The Graph Below Represents the Distance in Miles by Car a Travels in X Minutes Eureka Math

Engage NY Eureka Math 8th Grade Module 4 Lesson 24 Answer Cardinal

Eureka Math Grade 8 Module iv Lesson 24 Exercise Answer Key

Exercises

Do 1.
Derek scored xxx points in the basketball game he played, and not once did he become to the free throw line. That means that Derek scored 2-point shots and 3-point shots. List as many combinations of two- and three-pointers as you lot can that would full 30 points.

Answer:

Write an equation to describe the information.
Respond:
Let 10 correspond the number of 2-pointers and y represent the number of iii-pointers.
30 = 2x + 3y

Practice 2.
Derek tells you that the number of 2-point shots that he fabricated is five more than than the number of three-indicate shots. How many combinations can y'all come upwards with that fit this scenario? (Don't worry about the full number of points.)

Answer:

Write an equation to describe the data.
Respond:
Allow x correspond the number of two-pointers and y represent the number of 3-pointers.
x = 5 + y

Exercise iii.
Which pair of numbers from your table in Exercise ii would show Derek's bodily score of thirty points?
Answer:
The pair 9 and 4 would show Derek'southward actual score of xxx points.

Practise iv.
Efrain and Fernie are on a road trip. Each of them drives at a constant speed. Efrain is a safe commuter and travels 45 miles per hour for the unabridged trip. Fernie is not such a safe driver. He drives 70 miles per 60 minutes throughout the trip. Fernie and Efrain left from the same location, but Efrain left at 8:00 a.m., and Fernie left at eleven:00 a.1000. Assuming they take the same route, will Fernie ever grab upwards to Efrain? If so, approximately when?
a. Write the linear equation that represents Efrain's abiding speed. Make sure to include in your equation the extra fourth dimension that Efrain was able to travel.
Reply:
Efrain's rate is \(\frac{45}{1}\) miles per hour, which is the same as 45 miles per 60 minutes. If he drives y miles in x hours at that abiding rate, then y = 45x. To account for his boosted iii hours of driving fourth dimension that Efrain gets, we write the equation y = 45(x + 3).
y = 45x + 135

b. Write the linear equation that represents Fernie's abiding speed.
Answer:
Fernie'south rate is \(\frac{70}{one}\) miles per hour, which is the same as 70 miles per hour. If he drives y miles in x hours at that constant charge per unit, and then y = 70x.

c. Write the organisation of linear equations that represents this situation.
Reply:
y = 45x + 135
y = 70x

d. Sketch the graphs of the two linear equations.

Answer:

e. Volition Fernie ever take hold of up to Efrain? If so, approximately when?
Answer:
Yes, Fernie will catch up to Efrain after about four \(\frac{one}{2}\) hours of driving or after traveling about 325 miles.

f. At approximately what point do the graphs of the lines intersect?
Reply:
The lines intersect at approximately (4.5, 325).

Practise five.
Jessica and Karl run at constant speeds. Jessica can run 3 miles in 24 minutes. Karl can run 2 miles in 14 minutes. They decide to race each other. Equally presently every bit the race begins, Karl trips and takes 2 minutes to recover.
a. Write the linear equation that represents Jessica'south abiding speed. Make sure to include in your equation the extra fourth dimension that Jessica was able to run.
Answer:
Jessica'due south charge per unit is \(\frac{3}{24}\) miles per minute, which is equivalent to \(\frac{one}{viii}\) miles per infinitesimal. If Jessica runs y miles x minutes at that constant speed, so y = \(\frac{1}{8}\) x. To account for her additional ii minute of running that Jessica gets, nosotros write the equation
y = \(\frac{i}{viii}\) (ten + 2)
y = \(\frac{ane}{8}\) x + \(\frac{1}{4}\)

b. Write the linear equation that represents Karl's constant speed.
Answer:
Karl'south rate is \(\frac{ii}{14}\) miles per minute, which is the same as \(\frac{1}{vii}\) miles per minute. If Karl runs y miles in x minutes at that constant speed, and then y = \(\frac{one}{vii}\) 10.

c. Write the system of linear equations that represents this state of affairs.
Answer:
y = \(\frac{1}{8}\) x + \(\frac{1}{viii}\)
y = \(\frac{1}{7}\) ten

d. Sketch the graphs of the ii linear equations.

Answer:

e. Use the graph to answer the questions beneath.
i. If Jessica and Karl raced for 3 miles, who would win? Explain.
Answer:
If the race were 3 miles, and then Karl would win. It only takes Karl 21 minutes to run three miles, but it takes Jessica 24 minutes to run the altitude of 3 miles.

2. At approximately what point would Jessica and Karl exist tied? Explain.
Respond:
Jessica and Karl would exist tied after about 4 minutes or a distance of 1 mile. That is where the graphs of the lines intersect.

Eureka Math Form viii Module 4 Lesson 24 Problem Gear up Answer Fundamental

Question ane.
Jeremy and Gerardo run at abiding speeds. Jeremy can run 1 mile in viii minutes, and Gerardo tin run 3 miles in 33 minutes. Jeremy started running 10 minutes later on Gerardo. Bold they run the same path, when volition Jeremy grab up to Gerardo?
a. Write the linear equation that represents Jeremy's constant speed.
Reply:
Jeremy's rate is \(\frac{1}{8}\) miles per infinitesimal. If he runs y miles in x minutes, so y = \(\frac{1}{8}\) x.

b. Write the linear equation that represents Gerardo's abiding speed. Make sure to include in your equation the extra time that Gerardo was able to run.
Answer:
Gerardo'due south rate is \(\frac{three}{33}\) miles per infinitesimal, which is the same every bit \(\frac{1}{11}\) miles per minute. If he runs y miles in x minutes, then y = \(\frac{1}{11}\) x. To account for the extra time that Gerardo gets to run, we write the equation
y = \(\frac{1}{11}\) (x + 10)
y = \(\frac{1}{11}\) x + \(\frac{10}{11}\)

c. Write the organization of linear equations that represents this situation.
Answer:
y = \(\frac{1}{8}\) x
y = \(\frac{1}{11}\) ten + \(\frac{10}{11}\)

d. Sketch the graphs of the two equations.

Answer:

e. Will Jeremy ever take hold of upwardly to Gerardo? If so, approximately when?
Answer:
Yes, Jeremy will take hold of upwards to Gerardo afterward about 24 minutes or almost iii miles.

f. At approximately what signal exercise the graphs of the lines intersect?
Respond:
The lines intersect at approximately (24, 3).

Question 2.
2 cars drive from town A to town B at constant speeds. The blue auto travels 25 miles per hour, and the scarlet car travels sixty miles per hour. The blue auto leaves at 9:thirty a.m., and the blood-red machine leaves at noon. The distance between the ii towns is 150 miles.
a. Who will go there first? Write and graph the system of linear equations that represents this state of affairs.

Answer:
The linear equation that represents the distance traveled by the blue car is y = 25(10 + two.5), which is the same every bit y = 25x + 62.5. The linear equation that represents the distance traveled by the scarlet car is
y = 60x. The organisation of linear equations that represents this situation is
y = 25x + 62.v
y = 60x

The red car will go to town B first.

b. At approximately what betoken practice the graphs of the lines intersect?
Reply:
The lines intersect at approximately (1.eight, 110).

Eureka Math Course 8 Module 4 Lesson 24 Exit Ticket Answer Primal

Question 1.
Darnell and Hector ride their bikes at abiding speeds. Darnell leaves Hector's firm to bike home. He tin bike the 8 miles in 32 minutes. Five minutes after Darnell leaves, Hector realizes that Darnell left his phone. Hector rides to catch up. He tin can ride to Darnell's house in 24 minutes. Assuming they bike the same path, will Hector take hold of up to Darnell before he gets home?
a. Write the linear equation that represents Darnell'due south constant speed.
Answer:
Darnell's rate is \(\frac{1}{4}\) miles per minute. If he bikes y miles in ten minutes at that constant speed, then y = \(\frac{one}{4}\) x.

b. Write the linear equation that represents Hector's constant speed. Brand sure to have into account that Hector left after Darnell.
Answer:
Hector's charge per unit is \(\frac{1}{3}\) miles per minute. If he bikes y miles in ten minutes, and so y = \(\frac{1}{three}\) x. To account for the extra time Darnell has to cycle, nosotros write the equation
y = \(\frac{1}{three}\) (x – 5)
y = \(\frac{1}{3}\) x-\(\frac{v}{iii}\)

c. Write the system of linear equations that represents this situation.
Answer:
y = \(\frac{one}{4}\) ten
y = \(\frac{1}{iii}\) x-\(\frac{5}{iii}\)

d. Sketch the graphs of the ii equations.

Answer:

e. Will Hector catch up to Darnell before he gets home? If and so, approximately when?
Answer:
Hector will catch upwards 20 minutes subsequently Darnell left his house (or 15 minutes of biking by Hector) or approximately five miles.

f. At approximately what point do the graphs of the lines intersect?
Reply:
The lines intersect at approximately (twenty, 5).

Share this post