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Engage NY Eureka Math Grade 6 Module 5 Mid Module Assessment Answer Key

Eureka Math Grade 6 Module 5 Mid Module Assessment Answer Key

Question 1.
David is the groundskeeper at Triangle Park, scale shown below.

Answer:

a. David needs to cut the grass four times a month. How many square yards of grass will he cut altogether each month?
Answer:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) 300yd. 50yd
A = \(\frac{1}{2}\) . 15,000 yd²
A = 7,500 yd²
4 . 7,500 yd² = 30,000 yd²

b. During the winter the triangular park and adjacent square parking lot are flooded with water and allowed to freeze so that people can go ice skating. What is the area of the ice?

Answer:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) . 300yd. 50 yd
A = \(\frac{1}{2}\) . 15,000 yd²
A = 7,500 yd²

A = S²
A = (50 yd)²
A = 2,500 yd²

7,500 yd² + 2,500 yd² = 10,000 yd²

Question 2.
Mariska is looking for a new computer table. Below is a sketch of two computer tables she likes when looking at them from above. All measurements are in feet.

a. If Mariska needs to choose the one with the greater area, which one should she choose? Justify your answer with evidence, using coordinates to determine side lengths.
Answer:
Table A:
(7, 15) → (7, 18) = 3 ft.
(7, 18) → (16, 18) = 9 ft.
(16, 18) → (16, 11) = 7 ft.
(16, 11) → (13, 11) = 3 ft.
(13, 11) → (13, 15) = 4 ft.
(13, 15) → (7, 15) = 6 ft.

A = bh
A = 9 ft. . 3 ft.
A = 27 ft²

A = bh
A = 3 ft. . 4 ft.
A = 12 ft²

27 ft² + 12 ft² = 39 ft²

Table B:
(5, 8) → (5, 10) = 2 ft.
(5, 10) → (2, 10) = 7 ft.
(12, 10) → (12, 3) = 7 ft.
(12, 3) → (5, 3) = 7 ft.
(5, 3) → (5, 5) = 2 ft.
(5, 5) → (9, 5) = 4 ft.
(9, 5) → (9, 8) = 3 ft.
(9, 8) → (5, 8) = 4 ft.

A = bh
A = 7 ft. . 2 ft.
A = 14 ft²

A = bh
A = 3 ft. 3 ft
A = 9 ft²

A = bh
A = 7 ft. . 2 ft
A = 14 ft²

14 ft² + 9 ft² + 14 ft² = 37 ft²

Mariska will need to choose Table A because it is the table with the greatest area

b. If Mariska needs to choose the one with the greater perimeter, which one should she choose? Justify your answer with evidence, using coordinates to determine side lengths.

Answer:
Table A:
P = 3 ft. + 9 ft. + 7 ft. + 3 ft. + 4 ft. + 6 ft.
P = 32 ft.

Table B:
P = 2 ft. + 7 ft. + 7 ft. + 7 ft. + 2 ft. + 3 ft. + 4 ft. + 4 ft.
P = 36 ft.
Table B has a large perimeter.

Question 3.
Find the area of the triangular region.

Answer:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) . 5 in. 7 in.
A = \(\frac{1}{2}\) . 35 in²
A = 17.5 in²

Question 4.
The grid below shows a bird’s-eye view of a middle school.

a. Write the coordinates of each point in the table.
Answer:

b. Each space on the grid stands for 10 meters. Find the length of each wall of the school.
Answer:

c. Find the area of the entire building. Show your work.
Answer:
A = bh
A = 100 m . 60 m
A = 6,000 m²

A = bh
A = 30 m . 50 m
A = 1,500 m²

A = bh
A = 20 m . 40 m
A = 800 m²

6,000 m² + 1,500 m² + 800 m² = 8,300 m²

Engage NY Eureka Math Grade 6 Module 5 Lesson 19a Answer Key

Eureka Math Grade 6 Module 5 Lesson 19a Opening Exercise Answer Key

Opening Exercise:

Question 1.
Determine the volume of this aquarium.

Answer:
V = l × w × h
V = 20 in. × 10 in. × 12 in.
V = 2,400 in³

Eureka Math Grade 6 Module 5 Lesson 19a Mathematical Modeling Exercise Answer Key

Mathematical Modeling Exercise: Using Ratios and Unit Rate to Determine Volume

For his environmental science project, Jamie is creating habitats for various wildlife including fish, aquatic turtles, and aquatic frogs. For each of these habitats, he uses a standard aquarium with length, width, and height dimensions measured in inches, identical to the aquarium mentioned in the Opening Exercise. To begin his project, Jamie needs to determine the volume, or cubic inches, of water that can fill the aquarium.

Use the table below to determine the unit rate of gallons/cubic inches.

Answer:

There are 231 cubic inches for every 1 gallon of water. So, the unit rate is 231.

→ Since we determined that for every gallon of water, there are 231 cubic inches, determine how many cubic inches are in the 10 gallons of water that Jamie needs for the fish.
→ How can we determine how many cubic inches are in 10 gallons of water?
We could use a tape diagram or a double number line, or we could find equivalent ratios.
→ Using either of these representations, determine the volume of the aquarium.

Determine the volume of this aquarium.
Answer:
Answers will vary depending on student choice. An example of a tape diagram is below.

→ We determined the volume of this tank is 2,310 in³. This is not the same volume we calculated earlier in the Opening Exercise. Why do you think the volumes are different?
Answers will vary but should include a discussion that there needs to be room for a lid; also, the water level cannot go all the way to the top so that there is room for heaters, filters, and fish, etc., without the water spilling over.

→ Generally, it is suggested that the highest level of water in this tank should be approximately 11.55 inches.
Calculate the volume of the aquarium using this new dimension.
V = l × w × h; V = 20 in. × 10 in. × 11.55 in.; V = 2,310 in³

→ What do you notice about this volume?
This volume is the same as the volume we determined when we found the volume using ratio and unit rates.

→ Let’s use the dimensions 20 in. × 10 in. × 11.55 in. for our exploration.

Eureka Math Grade 6 Module 5 Lesson 19a Exercise Answer Key

Optional Exercise 1:

→ We have determined that the volume for the 10-gallon aquarium with dimensions 20 in. × 10 in. × 11.55 in.
is 2,310 in³.
→ Suppose Jamie needs to fill the aquarium to the top in order to prepare the tank for fish. According to our calculations, if Jamie pours 10 gallons of water into the tank, the height of the water is approximately 11.55 in.
→ Let’s test it. Begin pouring water into the aquarium 1 gallon at a time. Be sure to keep track of the number of gallons. Use a tally system.

→ Measure the height of the water with your ruler.
→ What did you find about our height calculation?
Our calculation was correct. The height is approximately 11.55 in.

Exercise 1:

a. Determine the volume of the tank when filled with 7 gallons of water.
Answer:
231 \(\frac{\text { cubic inches }}{\text { gallon }}\) = 1,617 in³
The volume for 7 gallons of water is 1,617 in³.

b. Work with your group to determine the height of the water when Jamie places 7 gallons of water in the aquarium.
Answer:
1,617 in³ = (20 in. )(10 in. ) h
\(\frac{1,617 \mathrm{in}^{3}}{200 \mathrm{in}^{2}}=\frac{200 \mathrm{in}^{2}}{200 \mathrm{in}^{2}} h\)
8.085 in. = h
The tank should have a water height of 8.085 inches.

Optional Exercise 2:

→ Let’s test it. Begin by pouring water into the aquarium 1 gallon at a time.
→ Be sure to keep track of the number of gallons poured. Use a tally system.
Or, have students mark the height of the water using a wax marker or a dry erase marker on the outside of the tank after each gallon is poured in. Then, students measure the intervals (distance between the marks). Students should notice that the intervals are equal.
→ Test the height at 7 gallons, and record the height measurement.

→ What did you find about our calculation?
Our calculation was correct. The height is about 8 inches.

Exercise 2:

a. Use the table from Example 1 to determine the volume of the aquarium when Jamie pours 3 gallons of water into the tank.
Answer:
The volume of the tank is 231 in³ × 3 = 693 in³.

b. Use the volume formula to determine the missing height dimension.
693 in³ = 20 in.(10 in.) h
\(\frac{693 \mathrm{in}^{3}}{200 \mathrm{in}^{2}}=\frac{200 \mathrm{in}^{2}}{200 \mathrm{in}^{2}} h\)
3.465 in. = h
The tank should have a water height of 3.465 in.

Optional Exercise 3:

→ Let’s test it. Begin by pouring water into the aquarium 1 gallon at a time.
→ Be sure to keep track of the number of gallons poured. Use a tally system.
→ Test the height at 3 gallons, and record the height measurement.

→ What did you find about our calculation?
Our calculation was correct. The height is about 3\(\frac{1}{2}\) inches.

Exercise 3:

a. Using the table of values below, determine the unit rate of liters to gallon.

Answer:

The unit rate is 3.785.

b. Using this conversion, determine the number of liters needed to fill the 10-gallon tank. liters.
Answer:
3.785 \(\frac{\text { liters }}{\text { gallon }}\) × 10 gallons = 37.85 liters

c. The ratio of the number of centimeters to the number of inches is 2.54: 1. What is the unit rate?
Answer:
2.54

d. Using this information, complete the table to convert the heights of the water in inches to the heights of the water in centimeters Jamie will need for his project at home.

Answer:

Exercise 4:

a. Determine the amount of plastic film the manufacturer uses to cover the aquarium faces. Draw a sketch of the aquarium to assist in your calculations. Remember that the actual height of the aquarium is inches.
Answer:
SA = (2lW) + (2lh) + (2wh)
SA = (2.20 in. . 10 in.) + (2.20 in. . 12 in.) + (2.10in. . 12 in.)
SA = 400 in² + 480 in² + 240 in²
SA = 1,120 in²

b. We do not include the measurement of the top of the aquarium since It is open without glass and does not need to be covered with film. Determine the area of the top of the aquarium, and find the amount of film the manufacturer uses to cover only the sides, front, back, and bottom.
Answer:
Area of the top of the aquarium = l . w
Area of the top of the aquarium = 20 in. . 10 in.
Area of the top of the aquarium = 200 in²
SA of aquarium without the top = 1,120 in² – 200 in² = 920 in²

c. Since Jamie needs three aquariums, determine the total surface area of the three aquariums.
Answer:
920 in² + 920 in² + 920 in² = 2.760 in² or 3920 in² = 2,760 in²

Eureka Math Grade 6 Module 5 Lesson 19a Problem Set Answer Key

This Problem Set is a culmination of skills learned in this module. Note that the figures are not drawn to scale.

Question 1.
Calculate the area of the figure below.

Answer:
A = bh
A = (40 ft.) (20 ft.)
A = 800 ft²

Question 2.
Calculate the area of the figure below.

Answer:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (1.3 m) (1.2 m)
A = 0.78 m²

Question 3.
Calculate the area of the figure below.

Answer:
Area of top rectangle:
A = lw
A = (25 in.)(12 in.)
A = 300 in²

Area of bottom rectangle:
A = lw
A= (7 in.)(20 in.)
A = 140 in²

Total Area = 300 in² + 140 in² = 440 in²

Question 4.
Complete the table using the diagram on the coordinate plane.

Answer:

Question 5.
Plot the points below, and draw the shape. Then, determine the area of the polygon. A(-3, 5), B(4, 3), C(0, -5)

Answer:

Area of Rectangle:
Area = lw
Area = (7 units)(10 units)
Area = 70 units²

Area of Triangle on Left:
Area = \(\frac{1}{2}\)bh
Area = \(\frac{1}{2}\)(3 units)(10 units)
Area = 15 units²

Area of Triangle on Top:
Area on top = \(\frac{1}{2}\)bh
Area = \(\frac{1}{2}\)(7 units) (2 units)
Area = 7 units²

Area of Triangle on Right:
Area = \(\frac{1}{2}\)bh
Area = \(\frac{1}{2}\)(4 units)(8 units)
Area = 16 units²

Total Area = 70 units² – 15 units² – 7 units² – 16 units²
Total Area = 32 units²

Question 6.
Determine the volume of the figure.

Answer:
V = l w h
V = (3\(\frac{1}{2}\) m) (\(\frac{7}{8}\) m) (1\(\frac{1}{4}\) m)
V = \(\frac{245}{64}\) m³
V = 3\(\frac{53}{64}\) m³

Question 7.
Give at least three more expressions that could be used to determine the volume of the figure in Problem 6.
Answer:
Answers will vary. Some examples include the following:
(\(\frac{35}{32}\) m²) (3\(\frac{53}{64}\) m)
(1\(\frac{1}{4}\) m) (\(\frac{7}{8}\) m) (3\(\frac{1}{2}\) m)
(\(\frac{49}{16}\) m²) (1\(\frac{1}{4}\) m)

Question 8.
Determine the volume of the irregular figure.

Answer:
Volume of the back Rectangular Prism:
V = l w h
V = (3\(\frac{5}{8}\) ft.) (1\(\frac{1}{3}\) ft) (1\(\frac{1}{4}\) ft.)
V = \(\frac{580}{96}\) ft³

Volume of the front Rectan guiar Prism:
V = l w h
V = (1\(\frac{1}{4}\) ft.) (1\(\frac{1}{6}\) ft.)(1\(\frac{1}{4}\) ft.)
V = \(\frac{175}{96}\) ft³

Total Volume = \(\frac{580}{96}\) ft³ + \(\frac{175}{96}\) ft³
= \(\frac{755}{96}\) ft³ = 7\(\frac{83}{96}\) ft³

Question 9.
Draw and label a net for the following figure. Then, use the net to determine the surface area of the figure.

Answer:

SA = 120 cm² + 84 cm² + 70 cm² + 84 cm² + 120 cm² + 70 cm²
= 548 cm²

Question 10.
Determine the surface area of the figure in Problem 9 using the formula SA = 21w + 21h + 2wh. Then, compare your answer to the solution in Problem 9.
Answer:
SA = 2lw + 2lh + 2wh
SA = 2(10 cm)(7 cm) + 2(10 cm)(12 cm) + 2 (7 cm)(12 cm)
SA = 140 cm² + 240 cm² + 168 cm²
SA= 548 cm²

The answer in Problem 10 is the same as in Problem 9. The formula finds the areas of each pair of equal faces and adds them together, like we did with the net.

Question 11.
A parallelogram has a base of 4.5 cm and an area of 9.495 cm². Tania wrote the equation 4.5x = 9.495 to represent this situation.

a. Explain what x represents in the equation.
Answer:
x represents the height of the parallelogram in centimeters.

b. Solve the equation for x and determine the height of the parallelogram.
Answer:
\(\frac{4.5 x}{4.5}=\frac{9.495}{4.5}\)
x = 2.11
The height of the parallelogram is 2.11 cm.

Question 12.
Triangle A has an area equal to one-third the area of Triangle B. Triangle A has an area of 3\(\frac{1}{2}\) square meters.

a. Gerard wrote the equation \(\frac{B}{3}\) = 3\(\frac{1}{2}\). Explain what B represents in the equation.
Answer:
B represents the area of Triangle B in square meters.

b. Determine the area of Triangle B.
Answer:
\(\frac{B}{3}\) . 3 = 3\(\frac{1}{2}\) . 3
The area of Triangle is 10\(\frac{1}{2}\) square meters.

Eureka Math Grade 6 Module 5 Lesson 19a Exit Ticket Answer Key

Question 1.
What did you learn today? Describe at least one situation in real life that would draw on the skills you used today.
Answer:
Answers will vary.

Engage NY Eureka Math Grade 6 Module 5 Lesson 19 Answer Key

Eureka Math Grade 6 Module 5 Lesson 19 Opening Exercise Answer Key

Opening Exercise:

Question 1.
A box needs to be painted. How many square inches need to be painted to cover the entire surface of the box?

Answer:
SA = 2(15 in.)(12 in.) + 2(15 in.)(6 in.) + 2(12 in.)(6 in.)
SA = 360 in² + 180 in² + 144 in²
SA = 684 in²

Question 2.
A juice box is 4 in. tall, 1 in. wide, and 2 in. long. How much juice fits inside the juice box?
Ans;
V = 1 in. × 2 in. × 4 in. = 8 in³

Question 3.
How did you decide how to solve each problem?
Answer:
I chose to use surface area to solve the first problem because you would need to know how much area the paint would need to cover. I chose to use volume to solve the second problem because you would need to know how much space is inside the juice box to determine how much juice it can hold.

Eureka Math Grade 6 Module 5 Lesson 19 Discussion Answer Key

Discussion:

Answer:

Eureka Math Grade 6 Module 5 Lesson 19 Example Answer Key

Example 1:

Vincent put logs in the shape of a rectangular prism outside his house. However, it is supposed to snow, and Vincent wants to buy a cover so the logs stay dry. If the pile of logs creates a rectangular prism with these measurements: 33 cm long, 12 cm wide, and 48 cm high, what is the minimum amount of material needed to cover the pile of logs?
Answer:

→ Where do we start?
We need to find the size of the cover for the logs, so we need to calculate the surface area. in order to find the surface area, we need to know the dimensions of the pile of logs.

→ Why do we need to find the surface area and not the volume?
We want to know the size of the cover Vincent wants to buy. If we calculated volume, we would not have the information Vincent needs when he goes shopping for a cover.

→ What are the dimensions of the pile of logs?
The length is 33 cm, the width is 12 cm, and the height is 48 cm.

→ How do we calculate the surface area to determine the size of the cover?
We can use the surface area formula for a rectangular prism.
SA = 2(33 cm)(12 cm) + 2(33 cm)(48 cm) + 2(12 cm)(48 cm)
SA = 792 cm2 + 3,168 cm² + 1,152 cm2²
SA = 5,112 cm²

→ What is different about this problem from other surface area problems of rectangular prisms you have encountered? How does this change the answer?
If Vincent just wants to cover the wood to keep it dry, he does not need to cover the bottom of the pile of logs. Therefore, the cover can be smaller.

→ How can we change our answer to find the exact size of the cover Vincent needs?
We know the area of the bottom of the pile of logs has the dimensions 33 cm and 12 cm. We can calculate the area and subtract this area from the total surface area.
The area of the bottom of the pile of logs is 396 cm²; therefore, the total surface area of the cover would need to be 5,112 cm² – 396 cm² = 4,716 cm².

Eureka Math Grade 6 Module 5 Lesson 19 Exercise Answer Key

Exercises:

Use your knowledge of volume and surface area to answer each problem.

Exercise 1.
Quincy Place wants to add a pool to the neighborhood. When determining the budget, Quincy Place determined that it would also be able to install a baby pool that requires less than 15 cubic feet of water. Quincy Place has three different models of a baby pool to choose from.
Choke One: 5 ft. × 5 ft. × 1 ft.
ChokeTwo: 4 ft. × 3 ft. × 1 fi
Choke Three: 4 ft. × 2 ft. × 2 ft.
Which of these chokes is best for the baby pool? Why are the others not good choices?
Answer:
Choice One Volume: 5 ft. × 5 ft. × 1 ft. = 25 ft³
Choice Two Volume: 4 ft. × 3 ft. × 1 ft. = 12 ft³
Choice Three Volume: 4 ft. × 2 ft. × 2 ft. = 16 ft³
Choice Two is within the budget because it holds less than 15 cubic feet of water. The other two choices do not work because they require too much water, and Quincy Place will not be able to afford the amount of water it takes to fill the baby pool.

Exercise 2.
A packaging firm has been hired to create a box for baby blocks. The firm was hired because it could save money by creating a box using the least amount of material. The packaging firm knows that the volume of the box must be 18 cm³.

a. What are possible dimensions for the box if the volume must be exactly 18 cm³?
Answer:
Choice 1: 1 cm × 1 cm × 18 cm
Choice 2: 1 cm × 2 cm × 9 cm
Choice 3: 1 cm × 3 cm × 6 cm
Choice 4: 2 cm × 3 cm × 3 cm

b. Which set of dimensions should the packaging firm choose in order to use the least amount of material? Explain.
Answer:
Choice 1: SA = 2(1 cm)(1 cm) + 2(1 cm)(18 cm) + 2(1 cm)(18 cm) = 74 cm²
Choice 2: SA = 2(1 cm)(2 cm) + 2(1 cm)(9 cm) + 2(2 cm)(9 cm) = 58 cm²
Choice 3: SA = 2(1 cm)(3 cm) + 2(1 cm)(6 cm) + 2(3 cm)(6 cm) = 54 cm²
Choice 4: SA = 2(2 cm)(3 cm) + 2(2 cm)(3 cm) + 2(3 cm)(3 cm) = 42 cm²
The packaging firm should choose Choice 4 because it requires the least amount of material. in order to find the amount of material needed to create a box,, the packaging firm would have to calculate the surface area of each box. The box with the smallest surface area requires the least amount of material.

Exercise 3.
A gift has the dimensions of 50 cm × 35 cm × 5 cm. You have wrapping paper with dimensions of 75 cm × 60 cm. Do you have enough wrapping paper to wrap the gift? Why or why not?
Answer:
Surface Area of the Present SA = 2(50 cm)(35 cm) + 2(50 cm)(5 cm) + 2(35 cm)(5 cm) = 3,500 cm² + 500 cm² + 350 cm² = 4,350cm²
Area of Wrapping Paper: A = 75 cm × 60 cm = 4,500 cm²
I do have enough paper to wrap the present because the present requires 4,350 square centimeters of paper, andi
have 4,500 square centimeters of wrapping paper.

Exercise 4.
Tony bought a flat-rate box from the post office to send a gift to his mother for Mother’s Day. The dimensions of the medium-size box are 14 inches × 12 inches × 3.5 inches. What is the volume of the largest gift he can send to his mother?
Answer:
Volume of the Box: V = 14 in. × 12 in. × 3.5 in. = 588 in³
Tony would have 588 cubic inches of space to fill with a gift for his mother.

Exercise 5.
A cereal company wants to change the shape of its cereal box in order to attract the attention of shoppers. The original cereal box has dimensions of 8 inches × 3 inches × 11 inches. The new box the cereal company is thinking of would have dimensions of 10 inches × 10 inches × 3 inches.

a. Which box holds more cereal?
Answer:
Volume of Original Box: V = 8 in. × 3 in. × 11 in. = 264 in³
Volume of New Box: V= 10 in. × 10 in. × 3 in.= 300 in³
The new box holds more cereal because it has a larger volume.

b. Which box requires more material to make?
Answer:
Surface Area of Original Box: SA = 2(8 in.)(3 in.) + 2(8 in.)(11 in.)+ 2(3 in.)(11 in.) = 48 in² + 176 in² + 66 in² = 290 in²

Surface Area of New Box: SA = 2(10 in.)(10 in.) + 2(10 in.)(3 in.) + 2(10 in.)(3 in.) = 200 in² + 60 in² + 60 in² = 320 in²

The new box requires more material than the original box because the new box has a larger surface area.

Exercise 6.
Cinema theaters created a new popcorn box in the shape of a rectangular prism. The new popcorn box has a length of 6 inches, a width of 3. 5 inches, and a height of 3.5 inches but does not include a lid.

a. How much material is needed to create the box?
Answer:
Surface Area of the Box: SA = 2(6 in.)(3. 5 in.) + 2(6 in.)(3.5 in.) + 2(3.5 in.)(3. 5 in.) = 42 in² + 42 in² + 24.5 in² = 108.5 in²
The box does not have a lid, so we have to subtract the area of the lid from the surface area.
Area of Lid: 6 in. × 3.5 in. = 21 in²
Total Surface Area: 108.5 in² – 21 in² = 87.5 in²
87.5 square inches of material is needed to create the new popcorn box.

b. How much popcorn does the box hold?
Answer:
Volume of the Box: V= 6 in. × 3.5 in. × 3.5 in. = 73.5 in³
The box holds 73.5 in³ of popcorn.

Eureka Math Grade 6 Module 5 Lesson 19 Problem Set Answer Key

Solve each problem below.

Question 1.
Dante built a wooden, cubic toy box for his son. Each side of the box measures 2 feet.

a. How many square feet of wood did he use to build the box?
Answer:
Surface Area of the Box: SA = 6(2 ft)² = 6(4 ft²) = 24 ft²
Dante used 24 square feet of wood to build the box.

b. How many cubic feet of toys will the box hold?
Answer:
Volume of the Box: V = 2 ft. × 21 ft. × 2 ft.= 8 ft³
The toy box would hold 8 cubic feet of toys.

Question 2.
A company that manufactures gift boxes wants to know how many different-sized boxes having a volume of 50 cubic centimeters it can make if the dimensions must be whole centimeters.

a. List all the possible whole number dimensions for the box.
Answer:
Choice One: 1 cm × 1 cm × 50 cm
Choice Two: 1 cm × 2 cm × 25 cm
Choice Three: 1 cm × 5 cm × 10 cm
Choice Four: 2 cm × 5 cm × 5 cm

b. Which possibility requires the least amount of material to make?
Answer:
Choice One: SA = 2(1 cm)(1 cm) + 2(1 cm)(50 cm) + 2(1 cm)(50 cm) = 2 cm² + 100 cm² + 100 cm² = 202 cm²

Choice Two: SA = 2(1 cm)(2 cm) + 2(1 cm)(25 cm) + 2(2 cm)(25 cm) = 4 cm² + 50 cm² + 100 cm² = 154 cm²

Choice Three: SA = 2(1 cm)(5 cm) + 2(1 cm)(10 cm) + 2(5 cm)(10 cm) = 10 cm² + 20 cm² + 100 cm² = 130 cm²

Choice Four: SA = 2(2 cm)(5 cm) + 2(2 cm)(5 cm) + 2(5 cm)(5 cm) = 20 cm² + 20 cm² + 50 cm² = 90 cm²

Choice Four requires the least amount of material because it has the smallest surface area.

c. Which box would you recommend the company use? Why?
Answer:
I would recommend the company use the box with dimensions of 2 cm × 5 cm × 5 cm (Choice Four) because it requires the least amount of material to make, so it would cost the company the least amount of money to make.

Question 3.
A rectangular box of rice is shown below. What is the greatest amount of rice, in cubic inches, that the box can hold?

Answer:
Volume of the Rice Box: V = 15\(\frac{1}{3}\) in. × 7\(\frac{2}{3}\) in. × 6\(\frac{1}{3}\) in. = \(\frac{20,102}{27}\) in³ = 744\(\frac{14}{27}\) in³

Question 4.
The Mars Cereal Company has two different cereal boxes for Mars Cereal. The large box is 8 inches wide, 11 inches high, and 3 inches deep. The small box is 6 inches wide, 10 inches high, and 2.5 inches deep.

a. How much more cardboard is needed to make the large box than the small box?
Answer:
SurfaceArea of the Large Box: SA = 2(8 in.)(11 in.) + 2(8 in.)(3 in.) + 2(11 in.)(3 in.) = 176 in² + 48 in² +66 in² = 290 in²

Surface Area of the Small Box: SA = 2(6 in. )(10 in.) + 2(6 in.)(2. 5 in.) + 2(10 in. )(2. 5 in.) = 120 in² + 30 in² + 50 in² = 200 in²

Difference: 290 in² – 200 in² = 90 in²
The large box requires 90 square inches more cardboard than the small box.

b. How much more cereal does the large box hold than the small box?
Answer:
Volume of the Large Box: V = 8 in. × 11 in. × 3 in. = 264 in³
Volume of the Small Box: V = 6 in. × 10 in. × 2.5 in. = 150 in³
Difference: 264 in³ – 150 in³ = 114 in³
The large box holds 114 cubic inches more cereal than the small box.

Question 5.
A swimming pool is 8 meters long, 6 meters wide, and 2 meters deep. The water-resistant paint needed for the pool costs $6 per square meter. How much will it cost to paint the pool?

a. How many faces of the poo1 do you have to paint?
Answer:
You have to point 5 faces.

b. How much paint (in square meters) do you need to paint the pool?
Answer:
SA = 2(8 m × 6 m) + 2(8 m × 2 m) + 2(6 m × 2 m) = 96 m² + 32 m² + 24 m² = 152 m²
Area of Top of Pool: 8 m × 6 m= 48 m²
Total Point Needed: 152 m² – 48 m² = 104 m²

c. How much will it cost to paint the pool?
Answer:
104 m² × $6/m² = $624
It will cost $624 to point the pool.

Question 6.
Sam is in charge of filling a rectangular hole with cement. The hole is 9 feet long, 3 feet wide, and 2 feet deep. How much cement will Sam need?
Answer:
V = 9 ft. × 3 ft. × 2 ft. = 54 ft³
Sam will need 54 cubic feet of cement to fill the hole.

Question 7.
The volume of Box D subtracted from the volume of Box C is 23.14 cubic centimeters. Box D has a volume of 10.115 cubic centimeters.

a. Let C be the volume of Box C in cubic centimeters. Write an equation that could be used to determine the volume of Box C.
Answer:
C – 10.115 cm³ = 23.14 cm³

b. Solve the equation to determine the volume of Box C.
Answer:
C – 10.115 cm³ + 10.115 cm³ = 23.14 cm³ + 10.115 cm³
C = 33. 255 cm³

c. The volume of Box C is one-tenth of the volume of another box, Box E. Let E represent the volume of Box E in cubic centimeters. Write an equation that could be used to determine the volume of Box E, using the result from part (b).
Answer:
33.255 cm³ = \(\frac{1}{10}\)E

d. Solve the equation to determine the volume of Box E.
Answer:
33.255cm³ ÷ \(\frac{1}{10}\) = \(\frac{1}{10}\)E ÷ \(\frac{1}{10}\)
332.55 cm³ = E

Eureka Math Grade 6 Module 5 Lesson 19 Exit Ticket Answer Key

Solve the word problem below.

Question 1.
Kelly has a rectangular fish aquarium that measures 18 inches long. 8 inches wide, and 12 inches tall.

a. What is the maximum amount of water the aquarium can hold?
Answer:
Volume of the Aquarium: V = 18 in. × 8 in. × 12 in. = 1,728 in³
The maximum amount of water the aquarium can hold is 1, 728 cubic inches.

b. If Kelly wanted to put a protective covering on the four glass walls of the aquarium, how big does the cover have to be?
Answer:
Surface Area of theAquarium: SA = 2(18 in.)(8 in.)+ 2(18 in.)(12 in.)+ 2(8 in.)(12 in.) = 288 in² + 432 in² + 192 in² = 912 in²

We only need to cover the four glass walls, so we can subtract the area of both the top and bottom of the aquarium.
Area of Top: A = 18 in. × 8 in. = 144 in²
Area of Bottom: A = 18 in. × 8 in. = 144 in²
Surface Area of the Four Walls: SA = 912 in² – 144 in² – 144 in² = 624 in²
Kelly would need 624 in² to cover the four walls of the aquarium.

Eureka Math Grade 6 Module 5 Lesson 19 Area of shapes Answer Key

Area of shapes:

Question 1.

Answer:
A = 80 ft²

Question 2.

Answer:
A = 30 m²

Question 3.

Answer:
A = 484 in²

Question 4.

Answer:
A = 1,029 cm²

Question 5.

Answer:
A = 72 ft²

Question 6.

Answer:
A = 156 km²

Question 7.

Answer:
A = 110 in²

Question 8.

Answer:
A = 192 cm²

Question 9.

Answer:
A = 576 m²

Question 10.

Answer:
A = 1,476 ft²

Engage NY Eureka Math Grade 6 Module 5 Lesson 18 Answer Key

Eureka Math Grade 6 Module 5 Lesson 18 Opening Exercise Answer Key

Opening Exercise:

a. What three-dimensional figure does the net create?
Answer:
Rectangular Prism

b. Measure (in inches) and label each side of the figure.
Answer:

c. Calculate the area of each face, and record this value inside the corresponding rectangle.
Answer:

d. How did we compute the surface area of solid figures in previous lessons?
Answer:
To determine surface area, we found the area of each of the faces and then added those areas.

e. Write an expression to show how we can calculate the surface area of the figure above.
Answer:
(4 in. × 1 in.) + (4 in. × 2 in.) + (4 in. × 1 in.) + (4 in. × 2 in.) + (2 in. × 1 in.) + (2 in. × 1 in.) OR
2(4 in. × 1 in.) + 2(4 in. × 2 in.) + 2(2 in. × 1 in.)

f. What does each part of the expression represent?
Answer:
Each part of the expression represents an area of one face of the given figure. We were able to write a more compacted form because there are three pairs of two faces that are identical.

g. What is the surface area of the figure?
Answer:
(4in.× 1 in) + (4 in. × 2 in.) + (4 in. × 1 in.) + (4 in. × 2 in.) + (2 in. × 1 in.) + (2 in. × 1 in.)
2(4 in. × 1 in.) + 2(4 in. × 2 in.) + 2(2 in. × 1 in.)
28 in²

Eureka Math Grade 6 Module 5 Lesson 18 Example Answer Key

Example 1:

Fold the net used in the Opening Exercise to make a rectangular prism. Have the two faces with the largest area be the bases of the prism. Fill in the first row of the table below.

Answer:

→ What do you notice about the areas of the faces?
Pairs of faces have equal areas.

→ What is the relationship between the faces having equal area?
The faces that have the same area are across from each other. The bottom and top have the same area, the front and the back have the same area, and the two sides have the same area.

→ How do we calculate the area of the two bases of the prism?
length × width

→ How do we calculate the area of the front and back faces of the prism?
length × height

→ How do we calculate the area of the right and left faces of the prism?
width × height

→ Using the name of the dimensions, fill in the third row of the table.

Examine the rectangular prism below. Complete the table.


Answer:

→ When comparing the methods to finding surface area of the two rectangular prisms, can you develop a general formula?
SA = l × w + l × w + l × h + l × h + w × h + w × h

→ Since we use the same expression to calculate the area of pairs of faces, we can use the distributive property to write an equivalent expression for the surface area of the figure that uses half as many terms.

→ We have determined that there are two l × w dimensions. Let’s record that as 2 times l times w, or simply 2(l × w). How can we use this knowledge to alter other parts of the formula?
We also have two l × h, so we can write that as 2(l × h), and we can write the two w × h as 2(w × h).

→ Writing each pair in a simpler way, what is the formula to calculate the surface area of a rectangular prism?
SA = 2(l × w) + 2(l × h) + 2(w × h)
→ Knowing the formula to calculate surface area makes it possible to calculate the surface area without a net.

Example 2:

Answer:
→ What are the dimensions of the rectangular prism?
The length is 20 cm, the width is 5 cm, and the height is 9 cm.

→ We use substitution in order to calculate the area. Substitute the given dimensions into the surface area formula.
SA = 2(20 cm)(5 cm) + 2(20 cm)(9 cm) + 2(5 cm)(9 cm)

→ Solve the equation. Remember to use order of operations.
SA = 200cm² + 360 cm² + 90 cm²
SA = 650 cm²

Eureka Math Grade 6 Module 5 Lesson 18 Exercise Answer Key

Exercises 1 – 3:

Exercise 1.
Calculate the surface area of each of the rectangular prisms below.

a.
Answer:
SA = 2(12 in.)(2 in.) + 2(12 in.)(3 in.) + 2(2 in.)(3 in.)
SA = 48 in² + 72 in² + 12 in²
SA = 132 in²

b.
Answer:
SA = 2(8 m)(6 m) + 2(8 m)(22 m) + 2(6 m)(22 m)
SA = 96 m² + 352 m² + 264 m²
SA = 712 m²

c.
Answer:
SA = 2(29 ft. ) (16 ft.) + 2(29 ft.) (23 ft.) + 2(16 ft.) (23 ft.)
SA = 928 ft² + 1,334 ft² + 736 ft²
SA = 2,998 ft²

d.
Answer:
SA = 2(4 cm) (1.2 cm) + 2(4 cm) (2.8 cm) + 2(1.2 cm) (2.8 cm)
SA = 9.6 cm² + 22.4 cm² + 6.72 cm²
SA = 38.72 cm²

Exercise 2.
Calculate the surface area of the cube.

Answer:
SA = 2(5 in.) (5 in.) + 2(5 in.) (5 in.) + 2(5 in.) (5 in.)
SA = 50 in² + 50 in² + 50 in²
SA = 150 in²

Exercise 3.
All the edges of a cube have the same length. Tony claims that the formula SA = 6s2, where s is the length of each side of the cube, can be used to calculate the surface area of a cube.

a. Use the dimensions from the cube in Problem 2 to determine if Tony’s formula is correct.
Answer:
Tony’s formula is correct because SA = 6(5 km)² = 150 km², which is the same surface area when we use the surface area formula for rectangular prisms.

b. Why does this formula work for cubes?
Answer:
Each face is a square, and to find the area of a square, you multiply the side lengths together. However, since the side lengths are the same, you can just square the side length. Also, a cube has 6 identical faces, so after calculating the area of one face, we can just multiply this area by 6 to determine the total surface area of the cube.

c. Becca does not want to try to remember two formulas for surface area, so she is only going to remember the formula for a cube. Is this a good idea? Why or why not?
Answer:
Becca’s idea is not a good idea. The surface area formula for cubes only works for cubes because rectangular prisms do not have 6 identical faces. Therefore, Becca also needs to know the surface area formula for rectangular prisms.

Eureka Math Grade 6 Module 5 Lesson 18 Problem Set Answer Key

Calculate the surface area of each figure below. Figures are not drawn to scale.

Question 1.

Answer:
SA = 2(15 in.)(15 in.) + 2(15 in.)(7 in.) + 2(15 in.)(7 in.)
SA = 450 in² + 210 in² + 210 in²
SA = 870 in²

Question 2.

Answer:
SA = 2(18.7 cm) (2. 3 cm) + 2(18.7 cm) (8. 4cm) + 2(2.3 cm) (8. 4 cm)
SA = 86.02 cm² + 314. 16 cm² + 38.64 cm²
SA = 438.82 cm²

Question 3.

Answer:
SA = 6(2\(\frac{1}{3}\) ft.)²

SA = 6(\(\frac{7}{3}\) ft.)²

SA = 6(\(\frac{49}{9}\) ft.)²

SA = \(\frac{294}{9}\) ft² = 32\(\frac{2}{3}\) ft²

Question 4.

Answer:
SA = 2(32.3 m) (24.7 m) + 2(32.3 m) (7.9 m) + 2(24.7 m) (7.9 m)
SA= 1,595.62 m² + 510.34 m² + 390.26 m²
SA = 2,496.22 m²

Question 5.
Write a numerical expression to show how to calculate the surface area of the rectangular prism. Explain each part of the expression.

Answer:
2(12 ft. × 3 ft.) + 2(12 ft. × 7 ft.) + 2(7 ft. × 3 ft.)
The first part of the expression shows the area of the top and bottom of the rectangular prism. The second part of the expression shows the area of the front and back of the 7 ft. rectangular prism. The third part of the expressin shows the area of the two sides of the rectangular prism. 3 ft. The surface area of the figure is 282 ft².

Question 6.
When Louie was calculating the surface area for Problem 4, he identified the following:
length = 24. 7 m, width = 32.3 m, and height = 7.9 m.
However, when Rocko was calculating the surface area for the same problem, he identified the following:
length = 32.3 m, width = 24.7 m, and height = 7.9 m.
Would Louie and Rocko get the same answer? Why or why not?
Answer:
Louie and Rocko would get the same answer because they are still finding the correct area of all six faces of the rectangular prism.

Eureka Math Grade 6 Module 5 Lesson 18 Exit Ticket Answer Key

Calculate the surface area of each figure below. Figures are not drawn to scale.

Question 1.

Answer:
SA = 2lw + 2lh + 2wh
SA = 2(12 ft.) (2 ft.) + 2(12 ft ) (10 ft.) + 2(2 ft) (10 ft.)
SA = 48 ft² + 240 ft² + 40 ft²
SA = 328 ft²

Question 2.

Answer:
SA = 6S²
SA = 6(8 cm)²
SA = 6(64 cm²)
SA = 384 cm²

Engage NY Eureka Math 8th Grade Module 4 Lesson 17 Answer Key

Eureka Math Grade 8 Module 4 Lesson 17 Exercise Answer Key

Exercise 1.
Find at least three solutions to the equation y=2x, and graph the solutions as points on the coordinate plane. Connect the points to make a line. Find the slope of the line.

Answer:
The slope of the line is 2; m=2.

Exercise 2.
Find at least three solutions to the equation y=3x-1, and graph the solutions as points on the coordinate plane. Connect the points to make a line. Find the slope of the line.

Answer:
The slope of the line is 3; m=3.

Exercise 3.
Find at least three solutions to the equation y=3x+1, and graph the solutions as points on the coordinate plane. Connect the points to make a line. Find the slope of the line.

Answer:
The slope of the line is 3; m=3.

Exercises 4–11
Students work independently or in pairs to identify the slope from an equation and to transform the standard form of an equation into slope-intercept form.

Exercise 4.
The graph of the equation y=7x-3 has what slope?
Answer:
The slope is 7.

Exercise 5.
The graph of the equation y=-\(\frac{3}{4}\) x-3 has what slope?
Answer:
The slope is –\(\frac{3}{4}\).

Exercise 6.
You have $20 in savings at the bank. Each week, you add $2 to your savings. Let y represent the total amount of money you have saved at the end of x weeks. Write an equation to represent this situation, and identify the slope of the equation. What does that number represent?
Answer:
y=2x+20
The slope is 2. It represents how much money is saved each week.

Exercise 7.
A friend is training for a marathon. She can run 4 miles in 28 minutes. Assume she runs at a constant rate. Write an equation to represent the total distance, y, your friend can run in x minutes. Identify the slope of the equation. What does that number represent?
Answer:
\(\frac{y}{x}\)=\(\frac{4}{28}\)
y=\(\frac{4}{28}\) x
y=\(\frac{1}{7}\) x
The slope is \(\frac{1}{7}\). It represents the rate at which my friend can run, one mile in seven minutes.

Exercise 8.
Four boxes of pencils cost $5. Write an equation that represents the total cost, y, for x boxes of pencils. What is the slope of the equation? What does that number represent?
Answer:
y=\(\frac{5}{4}\) x
The slope is \(\frac{5}{4}\). It represents the cost of one box of pencils, $1.25.

Exercise 9.
Solve the following equation for y, and then identify the slope of the line: 9x-3y=15.
Answer:
9x-3y=15
9x-9x-3y=15-9x
-3y=15-9x
\(\frac{-3}{-3}\) y=\(\frac{15-9 x}{-3}\)
y=\(\frac{15}{-3}\)–\(\frac{9 x}{-3}\)
y=-5+3x
y=3x-5
The slope of the line is 3.

Exercise 10.
Solve the following equation for y, and then identify the slope of the line: 5x+9y=8.
Answer:
5x+9y=8
5x-5x+9y=8-5x
9y=8-5x
\(\frac{9}{\mathbf{9}}\) y=\(\frac{8-5 x}{9}\)
y=\(\frac{8}{9}\)–\(\frac{5}{9}\)x
y=-\(\frac{5}{9}\)x+\(\frac{8}{9}\)
The slope of the line is –\(\frac{5}{9}\).

Question 11.
Solve the following equation for y, and then identify the slope of the line: ax+by=c.
Answer:
ax+by=c
ax-ax+by=c-ax
by=c-ax
\(\frac{b}{b}\) y=\(\frac{c-a x}{b}\)
y=\(\frac{c}{b}\)–\(\frac{ax}{b}\)
y=\(\frac{c}{b}\)–\(\frac{a}{b}\) x
y=-\(\frac{a}{b}\) x+\(\frac{c}{b}\)
The slope of the line is –\(\frac{a}{b}\).

Eureka Math Grade 8 Module 4 Lesson 17 Problem Set Answer Key

Students practice transforming equations from standard form into slope-intercept form and showing that the line joining two distinct points of the graph y=mx+b has slope m. Students graph the equation and informally note the y-intercept.

Question 1.
Solve the following equation for y: -4x+8y=24. Then, answer the questions that follow.
Answer:
-4x+8y=24
-4x+4x+8y=24+4x
8y=24+4x
\(\frac{8}{8}\) y=\(\frac{24}{8}\)+\(\frac{4}{8}\) x
y=3+\(\frac{1}{2}\) x
y=\(\frac{1}{2}\) x+3

a. Based on your transformed equation, what is the slope of the linear equation -4x+8y=24?
Answer:
The slope is \(\frac{1}{2}\).

b. Complete the table to find solutions to the linear equation.

Answer:

c. Graph the points on the coordinate plane.
Answer:

d. Find the slope between any two points.
Answer:
Using points (0,3) and (2,4):
m=\(\frac{4-3}{2-0}\)
=\(\frac{1}{2}\)

e. The slope you found in part (d) should be equal to the slope you noted in part (a). If so, connect the points to make the line that is the graph of an equation of the form y=mx+b that has slope m.

f. Note the location (ordered pair) that describes where the line intersects the y-axis.
Answer:
(0,3) is the location where the line intersects the y-axis.

Question 2.
Solve the following equation for y: 9x+3y=21. Then, answer the questions that follow.
Answer:
9x+3y=21
9x-9x+3y=21-9x
3y=21-9x
\(\frac{3}{3}\) y=\(\frac{21}{3}\)–\(\frac{9}{3}\) x
y=7-3x
y=-3x+7

a. Based on your transformed equation, what is the slope of the linear equation 9x+3y=21?
Answer:
The slope is -3.

b. Complete the table to find solutions to the linear equation.

Answer:

c. Graph the points on the coordinate plane.
Answer:

d. Find the slope between any two points.
Answer:
Using points (1,4) and (2,1):
m=\(\frac{4-1}{1-2}\)
=\(\frac{3}{-1}\)
=-3

e. The slope you found in part (d) should be equal to the slope you noted in part (a). If so, connect the points to make the line that is the graph of an equation of the form y=mx+b that has slope m.

f. Note the location (ordered pair) that describes where the line intersects the y-axis.
Answer:
(0,7) is the location where the line intersects the y-axis.

Question 3.
Solve the following equation for y: 2x+3y=-6. Then, answer the questions that follow.
Answer:
2x+3y=-6
2x-2x+3y=-6-2x
3y=-6-2x
\(\frac{3}{3}\) y=\(\frac{-6}{3}\)–\(\frac{2}{3}\) x
y=-2-\(\frac{2}{3}\) x
y=-\(\frac{2}{3}\) x-2

a. Based on your transformed equation, what is the slope of the linear equation 2x+3y=-6?
Answer:
The slope is –\(\frac{2}{3}\).

b. Complete the table to find solutions to the linear equation.

Answer:

c. Graph the points on the coordinate plane.
Answer:

d. Find the slope between any two points.
Answer:
Using points (-6,2) and (3,-4):
m=\(\frac{2-(-4)}{-6-3}\)
=\(\frac{6}{-9}\)
=-\(\frac{2}{3}\)

e. The slope you found in part (d) should be equal to the slope you noted in part (a). If so, connect the points to make the line that is the graph of an equation of the form y=mx+b that has slope m.

f. Note the location (ordered pair) that describes where the line intersects the y-axis.
Answer:
(0,-2) is the location where the line intersects the y-axis.

Question 4.
Solve the following equation for y: 5x-y=4. Then, answer the questions that follow.
Answer:
5x-y=4
5x-5x-y=4-5x
-y=4-5x
y=-4+5x
y=5x-4

a. Based on your transformed equation, what is the slope of the linear equation 5x-y=4?
Answer:
The slope is 5.

b. Complete the table to find solutions to the linear equation.

Answer:

c. Graph the points on the coordinate plane.

d. Find the slope between any two points.
Answer:
Using points (0,-4) and (1,1):
m=\(\frac{-4-1}{0-1}\)
=\(\frac{-5}{-1}\)
=5

e. The slope you found in part (d) should be equal to the slope you noted in part (a). If so, connect the points to make the line that is the graph of an equation of the form y=mx+b that has slope m.

f. Note the location (ordered pair) that describes where the line intersects the y-axis.
Answer:
(0,-4) is the location where the line intersects the y-axis.

Eureka Math Grade 8 Module 4 Lesson 17 Exit Ticket Answer Key

Question 1.
Solve the following equation for y: 35x-7y=49.
Answer:
35x-7y=49
35x-35x-7y=49-35x
-7y=49-35x
\(\frac{-7}{-7}\) y=\(\frac{49}{-7}\)–\(\frac{35}{-7}\) x
y=-7-(-5x)
y=5x-7

Question 2.
What is the slope of the equation in Problem 1?
Answer:
The slope of y=5x-7 is 5.

Question 3.
Show, using similar triangles, why the graph of an equation of the form y=mx is a line with slope m.

Answer:

Solutions will vary. A sample solution is shown below.
The line shown has slope 2. When we compare the corresponding side lengths of the similar triangles, we have the ratios \(\frac{2}{1}\)=\(\frac{4}{2}\)=2. In general, the ratios would be \(\frac{x}{1}\)=\(\frac{y}{m}\), equivalently y=mx, which is a line with slope m.

Engage NY Eureka Math 8th Grade Module 4 Lesson 18 Answer Key

Eureka Math Grade 8 Module 4 Lesson 19 Opening Exercise Answer Key

Examine each of the graphs and their equations. Identify the coordinates of the point where the line intersects the y-axis. Describe the relationship between the point and the equation y=mx+b.
a. y=\(\frac{1}{2}\) x+3

b. y=-3x+7

c. y=-\(\frac{2}{3}\) x-2

d. y=5x-4

Answer:
A point is noted in each graph above where the line intersects the y-axis:
y=\(\frac{1}{2}\) x+3, (0,3)
y=-3x+7, (0,7)
y=-\(\frac{2}{3}\) x-2, (0,-2)
y=5x-4, (0,-4)
In each equation, the number b was the y-coordinate of the point where the line intersected the y-axis.

Eureka Math Grade 8 Module 4 Lesson 19 Example Answer Key

Graph an equation in the form of y=mx+b.

Example 1.
Graph the equation y=\(\frac{2}{3}\) x+1. Name the slope and y-intercept point.
Answer:
The slope is m=\(\frac{2}{3}\), and the y-intercept point is (0,1).

→ To graph the equation, we must begin with the known point. In this case, that is the y-intercept point. We cannot begin with the slope because the slope describes the rate of change between two points. That means we need a point to begin with. On a graph, we plot the point (0,1).

→ Next, we use the slope to find the second point. We know that m=\(\frac{|Q R|}{|P Q|}\) =\(\frac{2}{3}\). The slope tells us exactly how many units to go to the right of P to find point Q and then how many vertical units we need to go from Q to find point R. How many units will we go to the right in order to find point Q? How do you know?
→ We need to go 3 units to the right of point P to find Q. We go 3 units because |PQ|=3.

→ How many vertical units from point Q must we go to find point R? How do you know?
→ We need to go 2 units from point Q to find R. We go 2 units because |QR|=2.
→ Will we go up from point Q or down from point Q to find R? How do you know?
→ We need to go up because the slope is positive. That means that the line will be left-to-right inclining.

→ Since we know that the line joining two distinct points of the form y=mx+b has slope m, and we specifically constructed points P and R with the slope in mind, we can join the points with a line.

Example 2.
Graph the equation y=-\(\frac{3}{4}\) x-2. Name the slope and y-intercept point.
Answer:
The slope is m=-\(\frac{3}{4}\), and the y-intercept point is (0,-2).

→ How do we begin?
→ We must begin by putting a known point on the graph, (0,-2).

→ We know that m=\(\frac{|Q R|}{|P Q|}\) =-\(\frac{3}{4}\). How many units will we go to the right in order to find point Q? How do you know?
→ We need to go 4 units to the right of point P to find Q. We go 4 units because |PQ|=4.

→ How many units from point Q must we go to find point R? How do you know?
→ We need to go 3 units from point Q to find R. We go 3 units because |QR|=3.
→ Will we go up from point Q or down from point Q to find R? How do you know?
→ We need to go down from point Q to point R because the slope is negative. That means that the line will be left-to-right declining.

→ Now we draw the line through the points P and R.

Example 3
Graph the equation y=4x-7. Name the slope and y-intercept point.
Answer:
The slope is m=4, and the y-intercept point is (0,-7).

→ Graph the equation y=4x-7. Name the slope and y-intercept point.
→ The slope is m=4, and the y-intercept point is (0,-7).
→ How do we begin?
→ We must begin by putting a known point on the graph, (0,-7).

→ Notice this time that the slope is the integer 4. In the last two examples, our slopes have been in the form of a fraction so that we can use the information in the numerator and denominator to determine the lengths of |PQ| and |QR|. Since m=\(\frac{|Q R|}{|P Q|}\) =4, what fraction can we use to represent slope to help us graph?
→ The number 4 is equivalent to the fraction \(\frac{4}{1}\).
→ Using m=\(\frac{|Q R|}{|P Q|}\) =\(\frac{4}{1}\), how many units will we go to the right in order to find point Q? How do you know?
→ We need to go 1 unit to the right of point P to find Q. We go 1 unit because |PQ|=1.

→ How many vertical units from point Q must we go to find point R? How do you know?
→ We need to go 4 units from point Q to find R. We go 4 units because |QR|=4.
→ Will we go up from point Q or down from point Q to find R? How do you know?
→ We need to go up from point Q to point R because the slope is positive. That means that the line will be left-to-right inclining.

→ Now we join the points P and R to make the line.

Eureka Math Grade 8 Module 4 Lesson 19 Exercise Answer Key

Exercises 1–4 (5 minutes)
Students complete Exercises 1–4 individually or in pairs.

Exercise 1.
Graph the equation y=\(\frac{5}{2}\) x -4.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=\(\frac{5}{2}\), and the y-intercept point is (0,-4).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Exercise 2.
Graph the equation y=-3x+6.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=-3, and the y-intercept point is (0,6).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 3.
The equation y=1x+0 can be simplified to y=x. Graph the equation y=x.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=1, and the y-intercept point is (0,0).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 4.
Graph the point (0,2).
Answer:

a. Find another point on the graph using the slope, m=\(\frac{2}{7}\).

b. Connect the points to make the line.

c. Draw a different line that goes through the point (0,2) with slope m=\(\frac{2}{7}\). What do you notice?
Answer:
Only one line can be drawn through the given point with the given slope.

Exercises 5–6
Students complete Exercises 5–6 individually or in pairs.

Exercise 5.
A bank put $10 into a savings account when you opened the account. Eight weeks later, you have a total of $24. Assume you saved the same amount every week.
a. If y is the total amount of money in the savings account and x represents the number of weeks, write an equation in the form y=mx+b that describes the situation.
Answer:
24=m(8)+10
14=8m
\(\frac{14}{8}\)=m
\(\frac{7}{4}\)=m
y=\(\frac{7}{4}\) x+10

b. Identify the slope and the y-intercept point. What do these numbers represent?
Answer:
The slope is \(\frac{7}{4}\), and the y-intercept point is (0,10). The y-intercept point represents the amount of money the bank gave me, in the amount of $10. The slope represents the amount of money I save each week, \(\frac{7}{4}\)=$1.75.

c. Graph the equation on a coordinate plane.
Answer:

d. Could any other line represent this situation? For example, could a line through point (0,10) with slope \(\frac{7}{5}\) represent the amount of money you save each week? Explain.
Answer:
No, a line through point (0,10) with slope \(\frac{7}{5}\) cannot represent this situation. That line would show that at the end of the 8 weeks I would have $21.20, but I was told that I would have $24 by the end of the 8 weeks.

Exercise 6.
A group of friends are on a road trip. After 120 miles, they stop to eat lunch. They continue their trip and drive at a constant rate of 50 miles per hour.
a. Let y represent the total distance traveled, and let x represent the number of hours driven after lunch. Write an equation to represent the total number of miles driven that day.
Answer:
y=50x+120

b. Identify the slope and the y-intercept point. What do these numbers represent?
Answer:
The slope is 50 and represents the rate of driving. The y-intercept point is 120 and represents the number of miles they had already driven before driving at the given constant rate.

c. Graph the equation on a coordinate plane.
Answer:

d. Could any other line represent this situation? For example, could a line through point (0,120) with slope 75 represent the total distance the friends drive? Explain.
Answer:
No, a line through point (0,120) with a slope of 75 could not represent this situation. That line would show that after an hour, the friends traveled a total distance of 195 miles. According to the information given, the friends would only have traveled 170 miles after one hour.

Eureka Math Grade 8 Module 4 Lesson 19 Exit Ticket Answer Key

Mrs. Hodson said that the graphs of the equations below are incorrect. Find the student’s errors, and correctly graph the equations.

Question 1.
Student graph of the equation y=\(\frac{1}{2}\) x+4:

Error:
Answer:
The student should have gone up 1 unit when finding |QR| since the slope is positive.
Correct graph of the equation:

Answer:

Question 2.
Student graph of the equation y=-\(\frac{3}{5}\) x-1:

Error:
Answer:
The student did not find the y-intercept point correctly. It should be the point (0,-1).
Correct graph of the equation:

Answer:

Eureka Math Grade 8 Module 4 Lesson 19 Problem Set Answer Key

Students practice graphing equations using y-intercept point and slope. Students need graph paper to complete the Problem Set. Optional Problem 11 has students show that there is only one line passing through a point with a given negative slope.

Graph each equation on a separate pair of x- and y-axes.

Question 1.
Graph the equation y=\(\frac{4}{5}\) x-5.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=\(\frac{4}{5}\), and the y-intercept point is (0,-5).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 2.
Graph the equation y=x+3.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=1, and the y-intercept point is (0,3).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 3.
Graph the equation y=-\(\frac{4}{3}\) x+4.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=-\(\frac{4}{3}\), and the y-intercept point is (0,4).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 4.
Graph the equation y=\(\frac{5}{2}\) x.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=\(\frac{5}{2}\), and the y-intercept point is (0,0).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 5.
Graph the equation y=2x-6.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=2, and the y-intercept point is (0,-6).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 6.
Graph the equation y=-5x+9.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=-5, and the y-intercept point is (0,9).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 7.
Graph the equation y=\(\frac{1}{3}\) x+1.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=\(\frac{1}{3}\), and the y-intercept point is (0,1).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 8.
Graph the equation 5x+4y=8. (Hint: Transform the equation so that it is of the form y=mx+b.)
a. Name the slope and the y-intercept point.
Answer:
5x+4y=8
5x-5x+4y=8-5x
4y=8-5x
\(\frac{4}{4}\) y=\(\frac{8}{4}\)–\(\frac{5}{4}\) x
y=2-\(\frac{5}{4}\) x
y=-\(\frac{5}{4}\) x+2
The slope is m=-\(\frac{5}{4}\), and the y-intercept point is (0,2).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 9.
Graph the equation -2x+5y=30.
a. Name the slope and the y-intercept point.
Answer:
-2x+5y=30
2x+2x+5y=30+2x
5y=30+2x
\(\frac{5}{5}\) y=\(\frac{30}{5}\)+\(\frac{2}{5}\) x
y=6+\(\frac{2}{5}\) x
y=\(\frac{2}{5}\) x+6
The slope is m=\(\frac{2}{5}\), and the y-intercept point is (0,6).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 10.
Let l and l’ be two lines with the same slope m passing through the same point P. Show that there is only one line with a slope m, where m<0, passing through the given point P. Draw a diagram if needed.
Answer:
First, assume that there are two different lines l and l’ with the same negative slope passing through P. From point P, I mark a point Q one unit to the right. Then, I draw a line parallel to the y-axis through point Q. The intersection of this line and line l and l’ are noted with points R and R’, respectively. By definition of slope, the lengths |QR| and |QR’ | represent the slopes of lines l and l’, respectively. We are given that the lines have the same slope, which means that lengths |QR| and |QR’| are equal. Since that is true, then points R and R’coincide and so do lines l and l’. Therefore, our assumption that they are different lines is false; l and l’ must be the same line. Therefore, there is only one line with slope m passing through the given point P.

Engage NY Eureka Math 8th Grade Module 4 Lesson 16 Answer Key

Eureka Math Grade 8 Module 4 Lesson 16 Example Answer Key

Example 1.
Using what you learned in the last lesson, determine the slope of the line with the following graph.

Answer:
The slope of the line is 3.

Example 2.
Using what you learned in the last lesson, determine the slope of the line with the following graph.

Answer:
The slope of this line is 2.

Example 3.
What is different about this line compared to the last two examples?

Answer:
This time, if we choose two points on the line that have a horizontal distance at 1, we cannot precisely determine the slope of the line because the vertical change is not an integer. It is some fractional amount.

→ Make a conjecture about how you could find the slope of this line.

Have students write their conjectures and share their ideas about how to find the slope of the line in this example; then, continue with the Discussion that follows.

Eureka Math Grade 8 Module 4 Lesson 16 Exercise Answer Key

Exercise
Let’s investigate concretely to see if the claim that we can find slope between any two points is true.

a. Select any two points on the line to label as P and R.
Answer:
Sample points are selected on the graph.

b. Identify the coordinates of points P and R.
Answer:
Sample points are labeled on the graph.

c. Find the slope of the line using as many different points as you can. Identify your points, and show your work below.
Answer:
Points selected by students will vary, but the slope should always equal 2. Students could choose to use points (0,5), (-1,3), (-2,1), (-3,-1), (-4,-3), and (-5,-5).

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key

Students practice finding slope between any two points on a line. Students also see that m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\) yields the same result as m=\(\frac{r_{2}-p_{2}}{r_{1}-p_{1}}\).

Question 1.
Calculate the slope of the line using two different pairs of points.

m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{30-(-10)}{-10-30}\)
=\(\frac{40}{-40}\)
=\(\frac{1}{1}\)
=-1
m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{10-(-10)}{10-30}\)
=\(\frac{20}{-20}\)
=-1

Question 2.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{3-2}{-2-2}\)
=\(\frac{1}{-4}\)
=-\(\frac{1}{4}\)

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{1-2}{6-2}\)
=\(\frac{-1}{4}\)
=-\(\frac{1}{4}\)

Question 3.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{1-5}{5-6}\)
=\(\frac{-4}{-1}\)
=\(\frac{4}{1}\)
=4

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{-3-5}{4-6}\)
=\(\frac{-8}{-2}\)
=4

Question 4.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{5-1}{3-5}\)
=\(\frac{4}{-2}\)
=\(-\frac{2}{1}\)
=-2

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{3-1}{4-5}\)
=\(\frac{2}{-1}\)
=-2

Question 5.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{2-1}{1-6}\)
=\(\frac{1}{-5}\)
=-\(\frac{1}{5}\)

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{0-1}{11-6}\)
=\(\frac{-1}{5}\)
=-\(\frac{1}{5}\)

Question 6.
Calculate the slope of the line using two different pairs of points.

a. Select any two points on the line to compute the slope.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{-5-(-2)}{-7-(-5)}\)
=\(\frac{-3}{-2}\)
=\(\frac{3}{2}\)

b. Select two different points on the line to calculate the slope.
Answer:
Let the two new points be (-3,1) and (-1,4).
m=\(\frac{q_{2}-s_{2}}{q_{1}-s_{1}}\)
=\(\frac{1-4}{-3-(-1)}\)
=\(\frac{-3}{-2}\)
=\(\frac{3}{2}\)

c. What do you notice about your answers in parts (a) and (b)? Explain.
Answer:
The slopes are equal in parts (a) and (b). This is true because of what we know about similar triangles.
The slope triangle that is drawn between the two points selected in part (a) is similar to the slope triangle that is drawn between the two points in part (b) by the AA criterion. Then, because the corresponding sides of similar triangles are equal in ratio, the slopes are equal.

Question 7.
Calculate the slope of the line in the graph below.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{-4-2}{-7-2}\)
=\(\frac{-6}{-9}\)
=\(\frac{2}{3}\)

Question 8.
Your teacher tells you that a line goes through the points (-6, \(\frac{1}{2}\)) and (-4,3).
a. Calculate the slope of this line.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{\frac{1}{2}-3}{-6-(-4)}\)
=\(\frac{-\frac{5}{2}}{-2}\)
= \(\frac{\frac{5}{2}}{2}\)
=\(\frac{5}{2}\)÷2
=\(\frac{5}{2}\)×\(\frac{1}{2}\)
= \(\frac{5}{4}\)

b. Do you think the slope will be the same if the order of the points is reversed? Verify by calculating the slope, and explain your result.
The slope should be the same because we are joining the same two points.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
= \(\frac{3-\frac{1}{2}}{-4-(-6)}\)
= \(\frac{\frac{5}{2}}{2}\)
= \(\frac{5}{4}\)
Since the slope of a line can be computed using any two points on the same line, it makes sense that it does not matter which point we name as P and which point we name as R.

Question 9.
Use the graph to complete parts (a)–(c).

a. Select any two points on the line to calculate the slope.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{1-(-3)}{-3-0}\)
=\(\frac{4}{-3}\)
=-\(\frac{4}{3}\)

b. Compute the slope again, this time reversing the order of the coordinates.
Answer:
m=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\)
=\(\frac{-3-1}{0-(-3)}\)
=\(\frac{-4}{3}\)
=-\(\frac{4}{3}\)

c. What do you notice about the slopes you computed in parts (a) and (b)?
Answer:
The slopes are equal.

d. Why do you think m=\(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\)=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\).?
Answer:
If I multiply the first fraction by \(\frac{-1}{-1}\), then I get the second fraction:
\(\frac{-1}{-1}\)×\(\left(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\right)\)=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\).
I can do the same thing to the second fraction to obtain the first:
\(\frac{-1}{-1}\)×\(\left(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\right)\)=\(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\)
Also, since I know that I can find the slope between any two points, it should not matter which point I pick first.

Question 10.
Each of the lines in the lesson was non-vertical. Consider the slope of a vertical line, x=2. Select two points on the line to calculate slope. Based on your answer, why do you think the topic of slope focuses only on non-vertical lines?

Answer:
Students can use any points on the line x=2 to determine that the slope is undefined. The computation of slope using the formula leads to a fraction with zero as its denominator. Since the slope of a vertical line is undefined, there is no need to focus on them.

Challenge:

Question 11.
A certain line has a slope of \(\frac{1}{2}\). Name two points that may be on the line.
Answer:
Answers will vary. Accept any answers that have a difference in y-values equal to 1 and a difference of x-values equal to 2. Points (6,4) and (4,3) may be on the line, for example.

Eureka Math Grade 8 Module 4 Lesson 16 Exit Ticket Answer Key

Find the rate of change of the line by completing parts (a) and (b).

a. Select any two points on the line to label as P and R. Name their coordinates.
Answer:
Answers will vary. Other points on the graph may have been chosen.
P(-1,0) and R(5,3)

b. Compute the rate of change of the line.
Answer:
m = \(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\)
m=\(\frac{0-3}{-1-5}\)
=\(\frac{-3}{-6}\)
=\(\frac{1}{2}\)

Engage NY Eureka Math 8th Grade Module 4 Lesson 15 Answer Key

Eureka Math Grade 8 Module 4 Lesson 15 Opening Exercise Answer Key

a. Which graph is steeper?
Answer:
It looks like Graph B is steeper.

b. Write directions that explain how to move from one point on the graph to the other for both Graph A and Graph B.
Answer:
For Graph A, move 2 units up and 3 units right. For Graph B, move 4 units up and 3 units right.

c. Write the directions from part (b) as ratios, and then compare the ratios. How does this relate to which graph was steeper in part (a)?
Answer:
\(\frac{2}{3}\)<\(\frac{4}{3}\).

Pair 1:

a. Which graph is steeper?
Answer:
It looks like Graph A is steeper.

b. Write directions that explain how to move from one point on the graph to the other for both Graph A and Graph B.
Answer:
For Graph A, move 4 units up and 5 units right. For Graph B, move 1 unit up and 12 units right.

c. Write the directions from part (b) as ratios, and then compare the ratios. How does this relate to which graph was steeper in part (a)?
Answer:
\(\frac{4}{5}\)>\(\frac{1}{12}\). Graph A was steeper and had the greater ratio.

Pair 2:

a. Which graph is steeper?
Answer:
It looks like Graph A is steeper.

b. Write directions that explain how to move from one point on the graph to the other for both Graph A and Graph B.
Answer:
For Graph A, move 7 units up and 1 unit right. For Graph B, move 3 units up and 6 units right.

c. Write the directions from part (b) as ratios, and then compare the ratios. How does this relate to which graph was steeper in part (a)?
Answer:
\(\frac{7}{1}\)>\(\frac{3}{6}\). Graph A was steeper and had the greater ratio.

Pair 3:

a. Which graph is steeper?
Answer:
It looks like Graph B is steeper.

b. Write directions that explain how to move from one point on the graph to the other for both Graph A and Graph B.
Answer:
For Graph A, move 1 unit up and 5 units right. For Graph B, move 4 units up and 1 unit right.

c. Write the directions from part (b) as ratios, and then compare the ratios. How does this relate to which graph was steeper in part (a)?
\(\frac{1}{5}\)<\(\frac{4}{1}\). Graph B was steeper and had the greater ratio.

Pair 4:

a. Which graph is steeper?
Answer:
They look about the same steepness.

b. Write directions that explain how to move from one point on the graph to the other for both Graph A and Graph B.
Answer:
For Graph A, move 4 units up and 4 units right. For Graph B, move 3 units up and 3 units right.

c. Write the directions from part (b) as ratios, and then compare the ratios. How does this relate to which graph was steeper in part (a)?
Answer:
\(\frac{4}{4}\)=\(\frac{3}{3}\). The graphs have equal ratios, which may explain why they look like the same steepness.

Eureka Math Grade 8 Module 4 Lesson 15 Exercise Answer Key

Use your transparency to find the slope of each line if needed.

Exercise 1.
What is the slope of this non-vertical line?

Answer:
The slope of this line is 4, m=4.

Exercise 2.
What is the slope of this non-vertical line?

Answer:
The slope of this line is 3, m=3.

Exercise 3.
Which of the lines in Exercises 1 and 2 is steeper? Compare the slopes of each of the lines. Is there a relationship between steepness and slope?
Answer:
The graph in Exercise 1 seems steeper. The slopes are 4 and 3. It seems like the greater the slope, the steeper the line.

Exercise 4.
What is the slope of this non-vertical line?

Answer:
The slope of this line is -1, m=-1.

Exercise 5.
What is the slope of this non-vertical line?

Answer:
The slope of this line is -4, m=-4.

Exercise 6.
What is the slope of this non-vertical line?

Answer:
The slope of this line is 0, m=0.

Eureka Math Grade 8 Module 4 Lesson 15 Problem Set Answer Key

Students practice identifying lines as having positive or negative slope. Students interpret the unit rate of a graph as the slope of the graph.

Question 1.
Does the graph of the line shown below have a positive or negative slope? Explain.

Answer:
The graph of this line has a negative slope. First of all, it is left-to-right declining, which is an indication of negative slope. Also, if we were to mark a point P and a point Q one unit to the right of P and then draw a line parallel to the y-axis through Q, then the intersection of the two lines would be below Q, making the number that represents slope negative.

Question 2.
Does the graph of the line shown below have a positive or negative slope? Explain.

Answer;
The graph of this line has a positive slope. First of all, it is left-to-right inclining, which is an indication of positive slope. Also, if we were to mark a point P and a point Q one unit to the right of P and then draw a line parallel to the y-axis through Q, then the intersection of the two lines would be above Q, making the number that represents slope positive.

Question 3.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is 1, m=1.

Question 4.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is –\(\frac{3}{2}\), m=-\(\frac{3}{2}\).

Question 5.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is \(\frac{5}{2}\), m=\(\frac{5}{2}\).

Question 6.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is 16, m=16.

Question 7.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is -10, m=-10.

Question 8.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is -5, m=-5.

Question 9.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is 2, m=2.

Question 10.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is -2, m=-2.

Question 11.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is 2, m=2.

Question 12.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is 5, m=5.

Question 13.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is \(\frac{3}{2}\), m=\(\frac{3}{2}\).

Question 14
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of this line is 0, m=0.

In Lesson 11, you did the work below involving constant rate problems. Use the table and the graphs provided to answer the questions that follow.

Question 15.
Suppose the volume of water that comes out in three minutes is 10.5 gallons.

a. How many gallons of water flow out of the faucet per minute? In other words, what is the unit rate of water flow?
Answer:
The unit rate of water flow is 3.5 gallons per minute.

b. Assume that the graph of the situation is a line, as shown in the graph. What is the slope of the line?
Answer:
The slope of the line is 3.5, m=3.5.

Question 16.
Emily paints at a constant rate. She can paint 32 square feet in five minutes.

a. How many square feet can Emily paint in one minute? In other words, what is her unit rate of painting?
Answer:
The unit rate at which Emily paints is 6.4 square feet per minute.

b. Assume that the graph of the situation is a line, as shown in the graph. What is the slope of the line?
Answer:
The slope of the line is 6.4, m=6.4.

Question 17.
A copy machine makes copies at a constant rate. The machine can make 80 copies in 2 1/2 minutes.


a. How many copies can the machine make each minute? In other words, what is the unit rate of the copy machine?
Answer:
The unit rate of the copy machine is 32 copies per minute.

b. Assume that the graph of the situation is a line, as shown in the graph. What is the slope of the line?
Answer:
The slope of the line is 32, m=32.

Eureka Math Grade 8 Module 4 Lesson 15 Exit Ticket Answer Key

Question 1.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of the line is 3, m=3.

Question 2.
What is the slope of this non-vertical line? Use your transparency if needed.

Answer:
The slope of the line is -0.6, which is equal to –\(\frac{3}{5}\), m=-\(\frac{3}{5}\).

Engage NY Eureka Math 8th Grade Module 4 Lesson 14 Answer Key

Eureka Math Grade 8 Module 4 Lesson 14 Exercise Answer Key

Exercise 1.
Find at least four solutions to graph the linear equation 1x+2y=5.
Answer:

Exercise 2.
Find at least four solutions to graph the linear equation 1x+0y=5.
Answer:

Exercise 3.
What was different about the equations in Exercises 1 and 2? What effect did this change have on the graph?
Answer:
In the first equation, the coefficient of y was 2. In the second equation, the coefficient of y was 0. The graph changed from being slanted to a vertical line.

Exercises 4–6
Students complete Exercises 4–6 independently. Students need graph paper to complete the exercises.

Exercise 4.
Graph the linear equation x=-2.
Answer:

Exercise 5.
Graph the linear equation x=3.
Answer:

Exercise 6.
What will the graph of x=0 look like?
Answer:
The graph of x=0 will look like a vertical line that goes through the point (0,0). It will be the same as the y-axis.

Exercises 7–9
Students complete Exercises 7–9 independently or in pairs in preparation for the discussion that follows. Students need graph paper to complete the exercises.

Exercise 7.
Find at least four solutions to graph the linear equation 2x+1y=2.
Answer:

Exercise 8.
Find at least four solutions to graph the linear equation 0x+1y=2.
Answer:

Exercise 9.
What was different about the equations in Exercises 7 and 8? What effect did this change have on the graph?
Answer:
In the first equation, the coefficient of x was 2. In the second equation, the coefficient of x was 0. The graph changed from being a slanted line to a horizontal line.

Exercises 10–12
Students complete Exercises 10–12 independently. Students need graph paper to complete the exercises.

Exercise 10.
Graph the linear equation y=-2.
Answer:

Exercise 11.
Graph the linear equation y=3.
Answer:

Exercise 12.
What will the graph of y=0 look like?
Answer;
The graph of y=0 will look like a horizontal line that goes through the point (0,0). It will be the same as the x-axis.

Eureka Math Grade 8 Module 4 Lesson 14 Exit Ticket Answer Key

Question 1.
Graph the linear equation ax+by=c, where a=0, b=1, and c=1.5.

Answer:

Question 2.
Graph the linear equation ax+by=c, where a=1, b=0, and c=-\(\frac{5}{2}\).

Answer:

Question 3.
What linear equation represents the graph of the line that coincides with the x-axis?
Answer:
y=0

Question 4.
What linear equation represents the graph of the line that coincides with the y-axis?
Answer:
x=0

Eureka Math Grade 8 Module 4 Lesson 14 Problem Set Answer Key

Students need graph paper to complete the Problem Set.

Question 1.
Graph the two-variable linear equation ax+by=c, where a=0, b=1, and c=-4.
Answer:

Question 2.
Graph the two-variable linear equation ax+by=c, where a=1, b=0, and c=9.
Answer:

Question 3.
Graph the linear equation y=7.
Answer:

Question 4.
Graph the linear equation x=1.
Answer:

Question 5.
Explain why the graph of a linear equation in the form of y=c is the horizontal line, parallel to the x-axis passing through the point (0,c).
Answer:
The graph of y=c passes through the point (0,c), which means the graph of y=c cannot be parallel to the y-axis because the graph intersects it. For that reason, the graph of y=c must be the horizontal line parallel to the
x-axis.

Question 6.
Explain why there is only one line with the equation y=c that passes through the point (0,c).
Answer:
There can only be one line parallel to another that goes through a given point. Since the graph of y=c is parallel to the x-axis and it goes through the point (0,c), then it must be the only line that does. Therefore, there is only one line that is the graph of the equation y=c that passes through (0,c).

Engage NY Eureka Math 8th Grade Module 4 Lesson 13 Answer Key

Eureka Math Grade 8 Module 4 Lesson 13 Exercise Answer Key

Exercise 1.
Find at least ten solutions to the linear equation 3x+y=-8, and plot the points on a coordinate plane.

Answer:

What shape is the graph of the linear equation taking?
Answer:
The graph appears to be the shape of a line.

Exercise 2.
Find at least ten solutions to the linear equation x-5y=11, and plot the points on a coordinate plane.

Answer:

What shape is the graph of the linear equation taking?
Answer:
The graph appears to be the shape of a line.

Exercise 3.
Compare the solutions you found in Exercise 1 with a partner. Add the partner’s solutions to your graph. Is the prediction you made about the shape of the graph still true? Explain.
Answer:
Yes. With the additional points, the graph still appears to be the shape of a line.

Exercise 4.
Compare the solutions you found in Exercise 2 with a partner. Add the partner’s solutions to your graph. Is the prediction you made about the shape of the graph still true? Explain.
Answer:
Yes. With the additional points, the graph still appears to be the shape of a line.

Exercise 5.
Joey predicts that the graph of -x+2y=3 will look like the graph shown below. Do you agree? Explain why or why not.

Answer:
No, I do not agree with Joey. The graph that Joey drew contains the point (0,0). If (0,0) is on the graph of the linear equation, then it will be a solution to the equation; however, it is not. Therefore, the point cannot be on the graph of the equation, which means Joey’s prediction is incorrect.

Exercise 6.
We have looked at some equations that appear to be lines. Can you write an equation that has solutions that do not form a line? Try to come up with one, and prove your assertion on the coordinate plane.
Answer:
Answers will vary. Any nonlinear equation that students write will graph as something other than a line. For example, the graph of y=x² or the graph of y=x³ will not be a line.

Eureka Math Grade 8 Module 4 Lesson 13 Problem Set Answer Key

In Problem 1, students graph linear equations by plotting points that represent solutions. For that reason, they need graph paper. Students informally explain why the graph of a linear equation is not curved by showing that a point on the curve is not a solution to the linear equation.

Question 1.
Find at least ten solutions to the linear equation \(\frac{1}{2}\) x+y=5, and plot the points on a coordinate plane.
Answer:

What shape is the graph of the linear equation taking?
Answer:
The graph appears to be the shape of a line.

Question 2.
Can the following points be on the graph of the equation x-y=0? Explain.

Answer:
The graph shown contains the point (0,-2). If (0,-2) is on the graph of the linear equation, then it will be a solution to the equation. It is not; therefore, the point cannot be on the graph of the equation, which means the graph shown cannot be the graph of the equation x-y=0.

Question 3.
Can the following points be on the graph of the equation x+2y=2? Explain.

Answer:
The graph shown contains the point (-4,1). If (-4,1) is on the graph of the linear equation, then it will be a solution to the equation. It is not; therefore, the point cannot be on the graph of the equation, which means the graph shown cannot be the graph of the equation x+2y=2.

Question 4.
Can the following points be on the graph of the equation x-y=7? Explain.

Answer:
Yes, because each point on the graph represents a solution to the linear equation x-y=7.

Question 5.
Can the following points be on the graph of the equation x+y=2? Explain.

Answer:
Yes, because each point on the graph represents a solution to the linear equation x+y=2.

Question 6.
Can the following points be on the graph of the equation 2x-y=9? Explain.

Answer:
Yes, because each point on the graph represents a solution to the linear equation 2x-y=9.

Question 7.
Can the following points be on the graph of the equation x-y=1? Explain.

Answer:
The graph shown contains the point (-2,-1). If (-2,-1) is on the graph of the linear equation, then it will be a solution to the equation. It is not; therefore, the point cannot be on the graph of the equation, which means the graph shown cannot be the graph of the equation x-y=1.

Eureka Math Grade 8 Module 4 Lesson 13 Exit Ticket Answer Key

Question 1.
Ethan found solutions to the linear equation 3x-y=8 and graphed them. What shape is the graph of the linear equation taking?

Answer:
It appears to take the shape of a line.

Question 2.
Could the following points be on the graph of -x+2y=5?

Answer:
Students may have chosen any point to make the claim that this is not the graph of the equation -x+2y=5.
Although the graph appears to be a line, the graph contains the point (0,3). The point (0,3) is not a solution to the linear equation; therefore, this is not the graph of -x+2y=5.
Note to teacher: Accept any point as not being a solution to the linear equation.

Engage NY Eureka Math 8th Grade Module 4 Lesson 12 Answer Key

Eureka Math Grade 8 Module 4 Lesson 12 Opening Exercise Answer Key

Emily tells you that she scored 32 points in a basketball game. Write down all the possible ways she could have scored 32 with only two- and three-point baskets. Use the table below to organize your work.

Answer:

Let x be the number of two-pointers and y be the number of three-pointers that Emily scored. Write an equation to represent the situation.
Answer:
2x+3y=32

Eureka Math Grade 8 Module 4 Lesson 12 Exercise Answer Key

Exploratory Challenge/Exercises

Question 1.
Find five solutions for the linear equation x+y=3, and plot the solutions as points on a coordinate plane.

Answer:

Question 2.
Find five solutions for the linear equation 2x-y=10, and plot the solutions as points on a coordinate plane.

Answer:

Question 3.
Find five solutions for the linear equation x+5y=21, and plot the solutions as points on a coordinate plane.

Answer:

Question 4.
Consider the linear equation \(\frac{2}{5}\) x+y=11.
a. Will you choose to fix values for x or y? Explain.
Answer:
If I fix values for x, it will make the computations easier. Solving for y can be done in one step.

b. Are there specific numbers that would make your computational work easier? Explain.
Answer:
Values for x that are multiples of 5 will make the computations easier. When I multiply \(\frac{2}{5}\) by a multiple of 5, I will get an integer.

c. Find five solutions to the linear equation \(\frac{2}{5}\) x+y=11, and plot the solutions as points on a coordinate plane.
x Linear Equation:

Answer:

Question 5.
At the store, you see that you can buy a bag of candy for $2 and a drink for $1. Assume you have a total of $35 to spend. You are feeling generous and want to buy some snacks for you and your friends.
a. Write an equation in standard form to represent the number of bags of candy, x, and the number of drinks, y, that you can buy with $35.
Answer:
2x+y=35

b. Find five solutions to the linear equation from part (a), and plot the solutions as points on a coordinate plane.

Answer:

Eureka Math Grade 8 Module 4 Lesson 12 Exit Ticket Answer Key

Question 1.
Is the point (1,3) a solution to the linear equation 5x-9y=32? Explain.
Answer:
No, (1,3) is not a solution to 5x-9y=32 because 5(1)-9(3)=5-27=-22, and -22≠32.

Question 2.
Find three solutions for the linear equation 4x-3y=1, and plot the solutions as points on a coordinate plane.

Answer:

Eureka Math Grade 8 Module 4 Lesson 12 Problem Set Answer Key

Students practice finding and graphing solutions for linear equations that are in standard form.

Question 1.
Consider the linear equation x-\(\frac{3}{2}\) y=-2.
a. Will you choose to fix values for x or y? Explain.
Answer:
If I fix values for y, it will make the computations easier. Solving for x can be done in one step.

b. Are there specific numbers that would make your computational work easier? Explain.
Answer:
Values for y that are multiples of 2 will make the computations easier. When I multiply \(\frac{3}{2}\) by a multiple of 2, I will get a whole number.

c. Find five solutions to the linear equation x-\(\frac{3}{2}\) y=-2, and plot the solutions as points on a coordinate plane.

Answer:

Question 2.
Find five solutions for the linear equation \(\frac{1}{3}\) x+y=12, and plot the solutions as points on a coordinate plane.
Answer:

Question 3.
Find five solutions for the linear equation -x+\(\frac{3}{4}\) y=-6, and plot the solutions as points on a coordinate plane.
Answer:

Question 4.
Find five solutions for the linear equation 2x+y=5, and plot the solutions as points on a coordinate plane.
Answer:

Question 5.
Find five solutions for the linear equation 3x-5y=15, and plot the solutions as points on a coordinate plane.
Answer:

Engage NY Eureka Math 8th Grade Module 4 Lesson 10 Answer Key

Eureka Math Grade 8 Module 4 Lesson 10 Example Answer Key

Example 1.
Consider the word problem below. We can do several things to answer this problem, but let’s begin to organize our work using a table for time and distance:

Example 1.
Paul walks 2 miles in 25 minutes. How many miles can Paul walk in 137.5 minutes?

Answer:

As students answer the questions below, fill in the table.
→ How many miles would Paul be able to walk in 50 minutes? Explain.
→ Paul could walk 4 miles in 50 minutes because 50 minutes is twice the time we were given, so we can calculate twice the distance, which is 4.
→ How many miles would Paul be able to walk in 75 minutes? Explain.
→ Paul could walk 6 miles in 75 minutes because 75 minutes is three times the number of minutes we were given, so we can calculate three times the distance, which is 6.
→ How many miles would Paul be able to walk in 100 minutes?
→ He could walk 8 miles.
→ How many miles would he walk in 125 minutes?
→ He could walk 10 miles.
→ How could we determine the number of miles Paul could walk in 137.5 minutes?

Provide students time to think about the answer to this question. They may likely say that they can write a proportion to figure it out. Allow them to share and demonstrate their solutions. Then, proceed with the discussion below, if necessary.
→ Since the relationship between the distance Paul walks and the time it takes him to walk that distance is proportional, we let y represent the distance Paul walks in 137.5 minutes and write the following:
Answer:
\(\frac{25}{2}\)=\(\frac{13.75}{y}\)
25y=137.5(2)
25y=275
y=11
Therefore, Paul can walk 11 miles in 137.5 minutes.

→ How many miles, y, can Paul walk in x minutes?
Provide students time to think about the answer to this question. Allow them to share their ideas, and then proceed with the discussion below, if necessary.
→ We know for a fact that Paul can walk 2 miles in 25 minutes, so we can write the ratio \(\frac{25}{2}\) as we did with the proportion. We can write another ratio for the number of miles, y, Paul walks in x minutes. It is \(\frac{x}{y}\). For the same reason we could write the proportion before, we can write one now with these two ratios:
\(\frac{25}{2}\)=\(\frac{x}{y}\).
Does this remind you of something we have done recently? Explain.
→ This is a linear equation in disguise. All we need to do is multiply each numerator by the other fraction’s denominator, and then we will have a linear equation.
25y=2x
→ Recall our original question: How many miles, y, can Paul walk in x minutes? We need to solve this equation for y.
25y=2x
y=\(\frac{2}{25}\) x
y=0.08x
Paul can walk 0.08x miles in x minutes. This equation will allow us to answer all kinds of questions about Paul with respect to any given number of minutes or miles.

→ Let’s go back to the table and look for y=0.08x or its equivalent y=\(\frac{2}{25}\) x. What do you notice?

→ The fraction \(\frac{2}{25}\) came from the first row in the table. It is the distance traveled divided by the time it took to travel that distance. It is also in between each row of the table. For example, the difference between 4 miles and 2 miles is 2, and the difference between the associated times 50 and 25 is 25. The pattern repeats throughout the table.

Show on the table the +2 between each distance interval and the +25 between each time interval. Remind students that they have done work like this before, specifically finding a unit rate for a proportional relationship. Make clear that the unit rate found in the table was exactly the same as the unit rate found using the proportion and that the unit rate is the rate at which Paul walks.
→ Let’s look at another problem where only a table is provided.

We want to know how many miles, y, can be traveled in any number of hours x. Using our previous work, what should we do?
→ We can write and solve a proportion that contains both x and y or use the table to help us determine the unit rate.
→ How many miles, y, can be traveled in any number of hours x?
→ Student work:
\(\frac{123}{3}\)=\(\frac{y}{x}\)
123x=3y
\(\frac{123}{3}\) x=y
41x=y
→ What does the equation y=41x mean?
→ It means that the distance traveled, y, is equal to the rate of 41 multiplied by the number of hours x traveled at that rate.

Example 2.
The point of this problem is to make clear to students that constant rate must be assumed in order to write linear equations in two variables and to use those equations to answer questions about distance, time, and rate.
→ Consider the following word problem: Alexxa walked from Grand Central Station (GCS) on 42^nd Street to Penn Station on 7^th Avenue. The total distance traveled was 1.1 miles. It took Alexxa 25 minutes to make the walk. How many miles did she walk in the first 10 minutes?
→ Give students a minute to think and/or work on the problem. Expect them to write a proportion and solve the problem. The next part of the discussion gets them to think about what is meant by “constant” speed or, rather, lack of it.
→ She walked 0.44 miles. (Assuming students used a proportion to solve.)
→ Are you sure about your answer? How often do you walk at a constant speed? Notice the problem did not even mention that she was walking at the same rate throughout the entire 1.1 miles. What if you have more information about her walk: Alexxa walked from GCS along 42^nd Street to an ATM 0.3 miles away in 8 minutes. It took her 2 minutes to get some money out of the machine. Do you think your answer is still correct?
→ Probably not since we now know that she had to stop at the ATM.
→ Let’s continue with Alexxa’s walk: She reached the 7^th Avenue junction 13 minutes after she left GCS, a distance of 0.6 miles. There, she met her friend Karen with whom she talked for 2 minutes. After leaving her friend, she finally got to Penn Station 25 minutes after her walk began.
→ Is this a more realistic situation than believing that she walked the exact same speed throughout the entire trip? What other events typically occur during walks in the city?
→ Stoplights at crosswalks, traffic, maybe a trip/fall, or running an errand
→ This is precisely the reason we need to take a critical look at what we call proportional relationships and constant speed, in general.
→ The following table shows an accurate picture of Alexxa’s walk:

With this information, we can answer the question. Alexxa walked exactly 0.3 miles in 10 minutes.
→ Now that we have an idea of what could go wrong when we assume a person walks at a constant rate or that a proportion can give us the correct answer all of the time, let’s define what is called average speed.
→ Suppose a person walks a distance of d (miles) in a given time interval t (minutes). Then, the average speed in the given time interval is \(\frac{d}{t}\) in miles per minute.
→ With this definition, we can calculate Alexxa’s average speed: The distance that Alexxa traveled divided by the time interval she walked is \(\frac{1.1}{25}\) miles per minute.
→ If we assume that someone can actually walk at the same average speed over any time interval, then we say that the person is walking at a constant speed.
Suppose the average speed of a person is the same constant C for any given time interval. Then, we say that the person is walking at a constant speed C.
→ If the original problem included information specifying constant speed, then we could write the following:
Alexxa’s average speed for 25 minutes is \(\frac{1.1}{25}\).
Let y represent the distance Alexxa walked in 10 minutes. Then, her average speed for 10 minutes is \(\frac{y}{10}\).
Since Alexxa is walking at a constant speed of C miles per minute, then we know that
\(\frac{1.1}{25}\)=C, and \(\frac{y}{10}\)=C.
Since both fractions are equal to C, then we can write
\(\frac{1.1}{25}\) = \(\frac{y}{10}\)
With the assumption of constant speed, we now have a proportional relationship, which would make the answer you came up with in the beginning correct.
We can go one step further and write a statement in general. If Alexxa walks y miles in x minutes, then
\(\)=C, and \(\)=\(\).
To find how many miles y Alexxa walks in x miles, we solve the equation for y:
\(\frac{1.1}{25}\)=\(\frac{y}{x}\)
25y=1.1x
\(\frac{25}{2}\)5 y=\(\frac{1.1}{25}\) x
y=\(\frac{1.1}{25}\) x,
where the last equation is an example of a linear equation in two variables x and y. With this general equation, we can find the distance y Alexxa walks in any given time x. Since we have more information about Alexxa’s walk, where and when she stopped, we know that the equation cannot accurately predict the distance she walks after a certain number of minutes. To do so requires us to assume that she walks at a constant rate. This is an assumption we generally take for granted when solving problems about rate.

Eureka Math Grade 8 Module 4 Lesson 10 Exercise Answer Key

Question 1.
Wesley walks at a constant speed from his house to school 1.5 miles away. It took him 25 minutes to get to school.
a. What fraction represents his constant speed, C?
Answer:
\(\frac{1.5}{25}\)=C

b. You want to know how many miles he has walked after 15 minutes. Let y represent the distance he traveled after 15 minutes of walking at the given constant speed. Write a fraction that represents the constant speed, C, in terms of y.
Answer:
\(\frac{y}{15}\)=C

c. Write the fractions from parts (a) and (b) as a proportion, and solve to find how many miles Wesley walked after 15 minutes.
Answer:
\(\frac{1.5}{25}\)=\(\frac{y}{15}\)
25y=22.5
\(\frac{25}{25}\) y=\(\frac{22.5}{25}\)
y=0.9
Wesley walks 0.9 miles in 15 minutes.

d. Let y be the distance in miles that Wesley traveled after x minutes. Write a linear equation in two variables that represents how many miles Wesley walked after x minutes.
Answer:
\(\frac{1.5}{25}\)=\(\frac{y}{x}\)
25y=1.5x
\(\frac{25}{25}\)y=\(\frac{1.5}{25}\) x
y=\(\frac{1.5}{25}\) x

Question 2.
Stefanie drove at a constant speed from her apartment to her friend’s house 20 miles away. It took her 45 minutes to reach her destination.
a. What fraction represents her constant speed, C?
Answer:
\(\frac{20}{45}\)=C

b. What fraction represents constant speed, C, if it takes her x number of minutes to get halfway to her friend’s house?
Answer:
\(\frac{10}{x}\)=C

c. Write and solve a proportion using the fractions from parts (a) and (b) to determine how many minutes it takes her to get to the halfway point.
Answer:
\(\frac{20}{45}\)=\(\frac{10}{x}\)
20x=450
\(\frac{20}{20}\) x=\(\frac{450}{20}\)
x=22.5
Stefanie gets halfway to her friend’s house, 10 miles away, after 22.5 minutes.

d. Write a two-variable equation to represent how many miles Stefanie can drive over any time interval.
Answer:
Let y represent the distance traveled over any time interval x. Then,
\(\frac{20}{45}\)=\(\frac{y}{x}\)
20x=45y
\(\frac{20}{45}\) x=\(\frac{45}{45}\) y
\(\frac{4}{9}\) x=y.

Exercise 3.
The equation that represents how many miles, y, Dave travels after x hours is y=50x+15. Use the equation to complete the table below.

Answer:

Eureka Math Grade 8 Module 4 Lesson 10 Exit Ticket Answer Key

Alex skateboards at a constant speed from his house to school 3.8 miles away. It takes him 18 minutes.
a. What fraction represents his constant speed, C?
Answer:
\(\frac{3.8}{18}\)=C

b. After school, Alex skateboards at the same constant speed to his friend’s house. It takes him 10 minutes. Write the fraction that represents constant speed, C, if he travels a distance of y.
Answer:
\(\frac{y}{10}\)=C

c. Write the fractions from parts (a) and (b) as a proportion, and solve to find out how many miles Alex’s friend’s house is from school. Round your answer to the tenths place.
\(\frac{3.8}{18}\)=\(\frac{y}{10}\)
3.8(10)=18y
38=18y
\(\frac{38}{18}\)=y
2.1≈y
Alex’s friend lives about 2.1 miles from school.

Eureka Math Grade 8 Module 4 Lesson 10 Problem Set Answer Key

Students practice writing and solving proportions to solve constant speed problems. Students write two variable equations to represent situations, generally.

Question 1.
Eman walks from the store to her friend’s house, 2 miles away. It takes her 35 minutes.
a. What fraction represents her constant speed, C?
Answer:
\(\frac{2}{35}\)=C

b. Write the fraction that represents her constant speed, C, if she walks y miles in 10 minutes.
Answer:
\(\frac{y}{10}\)=C

c. Write and solve a proportion using the fractions from parts (a) and (b) to determine how many miles she walks after 10 minutes. Round your answer to the hundredths place.
Answer:
\(\frac{2}{35}\)=\(\frac{y}{10}\)
35y=20
\(\frac{35}{35}\) y=\(\frac{20}{35}\)
y=0.57142…
Eman walks about 0.57 miles after 10 minutes.

d. Write a two-variable equation to represent how many miles Eman can walk over any time interval.
Answer:
Let y represent the distance Eman walks in x minutes.
\(\frac{2}{35}\)=\(\frac{y}{x}\)
35y=2x
\(\frac{35}{35}\) y=\(\frac{2}{35}\) x
y=\(\frac{2}{35}\) x

Question 2.
Erika drives from school to soccer practice 1.3 miles away. It takes her 7 minutes.
a. What fraction represents her constant speed, C?
Answer:
\(\frac{1.3}{7}\)=C

b. What fraction represents her constant speed, C, if it takes her x minutes to drive exactly 1 mile?
Answer:
\(\frac{1}{x}\)=C

c. Write and solve a proportion using the fractions from parts (a) and (b) to determine how much time it takes her to drive exactly 1 mile. Round your answer to the tenths place.
Answer:
\(\frac{1.3}{7}\)=\(\frac{1}{x}\)
1.3x=7
\(\frac{1.3}{1.3}\) x=7/(1.3)
x=5.38461…
It takes Erika about 5.4 minutes to drive exactly 1 mile.

d. Write a two-variable equation to represent how many miles Erika can drive over any time interval.
Answer:
Let y be the number of miles Erika travels in x minutes.
\(\frac{1.3}{7}\)=\(\frac{y}{x}\)
7y=1.3x
\(\frac{7}{7}\) y=\(\frac{1.3}{7}\) x
y=\(\frac{1.3}{7}\) x

Question 3.
Darla drives at a constant speed of 45 miles per hour.
a. If she drives for y miles and it takes her x hours, write the two-variable equation to represent the number of miles Darla can drive in x hours.
Answer;
\(\frac{y}{x}\)=45
y=45x

b. Darla plans to drive to the market 14 miles from her house, then to the post office 3 miles from the market, and then return home, which is 15 miles from the post office. Assuming she drives at a constant speed the entire time, how long will it take her to run her errands and get back home? Round your answer to the hundredths place.
Answer:
Altogether, Darla plans to drive 32 miles because 14+3+15=32.
32=45x
\(\frac{32}{45}\)=\(\frac{45}{45}\) x
0.71111…=x
It will take Darla about 0.71 hours to run her errands and get back home.

Question 4.
Aaron walks from his sister’s house to his cousin’s house, a distance of 4 miles, in 80 minutes. How far does he walk in 30 minutes?
Answer:
I cannot say for sure how far Aaron walks in 30 minutes because I do not know if he is walking at a constant speed. Maybe he stopped at his friend’s house for 20 minutes.

Question 5.
Carlos walks 4 miles every night for exercise. It takes him exactly 63 minutes to finish his walk.
a. Assuming he walks at a constant rate, write an equation that represents how many miles, y, Carlos can walk in x minutes.
Answer:
Since \(\frac{4}{63}\)=C and \(\frac{y}{x}\)=C, then
\(\frac{4}{63}\)=\(\frac{y}{x}\)
63y=4x
\(\frac{63}{63}\)y=\(\frac{4}{63}\) x
y=\(\frac{4}{63}\) x”.”

b. Use your equation from part (a) to complete the table below. Use a calculator, and round all values to the hundredths place.

Answer:

Engage NY Eureka Math 8th Grade Module 4 Lesson 11 Answer Key

Eureka Math Grade 8 Module 4 Lesson 11 Example Answer Key

Example 1.
Pauline mows a lawn at a constant rate. Suppose she mows a 35-square-foot lawn in 2.5 minutes. What area, in square feet, can she mow in 10 minutes? t minutes?

→ What is Pauline’s average rate in 2.5 minutes?
→ Pauline’s average rate in 2.5 minutes is \(\frac{35}{2.5}\) square feet per minute.
→ What is Pauline’s average rate in 10 minutes?
→ Let A represent the square feet of the area mowed in 10 minutes. Pauline’s average rate in 10 minutes is \(\frac{A}{10}\) square feet per mintute.
→ Let C be Pauline’s constant rate in square feet per minute; then, \(\frac{35}{2.5}\)=C, and \(\frac{A}{10}\)=C. Therefore,
Answer;
\(\frac{35}{2.5}\)=\(\frac{A}{10}\)
350=2.5A
\(\frac{350}{2.5}\)=\(\frac{2.5}{2.5}\) A
140=A
Pauline mows 140 square feet of lawn in 10 minutes.

→ If we let y represent the number of square feet Pauline can mow in t minutes, then Pauline’s average rate in t minutes is \(\frac{y}{t}\) square feet per minute.
→ Write the two-variable equation that represents the area of lawn, y, Pauline can mow in t minutes.
Answer:
\(\frac{35}{2.5}\)=\(\frac{y}{t}\)
2.5y=35t
\(\frac{2.5}{2.5}\) y=\(\frac{35}{2.5}\) t
y=\(\frac{35}{2.5}\) t

→ What is the meaning of \(\frac{35}{2.5}\) in the equation y=\(\frac{35}{2.5}\) t?
→ The number \(\frac{35}{2.5}\) represents the constant rate at which Pauline can mow a lawn.
→ We can organize the information in a table.

Answer:

→ On a coordinate plane, we will let the x-axis represent time t, in minutes, and the y-axis represent the area of mowed lawn in square feet. Then we have the following graph.

Answer:

→ Because Pauline mows at a constant rate, we would expect the square feet of mowed lawn to continue to rise as the time, in minutes, increases.

Example 2.
Water flows at a constant rate out of a faucet. Suppose the volume of water that comes out in three minutes is 10.5 gallons. How many gallons of water come out of the faucet in t minutes?
→ Write the linear equation that represents the volume of water, V, that comes out in t minutes.
Answer:
Let C represent the constant rate of water flow.
\(\frac{10.5}{3}\)=C, and \(\frac{V}{t}\)=C; then, \(\frac{10.5}{3}\)=\(\frac{V}{t}\).
\(\frac{10.5}{3}\)=\(\frac{V}{t}\)
3V=10.5t
\(\frac{3}{3}\) V=\(\frac{10.5}{3}\) t
V=\(\frac{10.5}{3}\) t

→ What is the meaning of the number \(\frac{10.5}{3}\) in the equation V=\(\frac{10.5}{3}\) t?
→ The number \(\frac{10.5}{3}\) represents the constant rate at which water flows from a faucet.
→ Using the linear equation V=\(\frac{10.5}{3}\) t, complete the table.

Answer:

→ On a coordinate plane, we will let the x-axis represent time t in minutes and the y-axis represent the volume of water. Graph the data from the table.

Answer:

→ Using the graph, about how many gallons of water do you think would flow after 1 \(\frac{1}{2}\) minutes? Explain.
→ After 1 \(\frac{1}{2}\) minutes, between 3 \(\frac{1}{2}\) and 7 gallons of water will flow. Since the water is flowing at a constant rate, we can expect the volume of water to rise between 1 and 2 minutes. The number of gallons that flow after 1 \(\frac{1}{2}\) minutes then would have to be between the number of gallons that flow out after 1 minute and 2 minutes.
→ Using the graph, about how long would it take for 15 gallons of water to flow out of the faucet? Explain.
→ It would take between 4 and 5 minutes for 15 gallons of water to flow out of the faucet. It takes 4 minutes for 14 gallons to flow; therefore, it must take more than 4 minutes for 15 gallons to come out. It must take less than 5 minutes because 3 \(\frac{1}{2}\) gallons flow out every minute.
→ Graphing proportional relationships like these last two constant rate problems provides us more information than simply solving an equation and calculating one value. The graph provides information that is not so obvious in an equation.

Eureka Math Grade 8 Module 4 Lesson 11 Exercise Answer Key

Exercise 1.
Juan types at a constant rate. He can type a full page of text in 3 \(\frac{1}{2}\) minutes. We want to know how many pages, p, Juan can type after t minutes.
a. Write the linear equation in two variables that represents the number of pages Juan types in any given time interval.
Let C represent the constant rate that Juan types in pages per minute. Then,
Answer:
\(\frac{1}{3.5}\)=C, and \(\frac{p}{t}\)=C; therefore, \(\frac{1}{3.5}\)=\(\frac{p}{t}\).
\(\frac{1}{3.5}\)=\(\frac{p}{t}\)
3.5p=t
\(\frac{3.5}{3.5}\) p=\(\frac{1}{3.5}\) t
p=\(\frac{1}{3.5}\) t

b. Complete the table below. Use a calculator, and round your answers to the tenths place.

Answer:

c. Graph the data on a coordinate plane.

Answer:

d. About how long would it take Juan to type a 5-page paper? Explain.
Answer;
It would take him between 15 and 20 minutes. After 15 minutes, he will have typed 4.3 pages. In 20 minutes, he can type 5.7 pages. Since 5 pages is between 4.3 and 5.7, then it will take him between 15 and 20 minutes.

Question 2.
Emily paints at a constant rate. She can paint 32 square feet in 5 minutes. What area, A, in square feet, can she paint in t minutes?
a. Write the linear equation in two variables that represents the number of square feet Emily can paint in any given time interval.
Answer:
Let C be the constant rate that Emily paints in square feet per minute. Then,
\(\frac{32}{5}\)=C, and \(\frac{A}{t}\)=C; therefore, \(\frac{32}{5}\)=\(\frac{A}{t}\).
\(\frac{32}{5}\)=\(\frac{A}{t}\)
5A=32t
\(\frac{5}{5}\) A=\(\frac{32}{5}\) t
A=\(\frac{32}{5}\) t

b. Complete the table below. Use a calculator, and round answers to the tenths place.

Answer:

c. Graph the data on a coordinate plane.

Answer:

d. About how many square feet can Emily paint in 2 \(\frac{1}{2}\) minutes? Explain.
Answer:
Emily can paint between 12.8 and 19.2 square feet in 2 \(\frac{1}{2}\) minutes. After 2 minutes, she paints 12.8 square feet, and after 3 minutes, she will have painted 19.2 square feet.

Question 3.
Joseph walks at a constant speed. He walked to a store that is one-half mile away in 6 minutes. How many miles, m, can he walk in t minutes?
a. Write the linear equation in two variables that represents the number of miles Joseph can walk in any given time interval, t.
Answer:
Let C be the constant rate that Joseph walks in miles per minute. Then,
\(\frac{0.5}{6}\)=C, and \(\frac{m}{t}\)=C; therefore, \(\frac{0.5}{6}\)=\(\frac{m}{t}\).
\(\frac{0.5}{6}\)=\(\frac{m}{t}\)
6m=0.5t
\(\frac{6}{6}\) m=\(\frac{0.5}{6}\) t
m=\(\frac{0.5}{6}\) t

b. Complete the table below. Use a calculator, and round answers to the tenths place.

Answer:

c. Graph the data on a coordinate plane.

Answer:

d. Joseph’s friend lives 4 miles away from him. About how long would it take Joseph to walk to his friend’s house? Explain.
Answer:
It will take Joseph a little less than an hour to walk to his friend’s house. Since it takes 30 minutes for him to walk 2.5 miles and 60 minutes to walk 5 miles, and 4 is closer to 5 than 2.5, it will take Joseph less than an hour to walk the 4 miles.

Eureka Math Grade 8 Module 4 Lesson 11 Exit Ticket Answer Key

Vicky reads at a constant rate. She can read 5 pages in 9 minutes. We want to know how many pages, p, Vicky can read after t minutes.

a. Write a linear equation in two variables that represents the number of pages Vicky reads in any given time interval.
Answer:
Let C represent the constant rate that Vicky reads in pages per minute. Then,
\(\frac{5}{9}\)=C, and \(\frac{p}{t}\)=C; therefore, \(\frac{5}{9}\)=\(\frac{p}{t}\).
\(\frac{5}{9}\)=\(\frac{p}{t}\)
9p=5t
\(\frac{9}{9}\) p=\(\frac{5}{9}\) t
p=\(\frac{5}{9}\) t

b. Complete the table below. Use a calculator, and round answers to the tenths place.

Answer:

c. About how long would it take Vicky to read 25 pages? Explain.
Answer:
It would take her a little over 40 minutes. After 40 minutes, she can read about 22.2 pages, and after 1 hour, she can read about 33.3 pages. Since 25 pages is between 22.2 and 33.3, it will take her between 40 and 60 minutes to read 25 pages.

Eureka Math Grade 8 Module 4 Lesson 11 Problem Set Answer Key

Students practice writing two-variable equations that represent a constant rate.

Question 1.
A train travels at a constant rate of 45 miles per hour.
a. What is the distance, d, in miles, that the train travels in t hours?
Answer:
Let C be the constant rate the train travels. Then, \(\frac{45}{1}\)=C, and \(\frac{d}{t}\)=C; therefore, \(\frac{45}{1}\)=\(\frac{d}{t}\).
\(\frac{45}{1}\)=\(\frac{d}{t}\)
d=45t

b. How many miles will it travel in 2.5 hours?
Answer:
d=45(2.5)
=112.5
The train will travel 112.5 miles in 2.5 hours.

Question 2.
Water is leaking from a faucet at a constant rate of \(\frac{1}{3}\) gallons per minute.
a. What is the amount of water, w, in gallons per minute, that is leaked from the faucet after t minutes?
Answer:
Let C be the constant rate the water leaks from the faucet in gallons per minute. Then,
\(\frac{\frac{1}{3}}{1}\)=C, and \(\frac{w}{t}\)=C; therefore, \(\frac{\frac{1}{3}}{1}\) = \(\frac{w}{t}\).
\(\frac{\frac{1}{3}}{1}\)=\(\frac{w}{t}\)
w=\(\frac{1}{3}\) t

b. How much water is leaked after an hour?
Answer:
w=\(\frac{1}{3}\) t
=\(\frac{1}{3}\) (60)
=20
The faucet will leak 20 gallons in one hour.

Question 3.
A car can be assembled on an assembly line in 6 hours. Assume that the cars are assembled at a constant rate.
a. How many cars, y, can be assembled in t hours?
Answer:
Let C be the constant rate the cars are assembled in cars per hour. Then,
\(\frac{1}{6}\)=C, and \(\frac{y}{t}\)=C; therefore, \(\frac{1}{6}\)=\(\frac{y}{t}\).
\(\frac{1}{6}\)=\(\frac{y}{t}\)
6y=t
\(\frac{6}{6}\) y=\(\frac{1}{6}\)t
y=\(\frac{1}{6}\) t

b. How many cars can be assembled in a week?
Answer:
A week is 24×7=168 hours. So, y=\(\frac{1}{6}\) (168)=28. Twenty-eight cars can be assembled in a week.

Question 4.
A copy machine makes copies at a constant rate. The machine can make 80 copies in 2 \(\frac{1}{2}\) minutes.
a. Write an equation to represent the number of copies, n, that can be made over any time interval in minutes, t.
Answer:
Let C be the constant rate that copies can be made in copies per minute. Then,
\(\frac{80}{2 \frac{1}{2}}\))=C, and \(\frac{n}{t}\)=C; therefore, \(\frac{80}{2 \frac{1}{2}}\)= \(\frac{n}{t}\).
\(\frac{80}{2 \frac{1}{2}}\)=\(\frac{n}{t}\)
2 \(\frac{1}{2}\) n=80t
\(\frac{5}{2}\) n=80t
\(\frac{2}{5}\)∙\(\frac{5}{2}\) n=\(\frac{2}{5}\)∙80t
n=32t

b. Complete the table below.

Answer:

c. Graph the data on a coordinate plane.

Answer:

d. The copy machine runs for 20 seconds and then jams. About how many copies were made before the jam occurred? Explain.
Answer:
Since 20 seconds is approximately 0.3 of a minute, then the number of copies made will be between 8 and 16 because 0.3 is between 0.25 and 0.5.

Question 5.
Connor runs at a constant rate. It takes him 34 minutes to run 4 miles.
a. Write the linear equation in two variables that represents the number of miles Connor can run in any given time interval in minutes, t.
Answer:
Let C be the constant rate that Connor runs in miles per minute, and let m represent the number of miles he ran in t minutes. Then,
\(\frac{4}{34}\)=C, and \(\frac{m}{t}\)=C; therefore, \(\frac{4}{34}\)=\(\frac{m}{t}\).
\(\frac{4}{34}\)=\(\frac{m}{t}\)
34m=4t
\(\frac{34}{34}\) m=\(\frac{4}{34}\) t
m=\(\frac{4}{34}\) t
m=\(\frac{2}{17}\) t

b. Complete the table below. Use a calculator, and round answers to the tenths place.

Answer:

c. Graph the data on a coordinate plane.

Answer:

d. Connor ran for 40 minutes before tripping and spraining his ankle. About how many miles did he run before he had to stop? Explain.
Answer:
Since Connor ran for 40 minutes, he ran more than 3.5 miles but less than 5.3 miles. Since 40 is between 30 and 45, then we can use those reference points to make an estimate of how many miles he ran in 40 minutes, probably about 5 miles.

Engage NY Eureka Math 8th Grade Module 4 Lesson 9 Answer Key

Eureka Math Grade 8 Module 4 Lesson 9 Exercise Answer Key

Exercise 1.
Write the equation for the 15^th step.
Answer:
S₁₅-7+7∙5¹⁵=5S₁₅

Exercise 2.
How many people would see the photo after 15 steps? Use a calculator if needed.
Answer:
S₁₅-7+7∙5¹⁵=5S₁₅
S₁₅-5S₁₅-7++7∙5¹⁵=5S₁₅-5S₁₅
S₁₅ (1-5)-7+7∙5¹⁵=0
S₁₅ (1-5)-7+7+7∙5¹⁵-7∙5¹⁵=7-7∙5¹⁵
S₁₅ (1-5)=7(1-5¹⁵ )
S₁₅=\(\frac{7\left(1-5^{15}\right)}{(1-5)}\)
S₁₅=53 405 761 717

Exercises 3–11 as an Alternative to Discussion
Students should be able to complete the following problems independently as they are an application of skills learned to this point, namely, transcription and solving linear equations in one variable. Have students work on the problems one at a time and share their work with the whole class, or assign the entire set and allow students to work at their own pace. Provide correct solutions at the end of the lesson.

Exercise 3.
Marvin paid an entrance fee of $5 plus an additional $1.25 per game at a local arcade. Altogether, he spent $26.25. Write and solve an equation to determine how many games Marvin played.
Let x represent the number of games he played.
5+1.25x=26.25
1.25x=21.25
x=\(\frac{21.25}{1.25}\)
x=17
Marvin played 17 games.

Exercise 4.
The sum of four consecutive integers is -26. What are the integers?
Answer:
Let x be the first integer.
x+(x+1)+(x+2)+(x+3)=-26
4x+6=-26
4x=-32
x=-8
The integers are -8, -7, -6, and -5.

Exercise 5.
A book has x pages. How many pages are in the book if Maria read 45 pages of a book on Monday, \(\frac{1}{2}\) the book on Tuesday, and the remaining 72 pages on Wednesday?
Answer:
Let x be the number of pages in the book.
x=45+\(\frac{1}{2}\) x+72
x=117+\(\frac{1}{2}\) x
\(\frac{1}{2}\) x=117
x=234
The book has 234 pages.

Exercise 6.
A number increased by 5 and divided by 2 is equal to 75. What is the number?
Answer:
Let x be the number.
\(\frac{x+5}{2}\)=75
x+5=150
x=145
The number is 145.

Exercise 7.
The sum of thirteen and twice a number is seven less than six times a number. What is the number?
Answer:
Let x be the number.
13+2x=6x-7
20+2x=6x
20=4x
5=x
The number is 5.

Exercise 8.
The width of a rectangle is 7 less than twice the length. If the perimeter of the rectangle is 43.6 inches, what is the area of the rectangle?
Answer:
Let x represent the length of the rectangle.
2x+2(2x-7)=43.6
2x+4x-14=43.6
6x-14=43.6
6x=57.6
x=\(\frac{57.6}{6}\)
x=9.6
The length of the rectangle is 9.6 inches, and the width is 12.2 inches, so the area is 117.12 in².

Exercise 9.
Two hundred and fifty tickets for the school dance were sold. On Monday, 35 tickets were sold. An equal number of tickets were sold each day for the next five days. How many tickets were sold on one of those days?
Answer:
Let x be the number of tickets sold on one of those days.
250=35+5x
215=5x
43=x
43 tickets were sold on each of the five days.

Exercise 10.
Shonna skateboarded for some number of minutes on Monday. On Tuesday, she skateboarded for twice as many minutes as she did on Monday, and on Wednesday, she skateboarded for half the sum of minutes from Monday and Tuesday. Altogether, she skateboarded for a total of three hours. How many minutes did she skateboard each day?
Answer:
Let x be the number of minutes she skateboarded on Monday.
x+2x+\(\frac{2 x+x}{2}\)=180
\(\frac{2 x}{2}\)+\(\frac{4 x}{2}\)+\(\frac{2 x+x}{2}\)=180
\(\frac{9 x}{2}\)=180
9x=360
x=40
Shonna skateboarded 40 minutes on Monday, 80 minutes on Tuesday, and 60 minutes on Wednesday.

Exercise 11.
In the diagram below, △ABC ~△A^’ B^’ C^’. Determine the length of \(\overline{A C}\) and \(\overline{B C}\).

Answer:
\(\frac{18}{10.3-x}\)=\(\frac{38}{5 x-0.3}\)
18(5x-0.3)=38(10.3-x)
90x-5.4=391.4-38x
128x-5.4=391.4
128x=396.8
x=\(\frac{396.8}{128}\)
x=3.1
The length of \(\overline{A C}\) is 7.2 mm, and the length of \(\overline{B C}\) is 15.2 mm.

Eureka Math Grade 8 Module 4 Lesson 9 Exit Ticket Answer Key

Question 1.
Rewrite the equation that would represent the sum in the fifth step of the Facebook problem:
S₅=7+7∙5+7∙5²+7∙5³+7∙5⁴
S₅=7+7∙5+7∙5²+7∙5³+7∙5⁴
S₅-7=7∙5+7∙5²+7∙5³+7∙5⁴
S₅-7+7∙5⁵=7∙5+7∙5²+7∙5³+7∙5⁴+7∙5⁵
S₅-7+7∙5⁵=5(7+7∙5+7∙5²+7∙5³+7∙5⁴ )
S₅-7+7∙5⁵=5(S₅ )
S₅-5S₅-7+7∙5⁵=0
S₅-5S₅=7-(7∙5⁵ )
(1-5) S₅=7-(7∙5⁵ )
(1-5) S₅=7(1-5⁵ )
S₅=\(\frac{7\left(1-5^{5}\right)}{1-5}\)

Question 2.
The sum of four consecutive integers is 74. Write an equation, and solve to find the numbers.
Answer:
Let x be the first number.
x+(x+1)+(x+2)+(x+3)=74
4x+6=74
4x=68
x=17
The numbers are 17, 18, 19, and 20.

Eureka Math Grade 8 Module 4 Lesson 9 Problem Set Answer Key

Assign the problems that relate to the elements of the lesson that were used with students.

Question 1.
You forward an e-card that you found online to three of your friends. They liked it so much that they forwarded it on to four of their friends, who then forwarded it on to four of their friends, and so on. The number of people who saw the e-card is shown below. Let S₅ represent the number of people who saw the e-card after one step, let S₂ represent the number of people who saw the e-card after two steps, and so on.
S₅=3
S₂=3+3∙4
S₃=3+3∙4+3∙4²
S₄=3+3∙4+3∙4²+3∙4³

a. Find the pattern in the equations.
Answer:
S₂=3+3∙4
S₂-3=3∙4
S₂-3+3∙4²=3∙4+3∙4²
S₂-3+3∙4²=4(3+3∙4)
S₂-3+3∙4²=4S₂

S₃=3+3∙4+3∙4²
S₃-3=3∙4+3∙4²
S₃-3+3∙4³=3∙4+3∙4²+3∙4³
S₃-3+3∙4³=4(3+3∙4+3∙4² )
S₃-3+3∙4³=4S₃

S₄=3+3∙4+3∙4²+3∙4³
S₄-3=3∙4+3∙4²+3∙4³
S₄-3+3∙4⁴=3∙4+3∙4²+3∙4³+3∙4⁴
S₄-3+3∙4⁴=4(3+3∙4+3∙4²+3∙4³ )
S₄-3+3∙4⁴=4S₄

b. Assuming the trend continues, how many people will have seen the e-card after 10 steps?
Answer:
S₅0-3+3∙4¹⁰=4S₁₀
S₁₀-4S₁₀-3+3∙4¹⁰=0
S₁₀(1-4)=3-3∙4¹⁰
S₁₀(1-4)=3(1-4¹⁰)
S₁₀=\(\frac{3\left(1-4^{10}\right)}{(1-4)}\)
S₁₀=1 048 575
After 10 steps, 1,048,575 people will have seen the e-card.

c. How many people will have seen the e-card after n steps?
Answer:
S_n=\(\frac{3\left(1-4^{n}\right)}{(1-4)}\)

For each of the following questions, write an equation, and solve to find each answer.

Question 2.
Lisa has a certain amount of money. She spent $39 and has \(\frac{3}{4}\) of the original amount left. How much money did she have originally?
Answer:
Let x be the amount of money Lisa had originally.
x-39=\(\frac{3}{4}\) x
-39=-\(\frac{1}{4}\) x
156=x
Lisa had $156 originally.

Question 3.
The length of a rectangle is 4 more than 3 times the width. If the perimeter of the rectangle is 18.4 cm, what is the area of the rectangle?
Answer:
Let x represent the width of the rectangle.
2(4+3x)+2x=18.4
8+6x+2x=18.4
8+8x=18.4
8x=10.4
x=\(\frac{10.4}{8}\)
x=1.3
The width of the rectangle is 1.3 cm, and the length is 7.9 cm, so the area is 10.27 cm².

Question 4.
Eight times the result of subtracting 3 from a number is equal to the number increased by 25. What is the number?
Answer:
Let x be the number.
8(x-3)=x+25
8x-24=x+25
7x-24=25
7x=49
x=7
The number is 7.

Question 5.
Three consecutive odd integers have a sum of 3. What are the numbers?
Answer:
Let x be the first odd number.
x+(x+2)+(x+4)=3
3x+6=3
3x=-3
x=-1
The three consecutive odd integers are -1, 1, and 3.

Question 6.
Each month, Liz pays $35 to her phone company just to use the phone. Each text she sends costs her an additional $0.05. In March, her phone bill was $72.60. In April, her phone bill was $65.85. How many texts did she send each month?
Answer:
Let x be the number of texts she sent in March.
35+0.05x=72.60
0.05x=37.6
x=\(\frac{37.6}{0.05}\)
x=752
She sent 752 texts in March.
Let y be the number of texts she sent in April.
35+0.05y=65.85
0.05y=30.85
y=\(\frac{30.85}{0.05}\)
y=617
She sent 617 texts in April.

Question 7.
Claudia is reading a book that has 360 pages. She read some of the book last week. She plans to read 46 pages today. When she does, she will be \(\frac{4}{5}\) of the way through the book. How many pages did she read last week?
Answer:
Let x be the number of pages she read last week.
x+46=\(\frac{4}{5}\) (360)
x+46=288
x=242
Claudia read 242 pages last week.

Question 8.
In the diagram below, △ABC ~△A^’ B^’ C^’. Determine the measure of ∠A.

Answer:
7x-4=x+32
6x-4=32
6x=36
x=6
The measure of ∠A is 38°.

Question 9.
In the diagram below, △ABC ~△A^’ B^’ C^’. Determine the measure of ∠A.

Answer:
10x-9=4x+57
6x-9=57
6x=66
x=11
The measure of ∠A is 101°.

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

Eureka Math Grade 8 Module 4 Lesson 8 Example Answer Key

Example 1.
→ Given a linear equation in disguise, we will try to solve it. To do so, we must first assume that the following equation is true for some number x.
\(\frac{x-1}{2}\) = \(\frac{x+\frac{1}{3}}{4}\)
We want to make this equation look like the linear equations we are used to. For that reason, we will multiply both sides of the equation by 2 and 4, as we normally do with proportions:
2(x+\(\frac{1}{3}\))=4(x-1).
→ Is this a linear equation? How do you know?
→ Yes, this is a linear equation because the expressions on the left and right of the equal sign are linear expressions.
→ Notice that the expressions that contained more than one term were put in parentheses. We do that so we do not make a mistake and forget to use the distributive property.
→ Now that we have a linear equation, we will use the distributive property and solve as usual.
2(x+\(\frac{1}{3}\))=4(x-1)
2x+\(\frac{2}{3}\)=4x-4
2x-2x+\(\frac{2}{3}\)=4x-2x-4
\(\frac{2}{3}\)=2x-4
\(\frac{2}{3}\)+4=2x-4+4
\(\frac{14}{3}\)=2x
\(\frac{1}{2}\)∙\(\frac{14}{3}\)=\(\frac{1}{2}\)∙2x
\(\frac{7}{3}\)=x
→ How can we verify that \(\frac{7}{3}\) is the solution to the equation?
→ We can replace x with \(\frac{7}{3}\) in the original equation.

Since \(\frac{7}{3}\) made the equation true, we know it is a solution to the equation.

Example 2.
→ Can we solve the following equation? Explain.
\(\frac{\frac{1}{5}-x}{7}\) = \(\frac{2 x+9}{3}\)
→ We need to multiply each numerator with the other fraction’s denominator.
→ So,
\(\frac{\frac{1}{5}-x}{7}\) = \(\frac{2 x+9}{3}\)
7(2x+9)=3(\(\frac{1}{5}\)-x).
→ What would be the next step?
→ Use the distributive property.
→ Now we have
7(2x+9)=3(\(\frac{1}{5}\)-x)
14x+63=\(\frac{3}{5}\)-3x
14x+3x+63=\(\frac{3}{5}\)-3x+3x
17x+63=\(\frac{3}{5}\)
17x+63-63=\(\frac{3}{5}\)-63
17x=\(\frac{3}{5}\)–\(\frac{315}{5}\)
17x=-\(\frac{312}{5}\)
\(\frac{1}{17}\) (17x)=(-\(\frac{312}{5}\))\(\frac{1}{17}\)
x=-\(\frac{312}{85}\).
→ Is this a linear equation? How do you know?
→ Yes, this is a linear equation because the left and right side are linear expressions.

Example 3.
Can this equation be solved?
\(\frac{6+x}{7 x+\frac{2}{3}}\)=\(\frac{3}{8}\)
Give students a few minutes to work. Provide support to students as needed.
→ Yes, we can solve the equation because we can multiply each numerator with the other fraction’s denominator and then use the distributive property to begin solving it.
Answer:
\(\frac{6+x}{7 x+\frac{2}{3}}\)=\(\frac{3}{8}\)
(6+x)8=(7x+\(\frac{2}{3}\))3
48+8x=21x+2
48+8x-8x=21x-8x+2
48=13x+2
48-2=13x+2-2
46=13x
\(\frac{46}{13}\)=x

Example 4.
Can this equation be solved?
\(\frac{7}{3 x+9}\)=\(\frac{1}{8}\)
Give students a few minutes to work. Provide support to students as needed.
→ Yes, we can solve the equation because we can multiply each numerator with the other fraction’s denominator and then use the distributive property to begin solving it.
Answer:
\(\frac{7}{3 x+9}\)=\(\frac{1}{8}\)
7(8)=(3x+9)1
56=3x+9
56-9=3x+9-9
47=3x
\(\frac{47}{3}\)=x

Example 5.
In the diagram below, △ABC~ △A’ B’ C’. Using what we know about similar triangles, we can determine the value of x.

→ Begin by writing the ratios that represent the corresponding sides.
Answer:
\(\frac{x-2}{9.5}\) = \(\frac{x+2}{12}\)

It is possible to write several different proportions in this case. If time, discuss this fact with students.
→ Now that we have the ratios, solve for x and find the lengths of \(\overline{A B}\) and \(\overline{A C}\).
Answer:
\(\frac{x-2}{9.5}\) = \(\frac{x+2}{12}\)
(x-2)12=9.5(x+2)
12x-24=9.5x+19
12x-24+24=9.5x+19+24
12x=9.5x+43
12x-9.5x=9.5x-9.5x+43
2.5x=43
x=\(\frac{43}{2.5}\)
x=17.2
|AB|=15.2 cm, and |AC|=19.2 cm.

Eureka Math Grade 8 Module 4 Lesson 8 Exercise Answer Key

Solve the following equations of rational expressions, if possible.

Question 1.
\(\frac{2 x+1}{9}\)=\(\frac{1-x}{6}\)
Answer:
\(\frac{2 x+1}{9}\)=\(\frac{1-x}{6}\)
9(1-x)=(2x+1)6
9-9x=12x+6
9-9x+9x=12x+9x+6
9=21x+6
9-6=21x+6-6
3=21x
\(\frac{3}{21}\)=\(\frac{21}{2}\)1 x
\(\frac{1}{7}\)=x

Question 2.
\(\frac{5+2 x}{3 x-1}\)=\(\frac{6}{7}\)
Answer:
\(\frac{5+2 x}{3 x-1}\)=\(\frac{6}{7}\)
(5+2x)7=(3x-1)6
35+14x=18x-6
35-35+14x=18x-6-35
14x=18x-41
14x-18x=18x-18x-41
-4x=-41
\(\frac{-4}{-4}\) x=\(\frac{-41}{-4}\)
x=\(\frac{41}{4}\)

Question 3.
\(\frac{x+9}{12}\)=\(\frac{-2 x-\frac{1}{2}}{3}\)
Answer:
\(\frac{x+9}{12}\)=\(\frac{-2 x-\frac{1}{2}}{3}\)
12(-2x-\(\frac{1}{2}\))=(x+9)3
-24x-6=3x+27
-24x+24x-6=3x+24x+27
-6=27x+27
-6-27=27x+27-27
-33=27x
\(\frac{-33}{27}\)=\(\frac{27}{27}\) x
–\(\frac{11}{9}\)=x

Question 8.
\(\frac{8}{3-4 x}\) = \(\frac{5}{2 x+\frac{1}{4}}\)
Answer:
\(\frac{8}{3-4 x}\) = \(\frac{5}{2 x+\frac{1}{4}}\)
8(2x+\(\frac{1}{4}\))=(3-4x)5
16x+2=15-20x
16x+2-2=15-2-20x
16x=13-20x
16x+20x=13-20x+20x
36x=13
\(\frac{36}{36}\) x=\(\frac{13}{36}\)
x=\(\frac{13}{36}\)

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

Students practice solving equations with rational expressions, if a solution is possible.

Solve the following equations of rational expressions, if possible. If an equation cannot be solved, explain why.

Question 1.
\(\frac{5}{6 x-2}\) = \(\frac{-1}{x+1}\)
Answer:
\(\frac{5}{6 x-2}\) = \(\frac{-1}{x+1}\)
5(x+1)=-1(6x-2)
5x+5=-6x+2
5x+5-5=-6x+2-5
5x=-6x-3
5x+6x=-6x+6x-3
11x=-3
x=-\(\frac{3}{11}\)

Question 2.
\(\frac{4-x}{8}\) = \(\frac{7 x-1}{3}\)
Answer:
\(\frac{4-x}{8}\) = \(\frac{7 x-1}{3}\)
8(7x-1)=(4-x)3
56x-8=12-3x
56x-8+8=12+8-3x
56x=20-3x
56x+3x=20-3x+3x
59x=20
\(\frac{59}{59}\) x=\(\frac{20}{59}\)
x=\(\frac{20}{59}\)

Question 3.
\(\frac{3 x}{x+2}\) = \(\frac{5}{9}\)
Answer:
\(\frac{3 x}{x+2}\) = \(\frac{5}{9}\)
9(3x)=(x+2)5
27x=5x+10
27x-5x=5x-5x+10
22x=10
\(\frac{22}{22}\) x=\(\frac{10}{22}\)
x=\(\frac{5}{11}\)

Question 4.
\(\frac{\frac{1}{2} x+6}{3}\) = \(\frac{x-3}{2}\)
Answer:
\(\frac{\frac{1}{2} x+6}{3}\) = \(\frac{x-3}{2}\)
3(x-3)=2(\(\frac{1}{2}\) x+6)
3x-9=x+12
3x-9+9=x+12+9
3x=x+21
3x-x=x-x+21
2x=21
x=\(\frac{21}{2}\)

Question 5.
\(\frac{7-2 x}{6}\) = \(\frac{x-5}{1}\)
Answer:
\(\frac{7-2 x}{6}\) = \(\frac{x-5}{1}\)
6(x-5)=(7-2x)1
6x-30=7-2x
6x-30+30=7+30-2x
6x=37-2x
6x+2x=37-2x+2x
8x=37
\(\frac{8}{8}\) x=\(\frac{37}{8}\)
x=\(\frac{37}{8}\)

Question 6.
\(\frac{2 x+5}{2}\) = \(\frac{3 x-2}{6}\)
Answer:
\(\frac{2 x+5}{2}\) = \(\frac{3 x-2}{6}\)
2(3x-2)=6(2x+5)
6x-4=12x+30
6x-4+4=12x+30+4
6x=12x+34
6x-12x=12x-12x+34
-6x=34
x=-\(\frac{34}{6}\)
x=-\(\frac{17}{3}\)

Question 7.
\(\frac{6 x+1}{3}\) = \(\frac{9-x}{7}\)
Answer:
\(\frac{6 x+1}{3}\) = \(\frac{9-x}{7}\)
(6x+1)7=3(9-x)
42x+7=27-3x
42x+7-7=27-7-3x
42x=20-3x
42x+3x=20-3x+3x
45x=20
\(\frac{45}{45}\)x=\(\frac{20}{45}\)
x=\(\frac{4}{9}\)

Question 8.
\(\frac{\frac{1}{3} x-8}{12}\) = \(\frac{-2-x}{15}\)
Answer:
\(\frac{\frac{1}{3} x-8}{12}\) = \(\frac{-2-x}{15}\)
12(-2-x)=(\(\frac{1}{3}\) x-8)15
-24-12x=5x-120
-24-12x+12x=5x+12x-120
-24=17x-120
-24+120=17x-120+120
96=17x
\(\frac{96}{17}\)=\(\frac{17}{17}\) x
\(\frac{96}{17}\)=x

Question 9.
\(\frac{3-x}{1-x}\)=\(\frac{3}{2}\)
Answer:
\(\frac{3-x}{1-x}\)=\(\frac{3}{2}\)
(3-x)2=(1-x)3
6-2x=3-3x
6-2x+2x=3-3x+2x
6=3-x
6-3=3-3-x
3=-x
-3=x

Question 10.
In the diagram below, △ABC~ △A’ B’ C’. Determine the lengths of \(\overline{A C}\) and \(\overline{B C}\).

Answer:
\(\frac{x+4}{4.5}\) = \(\frac{3 x-2}{9}\)
9(x+4)=4.5(3x-2)
9x+36=13.5x-9
9x+36+9=13.5x-9+9
9x+45=13.5x
9x-9x+45=13.5x-9x
45=4.5x
10=x
|AC|=14 cm, and |BC|=28 cm.

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

Solve the following equations for x.

Question 1.
\(\frac{5 x-8}{3}\) = \(\frac{11 x-9}{5}\)
Answer:
\(\frac{5 x-8}{3}\) = \(\frac{11 x-9}{5}\)
5(5x-8)=3(11x-9)
25x-40=33x-27
25x-25x-40=33x-25x-27
-40=8x-27
-40+27=8x-27+27
-13=8x
–\(\frac{13}{8}\)=x

Question 2.
\(\frac{x+11}{7}\)=\(\frac{2 x+1}{-8}\)
Answer:
\(\frac{x+11}{7}\)=\(\frac{2 x+1}{-8}\)
7(2x+1)=-8(x+11)
14x+7=-8x-88
14x+7-7=-8x-88-7
14x=-8x-95
14x+8x=-8x+8x-95
22x=-95
x=-\(\frac{95}{22}\)

Question 3.
\(\frac{-x-2}{-4}\) = \(\frac{3 x+6}{2}\)
Answer:
\(\frac{-x-2}{-4}\) = \(\frac{3 x+6}{2}\)
-4(3x+6)=2(-x-2)
-12x-24=-2x-4
-12x-24+24=-2x-4+24
-12x=-2x+20
-12x+2x=-2x+2x+20
-10x=20
x=-2