Eureka Math Grade 8 Module 6 Lesson 7 Answer Key

Engage NY Eureka Math 8th Grade Module 6 Lesson 7 Answer Key

Eureka Math Grade 8 Module 6 Lesson 7 Exercise Answer Key

Example 1.
In the previous lesson, you learned that scatter plots show trends in bivariate data.
When you look at a scatter plot, you should ask yourself the following questions:
a. Does it look like there is a relationship between the two variables used to make the scatter plot?
b. If there is a relationship, does it appear to be linear?
c. If the relationship appears to be linear, is the relationship a positive linear relationship or a negative linear relationship?

To answer the first question, look for patterns in the scatter plot. Does there appear to be a general pattern to the points in the scatter plot, or do the points look as if they are scattered at random? If you see a pattern, you can answer the second question by thinking about whether the pattern would be well described by a line. Answering the third question requires you to distinguish between a positive linear relationship and a negative linear relationship. A positive linear relationship is one that is described by a line with a positive slope. A negative linear relationship is one that is described by a line with a negative slope.

Exercises 1–9
Take a look at the following five scatter plots. Answer the three questions in Example 1 for each scatter plot.

Exercise 1.
Scatter Plot 1
Engage NY Math Grade 8 Module 6 Lesson 7 Exercise Answer Key 1
Is there a relationship?
Answer:
Yes

If there is a relationship, does it appear to be linear?
Answer:
Yes

If the relationship appears to be linear, is it a positive or a negative linear relationship?
Answer:
Negative

Exercise 2.
Scatter Plot 2
Engage NY Math Grade 8 Module 6 Lesson 7 Exercise Answer Key 2
Is there a relationship?
Answer:
Yes

If there is a relationship, does it appear to be linear?
Answer:
Yes

If the relationship appears to be linear, is it a positive or a negative linear relationship?
Answer:
Positive

Exercise 3.
Scatter Plot 3
Engage NY Math Grade 8 Module 6 Lesson 7 Exercise Answer Key 3
Is there a relationship?
Answer:
No

If there is a relationship, does it appear to be linear?
Answer:
Not applicable

If the relationship appears to be linear, is it a positive or a negative linear relationship?
Answer:
Not applicable

Exercise 4.
Scatter Plot 4
Engage NY Math Grade 8 Module 6 Lesson 7 Exercise Answer Key 4
Is there a relationship?
Answer:
Yes

If there is a relationship, does it appear to be linear?
Answer:
No

If the relationship appears to be linear, is it a positive or a negative linear relationship?
Answer:
Not applicable

Exercise 5.
Scatter Plot 5
Engage NY Math Grade 8 Module 6 Lesson 7 Exercise Answer Key 5
Is there a relationship?
Answer:
Yes

If there is a relationship, does it appear to be linear?
Answer:
Yes

If the relationship appears to be linear, is it a positive or a negative linear relationship?
Answer:
Negative

Exercise 6.
Below is a scatter plot of data on weight in pounds (x) and fuel efficiency in miles per gallon (y) for 13 cars. Using the questions at the beginning of this lesson as a guide, write a few sentences describing any possible relationship between x and y.
Engage NY Math Grade 8 Module 6 Lesson 7 Exercise Answer Key 6
Answer:
Possible response: There appears to be a negative linear relationship between fuel efficiency and weight. Students may note that this is a fairly strong negative relationship. The cars with greater weight tend to have lesser fuel efficiency.

Exercise 7.
Below is a scatter plot of data on price in dollars (x) and quality rating (y) for 14 bike helmets. Using the questions at the beginning of this lesson as a guide, write a few sentences describing any possible relationship between x and y.
Engage NY Math Grade 8 Module 6 Lesson 7 Exercise Answer Key 7
Answer:
Possible response: There does not appear to be a relationship between quality rating and price. The points in the scatter plot appear to be scattered at random, and there is no apparent pattern in the scatter plot.

Exercise 8.
Below is a scatter plot of data on shell length in millimeters (x) and age in years (y) for 27 lobsters of known age. Using the questions at the beginning of this lesson as a guide, write a few sentences describing any possible relationship between x and y.
Engage NY Math Grade 8 Module 6 Lesson 7 Exercise Answer Key 8
Answer:
Possible response: There appears to be a relationship between shell length and age, but the pattern in the scatter plot is curved rather than linear. Age appears to increase as shell length increases, but the increase is not at a constant rate.

Exercise 9.
Below is a scatter plot of data from crocodiles on body mass in pounds (x) and bite force in pounds (y). Using the questions at the beginning of this lesson as a guide, write a few sentences describing any possible relationship between x and y.
Engage NY Math Grade 8 Module 6 Lesson 7 Exercise Answer Key 9
Data Source: http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0031781#pone-0031781-t001
(Note: Body mass and bite force have been converted to pounds from kilograms and newtons, respectively.)
Answer:
Possible response: There appears to be a positive linear relationship between bite force and body mass. For crocodiles, the greater the body mass, the greater the bite force tends to be. Students may notice that this is a positive relationship but not quite as strong as the relationship noted in Exercise 6.

Example 2: Clusters and Outliers
In addition to looking for a general pattern in a scatter plot, you should also look for other interesting features that might help you understand the relationship between two variables. Two things to watch for are as follows:
CLUSTERS: Usually, the points in a scatter plot form a single cloud of points, but sometimes the points may form two or more distinct clouds of points. These clouds are called clusters. Investigating these clusters may tell you something useful about the data.
OUTLIERS: An outlier is an unusual point in a scatter plot that does not seem to fit the general pattern or that is far away from the other points in the scatter plot.
The scatter plot below was constructed using data from a study of Rocky Mountain elk (“Estimating Elk Weight from Chest Girth,” Wildlife Society Bulletin, 1996). The variables studied were chest girth in centimeters (x) and weight in kilograms (y).
Engage NY Math Grade 8 Module 6 Lesson 7 Exercise Answer Key 10

Exercises 10–12

Exercise 10.
Do you notice any point in the scatter plot of elk weight versus chest girth that might be described as an outlier? If so, which one?
Answer:
Possible response: The point in the lower left-hand corner of the plot corresponding to an elk with a chest girth of about 96 cm and a weight of about 100 kg could be described as an outlier. There are no other points in the scatter plot that are near this one.

Exercise 11.
If you identified an outlier in Exercise 10, write a sentence describing how this data observation differs from the others in the data set.
Answer:
Possible response: This point corresponds to an observation for an elk that is much smaller than the other elk in the data set, both in terms of chest girth and weight.

Exercise 12.
Do you notice any clusters in the scatter plot? If so, how would you distinguish between the clusters in terms of chest girth? Can you think of a reason these clusters might have occurred?
Answer:
Possible response: Other than the outlier, there appear to be three clusters of points. One cluster corresponds to elk with chest girths between about 105 cm and 115 cm. A second cluster includes elk with chest girths between about 120 cm and 145 cm. The third cluster includes elk with chest girths above 150 cm. It may be that age and sex play a role. Maybe the cluster with the smaller chest girths includes young elk. The two other clusters might correspond to females and males if there is a difference in size for the two sexes for Rocky Mountain elk. If we had data on age and sex, we could investigate this further.

Eureka Math Grade 8 Module 6 Lesson 7 Problem Set Answer Key

Question 1.
Suppose data was collected on size in square feet (x) of several houses and price in dollars (y). The data was then used to construct the scatterplot below. Write a few sentences describing the relationship between price and size for these houses. Are there any noticeable clusters or outliers?
Eureka Math 8th Grade Module 6 Lesson 7 Problem Set Answer Key 1
Answer:
Answers will vary. Possible response: There appears to be a positive linear relationship between size and price. Price tends to increase as size increases. There appear to be two clusters of houses—one that includes houses that are less than 3,000 square feet in size and another that includes houses that are more than 3,000 square feet
in size.

Question 2.
The scatter plot below was constructed using data on length in inches (x) of several alligators and weight in pounds (y). Write a few sentences describing the relationship between weight and length for these alligators. Are there any noticeable clusters or outliers?
Eureka Math 8th Grade Module 6 Lesson 7 Problem Set Answer Key 2
Data Source: Exploring Data, Quantitative Literacy Series, James Landwehr and Ann Watkins, 1987.
Answer:
Answers will vary. Possible response: There appears to be a positive relationship between length and weight, but the relationship is not linear. Weight tends to increase as length increases. There are three observations that stand out as outliers. These correspond to alligators that are much bigger in terms of both length and weight than the other alligators in the sample. Without these three alligators, the relationship between length and weight would look linear. It might be possible to use a line to model the relationship between weight and length for alligators that have lengths of fewer than 100 inches.

Question 3.
Suppose the scatter plot below was constructed using data on age in years (x) of several Honda Civics and price in dollars (y). Write a few sentences describing the relationship between price and age for these cars. Are there any noticeable clusters or outliers?
Eureka Math 8th Grade Module 6 Lesson 7 Problem Set Answer Key 3
Answer:
Answers will vary. Possible response: There appears to be a negative linear relationship between price and age. Price tends to decrease as age increases. There is one car that looks like an outlier—the car that is 10 years old. This car has a price that is lower than expected based on the pattern of the other points in the scatter plot.

Question 4.
Samples of students in each of the U.S. states periodically take part in a large-scale assessment called the National Assessment of Educational Progress (NAEP). The table below shows the percent of students in the northeastern states (as defined by the U.S. Census Bureau) who answered Problems 7 and 15 correctly on the 2011 eighth-grade test. The scatter plot shows the percent of eighth-grade students who got Problems 7 and 15 correct on the 2011 NAEP.
Eureka Math 8th Grade Module 6 Lesson 7 Problem Set Answer Key 4
Eureka Math 8th Grade Module 6 Lesson 7 Problem Set Answer Key 5
a. Why does it appear that there are only eight points in the scatter plot for nine states?
Answer:
Two of the states, New Hampshire and Rhode Island, had exactly the same percent correct on each of the questions, (29,52).

b. What is true of the states represented by the cluster of five points in the lower left corner of the graph?
Answer:
Answers will vary; those states had lower percentages correct than the other three states in the upper right.

c. Which state did the best on these two problems? Explain your reasoning.
Answer:
Answers will vary; some students might argue that Massachusetts at (35,56) did the best. Even though Vermont actually did a bit better on Problem 15, it was lower on Problem 7.

d. Is there a trend in the data? Explain your thinking.
Answer:
Answers will vary; there seems to be a positive linear trend, as a large percent correct on one question suggests a large percent correct on the other, and a low percent on one suggests a low percent on the other.

Question 5.
The plot below shows the mean percent of sunshine during the year and the mean amount of precipitation in inches per year for the states in the United States.
Eureka Math 8th Grade Module 6 Lesson 7 Problem Set Answer Key 6
Data source: www.currentresults.com/Weather/US/average-annual-state-sunshine.php
www.currentresults.com/Weather/US/average-annual-state-precipitation.php
a. Where on the graph are the states that have a large amount of precipitation and a small percent of sunshine?
Answer:
Those states will be in the lower right-hand corner of the graph.

b. The state of New York is the point (46,41.8). Describe how the mean amount of precipitation and percent of sunshine in New York compare to the rest of the United States.
Answer:
New York has a little over 40 inches of precipitation per year and is sunny about 45% of the time. It has a smaller percent of sunshine over the year than most states and is about in the middle of the states in terms of the amount of precipitation, which goes from about 10 to 65 inches per year.

c. Write a few sentences describing the relationship between mean amount of precipitation and percent of sunshine.
Answer:
There is a negative relationship, or the more precipitation, the less percent of sun. If you took away the three states at the top left with a large percent of sun and very little precipitation, the trend would not be as pronounced. The relationship is not linear.

Question 6.
At a dinner party, every person shakes hands with every other person present.
a. If three people are in a room and everyone shakes hands with everyone else, how many handshakes take place?
Answer:
Three handshakes

b. Make a table for the number of handshakes in the room for one to six people. You may want to make a diagram or list to help you count the number of handshakes.
Eureka Math 8th Grade Module 6 Lesson 7 Problem Set Answer Key 7
Answer:
Eureka Math 8th Grade Module 6 Lesson 7 Problem Set Answer Key 8

c. Make a scatter plot of number of people (x) and number of handshakes (y). Explain your thinking.
Eureka Math 8th Grade Module 6 Lesson 7 Problem Set Answer Key 9
Answer:
Eureka Math 8th Grade Module 6 Lesson 7 Problem Set Answer Key 10

d. Does the trend seem to be linear? Why or why not?
Answer:
The trend is increasing, but it is not linear. As the number of people increases, the number of handshakes also increases. It does not increase at a constant rate.

Eureka Math Grade 8 Module 6 Lesson 7 Exit Ticket Answer Key

Question 1.
Which of the following scatter plots shows a negative linear relationship? Explain how you know.
Eureka Math Grade 8 Module 6 Lesson 7 Exit Ticket Answer Key 1
Answer:
Only Scatter Plot 3 shows a negative linear relationship because the y-values tend to decrease as the value of x increases.

Question 2.
The scatter plot below was constructed using data from eighth-grade students on number of hours playing video games per week (x) and number of hours of sleep per night (y). Write a few sentences describing the relationship between sleep time and time spent playing video games for these students. Are there any noticeable clusters or outliers?
Eureka Math Grade 8 Module 6 Lesson 7 Exit Ticket Answer Key 2
Answer:
Answers will vary. Sample response: There appears to be a negative linear relationship between the number of hours per week a student plays video games and the number of hours per night the student sleeps. As video game time increases, the number of hours of sleep tends to decrease. There is one observation that might be considered an outlier—the point corresponding to a student who plays video games 32 hours per week. Other than the outlier, there are two clusters—one corresponding to students who spend very little time playing video games and a second corresponding to students who play video games between about 10 and 25 hours per week.

Question 3.
In a scatter plot, if the values of y tend to increase as the value of x increases, would you say that there is a positive relationship or a negative relationship between x and y? Explain your answer.
Answer:
There is a positive relationship. If the value of y increases as the value of x increases, the points go up on the scatter plot from left to right.

Eureka Math Grade 8 Module 6 Lesson 6 Answer Key

Engage NY Eureka Math 8th Grade Module 6 Lesson 6 Answer Key

Eureka Math Grade 8 Module 6 Lesson 6 Example Answer Key

Example 1.
A bivariate data set consists of observations on two variables. For example, you might collect data on 13 different car models. Each observation in the data set would consist of an (x,y) pair.
x: weight (in pounds, rounded to the nearest 50 pounds)
and
y: fuel efficiency (in miles per gallon, mpg)
The table below shows the weight and fuel efficiency for 13 car models with automatic transmissions manufactured in 2009 by Chevrolet.
Engage NY Math 8th Grade Module 6 Lesson 6 Example Answer Key 1

Eureka Math Grade 8 Module 6 Lesson 6 Exercise Answer Key

Exercises 1–8

Exercise 1.
In the Example 1 table, the observation corresponding to Model 1 is (3200,23). What is the fuel efficiency of this car? What is the weight of this car?
Answer:
The fuel efficiency is 23 miles per gallon, and the weight is 3,200 pounds.

Exercise 2.
Add the points corresponding to the other 12 observations to the scatter plot.
Engage NY Math Grade 8 Module 6 Lesson 6 Exercise Answer Key 1
Answer:
Engage NY Math Grade 8 Module 6 Lesson 6 Exercise Answer Key 2

Exercise 3.
Do you notice a pattern in the scatter plot? What does this imply about the relationship between weight (x) and fuel efficiency (y)?
Answer:
There does seem to be a pattern in the plot. Higher weights tend to be paired with lesser fuel efficiencies, so it looks like heavier cars generally have lower fuel efficiency.

Is there a relationship between price and the quality of athletic shoes? The data in the table below are from the Consumer Reports website.
x: price (in dollars)
and
y: Consumer Reports quality rating
The quality rating is on a scale of 0 to 100, with 100 being the highest quality.
Engage NY Math Grade 8 Module 6 Lesson 6 Exercise Answer Key 3

Exercise 4.
One observation in the data set is (110,57). What does this ordered pair represent in terms of cost and quality?
Answer:
The pair represents a shoe that costs $110 with a quality rating of 57.

Exercise 5.
To construct a scatter plot of these data, you need to start by thinking about appropriate scales for the axes of the scatter plot. The prices in the data set range from $30 to $110, so one reasonable choice for the scale of the x-axis would range from $20 to $120, as shown below. What would be a reasonable choice for a scale for the y-axis?
Engage NY Math Grade 8 Module 6 Lesson 6 Exercise Answer Key 4
Answer:
Sample response: The smallest y-value is 51, and the largest y-value is 71. So, the y-axis could be scaled from 50 to 75.
Engage NY Math Grade 8 Module 6 Lesson 6 Exercise Answer Key 5

Exercise 6.
Add a scale to the y-axis. Then, use these axes to construct a scatter plot of the data.

Exercise 7.
Do you see any pattern in the scatter plot indicating that there is a relationship between price and quality rating for athletic shoes?
Answer:
Answers will vary. Students may say that they do not see a pattern, or they may say that they see a slight downward trend.

Exercise 8.
Some people think that if shoes have a high price, they must be of high quality. How would you respond?
Answer:
Answers will vary. The data do not support this. Students will either respond that there does not appear to be a relationship between price and quality, or if they saw a downward trend in the scatter plot, they might even indicate that the higher-priced shoes tend to have lower quality. Look for consistency between the answer to this question and how students answered the previous question.

Exercises 9–10

Exercise 9.
Data were collected on
x: shoe size
and
y: score on a reading ability test
for 29 elementary school students. The scatter plot of these data is shown below. Does there appear to be a statistical relationship between shoe size and score on the reading test?
Engage NY Math Grade 8 Module 6 Lesson 6 Exercise Answer Key 6
Answer:
Possible response: The pattern in the scatter plot appears to follow a line. As shoe sizes increase, the reading scores also seem to increase. There does appear to be a statistical relationship because there is a pattern in the scatter plot.

Exercise 10.
Explain why it is not reasonable to conclude that having big feet causes a high reading score. Can you think of a different explanation for why you might see a pattern like this?
Answer:
Possible response: You cannot conclude that just because there is a statistical relationship between shoe size and reading score that one causes the other. These data were for students completing a reading test for younger elementary school children. Older children, who would have bigger feet than younger children, would probably tend to score higher on a reading test for younger students.

Eureka Math Grade 8 Module 6 Lesson 6 Problem Set Answer Key

Question 1.
The table below shows the price and overall quality rating for 15 different brands of bike helmets.
Data source: www.consumerreports.org
Eureka Math 8th Grade Module 6 Lesson 6 Problem Set Answer Key 1
Construct a scatter plot of price (x) and quality rating (y). Use the grid below.
Eureka Math 8th Grade Module 6 Lesson 6 Problem Set Answer Key 2
Answer:
Eureka Math 8th Grade Module 6 Lesson 6 Problem Set Answer Key 3

Question 2.
Do you think that there is a statistical relationship between price and quality rating? If so, describe the nature of the relationship.
Answer:
Sample response: No. There is no pattern visible in the scatter plot. There does not appear to be a relationship between price and the quality rating for bike helmets.

Question 3.
Scientists are interested in finding out how different species adapt to finding food sources. One group studied crocodilian species to find out how their bite force was related to body mass and diet. The table below displays the information they collected on body mass (in pounds) and bite force (in pounds).
Eureka Math 8th Grade Module 6 Lesson 6 Problem Set Answer Key 4
Data Source: http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0031781#pone-0031781-t001
(Note: Body mass and bite force have been converted to pounds from kilograms and newtons, respectively.)
Construct a scatter plot of body mass (x) and bite force (y). Use the grid below, and be sure to add an appropriate scale to the axes.
Eureka Math 8th Grade Module 6 Lesson 6 Problem Set Answer Key 5
Answer:
Eureka Math 8th Grade Module 6 Lesson 6 Problem Set Answer Key 6

Question 4.
Do you think that there is a statistical relationship between body mass and bite force? If so, describe the nature of the relationship.
Answer:
Sample response: Yes, because it looks like there is an upward pattern in the scatter plot. It appears that alligators with larger body mass also tend to have greater bite force.

Question 5.
Based on the scatter plot, can you conclude that increased body mass causes increased bite force? Explain.
Answer:
Sample response: No. Just because there is a statistical relationship between body mass and bite force does not mean that there is a cause-and-effect relationship.

Eureka Math Grade 8 Module 6 Lesson 6 Exit Ticket Answer Key

Energy is measured in kilowatt-hours. The table below shows the cost of building a facility to produce energy and the ongoing cost of operating the facility for five different types of energy.
Eureka Math Grade 8 Module 6 Lesson 6 Exit Ticket Answer Key 1
Question 1.
Construct a scatter plot of the cost to build the facility in dollars per kilowatt-hour (x) and the cost to operate the facility in cents per kilowatt-hour (y). Use the grid below, and be sure to add an appropriate scale to the axes.
Eureka Math Grade 8 Module 6 Lesson 6 Exit Ticket Answer Key 2
Answer:
Eureka Math Grade 8 Module 6 Lesson 6 Exit Ticket Answer Key 3

Question 2.
Do you think that there is a statistical relationship between building cost and operating cost? If so, describe the nature of the relationship.
Answer:
Answers may vary. Sample response: Yes, because it looks like there is a downward pattern in the scatter plot. It appears that the types of energy that have facilities that are more expensive to build are less expensive to operate.

Question 3.
Based on the scatter plot, can you conclude that decreased building cost is the cause of increased operating cost? Explain.
Answer:
Sample response: No. Just because there may be a statistical relationship between cost to build and cost to operate does not mean that there is a cause-and-effect relationship.

Eureka Math Grade 8 Module 6 Lesson 5 Answer Key

Engage NY Eureka Math 8th Grade Module 6 Lesson 5 Answer Key

Eureka Math Grade 8 Module 6 Lesson 5 Exercise Answer Key

Exercises 1–2

Exercise 1.
In your own words, describe what is happening as Aleph is running during the following intervals of time.
a. 0 to 15 minutes
Answer:
Aleph runs 2 miles in 15 minutes.

b. 15 to 30 minutes
Answer:
Aleph runs another 2 miles in 15 minutes for a total of 4 miles.

c. 30 to 45 minutes
Answer:
Aleph runs another 2 miles in 15 minutes for a total of 6 miles.

d. 45 to 60 minutes
Answer:
Aleph runs another 2 miles in 15 minutes for a total of 8 miles.

Exercise 2.
In your own words, describe what is happening as Shannon is running during the following intervals of time.
a. 0 to 15 minutes
Answer:
Shannon runs 1.5 miles in 15 minutes.

b. 15 to 30 minutes
Answer:
Shannon runs another 0.6 mile in 15 minutes for a total of 2.1 miles.

c. 30 to 45 minutes
Answer:
Shannon runs another 0.5 mile in 15 minutes for a total of 2.6 miles.

d. 45 to 60 minutes
Answer:
Shannon runs another 0.4 mile in 15 minutes for a total of 3.0 miles.

Exercises 3–5

Exercise 3.
Different breeds of dogs have different growth rates. A large breed dog typically experiences a rapid growth rate from birth to age 6 months. At that point, the growth rate begins to slow down until the dog reaches full growth around 2 years of age.
a. Sketch a graph that represents the weight of a large breed dog from birth to 2 years of age.
Engage NY Math Grade 8 Module 6 Lesson 5 Exercise Answer Key 1
Answer:
Answers will vary.
Engage NY Math Grade 8 Module 6 Lesson 5 Exercise Answer Key 2

b. Is the function represented by the graph linear or nonlinear? Explain.
Answer:
The function is nonlinear because the growth rate is not constant.

c. Is the function represented by the graph increasing or decreasing? Explain.
Answer:
The function is increasing but at a decreasing rate. There is rapid growth during the first 6 months, and then the growth rate decreases.

Exercise 4.
Nikka took her laptop to school and drained the battery while typing a research paper. When she returned home, Nikka connected her laptop to a power source, and the battery recharged at a constant rate.
a. Sketch a graph that represents the battery charge with respect to time.
Engage NY Math Grade 8 Module 6 Lesson 5 Exercise Answer Key 3
Answer:
Answers will vary.
Engage NY Math Grade 8 Module 6 Lesson 5 Exercise Answer Key 4

b. Is the function represented by the graph linear or nonlinear? Explain.
Answer:
The function is linear because the battery is recharging at a constant rate.

c. Is the function represented by the graph increasing or decreasing? Explain.
Answer:
The function is increasing because the battery is being recharged.

Exercise 5.
The long jump is a track-and-field event where an athlete attempts to leap as far as possible from a given point. Mike Powell of the United States set the long jump world record of 8.95 meters (29.4 feet) during the 1991 World Championships in Tokyo, Japan.
a. Sketch a graph that represents the path of a high school athlete attempting the long jump.
Engage NY Math Grade 8 Module 6 Lesson 5 Exercise Answer Key 5
Answer:
Answers will vary
Engage NY Math Grade 8 Module 6 Lesson 5 Exercise Answer Key 6

b. Is the function represented by the graph linear or nonlinear? Explain.
Answer:
The function is nonlinear. The rate of change is not constant.

c. Is the function represented by the graph increasing or decreasing? Explain.
Answer:
The function both increases and decreases over different intervals. The function increases as the athlete begins the jump and reaches a maximum height. The function decreases after the athlete reaches maximum height and begins descending back toward the ground.

Exercises 6–9

Exercise 6.
Use the graph from Example 3 to answer the following questions.
a. Is the function represented by the graph linear or nonlinear?
Answer:
The function is nonlinear. The rate of change is not constant.

b. Where is the function increasing? What does this mean within the context of the problem?
Answer:
The function is increasing during the following intervals of time: 0 to 4 seconds, 8 to 12 seconds, 16 to 20 seconds, 24 to 28 seconds, and 32 to 36 seconds. It means that Lamar and his sister are rising in the air.

c. Where is the function decreasing? What does this mean within the context of the problem?
Answer:
The function is decreasing during the following intervals of time: 4 to 8 seconds, 12 to 16 seconds, 20 to 24 seconds, 28 to 32 seconds, and 36 to 40 seconds. Lamar and his sister are traveling back down toward the ground.

Exercise 7.
How high above the ground is the platform for passengers to get on the Ferris wheel? Explain your reasoning.
Answer:
The lowest point on the graph, which is at 5 feet, can represent the platform where the riders get on the Ferris wheel.

Exercise 8.
Based on the graph, how many revolutions does the Ferris wheel complete during the 40-second time interval? Explain your reasoning.
Answer:
The Ferris wheel completes 5 revolutions. The lowest points on the graph can represent Lamar and his sister at the beginning of a revolution or at the entrance platform of the Ferris wheel. So, one revolution occurs between 0 and 8 seconds, 8 and 16 seconds, 16 and 24 seconds, 24 and 32 seconds, and 32 and 40 seconds.

Exercise 9.
What is the diameter of the Ferris wheel? Explain your reasoning.
Answer:
The diameter of the Ferris wheel is 40 feet. The lowest point on the graph represents the base of the Ferris wheel, and the highest point on the graph represents the top of the Ferris wheel. The difference between the two values is 40 feet, which is the diameter of the wheel.

Eureka Math Grade 8 Module 6 Lesson 5 Problem Set Answer Key

Question 1.
Read through the following scenarios, and match each to its graph. Explain the reasoning behind your choice.
a. This shows the change in a smartphone battery charge as a person uses the phone more frequently.
b. A child takes a ride on a swing.
c. A savings account earns simple interest at a constant rate.
d. A baseball has been hit at a youth baseball game.
Scenario: ____
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 1
Answer:
Scenario: c

The savings account is earning interest at a constant rate, which means that the function is linear.

Scenario: ____
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 2
Answer:
Scenario: d

The baseball is hit into the air, reaches a maximum height, and falls back to the ground at a variable rate.

Scenario: ____
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 3
Answer:
Scenario: b

The distance from the ground increases as the child swings up into the air and then decreases as the child swings back down toward the ground.

Scenario: ____
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 4
Answer:
Scenario: a

The battery charge is decreasing as a person uses the phone more frequently.

Question 2.
The graph below shows the volume of water for a given creek bed during a 24-hour period. On this particular day, there was wet weather with a period of heavy rain.
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 5
Describe how each part (A, B, and C) of the graph relates to the scenario.
Answer:
A: The rain begins, and the volume of water flowing in the creek bed begins to increase.
B: A period of heavy rain occurs, causing the volume of water to increase.
C: The heavy rain begins to subside, and the volume of water continues to increase.

Question 3.
Half-life is the time required for a quantity to fall to half of its value measured at the beginning of the time period. If there are 100 grams of a radioactive element to begin with, there will be 50 grams after the first half-life, 25 grams after the second half-life, and so on.
a. Sketch a graph that represents the amount of the radioactive element left with respect to the number of half-lives that have passed.
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 6
Answer:
Answers will vary.
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 7

b. Is the function represented by the graph linear or nonlinear? Explain.
Answer:
The function is nonlinear. The rate of change is not constant with respect to time.

c. Is the function represented by the graph increasing or decreasing?
Answer:
The function is decreasing.

Question 4.
Lanae parked her car in a no-parking zone. Consequently, her car was towed to an impound lot. In order to release her car, she needs to pay the impound lot charges. There is an initial charge on the day the car is brought to the lot. However, 10% of the previous day’s charges will be added to the total charge for every day the car remains in the lot.
a. Sketch a graph that represents the total charges with respect to the number of days a car remains in the impound lot.
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 8
Answer:
Answers will vary.
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 9

b. Is the function represented by the graph linear or nonlinear? Explain.
Answer:
The function is nonlinear. The function is increasing.

c. Is the function represented by the graph increasing or decreasing? Explain.
Answer:
The function is increasing. The total charge is increasing as the number of days the car is left in the lot increases.

Question 5.
Kern won a $50 gift card to his favorite coffee shop. Every time he visits the shop, he purchases the same coffee drink.
a. Sketch a graph of a function that can be used to represent the amount of money that remains on the gift card with respect to the number of drinks purchased.
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 10
Answer:
Answers will vary.
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 11

b. Is the function represented by the graph linear or nonlinear? Explain.
Answer:
The function is linear. Since Kern purchases the same drink every visit, the balance is decreasing by the same amount or, in other words, at a constant rate of change.

c. Is the function represented by the graph increasing or decreasing? Explain.
Answer:
The function is decreasing. With each drink purchased, the amount of money on the card decreases.

Question 6.
Jay and Brooke are racing on bikes to a park 8 miles away. The tables below display the total distance each person biked with respect to time.
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 12
a. Which person’s biking distance could be modeled by a nonlinear function? Explain.
Answer:
The distance that Jay biked could be modeled by a nonlinear function because the rate of change is not constant. The distance that Brooke biked could be modeled by a linear function because the rate of change is constant.

b. Who would you expect to win the race? Explain.
Answer:
Jay will win the race. The distance he bikes during each five-minute interval is increasing, while Brooke’s biking distance remains constant. If the trend remains the same, it is estimated that both Jay and Brooke will travel about 7.2 miles in 30 minutes. So, Jay will overtake Brooke during the last 5 minutes to win the race.

Question 7.
Using the axes in Problem 7(b), create a story about the relationship between two quantities.
a. Write a story about the relationship between two quantities. Any quantities can be used (e.g., distance and time, money and hours, age and growth). Be creative! Include keywords in your story such as increase and decrease to describe the relationship.
Answer:
Answers will vary.
A person in a car is at a red stoplight. The light turns green, and the person presses down on the accelerator with increasing pressure. The car begins to move and accelerate. The rate at which the car accelerates is
not constant.

b. Label each axis with the quantities of your choice, and sketch a graph of the function that models the relationship described in the story.
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 13
Answer:
Answers will vary based on the story from part (a).
Eureka Math 8th Grade Module 6 Lesson 5 Problem Set Answer Key 14

Eureka Math Grade 8 Module 6 Lesson 5 Exit Ticket Answer Key

Question 1.
Lamar and his sister continue to ride the Ferris wheel. The graph below represents Lamar and his sister’s distance above the ground with respect to time during the next 40 seconds of their ride.
Eureka Math Grade 8 Module 6 Lesson 5 Exit Ticket Answer Key 1
a. Name one interval where the function is increasing.
Answer:
The function is increasing during the following intervals of time: 40 to 44 seconds, 48 to 52 seconds, 56 to 60 seconds, and 72 to 76 seconds.

b. Name one interval where the function is decreasing.
Answer:
The function is decreasing during the following intervals of time: 44 to 48 seconds, 52 to 56 seconds, 64 to 66 seconds, 70 to 72 seconds, and 76 to 80 seconds.

c. Is the function linear or nonlinear? Explain.
Answer:
The function is both linear and nonlinear during different intervals of time. It is linear from 60 to 64 seconds and from 66 to 70 seconds. It is nonlinear from 40 to 60 seconds and from 70 to 80 seconds.

d. What could be happening during the interval of time from 60 to 64 seconds?
Answer:
The Ferris wheel is not moving during that time, so riders may be getting off or getting on.

e. Based on the graph, how many complete revolutions are made during this 40-second interval?
Answer:
Four revolutions are made during this time period.

Eureka Math Grade 8 Module 6 Lesson 4 Answer Key

Engage NY Eureka Math 8th Grade Module 6 Lesson 4 Answer Key

Eureka Math Grade 8 Module 6 Lesson 4 Exercise Answer Key

Exercise 1.
Read through each of the scenarios, and choose the graph of the function that best matches the situation. Explain the reason behind each choice.
a. A bathtub is filled at a constant rate of 1.75 gallons per minute.
b. A bathtub is drained at a constant rate of 2.5 gallons per minute.
c. A bathtub contains 2.5 gallons of water.
d. A bathtub is filled at a constant rate of 2.5 gallons per minute.
Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 1
Scenario:
Explanation:
Answer:
Scenario: c

Explanation: The amount of water in the tub does not change over time; it remains constant at 2.5 gallons.

Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 2
Scenario:
Explanation:
Answer:
Scenario: b

Explanation: The bathtub is being drained at a constant rate of 2.5 gallons per minute. So, the amount of water is decreasing, which means that the slope of the line is negative. The graph of the function also shows that there are initially 20 gallons of water in the tub.

Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 3
Scenario:
Explanation:
Answer:
Scenario: d

Explanation: The tub is being filled at a constant rate of 2.5 gallons per minute, which implies that the amount of water in the tub is increasing, so the line has a positive slope. Based on the graph, the amount of water is also increasing at a faster rate than in choice (a).

Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 4
Scenario:
Explanation:
Answer:
Scenario: d

Explanation: The tub is being filled at a constant rate of 2.5 gallons per minute, which implies that the amount of water in the tub is increasing, so the line has a positive slope. Based on the graph, the amount of water is also increasing at a faster rate than in choice (a).

Exercise 2.
Read through each of the scenarios, and sketch a graph of a function that models the situation.
a. A messenger service charges a flat rate of $4.95 to deliver a package regardless of the distance to the destination.
Answer:
Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 5
Answer:
The delivery charge remains constant regardless of the distance to the destination.
Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 6

b. At sea level, the air that surrounds us presses down on our bodies at 14.7 pounds per square inch (psi). For every 10 meters that you dive under water, the pressure increases by 14.7 psi.
Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 7
Answer:
The initial value is 14.7 psi. The function increases at a rate of 14.7 psi for every 10 meters, or 1.47 psi per meter.
Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 8

c. The range (driving distance per charge) of an electric car varies based on the average speed the car is driven. The initial range of the electric car after a full charge is 400 miles. However, the range is reduced by 20 miles for every 10 mph increase in average speed the car is driven.
Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 9
Answer:
The initial value of the function is 400. The function is decreasing by 20 miles for every 10 mph increase in speed. In other words, the function decreases by 2 miles for every 1 mph increase in speed.
Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 10

Exercise 3.
The graph below represents the total number of smartphones that are shipped to a retail store over the course of 50 days.
Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 11
Match each part of the graph (A, B, and C) to its verbal description. Explain the reasoning behind your choice.
i. Half of the factory workers went on strike, and not enough smartphones were produced for normal shipments.
Answer:
C; if half of the workers went on strike, then the number of smartphones produced would be less than normal. The rate of change for C is less than the rate of change for A.

ii. The production schedule was normal, and smartphones were shipped to the retail store at a constant rate.
Answer:
A; if the production schedule is normal, the rate of change of interval A is greater than the rate of change of interval C.

iii. A defective electronic chip was found, and the factory had to shut down, so no smartphones were shipped.
Answer:
B; if no smartphones are shipped to the store, the total number remains constant during that time.

Exercise 4.
The relationship between Jameson’s account balance and time is modeled by the graph below.
Engage NY Math Grade 8 Module 6 Lesson 4 Exercise Answer Key 12
a. Write a story that models the situation represented by the graph.
Answer:
Answers will vary.
Jameson was sick and did not work for almost a whole week. Then, he mowed several lawns over the next few days and deposited the money into his account after each job. It rained several days, so instead of working, Jameson withdrew money from his account each day to go to the movies and out to lunch with friends.

b. When is the function represented by the graph increasing? How does this relate to your story?
Answer:
It is increasing between 6 and 9 days. Jameson earned money mowing lawns and made a deposit to his account each day. The money earned for each day was constant for these days. This is represented by a straight line.

c. When is the function represented by the graph decreasing? How does this relate to your story?
Answer:
It is decreasing between 9 and 14 days. Since Days 9–14 are represented by a straight line, this means that Jameson spent the money constantly over these days. Jameson cannot work because it is raining. Perhaps he withdraws money from his account to spend on different activities each day because he cannot work.

Eureka Math Grade 8 Module 6 Lesson 4 Problem Set Answer Key

Question 1.
Read through each of the scenarios, and choose the graph of the function that best matches the situation. Explain the reason behind each choice.
a. The tire pressure on Regina’s car remains at 30 psi.
b. Carlita inflates her tire at a constant rate for 4 minutes.
c. Air is leaking from Courtney’s tire at a constant rate.
Eureka Math 8th Grade Module 6 Lesson 4 Problem Set Answer Key 1
Scenario:
Explanation:
Answer:
Scenario: c

Explanation: The tire pressure decreases each minute at a constant rate.

Eureka Math 8th Grade Module 6 Lesson 4 Problem Set Answer Key 2
Scenario:
Explanation:
Answer:
Scenario: a

Explanation: The tire pressure remains at 30 psi.

Eureka Math 8th Grade Module 6 Lesson 4 Problem Set Answer Key 3
Scenario:
Explanation:
Answer:
Scenario: b

Explanation: The tire pressure is increasing each minute at a constant rate for 4 minutes.

Question 2.
A home was purchased for $275,000. Due to a recession, the value of the home fell at a constant rate over the next 5 years.
a. Sketch a graph of a function that models the situation.
Eureka Math 8th Grade Module 6 Lesson 4 Problem Set Answer Key 4
Answer:
Graphs will vary; a sample graph is provided.
Eureka Math 8th Grade Module 6 Lesson 4 Problem Set Answer Key 5

b. Based on your graph, how is the home value changing with respect to time?
Answer:
Answers will vary; a sample answer is provided.
The value is decreasing by $25,000 over 5 years or at a constant rate of $5,000 per year.

Question 3.
The graph below displays the first hour of Sam’s bike ride.
Eureka Math 8th Grade Module 6 Lesson 4 Problem Set Answer Key 6
Match each part of the graph (A, B, and C) to its verbal description. Explain the reasoning behind your choice.
i. Sam rides his bike to his friend’s house at a constant rate.
Answer:
A; the distance from home should be increasing as Sam is riding toward his friend’s house.

ii. Sam and his friend bike together to an ice cream shop that is between their houses.
Answer:
C; Sam was at his friend’s house, but as they start biking to the ice cream shop, the distance from Sam’s home begins to decrease.

iii. Sam plays at his friend’s house.
Answer:
B; Sam remains at the same distance from home while he is at his friend’s house.

Question 4.
Using the axes below, create a story about the relationship between two quantities.
a. Write a story about the relationship between two quantities. Any quantities can be used (e.g., distance and time, money and hours, age and growth). Be creative. Include keywords in your story such as increase and decrease to describe the relationship.
Answer:
Answers will vary. Give students the freedom to write a basic linear story or a piecewise story.
A rock climber begins her descent from a height of 50 feet. She slowly descends at a constant rate for 4 minutes. She takes a break for 1 minute; she then realizes she left some of her gear on top of the rock and climbs more quickly back to the top at a constant rate.

b. Label each axis with the quantities of your choice, and sketch a graph of the function that models the relationship described in the story.
Eureka Math 8th Grade Module 6 Lesson 4 Problem Set Answer Key 7
Answer:
Answers will vary based on the story from part (a).
Eureka Math 8th Grade Module 6 Lesson 4 Problem Set Answer Key 8

Eureka Math Grade 8 Module 6 Lesson 4 Exit Ticket Answer Key

Question 1.
The graph below shows the relationship between a car’s value and time.
Eureka Math Grade 8 Module 6 Lesson 4 Exit Ticket Answer Key 1
Match each part of the graph (A, B, and C) to its verbal description. Explain the reasoning behind your choice.
i. The value of the car holds steady due to a positive consumer report on the same model.
Answer:
B; if the value is holding steady, there is no change in the car’s value between years.

ii. There is a shortage of used cars on the market, and the value of the car rises at a constant rate.
Answer:
C; if the value of the car is rising, it represents an increasing function.

iii. The value of the car depreciates at a constant rate.
Answer:
A; if the value depreciates, it represents a decreasing function.

Question 2.
Henry and Roxy both drive electric cars that need to be recharged before use. Henry uses a standard charger at his home to recharge his car. The graph below represents the relationship between the battery charge and the amount of time it has been connected to the power source for Henry’s car.
Eureka Math Grade 8 Module 6 Lesson 4 Exit Ticket Answer Key 2
Answer:
Eureka Math Grade 8 Module 6 Lesson 4 Exit Ticket Answer Key 3
a. Describe how Henry’s car battery is being recharged with respect to time.
Answer:
The battery charge is increasing at a constant rate of 10% every 10 minutes.

b. Roxy has a supercharger at her home that can charge about half of the battery in 20 minutes. There is no remaining charge left when she begins recharging the battery. Sketch a graph that represents the relationship between the battery charge and the amount of time on the axes above. Assume the relationship is linear.
Answer:
See the above graph.

c. Which person’s car will be recharged to full capacity first? Explain.
Answer:
Roxy’s car will be completely recharged first. Her supercharger has a greater rate of change compared to Henry’s charger.

Eureka Math Grade 8 Module 7 Lesson 1 Answer Key

Engage NY Eureka Math 8th Grade Module 7 Lesson 1 Answer Key

Eureka Math Grade 8 Module 7 Lesson 1 Example Answer Key

Example 1.
Write an equation that allows you to determine the length of the unknown side of the right triangle.
Engage NY Math 8th Grade Module 7 Lesson 1 Example Answer Key 1
Answer:
Write an equation that allows you to determine the length of the unknown side of the right triangle.
Note: Students may use a different symbol to represent the unknown side length.
Let b cm represent the unknown side length. Then, 52 + b2 = 132.
Verify that students wrote the correct equation; then, allow them to solve it. Ask them how they knew the correct answer was 12. They should respond that 132-52 = 144, and since 144 is a perfect square, they knew that the unknown side length must be 12 cm.

Example 2.
Write an equation that allows you to determine the length of the unknown side of the right triangle.
Engage NY Math 8th Grade Module 7 Lesson 1 Example Answer Key 2
Answer:
→ Write an equation that allows you to determine the length of the unknown side of the right triangle.
Let c cm represent the length of the hypotenuse. Then, 42 + 92 = c2.

→ There is something different about this triangle. What is the length of the missing side? If you cannot find the length of the missing side exactly, then find a good approximation.
Provide students time to find an approximation for the length of the unknown side. Select students to share their answers and explain their reasoning. Use the points below to guide their thinking as needed.

→ How is this problem different from the last one?
The answer is c2 = 97. Since 97 is not a perfect square, the exact length cannot be represented as an integer.

→ Since 97 is not a perfect square, we cannot determine the exact length of the hypotenuse as an integer; however, we can make an estimate. Think about all of the perfect squares we have seen and calculated in past discussions. The number 97 is between which two perfect squares?
The number 97 is between 81 and 100.
If c2 were 81, what would be the length of the hypotenuse?
The length would be 9 cm.
If c2 were 100, what would be the length of the hypotenuse?
The length would be 10 cm.

→ At this point, we know that the length of the hypotenuse is somewhere between 9 cm and 10 cm. Think about the length to which it is closest. The actual length of the hypotenuse is determined by the equation c2 = 97. To which perfect square number, 100 or 81, is 97 closer?
The number 97 is closer to the perfect square 100 than to the perfect square 81.

→ Now that we know that the length of the hypotenuse of this right triangle is between 9 cm and 10 cm, but closer to 10 cm, let’s try to get an even better estimate of the length. Choose a number between 9 and 10 but closer to 10. Square that number. Do this a few times to see how close you can get to the number 97.

Provide students time to check a few numbers between 9 and 10. Students should see that the length is somewhere between 9.8 cm and 9.9 cm because 9.82 = 96.04 and 9.92 = 98.01. Some students may even check 9.85; 9.852 = 97.0225. This activity shows students that an estimation of the length being between 9 cm and 10 cm is indeed accurate, and it helps students develop an intuitive sense of how to estimate square roots.

Example 3.
Write an equation to determine the length of the unknown side of the right triangle.
Engage NY Math 8th Grade Module 7 Lesson 1 Example Answer Key 3
Answer:
→ Write an equation to determine the length of the unknown side of the right triangle.
Let c cm represent the length of the hypotenuse. Then, 32 + 82 = c2.
Verify that students wrote the correct equation, and then allow them to solve it. Instruct them to estimate the length, if necessary. Then, let them continue to work. When most students have finished, ask the questions below.

→ Could you determine an answer for the length of the hypotenuse as an integer?
No. Since c2 = 73, the length of the hypotenuse is not a perfect square.
Optionally, you can ask, “Can anyone find the exact length of the hypotenuse as a rational number?” It is important that students recognize that no one can determine the exact length of the hypotenuse as a rational number at this point.

→ Since 73 is not a perfect square, we cannot determine the exact length of the hypotenuse as a whole number. Let’s estimate the length. Between which two whole numbers is the length of the hypotenuse? Explain.
Since 73 is between the two perfect squares 64 and 81, we know the length of the hypotenuse must be between 8 cm and 9 cm.
→ Is the length closer to 8 cm or 9 cm? Explain.
The length is closer to 9 cm because 73 is closer to 81 than it is to 64.
→ The length of the hypotenuse is between 8 cm and 9 cm but closer to 9 cm.

Example 4.
In the figure below, we have an equilateral triangle with a height of 10 Inches. What do we know about an equilateral triangle?
Engage NY Math 8th Grade Module 7 Lesson 1 Example Answer Key 4
Answer:
→ In the figure below, we have an equilateral triangle with a height of 10 inches. What do we know about an equilateral triangle?
Equilateral triangles have sides that are all of the same length and angles that are all of the same degree, namely 60°.
Let’s say the length of the sides is x inches. Determine the approximate length of the sides of the triangle.
Engage NY Math 8th Grade Module 7 Lesson 1 Example Answer Key 5
→ What we actually have here are two congruent right triangles.
Trace one of the right triangles on a transparency, and reflect it across the line representing the height of the triangle to convince students that an equilateral triangle is composed of two congruent right triangles.
Engage NY Math 8th Grade Module 7 Lesson 1 Example Answer Key 6
With this knowledge, we need to determine the length of the base of one of the right triangles. If we know that the length of the base of the equilateral triangle is x inches, then what is the length of the base of one of the right triangles? Explain.
Engage NY Math 8th Grade Module 7 Lesson 1 Example Answer Key 7

Eureka Math Grade 8 Module 7 Lesson 1 Exercise Answer Key

Exercise 1.
Use the Pythagorean theorem to find a whole number estimate of the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Engage NY Math Grade 8 Module 7 Lesson 1 Exercise Answer Key 1
Answer:
Let x cm be the length of the unknown side.
62 + x2 = 112
36 + x2 = 121
x2 = 85
The length of the unknown side of the triangle is approximately 9 cm. The number 85 is between the perfect squares 81 and 100. Since 85 is closer to 81 than 100, then the length of the unknown side of the triangle is closer to 9 cm than it is to 10 cm.

Exercise 2.
Use the Pythagorean theorem to find a whole number estimate of the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Engage NY Math Grade 8 Module 7 Lesson 1 Exercise Answer Key 2
Answer:
Let c in. be the length of the hypotenuse.
62 + 102 = c2
36 + 100 = c2
136 = c2
The length of the hypotenuse is approximately 12 in. The number 136 is between the perfect squares 121 and 144. Since 136 is closer to 144 than 121, the length of the unknown side of the triangle is closer to 12 in. than it is to 11 in.

Exercise 3.
Use the Pythagorean theorem to find a whole number estimate of the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Engage NY Math Grade 8 Module 7 Lesson 1 Exercise Answer Key 3
Answer:
Let x mm be the length of the unknown side.
92 + x2 = 112
81 + x2 = 121
x2 = 40
The length of the hypotenuse is approximately 6 mm. The number 40 is between the perfect squares 36 and 49. Since 40 is closer to 36 than 49, then the length of the unknown side of the triangle is closer to 6 mm than it is to 7 mm.

Eureka Math Grade 8 Module 7 Lesson 1 Problem Set Answer Key

Question 1.
Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Eureka Math 8th Grade Module 7 Lesson 1 Problem Set Answer Key 1
Answer:
Let x in. be the length of the unknown side.
132 + x2 = 152
169 + x2 = 225
x2 = 56
The number 56 is between the perfect squares 49 and 64. Since 56 is closer to 49 than it is to 64, the length of the unknown side of the triangle is closer to 7 in. than it is to 8 in.

Question 2.
Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Eureka Math 8th Grade Module 7 Lesson 1 Problem Set Answer Key 2
Answer:
Let x cm be the length of the unknown side.
x2 + 122 = 132
x2 + 144 = 169
x2 = 25
x = 5
The length of the unknown side is 5 cm. The Pythagorean theorem led me to the fact that the square of the value of the unknown length is 25. Since 25 is a perfect square, 25 is equal to 52; therefore, x = 5, and the unknown length of the triangle is 5 cm.

Question 3.
Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Eureka Math 8th Grade Module 7 Lesson 1 Problem Set Answer Key 3
Answer:
Let c in. be the length of the hypotenuse.
42 + 122 = c2
16 + 144 = c2
160 = c2
The number 160 is between the perfect squares 144 and 169. Since 160 is closer to 169 than it is to 144, the length of the hypotenuse of the triangle is closer to 13 in. than it is to 12 in.

Question 4.
Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Eureka Math 8th Grade Module 7 Lesson 1 Problem Set Answer Key 4
Answer:
Let x cm be the length of the unknown side.
x2 + 112 = 132
x2 + 121 = 169
x2 = 48
The number 48 is between the perfect squares 36 and 49. Since 48 is closer to 49 than it is to 36, the length of the unknown side of the triangle is closer to 7 cm than it is to 6 cm.

Question 5.
Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Eureka Math 8th Grade Module 7 Lesson 1 Problem Set Answer Key 5
Answer:
Let c in. be the length of the hypotenuse.
62 + 82 = c2
36 + 64 = c2
100 = c2
10 = c
The length of the hypotenuse is 10 in. The Pythagorean theorem led me to the fact that the square of the value of the unknown length is 100. We know 100 is a perfect square, and 100 is equal to 102; therefore, c = 10, and the length of the hypotenuse of the triangle is 10 in.

Question 6.
Determine the length of the unknown side of the right triangle. Explain how you know your answer is correct.
Eureka Math 8th Grade Module 7 Lesson 1 Problem Set Answer Key 6
Answer:
Let c cm be the length of the hypotenuse.
72 + 42 = c2
49 + 16 = c2
65 = c2
The number 65 is between the perfect squares 64 and 81. Since 65 is closer to 64 than it is to 81, the length of the hypotenuse of the triangle is closer to 8 cm than it is to 9 cm.

Question 7.
Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Eureka Math 8th Grade Module 7 Lesson 1 Problem Set Answer Key 7
Answer:
Let x mm be the length of the unknown side.
32 + x2 = 122
9 + x2 = 144
x2 = 135
The number 135 is between the perfect squares 121 and 144. Since 135 is closer to 144 than it is to 121, the length of the unknown side of the triangle is closer to 12 mm than it is to 11 mm.

Question 8.
The triangle below is an isosceles triangle. Use what you know about the Pythagorean theorem to determine the approximate length of the base of the isosceles triangle.
Eureka Math 8th Grade Module 7 Lesson 1 Problem Set Answer Key 8
Answer:
Let x ft. represent the length of the base of one of the right triangles of the isosceles triangle.
x2 + 72 = 92
x2 + 49 = 81
x2 = 32
Since 32 is between the perfect squares 25 and 36 but closer to 36, the approximate length of the base of the right triangle is 6 ft. Since there are two right triangles, the length of the base of the isosceles triangle is approximately 12 ft.

Question 9.
Give an estimate for the area of the triangle shown below. Explain why it is a good estimate.
Eureka Math 8th Grade Module 7 Lesson 1 Problem Set Answer Key 9
Answer:
Let x cm represent the length of the base of the right triangle.
x2 + 32 = 72
x2 + 9 = 49
x2 = 40
Since 40 is between the perfect squares 36 and 49 but closer to 36, the approximate length of the base is 6 cm.
A = \(\frac{1}{2}\)(6)(3) = 9
So, the approximate area of the triangle is 9 cm2. This is a good estimate because of the approximation of the length of the base. Further, since the hypotenuse is the longest side of the right triangle, approximating the length of the base as 6 cm makes mathematical sense because it has to be shorter than the hypotenuse.

Eureka Math Grade 8 Module 7 Lesson 1 Exit Ticket Answer Key

Question 1.
Determine the length of the unknown side of the right triangle. If you cannot determine the length exactly, then determine which two integers the length is between and the integer to which it is closest.
Eureka Math Grade 8 Module 7 Lesson 1 Exit Ticket Answer Key 1
Answer:
Let x in. be the length of the unknown side.
92 + x2 = 152
81 + x2 = 225
x2 = 144
x = 12
The length of the unknown side is 12 in. The Pythagorean theorem led me to the fact that the square of the value of the unknown length is 144. We know 144 is a perfect square, and 144 is equal to 122; therefore, x = 12, and the unknown length of the triangle is 12 in.

Question 2.
Determine the length of the unknown side of the right triangle. If you cannot determine the length exactly, then determine which two integers the length is between and the integer to which it is closest.
Eureka Math Grade 8 Module 7 Lesson 1 Exit Ticket Answer Key 2
Answer:
Let x mm be the length of the unknown side.
22 + 72 = x2
4 + 49 = x2
53 = x2
The number 53 is between the perfect squares 49 and 64. Since 53 is closer to 49 than 64, the length of the unknown side of the triangle is closer to 7 mm than 8 mm.

PEMDAS Rule Involving Decimals – Definition, Examples | How to Solve PEMDAS?

PEMDAS Rule Involving Decimals

PEMDAS Rule – Involving Decimals helps you to solve all the decimal operations easily in minutes. Improve your problem-solving skills using PEMDAS Rule while calculating decimal operations. The order of operations is also the same for the calculation of the decimal number using the PEMDAS Rule. The order of operations is Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction. Check out different examples and also the order of operation in the below article.

Do Refer:

How to Solve PEMDAS(Order of Operations)?

1. Parenthesis Terms
We have to give the first importance to the parenthesis terms while simplifying an arithmetic expression. For example
0.3 + (0.2 × 0.2).
In the above, expression the parenthesis term is (0.2 × 0.2). By simplifying the term. We will get
0.3 + 0.04.
0.34 (addition).

2. Exponent Terms
After solving the parenthesis terms, we have to give priority to exponent terms in the calculation. For example
0.5 × (0.6 + 0.2) + (0.2)².
First, simplify the parenthesis term. That is
0.5 × 0.8 + (0.2)².
In the above expression, the exponent term is (0.2)². By simplifying the exponent term, we will get
0.5 × 0.8 + 0.04.
0.4 + 0.04 (Multiplication 0.5 × 0.8 = 0.4).
0.44 (addition).

3. Multiplication Terms
While simplifying an arithmetic expression, we need to give the third priority for multiplication terms. For example
0.1 × [(0.2 + 0.5) + (0.8 – 0.4)] + (0.3 + 0.3)².
Simplify the parenthesis terms. That is
0.1×[ 0.7 + 0.4 ] + (0.6)².
0.1×1.1 + (0.6)². (again Parenthesis term simplification).
Next, simplify the Exponent terms. That is
0.1× 1.1 + 0.36.
Now, simplify the Multiplication term 0.1× 1.1. we will get
0.11 + 0.36 = 0.47.

4. Division Simplification.
The next priority while simplification of expression is division terms. For example
0.3 × (0.25 + 0.12) ÷ 0.1 + (0.24)².
First, simplify the parenthesis terms. That is
0.3 × 0.37 ÷ 0.1 + (0.24)².
The second priority for exponent terms. That is
0.3 × 0.37 ÷ 0.1 + 0.0576.
Simplify the Multiplication Terms. That is
0.111 ÷ 0.1 + 0.0576.
Now, Simplify the division terms. We will get
1.11 + 0.0576.
1.1676.

5. Addition Terms Simplification.
The fifth priority while calculating an arithmetic expression is addition terms. For example
0.5 + 0.25 ÷ 0.1 + (0.8)² + (0.2 + 0.6).
Simplify the parenthesis terms. We will get
0.5 + 0.25 ÷ 0.1 + (0.8)² + 0.8.
Exponent terms simplification.
0.5 + 0.25 ÷ 0.1 + 0.64 + 0.8.
Division terms simplification.
0.5 + 2.5 + 0.64 + 0.8.
Now, simplify the additional terms. That is
1.44.

6. Subtraction Terms Simplification.
The last priority while the simplification process is subtraction. For example
0.4 + (0.2)³ × 0.1 ÷ 0.2 – 0.25.
Simplify the exponent terms first. That is
0.4 + 0.008× 0.1 ÷ 0.2 – 0.25.
Multiplication terms simplification.
0.4 + 0.0008 ÷ 0.2 – 0.25.
Simplify the division terms.
0.4 + 0.004 – 0.25.
Addition terms simplification.
0.404 – 0.25.
Subtraction simplification.
0.154.

PEMDAS Rule Involving Decimals Worked Out Problems

1. Simplify the given expressions by using the PEMDAS Rule
(i) 10.5 – 0.5 × 0.25 + (0.5)² + (0.4 + 0.6).
(ii) 12.5 + 0.6 × 0.23 – 0.89 + (1.25 – 0.2).
(iii) 20.25 ÷ 0.5 + 0.26 – 6.45 × (0.3)².
(iv) 50.6 – 25.2 × 0.45 + 0.5 + (0.2 – 0.1)³.

(i) 10.5 – 0.5 × 0.25 + (0.5)² + (0.4 + 0.6).

Solution:

The given expression is 10.5 – 0.5 × 0.25 + (0.5)² + (0.4 + 0.6).
As per the PEMDAS rule, we have to follow the order of operations while simplifying the expression. That is, Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction.
10.5 – 0.5 × 0.25 + (0.5)² + (0.4 + 0.6).
10.5 – 0.5 × 0.25 + (0.5)² + 1.
10.5 – 0.5 × 0.25 + 0.25 + 1.
10.5 –0.125 + 0.25 + 1.
11.75 – 0.125.
11.625.
By simplifying the expression 10.5 – 0.5 × 0.25 + (0.5)² + (0.4 + 0.6), we will get the resultant value as 11.625.

(ii) 12.5 + 0.6 × 0.23 – 0.89 + (1.25 – 0.2).

Solution:

The given expression is 12.5 + 0.6 × 0.23 – 0.89 + (1.25 – 0.2).
As per the PEMDAS rule, we have to follow the order of operations while simplifying the expression. That is Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction.
12.5 + 0.6 × 0.23 – 0.89 + (1.25 – 0.2).
12.5 + 0.6 × 0.23 – 0.89 + 1.05.
12.5 + 0.138 – 0.89 + 1.05.
13.688 – 0.89.
12.798.
By simplifying the expression 12.5 + 0.6 × 0.23 – 0.89 + (1.25 – 0.2)
we will get the resultant value as 12.798.

(iii) 20.25 ÷ 0.5 + 0.26 – 6.45 × (0.3)².

Solution:

The given expression is 20.25 ÷ 0.5 + 0.26 – 6.45 × (0.3)².
As per the PEMDAS rule, we have to follow the order of operations while simplifying the expression. That is, Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction.
20.25 ÷ 0.5 + 0.26 – 6.45 × (0.3)².
20.25 ÷ 0.5 + 0.26 – 6.45 × 0.09.
20.25 ÷ 0.5 + 0.26 – 0.5805.
40.5 + 0.26 – 0.5805.
40.76 – 0.5805.
40.1795.
By simplifying the expression 20.25 ÷ 0.5 + 0.26 – 6.45 × (0.3)²
we will get the resultant value as 40.1795.

(iv) 50.6 – 25.2 × 0.45 + 0.5 + (0.2 – 0.1)³.

Solution:

The given expression is 50.6 – 25.2 × 0.45 + 0.5 + (0.2 – 0.1)³.
As per the PEMDAS rule, we have to follow the order of operations while simplifying the expression. That is, Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction.
50.6 – 25.2 × 0.45 + 0.5 + (0.2 – 0.1)³.
50.6 – 25.2 × 0.45 + 0.5 + (0.1)³.
50.6 – 25.2 × 0.45 + 0.5 + 0.001.
50.6 – 11.34 + 0.5 + 0.001.
51.101 – 11.34.
39.761.
By simplifying the expression 50.6 – 25.2 × 0.45 + 0.5 + (0.2 – 0.1)³ we will get the resultant value as 39.761.

PEMDAS Rules Involving Integers – Definition, Examples | How to Simplify PEMDAS Involving Integers?

PEMDAS Rule involving Integers

We have multiple operations in one arithmetic expression. To simplify the expression very easily, we have to follow the PEMDAS rules. The PEMDAS rule is an order of operations. By using the PEMDAS Rule, we can know the priority levels of the operations in mathematics. So, we can easily simplify the arithmetic expressions involves integers. Check out the PEMDAS Rules applied to the integers and solving methods to find the answers.

Also, Read:

How to do Order of Operations with Integers?

1. First priority for Parenthesis terms {}, (), [].
For example, 2 × (10+15).
2 × 25 (parenthesis term first).
50 (multiplication).

2. Second Priority for Exponent Terms a².
For example, 10 + (20 × 2) + 4².
10 + (20 × 2) + 4² = 10 + 40 + 4² (Parenthesis term 20 × 2 = 40).
= 10 + 40 + 16 (Exponent Term 4² = 16 second priority).
= 66 (addition).

3. Third Priority for Multiplication a × b.
For example, 25 × 2 + (10 + 2) + 2².
25 × 2 + 12 + 2² (Parenthesis first 10 +2 = 12).
25 × 2 + 12 + 4 (Exponent Second 2² = 4).
50 + 12 + 4 (Multiplication third 25 × 2 = 50).
66 (addition).

4. Fourth Priority for Division a ÷ b.
For example, 100 ÷ 2 × 25 + (20 × 2) + (10 + 2)².
100 ÷ 2 × 25 + 40 + 12² (Parenthesis first 20 × 2 = 40 and 10 + 2 = 12).
100 ÷ 2 × 25 + 40 + 144 (Exponent Second 12² = 144).
100 ÷ 50 + 40 + 144 (Multiplication Third 25 × 2 = 50).
2 + 40 + 144 (Division Fourth 100 ÷ 50 = 2).
186 (addition).

5. Next priority for Addition a + b.
For example, 14 × 2 ÷ 4 + (25 + 5) + 5².
14 × 2 ÷ 4 + 30 + 5² (Parenthesis First 25 + 5 = 30).
14 × 2 ÷ 4 + 30 + 25 (Exponent 5² = 25).
28 ÷ 4 + 30 + 25 (Multiplication Third 14 × 2 = 28).
7 + 30 + 25 (Division Fourth 28 ÷ 4 = 7).
62 (fifth priority for addition 7 + 30 + 25 = 62).

6. Next priority for Subtraction a – b.
For example, 10 × 2 + 15 ÷ 5 + (100 × 2) – 15 + 2².
10 × 2 + 15 ÷ 5 + 200 – 15 + 2² (Parenthesis First 100 × 2 = 200).
10 × 2 + 15 ÷ 5 + 200 – 15 + 4 (Exponent second priority 2² =4).
20 + 15 ÷ 5 + 200 – 15 + 4 (Multiplication Third 10 × 2 = 20).
20 + 3 + 200 – 15 + 4 (Division fourth priority 15 ÷ 5 = 3).
227 – 15 (Addition Fifth Priority 20 + 3+ 200 + 4 = 227).
212 (Subtraction 227 – 15 = 212).

Solved Examples on How to Simplify PEMDAS Involving Integers

1. Simplify the given Expressions by using the PEMDAS Rule
(i) 10 – 24 ÷ 6 + 20 × (30 + 5).
(ii) 25 – [(15 × 2) + (30 +10)] + 5².
(iii) 62 + 25 of (50 – 20) × 25².
(iv) 40 + 25- 46 × 30 + (10 + 35).

(i) 10 – 24 ÷ 6 + 20 × (30 + 5).

Solution:

The given expression is 10 – 24 ÷ 6 + 20 × (30 + 5).
Based on the PEMDAS rule, we need to follow the order of operations. That is Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction.
10 – 24 ÷ 6 + 20 × (30 + 5).
Simplify the Parenthesis terms first. That is
10 – 24 ÷ 6 + 20 × 35.
Simplify the Multiplication Term. That is,
10 – 24 ÷ 6 + 700.
Simplify the Division Term. That is,
10 – 4 + 700.
Simplify the Addition term. That is,
710 – 4.
Simplify the subtraction. That is,
706.
Therefore, 10 – 24 ÷ 6 + 20 × (30 + 5) is equal to 706.

(ii) 25 – [(15 × 2) + (30 +10)] + 5².

Solution:

The given expression is25 – [(15 × 2) + (30 +10)] + 5².
As per the PEMDAS rule, the order of operations is Parenthesis, Exponent, Multiplication, Division, Addition, Subtraction.
25 – [(15 × 2) + (30 +10)] + 5².
25 – [30 + 40] + 5² (parenthesis term simplification).
25 – 70 + 5² (again parenthesis term simplification).
25 – 70 + 25 (Exponent term simplification).
50 – 70 (addition simplification).
-20 (subtraction).
Therefore, 25 – [(15 × 2) + (30 +10)] + 5² is equal to – 20.

(iii) 62 + 25 of (50 – 20) × 25².

Solution:

The given expression is 62 + 25 of (50 – 20) × 25².
As per the PEMDAS rule, the order of operations is Parenthesis, Exponent, Multiplication, Division, Addition, Subtraction.
62 + 25 of (50 – 20) × 25².
Solve the Parenthesis term. That is
62 + 25 of (30) × 25².
We can write it as 62 + 25 × 30 × 25².
Next, simplify the exponent terms. That is
62 + 25 × 30 × 625.
The next priority for simplification is multiplication. That is
62 + 4,68,750.
4,68,812.
Therefore, 62 + 25 of (50 – 20) × 25² is equal to 4,68,812.

(iv) 40 + 25- 46 × 30 + (10 + 35).

Solution:

The given expression is 40 + 25- 46 × 30 + (10 + 35).
As per the PEMDAS rule, the order of operations is Parenthesis, Exponent, Multiplication, Division, Addition, Subtraction.
40 + 25- 46 × 30 + (10 + 35).
We need to simplify the parenthesis terms first. That is
40 + 25- 46 × 30 + 45.
Next, simplify the multiplication terms. That is
40 + 25 – 1380 + 45.
Simplify the additional terms. That is
110 – 1380.
Subtract the above terms. That is
-1270.
Therefore, 40 + 25- 46 × 30 + (10 + 35) is equal to – 1270.

PEMDAS Rule – Definition, Full Form, Examples with Answers | Steps to Simplify Order of Operations

PEMDAS Rule

In mathematics, we have different types of operations such as addition, subtraction, multiplication, division, etc. If we start with any one of the operations to solve an arithmetic expression, then we won’t get the exact result for the expression. For the exact result, we have to follow one basic rule in mathematics and that is PEMDAS Rule. PEMDAS Rule is the same as BODMAS Rule. An acronym for the PEMDAS rule is “Please Excuse My Dear Sally”.

PEMDAS Full Form

The PEMDAS stands for
P —- Parenthesis.
E —- Exponent.
M —- Multiplication.
D —- Division.
A —- Addition.
S —- Subtraction.

PEMDAS Rule – Order of Operations

The PEMDAS Rule states that the order of the operations for better and easy calculation and the exact result of the expression. The order of operations is

  • The parenthesis terms are the first priority terms in the expression. Which are {}, [], (), and etc..
  • Exponent terms are the second priority terms. They are a² or √a.
  • The third priority of the operations is multiplication operation like a b.
  • Next, we have to solve the division operations a ÷ b.
  • The Next priority in mathematics is for the addition operation that is a + b.
  • Then, solve the subtraction operations a – b.

An arithmetic expression is arranged with the number of operations but by following the above order of operations, we need to solve the parenthesis operations first. After the parenthesis operations only, we have to solve the exponent operation, and then multiplication, division, addition, subtraction.

Example for the order of operations 2 + 3× 6 =?
If we follow left to right method, then 2 + 3 × 6 = 5 6 = 30. (wrong calculation).
If we follow the PEMDAS Rule, 2 + 3 × 6 = 2 + 18 (multiplication first).
2 + 18= 20. (addition next).
2 + 3 × 6 = 20 is the correct answer.

Some people use the BODMAS Rule (Bracket, Order, Division, Multiplication, Addition, Subtraction) for simple calculations and some people will follow the PEMDAS Rule. But both are the same there is no difference. In Canada, BEDMAS Rule (Bracket, Exponent, Division, Multiplication, Addition, Subtraction) is used for better calculation.

Also, Check:

Common Errors While using PEMDAS Rule

If the multiple parenthesis operations are there in an expression, then it leads to an incorrect solution in most of the conditions. Why because, we will get confused, which operation we have to solve first. So, learn the PEMDAS Rule concept carefully and solve the solutions with perfect answers.

Steps to Simplify PEMDAS Rule(Order of Operations)

1. Solve Parenthesis first?
Example: (5 + 8) ×3?

Solution:

The given expression is (5 + 8) × 3.
Based on the PEMDAS Rule, solve the parenthesis terms first. That is
(5 + 8) × 3 = 13 × 3.
Multiplication next. That is,
13 × 3 = 39.
So, (5 + 8) × 3 is equal to 39.

2. Solve Exponent term.
Example: 5 + 2²?

Solution:

The given expression is 5 + 2².
Based on the PEMDAS Rule, we have to give the importance for exponent term first. That is,
5 + 2² = 5 + 4.
5 + 4 = 9(addition).
Therefore, 5 + 2² is equal to 9.

3. Third priority is for multiplication.
Example: 20 + 3 × 3 + 5?

Solution:

The given expression is 20 + 3 × 3 + 5.
Based on the PEMDAS Rule, we have to give the importance for multiplication term first. That is,
20 + 3 × 3 + 5 = 20 + 9 + 5 (multiplication 3 × 3 = 9).
20 + 9 + 5 = 34 (addition).
So, 20 + 3 × 3 + 5 is equal to 34.

4. Next priority for division.
Example: 20 ÷ 5 – 2 + 3?

Solution:

The given expression is 20 ÷ 5 – 2 + 3.
Based on the PEMDAS Rule, we have to give the importance for Division term first. That is,
20 ÷ 5 – 2 + 3 = 4 – 2 + 3 (division first 20 ÷ 5 = 4).
= 7 – 2 (addition 4 + 3 = 7).
= 5 (subtraction 7 – 2 = 5).

5. Solve the addition and subtraction.
Example: 15 + 25 – 10?

Solution:

The given expression is 15 + 25 – 10.
Based on the PEMDAS Rule, we have to give the importance for addition term first and then subtraction. That is,
15 + 25 – 10 = 40 – 10 (addition 15 + 25 = 40).
40 – 10 = 30 (subtraction).
So, 15 + 25 – 10 is equal to 30.

PEMDAS Examples with Answers

1. Simplify the expressions by using the PEMDAS Rule
(i) 125 ÷ 10 + 20 × 5 -10.
(ii) 20 × 10 + [(5 – 2) + (25 + 5)].
(iii) 100 ÷ 5 – [30 + 20 – 15] + 4².
(iv) 60 + 20 – {(56 – 24 × 10) + (10 – 2+ 6)}.

(i) 125 ÷ 10 + 20 × 5 -10.

Solution:

The given expression is 125 ÷ 10 + 20 × 5 -10.
As per the PEMDAS rule,
125 ÷ 10 + 20 × 5 -10 = 125 ÷ 10 + 100 – 10 (Multiplication 20 x 5 = 100).
= 12.5 + 100 -10 (division 125 ÷ 10 = 12.5).
= 112.5 – 10 (addition 12.5 + 100 = 112.5).
= 102.5 ( subtraction 112.5 – 10 = 102.5).
Therefore, 125 ÷ 10 + 20 × 5 -10 is equal to 102.5.
(ii) 20 × 10 + [(5 – 2) + (25 + 5)].

Solution:

The given expression is 20 × 10 + [(5 – 2) + (25 + 5)].
As per the PEMDAS Rule,
20 × 10 + [(5 – 2) + (25 + 5)] = 20 × 10 + [3 + 30] (parenthesis terms simplification 5 – 2 = 3 and 25 + 5 = 30).
= 20 × 10 + 33.
= 200 + 33 (multiplication 20 x 10 = 200).
= 233 (addition).
So, 20 × 10 + [(5 – 2) + (25 + 5)] is equal to 233.
(iii) 100 ÷ 5 – [30 + 20 – 15] + 4².

Solution:

The given expression is 100 ÷ 5 – [30 + 20 – 15] + 4².
Based on the PEMDAS Rule,
100 ÷ 5 – [30 + 20 – 15] + 4² = 100 ÷ 5 – [50 – 15] + 4²( parenthesis addition 30 + 20 = 50).
= 100 ÷ 5 – [35] + 4² (subtraction in parenthesis 50 – 15 = 35).
= 100 ÷ 5 – 35 + 16 (exponent 4² = 16).
= 20 – 35 + 16 (division 100 ÷ 5 = 20).
= 36 – 35 (addition 20 + 16 = 36).
= 1 (subtraction).
Therefore, by following the PEMDAS Rule 100 ÷ 5 – [30 + 20 – 15] + 4² is equal to 1.

(iv) 60 + 20 – {(56 – 24 × 10) + (10 – 2+ 6)}.

Solution:

The given expression is 60 + 20 – {(56 – 24 × 10) + (10 – 2+ 6)}.
As per the PEMDAS Rule
60 + 20 – {(56 – 24 × 10) + (10 – 2+ 6)} = 60 + 20 – {(56 – 240) + (16 – 2)}.
= 60 + 20 – { – 184 + 14 }.
= 60 + 20 – {-170}.
= 60 + 20 + 170.
= 250.
So, by following the PEMDAS rule, 60 + 20 – {(56 – 24 × 10) + (10 – 2+ 6)} is equal to 250.

FAQs on PEMDAS Rule

1. What is PEMDAS Rule?

PEMDAS rule is used for better calculation.

2. What is the Full form of PEMDAS?

The full form of PEMDAS is Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction.

3. What is ‘E’ in PEMDAS?

E stands for Exponent in PEMDAS Rule.

4. Is BODMAS and PEMDAS the same?

BODMAS stands for Brackets, Orders, Division, Multiplication, Addition, and Subtraction. PEMDAS stands for Parenthesis, Exponent, Multiplication, Division, Addition, Subtraction. BIDMAS and PEMDAS do exactly the same operations but using different words.

Indirect Variation – Definition, Formula, Equation, Graph, Examples

Indirect Variation

Know the indirect variation definition here. Check day-to-day usage of inverse variations and know various problems involved in it. Gather the complete material of inverse proportions or indirect proportions. Follow the step-by-step process which helps you in solving the complicated problems. Get the detailed preparation material of indirect variation and follow all the tips and tricks mentioned below. Scroll down to the below sections to know the various details like definition, word problems, examples, formulae, key ideas, tips, tricks, etc.

Also, Read:

Importance of Inverse Variation

We use variation in values of quantities in day-to-day life. In most cases, the variation of the value of some quantity depends on the value of another quantity. Inverse variation or inverse proportion is defined as the variable which varies inversely in respect to another variable.

For suppose, If the train is traveling for x distance at a constant speed, the time taken to travel x distance remains constant, if there is a change in the speed then the time taken will also change. Hence, the inverse proportion values work here.

Indirect Variation – Definition

As we know that direct variation implies the direct proportion of one quantity to another, an inverse proportion is vice-verse of direct proportion. The change in one quantity is inversely proportional to the other quantity. i.e., If there is an increase in the value of one quantity, then there will be a decrease in the value of another quantity. Thus, both the values of quantities are defined to be indirectly proportional.

Inverse Variation Equation

Quantities that are available in inverse variation are expressed as,

x ∝ 1/y

xy = k, where k is the constant of proportionality and x,y are the values of 2 quantities. To define the change in values of two quantities, suppose that the initial values are x1, y1 and the final values are x2, y2 which are in inverse variation. The equation can be expressed as,

x1/x2 = y1/y2

Inverse Variation Example Graph 

Inverse Variation Graphical Representation

How to Solve Inverse Variation Problems?

Go through the simple procedure listed below to solve the Problems on Inverse Variation. They are along the lines

Step 1: First of all, check the given equation. We use the formula y = k/x to solve indirect proportions. When you are working on the word problems, consider the variables given other than x and y and also use those variables which are relevant to the problem to be solved. Check all the values carefully to determine the changes in inverse variation equations like square roots, squares, and cubes.

Step 2:  Find the complete information in the problem and find the “k” value which is found in step 2.

Step 3: Rewrite the resultant equation which is obtained in step 1. Substitute the value of k which we got in step 2.

Step 4: With the equation we got in step 3 and also with the help of the remaining information in the question, we find the final solution in this step. While solving the word problems, include the units in the final answer.

How to Identify Inverse Variation?

  • We define it as yy inversely varies with xx if yy is defined as the product of a constant number kk and xx reciprocal
  • The value of kk will never be 0(zero), i.e, k≠0
  • kk is defined as the fixed or the constant product of xx and yy which means that multiplying both the values of xx and yy always gives the constant output of kk.

Inverse Variation Word Problems

Problem 1:

The number of hours constructing a deep well is inversely proportional to the number of men working doing it. It takes 10 hours for 2 men to construct. How many men are needed to complete the work in 4 hours?

Solution:

Let y be the number of men working

Let x be the number of hours

In the first case,

y = 10, x = 2

As we know the equation of inverse proportion,

yx = k

Substituting the values in the equation,

(10)(2) = k

k = 20

Therefore, the value of k = 20

In the second case,

y = ?, x = 4

As we already got the k value i.e., 20

Substitute the values in the above equation,

yx = k

(y)(4) = 20

4y = 20

y = 20/4

y = 5

Therefore, no of men required to finish the work in 4 hours = 5 men

Hence, the final solution is 5 men

Problem 2:

When riding a bus at 55km/hr average speed, it takes Marcel 3 hours to reach his destination. How long will it take him if he travels by van at 70 km/hr?

Solution:

Let y be the average speed

Let x be the number of hours

In the first case,

y = 55, x = 3

As we know the equation of inverse proportion,

yx = k

Substituting the values in the equation,

(55)(3) = k

k = 165

Therefore, the value of k is 165

In the second case,

y = 70, x = ?

As we already got the k value i.e., 165

Substitute the values in the above equation,

yx = k

70x = 165

x = 165/70

x = 2.36

Therefore, it takes 2.36 hours for him to travel by van at 70 km/hr

Thus, the final solution is 2.36 hours

Problem 3:

A group of 10 men decided to rent a house for $100 for the stay of one week. But two people of them got sick and couldn’t join them. How much would each man pay for the rent?

Solution:

Let no of men = y

Let payment of each man = x

In the first case,

y = 10

x = 10

As we know the equation of inverse proportion,

yx = k

Substituting the values in the equation,

(10)(10) = k

k = 100

Therefore, the value of k is 100

In the second case,

As 2 men dropped to join the rent, we get the value as

y = 10-2

y = 8

x =?

As we already got the k value i.e., 100

Substitute the values in the above equation,

yx = k

(8)(x) = 100

x = 100/8

x = 2.5

Therefore, each man would pay $2.5 for the rent.

Thus, the final solution is $2.5

Problem 4:

Clark on his first drive traveled from home to his destination or 6 hours with an average speed of 60 km/hr. What must be his average speed if he wants to get there within 5 hours?

Solution:

Let y be the no of hours

Let x be average speed

In the first case,

y = 6, x = 60

As we know the equation of inverse proportion,

yx = k

Substituting the values in the equation,

(6)(60) = k

k =360

In the second case,

y = 5, x = ?

As we already got the k value i.e., 360

Substitute the values in the above equation,

yx = k

(5)(x) = 360

x = 360/5

x = 72 km/hr

Therefore, he must maintain an average speed of 72km/hr to get there within 5 hours

Problem 5:

Three fishermen decided to contribute individually to purchase a boat worth $1500. They decided to let other fishermen join them so they can lower their payments. How many fishermen will they need so that each one will pay $300?

Solution:

Let y be the no of fishermen

Let x be the amount to be paid by each fisherman

In the first case,

y = 3, x = 1500/3 = 500

As we know the equation of inverse proportion,

yx = k

Substituting the values in the equation,

(3)(500) = k

k = 1500

Therefore, the value of k = 1500

In the second case,

y = ?, x = 300

As we already got the k value i.e., 1500

Substitute the values in the above equation,

yx = k

(y)(300) = 1500

y = 1500/300

y = 5

Therefore, they need 5 fishermen that each one will pay $300

Problem 6:

The volume of a gas t constant temperature varies inversely as the pressure. The volume of the gas is 75 milliliters when the pressure is 1.5 atmospheres. Find the volume of the gas when the pressure is increased to 2.5 atmospheres?

Solution:

Let the volume of gas be V

Let the pressure be P

As we know the equation,

V = k/P

75 = k/1.5

112.5 = k

The value of k is 112.5

Substitute the value of k in the above equation

V = 112.5/P

V = 112.5/2.5

V = 45 milliliters

Therefore, the volume of the gas is 45 milliliters when the pressure is increased to 2.5 atmospheres

Problem 7:

The bases of triangles having equal areas are inversely proportional to their altitudes. The base of a certain triangle is 24 cm and its altitude is 30 cm. Find the base of the triangle whose altitude is 40 cm?

Solution:

Let b be the bases of the triangle

Let a be the altitudes of the triangle

As we know the equation of inverse proportion

b = k/a

(24) = k/(30)

(30)(24) = k

k = 720

Therefore, the value of k is 720

Substitute the value of k and a in the equation,

b = 720/40

b = 18 cm

Therefore, the base of the triangle is 18 cm

Thus the final solution is 18 cm

Problem 8:

The number of hours required to do a job varies inversely as the number of people working together. If it takes 8 hours or 5 people to paint a house how long will it take 12 people to paint the house?

Solution:

Let h be the number of hours

Let p be the number of people

As we know the equation of inverse proportion

h = k/p

(8) = k/(5)

(5)(8) = k

k = 40

Substitute the value of k and p in the above equation

h = 40/12

h = 10/3

h ≡ 3.33 hours

Therefore, it takes 3.33 hours for 12 people to paint the house.

Sin Theta Equals 0 General Solution | How do you Solve Sin Theta = 0?

Sin Theta Equals 0

Sin Theta Equals 0 properties is here. Check Sine Definition in terms of Sin 0 value. Know the various Sine degrees and radians along with the formulae, tricks, and tips. Follow sin θ equals zero examples, frequently asked questions, steps to solve trigonometric equations, analysis of the solution, etc. Get the steps to solve trigonometric problems, formulae, examples, solutions, etc.

Do Read:

Sin 0 Definition

In trigonometric equations, there are 3 primary functions which are sine, cosine, and tangent. These functions are used to calculate the length and angles of the right-angled triangles. The sine function is something that defines the relationship between the hypotenuse side and the angle of the perpendicular side (or) sin θ is defined as the ratio of the hypotenuse and the perpendicular of the right-angled triangle.

Sin θ Formula

If we have to calculate the degree of sin 0 value, then find the coordinates points on the x and y plane. Sin 0 defines the x value where coordinates are 1 and the y coordinates value is 0, which is (x,y) = (1,0) which means that the value of the perpendicular or opposite side is 0 and the hypotenuse value is 1. Therefore, to place the sin ratio values for where θ=00 hypotenuse is 0 and perpendicular side is 1

Sin 0° = 0

or

Sin 0° =0/1

The relations of various trigonometric functions are

sin(θ) = Opposite/Hypotenues

tan(θ) = Opposite/Adjacent

cos(θ) = Adjacent/Hypotenues

From the above-written equations, sin 0 degrees value. Now have a look at radians or degree values for each revolution in the given table.

Sine Radians / DegreesSin Values
Sin (0°)0
Sin (30°) or Sin (Π/6)1/2
Sin (45°) or Sin (Π/4)1/√2
Sin (60°) or Sin (Π/3)3/√2
Sin (90°) or Sin (Π/2)1
Sin (180°) or Sin (Π)0
Sin (270°) or Sin (3Π/2)-1
Sin (360°) or Sin (2Π)0

As mentioned in the above table, we can determine the values of tan values

Tan(θ) = Sin(θ)/Cos(θ)

Hence,

Tan(0°)=Sin(0°)/Cos(0°) = 0

Tan(30°)=Sin(30°)/Cos(30°) = 3/√2

Tan(450)=Sin(45°)/Cos(45°) = 1

Tan(60°)=Sin(60°)/Cos(60°) = √3

Tan(90°)=Sin(90°)/Cos(90°) = Undefined

2Π is the period for both cosine and sine function. To find all the possible solutions, add 2Πk, where k is an integer to the initial solution. The period of the function is 2Π which states all the possible solutions for the given function.

The equation with the period 2Π for the function is

sinθ = sin(θ ± 2kπ)

For other trigonometric functions also, the possible solutions are indicated by the same rules. To solve the trigonometric equations, we must follow the same techniques that we use for the algebraic equations. We read and write a trigonometric equation from left to right, in the same way as we read the sentence. To make the straightforward process, we must look for the factors, patterns, find the common denominators, the substitution of certain expressions with the variable.

How to solve a Trigonometric Equation?

  • First of all, check for the pattern which helps you in minimizing the equation. Mostly the pattern will be of algebraic properties like factoring or a squares opportunity.
  • Now, use the single variable and substitute it in the trigonometric equation in such a way that u or x.
  • Follow the same pattern of the algebraic equation to solve trigonometric expressions.
  • Then, substitute the trigonometric expression in the resultant expression by using the variable.
  • Finally, solve the equation to find the angle of the equation.

Table of Trigonometric Ratios for Various Angles

Angles (In Degrees)30°45°60°90°180°270°360°
Angles (In Radians)π/6π/4π/3π/2π3π/2
sin01/21/√2√3/210-10
cos1√3/21/√21/20-101
tan01/√31√300
cot√311/√300
cosec2√2√2/31-1
sec1√2/3√22-11

Table of Trigonometric Ratios for Various Radians

Angle30°45°60°90°180°270°360°
Radian0Π/6Π/4Π/3Π/2Π3Π/2

Applications of Trigonometry

  • Trigonometric equations help us to find the missing sides and angles of the triangle.
  • These equations are mostly used by builders to measure the distance and height of the building from the viewpoint.
  • It is used by the students to solve trigonometry-based problems.

Problems on Sin Theta Equals 0

Problem 1:

If √3 sinθ- cosθ = 0 and 0 < θ < 90°, find the value of θ?

Solution:

As given in the question,

The equation is √3 sinθ- cosθ = 0

√3 sin θ = cos θ

sin θ = cos θ * 1/√3

sin θ / cos θ = 1/√3

tan θ = Tan 30

θ = 30°

Therefore, the value of θ is 30º

Problem 2:

If secθ.sinθ = 0, then find the value of θ?

Solution:

As given in the question,

The equation is secθ.sinθ = 0

As we know that sec θ = 1/cos θ

The equation will be

1/cos θ . sin θ = 0

tan θ = 0

tan θ = tan 45°

θ = 45°

Problem 3:

Find the values of θ in [0°,360) so that y/r = sin θ = 1/2?

Hint: Take y=1,r=2

Solution:

As given in the question,

y/r = sin θ = 1/2

r = 2, y = 1

From the given values, we use the hypotenuse theorem

Hence, we have to find the values of x

i.e., x = √3

Therefore, θ = 30°

As the side of the triangle is not mentioned, there is also another chance where the x can be negative

Hence, if the value of x is negative, then x = -√3

Therefore, θ = 150º

Thus, the values of θ in (0°,360) are 30° and 150°

Eureka Math Grade 8 Module 5 Lesson 8 Answer Key

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

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

Exercise 1.
Consider the function that assigns to each number x the value x2.
a. Do you think the function is linear or nonlinear? Explain.
Answer:
I think the function is nonlinear. The equation describing the function is not of the form y = mx + b.

b. Develop a list of inputs and outputs for this function. Organize your work using the table below. Then, answer the questions that follow.
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 1
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 2

c. Plot the inputs and outputs as ordered pairs defining points on the coordinate plane.
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 3
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 4

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

e. Find the rate of change using rows 1 and 2 from the table above.
Answer:
\(\frac{25 – 16}{ – 5 – ( – 4)}\) = \(\frac{9}{ – 1}\) = – 9

f. Find the rate of change using rows 2 and 3 from the table above.
Answer:
\(\frac{16 – 9}{ – 4 – ( – 3)} = \) = \(\frac{7}{ – 1}\) = – 7

g. Find the rate of change using any two other rows from the table above.
Answer:
Student work will vary.
\(\frac{16 – 25}{4 – 5}\) = \(\frac{ – 9}{ – 1}\) = 9

h. Return to your initial claim about the function. Is it linear or nonlinear? Justify your answer with as many pieces of evidence as possible.
Answer:
This is definitely a nonlinear function because the rate of change is not a constant for different intervals of inputs. Also, we would expect the graph of a linear function to be a set of points in a line, and this graph is not a line. As was stated before, the expression x2 is nonlinear.

Exercise 2.
Consider the function that assigns to each number x the value x3.
a. Do you think the function is linear or nonlinear? Explain.
Answer:
I think the function is nonlinear. The equation describing the function is not of the form y = mx + b.

b. Develop a list of inputs and outputs for this function. Organize your work using the table below. Then, answer the questions that follow.
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 5
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 6

c. Plot the inputs and outputs as ordered pairs defining points on the coordinate plane.
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 7
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 8

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

e. Find the rate of change using rows 2 and 3 from the table above.
Answer:
\(\frac{ – 8 – ( – 3.375)}{ – 2 – ( – 1.5)}\) = \(\frac{ – 4.625}{ – 0.5}\) = 9.25

f. Find the rate of change using rows 3 and 4 from the table above.
Answer:
\(\frac{ – 3.375 – ( – 1)}{ – 1.5 – ( – 1)}\) = \(\frac{ – 2.375}{ – 0.5}\) = 4.75

g. Find the rate of change using rows 8 and 9 from the table above.
Answer:
\(\frac{1 – 3.375}{1 – 1.5}\) = \(\frac{ – 2.375}{ – 0.5}\) = 4.75

h. Return to your initial claim about the function. Is it linear or nonlinear? Justify your answer with as many pieces of evidence as possible.
Answer:
This is definitely a nonlinear function because the rate of change is not a constant for any interval of inputs. Also, we would expect the graph of a linear function to be a line, and this graph is not a line. As was stated before, the expression x3 is nonlinear.

Exercise 3.
Consider the function that assigns to each positive number x the value \(\frac{1}{x}\).
a. Do you think the function is linear or nonlinear? Explain.
Answer:
I think the function is nonlinear. The equation describing the function is not of the form y = mx + b.

b. Develop a list of inputs and outputs for this function. Organize your work using the table below. Then, answer the questions that follow.
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 9
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 10

c. Plot the inputs and outputs as ordered pairs defining points on the coordinate plane.
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 11
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 12

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

e. Find the rate of change using rows 1 and 2 from the table above.
Answer:
\(\frac{10 – 5}{0.1 – 0.2}\) = \(\frac{5}{ – 0.1}\) = – 50

f. Find the rate of change using rows 2 and 3 from the table above.
Answer:
\(\frac{5 – 2.5}{0.2 – 0.4}\) = \(\frac{2.5}{ – 0.2}\) = – 12.5

g. Find the rate of change using any two other rows from the table above.
Answer:
Student work will vary.
\(\frac{1 – 0.625}{1 – 1.6}\) = \(\frac{0.375}{ – 0.6}\) = – 0.625

h. Return to your initial claim about the function. Is it linear or nonlinear? Justify your answer with as many pieces of evidence as possible.
Answer:
This is definitely a nonlinear function because the rate of change is not a constant for any interval of inputs. Also, we would expect the graph of a linear function to be a line, and this graph is not a line. As was stated before, the expression \(\frac{1}{x}\) is nonlinear.

Exercises 4–10
In each of Exercises 4–10, an equation describing a rule for a function is given, and a question is asked about it. If necessary, use a table to organize pairs of inputs and outputs, and then plot each on a coordinate plane to help answer the question.

Exercise 4.
What shape do you expect the graph of the function described by y = x to take? Is it a linear or nonlinear function?
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 13
I expect the shape of the graph to be a line. This function is a linear function described by the linear equation y = x. The graph of this function is a line.

Exercise 5.
What shape do you expect the graph of the function described by y = 2x2 – x to take? Is it a linear or nonlinear function?
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 14
I expect the shape of the graph to be something other than a line. This function is nonlinear because its graph is not a line. Also the equation describing the function is not of the form y = mx + b. It is not linear.

Exercise 6.
What shape do you expect the graph of the function described by 3x + 7y = 8 to take? Is it a linear or nonlinear function?
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 15
I expect the shape of the graph to be a line. This function is a linear function described by the linear equation 3x + 7y = 8. The graph of this function is a line. (We have y = – \(\frac{3}{7}\) x + \(\frac{8}{7}\).)

Exercise 7.
What shape do you expect the graph of the function described by y = 4x3 to take? Is it a linear or nonlinear function?
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 16
I expect the shape of the graph to be something other than a line. This function is nonlinear because its graph is not a line. Also the equation describing the function is not of the form y = mx + b. It is not linear.

Exercise 8.
What shape do you expect the graph of the function described by \(\frac{3}{x}\) = y to take? Is it a linear or nonlinear function? (Assume that an input of x = 0 is disallowed.)
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 17
I expect the shape of the graph to be something other than a line. This function is nonlinear because its graph is not a line. Also the equation describing the function is not of the form y = mx + b. It is not linear.

Exercise 9.
What shape do you expect the graph of the function described by \(\frac{4}{x^{2}}\) = y to take? Is it a linear or nonlinear function? (Assume that an input of x = 0 is disallowed.)
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 18
I expect the shape of the graph to be something other than a line. This function is nonlinear because its graph is not a line. Also the equation describing the function is not of the form y = mx + b. It is not linear.

Exercise 10.
What shape do you expect the graph of the equation x2 + y2 = 36 to take? Is it a linear or nonlinear function? Is it a function? Explain.
Answer:
Engage NY Math Grade 8 Module 5 Lesson 8 Exercise Answer Key 19
I expect the shape of the graph to be something other than a line. It is nonlinear because its graph is not a line. It is not a function because there is more than one output for any given value of x in the interval ( – 6, 6). For example, at x = 0 the y – value is both 6 and – 6. This does not fit the definition of function because functions assign to each input exactly one output. Since there is at least one instance where an input has two outputs, it is not a function.

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

Question 1.
Consider the function that assigns to each number x the value x2 – 4.
a. Do you think the function is linear or nonlinear? Explain.
Answer:
The equation describing the function is not of the form y = mx + b. It is not linear.

b. Do you expect the graph of this function to be a straight line?
Answer:
No

c. Develop a list of inputs and matching outputs for this function. Use them to begin a graph of the function.
Eureka Math 8th Grade Module 5 Lesson 8 Problem Set Answer Key 1
Answer:
Eureka Math 8th Grade Module 5 Lesson 8 Problem Set Answer Key 2

Question 2.
Consider the function that assigns to each number x greater than – 3 the value \(\frac{1}{x + 3}\).
a. Is the function linear or nonlinear? Explain.
Answer:
The equation describing the function is not of the form y = mx + b. It is not linear.

b. Do you expect the graph of this function to be a straight line?
Answer:
No

c. Develop a list of inputs and matching outputs for this function. Use them to begin a graph of the function.
Eureka Math 8th Grade Module 5 Lesson 8 Problem Set Answer Key 3
Answer:
Eureka Math 8th Grade Module 5 Lesson 8 Problem Set Answer Key 4

d. Was your prediction to (b) correct?
Answer:
Yes, the graph appears to be taking the shape of some type of curve.

Question 3.
a. Is the function represented by this graph linear or nonlinear? Briefly justify your answer.
Eureka Math 8th Grade Module 5 Lesson 8 Problem Set Answer Key 5
Answer:
The graph is clearly not a straight line, so the function is not linear.

b. What is the average rate of change for this function from an input of x = – 2 to an input of x = – 1?
Answer:
\(\frac{ – 2 – 1}{ – 2 – ( – 1)}\) = \(\frac{ – 3}{ – 1}\) = 3

c. What is the average rate of change for this function from an input of x = – 1 to an input of x = 0?
Answer:
\(\frac{1 – 2}{ – 1 – 0}\) = \(\frac{ – 1}{ – 1}\) = 1
As expected, the average rate of change of this function is not constant.
Eureka Math 8th Grade Module 5 Lesson 8 Problem Set Answer Key 6

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

Question 1.
The graph below is the graph of a function. Do you think the function is linear or nonlinear? Briefly justify your answer.
Eureka Math Grade 8 Module 5 Lesson 8 Exit Ticket Answer Key 1
Answer:
Student work may vary. The plot of this graph appears to be a straight line, and so the function is linear.

Question 2.
Consider the function that assigns to each number x the value \(\frac{1}{2}\) x2. Do you expect the graph of this function to be a straight line? Briefly justify your answer.
Answer:
The equation is nonlinear (not of the form y = mx + b), so the function is nonlinear. Its graph will not be a straight line.

Eureka Math Grade 8 Module 6 Lesson 2 Answer Key

Engage NY Eureka Math 8th Grade Module 6 Lesson 2 Answer Key

Eureka Math Grade 8 Module 6 Lesson 2 Example Answer Key

Example 1: Rate of Change and Initial Value
The equation of a line can be interpreted as defining a linear function. The graphs and the equations of lines are important in understanding the relationship between two types of quantities (represented in the following examples by x and y).
In a previous lesson, you encountered an MP3 download site that offers downloads of individual songs with the following price structure: a $3 fixed fee for a monthly subscription plus a fee of $0.25 per song. The linear function that models the relationship between the number of songs downloaded and the total monthly cost of downloading songs can be written as
y = 0.25x + 3,
where x represents the number of songs downloaded and y represents the total monthly cost (in dollars) for
MP3 downloads.
a. In your own words, explain the meaning of 0.25 within the context of the problem.
Answer:
In the example on the previous page, the value 0.25 means there is a cost increase of $0.25 for every 1 song downloaded.

b. In your own words, explain the meaning of 3 within the context of the problem.
Answer:
In the example on the previous page, the value of 3 represents an initial cost of $3 for downloading 0 songs. In other words, there is a fixed cost of $3 to subscribe to the site.

The values represented in the function can be interpreted in the following way:
Engage NY Math 8th Grade Module 6 Lesson 2 Example Answer Key 1

Eureka Math Grade 8 Module 6 Lesson 2 Exercise Answer Key

Exercises 1–6: Is It a Better Deal?
Another site offers MP3 downloads with a different price structure: a $2 fixed fee for a monthly subscription plus a fee of $0.40 per song.
Exercise 1.
Write a linear function to model the relationship between the number of songs downloaded and the total monthly cost. As before, let x represent the number of songs downloaded and y represent the total monthly cost (in dollars) of downloading songs.
Answer:
y = 0.4x + 2

Exercise 2.
Determine the cost of downloading 0 songs and 10 songs from this site.
Answer:
y = 0.4(0) + 2 = 2.00. For 0 songs, the cost is $2.00.
y = 0.4(10) + 2 = 6.00. For 10 songs, the cost is $6.00.

Exercise 3.
The graph below already shows the linear model for the first subscription site (Company 1): y = 0.25x + 3. Graph the equation of the line for the second subscription site (Company 2) by marking the two points from your work in Exercise 2 (for 0 songs and 10 songs) and drawing a line through those two points.
Engage NY Math Grade 8 Module 6 Lesson 2 Exercise Answer Key 1
Answer:
Engage NY Math Grade 8 Module 6 Lesson 2 Exercise Answer Key 2

Exercise 4.
Which line has a steeper slope? Which company’s model has the more expensive cost per song?
Answer:
The line modeled by the second subscription site (Company 2) is steeper. It has the larger slope value and the greater cost per song.

Exercise 5.
Which function has the greater initial value?
Answer:
The first subscription site (Company 1) has the greater initial value. Its monthly subscription fee is $3 compared to only $2 for the second site.

Exercise 6.
Which subscription site would you choose if you only wanted to download 5 songs per month? Which company would you choose if you wanted to download 10 songs? Explain your reasoning.
Answer:
For 5 songs: Company 1’s cost is $4.25 (y = 25(5) + 3); Company 2’s cost is $4.00 (y = 0.4(5) + 2). So, Company 2 would be the better choice. Graphically, Company 2’s model also has the smaller y – value when x = 5.
For 10 songs: Company 1’s cost is $5.50 (y = 0.25(10) + 3); Company 2’s cost is $6.00 (y = 0.4(10) + 2). So, Company 1 would be the better choice. Graphically, Company 1’s model also has the smaller y – value at x = 10.

Exercises 7–9: Aging Autos

Exercise 7.
When someone purchases a new car and begins to drive it, the mileage (meaning the number of miles the car has traveled) immediately increases. Let x represent the number of years since the car was purchased and y represent the total miles traveled. The linear function that models the relationship between the number of years since purchase and the total miles traveled is y = 15000x.
a. Identify and interpret the rate of change.
Answer:
The rate of change is 15,000. It means that the mileage is increasing by 15,000 miles per year.

b. Identify and interpret the initial value.
Answer:
The initial value is 0. This means that there were no miles on the car when it was purchased.

c. Is the mileage increasing or decreasing each year according to the model? Explain your reasoning.
Answer:
Since the rate of change is positive, it means the mileage is increasing each year.

Exercise 8.
When someone purchases a new car and begins to drive it, generally speaking, the resale value of the car (in dollars) goes down each year. Let x represent the number of years since purchase and y represent the resale value of the car (in dollars). The linear function that models the resale value based on the number of years since purchase is
y = 20000 – 1200x.
a. Identify and interpret the rate of change.
Answer:
The rate of change is – 1,200. The resale value of the car is decreasing $1,200 every year since purchase.

b. Identify and interpret the initial value.
Answer:
The initial value is $20,000. The car’s value at the time of purchase was $20,000.

c. Is the resale value increasing or decreasing each year according to the model? Explain.
Answer:
The slope is negative. This means that the resale value decreases each year.

Exercise 9.
Suppose you are given the linear function y = 2.5x + 10.
a. Write a story that can be modeled by the given linear function.
Answer:
Answers will vary. I am ordering cupcakes for a birthday party. The bakery is going to charge $2.50 per cupcake in addition to a $10 decorating fee.

b. What is the rate of change? Explain its meaning with respect to your story.
Answer:
The rate of change is 2.5, which means that the cost increases $2.50 for every additional cupcake ordered.

c. What is the initial value? Explain its meaning with respect to your story.
Answer:
The initial value is 10, which in this story means that there is a flat fee of $10 to decorate the cupcakes.

Eureka Math Grade 8 Module 6 Lesson 2 Problem Set Answer Key

Question 1.
A rental car company offers the following two pricing methods for its customers to choose from for a
one – month rental:
Method 1: Pay $400 for the month, or
Method 2: Pay $0.30 per mile plus a standard maintenance fee of $35.
a. Construct a linear function that models the relationship between the miles driven and the total rental cost for Method 2. Let x represent the number of miles driven and y represent the rental cost (in dollars).
Answer:
y = 35 + 0.30x

b. If you plan to drive 1,100 miles for the month, which method would you choose? Explain your reasoning.
Answer:
Method 1 has a flat rate of $400 regardless of miles. Using Method 2, the cost would be $365
(y = 35 + 0.3(1100)). So, Method 2 would be preferred.

Question 2.
Recall from a previous lesson that Kelly wants to add new music to her MP3 player. She was interested in a monthly subscription site that offered its MP3 downloading service for a monthly subscription fee plus a fee per song. The linear function that modeled the total monthly cost in dollars (y) based on the number of songs downloaded (x) is
y = 5.25 + 0.30x.
The site has suddenly changed its monthly price structure. The linear function that models the new total monthly cost in dollars (y) based on the number of songs downloaded (x) is y = 0.35x + 4.50.
a. Explain the meaning of the value 4.50 in the new equation. Is this a better situation for Kelly than before?
Answer:
The initial value is 4.50 and means that the monthly subscription cost is now $4.50. This is lower than before, which is good for Kelly.

b. Explain the meaning of the value 0.35 in the new equation. Is this a better situation for Kelly than before?
Answer:
The rate of change is 0.35. This means that the cost is increasing by $0.35 for every song downloaded. This is more than the download cost for the original plan.

c. If you were to graph the two equations (old versus new), which line would have the steeper slope? What does this mean in the context of the problem?
Answer:
The slope of the new line is steeper because the new linear function has a greater rate of change. It means that the total monthly cost of the new plan is increasing at a faster rate per song compared to the cost of the old plan.

d. Which subscription plan provides the better value if Kelly downloads fewer than 15 songs per month?
Answer:
If Kelly were to download 15 songs, both plans will cost the same ($9.75). Therefore, the new plan is cheaper if Kelly downloads fewer than 15 songs.

Eureka Math Grade 8 Module 6 Lesson 2 Exit Ticket Answer Key

In 2008, a collector of sports memorabilia purchased 5 specific baseball cards as an investment. Let y represent each card’s resale value (in dollars) and x represent the number of years since purchase. Each card’s resale value after 0, 1, 2, 3, and 4 years could be modeled by linear equations as follows:
Card A: y = 5 – 0.7x
Card B: y = 4 + 2.6x
Card C: y = 10 + 0.9x
Card D: y = 10 – 1.1x
Card E: y = 8 + 0.25x

Question 1.
Which card(s) are decreasing in value each year? How can you tell?
Answer:
Cards A and D are decreasing in value, as shown by the negative values for rate of change in each equation.

Question 2.
Which card(s) had the greatest initial value at purchase (at 0 years)?
Answer:
Since all of the models are in slope – intercept form, Cards C and D have the greatest initial values at $10 each.

Question 3.
Which card(s) is increasing in value the fastest from year to year? How can you tell?
Answer:
Card B is increasing in value the fastest from year to year. Its model has the greatest rate of change.

Question 4.
If you were to graph the equations of the resale values of Card B and Card C, which card’s graph line would be steeper? Explain.
Answer:
The Card B line would be steeper because the function for Card B has the greatest rate of change; the card’s value is increasing at a faster rate than the other values of other cards.

Question 5.
Write a sentence explaining the 0.9 value in Card C’s equation.
Answer:
The 0.9 value means that Card C’s value increases by 90 cents per year.

Eureka Math Grade 8 Module 6 Lesson 3 Answer Key

Engage NY Eureka Math 8th Grade Module 6 Lesson 3 Answer Key

Eureka Math Grade 8 Module 6 Lesson 3 Example Answer Key

Example 1: Rate of Change and Initial Value Given in the Context of the Problem
A truck rental company charges a $150 rental fee in addition to a charge of $0.50 per mile driven. Graph the linear function relating the total cost of the rental in dollars, C, to the number of miles driven, m, on the axes below.
Engage NY Math 8th Grade Module 6 Lesson 3 Example Answer Key 1
Answer:
Engage NY Math 8th Grade Module 6 Lesson 3 Example Answer Key 2

a. If the truck is driven 0 miles, what is the cost to the customer? How is this shown on the graph?
Answer:
$150, shown as the point (0, 150). This is the initial value. Some students might say “b.” Help them to use the term initial value.

b. What is the rate of change that relates cost to number of miles driven? Explain what it means within the context of the problem.
Answer:
The rate of change is 0.5. It means that the cost increases by $0.50 for every mile driven.

c. On the axes given, sketch the graph of the linear function that relates C to m.
Answer:
Students can plot the initial value (0, 150) and then use the rate of change to identify additional points as needed. A 1, 000-unit increase in m results in a 500-unit increase for C, so another point on the line is (1000, 650).

d. Write the equation of the linear function that models the relationship between number of miles driven and total rental cost.
Answer:
C = 0.5m + 150

Eureka Math Grade 8 Module 6 Lesson 3 Exercise Answer Key

Exercises
Jenna bought a used car for $18, 000. She has been told that the value of the car is likely to decrease by $2, 500 for each year that she owns the car. Let the value of the car in dollars be V and the number of years Jenna has owned the car be t.
Engage NY Math Grade 8 Module 6 Lesson 3 Exercise Answer Key 1
Answer:
Engage NY Math Grade 8 Module 6 Lesson 3 Exercise Answer Key 2

Exercise 1.
What is the value of the car when t = 0? Show this point on the graph.
Answer:
$18, 000. Shown by the point (0, 18000)

Exercise 2.
What is the rate of change that relates V to t? (Hint: Is it positive or negative? How can you tell?)
Answer:
-2, 500. The rate of change is negative because the value of the car is decreasing.

Exercise 3.
Find the value of the car when:
a. t = 1
Answer:
$18000 – $2500 = $15500

b. t = 2
Answer:
$18000 – 2($2500) = $13000

c. t = 7
Answer:
$18000 – 7($2500) = $500

Exercise 4.
Plot the points for the values you found in Exercise 3, and draw the line (using a straightedge) that passes through those points.
Answer:
See the graph above.

Exercise 5.
Write the linear function that models the relationship between the number of years Jenna has owned the car and the value of the car.
Answer:
V = 18000 – 2500t or V = -2500t + 18000

An online bookseller has a new book in print. The company estimates that if the book is priced at $15, then 800 copies of the book will be sold per day, and if the book is priced at $20, then 550 copies of the book will be sold per day.
Engage NY Math Grade 8 Module 6 Lesson 3 Exercise Answer Key 3
Answer:
Engage NY Math Grade 8 Module 6 Lesson 3 Exercise Answer Key 4

Exercise 6.
Identify the ordered pairs given in the problem. Then, plot both on the graph.
Answer:
The ordered pairs are (15, 800) and (20, 550). See the graph above.

Exercise 7.
Assume that the relationship between the number of books sold and the price is linear. (In other words, assume that the graph is a straight line.) Using a straightedge, draw the line that passes through the two points.
Answer:
See the graph above.

Exercise 8.
What is the rate of change relating number of copies sold to price?
Answer:
Between the points (15, 800) and (20, 550), the run is 5, and the rise is -(800-550) = -250. So, the rate of change is \(\frac{-250}{5}\) = -50.

Exercise 9.
Based on the graph, if the company prices the book at $18, about how many copies of the book can they expect to sell per day?
Answer:
650

Exercise 10.
Based on the graph, approximately what price should the company charge in order to sell 700 copies of the book per day?
Answer:
$17

Eureka Math Grade 8 Module 6 Lesson 3 Problem Set Answer Key

Question 1.
A plumbing company charges a service fee of $120, plus $40 for each hour worked. Sketch the graph of the linear function relating the cost to the customer (in dollars), C, to the time worked by the plumber (in hours), t, on the axes below.
Eureka Math 8th Grade Module 6 Lesson 3 Problem Set Answer Key 1
Answer:
Eureka Math 8th Grade Module 6 Lesson 3 Problem Set Answer Key 2

a. If the plumber works for 0 hours, what is the cost to the customer? How is this shown on the graph?
Answer:
$120 This is shown on the graph by the point (0, 120).

b. What is the rate of change that relates cost to time?
Answer:
40

c. Write a linear function that models the relationship between the hours worked and the cost to the customer.
Answer:
C = 40t + 120

d. Find the cost to the customer if the plumber works for each of the following number of hours.
i) 1 hour
Answer:
$160

ii) 2 hours
Answer:
$200

iii) 6 hours
Answer:
$360

e. Plot the points for these times on the coordinate plane, and use a straightedge to draw the line through the points.
Answer:
See the graph on the previous page.

Question 2.
An author has been paid a writer’s fee of $1, 000 plus $1.50 for every copy of the book that is sold.
a. Sketch the graph of the linear function that relates the total amount of money earned in dollars, A, to the number of books sold, n, on the axes below.
Eureka Math 8th Grade Module 6 Lesson 3 Problem Set Answer Key 3
Answer:
Eureka Math 8th Grade Module 6 Lesson 3 Problem Set Answer Key 4

b. What is the rate of change that relates the total amount of money earned to the number of books sold?
Answer:
1.5

c. What is the initial value of the linear function based on the graph?
Answer:
1, 000

d. Let the number of books sold be n and the total amount earned be A. Construct a linear function that models the relationship between the number of books sold and the total amount earned.
Answer:
A = 1.5n + 1000

Question 3.
Suppose that the price of gasoline has been falling. At the beginning of last month (t = 0), the price was $4.60 per gallon. Twenty days later (t = 20), the price was $4.20 per gallon. Assume that the price per gallon, P, fell at a constant rate over the twenty days.
Eureka Math 8th Grade Module 6 Lesson 3 Problem Set Answer Key 5
Answer:
Eureka Math 8th Grade Module 6 Lesson 3 Problem Set Answer Key 6

a. Identify the ordered pairs given in the problem. Plot both points on the coordinate plane above.
Answer:
(0, 4.60) and (20, 4.20); see the graph above.

b. Using a straightedge, draw the line that contains the two points.
Answer:
See the graph above.

c. What is the rate of change? What does it mean within the context of the problem?
Answer:
Using points (0, 4.60) and (20, 4.20), the rate of change is -0.02 because \(\frac{4.20-4.60}{20-0}\) = \(\frac{-0.4}{20}\) = -0.02. The price of gas is decreasing $0.02 each day.

d. What is the function that models the relationship between the number of days and the price per gallon?
Answer:
P = -0.02t + 4.6

e. What was the price of gasoline after 9 days?
Answer:
$4.42; see the graph above.

f. After how many days was the price $4.32?
Answer:
14 days; see the graph above.

Eureka Math Grade 8 Module 6 Lesson 3 Exit Ticket Answer Key

Question 1.
A car starts a journey with 18 gallons of fuel. Assuming a constant rate, the car consumes 0.04 gallon for every mile driven. Let A represent the amount of gas in the tank (in gallons) and m represent the number of miles driven.
Eureka Math Grade 8 Module 6 Lesson 3 Exit Ticket Answer Key 1
Answer:
Eureka Math Grade 8 Module 6 Lesson 3 Exit Ticket Answer Key 2

a. How much gas is in the tank if 0 miles have been driven? How would this be represented on the axes above?
Answer:
There are 18 gallons in the tank. This would be represented as (0, 18), the initial value, on the graph above.

b. What is the rate of change that relates the amount of gas in the tank to the number of miles driven? Explain what it means within the context of the problem.
Answer:
-0.04; the car consumes 0.04 gallon for every mile driven. It relates the amount of fuel to the miles driven.

c. On the axes above, draw the line that represents the graph of the linear function that relates A to m.
Answer:
See the graph above. Students can plot the initial value (0, 18) and then use the rate of change to identify additional points as needed. A 50-unit increase in m results in a 2-unit decrease for A, so another point on the line is (50, 16).

d. Write the linear function that models the relationship between the number of miles driven and the amount of gas in the tank.
Answer:
A = 18 – 0.04m or A = -0.04m + 18

Question 2.
Andrew works in a restaurant. The graph below shows the relationship between the amount Andrew earns in dollars and the number of hours he works.
Eureka Math Grade 8 Module 6 Lesson 3 Exit Ticket Answer Key 3
a. If Andrew works for 7 hours, approximately how much does he earn in dollars?
Answer:
$96

b. Estimate how long Andrew has to work in order to earn $64.
Answer:
3 hours

c. What is the rate of change of the function given by the graph? Interpret the value within the context of the problem.
Answer:
Using the ordered pairs (7, 96) and (3, 64), the slope is 8. It means that the amount Andrew earns increases by $8 for every hour worked.

Eureka Math Grade 8 Module 6 Lesson 1 Answer Key

Engage NY Eureka Math 8th Grade Module 6 Lesson 1 Answer Key

Eureka Math Grade 8 Module 6 Lesson 1 Example Answer Key

Example 2: Another Rate Plan
A second wireless access company has a similar method for computing its costs. Unlike the first company that Lenore was considering, this second company explicitly states its access fee is $0.15, and its usage rate is $0.04 per minute.
Total Session Cost = $0.15 + $0.04 (number of minutes)
Answer:
→ How is this plan presented differently?
In this case, we are given the access fee and usage rate with an equation. In the first example, just data points were given.

→ Based on the work with the first set of problems, how do you think the two plans are different?
The values for the access fee and usage charge per minute are different, or the initial value and the rate of change are different.

Eureka Math Grade 8 Module 6 Lesson 1 Exercise Answer Key

Exercises 1–6

Exercise 1.
Lenore makes a table of this information and a graph where number of minutes is represented by the horizontal axis and total session cost is represented by the vertical axis. Plot the three given points on the graph. These three points appear to lie on a line. What information about the access plan suggests that the correct model is indeed a linear relationship?
Engage NY Math Grade 8 Module 6 Lesson 1 Exercise Answer Key 1
Answer:
The amount charged for the minutes connected is based upon a constant usage rate in dollars per minute.
Engage NY Math Grade 8 Module 6 Lesson 1 Exercise Answer Key 2.1

Exercise 2.
The rate of change describes how the total cost changes with respect to time.
a. When the number of minutes increases by 10 (e.g., from 10 minutes to 20 minutes or from 20 minutes to 30 minutes), how much does the charge increase?
Answer:
When the number of minutes increases by 10 (e.g., from 10 minutes to 20 minutes or from 20 minutes to 30 minutes), the cost increases by $0.30 (30 cents).

b. Another way to say this would be the usage charge per 10 minutes of use. Use that information to determine the increase in cost based on only 1 minute of additional usage. In other words, find the usage charge per minute of use.
Answer:
If $0.30 is the usage charge per 10 minutes of use, then $0.03 is the usage charge per 1 minute of use (i.e., the usage rate). Since the usage rate is constant, students should use what they have learned in Module 4.

Exercise 3.
The company’s pricing plan states that the usage rate is constant for any number of minutes connected to the Internet. In other words, the increase in cost for 10 more minutes of use (the value that you calculated in Exercise 2) is the same whether you increase from 20 to 30 minutes, 30 to 40 minutes, etc. Using this information, determine the total cost for 40 minutes, 50 minutes, and 60 minutes of use. Record those values in the table, and plot the corresponding points on the graph in Exercise 1.
Answer:
Consider the following table and graphs.
Engage NY Math Grade 8 Module 6 Lesson 1 Exercise Answer Key 3

Exercise 4.
Using the table and the graph in Exercise 1, compute the hypothetical cost for 0 minutes of use. What does that value represent in the context of the values that Lenore is trying to figure out?
Answer:
Since there is a $0.30 decrease in cost for each decrease of 10 minutes of use, one could subtract $0.30 from the cost value for 10 minutes and arrive at the hypothetical cost value for 0 minutes. That cost would be $0.10. Students may notice that such a value follows the regular pattern in the table and would represent the fixed access fee for connecting. (This value could also be found from the graph after completing Exercise 6.)

Exercise 5.
On the graph in Exercise 1, draw a line through the points representing 0 to 60 minutes of use under this company’s plan. The slope of this line is equal to the constant rate of change, which in this case is the usage rate.
Answer:
Engage NY Math Grade 8 Module 6 Lesson 1 Exercise Answer Key 4

Exercise 6.
Using x for the number of minutes and y for the total cost in dollars, write a function to model the linear relationship between minutes of use and total cost.
Answer:
y = 0.03x + 0.10

Exercises 7–16

Exercise 7.
Let x represent the number of minutes used and y represent the total session cost in dollars. Construct a linear function that models the total session cost based on the number of minutes used.
Answer:
y = 0.04x + 0.15

Exercise 8.
Using the linear function constructed in Exercise 7, determine the total session cost for sessions of 0, 10, 20, 30, 40, 50, and 60 minutes, and fill in these values in the table below.
Engage NY Math Grade 8 Module 6 Lesson 1 Exercise Answer Key 5
Answer:
Engage NY Math Grade 8 Module 6 Lesson 1 Exercise Answer Key 6

Exercise 9.
Plot these points on the original graph in Exercise 1, and draw a line through these points. In what ways does the line that represents this second company’s access plan differ from the line that represents the first company’s access plan?
Answer:
The second company’s plan line begins at a greater initial value. The same plan also increases in total cost more quickly over time; in other words, the slope of the line for the second company’s plan is steeper.
Engage NY Math Grade 8 Module 6 Lesson 1 Exercise Answer Key 7

Exercises 10–12

MP3 download sites are a popular forum for selling music. Different sites offer pricing that depends on whether or not you want to purchase an entire album or individual songs à la carte. One site offers MP3 downloads of individual songs with the following price structure: a $3 fixed fee for a monthly subscription plus a charge of $0.25 per song.
Exercise 10.
Using x for the number of songs downloaded and y for the total monthly cost in dollars, construct a linear function to model the relationship between the number of songs downloaded and the total monthly cost.
Answer:
Since $3 is the initial cost and there is a 25 cent increase per song, the function would be
y = 3 + 0.25x or y = 0.25x + 3.

Exercise 11.
Using the linear function you wrote in Exercise 10, construct a table to record the total monthly cost (in dollars) for MP3 downloads of 10 songs, 20 songs, and so on up to 100 songs.
Engage NY Math Grade 8 Module 6 Lesson 1 Exercise Answer Key 8
Answer:
Engage NY Math Grade 8 Module 6 Lesson 1 Exercise Answer Key 9

Exercise 12.
Plot the 10 data points in the table on a coordinate plane. Let the x-axis represent the number of songs downloaded and the y-axis represent the total monthly cost (in dollars) for MP3 downloads.
Answer:
Engage NY Math Grade 8 Module 6 Lesson 1 Exercise Answer Key 10

A band will be paid a flat fee for playing a concert. Additionally, the band will receive a fixed amount for every ticket sold. If 40 tickets are sold, the band will be paid $200. If 70 tickets are sold, the band will be paid $260.
Exercise 13.
Determine the rate of change.
Answer:
The points (40,200) and (70,260) have been given.
So, the rate of change is 2 because \(\frac{260-200}{70-40}\) = 2.

Exercise 14.
Let x represent the number of tickets sold and y represent the amount the band will be paid in dollars. Construct a linear function to represent the relationship between the number of tickets sold and the amount the band will be paid.
Answer:
Using the rate of change and (40,200):
200 = 2(40) + b
200 = 80 + b
120 = b
Therefore, the function is y = 2x + 120.

Exercise 15.
What flat fee will the band be paid for playing the concert regardless of the number of tickets sold?
Answer:
The band will be paid a flat fee of $120 for playing the concert.

Exercise 16.
How much will the band receive for each ticket sold?
Answer:
The band receives $2 per ticket.

Eureka Math Grade 8 Module 6 Lesson 1 Problem Set Answer Key

Question 1.
Recall that Lenore was investigating two wireless access plans. Her friend in Europe says that he uses a plan in which he pays a monthly fee of 30 euro plus 0.02 euro per minute of use.
a. Construct a table of values for his plan’s monthly cost based on 100 minutes of use for the month, 200 minutes of use, and so on up to 1,000 minutes of use. (The charge of 0.02 euro per minute of use is equivalent to 2 euro per 100 minutes of use.)
Answer:
Eureka Math 8th Grade Module 6 Lesson 1 Problem Set Answer Key 1

b. Plot these 10 points on a carefully labeled graph, and draw the line that contains these points.
Eureka Math 8th Grade Module 6 Lesson 1 Problem Set Answer Key 3
Answer:

c. Let x represent minutes of use and y represent the total monthly cost in euro. Construct a linear function that determines the monthly cost based on minutes of use.
Answer:
y = 30 + 0.02x

d. Use the function to calculate the cost under this plan for 750 minutes of use. If this point were added to the graph, would it be above the line, below the line, or on the line?
Answer:
The cost for 750 minutes would be €45. The point (750,45) would be on the line.

Question 2.
A shipping company charges a $4.45 handling fee in addition to $0.27 per pound to ship a package.
a. Using x for the weight in pounds and y for the cost of shipping in dollars, write a linear function that determines the cost of shipping based on weight.
Answer:
y = 4.45 + 0.27x

b. Which line (solid, dotted, or dashed) on the following graph represents the shipping company’s pricing method? Explain.
Eureka Math 8th Grade Module 6 Lesson 1 Problem Set Answer Key 4
Answer:
The solid line would be the correct line. Its initial value is 4.45, and its slope is 0.27. The dashed line shows the cost decreasing as the weight increases, so that is not correct. The dotted line starts at an initial value that is too low.

Question 3.
Kelly wants to add new music to her MP3 player. Another subscription site offers its downloading service using the following: Total Monthly Cost = 5.25 + 0.30(number of songs).
a. Write a sentence (all words, no math symbols) that the company could use on its website to explain how it determines the price for MP3 downloads for the month.
Answer:
“We charge a $5.25 subscription fee plus 30 cents per song downloaded.”

b. Let x represent the number of songs downloaded and y represent the total monthly cost in dollars. Construct a function to model the relationship between the number of songs downloaded and the total monthly cost.
Answer:
y = 5.25 + 0.30x

c. Determine the cost of downloading 10 songs.
Answer:
5.25 + 0.30(10) = 8.25
The cost of downloading 10 songs is $8.25.

Question 4.
Li Na is saving money. Her parents gave her an amount to start, and since then she has been putting aside a fixed amount each week. After six weeks, Li Na has a total of $82 of her own savings in addition to the amount her parents gave her. Fourteen weeks from the start of the process, Li Na has $118.
a. Using x for the number of weeks and y for the amount in savings (in dollars), construct a linear function that describes the relationship between the number of weeks and the amount in savings.
Answer:
The points (6, 82) and (14, 118) have been given.
So, the rate of change is 4.5 because \(\frac{118-82}{14-6}\) = \(\frac{36}{8}\) = 4.5.
Using the rate of change and (6, 82):
82 = 4.5(6) + b
82 = 27 + b
55 = b
The function is y = 4.5x + 55.

b. How much did Li Na’s parents give her to start?
Answer:
Li Na’s parents gave her $55 to start.

c. How much does Li Na set aside each week?
Answer:
Li Na is setting aside $4.50 every week for savings.

d. Draw the graph of the linear function below (start by plotting the points for x = 0 and x = 20).
Eureka Math 8th Grade Module 6 Lesson 1 Problem Set Answer Key 5
Answer:
Eureka Math 8th Grade Module 6 Lesson 1 Problem Set Answer Key 6

Eureka Math Grade 8 Module 6 Lesson 1 Exit Ticket Answer Key

A rental car company offers a rental package for a midsize car. The cost comprises a fixed $30 administrative fee for the cleaning and maintenance of the car plus a rental cost of $35 per day.
Question 1.
Using x for the number of days and y for the total cost in dollars, construct a function to model the relationship between the number of days and the total cost of renting a midsize car.
Answer:
y = 35x + 30

Question 2.
The same company is advertising a deal on compact car rentals. The linear function y = 30x + 15 can be used to model the relationship between the number of days, x, and the total cost in dollars, y, of renting a compact car.
a. What is the fixed administrative fee?
Answer:
The administrative fee is $15.

b. What is the rental cost per day?
Answer:
It costs $30 per day to rent the compact car.

Eureka Math Grade 8 Module 5 End of Module Assessment Answer Key

Engage NY Eureka Math 8th Grade Module 5 End of Module Assessment Answer Key

Eureka Math Grade 8 Module 5 End of Module Assessment Task Answer Key

Question 1.
a. We define x as a year between 2008 and 2013 and y as the total number of smartphones sold that year, in millions. The table shows values of x and corresponding y values.
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 1
i) How many smartphones were sold in 2009?
Answer:
17.3 Million Smartphones were sold in 2009

ii) In which year were 90 million smartphones sold?
Answer:
90 Million Smartphones were sold in 2011

iii) Is y a function of x? Explain why or why not.
Answer:
Yes, It is a function because for each input there is exactly one output. Specifically only one number will be assigned to represent the number of smartphones sold in the given year.

b. Randy began completing the table below to represent a particular linear function. Write an equation to represent the function he was using and complete the table for him.
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 2
Answer:
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 9
y = 3x + 4

c. Create the graph of the function in part (b).
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 3
Answer:
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 10

d. At NYU in 2013, the cost of the weekly meal plan options could be described as a function of the number of meals. Is the cost of the meal plan a linear or nonlinear function? Explain.

8 meals: $125/week
10 meals: $135/week
12 meals: $155/week
21 meals: $220/week
Answer:
\(\frac{125}{8}\) = 15.625
\(\frac{135}{10}\) = 13.5
\(\frac{155}{12}\) = 12.917
\(\frac{220}{21}\) = 10.476
The cost of the meal plan is a nonlinear function. The cost per meal is different based on the plan. Chosen for example, one plan charges almost $16 per meal while another is about $10. Also, when the data is graphed, The points do not fall on a line.
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 11

Question 2.
The cost to enter and go on rides at a local water park, Wally’s Water World, is shown in the graph below.
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 4
A new water park, Tony’s Tidal Takeover, just opened. You have not heard anything specific about how much it costs to go to this park, but some of your friends have told you what they spent. The information is organized in the table below.
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 5
Each park charges a different admission fee and a different fee per ride, but the cost of each ride remains the same.
a. If you only have $14 to spend, which park would you attend (assume the rides are the same quality)? Explain.
Answer:
Let x represent the number of rides
Let w represent the total cost at wally’s water world
W = 2x + 8
Wally’s
W = 2x + 8
14 = 2x + 8
6 = 2x
3 = x

Tony’s
T = 0.75x + 12
14 = 0.75x + 12
2 = 0.75x
2.67 ≈ x
At wally’s you can go in 3 rides with $14, At tony’s just 2 rides. I would go to wally’s because i could go on more rides.

b. Another water park, Splash, opens, and they charge an admission fee of $30 with no additional fee for rides. At what number of rides does it become more expensive to go to Wally’s Water World than Splash? At what number of rides does it become more expensive to go to Tony’s Tidal Takeover than Splash?
Answer:
Let S represent total cost at splash, S = 30
Wally’s
30 = 2x + 8
22 = 2x
11 = x

Tony’s
30 = 0.75x + 12
18 = 0.75x
24 = x
At Wally’s you can go on 11 rides with $30. The 12th ride makes wally’s more expensive than splash.
At Tony’s you can go on 24 rides with $30. The 25th ride makes tony’s more expensive than splash.

c. For all three water parks, the cost is a function of the number of rides. Compare the functions for all three water parks in terms of their rate of change. Describe the impact it has on the total cost of attending each park.
Answer:
Wally’s rate of change is 2, $2 per ride.
Tony’s rate of change is 0.75, $0.75 per ride.
Splash’s rate of change is 0, $0 extra per ride.
Wally’s has the greatest rate of change that means that the total cost at wally’s will increase the fastest as we go on more rides. At tony’s the rate of change is just 0.75 so the total cost increases with the number of rides we go on, but not as quickly as wally’s. Splash has a rate of change of zero, The number of rides we go on does not impact the total cost at all.

Question 3.
For each part below, leave your answers in terms of π.
a. Determine the volume for each three-dimensional figure shown below.
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 6
Answer:
V = \(\frac{1}{3}\) π (16)(9)
= (16)(3)π
= 48 π
The volume is 48 π mm2

V = π (4)(5.3)
= π (21.2)
= 21.2 π
The volume is 21.2 π cm3.

V = \(\frac{4}{3}\) π (32)
= 4(9) π
= 36 π
The volume is 36 πin3

b. You want to fill the cylinder shown below with water. All you have is a container shaped like a cone with a radius of 3 inches and a height of 5 inches; you can use this cone-shaped container to take water from a faucet and fill the cylinder. How many cones will it take to fill the cylinder?
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 7
Answer:
Volume of cylinder = π (64) (3)
= 192 π
Volume of cone = \(\frac{1}{3}\) π (9) (5)
= \(\frac{45}{3}\) π
= 15 π
The volume of cylinder is 192 π in3
The volume of cone is 15 π in3
\(\frac{192 \pi}{15 \pi}\) = \(\frac{192}{15}\) = 12.8
It takes 12.8 cone of the given size to fill the cylinder.

c. You have a cylinder with a diameter of 15 inches and height of 12 inches. What is the volume of the largest sphere that will fit inside of it?
Engage NY Math 8th Grade Module 5 End of Module Assessment Answer Key 8
Answer:
The cylinder has radius of 7.5 cm, but the height is just 12 cm. That means the maximum radius for the sphere is 6 cm. Anything larger would not fit in the cylinder. Then the volume of the largest sphere that will fit in the cylinder is 288 in3.
V= \(\frac{4}{3}\) π (63)
= \(\frac{4}{3}\) π (216)
= 288 π