°

Exercise 1.
The following figure shows four lines intersecting at a point. In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x and y. Confirm your answers by measuring the angle with a protractor.

Answer:
The angles x°, 25°, y°, and 40° are angles on a line and have a sum of 180°; the angle marked y° is vertically opposite and equal to 96°.
y = 96, vert. ∠s
x + 25 + (96) + 40 = 180
x + 161 = 180
x + 161 – 161 = 180 – 161
x = 19

Exercise 2.
In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x and y. Confirm your answers by measuring the angles with a protractor.

Answer:
The measures of angles x and y have a sum of 90°; the measures of angles x and 27 have a sum of 90°.
x + 27 = 90
x + 27 – 27 = 90 – 27
x = 63
(63) + y = 90
63 – 63 + y = 90 – 63
y = 27

Exercise 3.
In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for x. Find the measure of ∠JKG. Confirm your answer by measuring the angle with a protractor.

Answer:
The sum of the degree measurements of ∠LKJ, ∠JKG, ∠GKM, and the arc that subtends ∠LKM is 360°.
5x + 24 + x + 90 = 360
6x + 114 = 360
6x + 114 – 114 = 360 – 114
6x = 246
(\(\frac{1}{6}\))6x = (\(\frac{1}{6}\))246
x = 41
m∠JKG = 41°

Eureka Math Grade 7 Module 3 Lesson 11 Problem Set Answer Key

In a complete sentence, describe the angle relationships in each diagram. Write an equation for the angle relationship(s) shown in the figure, and solve for the indicated unknown angle. You can check your answers by measuring each angle with a protractor.
Question 1.
Find the measures of ∠EAF, ∠DAE, and ∠CAD.

Answer:
∠GAF, ∠EAF, ∠DAE, and ∠CAD are angles on a line and their measures have a sum of 180°.
6x + 4x + 2x + 30 = 180
12x + 30 = 180
12x + 30 – 30 = 180 – 30
12x = 150
x = 12.5
m∠EAF = 2(12.5°) = 25°
m∠DAE = 4(12.5°) = 50°
m∠CAD = 6(12.5°) = 75°

Question 2.
Find the measure of a.

Answer:
Angles a°, 26°, a°, and 126° are angles at a point and have a sum of 360°.
a + 126 + a + 26 = 360
2a + 152 = 360
2a + 152 – 152 = 360 – 152
2a = 208
(\(\frac{1}{2}\))2a = (\(\frac{1}{2}\))208
a = 104

Question 3.
Find the measures of x and y.

Answer:
Angles y° and 65° and angles 25° and x° have a sum of 90°.
x + 25 = 90
x + 25 – 25 = 90 – 25
x = 65
65 + y = 90
65 + y = 90
65 – 65 + y = 90 – 65
y = 25

Question 4.
Find the measure of ∠HAJ.

Answer:
Adjacent angles x° and 15° together are vertically opposite from and are equal to angle 81°.
x + 15 = 81
x + 15 – 15 = 81 – 15
x = 66
m∠HAJ = 66°

Question 5.
Find the measures of ∠HAB and ∠CAB.

Answer:
The measures of adjacent angles ∠CAB and ∠HAB have a sum of the measure of ∠CAH, which is vertically opposite from and equal to the measurement of ∠DAE.
2x + 3x + 70 = 180
5x = 110
(\(\frac{1}{5}\))5x = (\(\frac{1}{5}\))110
x = 22
m∠HAB = 3(22°) = 66°
m∠CAB = 2(22°) = 44°

Question 6.
The measure of ∠SPT is b°. The measure of ∠TPR is five more than two times ∠SPT. The measure of ∠QPS is twelve less than eight times the measure of ∠SPT. Find the measures of ∠SPT, ∠TPR, and ∠QPS.

Answer:
∠QPS, ∠SPT, and ∠TPR are angles on a line and their measures have a sum of 180°.
(8b – 12) + b + (2b + 5) = 180
11b – 7 = 180
11b – 7 + 7 = 180 + 7
11b = 187
(\(\frac{1}{11}\))11b = (\(\frac{1}{11}\))187
b = 17
m∠SPT = (17°) = 17°
m∠TPR = 2(17°) + 5° = 39°
m∠QPS = 8(17°) – 12° = 124°

Question 7.
Find the measures of ∠HQE and ∠AQG.

Answer:
∠AQG, ∠AQH, and ∠HQE are adjacent angles whose measures have a sum of 90°.
2y + 21 + y = 90
3y + 21 = 90
3y + 21 – 21 = 90 – 21
3y = 69
(\(\frac{1}{3}\))3y = (\(\frac{1}{3}\))69
y = 23
m∠HQE = 2(23°) = 46°
m∠AQG = (23°) = 23°

Question 8.
The measures of three angles at a point are in the ratio of 2:3:5. Find the measures of the angles.
Answer:
∠A = 2x, ∠B = 3x, ∠C = 5x
2x + 3x + 5x = 360
10x = 360
(\(\frac{1}{10}\))10x = (\(\frac{1}{10}\))360
x = 36
∠A = 2(36°) = 72°
∠B = 3(36°) = 108°
∠C = 5(36°) = 180°

Question 9.
The sum of the measures of two adjacent angles is 72°. The ratio of the smaller angle to the larger angle is 1∶3. Find the measures of each angle.
Answer:
∠A = x, ∠B = 3x
x + 3x = 72
4x = 72
(\(\frac{1}{4}\))(4x) = (\(\frac{1}{4}\))(72)
x = 18
∠A = (18°) = 18°
∠B = 3(18°) = 54°

Question 10.
Find the measures of ∠CQA and ∠EQB.

Answer:
4x + 5x = 108
9x = 108
(\(\frac{1}{9}\))9x = (\(\frac{1}{9}\))108
x = 12
m∠CQA = 5(12°) = 60°
m∠EQB = 4(12°) = 48°

Eureka Math Grade 7 Module 3 Lesson 11 Exit Ticket Answer Key

Question 1.
Write an equation for the angle relationship shown in the figure and solve for x. Find the measures of ∠RQS and ∠TQU.

Answer:
3x + 90 + 4x + 221 = 360
7x + 311 = 360
7x + 311 – 311 = 360 – 311
7x = 49
(\(\frac{1}{7}\))7x = (\(\frac{1}{7}\))49
x = 7
m∠RQS = 3(7°) = 21°
m∠TQU = 4(7°) = 28°

Engage NY Eureka Math 7th Grade Module 3 Lesson 12 Answer Key

Eureka Math Grade 7 Module 3 Lesson 12 Example Answer Key

Example 1.
Preserves the inequality symbol:
Answer:
means the inequality symbol stays the same.

Reverses the inequality symbol:
Answer:
means the inequality symbol switches less than with greater than and less than or equal to with greater than or equal to.

Station 1
Add or Subtract a Number to Both Sides of the Inequality

Examine the results. Make a statement about what you notice, and justify it with evidence.
Answer:
When a number is added or subtracted to both numbers being compared, the symbol stays the same, and the inequality symbol is preserved.

Station 2
Multiply each term by – 1

Examine the results. Make a statement about what you notice and justify it with evidence.
Answer:
When both numbers are multiplied by – 1, the symbol changes, and the inequality symbol is reversed.

Station 3
Multiply or Divide Both Sides of the Inequality by a Positive Number

Examine the results. Make a statement about what you notice, and justify it with evidence.
Answer:
When both numbers being compared are multiplied by or divided by a positive number, the symbol stays the same, and the inequality symbol is preserved.

Station 4
Multiply or Divide Both Sides of the Inequality by a Negative Number

Examine the results. Make a statement about what you notice and justify it with evidence.
Answer:
When both numbers being compared are multiplied by or divided by a negative number, the symbol changes, and the inequality symbol is reversed.

Eureka Math Grade 7 Module 3 Lesson 12 Exercise Answer Key

Exercise
Complete the following chart using the given inequality, and determine an operation in which the inequality symbol is preserved and an operation in which the inequality symbol is reversed. Explain why this occurs.

Answer:
Solutions may vary. A sample student response is below.

Eureka Math Grade 7 Module 3 Lesson 12 Problem Set Answer Key

Question 1.
For each problem, use the properties of inequalities to write a true inequality statement.
The two integers are – 2 and – 5.
a. Write a true inequality statement.
Answer:
– 5 < – 2

b. Subtract – 2 from each side of the inequality. Write a true inequality statement.
Answer:
– 7 < – 4

c. Multiply each number by – 3. Write a true inequality statement.
Answer:
15 > 6

Question 2.
On a recent vacation to the Caribbean, Kay and Tony wanted to explore the ocean elements. One day they went in a submarine 150 feet below sea level. The second day they went scuba diving 75 feet below sea level.
a. Write an inequality comparing the submarine’s elevation and the scuba diving elevation.
Answer:
– 150 < – 75

b. If they only were able to go one – fifth of the capable elevations, write a new inequality to show the elevations they actually achieved.
Answer:
– 30 < – 15

c. Was the inequality symbol preserved or reversed? Explain.
Answer:
The inequality symbol was preserved because the number that was multiplied to both sides was NOT negative.

Question 3.
If a is a negative integer, then which of the number sentences below is true? If the number sentence is not true, give a reason.
a. 5 + a < 5
Answer:
True

b. 5 + a > 5
Answer:
False because adding a negative number to 5 will decrease 5, which will not be greater than 5.

c. 5 – a > 5
Answer:
True

d. 5 – a < 5
Answer:
False because subtracting a negative number is adding a positive number to 5, which will be larger than 5.

e. 5a < 5
Answer:
True

f. 5a > 5
Answer:
False because a negative number multiplied by a positive number is negative, which will be less than 5.

g. 5 + a > a
Answer:
True

h. 5 + a < a
Answer: False because adding 5 to a negative number is greater than the negative number itself.

i. 5 – a > a
Answer:
True

j. 5 – a < a
Answer:
False because subtracting a negative number is the same as adding a positive number, which is greater than the negative number itself.

k. 5a > a
Answer:
False because a negative number multiplied by a 5 is negative and will be 5 times smaller than a.

l. 5a < a
Answer:
True

Eureka Math Grade 7 Module 3 Lesson 12 Exit Ticket Answer Key

Question 1.
Given the initial inequality – 4 < 7, state possible values for c that would satisfy the following inequalities.
a. c( – 4) < c(7)
Answer:
c > 0

b. c( – 4) > c(7)
Answer:
c < 0

c. c(- 4) = c(7)
Answer:
c = 0

Question 2.
Given the initial inequality 2 > – 4, identify which operation preserves the inequality symbol and which operation reverses the inequality symbol. Write the new inequality after the operation is performed.
a. Multiply both sides by – 2.
Answer:
The inequality symbol is reversed.
2 > – 4
2( – 2) < – 4( – 2)
– 4 < 8

b. Add – 2 to both sides.
Answer:
The inequality symbol is preserved. 2 > – 4
2 + ( – 2) > – 4 + ( – 2)
0 > – 6

c. Divide both sides by 2.
Answer:
The inequality symbol is preserved.
2 > – 4
2 ÷ 2 > – 4 ÷ 2
1 > – 2

d. Multiply both sides by – \(\frac{1}{2}\).
Answer:
Inequality symbol is reversed.
2 > – 4
2( – \(\frac{1}{2}\) ) < – 4( – \(\frac{1}{2}\) )
– 1 < 2 e. Subtract – 3 from both sides. Answer: The inequality symbol is preserved. 2 > – 4
2 – ( – 3) > – 4 – ( – 3)
5 > – 1

Eureka Math Grade 7 Module 3 Lesson 12 Equations Answer Key

Progression of Exercises
Determine the value of the variable.

Set 1
Question 1.
x + 1 = 5
Answer:
x = 4

Question 2.
x + 3 = 5
Answer:
x = 2

Question 3.
x + 6 = 5
Answer:
x = – 1

Question 4.
x – 5 = 2
Answer:
x = 7

Question 5.
x – 5 = 8
Answer:
x = 13

Set 2
Question 1.
3x = 15
Answer:
x = 5

Question 2.
3x = 0
Answer:
x = 0

Question 3.
3x = – 3
Answer:
x = – 1

Question 4.
– 9x = 18
Answer:
x = – 2

Question 5.
– x = 18
Answer:
x = – 18

Set 3
Question 1.
\(\frac{1}{7}\) x = 5
Answer:
x = 35

Question 2.
\(\frac{2}{7}\) x = 10
Answer:
x = 35

Question 3.
\(\frac{3}{7}\) x = 15
Answer:
x = 35

Question 4.
\(\frac{4}{7}\) x = 20
Answer:
x = 35

Question 5.
– \(\frac{5}{7}\) x = – 25
Answer:
x = 35

Set 4
Question 1.
2x + 4 = 12
Answer:
x = 4

Question 2.
2x – 5 = 13
Answer:
x = 9

Question 3.
2x + 6 = 14
Answer:
x = 4

Question 4.
3x – 6 = 18
Answer:
x = 8

Question 5.
– 4x + 6 = 22
Answer:
x = – 4

Set 5
Question 1.
2x + 0.5 = 6.5
Answer:
x = 3

Question 2.
3x – 0.5 = 8.5
Answer:
x = 3

Question 3.
5x + 3 = 8.5
Answer:
x = 1.1

Question 4.
5x – 4 = 1.5
Answer:
x = 1.1

Question 5.
– 7x + 1.5 = 5
Answer:
x = – 0.5

Set 6
Question 1.
2(x + 3) = 4
Answer:
x = – 1

Question 2.
5(x + 3) = 10
Answer:
x = – 1

Question 3.
5(x – 3) = 10
Answer:
x = 5

Question 4.
– 2(x – 3) = 8
Answer:
x = – 1

Question 5.
– 3(x + 4) = 3
Answer:
x = – 5

Engage NY Eureka Math 7th Grade Module 3 Lesson 13 Answer Key

Eureka Math Grade 7 Module 3 Lesson 13 Example Answer Key

Example 1: Evaluating Inequalities—Finding a Solution
The sum of two consecutive odd integers is more than – 12. Write several true numerical inequality expressions.
Answer:

The sum of two consecutive odd integers is more than – 12. What is the smallest value that will make this true?
a. Write an inequality that can be used to find the smallest value that will make the statement true.
Answer:
x: an integer
2x + 1: odd integer
2x + 3: next consecutive odd integer
2x + 1 + 2x + 3 > – 12

b. Use if – then moves to solve the inequality written in part (a). Identify where the 0’s and 1’s were made using the if – then moves.
Answer:
4x + 4 > – 12
4x + 4 – 4 > – 12 – 4 If a > b, then a – 4 > b – 4.
4x + 0 > – 16 0 was the result.
(\(\frac{1}{4}\))(4x) > (\(\frac{1}{4}\))( – 16) If a > b, then a(\(\frac{1}{4}\)) > b(\(\frac{1}{4}\)).
x > – 4 1 was the result.

c. What is the smallest value that will make this true?
Answer:
To find the odd integer, substitute – 4 for x in 2x + 1.
2( – 4) + 1
– 8 + 1
– 7
The values that will solve the original inequality are all the odd integers greater than – 7. Therefore, the smallest values that will make this true are – 5 and – 3.

Eureka Math Grade 7 Module 3 Lesson 13 Exercise Answer Key

Opening Exercise: Writing Inequality Statements
Tarik is trying to save $265.49 to buy a new tablet. Right now, he has $40 and can save $38 a week from his allowance.
Write and evaluate an expression to represent the amount of money saved after …
2 weeks
Answer:
40 + 38(2)
40 + 76
116

3 weeks
Answer:
40 + 38(3)
40 + 114
154

4 weeks
Answer:
40 + 38(4)
40 + 152
192

5 weeks
Answer:
40 + 38(5)
40 + 190
230

6 weeks
Answer:
40 + 38(6)
40 + 228
268

7 weeks
Answer:
40 + 38(7)
40 + 266
306

8 weeks
Answer:
40 + 38(8)
40 + 304
344

When will Tarik have enough money to buy the tablet?
Answer:
From 6 weeks and onward

Write an inequality that will generalize the problem.
Answer:
38w + 40 ≥ 265.49 Where w represents the number of weeks it will take to save the money.

Exercise 1.
Connor went to the county fair with $22.50 in his pocket. He bought a hot dog and drink for $3.75 and then wanted to spend the rest of his money on ride tickets, which cost $1.25 each.
a. Write an inequality to represent the total spent where r is the number of tickets purchased.
Answer:
1.25r + 3.75 ≤ 22.50

b. Connor wants to use this inequality to determine whether he can purchase 10 tickets. Use substitution to show whether he will have enough money.
Answer:
1.25r + 3.75 ≤ 22.50
1.25(10) + 3.75 ≤ 22.50
12.5 + 3.75 ≤ 22.50
16.25 ≤ 22.50
True
He will have enough money since a purchase of 10 tickets brings his total spending to $16.25.

c. What is the total maximum number of tickets he can buy based upon the given information?
Answer:
1.25r + 3.75 ≤ 22.50
1.25r + 3.75 – 3.75 ≤ 22.50 – 3.75
1.25r + 0 ≤ 18.75
(\(\frac{1}{1.25}\))(1.25r) ≤ (\(\frac{1}{1.25}\))(18.75)
r ≤ 15
The maximum number of tickets he can buy is 15.

Exercise 2.
Write and solve an inequality statement to represent the following problem:
On a particular airline, checked bags can weigh no more than 50 pounds. Sally packed 32 pounds of clothes and five identical gifts in a suitcase that weighs 8 pounds. Write an inequality to represent this situation.
Answer:
x: weight of one gift
5x + 8 + 32 ≤ 50
5x + 40 ≤ 50
5x + 40 – 40 ≤ 50 – 40
5x ≤ 10
(\(\frac{1}{5}\))(5x) ≤ (\(\frac{1}{5}\))(10)
x ≤ 2
Each of the 5 gifts can weigh 2 pounds or less.

Eureka Math Grade 7 Module 3 Lesson 13 Problem Set Sample Answer Key

Question 1.
Match each problem to the inequality that models it. One choice will be used twice.
_________ The sum of three times a number and – 4 is greater than 17.         a. 3x + – 4 ≥ 17
_________ The sum of three times a number and – 4 is less than 17.               b. 3x + – 4 < 17
_________ The sum of three times a number and – 4 is at most 17.                 c. 3x + – 4 > 17
_________ The sum of three times a number and – 4 is no more than 17.       d. 3x + – 4 ≤ 17
_________ The sum of three times a number and – 4 is at least 17.
Answer:
c The sum of three times a number and – 4 is greater than 17.           a. 3x + – 4 ≥ 17
b The sum of three times a number and – 4 is less than 17.                b. 3x + – 4 < 17
d The sum of three times a number and – 4 is at most 17.                  c. 3x + – 4 > 17
d The sum of three times a number and – 4 is no more than 17.        d. 3x + – 4 ≤ 17
a The sum of three times a number and – 4 is at least 17.

Question 2.
If x represents a positive integer, find the solutions to the following inequalities.
a. x < 7
Answer:
x < 7 or 1, 2, 3, 4, 5, 6

b. x – 15 < 20
Answer:
x < 35

c. x + 3 ≤ 15
Answer:
x ≤ 12

d. – x > 2
Answer:
There are no positive integer solutions.

e. 10 – x > 2
Answer:
x < 8

f. – x ≥ 2
Answer:
There are no positive integer solutions.

g. \(\frac{x}{3}\) < 2
x < 6 Answer: h. – \(\frac{x}{3}\) > 2
Answer:
There are no positive integer solutions.

i. 3 – \(\frac{x}{4}\) > 2
Answer:
x < 4

Question 3.
Recall that the symbol ≠ means not equal to. If x represents a positive integer, state whether each of the following statements is always true, sometimes true, or false.
a. x > 0
Answer:
Always true

b. x < 0
Answer:
False

c. x > – 5
Answer:
Always true

d. x > 1
Answer:
Sometimes true

e. x ≥ 1
Answer:
Always true

f. x ≠ 0
Answer:
Always true

g. x ≠ – 1
Answer:
Always true

h. x ≠ 5
Answer:
Sometimes true

Question 4.
Twice the smaller of two consecutive integers increased by the larger integer is at least 25.
Model the problem with an inequality, and determine which of the given values 7, 8, and/or 9 are solutions. Then, find the smallest number that will make the inequality true.
Answer:
2x + x + 1 ≥ 25
The smallest integer would be 8.
For x = 7:
2x + x + 1 ≥ 25
2(7) + 7 + 1 ≥ 25
14 + 7 + 1 ≥ 25
22 ≥ 25
False

For x = 8:
2x + x + 1 ≥ 25
2(8) + 8 + 1 ≥ 25
16 + 8 + 1 ≥ 25
25 ≥ 25
True

For x = 9:
2x + x + 1 ≥ 25
2(9) + 9 + 1 ≥ 25
18 + 9 + 1 ≥ 25
28 ≥ 25
True
The smallest integer would be 8.

Question 5.
a. The length of a rectangular fenced enclosure is 12 feet more than the width. If Farmer Dan has 100 feet of fencing, write an inequality to find the dimensions of the rectangle with the largest perimeter that can be created using 100 feet of fencing.
Answer:
Let w represent the width of the fenced enclosure.
w + 12: length of the fenced enclosure
w + w + w + 12 + w + 12 ≤ 100
4w + 24 ≤ 100

b. What are the dimensions of the rectangle with the largest perimeter? What is the area enclosed by this rectangle?
Answer:
4w + 24 ≤ 100
4w + 24 – 24 ≤ 100 – 24
4w + 0 ≤ 76
(\(\frac{1}{4}\))(4w) ≤ (\(\frac{1}{4}\))(76)
w ≤ 19
Maximum width is 19 feet.
Maximum length is 31 feet.
Maximum area: A = lw
A = (19)(31)
A = 589
The area is 589 ft < sup > 2 < /sup > .

Question 6.
At most, Kyle can spend $50 on sandwiches and chips for a picnic. He already bought chips for $6 and will buy sandwiches that cost $4.50 each. Write and solve an inequality to show how many sandwiches he can buy. Show your work, and interpret your solution.
Answer:
Let s represent the number of sandwiches.
4.50s + 6 ≤ 50
4.50s + 6 – 6 ≤ 50 – 6
4.50s ≤ 44
(\(\frac{1}{4.50}\))(4.50s) ≤ (\(\frac{1}{4.50}\))(44)
s ≤ 9 \(\frac{7}{9}\)
At most, Kyle can buy 9 sandwiches with $50.

Eureka Math Grade 7 Module 3 Lesson 13 Exit Ticket Answer Key

Question 1.
Shaggy earned $7.55 per hour plus an additional $100 in tips waiting tables on Saturday. He earned at least $160 in all. Write an inequality and find the minimum number of hours, to the nearest hour, that Shaggy worked on Saturday.
Answer:
Let h represent the number of hours worked.
7.55h + 100 ≥ 160
7.55h + 100 – 100 ≥ 160 – 100
7.55h ≥ 60
(\(\frac{1}{7.55}\))(7.55h) ≥ (\(\frac{1}{7.55}\))(60)
h ≥ 7.9
If Shaggy earned at least $160, he would have worked at least 8 hours.

Engage NY Eureka Math 7th Grade Module 3 Lesson 14 Answer Key

Eureka Math Grade 7 Module 3 Lesson 14 Example Answer Key

Example 1.
A youth summer camp has budgeted $2,000 for the campers to attend the carnival. The cost for each camper is $17.95, which includes general admission to the carnival and two meals. The youth summer camp must also pay $250 for the chaperones to attend the carnival and $350 for transportation to and from the carnival. What is the greatest number of campers who can attend the carnival if the camp must stay within its budgeted amount?
Answer:
Let c represent the number of campers to attend the carnival.
17.95c + 250 + 350 ≤ 2000
17.95c + 600 ≤ 2000
17.95c + 600 – 600 ≤ 2000 – 600
17.95c ≤ 1400
(\(\frac{1}{17.95}\))(17.95c) ≤ (\(\frac{1}{17.95}\))(1400)
c ≤ 77.99
In order for the camp to stay in budget, the greatest number of campers who can attend the carnival is 77 campers.

Example 2.
The carnival owner pays the owner of an exotic animal exhibit $650 for the entire time the exhibit is displayed. The owner of the exhibit has no other expenses except for a daily insurance cost. If the owner of the animal exhibit wants to make more than $500 in profits for the 5 \(\frac{1}{2}\) days, what is the greatest daily insurance rate he can afford to pay?
Answer:
Let i represent the daily insurance cost.
650 – 5.5i > 500
– 5.5i + 650 – 650 > 500 – 650
– 5.5i + 0> – 150
(\(\frac{1}{ – 5.5}\))( – 5.5i)>(\(\frac{1}{ – 5.5}\))( – 150)
i<27.27 The maximum daily cost the owner can pay for insurance is $27.27. Example 3. Several vendors at the carnival sell products and advertise their businesses. Shane works for a recreational company that sells ATVs, dirt bikes, snowmobiles, and motorcycles. His boss paid him $500 for working all of the days at the carnival plus 5% commission on all of the sales made at the carnival. What was the minimum amount of sales Shane needed to make if he earned more than $1,500? Answer: Let s represent the sales, in dollars, made during the carnival. 500 + \(\frac{5}{100}\) s > 1,500
\(\frac{5}{100}\) s + 500 > 1,500
\(\frac{5}{100}\) s + 500 – 500 > 1,500 – 500
\(\frac{5}{100}\) s + 0 > 1,000
(\(\frac{100}{5}\))(\(\frac{5}{100}\) s) > (\(\frac{100}{5}\))(1,000)
s > 20,000
The sales had to be more than $20,000 for Shane to earn more than $1,500.

Eureka Math Grade 7 Module 3 Lesson 14 Exercise Answer Key

Opening Exercise
The annual County Carnival is being held this summer and will last 5 \(\frac{1}{2}\) days. Use this information and the other given information to answer each problem.

You are the owner of the biggest and newest roller coaster called the Gentle Giant. The roller coaster costs $6 to ride. The operator of the ride must pay $200 per day for the ride rental and $65 per day for a safety inspection. If you want to make a profit of at least $1,000 each day, what is the minimum number of people that must ride the roller coaster?
Write an inequality that can be used to find the minimum number of people, p, which must ride the roller coaster each day to make the daily profit.
Answer:
6p – 200 – 65 ≥ 1000

Solve the inequality.
Answer:
6p – 200 – 65 ≥ 1000
6p – 265 ≥ 1000
6p – 265 + 265 ≥ 1000 + 265
6p + 0 ≥ 1265
(\(\frac{1}{6}\))(6p) ≥ (\(\frac{1}{6}\))(1265)
p ≥ 210 \(\frac{5}{6}\)

Interpret the solution.
Answer:
There needs to be a minimum of 211 people to ride the roller coaster every day to make a daily profit of at least $1,000.

Eureka Math Grade 7 Module 3 Lesson 14 Problem Set Answer Key

Question 1.
As a salesperson, Jonathan is paid $50 per week plus 3% of the total amount he sells. This week, he wants to earn at least $100. Write an inequality with integer coefficients for the total sales needed to earn at least $100, and describe what the solution represents.
Answer:
Let the variable p represent the purchase amount.
50 + \(\frac{3}{100}\) p ≥ 100
\(\frac{3}{100}\) p + 50 ≥ 100
(100)(\(\frac{3}{100}\) p) + 100(50) ≥ 100(100)
3p + 5000 ≥ 10000
3p + 5000 – 5000 ≥ 10000 – 5000
3p + 0 ≥ 5000
(\(\frac{1}{3}\))(3p) ≥ (\(\frac{1}{3}\))(5000)
p ≥ 1666 \(\frac{2}{3}\)
Jonathan must sell $1,666.67 in total purchases.

Question 2.
Systolic blood pressure is the higher number in a blood pressure reading. It is measured as the heart muscle contracts. Heather was with her grandfather when he had his blood pressure checked. The nurse told him that the upper limit of his systolic blood pressure is equal to half his age increased by 110.
a. a is the age in years, and p is the systolic blood pressure in millimeters of mercury (mmHg). Write an inequality to represent this situation.
Answer:
p ≤ \(\frac{1}{2}\) a + 110

b. Heather’s grandfather is 76 years old. What is normal for his systolic blood pressure?
Answer:
p ≤ \(\frac{1}{2}\) a + 110, where a = 76.
p ≤ \(\frac{1}{2}\) (76) + 110
p ≤ 38 + 110
p ≤ 148
A systolic blood pressure for his age is normal if it is at most 148 mmHG.

Question 3.
Traci collects donations for a dance marathon. One group of sponsors will donate a total of $6 for each hour she dances. Another group of sponsors will donate $75 no matter how long she dances. What number of hours, to the nearest minute, should Traci dance if she wants to raise at least $1,000?
Answer:
Let the variable h represent the number of hours Traci dances.
6h + 75 ≥ 1000
6h + 75 – 75 ≥ 1000 – 75
6h + 0 ≥ 925
(\(\frac{1}{6}\))(6h) ≥ (\(\frac{1}{6}\))(925)
h ≥ 154 \(\frac{1}{6}\)
Traci would have to dance at least 154 hours and 10 minutes.

Question 4.
Jack’s age is three years more than twice the age of his younger brother, Jimmy. If the sum of their ages is at most 18, find the greatest age that Jimmy could be.
Answer:
Let the variable j represent Jimmy’s age in years.
Then, the expression 3 + 2j represents Jack’s age in years.
j + 3 + 2j ≤ 18
3j + 3 ≤ 18
3j + 3 – 3 ≤ 18 – 3
3j ≤ 15
(\(\frac{1}{3}\))(3j) ≤ (\(\frac{1}{3}\))(15)
j ≤ 5
Jimmy’s age is 5 years or less.

Question 5.
Brenda has $500 in her bank account. Every week she withdraws $40 for miscellaneous expenses. How many weeks can she withdraw the money if she wants to maintain a balance of a least $200?
Answer:
Let the variable w represent the number of weeks.
500 – 40w ≥ 200
500 – 500 – 40w ≥ 200 – 500
– 40w ≥ – 300
( – \(\frac{1}{40}\))( – 40w) ≤ ( – \(\frac{1}{40}\))( – 300)
w ≤ 7.5
$40 can be withdrawn from the account for seven weeks if she wants to maintain a balance of at least $200.

Question 6.
A scooter travels 10 miles per hour faster than an electric bicycle. The scooter traveled for 3 hours, and the bicycle traveled for 5 \(\frac{1}{2}\) hours. Altogether, the scooter and bicycle traveled no more than 285 miles. Find the maximum speed of each.
Answer:

3(x + 10) + 5 \(\frac{1}{2}\) x ≤ 285
3x + 30 + 5 \(\frac{1}{2}\) x ≤ 285
8 \(\frac{1}{2}\) x + 30 ≤ 285
8 \(\frac{1}{2}\) x + 30 – 30 ≤ 285 – 30
8 \(\frac{1}{2}\) x ≤ 255
\(\frac{17}{2}\) x ≤ 255
(\(\frac{2}{17}\))(\(\frac{17}{2}\) x) ≤ (255)(\(\frac{2}{17}\))
x ≤ 30
The maximum speed the bicycle traveled was 30 miles per hour, and the maximum speed the scooter traveled was 40 miles per hour.

Eureka Math Grade 7 Module 3 Lesson 14 Exit Ticket Answer Key

Question 1.
Games at the carnival cost $3 each. The prizes awarded to winners cost $145.65. How many games must be played to make at least $50?
Answer:
Let g represent the number of games played.
3g – 145.65 ≥ 50
3g – 145.65 + 145.65 ≥ 50 + 145.65
3g + 0 ≥ 195.65
(\(\frac{1}{3}\))(3g) ≥ (\(\frac{1}{3}\))(195.65)
g ≥ 65.217
There must be at least 66 games played to make at least $50.

Engage NY Eureka Math 7th Grade Module 3 Lesson 15 Answer Key

Eureka Math Grade 7 Module 3 Lesson 15 Example Answer Key

Example
A local car dealership is trying to sell all of the cars that are on the lot. Currently, there are 525 cars on the lot, and the general manager estimates that they will consistently sell 50 cars per week. Estimate how many weeks it will take for the number of cars on the lot to be less than 75.
Write an inequality that can be used to find the number of full weeks, w, it will take for the number of cars to be less than 75. Since w is the number of full or complete weeks, w = 1 means at the end of week 1.
Answer:
525 – 50w < 75

Solve and graph the inequality.
Answer:
525 – 50w < 75
– 50w + 525 – 525 < 75 – 525
– 50w + 0 < – 450 ( – \(\frac{1}{50}\))( – 50w) > ( – \(\frac{1}{50}\))( – 450)
w > 9

Interpret the solution in the context of the problem.
Answer:
The dealership can sell 50 cars per week for more than 9 weeks to have less than 75 cars remaining on the lot.

Verify the solution.
Answer:
w = 9:
525 – 50w < 75
525 – 50(9) < 75
525 – 450 < 75
75 < 75
False

w = 10:
525 – 50w < 75
525 – 50(10) < 75
525 – 500 < 75
25 < 75
True

Eureka Math Grade 7 Module 3 Lesson 15 Exercise Answer Key

Exercise 1.
Two identical cars need to fit into a small garage. The opening is 23 feet 6 inches wide, and there must be at least 3 feet 6 inches of clearance between the cars and between the edges of the garage. How wide can the cars be?
Answer:
Encourage students to begin by drawing a diagram to illustrate the problem. A sample diagram is as follows:

Have students try to find all of the widths that the cars could be. Challenge them to name one more width than the person next to them. While they name the widths, plot the widths on a number line at the front of the class to demonstrate the shading. Before plotting the widths, ask if the circle should be open or closed as a quick review of graphing inequalities. Ultimately, the graph should be

→ Describe how to find the width of each car.
To find the width of each car, I subtract the minimum amount of space needed on either side of each car and in between the cars from the total length. Altogether, the amount of space needed was 3(3.5 ft.) or 10.5 ft. Then, I divided the result, 23.5 – 10.5 = 13, by 2 since there were 2 cars. The answer would be no more than \(\frac{13}{2}\) ft. or 6.5 ft.

→ Did you take an algebraic approach to finding the width of each car or an arithmetic approach? Explain.
Answers will vary.

→ If arithmetic was used, ask, “If w is the width of one car, write an inequality that can be used to find all possible values of w.”
2w + 10.5 ≤ 23.5

→ Why is an inequality used instead of an equation?
Since the minimum amount of space between the cars and each side of the garage is at least 3 feet 6 inches, which equals 3.5 ft., the space could be larger than 3 feet 6 inches. If so, then the width of the cars would be smaller. Since the width in between the cars and on the sides is not exactly 3 feet 6 inches, and it could be more, then there are many possible car widths. An inequality will give all possible car widths.

→ If an algebraic approach was used initially, ask, “How is the work shown in solving the inequality similar to the arithmetic approach?”
The steps to solving the inequality are the same as in an arithmetic approach. First, determine the total minimum amount of space needed by multiplying 3 by 3.5. Then, subtract 10.5 from the total of 23.5 and divide by 2.

→ What happens if the width of each car is less than 6.5 feet?
The amount of space between the cars and on either side of the car and garage is more then 3 feet 6 inches.

→ What happens if the width of each car is exactly 6.5 feet?
The amount of space between the cars and on either side of the car and garage is exactly 3 feet 6 inches.

→ What happens if the width of each car is more than 6.5 feet?
The amount of space between the cars and on either side of the car and garage is less than 3 feet 6 inches.

→ How many possible car widths are there?
Any infinite number of possible widths.

→ What assumption is being made?
The assumption made is that the width of the car is greater than 0 feet. The graph illustrates all possible values less than 6.5 feet, but in the context of the problem, we know that the width of the car must be greater than 0 feet.

→ Since we have determined there is an infinite amount, how can we illustrate this on a number line?
Illustrate by drawing a graph with a closed circle on 6.5 and an arrow drawn to the left.

→ What if 6.5 feet could not be a width, but all other possible measures less than 6.5 can be a possible width; how would the graph be different?
The graph would have an open circle on 6.5 and an arrow drawn to the left.

Exercise 2.
The cost of renting a car is $25 per day plus a one – time fee of $75.50 for insurance. How many days can the car be rented if the total cost is to be no more than $525?
a. Write an inequality to model the situation.
Answer:
Let x represent the number of days the car is rented.
25x + 75.50 ≤ 525

b. Solve and graph the inequality.
Answer:
25x + 75.50 ≤ 525
25x + 75.50 – 75.50 ≤ 525 – 75.50
25x + 0 ≤ 449.50
(\(\frac{1}{25}\)(25x) ≤ (\(\frac{1}{25}\))(449.50)
x ≤ 17.98

OR

25x + 75.50 ≤ 525
2,500x + 7,550 ≤ 52,500
2,500x + 7,550 – 7,550 ≤ 52,500 – 7,550
(\(\frac{1}{2,500}\))(2,500x) ≤ (\(\frac{1}{2,500}\))(44,950)
x ≤ 17.98

c. Interpret the solution in the context of the problem.
Answer:
The car can be rented for 17 days or fewer and stay within the amount of $525. The number of days is an integer. The 18th day would put the cost over $525, and since the fee is charged per day, the solution set includes whole numbers.

Exercise 3.
Mrs. Smith decides to buy three sweaters and a pair of jeans. She has $120 in her wallet. If the price of the jeans is $35, what is the highest possible price of a sweater, if each sweater is the same price?
Answer:
Let w represent the price of one sweater.
3w + 35 ≤ 120
3w + 35 – 35 ≤ 120 – 35
3w + 0 ≤ 85
(\(\frac{1}{3}\))(3w) ≤ (\(\frac{1}{3}\))(85)
w ≤ 28.33
Graph:

Solution: The highest price Mrs. Smith can pay for a sweater and have enough money is $28.33.

Exercise 4.
The members of the Select Chorus agree to buy at least 250 tickets for an outside concert. They buy 20 fewer lawn tickets than balcony tickets. What is the least number of balcony tickets bought?
Answer:
Let b represent the number of balcony tickets.
Then b – 20 represents the number of lawn tickets.
b + b – 20 ≥ 250
2b – 20 ≥ 250
2b – 20 + 20 ≥ 250 + 20
2b + 0 ≥ 270
(\(\frac{1}{2}\))(2b) ≥ (\(\frac{1}{2}\))(270)
b ≥ 135
Graph:

Solution: The least number of balcony tickets bought is 135. The answers need to be integers.

Exercise 5.
Samuel needs $29 to download some songs and movies on his MP3 player. His mother agrees to pay him $6 an hour for raking leaves in addition to his $5 weekly allowance. What is the minimum number of hours Samuel must work in one week to have enough money to purchase the songs and movies?
Answer:
Let h represent the number of hours Samuel rakes leaves.
6h + 5 ≥ 29
6h + 5 – 5 ≥ 29 – 5
6h + 0 ≥ 24
(\(\frac{1}{6}\))(6h) ≥ (\(\frac{1}{6}\))(24)
h ≥ 4
Graph:

Solution: Samuel needs to rake leaves at least 4 hours to earn $29. Any amount of time over 4 hours will earn him extra money.

Eureka Math Grade 7 Module 3 Lesson 15 Problem Set Answer Key

Question 1.
Ben has agreed to play fewer video games and spend more time studying. He has agreed to play less than 10 hours of video games each week. On Monday through Thursday, he plays video games for a total of 5 \(\frac{1}{2}\) hours. For the remaining 3 days, he plays video games for the same amount of time each day. Find t, the amount of time he plays video games for each of the 3 days. Graph your solution.
Answer:
Let t represent the time in hours spent playing video games.
3t + 5 \(\frac{1}{2}\) < 10
3t + 5 \(\frac{1}{2}\) – 5 \(\frac{1}{2}\) < 10 – 5 \(\frac{1}{2}\)
3t + 0 < 4 \(\frac{1}{2}\)
(\(\frac{1}{3}\))(3t) < (\(\frac{1}{3}\))(4 \(\frac{1}{2}\))
t < 1.5
Graph:

Ben plays less than 1.5 hours of video games each of the three days.

Question 2.
Gary’s contract states that he must work more than 20 hours per week. The graph below represents the number of hours he can work in a week.

a. Write an algebraic inequality that represents the number of hours, h, Gary can work in a week.
Answer:
h > 20

b. Gary is paid $15.50 per hour in addition to a weekly salary of $50. This week he wants to earn more than $400. Write an inequality to represent this situation.
Answer:
15.50h + 50 > 400

c. Solve and graph the solution from part (b). Round your answer to the nearest hour.
Answer:
15.50h + 50 – 50 > 400 – 50
15.50h > 350
(\(\frac{1}{15.50}\))(15.50h) > 350(\(\frac{1}{15.50}\))
h > 22.58
Gary has to work 23 or more hours to earn more than $400.

Question 3.
Sally’s bank account has $650 in it. Every week, Sally withdraws $50 to pay for her dog sitter. What is the maximum number of weeks that Sally can withdraw the money so there is at least $75 remaining in the account? Write and solve an inequality to find the solution, and graph the solution on a number line.
Answer:
Let w represent the number of weeks Sally can withdraw the money.
650 – 50w ≥ 75
650 – 50w – 650 ≥ 75 – 650
– 50w ≥ – 575
(\(\frac{1}{ – 50}\))( – 50w) ≥ (\(\frac{1}{ – 50}\))( – 575)
w ≤ 11.5
The maximum number of weeks Sally can withdraw the weekly dog sitter fee is 11 weeks.

Question 4.
On a cruise ship, there are two options for an Internet connection. The first option is a fee of $5 plus an additional $0.25 per minute. The second option costs $50 for an unlimited number of minutes. For how many minutes, m, is the first option cheaper than the second option? Graph the solution.
Answer:
Let m represent the number of minutes of Internet connection.
5 + 0.25m < 50
5 + 0.25m – 5 < 50 – 5
0.25m + 0 < 45
(\(\frac{1}{0.25}\))(0.25m) < (\(\frac{1}{0.25}\))(45)
m < 180
If there are less than 180 minutes, or 3 hours, used on the Internet, then the first option would be cheaper. If 180 minutes or more are planned, then the second option is more economical.

Question 5.
The length of a rectangle is 100 centimeters, and its perimeter is greater than 400 centimeters. Henry writes an inequality and graphs the solution below to find the width of the rectangle. Is he correct? If yes, write and solve the inequality to represent the problem and graph. If no, explain the error(s) Henry made.

Answer:
Henry’s graph is incorrect. The inequality should be 2(100) + 2w > 400. When you solve the inequality, you get w > 100. The circle on 100 on the number line is correct; however, the circle should be an open circle since the perimeter is not equal to 400. Also, the arrow should be pointing in the opposite direction because the perimeter is greater than 400, which means the width is greater than 100. The given graph indicates an inequality of less than or equal to.

Eureka Math Grade 7 Module 3 Lesson 15 Exit Ticket Answer Key

Question 1.
The junior high art club sells candles for a fundraiser. The first week of the fundraiser, the club sells 7 cases of candles. Each case contains 40 candles. The goal is to sell at least 13 cases. During the second week of the fundraiser, the club meets its goal. Write, solve, and graph an inequality that can be used to find the possible number of candles sold the second week.
Answer:
Let n represent the number candles sold the second week.
\(\frac{n}{40}\) + 7 ≥ 13
\(\frac{n}{40}\) + 7 – 7 ≥ 13 – 7
\(\frac{n}{40}\) ≥ 6
(40)(\(\frac{n}{40}\)) ≥ 6(40)
n ≥ 240
The minimum number of candles sold the second week was 240.

OR

Let n represent the number of cases of candles sold the second week.
40n + 280 ≥ 520
40n + 280 – 280 ≥ 520 – 280
40n + 0 ≥ 240
(\(\frac{1}{40}\))(40n) ≥ 240(\(\frac{1}{40}\))
n ≥ 6

The minimum number of cases sold the second week was 6. Since there are 40 candles in each case, the minimum number of candles sold the second week would be (40)(6) = 240.

Eureka Math Grade 7 Module 3 Lesson 15 Inequalities Answer Key

Progression of Exercises
Determine the value(s) of the variable.
Set 1
Question 1.
x + 1 > 8
Answer:
x > 7

Question 2.
x + 3 > 8
Answer:
x > 5

Question 3.
x + 10 > 8
Answer:
x > – 2

Question 4.
x – 2 > 3
Answer:
x > 5

Question 5.
x – 4 > 3
Answer:
x > 7

Set 2
Question 1.
3x ≤ 15
Answer:
x ≤ 5

Question 2.
3x ≤ 21
Answer:
x ≤ 7

Question 3.
– x ≤ 4
Answer:
x ≥ – 4

Question 4.
– 2x ≤ 4
Answer:
x ≥ – 2

Question 5.
– x ≤ – 4
Answer:
x ≥ 4

Set 3
Question 1.
\(\frac{1}{2}\) x < 1
Answer:
x < 2

Question 2.
\(\frac{1}{2}\)x < 3
Answer:
x < 6

Question 3.
– \(\frac{1}{5}\)x < 2 Answer: x > – 10

Question 4.
– \(\frac{2}{5}\) x < 2 Answer: x > – 5

Question 5.
– \(\frac{3}{5}\) x < 3 Answer: x > – 5

Set 4
Question 1.
2x + 4 ≥ 8
Answer:
x ≥ 2

Question 2.
2x – 3 ≥ 5
Answer:
x ≥ 4

Question 3.
– 2x + 1 ≥ 7
Answer:
x ≤ – 3

Question 4.
– 3x + 1 ≥ – 8
Answer:
x ≤ 3

Question 5.
– 3x – 5 ≥ 10
Answer:
x ≤ – 5

Set 5
Question 1.
2x – 0.5 > 5.5
Answer:
x > 3

Question 2.
3x + 1.5 > 4.5
Answer:
x > 2

Question 3.
5x – 3 > 4.5
Answer:
x > 1.5

Question 4.
– 5x + 2 > 8.5
Answer:
x < – 1.3 Question 5. – 9x – 3.5 > 1
Answer:
x < – 0.5

Set 6
Question 1.
2(x + 3) ≤ 4
Answer:
x ≤ – 1

Question 2.
3(x + 3) ≤ 6
Answer:
x ≤ – 1

Question 3.
4(x + 3) ≤ 8
Answer:
x ≤ – 1

Question 4.
– 5(x – 3) ≤ – 10
Answer:
x ≥ 5

Question 5.
– 2(x + 3) ≤ 8
Answer:
x ≥ – 7

Engage NY Eureka Math 7th Grade Module 3 Lesson 16 Answer Key

Eureka Math Grade 7 Module 3 Lesson 16 Example Answer Key

Example
a. The following circles are not drawn to scale. Find the circumference of each circle. (Use \(\frac{22}{7}\) as an approximation for π.)

Answer:
66 cm; 286 ft.; 110 m; Ask students if these numbers are roughly three times the diameters.

b. The radius of a paper plate is 11.7 cm. Find the circumference to the nearest tenth. (Use 3.14 as an approximation for π.)
Answer:
Diameter: 23.4 cm; circumference: 73.5 cm

c. The radius of a paper plate is 11.7 cm. Find the circumference to the nearest hundredth. (Use the π button on your calculator as an approximation for π.)
Answer:
Circumference: 73.51 cm

d. A circle has a radius of r cm and a circumference of C cm. Write a formula that expresses the value of C in terms of r and π.
Answer:
C = π ∙ 2r, or C = 2πr.

e. The figure below is in the shape of a semicircle. A semicircle is an arc that is half of a circle. Find the perimeter of the shape. (Use 3.14 for π.)

Answer:
8 m + \(\frac{8(3.14)}{2}\) m = 20.56 m

Eureka Math Grade 7 Module 3 Lesson 16 Exercise Answer Key

Opening Exercise
a. Using a compass, draw a circle like the picture to the right.

C is the center of the circle.
The distance between C and B is the radius of the circle.

b. Write your own definition for the term circle.
Answer:
Student responses will vary. Many might say, “It is round.” “It is curved.” “It has an infinite number of sides.” “The points are always the same distance from the center.” Analyze their definitions, showing how other figures such as ovals are also “round” or “curved.” Ask them what is special about the compass they used. (Answer: The distance between the spike and the pencil is fixed when drawing the circle.) Let them try defining a circle again with this new knowledge.

c. Extend segment CB to a segment AB, where A is also a point on the circle.
Answer:

The length of the segment AB is called the diameter of the circle.
d. The diameter is ______________________ as long as the radius.
Answer:
The diameter is twice, or 2 times, as long as the radius.

e. Measure the radius and diameter of each circle. The center of each circle is labeled C.

Answer:
CB = 1.5 cm, AB = 3 cm, CF = 2 cm, EF = 4 cm
The radius of the largest circle is 3 cm. The diameter is 6 cm.

f. Draw a circle of radius 6 cm.
Answer:
Part (f) may not be as easy as it seems. Let students grapple with how to measure 6 cm with a compass. One difficulty they might encounter is trying to measure 6 cm by putting the spike of the compass on the edge of the ruler (i.e., the
0 cm mark). Suggest either of the following: (1) Measure the compass from the 1 cm mark to the 7 cm mark, or (2) Mark two points 6 cm apart on the paper first; then, use one point as the center.

Eureka Math Grade 7 Module 3 Lesson 16 Problem Set Answer Key

Question 1.
Find the circumference.

a. Give an exact answer in terms of π.
Answer:
C = 2πr
C = 2π ∙ 14 cm
C = 28π cm

b. Use π ≈ \(\frac{22}{7}\) , and express your answer as a fraction in lowest terms.
Answer:
C ≈ 2 ∙ \(\frac{22}{7}\) ∙ 14 cm
C ≈ 88 cm

c. Use the π button on your calculator, and express your answer to the nearest hundredth.
Answer:
C = 2 ∙ π ∙ 14 cm
C ≈ 87.96 cm

Question 2.
Find the circumference.

a. Give an exact answer in terms of π.
Answer:
d = 42 cm
C = πd
C = 42π cm

b. Use π ≈ \(\frac{22}{7}\) , and express your answer as a fraction in lowest terms.
Answer:
C ≈ 42 cm ∙ \(\frac{22}{7}\)
C ≈ 132 cm

Question 3.
The figure shows a circle within a square. Find the circumference of the circle. Let π ≈ 3.14.

Answer:
The diameter of the circle is the same as the length of the side of the square.
C = πd
C = π ∙ 16 in.
C ≈ 3.14 ∙ 16 in.
C ≈ 50.24 in.

Question 4.
Consider the diagram of a semicircle shown.

a. Explain in words how to determine the perimeter of a semicircle.
Answer:
The perimeter is the sum of the length of the diameter and half of the circumference of a circle with the same diameter.

b. Using d to represent the diameter of the circle, write an algebraic equation that will result in the perimeter of a semicircle.
Answer:
P = d + \(\frac{1}{2}\) πd

c. Write another algebraic equation to represent the perimeter of a semicircle using r to represent the radius of a semicircle.
Answer:
P = 2r + \(\frac{1}{2}\) π ∙ 2r
P = 2r + πr

Question 5.
Find the perimeter of the semicircle. Let π ≈ 3.14.

Answer:
P = d + \(\frac{1}{2}\) πd
P ≈ 17 in. + \(\frac{1}{2}\) ∙ 3.14 ∙ 17 in.
P ≈ 17 in. + 26.69 in.
P ≈ 43.69 in.

Question 6.
Ken’s landscape gardening business makes odd-shaped lawns that include semicircles. Find the length of the edging material needed to border the two lawn designs. Use 3.14 for π.
a. The radius of this flowerbed is 2.5 m.

Answer:
A semicircle has half of the circumference of a circle. If the circumference of the semicircle is C = \(\frac{1}{2}\)(π ∙ 2 ∙ 2.5 m), then the circumference approximates 7.85 m. The length of the edging material must include the circumference and the diameter; 7.85 m + 5 m = 12.85 m. Ken needs 12.85 meters of edging to complete his design.

b. The diameter of the semicircular section is 10 m, and the lengths of the two sides are 6 m.

Answer:
The circumference of the semicircular part has half of the circumference of a circle. The circumference of the semicircle is C = \(\frac{1}{2}\) π ∙ 10 m, which is approximately 15.7 m. The length of the edging material must include the circumference of the semicircle and the perimeter of two sides of the triangle;
15.7 m + 6 m + 6 m = 27.7 m. Ken needs 27.7 meters of edging to complete his design.

Question 7.
Mary and Margaret are looking at a map of a running path in a local park. Which is the shorter path from E to F, along the two semicircles or along the larger semicircle? If one path is shorter, how much shorter is it? Let π ≈ 3.14.

Answer:
A semicircle has half of the circumference of a circle. The circumference of the large semicircle is C = \(\frac{1}{2}\) π ∙ 4 km, or 6.28 km. The diameter of the two smaller semicircles is 2 km. The total circumference would be the same as the circumference for a whole circle with the same diameter. If C = π ∙ 2 km, then C = 6.28 km. The distance around the larger semicircle is the same as the distance around both of the semicircles. So, both paths are equal in distance.

Question 8.
Alex the electrician needs 34 yards of electrical wire to complete a job. He has a coil of wiring in his workshop. The coiled wire is 18 inches in diameter and is made up of 21 circles of wire. Will this coil be enough to complete the job? Let π ≈ 3.14.

Answer:
The circumference of the coil of wire is C = π ∙ 18 in., or approximately 56.52 in. If there are 21 circles of wire, then the number of circles times the circumference will yield the total number of inches of wire in the coil. If 56.52 in. ∙ 21 ≈ 1186.92 in., then \(\frac{1186.92 \mathrm{in.}}{36 \mathrm{in.}}\) ≈ 32.97 yd. (1 yd. = 3 ft. = 36 in. When converting inches to yards, you must divide the total inches by the number of inches in a yard, which is 36 inches.) Alex will not have enough wire for his job in this coil of wire.

Eureka Math Grade 7 Module 3 Lesson 16 Exit Ticket Answer Key

Brianna’s parents built a swimming pool in the backyard. Brianna says that the distance around the pool is 120 feet.
Question 1.
Is she correct? Explain why or why not.

Answer:
Brianna is incorrect. The distance around the pool is 131.4 ft. She found the distance around the rectangle only and did not include the distance around the semicircular part of the pool.

Question 2.
Explain how Brianna would determine the distance around the pool so that her parents would know how many feet of stone to buy for the edging around the pool.
Answer:
In order to find the distance around the pool, Brianna must first find the circumference of the semicircle, which is C = \(\frac{1}{2}\) ∙ π ∙ 20 ft., or 10π ft., or about 31.4 ft. The sum of the three other sides is
20 ft. + 40 ft. + 40 ft. = 100 ft.; the perimeter is 100 ft. + 31.4 ft. = 131.4 ft.

Question 3.
Explain the relationship between the circumference of the semicircular part of the pool and the width of the pool.
Answer:
The relationship between the circumference of the semicircular part and the width of the pool is the same as half of π because this is half the circumference of the entire circle.