Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions

Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions

Want to master in Algebra 2 Ch 6 Concepts like Exponential Growth and Decay Functions, Natural Base e, Logarithms, and its functions, properties of logarithms, etc.? Then, accessing this guide is the perfect solution for your concerns. Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions make you understand the topics quickly and clear all your subject queries. Moreover, the BIM Algebra 2 Ch 6 Solutions prevailing are written by the subject experts based on the guidelines of the latest common core curriculum. Hence, download the Big Ideas Math Algebra 2 Ch 6 Exponential and Logarithmic Functions & enhance your maths skills.

Big Ideas Math Book Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions

BIM Algebra 2 Ch 6 Exponential and Logarithmic Functions covers Questions from Exercises 6.1 to 6.7, Review Tests, Chapter Tests, Cumulative Assessments, etc. With the help of this ultimate preparation guide provided performance task test, students can check their knowledge on Chapter 6 Exponential and Logarithmic Functions concepts. Tap on the link directly and practice each & every lesson in Ch 6 Big Ideas Math Algebra 2 Solution Key Pdf. Access online Big Ideas Math Book Algebra 2 Ch 6 Exponential and Logarithmic Functions Answers pdf without paying a single penny.

Exponential and Logarithmic Functions Maintaining Mathematical Proficiency

Evaluate the expression.
Question 1.
3 • 24
Answer:

Question 2.
(−2)5
Answer:

Question 3.
−(\(\frac{5}{6}\))2
Answer:

Question 4.
(\(\frac{3}{4}\))3
Answer:

Find the domain and range of the function represented by the graph.
Question 5.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 1
Answer:

Question 6.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 2
Answer:

Question 7.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 3
Answer:

Question 8.
ABSTRACT REASONING
Consider the expressions −4n and (−4)n, where n is an integer. For what values of n is each expression negative? positive? Explain your reasoning.
Answer:

Exponential and Logarithmic Functions Mathematical Practices

Mathematically proficient students know when it is appropriate to use general methods and shortcuts.

Monitoring Progress

Determine whether the data can be modeled by an exponential or linear function. Explain your reasoning. Then write the appropriate model and find y when x = 10
Question 1.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 4.1
Answer:

Question 2.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 4.2
Answer:

Question 3.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 4.3
Answer:

Question 4.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 4.4
Answer:

Lesson 6.1 Exponential Growth and Decay Functions

Essential Question What are some of the characteristics of the graph of an exponential function?You can use a graphing calculator to evaluate an exponential function. For example, consider the exponential function f(x) = 2x
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 1

EXPLORATION 1

Identifying Graphs of Exponential Functions
Work with a partner. Match each exponential function with its graph. Use a table of values to sketch the graph of the function, if necessary.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 2

EXPLORATION 2

Characteristics of Graphs of Exponential Functions
Work with a partner. Use the graphs in Exploration 1 to determine the domain, range, and y-intercept of the graph of f(x) = bx, where b is a positive real number other than 1. Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 3

Communicate Your Answer

Question 3.
What are some of the characteristics of the graph of an exponential function?
Answer:

Question 4.
In Exploration 2, is it possible for the graph of f (x) =b x to have an x-intercept? Explain your reasoning.
Answer:

Monitoring Progress

Tell whether the function represents exponential growth or exponential decay. Then graph the function.
Question 1.
y = 4x
Answer:

Question 2.
y = (\(\frac{2}{3}\))x
Answer:

Question 3.
f(x) = (0.25)x
Answer:

Question 4.
f(x) = (1.5)x
Answer:

Question 5.
WHAT IF?
In Example 2, the value of the car can be approximated by the model y = 25(0.9)t. Identify the annual percent decrease in the value of the car. Estimate when the value of the car will be $8000.
Answer:

Question 6.
WHAT IF?
In Example 3, assume the world population increased by 1.5% each year. Write an equation to model this situation. Estimate the year when the world population was 7 billion.
Answer:

Question 7.
The amount y (in grams) of the radioactive isotope iodine-123 remaining after t hours is y = a(0.5)t/13, where a is the initial amount (in grams). What percent of the iodine-123 decays each hour?
Answer:

Question 8.
WHAT IF?
In Example 5, find the balance after 3 years when the interest is compounded daily.
Answer:

Exponential Growth and Decay Functions 6.1 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
In the exponential growth model y = 2.4(1.5)x, identify the initial amount, the growth factor, and the percent increase.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which characteristic of an exponential decay function does not belong with the other three? Explain your reasoning
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 4
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, evaluate the expression for (a) x = -2 and (b) x = 3.
Question 3.
2x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 3

Question 4.
4x
Answer:

Question 5.
8 • 3x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 5

Question 6.
6 • 2x
Answer:

Question 7.
5 + 3x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 7

Question 8.
2x − 2
Answer:

In Exercises 9–18, tell whether the function represents exponential growth or exponential decay. Then graph the function.
Question 9.
y = 6x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 9

Question 10.
y = 7x
Answer:

Question 11.
y =(\(\frac{1}{6}\))x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 11

Question 12.
y =(\(\frac{1}{8}\))x
Answer:

Question 13.
y = (\(\frac{4}{3}\))x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 13

Question 14.
y = (\(\frac{2}{5}\))x
Answer:

Question 15.
y = (1.2)x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 15

Question 16.
y = (0.75)x
Answer:

Question 17.
y = (0.6)x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 17

Question 18.
y = (1.8)x
Answer:

ANALYZING RELATIONSHIPS In Exercises 19 and 20, use the graph of f(x) = bx to identify the value of the base b.
Question 19.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 19

Question 20.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 6
Answer:

Question 21.
MODELING WITH MATHEMATICS
The value of a mountain bike y (in dollars) can be approximated by the model y = 200(0.75)t, where t is the number of years since the bike was new.
a. Tell whether the model represents exponential growth or exponential decay.
b. Identify the annual percent increase or decrease in the value of the bike.
c. Estimate when the value of the bike will be $50.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 21

Question 22.
MODELING WITH MATHEMATICS
The population P(in thousands) of Austin, Texas, during a recent decade can be approximated by y = 494.29(1.03)t, where t is the number of years since the beginning of the decade.
a. Tell whether the model represents exponential growth or exponential decay.
b. Identify the annual percent increase or decrease in population.
c. Estimate when the population was about 590,000.
Answer:

Question 23.
MODELING WITH MATHEMATICS
In 2006, there were approximately 233 million cell phone subscribers in the United States. During the next 4 years, the number of cell phone subscribers increased by about 6% each year.
a. Write an exponential growth model giving the number of cell phone subscribers y (in millions) t years after 2006. Estimate the number of cell phone subscribers in 2008.
b. Estimate the year when the number of cell phone subscribers was about 278 million.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 23

Question 24.
MODELING WITH MATHEMATICS
You take a 325 milligram dosage of ibuprofen. During each subsequent hour, the amount of medication in your bloodstream decreases by about 29% each hour.
a. Write an exponential decay model giving the amount y (in milligrams) of ibuprofen in your bloodstream t hours after the initial dose.
b. Estimate how long it takes for you to have 100milligrams of ibuprofen in your bloodstream.
Answer:

JUSTIFYING STEPS In Exercises 25 and 26, justify each step in rewriting the exponential function.
Question 25.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 25

Question 26.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 8
Answer:

Question 27.
PROBLEM SOLVING
When a plant or animal dies, it stops acquiring carbon-14 from the atmosphere. The amount y (in grams) of carbon-14 in the body of an organism after t years is y = a(0.5)t/5730, where a is the initial amount (in grams). What percent of the carbon-14 is released each year?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 27

Question 28.
PROBLEM SOLVING
The number y of duckweed fronds in a pond after t days is y = a(1230.25)t/16, where a is the initial number of fronds. By what percent does the duckweed increase each day?
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 9
Answer:

In Exercises 29–36, rewrite the function in the form y = a(1 + r)t or y = a(1 – r)t. Then state the growth or decay rate.
Question 29.
y = a(2)t/3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 29

Question 30.
y = a(4)t/6
Answer:

Question 31.
y = a(0.5)t/12
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 31

Question 32.
y = a(0.25)t/9
Answer:

Question 33.
y = a(\(\frac{2}{3}\))t/10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 33

Question 34.
y = a(\(\frac{5}{4}\))t/22
Answer:

Question 35.
y = a(2)8t
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 35

Question 36.
y = a(\(\frac{1}{3}\))3t
Answer:

Question 37.
PROBLEM SOLVING
You deposit $5000 in an account that pays 2.25% annual interest. Find the balance after 5 years when the interest is compounded quarterly.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 37

Question 38.
DRAWING CONCLUSIONS
You deposit $2200 into three separate bank accounts that each pay 3% annual interest. How much interest does each account earn after 6 years?
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 10
Answer:

Question 39.
ERROR ANALYSIS
You invest $500 in the stock of a company. The value of the stock decreases 2% each year. Describe and correct the error in writing a model for the value of the stock after t years.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 11
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 39

Question 40.
ERROR ANALYSIS
You deposit $250 in an account that pays 1.25% annual interest. Describe and correct the error in finding the balance after 3 years when the interest is compounded quarterly.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 12
Answer:

In Exercises 41– 44, use the given information to find the amount A in the account earning compound interest after 6 years when the principal is $3500.
Question 41.
r = 2.16%, compounded quarterly
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 41

Question 42.
r = 2.29%, compounded monthly
Answer:

Question 43.
r = 1.83%, compounded daily
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 43

Question 44.
r = 1.26%, compounded monthly
Answer:

Question 45.
USING STRUCTURE
A website recorded the number y of referrals it received from social media websites over a 10-year period. The results can be modeled by y = 2500(1.50)t, where t is the year and 0 ≤ t≤ 9. Interpret the values of a and b in this situation. What is the annual percent increase? Explain.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 45

Question 46.
HOW DO YOU SEE IT?
Consider the graph of an exponential function of the form f (x) = abx.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 13
a. Determine whether the graph of f represents exponential growth or exponential decay.
b. What are the domain and range of the function? Explain.
Answer:

Question 47.
MAKING AN ARGUMENT
Your friend says the graph of f(x) = 2x increases at a faster rate than the graph of g (x) = x2 when x ≥ 0. Is your friend correct? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 14
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 47

Question 48.
THOUGHT PROVOKING
The function f(x) = bx represents an exponential decay function. Write a second exponential decay function in terms of b and x.
Answer:

Question 49.
PROBLEM SOLVING
The population p of a small town after x years can be modeled by the function p = 6850(1.03)x. What is the average rate of change in the population over the first 6 years? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 49

Question 50.
REASONING
Consider the exponential function f(x) = abx.
a. Show that \(\frac{f(x+1)}{f(x)}\) = b.
b. Use the equation in part (a) to explain why there is no exponential function of the form f(x) = abx whose graph passes through the points in the table below.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 15
Answer:

Question 51.
PROBLEM SOLVING
The number E of eggs a Leghorn chicken produces per year can be modeled by the equation E = 179.2(0.89)w/52, where w is the age (in weeks) of the chicken and w ≥ 22.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 16
a. Identify the decay factor and the percent decrease.
b. Graph the model.
c. Estimate the egg production of a chicken that is 2.5 years old.d. Explain how you can rewrite the given equation so that time is measured in years rather than in weeks.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 51

Question 52.
CRITICAL THINKING
You buy a new stereo for $1300 and are able to sell it 4 years later for $275. Assume that the resale value of the stereo decays exponentially with time. Write an equation giving the resale value V(in dollars) of the stereo as a function of the time t (in years) since you bought it.
Answer:

Maintaining Mathematical Proficiency

Simplify the expression.
Question 53.
x9 • x2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 53

Question 54.
\(\frac{x^{4}}{x^{3}}\)
Answer:

Question 55.
4x • 6x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 55

Question 56.
\(\left(\frac{4 x^{8}}{2 x^{6}}\right)4\)
Answer:

Question 57.
\(\frac{x+3 x}{2}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 57

Question 58.
\(\frac{6x}{2}\) + 4x
Answer:

Question 59.
\(\frac{12x}{4x}\) + 5x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.1 a 59

Question 60.
(2x • 3x5)3
Answer:

Lesson 6.2 The Natural Base e

Essential Question What is the natural base e?
So far in your study of mathematics, you have worked with special numbers such as π and i. Another special number is called the natural base and is denoted by e. The natural base e is irrational, so you cannot find its exact value.

EXPLORATION 1

Approximating the Natural Base e
Work with a partner. One way to approximate the natural base e is to approximate the sum
1 + \(\frac{1}{1}+\frac{1}{1 \cdot 2}+\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{1 \cdot 2 \cdot 3 \cdot 4}+\cdots\)
Use a spreadsheet or a graphing calculator to approximate this sum. Explain the steps you used. How many decimal places did you use in your approximation?
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 1

EXPLORATION 2

Approximating the Natural Base e
Work with a partner. Another way to approximate the natural base e is to consider the expression(1 + \(\frac{1}{x}\))x .
As x increases, the value of this expression approaches the value of e. Copy and complete the table. Then use the results in the table to approximate e. Compare this approximation to the one you obtained in Exploration 1.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 2

EXPLORATION 3

Graphing a Natural Base Function
Work with a partner. Use your approximate value of e in Exploration 1 or 2 to complete the table. Then sketch the graph of the natural base exponential function y =ex. You can use a graphing calculator and the ex key to check your graph. What are the domain and range of y = ex? Justify your answers.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 3

Communicate Your Answer

Question 4.
What is the natural base e?
Answer:

Question 5.
Repeat Exploration 3 for the natural base exponential function y = e-x. Then compare the graph of y = ex to the graph of y = e-x.
Answer:

Question 6.
The natural base e is used in a wide variety of real-life applications. Use the Internet or some other reference to research some of the real-life applications of e.
Answer:

Monitoring Progress

Simplify the expression.
Question 1.
e7 • e4
Answer:

Question 2.
\(\frac{24 e^{8}}{8 e^{5}}\)
Answer:

Question 3.
(10e-3x)3
Answer:

Tell whether the function represents exponential growth or exponential decay. Then graph the function.
Question 4.
y = \(\frac{1}{2}\)ex
Answer:

Question 5.
y = 4e-x
Answer:

Question 6.
f(x) = 2e2x
Answer:

Question 7.
You deposit $4250 in an account that earns 5% annual interest compounded continuously. Compare the balance after 10 years with the accounts in Example 3.
Answer:

The Natural Base e 6.2 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
What is the natural base e?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 1

Question 2.
WRITING
Tell whether the function f(x) = \(\frac{1}{3}\)e4x represents exponential growth or exponential decay. Explain.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, simplify the expression.
Question 3.
e3 • e5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 3

Question 4.
e-4 • e6
Answer:

Question 5.
\(\frac{11 e^{9}}{22 e^{10}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 5

Question 6.
\(\frac{27 e^{7}}{3 e^{4}}\)
Answer:

Question 7.
(5e7x)4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 7

Question 8.
(4e-2x)3
Answer:

Question 9.
\(\sqrt{9 e^{6 x}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 9

Question 10.
\(\sqrt[3]{8 e^{12 x}}\)
Answer:

Question 11.
ex • e-6x • e8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 11

Question 12.
ex • e4 • ex+3
Answer:

ERROR ANALYSIS In Exercises 13 and 14, describe and correct the error in simplifying the expression.
Question 13.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 13

Question 14.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 5
Answer:

In Exercises 15–22, tell whether the function represents exponential growth or exponential decay. Then graph the function.
Question 15.
y = e3x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 15

Question 16.
y = e-2x
Answer:

Question 17.
y = 2e-x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 17

Question 18.
y = 3e2x
Answer:

Question 19.
y = 0.5ex
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 19

Question 20.
y = 0.25e-3x
Answer:

Question 21.
y = 0.4e-0.25x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 21

Question 22.
y = 0.6e0.5x
Answer:

ANALYZING EQUATIONS In Exercises 23–26, match the function with its graph. Explain your reasoning.
Question 23.
y = e2x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 23

Question 24.
y = e-2x
Answer:

Question 25.
y = 4e-0.5x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 25

Question 26.
y = 0.75ex
Answer:

Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 6

USING STRUCTURE In Exercises 27–30, use the properties of exponents to rewrite the function in the form y = a(1 + r)t or y = a(1 – r)t. Then find the percent rate of change.
Question 27.
y = e-0.25t
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 27

Question 28.
y = e-0.75t
Answer:

Question 29.
y = 2e0.4t
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 29

Question 30.
y = 0.5e0.8t
Answer:

USING TOOLS In Exercises 31–34, use a table of values or a graphing calculator to graph the function. Then identify the domain and range.
Question 31.
y = ex-2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 31

Question 32.
y = ex+1
Answer:

Question 33.
y = 2ex + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 33

Question 34.
y = 3ex − 5
Answer:

Question 35.
MODELING WITH MATHEMATICS
Investment accounts for a house and education earn annual interest compounded continuously. The balance H(in dollars) of the house fund after t years can be modeled by H = 3224e0.05t. The graph shows the balance in the education fund over time. Which account has the greater principal? Which account has a greater balance after 10 years?
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 35

Question 36.
MODELING WITH MATHEMATICS
Tritium and sodium-22 decay over time. In a sample of tritium, the amount y (in milligrams) remaining after t years is given by y = 10e−0.0562t. The graph shows the amount of sodium-22 in a sample over time. Which sample started with a greater amount? Which has a greater amount after 10 years?
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 8
Answer:

Question 37.
OPEN-ENDED
Find values of a, b, r, and q such that f(x) = aerx and g(x) = beqx are exponential decay functions, but \(\frac{f(x)}{g(x)}\) represents exponential growth.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 37

Question 38.
THOUGHT PROVOKING
Explain why A = P(1 + \(\frac{r}{n}\))nt approximates A =Pert as n approaches positive infinity.
Answer:

Question 39.
WRITING
Can the natural base e be written as a ratio of two integers? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 39

Question 40.
MAKING AN ARGUMENT
Your friend evaluates f(x) = e-x when x = 1000 and concludes that the graph of y = fx) has an x-intercept at (1000, 0). Is your friend correct? Explain your reasoning.
Answer:

Question 41.
DRAWING CONCLUSIONS
You invest $2500 in an account to save for college. Account 1 pays 6% annual interest compounded quarterly. Account 2 pays 4% annual interest compounded continuously. Which account should you choose to obtain the greater amount in 10 years? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 41

Question 42.
HOW DO YOU SEE IT?
Use the graph to complete each statement.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 9
a. f(x) approaches ____ as x approaches +∞.
b. f(x) approaches ____ as x approaches −∞.
Answer:

Question 43.
PROBLEM SOLVING
The growth of Mycobacterium tuberculosis bacteria can be modeled by the function N(t) = ae0.166t, where N is the number of cells after t hours and a is the number of cells when t = 0.
a. At 1:00 P.M., there are 30 M. tuberculosis bacteria in a sample. Write a function that gives the number of bacteria after 1:00 P.M.
b. Use a graphing calculator to graph the function in part (a).
c. Describe how to find the number of cells in the sample at 3:45 P.M.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 43

Maintaining Mathematical Proficiency

Write the number in scientific notation.
Question 44.
0.006
Answer:

Question 45.
5000
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 45

Question 46.
26,000,000
Answer:

Question 47.
0.000000047
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 47

Find the inverse of the function. Then graph the function and its inverse.
Question 48.
y = 3x + 5
Answer:

Question 49.
y = x2 − 1, x ≤ 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 49

Question 50.
y = \(\sqrt{x+6}\)
Answer:

Question 51.
y = x3 − 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.2 a 51

Lesson 6.3 Logarithms and Logarithmic Functions

Essential Question What are some of the characteristics of the graph of a logarithmic function?
Every exponential function of the form f(x) = bx, where b is a positive real number other than 1, has an inverse function that you can denote by g(x) = logbx. This inverse function is called a logarithmic function with base b.

EXPLORATION 1

Rewriting Exponential Equations
Work with a partner. Find the value of x in each exponential equation. Explain your reasoning. Then use the value of x to rewrite the exponential equation in its equivalent logarithmic form, x = logby.
a. 2x = 8
b. 3x = 9
c. 4x = 2
d. 5x = 1
e. 5x = \(\frac{1}{5}\)
f. 8x = 4

EXPLORATION 2

Graphing Exponential and Logarithmic Functions
Work with a partner. Complete each table for the given exponential function. Use the results to complete the table for the given logarithmic function. Explain your reasoning. Then sketch the graphs of f and g in the same coordinate plane.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 1

EXPLORATION 3

Characteristics of Graphs of LogarithmicFunctions
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 2
Work with a partner. Use the graphs you sketched in Exploration 2 to determine the domain, range, x-intercept, and asymptote of the graph of g(x) = logbx, where b is a positive real number other than 1. Explain your reasoning.

Communicate Your Answer

Question 4.
What are some of the characteristics of the graph of a logarithmic function?
Answer:

Question 5.
How can you use the graph of an exponential function to obtain the graph of a logarithmic function?
Answer:

Monitoring Progress

Rewrite the equation in exponential form.
Question 1.
log3 81 = 4
Answer:

Question 2.
log7 7 = 1
Answer:

Question 3.
log14 1 = 0
Answer:

Question 4.
log1/2 32 = −5
Answer:

Rewrite the equation in logarithmic form.
Question 5.
72 = 49
Answer:

Question 6.
500 = 1
Answer:

Question 7.
4-1 = \(\frac{1}{4}\)
Answer:

Question 8.
2561/8 = 2
Answer:

Evaluate the logarithm. If necessary, use a calculator and round your answer to three decimal places.
Question 9.
log2 32
Answer:

Question 10.
log27 3
Answer:

Question 11.
log 12
Answer:

Question 12.
ln 0.75
Answer:

Simplify the expression.
Question 13.
8log8 x
Answer:

Question 14.
log7 7−3x
Answer:

Question 15.
log2 64x
Answer:

Question 16.
eln 20
Answer:

Question 17.
Find the inverse of y = 4x.
Answer:

Question 18.
Find the inverse of y = ln(x− 5).
Answer:

Graph the function.
Question 19.
y = log2 x
Answer:

Question 20.
f(x) = log5 x
Answer:

Question 21.
y = log1/2 x
Answer:

Logarithms and Logarithmic Functions 6.3 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
A logarithm with base 10 is called a(n) ___________ logarithm.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 1

Question 2.
COMPLETE THE SENTENCE
The expression log3 9 is read as ______________.
Answer:

Question 3.
WRITING
Describe the relationship between y = 7x and y = log7 x.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 3

Question 4.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 3
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–10, rewrite the equation in exponential form.
Question 5.
log3 9 = 2
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 5

Question 6.
log4 4 = 1
Answer:

Question 7.
log6 1 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 7

Question 8.
log7 343 = 3
Answer:

Question 9.
log1/2 16 =−4
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 9

Question 10.
log3 \(\frac{1}{3}\) = −1
Answer:

In Exercises 11–16, rewrite the equation in logarithmic form.
Question 11.
62 = 36
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 11

Question 12.
120 = 1
Answer:

Question 13.
16-1 = \(\frac{1}{16}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 13

Question 14.
5-2 = \(\frac{1}{25}\)
Answer:

Question 15.
1252/3 = 25
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 15

Question 16.
491/2 = 7
Answer:

In Exercises 17–24, evaluate the logarithm.
Question 17.
log3 81
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 17

Question 18.
log7 49
Answer:

Question 19.
log3 3
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 19

Question 20.
log1/2 1
Answer:

Question 21.
log5 \(\frac{1}{625}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 21

Question 22.
log8 \(\frac{1}{512}\)
Answer:

Question 23.
log4 0.25
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 23

Question 24.
log10 0.001
Answer:

Question 25.
NUMBER SENSE
Order the logarithms from least value to greatest value.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 4
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 25

Question 26.
WRITING
Explain why the expressions log2(−1) and log1 1 are not defined.
Answer:

In Exercises 27–32, evaluate the logarithm using a calculator. Round your answer to three decimal places.
Question 27.
log 6
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 27

Question 28.
ln 12
Answer:

Question 29.
ln \(\frac{1}{3}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 29

Question 30.
log \(\frac{2}{7}\)
Answer:

Question 31.
3 ln 0.5
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 31

Question 32.
log 0.6 + 1
Answer:

Question 33.
MODELING WITH MATHEMATICS
Skydivers use an instrument called an altimeter to track their altitude as they fall. The altimeter determines altitude by measuring air pressure. The altitude h (in meters) above sea level is related to the air pressure P(in pascals) by the function shown in the diagram. What is the altitude above sea level when the air pressure is 57,000 pascals?
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 5
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 33

Question 34.
MODELING WITH MATHEMATICS
The pH value for a substance measures how acidic or alkaline the substance is. It is given by the formula pH = −log[H+], where H+ is the hydrogen ion concentration (in moles per liter). Find the pH of each substance.
a. baking soda: [H+] = 10-8 moles per liter
b. vinegar: [H+] = 10-3 moles per liter
Answer:

In Exercises 35–40, simplify the expression.
Question 35.
7log7x
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 35

Question 36.
3log3 5x
Answer:

Question 37.
eln4
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 37

Question 38.
10log 15
Answer:

Question 39.
log3 32x
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 39

Question 40.
ln ex + 1
Answer:

Question 41.
ERROR ANALYSIS
Describe and correct the error in rewriting 4−3 = \(\frac{1}{64}\) in logarithmic form.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 6
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 41

Question 42.
ERROR ANALYSIS
Describe and correct the error in simplifying the expression log4 64x.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 7
Answer:

In Exercises 43–52, find the inverse of the function.
Question 43.
y = 0.3x
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 43

Question 44.
y = 11x
Answer:

Question 45.
y = log2 x
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 45

Question 46.
y = log1/5 x
Answer:

Question 47.
y = ln(x − 1)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 47

Question 48.
y = ln 2x
Answer:

Question 49.
y = e3x
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 49

Question 50.
y = ex-4
Answer:

Question 51.
y = 5x − 9
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 51

Question 52.
y = 13 + log x
Answer:

Question 53.
PROBLEM SOLVING
The wind speed s (in miles per hour) near the center of a tornado can be modeled by s = 93 log d + 65, where d is the distance (in miles) that the tornado travels.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 8
a. In 1925, a tornado traveled 220 miles through three states. Estimate the wind speed near the center of the tornado.
b. Find the inverse of the given function. Describe what the inverse represents.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 53

Question 54.
MODELING WITH MATHEMATICS
The energy magnitude M of an earthquake can be modeled by M = \(\frac{2}{3}\)log E − 9.9, where E is the amount of energy released (in ergs).
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 9
a. In 2011, a powerful earthquake in Japan, caused by the slippage of two tectonic plates along a fault, released 2.24 × 1028 ergs. What was the energy magnitude of the earthquake?
b. Find the inverse of the given function. Describe what the inverse represents.
Answer:

In Exercises 55–60, graph the function.
Question 55.
y = log4 x
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 55

Question 56.
y = log6 x
Answer:

Question 57.
y = log1/3 x
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 57.1
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 57.2

Question 58.
y = log1/4 x
Answer:

Question 59.
y = log2 x − 1
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 59

Question 60.
y = log3(x + 2)
Answer:

USING TOOLS In Exercises 61–64, use a graphing calculator to graph the function. Determine the domain, range, and asymptote of the function.
Question 61.
y = log(x+ 2)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 61

Question 62.
y = −ln x
Answer:

Question 63.
y = ln(−x)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 63

Question 64.
y = 3 − log x
Answer:

Question 65.
MAKING AN ARGUMENT
Your friend states that every logarithmic function will pass through the point (1, 0). Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 65

Question 66.
ANALYZING RELATIONSHIPS
Rank the functions in order from the least average rate of change to the greatest average rate of change over the interval 1 ≤ x ≤ 10.
a. y = log6 x
b. y = log3/5 x
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 10
Answer:

Question 67.
PROBLEM SOLVING
Biologists have found that the length ℓ (in inches) of an alligator and its weight w (in pounds) are related by the function ℓ = 27.1 ln w − 32.8.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 11
a. Use a graphing calculator to graph the function.
b. Use your graph to estimate the weight of an alligator that is 10 feet long.
c. Use the zero feature to find the x-intercept of the graph of the function. Does this x-value make sense in the context of the situation? Explain.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 67

Question 68.
HOW DO YOU SEE IT?
The figure shows the graphs of the two functions f and g.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 12
a. Compare the end behavior of the logarithmic function g to that of the exponential function f.
b. Determine whether the functions are inverse functions. Explain.
c. What is the base of each function? Explain.
Answer:

Question 69.
PROBLEM SOLVING
A study in Florida found that the number s of fish species in a pool or lake can be modeled by the function s = 30.6 − 20.5 log A + 3.8(log A)2
where A is the area (in square meters) of the pool or lake.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 13
a. Use a graphing calculator to graph the function on the domain 200 ≤ A ≤ 35,000.
b. Use your graph to estimate the number of species in a lake with an area of 30,000 square meters.
c. Use your graph to estimate the area of a lake that contains six species of fish.
d. Describe what happens to the number of fish species as the area of a pool or lake increases. Explain why your answer makes sense.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 69

Question 70.
THOUGHT PROVOKING
Write a logarithmic function that has an output of −4. Then sketch the graph of your function.
Answer:

Question 71.
CRITICAL THINKING
Evaluate each logarithm. (Hint: For each logarithm logb x, rewrite b and x as powers of the same base.)
a. log125 25
b. log8 32
c. log27 81
d. log4 128
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 71

Maintaining Mathematical Proficiency

Let f(x) = \(\sqrt [ 3 ]{ x }\). Write a rule for g that represents the indicated transformation of the graph of f.
Question 72.
g(x) = −f(x)
Answer:

Question 73.
g(x) = f (\(\frac{1}{2}\)x )
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 73

Question 74.
g(x) = f(−x) + 3
Answer:

Question 75.
g(x) = f(x+ 2)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 75

Identify the function family to which f belongs. Compare the graph of f to the graph of its parent.
Question 76.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 14
Answer:

Question 77.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 15
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 a 77

Question 78.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions 6.3 16
Answer:

Lesson 6.4 Transformations of Exponential and Logarithmic Functions

Essential Question How can you transform the graphs of exponential and logarithmic functions?

EXPLORATION 1

Identifying Transformations
Work with a partner. Each graph shown is a transformation of the parent function
f(x) = ex or f(x) = ln x.
Match each function with its graph. Explain your reasoning. Then describe the transformation of f represented by g.
a. g(x) = ex+2 − 3
b. g(x) = −ex+2 + 1
c. g(x) = ex-2 − 1
d. g(x) = ln(x + 2)
e. g(x) = 2 + ln x
f. g(x) = 2 + ln(−x)
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 1

EXPLORATION 2

Characteristics of Graphs
Work with a partner. Determine the domain, range, and asymptote of each function in Exploration 1. Justify your answers.

Communicate Your Answer

Question 3.
How can you transform the graphs of exponential and logarithmic functions?
Answer:

Question 4.
Find the inverse of each function in Exploration 1. Then check your answer by using a graphing calculator to graph each function and its inverse in the same viewing window.
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 2
Answer:

Monitoring Progress

Describe the transformation of f represented by g. Then graph each function.
Question 1.
f(x) = 2x, g(x) = 2x-3 + 1
Answer:

Question 2.
f(x) = e-x, g(x) = e-x − 5
Answer:

Question 3.
f(x) = 0.4x, g(x) = 0.4-2x
Answer:

Question 4.
f(x) = ex, g(x) = −ex+6
Answer:

Describe the transformation of f represented by g. Then graph each function.
Question 5.
f(x) = log2x, g(x) = −3 log2x
Answer:

Question 6.
f(x) = log1/4x, g(x) = log1/4(4x) − 5
Answer:

Question 7.
Let the graph of g be a horizontal stretch by a factor of 3, followed by a translation 2 units up of the graph of f(x) = e-x. Write a rule for g.
Answer:

Question 8.
Let the graph of g be a reflection in the y-axis, followed by a translation 4 units to the left of the graph of f(x) = log x. Write a rule for g.
Answer:

Transformations of Exponential and Logarithmic Functions 6.4 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Given the function f(x) = abx-h + k, describe the effects of a, h, and k on the graph of the function.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 1

Question 2.
COMPLETE THE SENTENCE
The graph of g(x) = log4(−x) is a reflection in the __________ of the graph of f(x) = log4x,
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, match the function with its graph. Explain your reasoning.
Question 3.
f(x) = 2x+2 − 2
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 3

Question 4.
g(x) = 2x+2 + 2
Answer:

Question 5.
h(x) = 2x-2 − 2
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 5

Question 6.
k(x) = 2x-2 + 2
Answer:

Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 3

In Exercises 7–16, describe the transformation of f represented by g. Then graph each function.
Question 7.
f(x) = 3x, g(x) = 3x + 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 7

Question 8.
f(x) = 4x, g(x) = 4x − 8
Answer:

Question 9.
f(x) = ex, g(x) = ex − 1
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 9

Question 10.
f(x) = ex, g(x) = ex + 4
Answer:

Question 11.
f(x) = 2x, g(x) = 2x-7
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 11

Question 12.
f(x) = 5x, g(x) = 5x+1
Answer:

Question 13.
f(x) = e-x, g(x) = e-x + 6
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 13

Question 14.
f(x) = e-x, g(x) = e-x − 9
Answer:

Question 15.
f(x) = (\(\frac{1}{4}\))x, g (x) = (\(\frac{1}{4}\))x-3 + 12
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 15

Question 16.
f(x) = (\(\frac{1}{3}\))x, g(x) = (\(\frac{1}{3}\))x+2 − \(\frac{2}{3}\)
Answer:

In Exercises 17–24, describe the transformation of f represented by g. Then graph each function.
Question 17.
f(x) = ex, g(x) = e2x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 17

Question 18.
f(x) = ex, g (x) =4—3 ex
Answer:

Question 19.
f(x) = 2x, g(x) = −2x-3
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 19

Question 20.
f(x) = 4x, g(x) = 40.5x-5
Answer:

Question 21.
f(x) = e-x, g(x) = 3e-6x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 21

Question 22.
f(x) = e-x, g(x) = e-5x + 2
Answer:

Question 23.
f(x) = (\(\frac{1}{2}\))x, g(x) = 6 (\(\frac{1}{2}\))x+5 − 2
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 23

Question 24.
f(x) = (\(\frac{3}{4}\))x, g(x) = −(\(\frac{3}{4}\)) -7 + 1
Answer:

ERROR ANALYSIS In Exercises 25 and 26, describe and correct the error in graphing the function.
Question 25.
f(x) = 2x + 3
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 4
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 25

Question 26.
f(x) = 3-x
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 5
Answer:

In Exercises 27–30, describe the transformation of f represented by g. Then graph each function.
Question 27.
f(x) = log4 x, g(x) = 3 log4 x − 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 27

Question 28.
f(x) = log1/3 x, g(x) = log1/3(−x) + 6
Answer:

Question 29.
f(x) = log1/5 x, g(x) = − log1/5(x − 7)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 29

Question 30.
f(x) = log2 x, g(x) = log2(x + 2) − 3
Answer:

ANALYZING RELATIONSHIPS In Exercises 31–34, match the function with the correct transformation of the graph of f. Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 6
Question 31.
y = f(x− 2)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 31

Question 32.
y = f(x + 2)
Answer:

Question 33.
y = 2f(x)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 33

Question 34.
y = f(2x)
Answer:

Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 7

In Exercises 35–38, write a rule for g that represents the indicated transformations of the graph of f.
Question 35.
f(x) = 5x; translation 2 units down, followed by a reflection in the y-axis
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 35

Question 36.
f(x) = (\(\frac{2}{3}\))x; reflection in the x-axis, followed by a vertical stretch by a factor of 6 and a translation 4 units left
Answer:

Question 37.
f(x) = ex; horizontal shrink by a factor of \(\frac{1}{2}\), followed by a translation 5 units up
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 37

Question 38.
f(x) =e-x; translation 4 units right and 1 unit down, followed by a vertical shrink by a factor of \(\frac{1}{3}\)
Answer:

In Exercises 39–42, write a rule for g that represents the indicated transformation of the graph of f.
Question 39.
f(x) = log6 x; vertical stretch by a factor of 6, followed by a translation 5 units down
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 39

Question 40.
f(x) = log2 x; reflection in the x-axis, followed by a translation 9 units left
Answer:

Question 41.
f(x) = log1/2 x; translation 3 units left and 2 units up, followed by a reflection in the y-axis
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 41

Question 42.
f(x) = ln x; translation 3 units right and 1 unit up, followed by a horizontal stretch by a factor of 8
Answer:

JUSTIFYING STEPS In Exercises 43 and 44, justify each step in writing a rule for g that represents the indicated transformations of the graph of f.
Question 43.
f(x) = log7 x; reflection in the x-axis, followed by a translation 6 units down
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 8
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 43

Question 44.
f(x) = 8x; vertical stretch by a factor of 4, followed by a translation 1 unit up and 3 units left
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 9
Answer:

USING STRUCTURE In Exercises 45–48, describe the transformation of the graph of f represented by the graph of g. Then give an equation of the asymptote.
Question 45.
f(x) = ex, g(x) = ex + 4
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 45

Question 46.
f(x) = 3x, g(x) = 3x-9
Answer:

Question 47.
f(x) = ln x, g(x) = ln(x + 6)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 47

Question 48.
f(x) = log1/5 x, g(x) = log1/5 x + 13
Answer:

Question 49.
MODELING WITH MATHEMATICS
The slope Sof a beach is related to the average diameter d(in millimeters) of the sand particles on the beach by the equation S = 0.159 + 0.118 log d. Describe the transformation of f(d ) = log d represented by S. Then use the function to determine the slope of a beach for each sand type below.
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 10
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 49

Question 50.
HOW DO YOU SEE IT?
The graphs of f(x) = bx and g(x) = (\(\frac{1}{b}\))x are shown for b = 2.
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 11
a. Use the graph to describe a transformation of the graph of f that results in the graph of g.
b. Does your answer in part (a) change when 0 < b < 1? Explain.
Answer:

Question 51.
MAKING AN ARGUMENT
Your friend claims a single transformation of f(x) = log x can result in a function g whose graph never intersects the graph of f. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 51

Question 52.
THOUGHT PROVOKING
Is it possible to transform the graph of f(x) = ex to obtain the graph of g(x) = ln x? Explain your reasoning.
Answer:

Question 53.
ABSTRACT REASONING
Determine whether each statement is always, sometimes, or never true. Explain your reasoning.
a. A vertical translation of the graph of f(x) = log x changes the equation of the asymptote.
b. A vertical translation of the graph of f(x) = ex changes the equation of the asymptote.
c. A horizontal shrink of the graph of f(x) = log x does not change the domain.
d. The graph of g(x) = abx-h + k does not intersect the x-axis.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 53

Question 54.
PROBLEM SOLVING
The amount P (in grams) of 100 grams of plutonium-239 that remains after t years can be modeled by P= 100(0.99997)t.
a. Describe the domain and range of the function.
b. How much plutonium-239 is present after 12,000 years?
c. Describe the transformation of the function if the initial amount of plutonium were 550 grams.
d. Does the transformation in part (c) affect the domain and range of the function? Explain your reasoning.
Answer:

Question 55.
CRITICAL THINKING
Consider the graph of the function h(x) = e-x-2. Describe the transformation of the graph of f(x) = e-x represented by the graph of h. Then describe the transformation of the graph of g(x) = ex represented by the graph of h. Justify your answers.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 55

Question 56.
OPEN-ENDED
Write a function of the form y = abx-h + k whose graph has a y-intercept of 5 and an asymptote of y = 2.
Answer:

Maintaining Mathematical Proficiency

Perform the indicated operation.
Question 57.
Let f(x) = x4 and g(x) = x2. Find ( fg)(x). Then evaluate the product when x = 3.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 57

Question 58.
Let f(x) = 4x6 and g(x) = 2x3. Find (\(\frac{f}{g}\)) (x). Then evaluate the quotient when x = 5.
Answer:

Question 59.
Let f(x) = 6x3 and g(x) = 8x3. Find ( f + g)(x). Then evaluate the sum when x = 2.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 a 59

Question 60.
Let f(x) = 2xx and g(x) = 3xx. Find ( f – g)(x). Then evaluate the difference when x = 6.
Answer:

Exponential and Logarithmic Functions Study Skills: Forming a Weekly Study Group

6.1–6.4 What Did You Learn?

Core Vocabulary
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 12

Core Concepts
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 13

Mathematical Practices
Question 1.
How did you check to make sure your answer was reasonable in Exercise 23 on page 300?
Answer:

Question 2.
How can you justify your conclusions in Exercises 23–26 on page 307?
Answer:

Question 3.
How did you monitor and evaluate your progress in Exercise 66 on page 315?
Answer:

Study Skills: Forming a Weekly Study Group

  • Select students who are just as dedicated to doing well in the math class as you are.
  • ind a regular meeting place that has minimal distractions.
  • Compare schedules and plan at least one time a week to meet, allowing at least 1.5 hours for study time.
    Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.4 14

Exponential and Logarithmic Functions 6.1–6.4 Quiz

Tell whether the function represents exponential growth or exponential decay. Explain your reasoning.
Question 1.
f(x) = (4.25)x
Answer:

Question 2.
y = (\(\frac{3}{8}\))x
Answer:

Question 3.
y = e0.6x
Answer:

Question 4.
f(x) = 5e-2x
Answer:

Simplify the expression.
Question 5.
e8 • e4
Answer:

Question 6.
\(\frac{15 e^{3}}{3 e}\)
Answer:

Question 7.
(5e4x)3
Answer:

Question 8.
elln 9
Answer:

Question 9.
log7 49x
Answer:

Question 10.
log3 81-2x
Answer:

Rewrite the expression in exponential or logarithmic form.
Question 11.
log4 1024 = 5
Answer:

Question 12.
log1/3 27 = −3
Answer:

Question 13.
74 = 2401
Answer:

Question 14.
4-2 = 0.0625
Answer:

Evaluate the logarithm. If necessary, use a calculator and round your answer to three decimal places.
Question 15.
log 45
Answer:

Question 16.
ln 1.4
Answer:

Question 17.
log2 32
Answer:

Graph the function and its inverse.
Question 18.
f(x) = (\(\frac{1}{9}\))x
Answer:

Question 19.
y = ln(x − 7)
Answer:

Question 20.
f(x) = log5 (x+ 1)
Answer:

The graph of g is a transformation of the graph of f. Write a rule for g.
Question 21.
f(x) = log3 x
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions q 1
Answer:

Question 22.
f(x) = 3x
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions q 2
Answer:

Question 23.
f(x)= log1/2 x
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions q 3
Answer:

Question 24.
You purchase an antique lamp for $150. The value of the lamp increases by 2.15% each year. Write an exponential model that gives the value y (in dollars) of the lamp t years after you purchased it.
Answer:

Question 25.
A local bank advertises two certificate of deposit (CD) accounts that you can use to save money and earn interest. The interest is compounded monthly for both accounts.
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions q 4
a. You deposit the minimum required amounts in each CD account. How much money is in each account at the end of its term? How much interest does each account earn? Justify your answers.
b. Describe the benefits and drawbacks of each account.
Answer:

Question 26.
The Richter scale is used for measuring the magnitude of an earthquake. The Richter magnitude R is given by R = 0.67 ln E + 1.17, where E is the energy (in kilowatt-hours) released by the earthquake. Graph the model. What is the Richter magnitude for an earthquake that releases 23,000 kilowatt-hours of energy?
Answer:

Lesson 6.5 Properties of Logarithms

Essential Question How can you use properties of exponents to derive properties of logarithms?
Let x = logb m and y = logb n.
The corresponding exponential forms of these two equations are
bx = m and by = n.

EXPLORATION 1

Product Property of Logarithms
Work with a partner. To derive the Product Property, multiply m and n to obtain mn =bxby = bx+y.
The corresponding logarithmic form of mn = bx+y is logb mn = x + y. So,

Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 1

EXPLORATION 2

Quotient Property of Logarithms
Work with a partner. To derive the Quotient Property, divide m by n to obtain
\(\frac{m}{n}=\frac{b^{x}}{b^{y}}\) = bx-y.
The corresponding logarithmic form of \(\frac{m}{n}\) = bx-y is logb \(\frac{m}{n}\) = x − y. So,
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 2

EXPLORATION 3

Power Property of Logarithms
Work with a partner. To derive the Power Property, substitute bx for m in the expression logb mn, as follows.
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 3

Communicate Your Answer

Question 4.
How can you use properties of exponents to derive properties of logarithms?
Answer:

Question 5.
Use the properties of logarithms that you derived in Explorations 1–3 to evaluate each logarithmic expression.
a. log4 163
b. log3 81-3
c. ln e2 + ln e5
d. 2 ln e6 − ln e5
e. log5 75 − log5 3
f. log4 2 + log4 32
Answer:

Monitoring Progress

Use log6 5 ≈ 0.898 and log6 8 ≈ 1.161 to evaluate the logarithm.
Question 1.
log6 \(\frac{5}{8}\)
Answer:

Question 2.
log6 40
Answer:

Question 3.
log6 64
Answer:

Question 4.
log6 125
Answer:

Expand the logarithmic expression.
Question 5.
log6 3x4
Answer:

Question 6.
ln \(\frac{5}{12x}\)
Answer:

Condense the logarithmic expression.
Question 7.
log x − log 9
Answer:

Question 8.
ln 4 + 3 ln 3 − ln 12
Answer:

Use the change-of-base formula to evaluate the logarithm.
Question 9.
log5 8
Answer:

Question 10.
log8 14
Answer:

Question 11.
log26 9
Answer:

Question 12.
log12 30
Answer:

Question 13.
WHAT IF?
In Example 6, the artist turns up the volume so that the intensity of the sound triples. By how many decibels does the loudness increase?
Answer:

Properties of Logarithms 6.5 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
To condense the expression log3 2x + log3 y, you need to use the __________ Property of Logarithms.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 1

Question 2.
WRITING
Describe two ways to evaluate log7 12 using a calculator.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, use log7 4 ≈ 0.712 and log7 12 ≈ 1.277 to evaluate the logarithm.
Question 3.
log7 3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 3

Question 4.
log7 48
Answer:

Question 5.
log7 16
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 5

Question 6.
log7 64
Answer:

Question 7.
log7 \(\frac{1}{4}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 7

Question 8.
log7 \(\frac{1}{3}\)
Answer:

In Exercises 9–12, match the expression with the logarithm that has the same value. Justify your answer.
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 9
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 11

In Exercises 13–20, expand the logarithmic expression.
Question 13.
log3 4x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 13

Question 14.
log8 3x
Answer:

Question 15.
log 10x5
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 15

Question 16.
ln 3x4
Answer:

Question 17.
ln \(\frac{x}{3 y}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 17

Question 18.
ln \(\frac{6 x^{2}}{y^{4}}\)
Answer:

Question 19.
log7 5\(\sqrt{x}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 19

Question 20.
log5 \(\sqrt[3]{x^{2} y}\)
Answer:

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in expanding the logarithmic expression.
Question 21.
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 5
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 21

Question 22.
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 6
Answer:

In Exercises 23–30, condense the logarithmic expression.
Question 23.
log4 7 − log4 10
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 23

Question 24.
ln 12 − ln 4
Answer:

Question 25.
6 ln x+ 4 ln y
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 25

Question 26.
2 log x+ log 11
ans;

Question 27.
log5 4 + \(\frac{1}{3}\) log5 x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 27

Question 28.
6 ln 2 − 4 ln y
Answer:

Question 29.
5 ln 2 + 7 ln x + 4 ln y
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 29

Question 30.
log3 4 + 2 log3 \(\frac{1}{2}\) + log3 x
Answer:

Question 31.
REASONING
Which of the following is not equivalent to log3 \(\frac{y^{4}}{3 x}\)? Justify your answer.
A. 4 log5 y − log5 3x
B. 4 log5 y − log5 3 + log5 x
C. 4 log5 y− log5 3 − log5 x
D. log5 y4 − log5 3 − log5 x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 31

Question 32.
REASONING
Which of the following equations is correct? Justify your answer.
A. log7 x + 2 log7 y = log7 (x + y2)
B. 9 log x − 2 log y = log \(\frac{x^{9}}{y^{2}}\)
C. 5 log4 x + 7 log2 y = log6x5y7
D. log9x − 5 log9 y = log9 \(\frac{x}{5 y}\)
Answer:

In Exercises 33–40, use the change-of-base formula to evaluate the logarithm.
Question 33.
log4 7
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 33

Question 34.
log5 13
Answer:

Question 35.
log9 15
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 35

Question 36.
log8 22
Answer:

Question 37.
log6 17
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 37

Question 38.
log2 28
Answer:

Question 39.
log7 \(\frac{3}{16}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 39

Question 40.
log3 \(\frac{9}{40}\)
Answer:

Question 41.
MAKING AN ARGUMENT
Your friend claims you can use the change-of-base formula to graph y = log3x using a graphing calculator. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 41

Question 42.
HOW DO YOU SEE IT?
Use the graph to determine the value of \(\frac{\log 8}{\log 2}\).
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 7
Answer:

MODELING WITH MATHEMATICS In Exercises 43 and 44, use the function L(I ) given in Example 6.
Question 43.
The blue whale can produce sound with an intensity that is 1 million times greater than the intensity of the loudest sound a human can make. Find the difference in the decibel levels of the sounds made by a blue whale and a human.
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 8
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 43

Question 44.
The intensity of the sound of a certain television advertisement is 10 times greater than the intensity of the television program. By how many decibels does the loudness increase?
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 9
Answer:

Question 45.
REWRITING A FORMULA
Under certain conditions, the wind speed s (in knots) at an altitude of h meters above a grassy plain can be modeled by the function s(h) = 2 ln 100h.
a. By what amount does the wind speed increase when the altitude doubles?
b. Show that the given function can be written in terms of common logarithms as
s(h) = \(\frac{2}{\log e}\)(log h+ 2).
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 45

Question 46.
THOUGHT PROVOKING
Determine whether the formula
logb (M + N) = logb M + logbN
is true for all positive, real values of M, N, and b(with b≠ 1). Justify your answer.
Answer:

Question 47.
USING STRUCTURE
Use the properties of exponents to prove the change-of-base formula. (Hint: Let x= logb a, y = logb c, and z = logb a.)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 47

Question 48.
CRITICAL THINKING
Describe three ways to transform the graph of f(x) = log x to obtain the graph of g(x) = log 100x − 1. Justify your answers.
Answer:

Maintaining Mathematical Proficiency

Solve the inequality by graphing.
Question 49.
x2 − 4 > 0
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 49

Question 50.
2(x − 6)2 − 5 ≥ 37
Answer:

Question 51.
x2 + 13x + 42 < 0
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 51

Question 52.
−x2 − 4x + 6 ≤ −6
Answer:

Solve the equation by graphing the related system of equations.
Question 53.
4x2 − 3x − 6 = −x2 + 5x + 3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 53

Question 54.
−(x + 3)(x − 2) = x2 − 6x
Answer:

Question 55.
2x2 − 4x − 5 = −(x + 3)2 + 10
Answer:
Big Ideas Math Answers Algebra 2 Chapter 6 Exponential and Logarithmic Functions 6.5 a 55

Question 56.
−(x + 7)2 + 5 = (x + 10)2 − 3
Answer:

Lesson 6.6 Solving Exponential and Logarithmic Equations

Essential Question How can you solve exponential and logarithmic equations?

EXPLORATION 1

Solving Exponential and Logarithmic Equations
Work with a partner. Match each equation with the graph of its related system of equations. Explain your reasoning. Then use the graph to solve the equation.
a. ex = 2
b. ln x = −1
c. 2x = 3-x
d. log4x = 1
e. log5 x = \(\frac{1}{2}\)
f. 4x = 2
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 1

EXPLORATION 2

Solving Exponential and Logarithmic Equations
Work with a partner. Look back at the equations in Explorations 1(a) and 1(b). Suppose you want a more accurate way to solve the equations than using a graphical approach.
a. Show how you could use a numerical approach by creating a table. For instance, you might use a spreadsheet to solve the equations.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 2
b. Show how you could use an analytical approach. For instance, you might try solving the equations by using the inverse properties of exponents and logarithms.

Communicate Your Answer

Question 3.
How can you solve exponential and logarithmic equations?
Answer:

Question 4.
Solve each equation using any method. Explain your choice of method.
a. 16x = 2
b. 2x = 42x+1
c. 2x = 3x+1
d. log x = \(\frac{1}{2}\)
e. ln x = 2
f. log3 x = \(\frac{3}{2}\)
Answer:

Monitoring Progress

Solve the equation.
Question 1.
2x = 5
Answer:

Question 2.
79x = 15
Answer:

Question 3.
4e-0.3x − 7 = 13
Answer:

Question 4.
WHAT IF?
In Example 2, how long will it take to cool the stew to 100ºF when the room temperature is 75ºF?
Answer:

Solve the equation. Check for extraneous solutions.
Question 5.
ln(7x − 4) = ln(2x + 11)
Answer:

Question 6.
log2(x − 6) = 5
Answer:

Question 7.
log 5x + log(x − 1) = 2
Answer:

Question 8.
log4 (x + 12) + log4 x = 3
Answer:

Solve the inequality.
Question 9.
ex < 2
Answer:

Question 10.
102x-6 > 3
Answer:

Question 11.
log x + 9 < 45 Answer: Question 12. 2 ln x − 1 > 4
Answer:

Solving Exponential and Logarithmic Equations 6.6 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
The equation 3x-1 = 34 is an example of a(n) ___________ equation.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 1

Question 2.
WRITING
Compare the methods for solving exponential and logarithmic equations.
Answer:

Question 3.
WRITING
When do logarithmic equations have extraneous solutions?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 3

Question 4.
COMPLETE THE SENTENCE
If b is a positive real number other than 1, then bx = by if and only if _________
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–16, solve the equation.
Question 5.
73x+5 = 71-x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 5

Question 6.
e2x = e3x-1
Answer:

Question 7.
5x-3 = 25x-5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 7

Question 8.
62x-6 = 363x-5
Answer:

Question 9.
3x = 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 9

Question 10.
5x = 33
Answer:

Question 11.
495x+2 = (\(\frac{1}{7}\))11-x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 11

Question 12.
5125x-1 = (\(\frac{1}{8}\))-4-x
Answer:

Question 13.
75x = 12
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 13

Question 14.
116x = 38
Answer:

Question 15.
3e4x+9 = 15
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 15

Question 16.
2e2x − 7 = 5
Answer:

Question 17.
MODELING WITH MATHEMATICS
The length ℓ(in centimeters) of a scalloped hammerhead shark can be modeled by the function ℓ = 266 − 219-0.05t
where t is the age (in years) of the shark. How old is a shark that is 175 centimeters long?
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 17

Question 18.
MODELING WITH MATHEMATICS
One hundred grams of radium are stored in a container. The amount R(in grams) of radium present after t years can be modeled by R= 100e−0.00043t. After how many years will only 5 grams of radium be present?
Answer:

In Exercises 19 and 20, use Newton’s Law of Cooling to solve the problem.
Question 19.
You are driving on a hot day when your car overheats and stops running. The car overheats at 280°F and can be driven again at 230°F. When it is 80°F outside, the cooling rate of the car is r = 0.0058. How long do you have to wait until you can continue driving?
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 19

Question 20.
You cook a turkey until the internal temperature reaches 180°F. The turkey is placed on the table until the internal temperature reaches 100°F and it can be carved. When the room temperature is 72°F, the cooling rate of the turkey is r = 0.067. How long do you have to wait until you can carve the turkey?
Answer:

In Exercises 21–32, solve the equation.
Question 21.
ln(4x − 7) = ln(x + 11)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 21

Question 22.
ln(2x − 4) = ln(x + 6)
Answer:

Question 23.
log2(3x − 4) = log2 5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 23

Question 24.
log(7x + 3) = log 38
Answer:

Question 25.
log2(4x + 8) = 5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 25

Question 26.
log3(2x + 1) = 2
Answer:

Question 27.
log7(4x + 9) = 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 27

Question 28.
log5(5x + 10) = 4
Answer:

Question 29.
log(12x − 9) = log 3x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 29

Question 30.
log6(5x + 9) = log6 6x
Answer:

Question 31.
log2(x2 − x − 6) = 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 31

Question 32.
log3(x2 + 9x + 27) = 2
Answer:

In Exercises 33–40, solve the equation. Check for extraneous solutions.
Question 33.
log2 x + log2 (x − 2) = 3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 33

Question 34.
log6 3x + log6 (x − 1) = 3
Answer:

Question 35.
ln x + ln(x + 3) = 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 35

Question 36.
ln x + ln(x − 2) = 5
Answer:

Question 37.
log3 3x2 + log3 3 = 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 37

Question 38.
log4 (−x) + log4 (x + 10) = 2
Answer:

Question 39.
log3(x − 9) + log3(x − 3) = 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 39

Question 40.
log5 (x + 4) + log5 (x + 1) = 2
Answer:

ERROR ANALYSIS In Exercises 41 and 42, describe and correct the error in solving the equation.
Question 41.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 41

Question 42.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 6
Answer:

Question 43.
PROBLEM SOLVING
You deposit $100 in an account that pays 6% annual interest. How long will it take for the balance to reach $1000 for each frequency of compounding?
a. annual
b. quarterly
c. daily
d. continuously
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 43.1
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 43.2

Question 44.
MODELING WITH MATHEMATICS
The apparent magnitude of a star is a measure of the brightness of the star as it appears to observers on Earth. The apparent magnitude M of the dimmest star that can be seen with a telescope is M = 5 log D+ 2, where D is the diameter (in millimeters) of the telescope’s objective lens. What is the diameter of the objective lens of a telescope that can reveal stars with a magnitude of 12?
Answer:

Question 45.
ANALYZING RELATIONSHIPS
Approximate the solution of each equation using the graph.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 45

Question 46.
MAKING AN ARGUMENT
Your friend states that a logarithmic equation cannot have a negative solution because logarithmic functions are not defined for negative numbers. Is your friend correct? Justify your answer.
Answer:

In Exercises 47–54, solve the inequality.
Question 47.
9x > 54
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 47

Question 48.
4x ≤ 36
Answer:

Question 49.
ln x ≥ 3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 49

Question 50.
log4 x< 4
Answer:

Question 51.
34x-5 < 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 51

Question 52.
e3x+4 > 11
Answer:

Question 53.
−3 log5 x + 6 ≤ 9
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 53

Question 54.
−4 log5 x − 5 ≥ 3
Answer:

Question 55.
COMPARING METHODS
Solve log5 x< 2 algebraically and graphically. Which method do you prefer? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 55

Question 56.
PROBLEM SOLVING
You deposit $1000 in an account that pays 3.5% annual interest compounded monthly. When is your balance at least $1200? $3500?
Answer:

Question 57.
PROBLEM SOLVING
An investment that earns a rate of return r doubles in value in t years, where t = \(\frac{\ln 2}{\ln (1+r)}\) and r is expressed as a decimal. What rates of return will double the value of an investment in less than 10 years?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 57

Question 58.
PROBLEM SOLVING
Your family purchases a new car for $20,000. Its value decreases by 15% each year. During what interval does the car’s value exceed $10,000?
Answer:

USING TOOLS In Exercises 59–62, use a graphing calculator to solve the equation.
Question 59.
ln 2x = 3-x+2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 59

Question 60.
log x = 7-x
Answer:

Question 61.
log x = 3x-3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 61

Question 62.
ln 2x = ex-3
Answer:

Question 63.
REWRITING A FORMULA
A biologist can estimate the age of an African elephant by measuring the length of its footprint and using the equation ℓ = 45 − 25.7e−0.09a, where ℓis the length (in centimeters) of the footprint and a is the age (in years).
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 8
a. Rewrite the equation, solving for a in terms of ℓ.
b. Use the equation in part (a) to find the ages of the elephants whose footprints are shown.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 63

Question 64.
HOW DO YOU SEE IT?
Use the graph to solve the inequality 4 ln x + 6 > 9. Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 9
Answer:

Question 65.
OPEN-ENDED
Write an exponential equation that has a solution of x = 4. Then write a logarithmic equation that has a solution of x = −3.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 65

Question 66.
THOUGHT PROVOKING
Give examples of logarithmic or exponential equations that have one solution, two solutions, and no solutions.
Answer:

CRITICAL THINKING In Exercises 67–72, solve the equation.
Question 67.
2x+3 = 53x-1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 67

Question 68.
103x-8 = 25-x
Answer:

Question 69.
log3 (x − 6) = log9 2x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 69

Question 70.
log4 x = log8 4x
Answer:

Question 71.
22x − 12 • 2x + 32 = 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 71

Question 72.
52x + 20 • 5x − 125 = 0
Answer:

Question 73.
WRITING
In Exercises 67–70, you solved exponential and logarithmic equations with different bases. Describe general methods for solving such equations.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 73

Question 74.
PROBLEM SOLVING
When X-rays of a fixed wavelength strike a material x centimeters thick, the intensity I(x) of the X-rays transmitted through the material is given by I(x) = I0e−μx, where I0 is the initial intensity and μ is a value that depends on the type of material and the wavelength of the X-rays. The table shows the values of μ for various materials and X-rays of medium wavelength.
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 10
a. Find the thickness of aluminum shielding that reduces the intensity of X-rays to 30% of their initial intensity. (Hint: Find the value of x for which I(x) = 0.3I0.)
b. Repeat part (a) for the copper shielding.
c. Repeat part (a) for the lead shielding.
d. Your dentist puts a lead apron on you before taking X-rays of your teeth to protect you from harmful radiation. Based on your results from parts (a)–(c), explain why lead is a better material to use than aluminum or copper.
Answer:

Maintaining Mathematical Proficiency

Write an equation in point-slope form of the line that passes through the given point and has the given slope.
Question 75.
(1, −2); m = 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 75

Question 76.
(3, 2); m = −2
Answer:

Question 77.
(3, −8); m = − \(\frac{1}{3}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 77

Question 78.
(2, 5); m = 2
Answer:

Use finite differences to determine the degree of the polynomial function that fits the data. Then use technology to find the polynomial function.
Question 79.
(−3, −50), (−2, −13), (−1, 0), (0, 1), (1, 2), (2, 15), (3, 52), (4, 125)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 79

Question 80.
(−3, 139), (−2, 32), (−1, 1), (0, −2), (1, −1), (2, 4), (3, 37), (4, 146)
Answer:

Question 81.
(−3, −327), (−2, −84), (−1, −17), (0, −6), (1, −3), (2, −32), (3, −189), (4, −642)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 6 Exponential and Logarithmic Functions 6.6 a 81

Lesson 6.7 Modeling with Exponential and Logarithmic Functions

Essential Question How can you recognize polynomial, exponential, and logarithmic models?

EXPLORATION 1

Recognizing Different Types of Models
Work with a partner. Match each type of model with the appropriate scatter plot. Use a regression program to find a model that fits the scatter plot.
a. linear (positive slope)
b. linear (negative slope)
c. quadratic
d. cubic
e. exponential
f. logarithmic
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 1

EXPLORATION 2

Exploring Gaussian and Logistic Models
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 2
Work with a partner. Two common types of functions that are related to exponential functions are given. Use a graphing calculator to graph each function. Then determine the domain, range, intercept, and asymptote(s) of the function.
a. Gaussian Function: f(x) = e−x2
b. Logistic Function: f(x) = \(\frac{1}{1+e^{-x}}\)

Communicate Your Answer

Question 3.
How can you recognize polynomial, exponential, and logarithmic models?
Answer:

Question4 .
Use the Internet or some other reference to find real-life data that can be modeled using one of the types given in Exploration 1. Create a table and a scatter plot of the data. Then use a regression program to find a model that fits the data.
Answer:

Monitoring Progress

Determine the type of function represented by the table. Explain your reasoning.
Question 1.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 3
Answer:

Question 2.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 4
Answer:

Write an exponential function y = abx whose graph passes through the given points.
Question 3.
(2, 12), (3, 24)
Answer:

Question 4.
(1, 2), (3, 32)
Answer:

Question 5.
(2, 16), (5, 2)
Answer:

Question 6.
WHAT IF?
Repeat Examples 3 and 4 using the sales data from another store.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 5
Answer:

Question 7.
Use a graphing calculator to find an exponential model for the data in Monitoring Progress Question 6.
Answer:

Question 8.
Use a graphing calculator to find a logarithmic model of the form p = a + b ln h for the data in Example 6. Explain why the result is an error message.
Answer:

Modeling with Exponential and Logarithmic Functions 6.7 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
Given a set of more than two data pairs (x, y), you can decide whether a(n) __________ function fits the data well by making a scatter plot of the points (x, ln y).
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 1

Question 2.
WRITING
Given a table of values, explain how you can determine whether an exponential function is a good model for a set of data pairs (x, y).
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, determine the type of function represented by the table. Explain your reasoning.
Question 3.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 3

Question 4.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 7
Answer:

Question 5.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 5

Question 6.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 9
Answer:

In Exercises 7–16, write an exponential function y = abx whose graph passes through the given points.
Question 7.
(1, 3), (2, 12)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 7

Question 8.
(2, 24), (3, 144)
Answer:

Question 9.
(3, 1), (5, 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 9

Question 10.
(3, 27), (5, 243)
Answer:

Question 11.
(1, 2), (3, 50)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 11

Question 12.
(1, 40), (3, 640)
Answer:

Question 13.
(−1, 10), (4, 0.31)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 13

Question 14.
(2, 6.4), (5, 409.6)
Answer:

Question 15.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 15

Question 16.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 11
Answer:

ERROR ANALYSIS In Exercises 17 and 18, describe and correct the error in determining the type of function represented by the data.
Question 17.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 17

Question 18.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 13
Answer:

Question 19.
MODELING WITH MATHEMATICS
A store sells motorized scooters. The table shows the numbers y of scooters sold during the xth year that the store has been open. Write a function that models the data.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 14
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 19

Question 20.
MODELING WITH MATHEMATICS
The table shows the numbers y of visits to a website during the xth month. Write a function that models the data. Then use your model to predict the number of visits after 1 year.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 15
Answer:

In Exercises 21–24, determine whether the data show an exponential relationship. Then write a function that models the data.
Question 21.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 16
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 21

Question 22.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 17
Answer:

Question 23.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 18
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 23

Question 24.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 19
Answer:

Question 25.
MODELING WITH MATHEMATICS
Your visual near point is the closest point at which your eyes can see an object distinctly. The diagram shows the near point y (in centimeters) at age x (in years). Create a scatter plot of the data pairs (x, ln y) to show that an exponential model should be a good fit for the original data pairs (x, y). Then write an exponential model for the original data.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 20
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 25.1
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 25.2

Question 26.
MODELING WITH MATHEMATICS
Use the data from Exercise 19. Create a scatter plot of the data pairs (x, ln y) to show that an exponential model should be a good fit for the original data pairs (x, y). Then write an exponential model for the original data.
Answer:

In Exercises 27–30, create a scatter plot of the points (x, ln y) to determine whether an exponential model fits the data. If so, find an exponential model for the data.
Question 27.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 21
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 27.1
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 27.2

Question 28.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 22
Answer:

Question 29.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 23
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 29

Question 30.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 24
Answer:

Question 31.
USING TOOLS
Use a graphing calculator to find an exponential model for the data in Exercise 19. Then use the model to predict the number of motorized scooters sold in the tenth year.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 31

Question 32.
USING TOOLS
A doctor measures an astronaut’s pulse rate y (in beats per minute) at various times x(in minutes) after the astronaut has finished exercising. The results are shown in the table. Use a graphing calculator to find an exponential model for the data. Then use the model to predict the astronaut’s pulse rate after 16 minutes.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 25
Answer:

Question 33.
USING TOOLS
An object at a temperature of 160°C is removed from a furnace and placed in a room at 20°C. The table shows the temperatures d (in degrees Celsius) at selected times t (in hours) after the object was removed from the furnace. Use a graphing calculator to find a logarithmic model of the form t = a +b ln d that represents the data. Estimate how long it takes for the object to cool to 50°C.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 26
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 33

Question 34.
USING TOOLS
The f-stops on a camera control the amount of light that enters the camera. Let s be a measure of the amount of light that strikes the film and let f be the f-stop. The table shows several f-stops on a 35-millimeter camera. Use a graphing calculator to find a logarithmic model of the form s = a + b ln f that represents the data. Estimate the amount of light that strikes the film when f = 5.657.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 27
Answer:

Question 35.
DRAWING CONCLUSIONS
The table shows the average weight (in kilograms) of an Atlantic cod that is x years old from the Gulf of Maine.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 28
a. Show that an exponential model fits the data. Then find an exponential model for the data.
b. By what percent does the weight of an Atlantic cod increase each year in this period of time? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 35

Question 36.
HOW DO YOU SEE IT?
The graph shows a set of data points (x, ln y). Do the data pairs (x, y) fit an exponential pattern? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 29
Answer:

Question 37.
MAKING AN ARGUMENT
Your friend says it is possible to find a logarithmic model of the form d = a + b ln t for the data in Exercise 33. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 37

Question 38.
THOUGHT PROVOKING
Is it possible to write y as an exponential function of x? Explain your reasoning. (Assume p is positive.)
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 30
Answer:

Question 39.
CRITICAL THINKING
You plant a sunflower seedling in your garden. The height h (in centimeters) of the seedling after t weeks can be modeled by the logistic function
h(t) = \(\frac{256}{1+13 e^{-0.65 t}}\)
a. Find the time it takes the sunflower seedling to reach a height of 200 centimeters.
b. Use a graphing calculator to graph the function. Interpret the meaning of the asymptote in the context of this situation.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 39

Maintaining Mathematical Proficiency

Tell whether x and y are in a proportional relationship. Explain your reasoning.
Question 40.
y = \(\frac{x}{2}\)
Answer:

Question 41.
y = 3x − 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 41

Question 42.
y = \(\frac{5}{x}\)
Answer:

Question 43.
y = −2x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 43

Identify the focus, directrix, and axis of symmetry of the parabola. Then graph the equation.
Question 44.
x = \(\frac{1}{8}\)y2
Answer:

Question 45.
y = 4x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 45

Question 46.
x2 = 3y
Answer:

Question 47.
y2 = \(\frac{2}{5}\)x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 a 47

Exponential and Logarithmic Functions Performance Task: Measuring Natural Disasters

6.5–6.7 What Did You Learn?

Core Vocabulary
exponential equations, p. 334
logarithmic equations, p. 335

Core Concepts
Section 6.5
Properties of Logarithms, p. 328
Change-of-Base Formula, p. 329

Section 6.6
Property of Equality for Exponential Equations, p. 334
Property of Equality for Logarithmic Equations, p. 335

Section 6.7
Classifying Data, p. 342
Writing Exponential Functions, p. 343
Using Exponential and Logarithmic Regression, p. 345

Mathematical Practices
Question 1.
Explain how you used properties of logarithms to rewrite the function in part (b) of Exercise 45 on page 332.
Answer:

Question 2.
How can you use cases to analyze the argument given in Exercise 46 on page 339?
Answer:

Performance Task Measuring Natural Disasters
In 2005, an earthquake measuring 4.1 on the Richter scale barely shook the city of Ocotillo, California, leaving virtually no damage. But in 1906, an earthquake with an estimated 8.2 on the same scale devastated the city of San Francisco. Does twice the measurement on the Richter scale mean twice the intensity of the earthquake?
To explore the answers to these questions and more, go to BigIdeasMath.com.
Big Ideas Math Algebra 2 Answers Chapter 6 Exponential and Logarithmic Functions 6.7 31

Exponential and Logarithmic Functions Chapter Review

6.1 Exponential Growth and Decay Functions (pp. 295–302)

Tell whether the function represents exponential growth or exponential decay. Identify the percent increase or decrease. Then graph the function.
Question 1.
f (x) = (\(\frac{1}{3}\))x
Answer:

Question 2.
y = 5x
Answer:

Question 3.
f(x) = (0.2)x
Answer:

Question 4.
You deposit $1500 in an account that pays 7% annual interest. Find the balance after 2 years when the interest is compounded daily.
Answer:

6.2 The Natural Base e (pp. 303–308)

Simplify the expression.
Question 5.
e4 • e11
Answer:

Question 6.
\(\frac{20 e^{3}}{10 e^{6}}\)
Answer:

Question 7.
(−3e-5x)2
Answer:

Tell whether the function represents exponential growth or exponential decay. Then graph the function.
Question 8.
f(x) = \(\frac{1}{3}\)ex
Answer:

Question 9.
y = 6e-x
Answer:

Question 10.
y = 3e-0.75x
Answer:

6.3 Logarithms and Logarithmic Functions (pp. 309–316)

Evaluate the logarithm.
Question 11.
log2 8
Answer:

Question 12.
log6 \(\frac{1}{36}\)
Answer:

Question 13.
log2 1
Answer:

Find the inverse of the function.
Question 14.
f(x) = 8x
Answer:

Question 15.
y = ln(x − 4)
Answer:

Question 16.
y = log(x+ 9)
Answer:

Question 17.
Graph y = log1/5 x.
Answer:

6.4 Transformations of Exponential and Logarithmic Functions (pp. 317–324)

Describe the transformation of f represented by g. Then graph each function.
Question 18.
f(x) = e-x, g(x) = -5x − 8
Answer:

Question 19.
f(x) = log4 x, g(x) = \(\frac{1}{2}\)log4 (x + 5)
Answer:

Write a rule for g.
Question 20.
Let the graph of g be a vertical stretch by a factor of 3, followed by a translation 6 units left and 3 units up of the graph of f(x) = ex.
Answer:

Question 21.
Let the graph of g be a translation 2 units down, followed by a reflection in the y-axis of the graph of f(x) = log x.
Answer:

6.5 Properties of Logarithms (pp. 327–332)

Expand or condense the logarithmic expression.
Question 22.
log8 3xy
Answer:

Question 23.
log 10x3y
Answer:

Question 24.
ln \(\frac{3 y}{x^{5}}\)
Answer:

Question 25.
3 log7 4 + log7 6
Answer:

Question 26.
log2 12 − 2 log2 x
Answer:

Question 27.
2 ln x + 5 ln 2 − ln 8
Answer:

Use the change-of-base formula to evaluate the logarithm.
Question 28.
log2 10
Answer:

Question 29.
log7 9
Answer:

Question 30.
log23 42
Answer:

6.6 Solving Exponential and Logarithmic Equations (pp. 333–340)

Solve the equation. Check for extraneous solutions.
Question 31.
5x = 8
Answer:

Question 32.
log3 (2x − 5) = 2
Answer:

Question 33.
ln x + ln(x + 2) = 3
Answer:

Solve the inequality.
Question 34.
6x > 12
Answer:

Question 35.
ln x ≤ 9
Answer:

Question 36.
e4x-2 ≥ 16
Answer:

6.7 Modeling with Exponential and Logarithmic Functions (pp. 341–348)

Write an exponential model for the data pairs (x, y).
Question 37.
(3, 8), (5, 2)
Answer:

Question 38.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions cr 38
Answer:

Question 39.
A shoe store sells a new type of basketball shoe. The table shows the pairs sold s over time t (in weeks). Use a graphing calculator to find a logarithmic model of the form s = a +b ln t that represents the data. Estimate how many pairs of shoes are sold after 6 weeks.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions cr 39
Answer:

Exponential and Logarithmic Functions Chapter Test

Graph the equation. State the domain, range, and asymptote.
Question 1.
y = (\(\frac{1}{2}\))x
Answer:

Question 2.
y = log1/5 x
Answer:

Question 3.
y = 4e-2x
Answer:

Describe the transformation of f represented by g. Then write a rule for g.
Question 4.
f(x) = log x
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions ct 4
Answer:

Question 5.
f(x) = ex
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions ct 5
Answer:

Question 6.
f(x) = \(\frac{1}{4}\))x
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions ct 6
Answer:

Evaluate the logarithm. Use log3 4 ≈ 1.262 and log3 13 ≈ 2.335, if necessary.
Question 7.
log2 52
Answer:

Question 8.
log3 \(\frac{13}{9}\)
Answer:

Question 9.
log3 16
Answer:

Question 10.
log3 8 + log3 \(\frac{1}{2}\)
Answer:

Question 11.
Describe the similarities and differences in solving the equations 45x-2 = 16 and log4(10x + 6) = 1. Then solve each equation.
Answer:

Question 12.
Without calculating, determine whether log511, \(\frac{\log 11}{\log 5}\), and \(\frac{\ln 11}{\ln 5}\) are equivalent expressions. Explain your reasoning.
Answer:

Question 13.
The amount y of oil collected by a petroleum company drilling on the U.S. continental shelf can be modeled by y = 12.263 ln x − 45.381, where y is measured in billions of barrels and x is the number of wells drilled. About how many barrels of oil would you expect to collect after drilling 1000 wells? Find the inverse function and describe the information you obtain from finding the inverse.
Answer:

Question 14.
The percent L of surface light that filters down through bodies of water can be modeled by the exponential function L(x) = 100ekx, where k is a measure of the murkiness of the water and x is the depth (in meters) below the surface.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions ct 14
a. A recreational submersible is traveling in clear water with a k-value of about −0.02. Write a function that gives the percent of surface light that filters down through clear water as a function of depth.
b. Tell whether your function in part (a) represents exponential growth or exponential decay. Explain your reasoning.
c. Estimate the percent of surface light available at a depth of 40 meters.
Answer:

Question 15.
The table shows the values y (in dollars) of a new snowmobile after x years of ownership. Describe three different ways to find an exponential model that represents the data. Then write and use a model to find the year when the snowmobile is worth $2500.
Big Ideas Math Algebra 2 Solutions Chapter 6 Exponential and Logarithmic Functions ct 15
Answer:

Exponential and Logarithmic Functions Cumulative Assessment

Question 1.
Select every value of b for the equation y = bx that could result in the graph shown.
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions ca 1
Answer:

Question 2.
Your friend claims more interest is earned when an account pays interest compounded continuously than when it pays interest compounded daily. Do you agree with your friend? Justify your answer.
Answer:

Question 3.
You are designing a rectangular picnic cooler with a length four times its width and height twice its width. The cooler has insulation that is 1 inch thick on each of the four sides and 2 inches thick on the top and bottom.
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions ca 3
a. Let x represent the width of the cooler. Write a polynomial function T that gives the volume of the rectangular prism formed by the outer surfaces of the cooler.
b.Write a polynomial function C for the volume of the inside of the cooler.
c. Let I be a polynomial function that represents the volume of the insulation. How is I related to T and C?
d. Write I in standard form. What is the volume of the insulation when the width of the cooler is 8 inches?
Answer:

Question 4.
What is the solution to the logarithmic inequality −4 log2 x ≥ −20?
A. x ≤ 32
B. 0 ≤ x ≤ 32
C. 0 < x ≤ 32
D. x ≥ 32
Answer:

Question 5.
Describe the transformation of f(x) = log2 x represented by the graph of g.
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions ca 5
Answer:

Question 6.
Let f(x) = 2x3 − 4x2 + 8x− 1, g(x) = 2x − 3x4 − 6x3+ 5, and h(x) = −7 + x2 + x. Order the following functions from least degree to greatest degree.
A. (f + g)(x)
B. (hg)(x)
C. (h − f)(x)
D. (fh)(x)
Answer:

Question 7.
Write an exponential model that represents each data set. Compare the two models.
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions ca 7
Answer:

Question 8.
Choose a method to solve each quadratic equation. Explain your choice of method.
a. x2 + 4x = 10
b. x2 = −12
c. 4(x − 1)2 = 6x + 2
d. xx − 3x − 18 = 0
Answer:

Question 9.
At the annual pumpkin-tossing contest, contestants compete to see whose catapult will send pumpkins the longest distance. The table shows the horizontal distances y (in feet) a pumpkin travels when launched at different angles x (in degrees). Create a scatter plot of the data. Do the data show a linear, quadratic, or exponential relationship? Use technology to find a model for the data. Find the angle(s) at which a launched pumpkin travels 500 feet.
Big Ideas Math Answer Key Algebra 2 Chapter 6 Exponential and Logarithmic Functions ca 9
Answer:

Big Ideas Math Answers Grade 3 | Big Ideas Math Book 3rd Grade Answer Key

Big Ideas Math Answers Grade 3

Do you want to clear all your doubts about Big Ideas Math Grade 3 Questions? Then get the perfect study material which helps you to score good marks in the exam. One of the best Big Ideas Math Grade 3 Answer Key will guide you in solving all types of problems. Review all 3rd grade concepts and have a look at the handy material to make your practice more easy and simple. Find comprehensive and well-organized Big Ideas Math Book Grade 3 Answers which are given with step-by-step explanations in the below material. In addition to the answer key, you can also check BIM Book Grade 3 Solution Key in the following sections.

Big Ideas Math Answers 3rd Grade Solutions Pdf | Big Ideas Math Book Grade 3 Answer Key

Big Ideas Math Modeling Real Life Grade 3 Textbook Solutions have chapter tests, performance tasks, vocabulary, practice test, cumulative practice, and others. Have a look at the complete practice guide and solve the questions included in it. Become an expert in maths by overcoming all weak areas by referring to Big Ideas Math Answers Grade 3. The common core curriculum helps you to know in detail information about all concepts in the syllabus. In the following section, you will get the quick links for all the 15 chapters of Big Ideas Math Book Grade 3 Solutions.

Benefits Common Core Big Ideas Math Grade 3 Answer Key?

We have outlined the major benefits of referring to BIM Book Grade 3 Answers. They are as under

  • BIM 3rd Grade Answers are prepared based on the common core 2019 curriculum.
  • Students can easily find the problems and answers for Performance Task, Activity, Cumulative Practice, and Chapter Practice from Big Ideas Math Book 3rd Grade Answer Key which improves your mathematics skills and regulates better practice.
  • Step by step detailed explanation solutions for all the questions of all chapters is provided by the subject experts which enables the students to learn and practice subjects perfectly.
  • By using BIM Grade 3 Answers, students can get on track in the subject and can become a pro in all math concepts.
  • Follow our well-organised Big Ideas Math Grade 3 Solutions PDF which is the best study material and helps to solve many difficult problems.
  • With the help of mock tests or practice tests, you can increase the ability to solve math problems quickly and easily.

FAQs on BIM 3rd Grade Answer Key

1. Where do I find BigIdeas Math Grade 3 Answers for All Chapters?

You can find the Chapterwise Big Ideas Math Book 3rd Grade Answer Key at www.bigideasmathanswers.com with a detailed explanation.

2. How to Solve Big Ideas Math Textbook Questions of 3rd Grade Easily?

Practice various concepts regularly taking the help of the Big Ideas Math Grade 3 Answer Key and solve the textbook questions easily.

3. How to download Elementary School Big Ideas Math 3rd Grade Solutions PDF?

You can download Elementary School Big Ideas Math Textbook 3rd Grade Solutions PDF via quick links available for all Chapters on our page. Simply tap on the chapter you want and kick start your preparation.

4. Can I download Big Ideas Math Grade 3 Book Answer Key Pdf freely?

The solution for the above answer is Yes. It is possible to download BIM Grade 3 Answer Key Pdf freely by visiting Big Ideas Math Answer site. You can click on the direct links available in the above section which directs to the particular chapter answer key pdf.

Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors

Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors

Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors: Before knowing the numerical expressions, Grade 6 students must be familiar with composite numbers, factor pairs, and prime numbers. Refer to the concept of vocabulary terms, subtracting and adding the fractions, mixed fractions in the below sections.

Know the various terms like what is exponent, perfect square, power, composite numbers, prime numbers with the help of Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors. You can also get all the answers to the questions which are available in the Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors pdf format. Scroll to the below sections to check example problems, answer PDFs, etc.

Big Ideas Math Book 6th Grade Answer Key Chapter 1 Numerical Expressions and Factors

BIM 6th Grade Chapter 1 Numerical Expression and Factors Answer key helps you in easy and quick learning. Download Big Ideas Math Book 6th Grade Answer Key Chapter 1 Numerical Expressions and Factors pdf to kickstart your preparation. Get the solutions for all the questions in this article. There are various topics included in numerical expressions like Least Common Multiple, Prime Factorisation, Powers and Exponents, Order of Operations, Greatest Common Factor, Exponents and Powers. Click on the below links and prepare for the exam as per the topics.

Performance Task

Lesson 1: Powers and Exponents

Lesson 2: Order of Operations

Lesson 3: Prime Factorization

Lesson 4: Greatest Common Factor

Lesson 5: Least Common Multiple

Chapter: 1 – Numerical Expressions and Factors

Numerical Expressions and Factors Steam Video/Performance Task

Filling Piñatas

Common factors can be used to make identical groups of objects. Can you think of any situations in which you would want to separate objects into equal groups? Are there any common factors that may be more useful than others? Can you think of any other ways to use common factors?
watch the STEAM Video “Filling Piñatas.” Then answer the following questions. The table below shows the numbers of party favors that Alex and Enid use to make piñatas.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 1
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 1.1

Question 1.
When finding the number of identical piñatas that can be made, why is it helpful for Alex and Enid to list the factors of each number given in the table?

Answer: By using the list of the factors of all the numbers Alex and Enid can make identical groups of the objects.

Question 2.
You want to create 6 identical piñatas. How can you change the numbers of party favors in the table to make this happen? Can you do this without changing the total number of party favors?

Answer: You can change the number of party favors to create 6 identical pinatas.
There are 100 Mints. So divide it into two identical groups.
Change the number of mints to 50. And add 50 to new identical pinatas.

Performance Task

Setting the Table

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be asked to plan a fundraising event with the items below.
72 chairs
48 balloons
24 flowers
36 candles
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 2
You will find the greatest number of identical tables that can be prepared, and what will be in each centerpiece. When making arrangements for a party, should a party planner always use the greatest number of identical tables possible? Explain why or why not.

Answer:
72 chairs = 2 × 36
= 2 × 2 × 18
= 2 × 2 × 2 × 9
= 2 × 2 × 2 × 3 × 3
2, 2, 2, 3, 3
Therefore, 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72 are the factors of 72.
48 balloons = 2 × 24
= 2 × 2 × 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
The positive Integer factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
The factors of number 24 are 1, 2, 3, 4, 6, 8, 12, 24.
The factors of number 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36
The greatest number of identical tables possible are 1, 2, 3, 4, 6, 12.

Numerical Expressions and Factors Getting Ready for Chapter 1

Chapter Exploration

Work with a partner. In Exercises 1–3, use the table.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 3

  1. Cross out the multiples of 2 that are greater than 2. Do the same for 3, 5, and 7.
  2. The numbers that are not crossed out are called prime numbers. The numbers that are crossed out are called composite numbers. In your own words, describe the characteristics of prime numbers and composite numbers.
  3. MODELING REAL LIFE Work with a partner. Cicadas are insects that live underground and emerge from the ground after x or x + 4 years. Is it possible that both x and x +4 are prime? Give some examples.

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-3
The numbers that are not crossed are 2, 3, 5, 7, 9, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,61, 67, 71, 73, 79, 83, 89, 97.
These are not multiples of any numbers. So, the above numbers are the prime numbers.

Vocabulary

The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 4

Answer:
i. First, we solve any operations inside of parentheses or brackets. Second, we solve any exponents. Third, we solve all multiplication and division from left to right. Fourth, we solve all addition and subtraction from left to right.
ii. “Factors” are numbers we can multiply together to get another number. When we find the factors of two or more numbers, and then find some factors are the same, then they are the “common factors”.
iii. A common multiple is a whole number that is a shared multiple of each set of numbers. The multiples that are common to two or more numbers are called the common multiples of those numbers. The smallest positive number is a multiple of two or more numbers.

Lesson 1.1 Powers and Exponents

Exploration 1
Writing Expressions Using Exponents
Work with a partner. Copy and complete the table.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 5

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-5
i. In your own words, describe what the two numbers in the expression 35 mean.

Answer: 35 means the number 3 repeats 5 times.
3 × 3 × 3 × 3 × 3 = 243

EXPLORATION 2
Using a Calculator to Find a Pattern

Work with a partner. Copy the diagram. Use a calculator to find each value. Write one digit of the value in each box. Describe the pattern in the digits of the values.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 6

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-6

1.1 Lesson

A power is a product of repeated factors. The base of a power is the repeated factor. The exponent of a power indicates the number of times the base is used as a factor.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 7
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 8

Try It
Write the product as a power.
Question 1.
2 × 2 × 2

Answer: 2³  = 8
Two cubed or three to the two. Here 2 is repeated three times.

Question 2.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 13

Answer: 46656 = 66

Six to the six. Here 6 is repeated six times.

Question 3.
15 × 15 × 15 × 15

Answer: 56025 = 154

15 to the power 4. Here 15 is repeated four times.

Question 4.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 14

Answer: 1280000000 = 207

20 to the power 7. Here 20 is repeated seven times.

Try It
Find the value of the power.

Question 5.
63

Answer: 6 × 6 × 6 = 216
The value of the power 63 is 216.

Question 6.
92

Answer: 9 × 9 = 81
The value of the power 92 is 81

Question 7.
34

Answer: 3 × 3 × 3 × 3
The value of the power 34 is 81.

Question 8.
182

Answer: 18 × 18
The value of the power 182 is 324.

Try It
Determine whether the number is a perfect square.

Question 9.
25

Answer: 5²
Yes, 25 is the perfect square.
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Question 10.
2

Answer: 2 is not a perfect square. 2 cannot be expressed as the square of a number from the same number system.

Question 11.
99

Answer: 99 is not a perfect square. 99 cannot be expressed as the square of a number from the same number system.

Question 12.
36

Answer: 6²
36 is a perfect square
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
FINDING VALUES OF POWERS
Find the value of the power.

Question 13.
82

Answer: 8 × 8 = 64
The value of the power 82 is 64.

Question 14.
35

Answer: 3 × 3 × 3 × 3 × 3 = 243
The value of the power 35 is 243.

Question 15.
113

Answer: 11 × 11 × 11 = 1331
The value of the power 113 is 1331.

Question 16.
VOCABULARY
How are exponents and powers different?

Answer:
An expression that represents repeated multiplication of the same factor is called a power. The number 5 is called the base, and the number 2 is called the exponent. The exponent corresponds to the number of times the base is used as a factor.

Question 17.
VOCABULARY
Is 10 a perfect square? Is 100 a perfect square? Explain.

Answer: 10 is not a perfect square.
A perfect square is a number that is generated by multiplying two equal integers by each other.
100 is a perfect square. Because 10 × 10 = 100.

Question 18.
WHICH ONE DOESN’T BELONG?
Which one does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 15

Answer:
24 = 2 × 2 × 2 × 2 = 16
32 = 3 × 3 = 9
3 + 3 + 3 + 3 = 3 × 4
5.5.5 = 125
The 3rd option does not belong to the other three expressions.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 19.
A square solar panel has an area of 16 square feet. Write the area as a power. Then find the side lengths of the panel.

Answer: 4 feet

Explanation:
Given that,
A square solar panel has an area of 16 square feet.
A = s × s
16 = s²
4² = s²
s = 4
Thus the side length of the panel is 4 feet.

Question 20.
The four-square court shown is a square made up of four identical smaller squares. What is the area of the court?

Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 16

Answer:
Given,
The four-square court shown is a square made up of four identical smaller squares.
The side of each square is 6 feet.
6 + 6 = 12 feet
The area of the court is 12 ft × 12 ft
A = 144 square feet
Thus the area of the court is 144 square feet.

Question 21.
DIG DEEPER!
Each face of a number cube is a square with a side length of 16 millimeters. What is the total area of all of the faces of the number cube?

Answer:
Given that,
Each face of a number cube is a square with a side length of 16 millimeters.
Area of the cube = 6 a²
A = 6 × 16 × 16
A = 1536 sq. mm

Powers and Exponents Practice 1.1

Review & Refresh

Multiply.

Question 1.
150 × 2

Answer: 300

Explanation:
Multiply the two numbers 150 and 2.
First multiply 2 with ones place 2 × 0 = 0
Next multiply with tens place 2 × 50 = 100
Next multiply with hundreds place 2 × 100 = 200
200 + 100 = 300

Question 2.
175 × 8

Answer: 1400

Explanation:
Multiply the two numbers 175 and 8.
First, multiply 2 with ones place 8 × 5 = 40
Next multiply with tens place 8 × 70= 560
Next multiply with hundreds place 8 × 100 = 800
800 + 560 + 40 = 1400

Question 3.
123 × 3

Answer: 369

Explanation:
Multiply the two numbers 123 and 3.
First multiply 2 with ones place 3 × 3 = 9
Next multiply with tens place 3 × 20 = 60
Next multiply with hundreds place 3 ×100 = 300
300 + 60 + 9 = 369

Question 4.
151 × 9

Answer: 1359

Explanation:
Multiply the two numbers 151 and 9.
First multiply 2 with ones place 9 × 1 = 9
Next multiply with tens place 9 × 50 = 450
Next multiply with hundreds place 9 × 100 = 900
900 + 450 + 9 = 1359

Write the sentence as a numerical expression.

Question 5.
Add 5 and 8, then multiply by 4.

Answer: The numerical expression for the above sentence is 5 + 8 × 4

Question 6.
Subtract 7 from 11, then divide by 2.

Answer: The numerical expression for the above sentence is 11 – 7 ÷ 2
Round the number to the indicated place value.

Question 7.
4.03785 to the tenths

Answer: The number 4.03785 nearest to the tenths is 4.0

Question 8.
12.89503 to the hundredths

Answer: The number 12.89503 nearest to the hundredths is 12.90

Complete the sentence.

Question 9.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 17

Answer: 3

Explanation:
(1/10) × 30 = 30/10 = 3
The product of 1/10 and 30 is 3.

Question 10.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 18

Answer: 20

Explanation:
(4/5) × 25 = 4 × 5 = 20
The product of 4/5 and 25 is 20.

Concepts, Skills, & Problem Solving

WRITING EXPRESSIONS USING EXPONENTS
Copy and complete the table. (See Exploration 1, p. 3.)
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 19

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-19
WRITING EXPRESSIONS AS POWERS
Write the product as a power.

Question 15.
9 × 9

Answer: The exponential form of the given expression is 9²

Question 16.
13 × 13

Answer: The exponential form of the given expression is 13²

Question 17.
15 × 15 × 15

Answer: The exponential form of the given expression is 15³

Question 18.
2.2.2.2.2

Answer: The exponential form of the given expression is 25

Question 19.
14 × 14 × 14

Answer: The exponential form of the given expression is 14³

Question 20.
8.8.8.8

Answer: The exponential form of the given expression is 84

Question 21.
11 × 11 × 11 × 11 × 11

Answer: The exponential form of the given expression is 115

Question 22.
7.7.7.7.7.7

Answer: The exponential form of the given expression is 76

Question 23.
16.16.16.16

Answer: The exponential form of the given expression is 164

Question 24.
43 × 43 × 43 × 43 × 43

Answer: The exponential form of the given expression is 435

Question 25.
167 × 167 × 167

Answer: The exponential form of the given expression is 167³

Question 26.
245.245.245.245

Answer: The exponential form of the given expression is 2454

FINDING VALUES OF POWERS
Find the value of the power.

Question 27.
52

Answer: The value of the powers 52 is 5 × 5 = 25

Question 28.
43

Answer: The value of the powers 43 is 4 × 4 × 4 = 64

Question 29.
62

Answer: The value of the powers 62 is 6 × 6 = 36

Question 30.
17

Answer: The value of the powers 17 is 1 × 1 × 1 × 1 × 1 × 1 × 1 = 1

Question 31.
03

Answer: The value of the powers 03 is 0 × 0 × 0 = 0

Question 32.
84

Answer: The value of the powers 84 is 8 × 8 × 8 × 8 = 4096

Question 33.
24

Answer: The value of the powers 24 is 2 × 2 × 2 × 2 = 64

Question 34.
122

Answer: The value of the powers 122 is 12 × 12 = 144

Question 35.
73

Answer: The value of the powers 73 is 7 × 7 × 7 = 343

Question 36.
54

Answer: The value of the powers 54 is 5 × 5 × 5 × 5 = 625

Question 37.
25

Answer: The value of the powers 25 is 2 × 2 × 2 × 2 × 2 = 32

Question 38.
142

Answer: The value of the powers 142 is 14 × 14 = 196

USING TOOLS
Use a calculator to find the value of the power.

Question 39.
76

Answer: 7 × 7 × 7 × 7 × 7 × 7 = 117649

Question 40.
48

Answer: 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 = 256

Question 41.
124

Answer: 12 × 12 × 12 × 12 = 20736

Question 42.
175

Answer: 17 × 17 × 17 × 17 × 17 = 1419857

Question 43.
YOU BE THE TEACHER
Your friend finds the value of 83. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 20

Answer: Your friend is incorrect
83 is nothing but 8 repeats 3 times.
83 = 8 × 8 × 8 = 512

IDENTIFYING PERFECT SQUARES
Determine whether the number is a perfect square.

Question 44.
8

Answer: 8 is not the perfect square. 8 cannot be expressed as the square of a number from the same number system.

Question 45.
4

Answer: 4 is a perfect square.
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Question 46.
81

Answer: 81 perfect square
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Question 47.
44

Answer: 44 is not the perfect square. 44 cannot be expressed as the square of a number from the same number system

Question 48.
49

Answer: 49 is a perfect square
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Question 49.
125

Answer: 125 is not the perfect square. 125 cannot be expressed as the square of a number from the same number system

Question 50.
150

Answer: 150 is not the perfect square. 150 cannot be expressed as the square of a number from the same number system

Question 51.
144

Answer: 144 is the perfect square
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Question 52.
MODELING REAL LIFE
On each square centimeter of a person’s skin, there are about 392 bacteria. How many bacteria does this expression represent?

Answer:
Given,
On each square centimeter of a person’s skin, there are about 392 bacteria.
392  = 39 × 39 = 1521 centimeters
Thus the bacteria represents 1521 centimeters.

Question 53.
REPEATED REASONING
The smallest figurine in a gift shop is 2 inches tall. The height of each figurine is twice the height of the previous figurine. What is the height of the tallest figurine?

Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 21

Answer:
Given that,
The smallest figurine in a gift shop is 2 inches tall. The height of each figurine is twice the height of the previous figurine.
The second figurine is twice that of the first figurine = 2 × 2 = 4 inches
The third figurine is twice that of the second figurine = 4 × 4 = 16 inches
The fourth figurine is twice that of the third figurine = 16 × 16 = 256 inches
Thus the height of the tallest figurine is 256 inches.

Question 54.
MODELING REAL LIFE
A square painting measures 2 meters on each side. What is the area of the painting in square centimeters?

Answer:
Given that,
A square painting measures 2 meters on each side.
Area of the square = s × s
A = 2 m × 2 m = 4 sq. meters
Thus the area of the painting in square centimeters is 4.

Question 55.
NUMBER SENSE
Write three powers that have values greater than 120 and less than 130.

Answer:
11² = 11(11) = 121; this is between 120 and 130.
5³ = 5(5)(5) = 25(5) = 125; this is between 120 and 130.
2⁷ = 2(2)(2)(2)(2)(2)(2) = 4(2)(2)(2)(2)(2) = 8(2)(2)(2)(2) = 16(2)(2)(2) = 32(2)(2) = 64(2) = 128; this is between 120 and 130.

Question 56.
DIG DEEPER!
A landscaper has 125 tiles to build a square patio. The patio must have an area of at least 80 square feet.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 22
a. What are the possible arrangements for the patio?

Answer:
Given that a square patio of at least 80 square feet is to be built from 125 tiles of length 12 inches or 1 foot.
Since there are 125 tiles and the patio has a shape of a square of at least 80 square feet, then the possible dimensions of the patio are
9 ft × 9 ft = 81 ft
10 ft × 10 ft = 100 ft, and
11 ft × 11 ft = 121 ft.

b. How many tiles are not used in each arrangement?

Answer:
For a patio of dimensions, 9ft by 9ft, the number of tiles that will not be used is given by 125 – 81 = 44
For a patio of dimensions, 10ft by 10ft, the number of tiles that will not be used is given by 125 – 100 = 25
For a patio of dimensions, 11ft by 11ft, the number of tiles that will not be used is given by 125 – 121 = 4

Question 57.
PATTERNS
Copy and complete the table. Describe what happens to the value of the power as the exponent decreases. Use this pattern to find the value of 40.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 23

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-23
40 = 1
Thus the value of 40 is 1.

Question 58.
REPEATED REASONING
How many blocks do you need to add to Square 6 to get Square 7? to Square 9 to get Square 10? to Square 19 to get Square 20? Explain.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 24

Answer:
You need to add 14 blocks to get square 7. The square 7 contains 7 × 7 = 49 blocks
You need to add 32 blocks to get square 9. The square 9 contains 9 × 9 = 81 blocks
You need to add 19 blocks to get square 10. The square 10 contains 10 × 10 = 100 blocks
You need to add 261 blocks to get square 19. The square 19 contains 19 × 19 = 361 blocks
You need to add 39 blocks to get square 20. The square 20 contains 20 × 20 = 400 blocks

Lesson 1.2 Order of Operations

Order of Operations

EXPLORATION 1
Comparing Different Orders

Work with a partner. Find the value of each expression by using different orders of operations. Are your answers the same?

Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 25

Answer:
The answers for all the expressions are not the same. The values of each expression will change if you change the order of operations.
a. 3 + 2 × 2
5 × 2 = 10
Multiply, then add
3 + 2 × 2
3 + 4 = 7
b. Subtract then multiply
18 – 3 × 3
15 × 3 = 45
Multiply, then subtract
18 – 3 × 3
18 – 9 = 9
c. Multiply, then subtract
8 × 8 – 2
64 – 2 = 62
Subtract, then Multiply
8 × 8 – 2
8 × 6 = 48
d. Multiply, then add
6 × 6 + 2
36 + 2 = 38
Add, then multiply
6 × 6 + 2
6 × 8 = 48

EXPLORATION 2
Determining Order of Operations
Work with a partner.
a. Scientific calculators use a standard order of operations when evaluating expressions. Why is a standard order of operations needed?

Answer: The order of operations is a rule that tells you the right order in which to solve different parts of a math problem. The order of operations is important because it guarantees that people can all read and solve a problem in the same way.

b. Use a scientific calculator to evaluate each expression in Exploration 1. Enter each expression exactly as written. For each expression, which order of operations is correct?

Answer:
a. 3 + 2 × 2  – Multiply, then add
b. 18 – 3 × 3 – Multiply, then subtract
c. 8 × 8 – 2 – Multiply, then subtract
d. 6 × 6 + 2 – Multiply, then add

c. What order of operations should be used to evaluate 3 + 22, 18 − 32, 82 − 2, and 62 + 2?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 26

Answer:
Solve the expressions by using the calculator.
a. 3 + 2 × 2
3 + 4 = 7
b. 18 – 3 × 3
18 – 9 = 9
c. 8 × 8 – 2
64 – 2 = 62
d. 6 × 6 + 2
36 + 2 = 38
d. Do 18 ÷ 3.3 and 18 ÷ 32 have the same value? Justify your answer.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 27

Answer: No

Explanation:
18 ÷ 3.3
(18 ÷ 3) × 3
6 × 3 = 18
18 ÷ 32  = 2
By using the calculator you can find the difference.
e. How does evaluating powers fit into the order of operations?

Answer:
When an expression has parentheses and powers, evaluate it in the following order: contents of parentheses, powers from left to right, multiplication and division from left to right, and addition and subtraction from left to right.

1.2 Lesson

A numerical expression is an expression that contains numbers and operations. To evaluate, or find the value of, a numerical expression, use a set of rules called the order of operations.
Key Idea
order of operations

  1. Perform operations in grouping symbols.
  2. Evaluate numbers with exponents.
  3. Multiply and divide from left to right.
  4. Add and subtract from left to right.

Try It
a. Evaluate the expression.

Question 1.
7.5 + 3

Answer: 56

Explanation:
You have to evaluate the expression from left to right.
7(5 + 3) = 7 × 8
= 56

Question 2.
(28 – 20) ÷ 4

Answer: 2

Explanation:
You have to evaluate the expression from left to right.
28 – 20 = 8
8 ÷ 4 = 2

Question 3.
[6 + (15 – 10)] × 5

Answer: 55

Explanation:
You have to evaluate the expression from left to right.
[6 + (15 – 10)] × 5
[6 + 5] × 5
11 × 5 = 55

Try It
Evaluate the expression.

Question 4.
6 + 24 – 1

Answer: 21

Explanation:
You have to evaluate the expression from left to right.
6 + 24 – 1
6 + (16 – 1)
6 + 15 = 21
6 + 24 – 1 = 21

Question 5.
4.32 + 18 – 9

Answer: 45

Explanation:
You have to evaluate the expression from left to right.
4.32 + (18 – 9)
4.32 + 9
4 × 9 + 9
36 + 9 = 45

Question 6.
16 + (52 – 7) ÷ 3

Answer:

Explanation:
You have to evaluate the expression from left to right.
16 + (52 – 7) ÷ 3
16 + (25 – 7) ÷ 3
16 + (18) ÷ 3
16 + (18 ÷ 3)
16 + 6 = 22

Thee symbols × and . are used to indicate multiplication. You can also use parentheses to indicate multiplication. For example, 3(2 +7) is the same as 3 × (2 + 7).

Try It
Evaluate the expression.

Question 7.
50 + 6(12 ÷ 4) – 82

Answer: 4

Explanation:
You have to evaluate the expression from left to right.
50 + 6(12 ÷ 4) – 82
50 + 6(3) – 82
50 + 18 – 82
50 + 18 – 64
68 – 64
4

Question 8.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 28

Answer: 24

Explanation:
You have to evaluate the expression from left to right.
5² – 1/5 (10 – 5)
5² – 1/5 (5)
5² – 1
25 – 1
24

Question 9.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 29

Answer:

Explanation:
You have to evaluate the expression from left to right.
8(2+5) = 8 × 7
(8 × 7)/7 = 8

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

USING ORDER OF OPERATIONS
Evaluate the expression.

Question 10.
7 + 2.4

Answer: 15

Explanation:
You have to evaluate the expression from left to right.
7 + 2 × 4 = 7 + 8 = 15

Question 11.
8 ÷ 4 × 2

Answer: 4

Explanation:
You have to evaluate the expression from left to right.
8 ÷ 4 = 2
2 × 2 = 4

Question 12.
3(5 + 1) ÷ 32

Answer: 2

Explanation:
You have to evaluate the expression from left to right.
3(5 + 1) ÷ 32
3 × 6 ÷ 32
18 ÷ 9 =2

Question 13.
WRITING
Why does 12 − 8 ÷ 2 = 2?

Answer:
12 − 8 ÷ 2
4 ÷ 2 = 2

Question 14.
REASONING
Describe the steps in evaluating the expression 8 ÷ (6 − 4) + 32.

Answer:
8 ÷ (6 − 4) + 32
8 ÷ 2 + 32
4 + 9 = 13

Question 15.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 30

Answer: (52 – 8) × 2 does not belong to the other three. Because the order of operations and expressions are different for the fourth option.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 16.
A square plot of land has side lengths of 40 meters. An archaeologist divides the land into 64 equal parts. What is the area of each part?

Answer:
Given that,
A square plot of land has side lengths of 40 meters.
An archaeologist divides the land into 64 equal parts.
Side of the square field = 40m
Area of the square field = s × s
A = 40m × 40m
A = 1600 sq.m
Area of each part of the square = 1600/64 = 25 sq.m

Question 17.
A glass block window is made of two different-sized glass squares. The window has side lengths of 40 inches. The large glass squares have side lengths of 10 inches. Find the total area of the small glass squares.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 31

Answer:
Given,
A glass block window is made of two different-sized glass squares.
The window has side lengths of 40 inches. The large glass squares have side lengths of 10 inches.
40 × 10 = 400

Question 18.
DIG DEEPER!
A square vegetable garden has side lengths of 12 feet. You plant flowers in the center portion as shown. You divide the remaining space into 4 equal sections and plant tomatoes, onions, zucchini, and peppers. What is the area of the onion section?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 32

Answer:
A square vegetable garden has side lengths of 12 feet.
You plant flowers in the center portion of the garden, a square that has side lengths of 4 feet.
You divide the remaining space into 4 equal sections and plant tomatoes, onions, zucchini, and peppers.
Given that,
→ side of the square vegetable garden is = 12 feet.
So,
→ Area of square vegetable garden = (side)² = (12)² = 144 feet².
Now, given that, inside this area, there is a square of side 4 feet reserved for flowers.
So,
→ The area of the flower section = (side)² = (4)² = 16 feet².
Therefore,
→ The rest of the garden that is intended for vegetables is = The total garden area – The flower section area = 144 – 16 = 128 feet².
Now, this remaining area is to be divided into four equal sections.
So,
→ The area of the onion section = (1/4) of remaining area = (1/4) × 128 = 32 feet².

Order of Operations Practice 1.2

Review & Refresh

Write the product as a power.

Question 1.
11 × 11 × 11 × 11

Answer: The exponent for the product 11 × 11 × 11 × 11 is 114

Question 2.
13 × 13 × 13 × 13 × 13

Answer: The exponent for the product 13 × 13 × 13 × 13 × 13 is 135
Find the missing dimension of the rectangular prism.

Question 3.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 33

Answer:
Given that,
l = 6 in.
b = 4 in.
h = ?
v = 192 cu. in
Volume of the rectangular prism = lbh
192 = 6 × 4 × h
h = 192/24
h = 8
Thus the height of the rectangular prism is 8 inches.

Question 4.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 34

Answer:
Given that,
h = 9m
b = 3m
v = 135 cu. m
l = ?
The volume of the rectangular prism = lbh
135 = l × 3 × 9
135 = l × 27
l = 5m

Tell whether the number is prime or composite.

Question 5.
9

Answer: Composite Number
A natural number greater than 1 that is not prime is called a composite number.

Question 6.
11

Answer: Prime Number
A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers.

Question 7.
23

Answer: Prime Number
A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers.

Concepts, Skills, & Problem Solving

COMPARING DIFFERENT ORDERS
Find the value of the expression by using different orders of operations. Are your answers the same? (See Exploration 1, p. 9.)

Question 8.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 35

Answer:
4 + 6 × 6
10 × 6 = 60
4 + 6 × 6
4 + 36 = 40

Question 9.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 36

Answer:
5 × 5 – 3
5 × 2 = 10
5 × 5 – 3
25 – 3 = 22

USING ORDER OF OPERATIONS
Evaluate the expression.

Question 10.
5 + 18 ÷ 6

Answer: 8

Explanation:
First, divide then divide.
5 + 3 = 8

Question 11.
(11 – 3) ÷ 2 + 1

Answer:

Explanation:
First, subtract and divide.
(11 – 3) ÷ 2 + 1
8 ÷ 2 + 1
4 + 1 = 5

Question 12.
45 ÷ 9 × 2

Answer: 10

Explanation:
The first divide then multiply.
45 ÷ 9 × 2
5 × 2 = 10

Question 13.
62 – 3.4

Answer: 24

Explanation:
Multiply then subtract
62 – 3.4
36 – 12
24

Question 14.
42 ÷ (15 – 23)

Answer: 6

Explanation:
Subtract then divide.
42 ÷ (15 – 8)
42 ÷ 7
6

Question 15.
42.2 + 8.7

Answer: 88

Explanation:
Multiply then add
42.2 + 8.7
16 × 2 + 56
32 + 56 = 88

Question 16.
(52 – 2) × 15 + 4

Answer: 27

Explanation:
(52 – 2) × 15 + 4
(25 – 2) × 1 + 4
Add, subtract then multiply
23 + 4 = 27

Question 17.
4 + 2 × 32 – 9

Answer: 13

Explanation:
4 + 2 × 32 – 9
4 + 18 – 9
4 + 9 = 13

Question 18.
8 ÷ 2 × 3 + 42 ÷ 4

Answer: 16

Explanation:
8 ÷ 2 × 3 + 42 ÷ 4
(4 × 3) + (16 ÷ 4)
12 + 4
16

Question 19.
32 + 12 ÷ (6 – 3) × 8

Answer: 41

Explanation:
32 + 12 ÷ (6 – 3) × 8
9 + (12 ÷ (6 – 3)) × 8
9 + (12 ÷ 3) × 8
9 + 4 × 8
9 + 32
41

Question 20.
(10 + 4) ÷ (26 – 19)

Answer: 2

Explanation:
Add, subtract then divide
(10 + 4) ÷ (26 – 19)
14 ÷ 7
2

Question 21.
(52 – 4).2 – 18

Answer: 24

Explanation:
((52 – 4).2) – 18
((25 – 4) × 2) – 18
(21 × 2) – 18
42 – 18
24

Question 22.
2 × [(16 – 8) × 2]

Answer: 32

Explanation:
2 × [(16 – 8) × 2]
2 × [8 × 2]
2 × 16
32

Question 23.
12 + 8 × 33 – 24

Answer: 204

Explanation:
12 + 8 × 33 – 24
12 + (8 × 27) – 24
12 + 216 – 24
12 + 192 = 204

Question 24.
62 ÷ [(2 + 4) × 23]

Answer: 48

Explanation:
62 ÷ [(2 + 4) × 23]
36 ÷ [(2 + 4) × 23]
36 ÷ 6 × 8
6 × 8
48

YOU BE THE TEACHER
Your friend evaluates the expression. Is your friend correct? Explain your reasoning.

Question 25.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 37

Answer: Your friend is incorrect.
9 + 3 × 3²
9 + (27)
36

Question 26.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 38

Answer:
19 – 6 + 12
13 + 12
25

Question 27.
PROBLEM SOLVING
You need to read 20 poems in 5 days for an English project. Each poem is 2 pages long. Evaluate the expression 20 × 2 ÷ 5 to find how many pages you need to read each day.

Answer:
Given,
You need to read 20 poems in 5 days for an English project. Each poem is 2 pages long.
20 × 2 ÷ 5
40 ÷ 5 = 8
Thus you need to read 8 pages each day.

USING ORDER OF OPERATIONS
Evaluate the expression.

Question 28.
12 – 2(7 – 4)

Answer:
12 -(2 × (7 – 4))
12 – (2 × 3)
12 – 6 = 6

Question 29.
4(3 + 5) – 3(6 -2)

Answer:
4(3 + 5) – 3(6 -2)
4 × 8 – 3 × 4
32 – 12
20

Question 30.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 39

Answer:
6 + 1/4 (12 -8)
6 + 1/4(4)
6 + 1
7

Question 31.
92 – 8(6 + 2)

Answer:
81 – (8(6 + 2))
81 – (8 × 8)
81 – 64
17

Question 32.
4(3 – 1)3 + 7(6) – 52

Answer:
4(3 – 1)3 + 7(6) – 52
4(2)3 + 7(6) – 52
4 × 8 + 42 – 25
32 + 42 – 25 = 49

Question 33.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 40

Answer:
8[(1 1/6 + 5/6) ÷ 4]
[8[7/6 + 5/6] ÷ 4]
8[12/6] ÷ 4
8[2 ÷ 4]
8(1/2)
4

Question 34.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 41

Answer:
49 – 2((11-3)/8)
49 – 2 (8/8)
49 – 2
47

Question 35.
8(7.3 + 3.7 – 8) ÷ 2

Answer:
8(7.3 + 3.7 – 8) ÷ 2
(8(7.3 + 3.7 – 8)) ÷ 2
8 (11 – 8) ÷ 2
8 × 3 ÷ 2
24 ÷ 2
12

Question 36.
24(5.2 – 3.2) ÷ 4

Answer:
24(5.2 – 3.2) ÷ 4
16 (5.2 – 3.2) ÷ 4
16 (2) ÷ 4
32 ÷ 4
8

Question 37.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 42

Answer:
36(3+5)/4
36 × 8/4
36 × 2
72

Question 38.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 43

Answer:
(144 – 24 + 1)/121
121/121
1

Question 39.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 44

Answer:
26 ÷ 2 + 5 = 18
18/6 = 3

Question 40.
PROBLEM SOLVING
Before a show, there are 8 people in a theater. Five groups of 4 people enter, and then three groups of 2 people leave. Evaluate the expression 8 + 5(4) − 3(2) to find how many people are in the theater.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 45

Answer:
Given,
Before a show, there are 8 people in a theater. Five groups of 4 people enter, and then three groups of 2 people leave.
8 + (5 × 4) – (3 × 2)
8 + 20 – 6
28 – 6
22

Question 41.
MODELING REAL LIFE
The front door of a house is painted white and blue. Each window is a square with a side length of 7 inches. What is the area of the door that is painted blue?

Answer:
Given,
The front door of a house is painted white and blue. Each window is a square with a side length of 7 inches.
Area of the square = s × s
A = 7 in × 7 in
A = 49 sq. inches
Therefore the area of the door that is painted blue is 49 sq. inches.

Question 42.
PROBLEM SOLVING
You buy 6 notebooks, 10 folders, 1 pack of pencils, and 1 lunch box for school. After using a $10 gift card, how much do you owe? Explain how you solved the problem.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 46

Answer:
Given,
You buy 6 notebooks, 10 folders, 1 pack of pencils, and 1 lunch box for school.
Cost of 1 notebook = $2
6 notebooks = 6 × $2 = $12
Cost of 1 folder = $1
10 folders = 10 × $1 = $10
Cost of 1 pack of pencils = $3
Cost of 1 lunch box = $8
So the total cost is $11 + $10 + $3 + $8 = $31
You used $10 gift card.
$31 – $10 = $21
Thus you ow $21.

Question 43.
OPEN-ENDED
Use all four operations and at least one exponent to write an expression that has a value of 100.

Answer: You need to use +, -, ×, ÷  operations to write the expressions that have the value of 100.
(34 – 1) × 3 + 3² ÷ 9 = 100

Question 44.

Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 47

REPEATED REASONING
A Petri dish contains 35 cells. Every day, each cell in the Petri dish divides into 2 cells in a process called mitosis. How many cells are there after 14 days? Justify your answer.

Answer:
Given,
A Petri dish contains 35 cells. Every day, each cell in the Petri dish divides into 2 cells in a process called mitosis
35 ÷ 2 = 17.5
1 day = 0.5 + 0.5 cells
14 days = 14 × 1 = 14 cells
17.5 – 14 = 3.5
Thus there are 3.5 cells after 14 days.

Question 45.
REASONING
Two groups collect litter along the side of a road. It takes each group 5 minutes to clean up a 200-yard section. How long does it take both groups working together to clean up 2 miles? Explain how you solved the problem.

Answer:
Given,
Two groups collect litter along the side of a road. It takes each group 5 minutes to clean up a 200-yard section.
To convert 2 miles to yards, you have to multiply 2 by 1760, because 1 mile equals to 1760 yards:
2 × 1760 = 3520 yards.
If you would like to know how long does it take to clean up 2 miles, you can calculate this using the following steps:
5 × 3520 = 200 × x
17600 = 200 × x /200
x = 17600 / 200
x = 88 minutes

Question 46.
NUMBER SENSE
Copy each statement. Insert +, −, ×, or ÷ symbols to make each statement true.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 48

Answer:
You can find the value by using the calculator by inserting the suitable operations.
Big-Ideas-Math-Answers-6th-Grade-Chapter-1-Numerical-Numerical-Expressions-and-Factors-48

Lesson 1.3 Prime Factorization

Prime Factorization

EXPLORATION 1
Rewriting Numbers as Products of Factors

Work with a partner. Two students use factor trees to write 108 as a product of factors, as shown below.

Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 49
a. Without using 1 as a factor, can you write 108 as a product with more factors than each student used? Justify your answer. Math Practice
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 50

Answer: Yes you can find the factors by using the prime factorization.
108 = 2 × 54
= 2 × 2 × 27
= 2 × 2 × 3 × 9
= 2 × 2 × 3 × 3 × 3

b. Use factor trees to write 80, 162, and 300 as products of as many factors as possible. Do not use 1 as a factor.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 50.1

Answer:
80 = 2 × 40
= 2 × 2 × 20
= 2 × 2 × 2 × 10
= 2 × 2 × 2 × 2 × 5
162 = 2 × 81
= 2 × 3 × 27
= 2 × 3 × 3 × 9
= 2 × 3 × 3 × 3 × 3
300 = 2 × 150
= 2 × 2 × 75
= 2 × 2 × 3 × 25
= 2 × 2 × 3 × 5 × 5

c. Compare your results in parts (a) and (b) with other groups. For each number, identify the product with the greatest number of factors. What do these factors have in common?

Answer: 300 contains the greatest number of factors. (2, 2, 3, 5, 5)

1.3 Lesson

Because 2 is a factor of 10 and 2 . 5 =10, 5 is also a factor of 10. The pair 2, 5 is called a factor pair of 10.

Try It
List the factor pairs of the number.

Question 1.
18

Answer: The factor pairs of 18 are 1, 2, 3, 6, 9, 18

Explanation:
1 × 18 = 18
2 × 9 = 18
3 × 6 = 18
6 × 3 = 18
9 × 2 = 18
18 × 1 = 18

Question 2.
24

Answer: The factor pairs of 1, 2, 3, 4, 6, 8, 12, 24

Explanation:
1 × 24 = 24
2 × 12 = 24
3 × 8 = 24
4 × 6 = 24
6 × 4 = 24
8 × 3 = 24
12 × 2 = 24
24 × 1 = 24

Question 3.
51

Answer: The factor pairs of 1, 3, 17, 51

Explanation:
1 × 51 = 51
3 × 17 = 51
17 × 3 = 51
51 × 1 = 15

Question 4.
WHAT IF?
The woodwinds section of the marching band has 38 members. Which has more possible arrangements, the brass section or the woodwinds section? Explain.

Answer: Brass section. 38 has only two-factor pairs.
38 = 1 × 38
= 2 × 19

Key Idea

Prime Factorization
The prime factorization of a composite number is the number written as a product of its prime factors.
You can use factor pairs and a factor tree to help find the prime factorization of a number. The factor tree is complete when only prime factors appear in the product. A factor tree for 60 is shown.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 51
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 52

Try It
Write the prime factorization of the number.

Question 5.
20

Answer:
The Prime Factorization is:
2 x 2 x 5
In Exponential Form:
2² x 51
CSV Format:
2, 2, 5

Question 6.
88

Answer:
88 = 2 × 44
= 2 × 2 × 22
= 2 × 2 × 2 × 11
The Prime Factorization is: 2 × 2 × 2 × 11

Question 7.
90

Answer:
90 = 2 × 45
= 2 × 3 × 15
= 2 × 3 × 3 × 5
The Prime Factorization is: 2 × 3 × 3 × 5

Question 8.
462

Answer:
= 2 × 231
= 2 × 3 × 77
= 2 × 3 × 7 × 11
The Prime Factorization is: 2 × 3 × 7 × 11

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

WRITING A PRIME FACTORIZATION
Write the prime factorization of the number.

Question 9.
14

Answer:
14 = 2 × 7
The Prime Factorization is: 2 × 7

Question 10.
86

Answer:
86 = 2 × 43
The Prime Factorization is: 2 × 43

Question 11.
40

Answer:
40 = 2 × 20
= 2 × 2 × 10
= 2 × 2 × 2 × 5
The Prime Factorization is: 2 × 2 × 2 × 5

Question 12.
516

Answer:
516 = 2 × 258
= 2 × 2 × 129
= 2 × 2 × 3 × 43
The Prime Factorization is: 2 × 2 × 3 × 43

Question 13.
WRITING
Explain the difference between prime numbers and composite numbers.

Answer:
A prime number is a number that has exactly two factors i.e. ‘1’ and the number itself. A composite number has more than two factors, which means apart from getting divided by number 1 and itself, it can also be divided by at least one integer or number.

Question 14.
STRUCTURE
Your friend lists the following factor pairs and concludes that there are 6 factor pairs of 12. Explain why your friend is incorrect.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 53

Answer: Your friend is incorrect. Because there are 5-factor pairs of 12.
The factor pairs of 12 are
1 × 12 =12
2 × 6 = 12
3 × 4 = 12
4 × 3 = 12
6 × 2 = 12

Question 15.
WHICH ONE DOESN’T BELONG?
Which factor pair does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 54

Answer:
2, 28 = 2 × 28 = 56
4, 14 = 4 × 14 = 56
6, 9 = 6 × 9 = 54
7, 8 = 7 × 56
By this we can say that 6, 9 does not belong to the other three expressions.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 16.
A group of 20 friends plays a card game. The game can be played with 2 or more teams of equal size. Each team must have atleast 2 members. List the possible numbers and sizes of teams.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 55

Answer:
Given,
A group of 20 friends plays a card game. The game can be played with 2 or more teams of equal size. Each team must have at least 2 members.
20 = 1 × 20
2 × 10
4 × 5
5 × 4
10 × 2
Thus there are 5 possible numbers and size of teams.

Question 17.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 56
You arrange 150 chairs in rows for a school play. You want each row to have the same number of chairs. How many possible arrangements are there? Are all of the possible arrangements appropriate for the play? Explain.

Answer:
You arrange 150 chairs in rows for a school play. You want each row to have the same number of chairs.
150 = 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150
Thus there are 12 possible arrangements appropriate for the play.

Question 18.
What is the least perfect square that is a factor of 4536? What is the greatest perfect square that is a factor of 4536?

Answer: least perfect square that is a factor of 4536

Explanation:
What is the last number of 4,536? It is this number: 4536. The answer is 6. Is 6 in the list of numbers that are never perfect squares (2, 3, 7 or 8)?
Answer: No, 6 is not in the list of numbers that are never perfect squares. Let’s continue to the next step.
Step 2:
We now need to obtain the digital root of the number. Here’s how you do it:
Split the number up and add each digit together:
4 + 5 + 3 + 6 = 18
If the answer is more than one digit, you would add each digit of the answer together again:
1 + 8 = 9
1 x 4,5362 x 2,2683 x 1,5124 x 1,1346 x 7567 x 6488 x 5679 x 50412 x 37814 x 32418 x 25221 x 21624 x 18927 x 16828 x 16236 x 12642 x 10854 x 8456 x 8163 x 72
We’re looking for a factor combination with equal numbers for X and Y (like 3×3) above. Notice there isn’t an equal factor combination, that when multiplied together, produce the number 4,536. That means 4,536 is NOT a perfect square.

Question 19.
DIG DEEPER!
The prime factorization of a number is 24 × 34 × 54 × 72. Is the number a perfect square? Explain your reasoning.

Answer:
The prime factorization of a number is 24 × 34 × 54 × 72.
16 × 81 × 625 × 49 = 39690000
Yes, 39690000 is a perfect square.

Prime Factorization Practice 1.3

Review & Refresh

Evaluate the expression.

Question 1.
2 + 42(5 – 3)

Answer: 34

Explanation:
2 + 42(5 – 3)
2 + 42(2)
2 + 16(2) = 2 + 32 = 34

Question 2.
23 + 4 × 32

Answer: 44

Explanation:
23 + 4 × 32
8 + 4 × 9
8 + 36
44

Question 3.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 57

Answer:

Explanation:
9 × 5 – 16(5/2 – 1/2)
9 × 5 – 16(4/2)
45 – 16(2)
45 – 32
13

Plot the points in a coordinate plane. Draw a line segment connecting the points.

Question 4.
(1, 1) and (4, 3)

Answer:
Big Ideas Math Grade 6 Chapter 1 img_1

Question 5.
(2, 3) and (5, 9)

Answer:
Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_2

Question 6.
(2, 5) and (4, 8)

Answer:
Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_2

Use the Distributive Property to find the quotient. Justify your answer.

Question 7.
408 ÷ 4

Answer: 120
Write 408 as 204 and 204
204 ÷ 4 = 51
204 ÷ 4 = 51
51 + 51 = 102
408 ÷ 4 = 102

Question 8.
628 ÷ 2

Answer: 314
608 can be written as 314 and 314
314 ÷ 2 = 157
314 ÷ 2 = 157
157 + 157 = 314
628 ÷ 2 = 314

Question 9.
969 ÷ 3

Answer: 323
969 can be written as 900 and 69
900 ÷ 3 = 300
69 ÷ 3 = 23
300 + 23 = 323
969 ÷ 3 = 323

Classify the triangle in as many ways as possible.

Question 10.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 58

Answer: Acute
Acute angles measure less than 90 degrees. Right angles measure 90 degrees. Obtuse angles measure more than 90 degrees.

Question 11.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 59

Answer: Obtuse
An obtuse angle has a measurement greater than 90 degrees but less than 180 degrees. However, A reflex angle measures more than 180 degrees but less than 360 degrees.

Question 12.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 60

Answer: Right angle
In geometry and trigonometry, a right angle is an angle of exactly 90° (degrees), corresponding to a quarter turn. If a ray is placed so that its endpoint is on a line and the adjacent angles are equal, then they are right angles.

Concepts, Skills, & Problem Solving

REWRITING A NUMBER
Write the number as a product of as many factors as possible. (See Exploration 1, p. 15.)

Question 13.
60

Answer: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

The factors of 60 are:
1 × 60 = 60
2 × 30 = 60
3 × 20 = 60
4 × 15 = 60
5 × 12 = 60
6 × 10 = 60
10 × 6 = 60
12 × 5 = 60
15 × 4 = 60
20 × 3 = 60
30 × 2 = 60
60 × 1 = 60

Question 14.
63

Answer: The factors of 63 are 1, 3, 7, 9, 21, 63

The factors of 63 are:
1 × 63 = 63
3 × 21 = 63
7 × 9 = 63
9 × 7 = 63
21 × 3 = 63
63 × 1 = 63

Question 15.
120

Answer: The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

The factors of 120 are:
1 × 120 = 120
2 × 60 = 120
3 × 40 = 120
4 × 30 = 120
5 × 24 = 120
6 × 20 = 120
8 × 15 = 120
10 × 12 = 120
12 × 10 = 120
15 × 8 = 120
20 × 6 = 120
24 × 5 = 120
30 × 4 = 120
40 × 3 = 120
60 × 2 = 120
120 × 1 = 120

Question 16.
150

Answer: The factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150

The factors of 150 are:
1 × 150 = 150
2 × 75 = 150
3 × 50 = 150
5 × 30 = 150
6 × 25 = 150
10 × 15 = 150

FINDING FACTOR PAIRS
List the factor pairs of the number.

Question 17.
15

Answer: The factor pairs of 15 are (1, 15), (3, 5)
1 × 15 = 15
3 × 5 = 15
5 × 3 = 15
15 × 1 = 15

Question 18.
22

Answer: The factor pairs of 22 are (1, 22) (2, 11)
1 × 22 = 22
2 × 11 = 22
11 × 2 = 22
22 × 1 = 22

Question 19.
34

Answer: (1, 34) (2,17)

The factor pairs of 34 are
1 × 34 = 34
2 × 17 = 34
17 × 2 = 34
34 × 1 = 34

Question 20.
39

Answer: (1, 39) (3, 13)

The factor pairs of 39 are
1 × 39 = 39
3 × 13 = 39
13 × 3 = 39
39 × 1 = 39

Question 21.
45

Answer: (1, 45) (3, 15) (5, 9)

The factor pairs of 45 are
1 × 45 = 45
3 × 15 = 45
5 × 9 = 45

Question 22.
54

Answer: (1, 54) (2, 27) (3, 18) (6, 9)

The factor pairs of 54 are
1 × 54 = 54
2 × 27 = 54
3 × 18 = 54
6 × 9 = 54

Question 23.
59

Answer: (1, 59)
The factor pairs of 59 are
1 × 59 = 59
59 × 1 = 59

Question 24.
61

Answer: (1, 61)
The factor pairs of 61 are
1 × 61 = 61
61 × 1 = 61

Question 25.
100

Answer: (1, 100) (2, 50) (4, 25) (5, 20) (10, 10)
The factor pairs of 100 are
1 × 100 = 100
2 × 50 = 100
4 × 25 = 100
5 × 20 = 100
10 × 10 = 100

Question 26.
58

Answer: (1, 58) (2, 29)
The factor pairs of 58 are
1 × 58 = 58
2 × 29 = 58

Question 27.
25

Answer: (1, 25) (5, 5)
The factor pairs of 25 are
1 × 25 = 25
5 × 5 = 25

Question 28.
76

Answer: (1, 76) (2, 38) (4, 19)
The factor pairs of 76 are
1 × 76 = 76
2 × 38 = 76
4 × 19 = 76

Question 29.
52

Answer: (1, 52) (2, 26) (4, 13)
The factor pairs of 52 are
1 × 52 = 52
2 × 26 = 52
4 × 13 = 52

Question 30.
88

Answer: (1,88) (2,44) (4, 22) (8, 11)
The factor pairs of 88 are
1 × 88 = 88
2 × 44 = 88
4 × 22 = 88
8 × 11 = 88

Question 31.
71

Answer: (1,71)
The factor pairs of 71 are
1 × 71 = 71

Question 32.
91

Answer: (1, 91) (7, 13)
The factor pairs of 91 are
1 × 91 = 91
7 × 13 = 91

WRITING A PRIME FACTORIZATION
Write the prime factorization of the number.

Question 33.
16

Answer:
16 = 2 × 8
2 × 2 × 4
2 × 2 × 2 × 2

Question 34.
25

Answer:
25 = 5 × 5

Question 35.
30

Answer:
30 = 2 × 15
= 2 × 3 × 5

Question 36.
26

Answer:
26 = 2 × 13

Question 37.
84

Answer:
84 = 2 × 42
2 × 2 × 21
2 × 2 × 3 × 7

Question 38.
54

Answer:
54 = 2 × 27
2 × 3 × 9
2 × 3 × 3 × 3

Question 39.
65

Answer:
65 = 5 × 13

Question 40.
77

Answer:
77 = 7 × 11

Question 41.
46

Answer:
46 = 2 × 23

Question 42.
39

Answer:
39 = 3 × 13

Question 43.
99

Answer:
99 = 3 × 33
3 × 3 × 11

Question 44.
24

Answer:
24 = 2 × 12
2 × 2 × 6
2 × 2 × 2 × 3

Question 45.
315

Answer:
315 = 3 × 105
3 × 3 × 35
3 × 3 × 5 × 7

Question 46.
490

Answer:
490 = 2 × 245
2 × 5 × 49
2 × 5 × 7 × 7

Question 47.
140

Answer:
2 × 70
2 × 2 × 35
2 × 2 × 5 × 7

Question 48.
640

Answer:
640 = 2 × 320
2 × 2 × 160
2 × 2 × 2 × 80
2 × 2 × 2 × 2 × 40
2 × 2 × 2 × 2 × 2 × 20
2 × 2 × 2 × 2 × 2 × 2 × 10
2 × 2 × 2 × 2 × 2 × 2 × 2 × 5

USING A PRIME FACTORIZATION
Find the number represented by the prime factorization.

Question 49.
22.32.5

Answer:
4 × 9 × 5 = 180
We have to find the prime factorization for 180.
180 = 2 × 90
2 × 2 × 45
2 × 2× 3 × 15
2 × 2 × 3 × 3 × 5

Question 50.
32.52.7

Answer:
9 × 25 × 7 = 1575
We have to find the prime factorization for 1575.
1575 = 3 × 525
3 × 3 × 175
3 × 3 × 5 × 35
3 × 3 × 5 × 5 × 7

Question 51.
23.112.13

Answer:
8 × 11 × 13 = 1144
We have to find the prime factorization for 1144.
1144 = 2 × 572
2 × 2 × 286
2 × 2 × 2 × 143
2 × 2 × 2 × 11 × 13

Question 52.
YOU BE THE TEACHER
Your friend finds the prime factorization of 72. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 61

Answer:
72 = 2 × 36
2 × 2 × 18
2 × 2 × 2 × 9
2 × 2 × 2 × 3 × 3
Your friend is incorrect because you have to write the prime factorization for 9 also.
Thus the prime factorization for 72 is 2 × 2 × 2 × 3 × 3

USING A PRIME FACTORIZATION
Find the greatest perfect square that is a factor of the number.

Question 53.
250

Answer:
A = Calculate the square root of the greatest perfect square from the list of all factors of 250. The factors of 250 are 1, 2, 5, 10, 25, 50, 125, and 250. Furthermore, the greatest perfect square on this list is 25 and the square root of 25 is 5. Therefore, A equals 5.
B = Calculate 250 divided by the greatest perfect square from the list of all factors of 250. We determined above that the greatest perfect square from the list of all factors of 250 is 25. Furthermore, 250 divided by 25 is 10, therefore B equals 10.
Now we have A and B and can get our answer to 250 in its simplest radical form as follows:
√250 = A√B
√250 = 5√10

Question 54.
275

Answer:
A = Calculate the square root of the greatest perfect square from the list of all factors of 275. The factors of 275 are 1, 5, 11, 25, 55, and 275. Furthermore, the greatest perfect square on this list is 25 and the square root of 25 is 5. Therefore, A equals 5.
B = Calculate 275 divided by the greatest perfect square from the list of all factors of 275. We determined above that the greatest perfect square from the list of all factors of 275 is 25. Furthermore, 275 divided by 25 is 11, therefore B equals 11.
Now we have A and B and can get our answer to 275 in its simplest radical form as follows:
√275 = A√B
√275 = 5√11

Question 55.
392

Answer:
A = Calculate the square root of the greatest perfect square from the list of all factors of 392. The factors of 392 are 1, 2, 4, 7, 8, 14, 28, 49, 56, 98, 196, and 392. Furthermore, the greatest perfect square on this list is 196 and the square root of 196 is 14. Therefore, A equals 14.
B = Calculate 392 divided by the greatest perfect square from the list of all factors of 392. We determined above that the greatest perfect square from the list of all factors of 392 is 196. Furthermore, 392 divided by 196 is 2, therefore B equals 2.
Now we have A and B and can get our answer to 392 in its simplest radical form as follows:
√392 = A√B
√392 = 14√2

Question 56.
338

Answer:
A = Calculate the square root of the greatest perfect square from the list of all factors of 338. The factors of 338 are 1, 2, 13, 26, 169, 338. Furthermore, the greatest perfect square on this list is 324 and the square root of 324 is 18. Therefore, A equals 18.
B = Calculate 338 divided by the greatest perfect square from the list of all factors of 338. We determined above that the greatest perfect square from the list of all factors of 338 is 324.
Now we have A and B and can get our answer to 338 in its simplest radical form as follows:
√338= A√B
√392 = 18√14

Question 57.
244

Answer:
Our first step would be to find out all the factors of 244
Since this is an even number, we divide 2 we get the factors as 244/2 = 122
Next, we again divide by 2 we get the factor as 122/2 = 61
We cannot go down any further since 61 is a prime number
Since we divided by 2, 2 is itself a factor.
Lastly, we divided by 2, twice; hence 2*2 = 4 is also a factor
The factors of 244 are 2,4,61 and 122
Out of 2,4,61 and 122, the only perfect square is 4
So, the greatest perfect square that is a factor of the number 244 should be 4
Therefore, the answer is: 4

Question 58.
650

Answer:
factor 650 and find the pairs
650=2×5×5×13
the pair is 5×5 which is 5² or 25
the greatest perfect square that is a factor of 650 is 25.

Question 59.
756

Answer:
A = Calculate the square root of the greatest perfect square from the list of all factors of 756. The factors of 756 are 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 27, 28, 36, 42, 54, 63, 84, 108, 126, 189, 252, 378, and 756. Furthermore, the greatest perfect square on this list is 36 and the square root of 36 is 6. Therefore, A equals 6.
B = Calculate 756 divided by the greatest perfect square from the list of all factors of 756. We determined above that the greatest perfect square from the list of all factors of 756 is 36. Furthermore, 756 divided by 36 is 21, therefore B equals 21.
Now we have A and B and can get our answer to 756 in its simplest radical form as follows:
√756 = A√B
√756 = 6√21

Question 60.
1290

Answer:
There is no greatest perfect square that is a factor of 1290.
The factors of 1290 are 1, 2, 3, 5, 6, 10, 15, 30, 43, 86, 129, 215, 258, 430, 645, 1290. There are no perfect squares as factors.

Question 61.
2205

Answer: No, the number 2,205 is not a perfect square.
The factors of 2205 are 1, 3, 5, 7, 9, 15, 21, 35, 45, 49, 63, 105, 147, 245, 315, 441, 735. There are no perfect squares as factors.

Question 62.
1890

Answer: 1890 is not the perfect square.
The factors of 1890 are 1, 3, 9, 27, 67, 201, 603, 1809. There are no perfect squares as factors.

Question 63.
495

Answer: 495 is not the perfect square.
The factors 495 are 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495. There are no perfect squares as factors.

Question 64.
4725

Answer: 4725 is not the perfect square.
The factors of 4725 are 1, 3, 5, 7, 9, 15, 21, 25, 27, 35, 45, 63, 75, 105, 135, 175, 189, 225, 315, 525, 675, 945, 1575, 4725. There are no perfect squares as factors.

Question 65.
VOCABULARY
A botanist separates plants into equal groups of 5 for an experiment. Is the total number of plants in the experiment prime or composite? Explain.

Answer: The total number of plants will be a composite.

Explanation:
She is separating them into equal groups of five. So the total number will be a multiple of 5. Five is a prime number. Multiples of prime numbers are composite numbers.

Question 66.
REASONING
A teacher divides 36 students into equal groups for a scavenger hunt. Each group should have at least 4 students but no more than 8 students. What are the possible group sizes?

Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 62

Answer:
Given,
A teacher divides 36 students into equal groups for a scavenger hunt.
Each group should have at least 4 students but no more than 8 students.
The factors of 36 are 1, 2, 3, 4, 6, 9
So, there are 6 groups of 6, 9 groups of 4.

Question 67.
CRITICAL THINKING
Is 2 the only even prime number? Explain.

Answer:
The definition of a prime number is a positive integer that has exactly two distinct divisors. Since the divisors of 2 are 1 and 2, there are exactly two distinct divisors, so 2 is prime.

Question 68.
LOGIC
One table at a bake sale has 75 cookies. Another table has 60 cupcakes. Which table allows for more rectangular arrangements? Explain.

Answer:
Given,
One table at a bake sale has 75 cookies. Another table has 60 cupcakes.
75 = 3·5², so has 6 divisors. 6 rectangles are possible if you make the distinction between 1×75 and 75×1.
60 = 2²·3·5, so has 12 divisors. 12 rectangles are possible under the same conditions.

Question 69.
PERFECT NUMBERS
A perfect number is a number that equals the sum of its factors, not including itself. For example, the factors of 28 are 1, 2, 4, 7, 14, and 28. Because 1 + 2 + 4 + 7 + 14 = 28, 28 is a perfect number. What are the perfect numbers between 1 and 27?

Answer: Perfect number, a positive integer that is equal to the sum of its proper divisors. The smallest perfect number is 6, which is the sum of 1, 2, and 3. Other perfect numbers are 28, 496, and 8,128.

Question 70.
REPEATED REASONING
Choose any two perfect squares and find their product. Then multiply your answer by another perfect square. Continue this process. Are any of the products perfect squares? What can you conclude?

Answer:
2² × 3² = 4×9=36, which is a square number

Question 71.
PROBLEM SOLVING
The stage manager of a school play creates a rectangular stage that has whole number dimensions and an area of 42 square yards. String lights will outline the stage. What is the least number of yards of string lights needed to enclose the stage?

Answer:
Given that the stage manager of a school play creates a rectangular acting area of 42 square yards.
Let the length of the rectangular acting area be x, then the width is given by 42 / x.
The number of yards of string lights that the manager needs to enclose the area is given by the perimeter of the rectangular area.
Recall that the perimeter of a rectangle is given by
P = 2(length + width) = 2(x + 42/x) = 2x + 84/x
The perimeter is minimum when the differentiation of 2x + 84/x is equal to 0.
Therefore, the minimum number of yards of string lights the manager need to enclose in this area is given by
2x – 84/x = 0
2x² – 84 = 0
2x² = 84
x² = 84/2
x² = 42
x ≈ 6.48

Question 72.
DIG DEEPER!
Consider the rectangular prism shown. Using only whole number dimensions, how many different prisms are possible? Explain.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 63

Answer:
The volume of the rectangular prism is lbh
v = 40 cubic inches
Let the length be 5 inches
breadth be 2 inches
height be 4 inches
V = 5 × 2 × 4
V = 40 cu. inches

Lesson 1.4 Greatest Common Factor

A Venn diagram uses circles to describe relationships between two or more sets. The Venn diagram shows the factors of 12 and 15. Numbers that are factors of both 12 and 15 are represented by the overlap of the two circles.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 64

Answer: 1 and 3 are overlapped between the two circles.
So, 1 and 3 are the greatest common factors of 12 and 15.
The factors of 12 are 1, 2, 3, 4, 6, 12
The factors of 15 are 1, 3, 5, 15.

EXPLORATION 1

Identifying Common Factors

Work with a partner. In parts (a) – (d), create a Venn diagram that represents the factors of each number and identify any common factors.
a. 36 and 48
b. 16 and 56
c. 30 and 75
d. 54 and 90
e. Look at the Venn diagrams in parts (a)–(d). Explain how to identify the greatest common factor of each pair of numbers. Then circle it in each diagram.

Answer:
a. Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_3

b. Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_4

c. Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_5

d. Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_6

e. 36 and 48 have the greatest common factors.

EXPLORATION 2

Using Prime Factors
Work with a partner
a. Each Venn diagram represents the prime factorizations of two numbers. Identify each pair of numbers. Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 65

Answer:
i. Red 2 × 3 × 3 = 18
Green 3 × 3 × 3 = 27
GCF = 9
ii. Yellow – 2 × 2 × 3 × 3 × 5 = 180
Purple – 5 × 11 = 55
GCF = 5
b. Create a Venn diagram that represents the prime factorizations of 36 and 48.

Answer:
Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_3
c. Repeat part(b) for the remaining number pairs in Exploration 1.

Answer:
Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_4

d. STRUCTURE
Make a conjecture about the relationship between the greatest common factors you found in Exploration 1 and the numbers in the overlaps of the Venn diagrams you just created.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 66

Answer:
a.Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_3
The GCF between the two numbers 36 and 48 are 1,2,3,4,6,12

b.Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_4
The GCF between the two numbers 16 and 56 are 1,2,4,8
c. Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_5
The GCF between the two numbers 30 and 75 is 15.
d. Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_6

1.4 Lesson

Try It

Find the GCF of the numbers using lists of factors.

Question 1.
8, 36

Answer:
The factors of 8 are: 1, 2, 4, 8
The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36
Then the greatest common factor is 4.

Question 2.
18, 72

Answer:
The factors of 18 are: 1, 2, 3, 6, 9, 18
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Then the greatest common factor is 18.

Question 3.
14, 28, 49

Answer:
The factors of 14 are: 1, 2, 7, 14
The factors of 28 are: 1, 2, 4, 7, 14, 28
The factors of 49 are: 1, 7, 49
Then the greatest common factor is 7.

Another way to find the GCF of two or more numbers is by using prime factors. The GCF is the product of the common prime factors of the numbers.

Try It
Find the GCF of the numbers using prime factorizations.

Question 4.
20,45

Answer:
Find the prime factorization of 20
20 = 2 × 2 × 5
Find the prime factorization of 45
45 = 3 × 3 × 5
To find the GCF, multiply all the prime factors common to both numbers:
Therefore, GCF = 5

Question 5.
32,90

Answer:
Find the prime factorization of 32
32 = 2 × 2 × 2 × 2 × 2
Find the prime factorization of 90
90 = 2 × 3 × 3 × 5
To find the GCF, multiply all the prime factors common to both numbers:
Therefore, GCF = 2

Question 6.
45,75,120

Answer:
45= 1,3,5,9,15,45
75= 1,3,5,15,25,75
120=1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120
GCF is 15

Try It

Question 7.
Write a pair of numbers whose greatest common factor is 10.

Answer:
Let’s first find the greatest common factor (GCF) of two whole numbers. The GCF of two numbers is the greatest number that is a factor of both of the numbers. Take the numbers 50 and 30.
50 = 10 × 5
30 = 10 × 3
Their greatest common factor is 10. since 10 is the greatest factor that both numbers have in common.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

FINDING THE GCF
Find the GCF of the numbers.

Question 8.
16, 40

Answer:
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40
Then the greatest common factor is 8.

Question 9.
35, 63

Answer:
The factors of 35 are: 1, 5, 7, 35
The factors of 63 are: 1, 3, 7, 9, 21, 63
Then the greatest common factor is 7.

Question 10.
18, 72, 144

Answer:
The factors of 18 are: 1, 2, 3, 6, 9, 18
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
The factors of 144 are: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
Then the greatest common factor is 18.

Question 11.
MULTIPLE CHOICE
Which number is not a factor of 10? Explain.
A. 1
B. 2
C. 4
D. 5

Answer: 4

Explanation:
Factors of 10 are: 1, 2, 5, 10
Thus the correct answer is option C.

Question 12.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 66.1

Answer:
The Greatest Common Factor of 24 and 32 is 8
The Greatest Common Divisor of 24 and 32 is 4
The Greatest Common Prime Factor of 24 and 32 is 8
The product of common prime factors of 24 and 32 is 8.
The Greatest Common Divisor of 24 and 32 are different from others.

Self – Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 13.
You use 30 sandwiches and 42 granola bars to make identical picnic baskets. You make the greatest number of picnic baskets with no food left over. How many sandwiches and how many granola bars are in each basket?
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 67

Answer: 5 sandwiches and 7 granola in each basket

Explanation:
Given
Represent Sandwiches with S and Granola with G
S = 30
G = 42
To do this, we simply need to determine the ratio of S to G
S:G = 30:42
S:G = 5:7

Question 14.
You fill bags with cookies to give to your friends. You bake 45 chocolate chip cookies, 30 peanut butter cookies, and 15 oatmeal cookies. You want identical groups of cookies in each bag with no cookies left over. What is the greatest number of bags you can make?

Answer: 15 bags

Explanation:
Given,
You fill bags with cookies to give to your friends. You bake 45 chocolate chip cookies, 30 peanut butter cookies, and 15 oatmeal cookies.
You want identical groups of cookies in each bag with no cookies leftover.
45 chocolate chip cookies
30 peanut butter cookies
15 oatmeal cookies
So, the GCF is 15.

Greatest Common Factor Practice 1.4

Review & Refresh

List the factor pairs of the number.

Question 1.
20

Answer: The factor pairs of 20 are (1, 20) (4,5) (2,10)

Explanation:
1 × 20 = 20
4 × 5 = 20
2 × 10 = 20

Question 2.
16

Answer: The factor pairs of 16 are (1,16), (2, 8) (4,4)

Explanation:
1 × 16 = 16
2 × 8 = 16
4 × 4 = 16

Question 3.
56

Answer: The factor pairs of 56 are (1,56) (7,8) (28,2) (14,4)

Explanation:
1 × 56 = 56
7 × 8 = 56
28 × 2 = 56
14 × 4 = 56

Question 4.
87

Answer: The factor pairs of 87 are (1,87) (3,29)

Explanation:
1 × 87 = 87
3 × 29 = 87

Tell whether the statement is always, sometimes, or never true.

Question 5.
A rectangle is a rhombus.

Answer: sometimes

No, because all four sides of a rectangle don’t have to be equal. However, the sets of rectangles and rhombuses do intersect, and their intersection is the set of square all squares are both a rectangle and a rhombus.

Question 6.
A rhombus is a square.

Answer: true

A square is a special case of a rhombus because it has four equal-length sides and goes above and beyond that to also have four right angles. Every square you see will be a rhombus, but not every rhombus you meet will be a square.

Question 7.
A square is a rectangle.

Answer: not always

A square also fits the definition of a rectangle.

Question 8.
A trapezoid is a parallelogram.

Answer: never true

A trapezoid has one pair of parallel sides and a parallelogram has two pairs of parallel sides. So a parallelogram is also a trapezoid.

Concepts, Skills, & Problem Solving

USING A VENN DIAGRAM
Use a Venn diagram to find the greatest common factor of the numbers. (See Exploration 1, p. 21.)

Question 9.
12,30

Answer: 6

Big-ideas-math-answers-grade-6-chapter-1-img-3

Question 10.
32,54

Answer: 2
Big-ideas-math-grade-6-chapter-1-answer-key-img-4

Question 11.
24,108

Answer: 12

BIM 6th Grade Chapter 1 Answer Key Numerical Expressions and Factors img_5

FINDING THE GCF
Find the GCF of the numbers using lists of factors.

Question 12.
6, 15

Answer: GCF is 3

Explanation:
The factors of 6 are: 1,2,3,6
The factors of 15 are: 1,3,5,15
The common Factors in 6 and 15 is 3.
Thus the greatest common factor is 3.

Question 13.
14, 84

Answer: GCF is 14

Explanation:
The factors of 14 are: 1,2,7,14
The factors of 84 are: 1,2,3,4,6,7,12,14,21,28,42 84
The greatest common factor is 14.

Question 14.
45, 108

Answer: GCF is 9

Explanation:
The factors of 45 are: 1,3,5,9,15,45
The factors of 108 are: 1,2,3,4,6,9,12,18,27,36,54,108
The greatest common factor is 9.

Question 15.
39, 65

Answer: GCF is 13

Explanation:
The factors of 39 are: 1,3,13,39
The factors of 65 are: 1,5,13,65
Thus the greatest common factor is 13.

Question 16.
51, 85

Answer: GCF is 17

Explanation:
The factors of 51 are: 1,3,17,51
The factors of 1,5,17,85
Thus the greatest common factor is 17

Question 17.
40, 63

Answer: GCF is 1

Explanation:
The factors of 40 are: 1,2,4,5,8,10,20,40
The factors of 63 are: 1,3,7,9,21,63
Thus the greatest common factor is 1.

Question 18.
12, 48

Answer: GCF is 12

Explanation:
The factors of 12 are: 1,2,3,4,6,12
The factors of 48 are: 1,2,3,4,6,8,12,16,24,48
Thus the greatest common factor is 12.

Question 19.
24, 52

Answer: GCF is 4

Explanation:
The factors of 24 are: 1,2,3,4,6,8,12,24
The factors of 1,2,4,13,36,52
Thus the greatest common factor is 4.

Question 20.
30, 58

Answer: GCF is 2

Explanation:
The factors of 30 are: 1,2,3,5,6,10,15,30
The factors of 58 are: 1,2,29,58
Thus the greatest common factor is 2.

FINDING THE GCF
Find the GCF of the numbers using prime factorizations.

Question 21.
45, 60

Answer:
The prime factorization of 45 is 3 x 3 x 5
The prime factorization 60 is 2 x 2 x 3 x 5
GCF of 45, 60 is 3 × 5 = 15

Question 22.
27, 63

Answer: 9
The prime factorization of 27 is 3 x 3 x 3
The prime factorization of 63 is 3 x 3 x 7
Thus GCF of 27, 63 is 9

Question 23.
36, 81

Answer: 9
The prime factorization of 36 is 2 x 2 x 3 x 3
The prime factorization of 81 is 3 x 3 x 3 x 3
Thus the GCF of 36, 81 is 9.

Question 24.
72, 84

Answer: 12
The prime factorization of 72 is 2 x 2 x 2 x 3 x 3
The prime factorization of 84 is 2 x 2 x 3 x 7
Thus the GCF of 72, 84 is 12.

Question 25.
61, 73

Answer: 1
The prime factorization of 61 is 1 × 61
The prime factorization 73 is 1 × 73
Thus the GCF of 61, 73 is 1

Question 26.
38, 95

Answer: 19
The prime factorization of 38 is 2 x 19
The prime factorization of 95 is 5 x 19
Thus the GCF of 38, 95 is 19

Question 27.
60, 75

Answer: 15
The prime factorization of 60 is 2 x 2 x 3 x 5
The prime factorization of 75 is 3 x 5 x 5
Thus the GCF of 60, 75 is 15.

Question 28.
42, 60

Answer: 6
The prime factorization 42 is 2 × 3 × 7
The prime factorization 60 is 2 × 2 × 3 × 5
Thus the GCF of 42, 60 is 6

Question 29.
42, 63

Answer: 21
The prime factorization of 42 is 2 × 3 × 7
The prime factorization of 63 is 3 × 3 ×7 
Thus the GCF of 42, 63 is 21

Question 30.
24, 96

Answer: 24
The prime factorization of 24 is 2 × 2 × 2 × 3
The prime factorization of 96 is 2 x 2 x 2 x 2 x 2 x 3
Thus the GCF of 24, 96 is 24.

Question 31.
189, 200

Answer: 24
The prime factorization of 189 is 3 x 3 x 3 x 7
The prime factorization of 200 is 2 x 2 x 2 x 5 x 5
Thus the GCF of 189, 200 is 24.

Question 32.
90, 108

Answer: 18
The prime factorization of 90 is 2 x 3 x 3 x 5
The prime factorization of 108 is 2 x 2 x 3 x 3 x 3
Thus the GCF of 90, 108 is 18.

OPEN-ENDED
Write a pair of numbers with the indicated GCF.

Question 33.
5

Answer: 10, 15
The factors of 10 are: 1,2,5,10
The factors of 15 are: 1,3,5,15
Thus 10, 15 are pairs of numbers with the indicated GCF 5.

Question 34.
12

Answer: 72, 84
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
The factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Then the greatest common factor is 12.

Question 35.
37

Answer: 37,74
The factors of 37 are: 1,37
The factors of 74 are: 1,2,37,74
Thus 37,74 are pairs of numbers with the indicated GCF 37.

Question 36.
MODELING REAL LIFE
A teacher is making identical activity packets using 92 crayons and 23 sheets of paper. What is the greatest number of packets the teacher can make with no items left over?

Answer: 23 packets

Explanation:
Factor both numbers:
1. 92=2·46=2·2·23;
2. 23 is a prime number.
Then the greatest common factor GCF (92, 23)=23.
This means that teacher can make 23 packets, each of them will contain 4 crayons and 1 sheet.

Question 37.
MODELING REAL LIFE
You are making balloon arrangements for a birthday party. There are 16 white balloons and 24 red balloons. Each arrangement must be identical. What is the greatest number of arrangements you can make using every balloon?
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 67.1

Answer:
This is a GCF problem. To find the GCF, list the factors.
24
1×24, 2×12, 4×6, 8×3 …
16
1×16, 2×8, 4×4, …
The greatest common factor is 8 because it is the greatest factor of both numbers.
so 8 arrangements, 2 white balloons and 3 red balloons each.
Because- 8 is the number of groups, and 8×3 =24 so 3 reds, and 2 whites because 8×2=16

YOU BE THE TEACHER
Your friend finds the GCF of the two numbers. Is your friend correct? Explain your reasoning.

Question 38.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 68

Answer:
No, your friend is incorrect.
42 = 2 × 3 × 7
154 = 2 × 7 × 11
Thus the GCF is 14.

Question 39.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 69

Answer: Yes your friend is correct.
Thus the GCF of 36 and 60 is 12.

FINDING THE GCF
Find the GCF of the numbers.

Question 40.
35, 56, 63

Answer: GCF is 7

Explanation:
The factors of 35 are: 1,5,7,35
The factors of 56 are: 1,2,4,7,8,14,28,56
The factors of 63 are: 1,3,7,9,21,63
Thus the greatest common factor is 7.

Question 41.
30, 60, 78

Answer: GCF is 6

Explanation:
The factors of 30 are: 1,2,3,5,6,10,15,30
The factors of 60 are: 1,2,3,4,5,6,10,12,15,20,30,60
The factors of 78 are: 1,2,3,6,13,26,39,78
Thus the greatest common factor is 6.

Question 42.
42, 70, 84

Answer: GCF is 14

The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42
The factors of 70 are: 1, 2, 5, 7, 10, 14, 35, 70
The factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Then the greatest common factor is 14.

Question 43.
40, 55, 72

Answer: GCF is 1

The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40
The factors of 55 are: 1, 5, 11, 55
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Then the greatest common factor is 1.

Question 44.
18, 54, 90

Answer: GCF is 18

The factors of 18 are: 1, 2, 3, 6, 9, 18
The factors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54
The factors of 90 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
Then the greatest common factor is 18.

Question 45.
16, 48, 88

Answer: GCF is 8

The factors of 16 are: 1, 2, 4, 8, 16
The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
The factors of 88 are: 1, 2, 4, 8, 11, 22, 44, 88
Then the greatest common factor is 8.

Question 46.
52, 78, 104

Answer: GCF is 26

The factors of 52 are: 1, 2, 4, 13, 26, 52
The factors of 78 are: 1, 2, 3, 6, 13, 26, 39, 78
The factors of 104 are: 1, 2, 4, 8, 13, 26, 52, 104
Then the greatest common factor is 26.

Question 47.
96, 120, 156

Answer: GCF is 12

The factors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
The factors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
The factors of 156 are: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156
Then the greatest common factor is 12.

Question 48.
280, 300, 380

Answer: GCF is 20

The factors of 280 are: 1, 2, 4, 5, 7, 8, 10, 14, 20, 28, 35, 40, 56, 70, 140, 280
The factors of 300 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300
The factors of 380 are: 1, 2, 4, 5, 10, 19, 20, 38, 76, 95, 190, 380
Then the greatest common factor is 20.

Question 49.
OPEN-ENDED
Write three numbers that have a GCF of 16. What method did you use to find your answer?

Answer: 16, 32, 48

The factors of 16 are: 2 × 2 × 2 × 2
The factors of 32 are : 2 × 2 × 2 × 2 × 2
The factors of 48 are: 3 × 2 × 2 × 2 × 2

CRITICAL THINKING
Tell whether the statement is always, sometimes, or never true. Explain your reasoning.

Question 50.
The GCF of two even numbers is 2.

Answer: Always

Explanation:
Example:
The factors of 14 are: 1, 2, 7, 14
The factors of 16 are: 1, 2, 4, 8, 16
Then the greatest common factor is 2.

Question 51.
The GCF of two prime numbers is 1.

Answer: Always

Explanation:
Example:
The factors of 3 are: 1, 3
The factors of 5 are: 1, 5
Then the greatest common factor is 1.

Question 52.
When one number is a multiple of another, the GCF of the numbers is the greater of the numbers.

Answer:
When one number is a multiple of another, the GCF of the numbers is the greater of the numbers. This is never true since the GCF is a factor of both numbers. So the GCF is the smaller of the two numbers.

Question 53.
PROBLEM SOLVING
A science museum makes gift bags for students using 168 magnets, 48 robot figurines, and 24 packs of freeze-dried ice cream. What is the greatest number of gift bags that can be made using all of the items? How many of each item are in each gift bag?
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 70

Answer:
The greatest common factor of 24, 48, and 168 is 24, so 24 gift bags can be made. Each will have 1/24 of the number of gift items of each type that are available.
In each bag are
1/24 × 168 magnets = 7 magnets
1/24 × 48 robot figurines = 2 robot figurines
1/24 × 24 packs of ice cream = 1 pack of ice cream

Question 54.
VENN DIAGRAM
Consider the numbers 252, 270, and 300.
a. Create a Venn diagram using the prime factors of the numbers.

Answer:
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors img_6
b. Use the Venn diagram to find the GCF of 252, 270, and 300.

Answer:
The factors of 252 are: 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252
The factors of 270 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270
The factors of 300 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300
Then the greatest common factor is 6.
c. What is the GCF of 252 and 270? 252 and 300? 270 and 300? Explain how you found your answers.

Answer:
The factors of 252 are: 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252
The factors of 270 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270
The GCF of 252 and 270 is 18.
252 and 300:
The factors of 252 are: 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252
The factors of 300 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300
Then the greatest common factor is 12.
270 and 300:
The factors of 270 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270
The factors of 300 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300
Then the greatest common factor is 30.

Question 55.
REASONING
You are making fruit baskets using 54 apples, 36 oranges, and 73 bananas.
a. Explain why you cannot make identical fruit baskets without leftover fruit.

Answer: 73 is a prime number. It can only be divided by 1 and by itself.
b. What is the greatest number of identical fruit baskets you can make with the least amount of fruit left over? Explain how you found your answer.

Answer:
The GCF of the three numbers:
54 36 73
1×54 1×36 1×73
2×27 2×18
3×18 3×12
6×9 4×9
6×6
GCF of 54, 36, and 73 is 1
GCF of 54 and 36 is 18
If we divide 54 apples into 18 baskets, we have 3 apples in each basket
If we divide 36 oranges into 18 baskets, we have 2 oranges in each basket
If we divide 73 bananas into 18 baskets, we have 4 bananas in each basket + one banana left over.
So the greatest number of identical fruit baskets we can make with the least amount of fruit left over is 18 baskets

Question 56.
DIG DEEPER!
Two rectangular, adjacent rooms share a wall. One-foot-by-one-foot tiles cover the floor of each room. Describe how the greatest possible length of the adjoining wall is related to the total number of tiles in each room. Draw a diagram that represents one possibility.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 71

Answer:
Consider two adjacent rectangular rooms having Length=L, and, Breadth = B
Now Suppose the wall which is in between two rooms has a height or length =H.
The breadth of wall = B [ if the wall doesn’t exceed the breadth of the room]
Considering two rooms to be identical,
Area of each room= L × B square unit
Area of each tile = 1×1=1 square unit
Number of tiles required= L B ÷ 1= LB tiles( product of length and breadth of the room is the number of tiles required)
Suppose if,LB= N
B= N/LArea of the wall(W) = B×H= B H square unit
B =W/H
Equating (1) and (2)
N/L = W/ H
H = WL/N
H = WL/LB
H = W/B
H = Area of wall/Breadth of room or wall

Lesson 1.5 Least Common Multiple

EXPLORATION 1
Identifying Common Multiples

Work with a partner. In parts (a)–(d), create a Venn diagram that represents the first several multiples of each number and identify any common multiples.
a. 8 and 12
b. 4 and 14
c. 10 and 15
d. 20 and 35
e. Look at the Venn diagrams in parts (a)–(d). Explain how to identify the least common multiple of each pair of numbers. Then circle it in each diagram.

Answer:
a. BIM Grade 6 Chapter 1 Answer Key img_7
b. Big Ideas math 6th grade answers chapter 1 img_8
c.Big Ideas math 6th grade answers chapter 1 img_9

d. Big Ideas math 6th grade answers chapter 1 img_10

EXPLORATION 2
Using Prime Factors
Work with a partner.
a. Create a Venn diagram that represents the prime factorizations of 8 and 12.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 72

Answer:
BIM Grade 6 Chapter 1 Answer Key img_7
b. Repeat part (a) for the remaining number pairs in Exploration 1.

Answer:
Big Ideas math 6th grade answers chapter 1 img_8
c. STRUCTURE
Make a conjecture about the relationship between the least common multiples you found in Exploration 1 and the numbers in the Venn diagrams you just created.

Answer:
The numbers which are overlapped are the least common multiples of the numbers.
d. The Venn diagram shows the prime factors of two numbers.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 73
Use the diagram to complete the following tasks.

  • Identify the two numbers.
  • Find the greatest common factor.
  • Find the least common multiple

Answer:

  • 120, 180
  • The factors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120The factors of 180 are: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180Then the greatest common factor is 60.
  • Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
    Multiples of 120:
    120, 240, 360, 480, 600
    Multiples of 180:
    180, 360, 540, 720
    Therefore,
    LCM(120, 180) = 360

1.5 Lesson

Try It

Find the LCM of the numbers using lists of multiples.

Question 1.
3, 8

Answer: 24
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 3:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Multiples of 8:
8, 16, 24, 32, 40
Therefore,
LCM(3, 8) = 24

Question 2.
9, 12

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 9:
9, 18, 27, 36, 45, 54
Multiples of 12:
12, 24, 36, 48, 60
Therefore,
LCM(9, 12) = 36

Question 3.
6, 10

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 6:
6, 12, 18, 24, 30, 36, 42
Multiples of 10:
10, 20, 30, 40, 50
Therefore,
LCM(6, 10) = 30

Try It
Find the LCM of the numbers using prime factorizations.

Question 4.
14, 18

Answer:
List all prime factors for each number.
Prime Factorization of 14 is:
2 x 7
Prime Factorization of 18 is:
2 x 3 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 3, 3, 7
Multiply these factors together to find the LCM.
LCM = 2 x 3 x 3 x 7 = 126
LCM(14, 18) = 126

Question 5.
28, 36

Answer:
List all prime factors for each number.
Prime Factorization of 28 is:
2 x 2 x 7
Prime Factorization of 36 is:
2 x 2 x 3 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 3, 3, 7
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 3 x 3 x 7 = 252
LCM = 252
Therefore,
LCM(28, 36) = 252

Question 6.
24, 90

Answer:
List all prime factors for each number.
Prime Factorization of 24 is:
2 x 2 x 2 x 3
Prime Factorization of 90 is:
2 x 3 x 3 x 5
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 2, 3, 3, 5
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 2 x 3 x 3 x 5 = 360
LCM = 360
Therefore,
LCM(24, 90) = 360

Try It

Find the LCM of the numbers.

Question 7.
2, 5, 8

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 2:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44
Multiples of 5:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50
Multiples of 8:
8, 16, 24, 32, 40, 48, 56
Therefore,
LCM(2, 5, 8) = 40

Question 8.
6, 10, 12

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 6:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72
Multiples of 10:
10, 20, 30, 40, 50, 60, 70, 80
Multiples of 12:
12, 24, 36, 48, 60, 72, 84
Therefore,
LCM(6, 10, 12) = 60

Question 9.
Write three numbers that have a least common multiple of 100.

Answer: 1, 10, 100

Explanation:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 1:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102
Multiples of 10:
10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120
Multiples of 100:
100, 200, 300
Therefore,
LCM(1, 10, 100) = 100

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

FINDING THE LCM
Find the LCM of the numbers.

Question 10.
6, 9

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 6:
6, 12, 18, 24, 30
Multiples of 9:
9, 18, 27, 36
Therefore,
LCM(6, 9) = 18

Question 11.
30, 40

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 30:
30, 60, 90, 120, 150, 180
Multiples of 40:
40, 80, 120, 160, 200
Therefore,
LCM(30, 40) = 120

Question 12.
5, 11

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 5:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65
Multiples of 11:
11, 22, 33, 44, 55, 66, 77
Therefore,
LCM(5, 11) = 55

Question 13.
Reasoning
Write two numbers such that 18 and 30 are multiples of the numbers. Justify your answer.

Answer: 3, 6

Explanation:
Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Multiples of 6: 6,12,18,24,30,36,42,48
Thus 18 and 30 are the multiples of 3 and 6.

Question 14.
REASONING
You need to find the LCM of 13 and 14. Would you rather list their multiples or use their prime factorizations? Explain.

Answer:
List all prime factors for each number.
Prime Factorization of 13 shows:
13 is prime = 13
Prime Factorization of 14 is:
2 x 7
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 7, 13
Multiply these factors together to find the LCM.
LCM = 2 x 7 x 13 = 182
In exponential form:
LCM = 2 x 7 x 13 = 182
LCM = 182
Therefore,
LCM(13, 14) = 182

Question 15.
CHOOSE TOOLS
A student writes the prime factorizations of 8 and 12 in a table as shown. She claims she can use the table to find the greatest common factor and the least common multiple of 8 and 12. How is this possible?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 74
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 75

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-74
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 2, 3
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 2 x 3 = 24
The least common multiple of 8 and 12 is 24.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.
A geyser erupts every fourth day. Another geyser erupts every sixth day. Today both geysers erupted. In how many days will both geysers erupt on the same day again?

Answer:
Given,
A geyser erupts every fourth day. Another geyser erupts every sixth day. Today both geysers erupted.
The geyser erupts on the same day after 12 days.

Question 18.
A water park has two large buckets that slowly fill with water. One bucket dumps water every 12 minutes. The other bucket dumps water every 10 minutes. Five minutes ago, both buckets dumped water. When will both buckets dump water at the same time again?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 76

Answer:
Given,
A water park has two large buckets that slowly fill with water.
One bucket dumps water every 12 minutes.
The other bucket dumps water every 10 minutes.
Five minutes ago, both buckets dumped water.
Both buckets will dump again at the same time in 60 minutes (1 hour.)

Question 19.
DIG DEEPER!
You purchase disposable plates, cups, and forks for a cookout. Plates are sold in packages of 24, cups in packages of 32, and forks in packages of 48. What are the least numbers of packages you should buy in order to have the same number of plates, cups, and forks?

Answer:
Given,
You purchase disposable plates, cups, and forks for a cookout. Plates are sold in packages of 24, cups in packages of 32, and forks in packages of 48.
We solve this question using the Lowest Common Multiple (LCM) method/
Step 1
We list multiples of each number until the first common multiple is found. This is referred to as the lowest common multiple.
Plates are sold in packages of 24
Cups in packages of 32
Forks in packages of 48
Multiples of 24:
24, 48, 72, 96, 120, 144
Multiples of 32:
32, 64, 96, 128, 160
Multiples of 48:
48, 96, 144, 192
Therefore,
LCM(24, 32, 48) = 96
Hence, the least numbers of packages you should buy in order to have the same number of plates, cups, and forks is 96 packages

Least Common Multiple Practice 1.5

Review & Refresh

Find the GCF of the numbers.

Question 1.
18, 42

Answer:
The factors of 18 are: 1, 2, 3, 6, 9, 18
The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42
Then the greatest common factor is 6.

Question 2.
72, 96

Answer:
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
The factors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
Then the greatest common factor is 24.

Question 3.
38, 76, 114

Answer:
The factors of 38 are: 1, 2, 19, 38
The factors of 76 are: 1, 2, 4, 19, 38, 76
The factors of 114 are: 1, 2, 3, 6, 19, 38, 57, 114
Then the greatest common factor is 38.

Divide.

Question 4.
900 ÷ 6

Answer: 150

Explanation:
Divide the two numbers 900 and 6.
900/6 = 150
It means 6 divides 900 150 times.
Thus the quotient is 150.

Question 5.
1944 ÷ 9

Answer: 216

Explanation:
Divide the two numbers 1944 and 9
1944/9 = 216
It means that 9 divides 1944 216 times.
Thus the quotient is 216

Question 6.
672 ÷ 12

Answer: 56

Explanation:
Divide the two numbers 672 and 12.
672/12 = 56
It means 12 divides 672 56 times.
Thus the quotient is 56.

Write an ordered pair that corresponds to the point.

Question 7.
Point A

Answer: (2,4)
By seeing the below graph we can find the ordered pairs.
2 lies on the x-axis and 4 lies on the y-axis.
So, the ordered pairs to the point A is (2,4)

Question 8.
Point B

Answer: (3,1)
By seeing the below graph we can find the ordered pairs.
3 lies on the x-axis and 1 lies on the y-axis.
So, the ordered pairs to the point A is (3,1)

Question 9.
Point C

Answer: (4,7)
By seeing the below graph we can find the ordered pairs.
4 lies on the x-axis and 7 lies on the y-axis.
So, the ordered pairs to the point A is (4,7)

Question 10.
Point D

Answer: (9,6)
By seeing the below graph we can find the ordered pairs.
9 lies on the x-axis and 6 lies on the y-axis.
So, the ordered pairs to the point A is (9,6)

Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 77

Concepts, Skills, & Problem Solving

USING A VENN DIAGRAM
Use a Venn diagram to find the least common multiple of the numbers. (See Exploration 1, p. 27.)

Question 11.
3, 7

Answer: 21

Big Ideas Math Grade 6 Chapter 1 img_10

Question 12.
6, 8

Answer: 24

Big Ideas Math 6th Grade Answers Chapter 1 img_11

Question 13.
4, 5

Answer:
Big Ideas Math 6th Grade Answers Chapter 1 img_12

FINDING THE LCM
Find the LCM of the numbers using lists of multiples.

Question 14.
1, 5

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 1:
1, 2, 3, 4, 5, 6, 7
Multiples of 5:
5, 10, 15
Therefore,
LCM(1, 5) = 5

Question 15.
2, 6

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 2:
2, 4, 6, 8, 10
Multiples of 6:
6, 12, 18
Therefore,
LCM(2, 6) = 6

Question 16.
2, 3

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 2:
2, 4, 6, 8, 10
Multiples of 3:
3, 6, 9, 12
Therefore,
LCM(2, 3) = 6

Question 17.
2, 9

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 2:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
Multiples of 9:
9, 18, 27, 36
Therefore,
LCM(2, 9) = 18

Question 18.
3, 4

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 3:
3, 6, 9, 12, 15, 18
Multiples of 4:
4, 8, 12, 16, 20
Therefore,
LCM(3, 4) = 12

Question 19.
8, 9

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88
Multiples of 9:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90
Therefore,
LCM(8, 9) = 72

Question 20.
5, 8

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 5:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50
Multiples of 8:
8, 16, 24, 32, 40, 48, 56
Therefore,
LCM(5, 8) = 40

Question 21.
11, 12

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 11:
11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143, 154
Multiples of 12:
12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156
Therefore,
LCM(11, 12) = 132

Question 22.
12, 18

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 12:
12, 24, 36, 48, 60
Multiples of 18:
18, 36, 54, 72
Therefore,
LCM(12, 18) = 36

FINDING THE LCM
Find the LCM of the numbers using prime factorizations.

Question 23.
7, 12

Answer:
List all prime factors for each number.
Prime Factorization of 7 shows:
7 is prime
Prime Factorization of 12 is:
2 x 2 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 3, 7
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 3 x 7 = 84
LCM = 84
Therefore,
LCM(7, 12) = 84

Question 24.
5, 9 4

Answer:
List all prime factors for each number.
Prime Factorization of 4 is:
2 x 2
Prime Factorization of 5 shows:
5 is prime
Prime Factorization of 9 is:
3 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 3, 3, 5
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 3 x 3 x 5 = 180
LCM = 180
Therefore,
LCM(4, 5, 9) = 180

Question 25.
4, 11

Answer:
List all prime factors for each number.
Prime Factorization of 4 is:
2 x 2
Prime Factorization of 11 shows:
11 is prime
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 11
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 11 = 44
LCM = 44
Therefore,
LCM(4, 11) = 44

Question 26.
9, 10

Answer:
List all prime factors for each number.
Prime Factorization of 9 is:
3 x 3
Prime Factorization of 10 is:
2 x 5
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 3, 3, 5
Multiply these factors together to find the LCM.
LCM = 2 x 3 x 3 x 5 = 90
LCM = 90
Therefore,
LCM(9, 10) = 90

Question 27.
12, 27

Answer:
List all prime factors for each number.
Prime Factorization of 12 is:
2 x 2 x 3
Prime Factorization of 27 is:
3 x 3 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 3, 3, 3
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 3 x 3 x 3 = 108
LCM = 108
Therefore,
LCM(12, 27) = 108

Question 28.
18, 45

Answer:
List all prime factors for each number.
Prime Factorization of 18 is:
2 x 3 x 3
Prime Factorization of 45 is:
3 x 3 x 5
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 3, 3, 5
Multiply these factors together to find the LCM.
LCM = 2 x 3 x 3 x 5 = 90
LCM = 90
Therefore,
LCM(18, 45) = 90

Question 29.
22, 23

Answer:
List all prime factors for each number.
Prime Factorization of 22 is:
2 x 11
Prime Factorization of 23 shows:
23 is prime
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 11, 23
Multiply these factors together to find the LCM.
LCM = 2 x 11 x 23 = 506
LCM = 506
Therefore,
LCM(22, 23) = 506

Question 30.
36, 60

Answer:
List all prime factors for each number.
Prime Factorization of 36 is:
2 x 2 x 3 x 3
Prime Factorization of 60 is:
2 x 2 x 3 x 5
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 3, 3, 5
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 3 x 3 x 5 = 180
LCM = 180
Therefore,
LCM(36, 60) = 180

Question 31.
35, 50

Answer:
List all prime factors for each number.
Prime Factorization of 35 is:
5 x 7
Prime Factorization of 50 is:
2 x 5 x 5
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 5, 5, 7
Multiply these factors together to find the LCM.
LCM = 2 x 5 x 5 x 7 = 350
LCM = 350
Therefore,
LCM(35, 50) = 350

Question 32.
YOU BE THE TEACHER
Your friend finds the LCM of 6 and 9. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 78

Answer: No friend is incorrect.
List all prime factors for each number.
Prime Factorization of 6 is:
2 x 3
Prime Factorization of 9 is:
3 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 3, 3
Multiply these factors together to find the LCM.
LCM = 2 x 3 x 3 = 18
LCM = 18
Therefore,
LCM(6, 9) = 18

Question 33.
MODELING REAL LIFE
You have diving lessons every fifth day and swimming lessons every third day. Today you have both lessons. In how many days will you have both lessons on the same day again?

Answer:
After 15 days
diving is on 0 , 5 , 10 , 15 , 20 , … days
swimming is on 0 , 3, 6 , 9, 12 , 15 , 18 , … days
Thus the next will be in 15 days

Question 34.
REASONING
Which model represents an LCM that is different from the other three? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 79

Answer: The number line B is different from the other three number lines.

FINDING THE LCM
Find the LCM of the numbers.

Question 35.
2, 3, 7

Answer:
LCM(2,3) = (2 × 3) / GCF(2,3)
= (2 × 3) / 1
= 6 / 1
= 6
LCM(6,7) = (6 × 7) / GCF(6,7)
= (6 × 7) / 1
= 42 / 1
= 42
Therefore,
LCM(2, 3, 7) = 42

Question 36.
3, 5, 11

Answer:
LCM(3,5) = (3 × 5) / GCF(3,5)
= (3 × 5) / 1
= 15 / 1
= 15
LCM(15,11) = (15 × 11) / GCF(15,11)
= (15 × 11) / 1
= 165 / 1
= 165
Therefore,
LCM(3, 5, 11) = 165

Question 37.
4, 9, 12

Answer:
LCM(4,9) = (4 × 9) / GCF(4,9)
= (4 × 9) / 1
= 36 / 1
= 36
LCM(36,12) = (36 × 12) / GCF(36,12)
= (36 × 12) / 12
= 432 / 12
= 36
Therefore,
LCM(4, 9, 12) = 36

Question 38.
6, 8, 15

Answer:
LCM(6,8) = (6 × 8) / GCF(6,8)
= (6 × 8) / 2
= 48 / 2
= 24
LCM(24,15) = (24 × 15) / GCF(24,15)
= (24 × 15) / 3
= 360 / 3
= 120
Therefore,
LCM(6, 8, 15) = 120

Question 39.
7, 18, 21

Answer:
LCM(7,18) = (7 × 18) / GCF(7,18)
= (7 × 18) / 1
= 126 / 1
= 126
LCM(126,21) = (126 × 21) / GCF(126,21)
= (126 × 21) / 21
= 2646 / 21
= 126
Therefore,
LCM(7, 18, 21) = 126

Question 40.
9, 10, 28

Answer:
LCM(9,10) = (9 × 10) / GCF(9,10)
= (9 × 10) / 1
= 90 / 1
= 90
LCM(90,28) = (90 × 28) / GCF(90,28)
= (90 × 28) / 2
= 2520 / 2
= 1260
Therefore,
LCM(9, 10, 28) = 1260

Question 41.
PROBLEM SOLVING
At Union Station, you notice that three subway lines just arrived at the same time. How long must you wait until all three lines arrive at Union Station at the same time again?
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 80

Answer: 60 minutes

Explanation:
The complete question in the attached figure
Step 1
Find the least common multiple (LCM) of the three numbers
List the prime factors of each number
10 = 2 × 5
12 = 2 × 2 × 3
15 = 3 × 5
Multiply each factor the greatest number of times it occurs in any of the numbers to find out the LCM
The LCM is equal to
4 × 3 × 5 = 60
Thus You must wait 60 minutes for all three lines to arrive at Union Station at the same time again.

Question 42.
DIG DEEPER!
A radio station gives away $15 to every 15th caller, $25 to every 25th caller, and a free concert ticket to every 100th caller. When will the station first give away all three prizes to one caller? When this happens, how much money and how many tickets are given away?

Answer:
Given,
Radio Station gives :
1st prize: $15 to 15th caller
2nd prize: $25 to 25th caller
3rd prize: free concert tickets to 100th caller
So, in order to get all three prizes the caller must be 15th, 25th, and 100th caller at the same time. But to find when the radio station will give first all three prizes we calculate L.C.M. of ( 15, 25, 100 ) that is 300
Hence, the station first gives away all three prizes to the 300th caller.

Question 43.
LOGIC
You and a friend are running on treadmills. You run 0.5 mile every 3 minutes, and your friend runs 2 miles every 14 minutes. You both start and stop running at the same time and run a whole number of miles. What are the least possible numbers of miles you and your friend can run?

Answer:
If you run 0.5 miles every 3 minutes then you run 1 mile every 6 minutes.
If your friend runs 2 miles every 14 minutes then your friend runs 1 mile every 7 minutes.
You will both then both run a whole number of minutes for a time that is a multiple of 6 and 7.
The least common multiple of 6 and 7 is 42 so the least possible time you and your friend could run for and both run a whole number of miles is then 42 minutes.
Since you run 1 mile every 6 minutes, in 42 minutes you will run 42/6=7 miles.
Since your friend runs 1 mile every 7 minutes, in 42 minutes your friend will run 42/7=6 miles.

Question 44.
VENN DIAGRAM
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 81
Refer to the Venn diagram.
a. Copy and complete the Venn diagram.
b. What is the LCM of 16, 24, and 40?
c. What is the LCM of 16 and 40? 24 and 40? 16 and 24? Explain how you found your answers.

Answer:
Big-Ideas-Math-Answers-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-81
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
LCM of 16, 40:
Multiples of 16:
16, 32, 48, 64, 80, 96, 112
Multiples of 40:
40, 80, 120, 160
Therefore,
LCM(16, 40) = 80
LCM of 24, 40:
Multiples of 24:
24, 48, 72, 96, 120, 144, 168
Multiples of 40:
40, 80, 120, 160, 200
Therefore,
LCM(24, 40) = 120
LCM of 16, 24:
Multiples of 16:
16, 32, 48, 64, 80
Multiples of 24:
24, 48, 72, 96
Therefore,
LCM(16, 24) = 48

CRITICAL THINKING
Tell whether the statement is always, sometimes, or never true. Explain your reasoning.

Question 45.
The LCM of two different prime numbers is their product.

Answer: Always true

Example:
3 × 5 = 15

Question 46.
The LCM of a set of numbers is equal to one of the numbers in the set.

Answer:
The LCM of a set of numbers is equal to one of the numbers in the set. Always Sometimes Never true. Question 909193: The LCM of a set of numbers is equal to one of the numbers in the set. This is sometimes true.

Question 47.
The GCF of two different numbers is the LCM of the numbers.

Answer:
Another way to find the LCM of two numbers is to divide their product by their greatest common factor ( GCF ). Example 2: Find the least common multiple of 18 and 20. The common factors are 2 and 3 .

Numerical Expressions and Factors Connecting Concepts

Getting Ready for Chapter Connecting Concepts

Using the Problem-Solving Plan

Question 1.
A sports team gives away shirts at the stadium. There are 60 large shirts, 1.6 times as many small shirts as large shirts, and 1.5 times as many medium shirts as small shirts. The team wants to divide the shirts into identical groups to be distributed throughout the stadium. What is the greatest number of groups that can be formed using every shirt?

Understand the Problem
You know the number of large shirts and two relationships among the numbers of small, medium, and large shirts. You are asked to find the greatest number of identical groups that can be formed using every shirt.
Make a plan
Break the problem into parts. First use multiplication to find the number of each size shirt. Then find the GCF of these numbers.
Solve and Check
Use the plan to solve the problem. Then check your solution.

Answer:
Given,
A sports team gives away shirts at the stadium.
There are 60 large shirts, 1.6 times as many small shirts as large shirts, and 1.5 times as many medium shirts as small shirts. The team wants to divide the shirts into identical groups to be distributed throughout the stadium.
60 × 1.6 = 96
60 × 1.5 = 90
The factors of 90 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
The factors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
Then the greatest common factor is 6.

Question 2.
An escape artist fills the tank shown with water. Find the number of cubic feet of water needed to fill the tank. Then find the number of cubic yards of water that are needed to fill the tank. Justify your answer.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 75.1

Answer:
Given,
An escape artist fills the tank shown with water.
side = 6 ft
We know that
The volume of the cube = s³
V = 6ft × 6ft × 6ft
V = 216 cubic ft.

Performance Task

Setting the Table

At the beginning of this chapter, you watched a STEAM video called “Filling Piñatas.” You are now ready to complete the performance task for this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 82

Answer:
Factors of 50 – 1, 2, 5, 10, 25, and 50
Factors of 12 – 1, 2, 3, 4, 6, 12
Factors of 16 – 1, 2, 4, 8, 16
The factors of number 24 are 1, 2, 3, 4, 6, 8, 12, 24.
The factors of 100 are 1,2,4,5,10,20,25,50 and 100.

Numerical Expressions and Factors Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 83

Graphic Organizers

You can use an Information Frame to organize and remember concepts. Here is an example of an Information Frame for the vocabulary term power.

Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 84

Choose and complete a graphic organizer to help you study the concept.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 85

  1. perfect square
  2. numerical expression
  3. order of operations
  4. prime factorization
  5. greatest common factor (GCF)
  6. least common multiple (LCM)

Answer:
1. A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system. Examples of Numbers that are Perfect Squares. 25 is a perfect square.
2. A numerical expression is a mathematical sentence involving only numbers and one or more operation symbols. Examples of operation symbols are the ones for addition, subtraction, multiplication, and division.
3. Order of operations refers to which operations should be performed in what order, but it’s just convention.
4. “Prime Factorization” is finding which prime numbers multiply together to make the original number.
5. Greatest Common Factor. The highest number that divides exactly into two or more numbers.
6. Least Common Multiple. The smallest positive number that is a multiple of two or more numbers.

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 86

1.1 Powers and Exponents (pp. 3-8)

Write the product as a power.

Question 1.
3 × 3 × 3 × 3 × 3 × 3

Answer: The product of 3 × 3 × 3 × 3 × 3 × 3 is 36

Question 2.
5 × 5 × 5

Answer: The product of 5 × 5 × 5 is 53

Question 3.
17 . 17 . 17 . 17 . 17

Answer: The product of 17 . 17 . 17 . 17 . 17 is 175

Question 4.
33

Answer: 3 × 3 × 3

Question 5.
26

Answer: 2 × 2 × 2 × 2 × 2 × 2

Question 6.
44

Answer: 4 × 4 × 4 × 4

Question 7.
Write a power that has a value greater than 23 and less than 33.

Answer: The power that has a value greater than 23 and less than 33 is 4²

Question 8.
Without evaluating, determine whether 25 or 42 is greater. Explain.

Answer: 25 > 42

Explanation:
The exponent with the highest number will be greater.

Question 9.
The bases on a softball field are square. What is the area of each base?
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 87
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 88

Answer:
Given,
The bases on a softball field are square.
s = 15 inches
We know that,
Area of the square = s × s
A = 15 × 15
A = 225 sq. in
Thus the area of each base is 225 sq. in.

1.2 Order of Operations (pp. 9–14)

Evaluate the expression.

Question 10.
3 × 6 – 12 ÷ 6

Answer: 16

Explanation:
You have to evaluate from left to right.
(3 × 6) – (12 ÷ 6)
18 – 2 = 16

Question 11.
30 ÷ (14 – 22) × 5

Answer: 15

Explanation:
You have to evaluate from left to right.
30 ÷ (14 – 4) × 5
30 ÷ 10 × 5
3 × 5 = 15

Question 12.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 89

Answer: 15

Explanation:
You have to evaluate from left to right.
2.3 + 3.7 = 6
5(6)/2 = 30/2 = 15

Question 13.
Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 90

Answer: 37

Explanation:
You have to evaluate from left to right.
7² + 5 = 49 + 5 = 54
1/2 × 54 = 27
4³ – 27 = 64 – 27 = 37

Question 14.
20 (32 – 4) ÷ 50

Answer: 2

Explanation:
You have to evaluate from left to right.
(32 – 4) = 9 – 4 = 5
20 × 5 ÷ 50
100 ÷ 50 = 2

Question 15.
5 + 3(42 – 2) ÷ 6

Answer: 12

Explanation:
You have to evaluate from left to right.
(42 – 2) = 16 – 2 = 14
5 + 3(14) ÷ 6
5 + 42 ÷ 6
5 + 7 = 12

Question 16.
Use grouping symbols and at least one exponent to write a numerical expression that has a value of 80.

Answer: 6 + (9² – 7) = 80

1.3 Prime Factorization (pp. 15–20)

List the factor pairs of the number.

Question 17.
28

Answer: The factor pairs of the number 28 are 1, 2, 4, 7, 14, 28

Explanation:
28 = 1 × 28
2 × 14
4 × 7
7 × 4
14 × 2
28 × 1

Question 18.
44

Answer: The factor pairs of the number 44 are 1, 2, 4, 11, 44.

Explanation:
44 = 1 × 44
2 × 22
4 × 11
11 × 4
44 × 1

Question 19.
96

Answer: The factor pairs of the number 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

Explanation:
1 and 96 are a factor pair of 96 since 1 x 96= 96
2 and 48 are a factor pair of 96 since 2 x 48= 96
3 and 32 are a factor pair of 96 since 3 x 32= 96
4 and 24 are a factor pair of 96 since 4 x 24= 96
6 and 16 are a factor pair of 96 since 6 x 16= 96
8 and 12 are a factor pair of 96 since 8 x 12= 96
12 and 8 are a factor pair of 96 since 12 x 8= 96
16 and 6 are a factor pair of 96 since 16 x 6= 96
24 and 4 are a factor pair of 96 since 24 x 4= 96
32 and 3 are a factor pair of 96 since 32 x 3= 96
48 and 2 are a factor pair of 96 since 48 x 2= 96
96 and 1 are a factor pair of 96 since 96 x 1= 96

Question 20.
There are 36 graduated cylinders to put away on a shelf after science class. The shelf can fit a maximum of 20 cylinders across and 4 cylinders deep. The teacher wants each row to have the same number of cylinders. List the possible arrangements of the graduated cylinders on the shelf.
Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 91

Answer:
Given,
There are 36 graduated cylinders to put away on a shelf after science class.
The shelf can fit a maximum of 20 cylinders across and 4 cylinders deep.
The teacher wants each row to have the same number of cylinders.
There are three possible arrangements of the graduated cylinders on the shelf.
1. 4 rows of 9 graduated cylinders
2. 3 rows of 12 graduated cylinders.
3. 2 rows of 18 graduated cylinders.

Write the prime factorization of the number.

Question 21.
42

Answer: The prime factorization of the number 42 is 2 × 3 × 7

Explanation:
42 = 2 × 21
= 2 × 3 × 7

Question 22.
50

Answer: The prime factorization of the number  2 × 5 × 5

Explanation:
50 = 2 × 25
= 2 × 5 × 5

Question 23.
66

Answer: The prime factorization of the number 2 × 3 × 11

Explanation:
66 = 2 × 33
= 2 × 3 × 11

1.4 Greatest Common Factor (pp. 21–26)

Find the GCF of the numbers using lists of factors.

Question 24.
27, 45

Answer: GCF is 9

Explanation:
The factors of 27 are: 1, 3, 9, 27
The factors of 45 are: 1, 3, 5, 9, 15, 45
Then the greatest common factor is 9.

Question 25.
30, 48

Answer: GCF is 6

Explanation:
The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30
The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Then the greatest common factor is 6.

Question 26.
28, 48

Answer: GCF is 4

Explanation:
The factors of 28 are: 1, 2, 4, 7, 14, 28
The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Then the greatest common factor is 4.

Find the GCF of the numbers using prime factorizations.

Question 27.
24, 80

Answer:
The prime factorization is the product of the circled primes. So the prime factorization of 24 is 24 = 2 · 2 · 2 · 3 = 2³ . 3
The prime factorization is the product of the circled primes. So the prime factorization of 80 is 80 = 2 x 2 x 2 x 2 x 5 = 2² . 2² . 5

Question 28.
52, 68

Answer:
The prime factorization is the product of the circled primes. So the prime factorization of 52 is 2 x 2 x 13 = 2² . 13
The prime factorization is the product of the circled primes. So the prime factorization of 68 is 2 × 2 × 17 = 2². 17

Question 29.
32, 56

Answer:
The prime factorization is the product of the circled primes. So the prime factorization of 32 is 2 x 2 x 2 x 2 x 2 = 25
The prime factorization is the product of the circled primes. So the prime factorization of 56 is 2 x 2 x 2 x 7 = 2³ . 7

Question 30.
Write a pair of numbers that have a GCF of 20.

Answer: The prime factors of 20 are 2 x 2 x 5. The GCF of 20 is 5.

Question 31.
What is the greatest number of friends you can invite to an arcade using the coupon such that the tokens and slices of pizza are equally split between you and your friends with none left over? How many slices of pizza and tokens will each person receive?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 92

Answer: (n/4)-1

Explanation:
Total slices = n
Total number of people =4
Each people may be eat = n/4 slices
Here Harris eats 1 slice fewer
Then Harris eats (n/4)-1 slices

1.5 Least Common Multiple (pp. 27–32)

Find the LCM of the numbers using lists of multiples.

Question 32.
4, 14

Answer: 28

Explanation:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 4:
4, 8, 12, 16, 20, 24, 28, 32, 36
Multiples of 14:
14, 28, 42, 56
Therefore,
LCM(4, 14) = 28

Question 33.
6, 20

Answer: 60

Explanation:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 6:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60.
Multiples of 20:
20, 40, 60, 80, 100
The LCM of 6, 20 is 60

Question 34.
12, 28

Answer: 84

Explanation:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 12:
12, 24, 36, 48, 60, 72, 84, 96
Multiples of 28:
28, 56, 84, 112, 140, 168, 196
LCM of 12, 28 is 84

Find the LCM of the numbers using prime factorizations.

Question 35.
6, 45

Answer:
Prime Factorisation of 6: 2 × 3
Prime Factorisation of 45: 3 × 3 × 5
LCM is 2 × 3 × 3 × 3 × 5 = 60

Question 36.
10, 12

Answer: 60
Prime factorization of 10: 2 × 5
Prime factorization of 12: 2 × 2 × 3
LCM is 5 × 2 × 2 × 3 = 60

Question 37.
18, 27

Answer:
Prime factorization of 18: 2 × 3 × 3
Prime factorization of 27: 3 × 3 × 3
LCM is 2 × 3 × 3 × 3 = 54

Question 38.
Find the LCM of 8, 12, and 18.

Answer: 72
Prime Factorisation of 8: 2 × 2 × 2
Prime Factorisation of 12: 2 × 2 × 3
Prime factorization of 18: 2 × 3 × 3
LCM = 72

Question 39.
Write a pair of numbers that have an LCM of 84.

Answer: 84 and 12

Explanation:
The LCM of 84 and 12 is 84.
Prime factorization of 12: 2 × 2 × 3
Prime factorization of 84: 2 × 2 × 3 × 7
The Least Common Multiple is 2 × 2 × 3 × 7 = 84

Question 40.
Write three numbers that have an LCM of 45.

Answer: 3, 15, 45

Explanation:
The prime factorization of 15: 3 × 5
The prime factorization of 45: 3 × 3 × 5
The LCM of 3, 15, 45 is 45.

Question 41.
You water your roses every sixth day and your hydrangeas every fifth day. Today you water both plants. In how many days will you water both plants on the same day again?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 93

Answer: 30

Explanation:
Given,
You water your roses every sixth day and your hydrangeas every fifth day. Today you water both plants.
6 × 5 = 30
You water both plants for 30 days on the same day again.

Question 42.
Hamburgers are sold in packages of 20, while buns are sold in packages of 12. What are the least numbers of packages you should buy in order to have the same number of hamburgers and buns?

Answer:
Given,
Hamburgers are sold in packages of 20, while buns are sold in packages of 12.
At least 5 packages of buns and 3 packages of hamburgers.
20×3=60
12×5=60
So that is how you get the answer by seeing if they have any integers in common.

Question 43.
A science museum is giving away a magnetic liquid kit to every 50th guest and a plasma ball to every 35th guest until someone receives both prizes.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 94
a. Which numbered guest will receive both a magnetic liquid kit and a plasma ball?

Answer:
A magnetic liquid kit prize every 50 guests and a plasma ball every 35 guests.
1.Guest 50th
2.Guest 100th
3.Guest 150th
4.Guest 200th
5.Guest 250th
6.Guest 300th
7.Guest 350th
8.Guest 400th and so on, in case no coincidence would happen.

b. How many people will receive a plasma ball?

Answer:
1.Guest 35th
2.Guest 70th
3.Guest 105th
4.Guest 140th
5.Guest 175th
6.Guest 210th
7.Guest 245th
8.Guest 280th
9.Guest 315th
10.Guest 350th.
As you can see, Guest 350th will be the first one to receive both prizes, and including him or her, a total of ten guests will receive the plasma ball until that moment. There wasn’t any coincidence before Guest 350th.

Numerical Expressions and Factors Practice Test

Question 1.
Find the value of 23.

Answer:
23 can be written as 2 × 2 × 2 = 8
Thus the value of 23 is 8.

Question 2.
Evaluate Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 95

Answer:
5 + 4(12 – 2) = 5 + 4(10) = 5 + 40
(5 + 40)/3² = 45/9 = 5
Thus the value of Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 95is 5.

Question 3.
Write 264.264.264 as a power

Answer: 264.264.264 can be written as 264³

Question 4.
List the factor pairs of 66.

Answer: The factor pairs of 66 are (1,66) (2, 33) (6, 11)
66 = 1 × 66
66 = 2 × 33
66 = 6 × 11

Question 5.
Write the prime factorization of 56.

Answer:
56 = 2 × 28
= 2 × 2 × 14
= 2 × 2 × 2 × 7
Thus the prime factorization of 56 is 2 × 2 × 2 × 7

Find the GCF of the numbers.

Question 6.
24, 54

Answer: GCF is 6

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
The factors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54
Then the greatest common factor is 6.

Question 7.
16, 32, 72

Answer: GCF is 8

The factors of 16 are: 1, 2, 4, 8, 16
The factors of 32 are: 1, 2, 4, 8, 16, 32
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Then the greatest common factor is 8.

Question 8.
52, 65

Answer: GCF is 13

The factors of 52 are: 1, 2, 4, 13, 26, 52
The factors of 65 are: 1, 5, 13, 65
Then the greatest common factor is 13.

Find the LCM of the numbers.

Question 9.
9, 24

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 9:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90
Multiples of 24:
24, 48, 72, 96, 120
Therefore,
LCM(9, 24) = 72

Question 10.
26, 39

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 26:
26, 52, 78, 104, 130
Multiples of 39:
39, 78, 117, 156
Therefore,
LCM(26, 39) = 78

Question 11.
6, 12, 14

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 6:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96
Multiples of 12:
12, 24, 36, 48, 60, 72, 84, 96, 108
Multiples of 14:
14, 28, 42, 56, 70, 84, 98, 112
Therefore,
LCM(6, 12, 14) = 84

Question 12.
You have 16 yellow beads, 20 red beads, and 24 orange beads to make identical bracelets. What is the greatest number of bracelets that you can make using all of the beads?

Answer:
To find how many identical bracelets you can make, you need to find a common denominator.
In this case, all three numbers; 16, 20, and 24, can be divided by four.
So you now know you can have four bracelets.
Then you take your numbers of each color beads and divide them by four so you know how many of each color will be on the bracelets.
In the end, you have four bracelets, each with 4 yellow beads, 5 red beads and 6 orange beads

Question 13.
A bag contains equal numbers of green marbles and blue marbles. You can divide all of the green marbles into groups of 12 and all the blue marbles into groups of 16. What is the least number of each color of marble that can be in the bag?

Answer:
Given,
A bag contains equal numbers of green marbles and blue marbles.
You can divide all of the green marbles into groups of 12 and all the blue marbles into groups of 16.
To solve this problem, we need to find for the LCM of each number. That is:
12: 12, 24, 36, 48, 60
16: 16, 32, 48, 64, 80
So we can see that the LCM is 48.
Therefore the least number of each color of marble must be 48.

Question 14.
The ages of the members of a family are 65, 58, 27, 25, 5, and 2 years old. What is the total admission price for the family to visit the zoo?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 96

Answer:
The ages of the members of a family are 65, 58, 27, 25, 5, and 2 years old.
We can find the total admission price for the family to visit the zoo by following the above table.
$10 + $12 + $27 + $12 + $8 + $8 = $77

Question 15.
A competition awards prizes for fourth, third, second, and first place. The fourth place winner receives $5. Each place above that receives a prize that is five times the amount of the previous prize. How much prize money is awarded?

Answer:
A competition awards prizes for fourth, third, second, and first place. The fourth place winner receives $5. Each place above that receives a prize that is five times the amount of the previous prize.
Each place above that receives a prize that is five times the amount of the previous prize
So we can say that;
Pn = 5 ×  P(n+1)
Where n ⇒ Number Place
Pn = Price received by Number place
Substituting the values of n as 3,2,1 to find the price of third second first place winner.
P3 = 5 ×  P(3+1) = 5 × P4 = 5 × 5 = 25
P2 = 5 ×  P(2+1) = 5 × P3 = 5 × 25 = 125
P1 = 5 ×  P(1+1) = 5 × P2 = 5 × 125 = 625
Now We will find the Total Prize money awarded.
Total Prize money awarded = 625 + 125 + 25 + 5 = 780
Hence A total of $780 price money was awarded.

Question 16.
You buy tealight candles and mints as party favors for a baby shower. The tealight candles come in packs of 12 for $3.50. The mints come in packs of 50 for $6.25. What is the least amount of money you can spend to buy the same number of candles and mints?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 96.1

Answer: The least amount of money it can be spent is $125.

Explanation:
First, we write the prime factorization of each number:
12= 2·2·3
15= 2·5·5
Then, we search for each different factor which appears the greater number of times. The factor 2 appears in both factorizations so the least common multiple is:
LCM= 2·2·3·5·5=300
Hence, the total quantity of packs of each thing is:
Candles: 300÷12=25
Mints: 300÷50=6
The least amount of money it can be spent is:
T=25×$3.50 + 6×$6.25= $87.5 + $37.5= $125

Numerical Expressions and Factors Cumulative Practice

Question 1.
Find the value of 8 × 135?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 96.2
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 97

Answer: Multiply the two numbers 8 and 135
8 × 135 = 1080

Question 2.
Which number is equivalent to the expression blow?
3.23 – 8 ÷ 4

Answer: 22

Explanation:
Given the expression 3.23 – 8 ÷ 4
3 × 8 -(8÷4)
24 – (2)
24 – 2 = 22

Question 3.
The top of an end table is a square with a side length of 16 inches. What is the area of the tabletop?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 98

Answer: I

Explanation:
Given that
The top of an end table is a square with a side length of 16 inches.
Area of the square = s × s
A = 16 × 16
A = 256 in²
Thus the correct answer is option I.

Question 4.
You are filling baskets using 18 green eggs, 36 red eggs, and 54 blue eggs. What is the greatest number of baskets that you can fill so that the baskets are identical and there are no eggs left over?
A. 3
B. 6
C. 9
D. 18

Answer: D

Explanation:
Given,
You are filling baskets using 18 green eggs, 36 red eggs, and 54 blue eggs.
18/n = 36/n = 54/n
Factors of 18 are 2,3,6,9,18
Factors of 36 are 2,3,4,6,9,12,18.
Factors of 54 are 2,3,6,9,18,27,54.
The common multiples of 18,36 and 54 are 2,3,6,9,18.
Thus the greatest among them is 18.
Thus the correct answer is option D.

Question 5.
What is the value of 23.32.5?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 99

Answer:
23.32.5
2³ = 8
3² = 9
8 × 9 × 5 = 360

Question 6.
You hang the two strands of decorative lights shown below.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 100
Both strands just changed color. After how many seconds will the strands change color at the same time again?
F. 3 seconds
G. 30 seconds
H. 90 seconds
I. 270 seconds

Answer: 3 seconds

Explanation:

Strand I: Changes between red and blue every 15 seconds
Strand II: Changes between green and gold every 18 seconds
18 – 15 = 3 seconds
Thus the correct answer is option F.

Question 7.
Point P is plotted in the coordinate plane below.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 101
What are the coordinates of Point P ?
A. (5, 3)
B. (4, 3)
C. (3, 5)
D. (3, 4)

Answer: C
By seeing the above graph we can find the coordinates of point p.
The X-axis is on 3 and Y-axis is on 5.
Thus the correct answer is option c.

Question 8.
What is the prime factorization of 1100?
F. 2 × 5 × 11
G. 22 × 52 × 11
H. 4 × 52 × 11
I. 22 × 5 × 55

Answer: G
Prime factorization of 1100: 2 × 550
2 × 2 × 275
2 × 2 × 5 × 55
2 × 2 × 5 × 5 × 11
Thus the correct answer is option b.

Question 9.
What is the least common multiple of 3, 8, and 10?
A. 24
B. 30
C. 80
D. 120

Answer: D

Multiples of 3:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126
Multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136
Multiples of 10:
10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 120, 140.
The common multiple among the three is 120.
Thus the correct answer is option D.

Question 10.
What is the area of the shaded region of the figure below?
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 102

Answer:
The above figure is square.
s = 4 yd
Area of the square = s × s
A = 4 yd × 4 yd
A = 16 sq. yd
The area of the outer box.
s = 9 yd
Area of the square = s × s
A = 9 yd × 9 yd
A = 81 sq. yd
The area of the shaded region is 81 – 16 = 65 sq. yd
Thus the correct answer is option G.

Question 11.
Which expression represents a prime factorization?
A. 4 × 4 × 7
B. 22 × 21 × 23
C. 34 × 5 × 7
D. 5 × 5 × 9 × 11

Answer: B

Prime factorization:
22 × 21 × 23
2, 21, 23 is a prime number.
Thus the correct answer is option B.

Question 12.
Find the greatest common factor for each pair of numbers.
Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 103
What can you conclude about the greatest common factor of 10, 15, and 21? Explain your reasoning.

Answer: 1

Explanation:
The factors of 10 are: 1, 2, 5, 10
The factors of 15 are: 1, 3, 5, 15
The factors of 21 are: 1, 3, 7, 21
Then the greatest common factor is 1.

Final Words:
I wish that the details prevailed in this article regarding Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors Answer Key is helpful for all the students and also teachers. Make use of the solutions and score good marks in the exams. Best Of Luck!!

Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines

Big Ideas Math Geometry Answers Chapter 3

Seeking help regarding the concepts of Big Ideas Geometry Answer Key Ch 3 Parallel and Perpendicular Lines? If so, don’t bother as you will get a complete idea through our BIM Geometry Chapter 3 Parallel and Perpendicular Lines Answer Key. Make the most out of these preparation resources and stand out from the rest of the crowd. All the Questions prevailing here in Big Ideas Math Geometry Answers Chapter 3 adhere and meets the Common Core Curriculum Standards. You will find Solutions to all the BIM Book Geometry Ch 3 Parallel and Perpendicular Concepts aligned as per the BIM Textbooks.

Big Ideas Math Book Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines

Solving the concepts from the Big Ideas Math Book Geometry Ch 3 Parallel and Perpendicular Lines Answers on a regular basis boosts the problem-solving ability in you. Embedded mathematical practices, exercises provided make it easy for you to understand the concepts quite quickly. Begin your preparation right away and clear the exams with utmost confidence. Simply click on the below available and learn the respective topics in no time.

Parallel and Perpendicular Lines Maintaining Mathematical Proficiency

Find the slope of the line.

Question 1.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 1
Answer:
From the given coordinate plane,
Let the given points are:
A (-1, 2), and B (3, -1)
Compare the given points with
A (x1, y1), B (x2, y2)
We know that,
Slope of the line (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of the line (m) = \(\frac{-1 – 2}{3 + 1}\)
= \(\frac{-3}{4}\)
Hence, from the above,
We can conclude that the slope of the given line is: \(\frac{-3}{4}\)

Question 2.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 2
Answer:
From the given coordinate plane,
Let the given points are:
A (-2, 2), and B (-3, -1)
Compare the given points with
A (x1, y1), B (x2, y2)
We know that,
Slope of the line (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of the line (m) = \(\frac{-1 – 2}{-3 + 2}\)
= \(\frac{-3}{-1}\)
= 3
Hence, from the above,
We can conclude that the slope of the given line is: 3

Question 3.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3
Answer:
From the given coordinate plane,
Let the given points are:
A (-3, -2), and B (1, -2)
Compare the given points with
A (x1, y1), B (x2, y2)
We know that,
Slope of the line (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of the line (m) = \(\frac{-2 + 2}{3 + 1}\)
= \(\frac{0}{4}\)
= 0
Hence, from the above,
We can conclude that the slope of the given line is: 0

Write an equation of the line that passes through the given point and has the given slope.

Question 4.
(6, 1); m = – 3
Answer:
The given point is: (6, 1)
The slope of the given line is: m = -3
We know that,
The equation of a line is:
y = mx + b
Where,
m is the slope
b is the y-intercept
So,
y = -3x + b ——–(1)
To find the value of b,
Substitute the given point in eq. (1)
So,
1 = -3 (6) + b
1 = -18 + b
1 + 18 = b
b = 19
Hence,
The equation of the line along with y-intercept is:
y = -3x + 19

Question 5.
(-3, 8); m = – 2
Answer:
The given point is: (-3, 8)
The slope of the given line is: m = -2
We know that,
The equation of a line is:
y = mx + b
Where,
m is the slope
b is the y-intercept
So,
y = -2x + b ——–(1)
To find the value of b,
Substitute the given point in eq. (1)
So,
8 = -2 (-3) + b
8 = 6 + b
8 – 6 = b
b = 2
Hence,
The equation of the line along with y-intercept is:
y = -2x + 2

Question 6.
(- 1, 5); m = 4
Answer:
The given point is: (-1, 5)
The slope of the given line is: m = 4
We know that,
The equation of a line is:
y = mx + b
Where,
m is the slope
b is the y-intercept
So,
y = 4x + b ——–(1)
To find the value of b,
Substitute the given point in eq. (1)
So,
5 = 4 (-1) + b
5 = -4 + b
5 + 4 = b
b = 9
Hence,
The equation of the line along with y-intercept is:
y = 4x + 9

Question 7.
(2, – 4); m = \(\frac{1}{2}\)
Answer:
The given point is: (2, -4)
The slope of the given line is: m = \(\frac{1}{2}\)
We know that,
The equation of a line is:
y = mx + b
Where,
m is the slope
b is the y-intercept
So,
y = \(\frac{1}{2}\)x + b ——–(1)
To find the value of b,
Substitute the given point in eq. (1)
So,
-4 = \(\frac{1}{2}\) (2) + b
-4 = 1 + b
-4 – 1 = b
b = -5
Hence,
The equation of the line along with y-intercept is:
y = \(\frac{1}{2}\)x – 5

Question 8.
(- 8, – 5); m = –\(\frac{1}{4}\)
Answer:
The given point is: (-8, -5)
The slope of the given line is: m = –\(\frac{1}{4}\)
We know that,
The equation of a line is:
y = mx + b
Where,
m is the slope
b is the y-intercept
So,
y = –\(\frac{1}{4}\)x + b ——–(1)
To find the value of b,
Substitute the given point in eq. (1)
So,
-5 = –\(\frac{1}{4}\) (-8) + b
-5 = 2 + b
-5 – 2 = b
b = -7
Hence,
The equation of the line along with y-intercept is:
y = –\(\frac{1}{4}\)x – 7

Question 9.
(0, 9); m = \(\frac{2}{3}\)
Answer:
The given point is: (0, 9)
The slope of the given line is: m = \(\frac{2}{3}\)
We know that,
The equation of a line is:
y = mx + b
Where,
m is the slope
b is the y-intercept
So,
y = \(\frac{2}{3}\)x + b ——–(1)
To find the value of b,
Substitute the given point in eq. (1)
So,
9 = \(\frac{2}{3}\) (0) + b
9 = 0 + b
9 – 0 = b
b = 9
Hence,
The equation of the line along with y-intercept is:
y = \(\frac{2}{3}\)x + 9

Question 10.
ABSTRACT REASONING
Why does a horizontal line have a slope of 0, but a vertical line has an undefined slope?
Answer:
We know that,
Slope of the line (m) = \(\frac{y2 – y1}{x2 – x1}\)
We know that,
For a horizontal line,
The coordinates of y are the same. i.e.,
y1 = y2 = y3  ………
For a vertical line,
The coordinates of x are the same. i.e.,
x1 = x2 = x3 …….
So,
The slope of the horizontal line (m) = \(\frac{y2 – y2}{x2 – x1}\)
We know that,
Any fraction that contains 0 in the numerator has its value equal to 0
So,
The slope of horizontal line (m) = 0
The slope of vertical line (m) = \(\frac{y2 – y1}{x2 – x1}\)
We know that,
Any fraction that contains 0 in the denominator has its value undefined
So,
The slope of the vertical line (m) = Undefined

Parallel and Perpendicular Lines Mathematical Practices

Use a graphing calculator to graph the pair of lines. Use a square viewing window. Classify the lines as parallel, perpendicular, coincident, or non-perpendicular intersecting lines. Justify your answer.

Question 1.
x + 2y = 2
2x – y = 4
Answer:
The given pair of lines are:
x + 2y = 2
2x – y = 4
Hence,
The representation of the given pair of lines in the coordinate plane is:

We know that,
For a pair of lines to be perpendicular, the product of the slopes i.e., the product of the slope of the first line and the slope of the second line will be equal to -1
So,
By comparing the given pair of lines with
y = mx + b
We get
The slope of first line (m1) = –\(\frac{1}{2}\)
The slope of second line (m2) = 2
So,
m1 ×m2 = –\(\frac{1}{2}\) × 2
= -1
Hence, from the above,
We can conclude that the given pair of lines are perpendicular lines

Question 2.
x + 2y = 2
2x + 4y = 4
Answer:
The give pair of lines are:
x + 2y = 2
2x + 4y = 4
Hence,
The representation of the given pair of lines in the coordinate plane is:

We know that,
For a pair of lines to be coincident, the pair of lines have the same slope and the same y-intercept
So,
By comparing the given pair of lines with
y = mx + b
We get
m1 = –\(\frac{1}{2}\), b1 = 1
m2 = –\(\frac{1}{2}\), b2 = 1
Hence, from the above,
We can conclude that the given pair of lines are coincident lines

Question 3.
x + 2y = 2
x + 2y = – 2
Answer:
The given pair of lines are:
x + 2y = 2
x + 2y = -2
Hence,
The representation of the given pair of lines in the coordinate plane is:

We know that,
For a pair of lines to be parallel, the pair of lines have the same slope but different y-intercepts
So,
By comparing the given pair of lines with
y = mx + b
We get
m1 = –\(\frac{1}{2}\), b1 = 1
m2 = \(\frac{1}{2}\), b2 = -1
Hence, from the above,
We can conclude that the given pair of lines are parallel lines

Question 4.
x – 2y = 2
x – y = – 4
Answer:
The given pair of lines are:
x – 2y = 2
x – y = -4
Hence,
The representation of the given pair of lines in the coordinate plane is:

We know that,
For a pair of lines to be non-perpendicular, the product of the slopes i.e., the product of the slope of the first line and the slope of the second line will not be equal to -1
So,
By comparing the given pair of lines with
y = mx + b
We get
The slope of first line (m1) = \(\frac{1}{2}\)
The slope of second line (m2) = 1
So,
m1 ×m2  = \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the given pair of lines are non-perpendicular lines

3.1 Pairs of Lines and Angles

Exploration 1

Points of intersection

work with a partner: Write the number of points of intersection of each pair of coplanar lines.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 4
Answer:
The given coplanar lines are:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 4
a. The points of intersection of parallel lines:
We know that,
The “Parallel lines” have the same slope but have different y-intercepts
So,
We can say that any parallel line do not intersect at any point
Hence, from the above,
We can conclude that the number of points of intersection of parallel lines is: 0

a. The points of intersection of intersecting lines:
We know that,
The “Intersecting lines” have a common point to intersect
So,
We can say that any intersecting line do intersect at 1 point
Hence, from the above,
We can conclude that the number of points of intersection of intersecting lines is: 1

c. The points of intersection of coincident lines:
We know that,
The “Coincident lines” may be intersecting or parallel
So,
We can say that any coincident line do not intersect at any point or intersect at 1 point
Hence, from the above,
We can conclude that the number of points of intersection of coincident lines is: 0 or 1

Exploration 2

Classifying Pairs of Lines

Work with a partner: The figure shows a right rectangular prism. All its angles are right angles. Classify each of the following pairs of lines as parallel, intersecting, coincident, or skew. Justify your answers. (Two lines are skew lines when they do not intersect and are not coplanar.)
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 5
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 6
Answer:
The given rectangular prism is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 5
We know that,
The “Parallel lines” are the lines that do not intersect with each other and present in the same plane
The “Intersecting lines” are the lines that intersect with each other and in the same plane
The “Coincident lines” are the lines that lie on one another and in the same plane
The “Skew lines” are the lines that do not present in the same plane and do not intersect
Hence,
The completed table of the nature of the given pair of lines is:

Exploration 3

Identifying Pairs of Angles

Work with a partner: In the figure, two parallel lines are intersected by a third line called a transversal.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 7
a. Identify all the pairs of vertical angles. Explain your reasoning.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results.
Answer:
We know that,
The angles that are opposite to each other when two lines cross are called “Vertical angles”
Hence, from the given figure,
We can conclude that the vertical angles are:
∠1 and ∠3; ∠2 and ∠4; ∠5 and ∠7; ∠6 and ∠8

b. Identify all the linear pairs of angles. Explain your reasoning.
Answer:
We know that,
A “Linear pair” is a pair of adjacent angles formed when two lines intersect
Hence, from the given figure,
We can conclude that the linear pair of angles is:
∠1 and ∠2; ∠4 and ∠3; ∠5 and ∠6; ∠8 and ∠7

Communicate Your Answer

Question 4.
What does it mean when two lines are parallel, intersecting, coincident, or skew?
Answer:
The two lines are “Parallel” when they do not intersect each other and are coplanar
The two lines are “Intersecting” when they intersect each other and are coplanar
The two lines are “Coincident” when they lie on each other and are coplanar
The two lines are “Skew” when they do not intersect each other and are not coplanar

Question 5.
In Exploration 2. find more pairs of lines that are different from those given. Classify the pairs of lines as parallel, intersecting, coincident, or skew. Justify your answers.
Answer:
The given rectangular prism of Exploration 2 is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 5
The pair of lines that are different from the given pair of lines in Exploration 2 are:
a. \(\overline{C D}\) and \(\overline{A E}\)
b. \(\overline{D H}\) and \(\overline{F G}\)
Hence, from the above,
We can conclude that
a. \(\overline{C D}\) and \(\overline{A E}\) are “Skew lines” because they are not intersecting and are non coplanar
b. \(\overline{D H}\) and \(\overline{F G}\) are “Skew lines” because they are not intersecting and are non coplanar

Lesson 3.1 Pairs of Lines and Angles

Monitoring Progress

Question 1.
Look at the diagram in Example 1. Name the line(s) through point F that appear skew to Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 8.
Answer:
From Example 1,
We can observe that
The line that passes through point F that appear skew to \(\overline{E H}\) is: \(\overline{F C}\)

Question 2.
In Example 2, can you use the Perpendicular Postulate to show that Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 9 is not perpendicular to Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 10? Explain why or why not.
Answer:
Perpendicular Postulate:
According to this Postulate,
If there is a line and a point not on the line, then there is exactly one line through the point perpendicular to the given line
Now,
In Example 2,
We can observe that \(\overline{A C}\) is not perpendicular to \(\overline{B F}\) because according to the perpendicular Postulate, \(\overline{A C}\) will be a straight line but it is not a straight line when we observe Example 2
Hence, from the above,
We can conclude that we can use “Perpendicular Postulate” to show that \(\overline{A C}\) is not perpendicular to \(\overline{B F}\)

Classify the pair of numbered angles.

Question 3.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 11
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 11
We know that,
The angles that have the same corner are called “Adjacent angles”
Hence, from the above,
We can conclude that ∠1 and ∠5 are the adjacent angles

Question 4.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 12
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 12
We know that,
The angles that have the opposite corners are called “Vertical angles”
Hence, from the above,
We can conclude that ∠2 and ∠7 are the “Vertical angles”

Question 5.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 13
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 13
We know that,
The angles that have the opposite corners are called “Vertical angles”
Hence, from the above,
We can conclude that ∠4 and ∠5 are the “Vertical angles”

Exercise 3.1 Pairs of Lines and Angles

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Two lines that do not intersect and are also not parallel are ________ lines.
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 1

Question 2.
WHICH ONE did DOESN’T BELONG?
Which angle pair does not belong with the other three? Explain our reasoning.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 14
∠2 and ∠3
∠4 and ∠5
∠1 and ∠8
∠2 and∠7
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 14
We know that,
The angles that have the common side are called “Adjacent angles”
The angles that are opposite to each other when 2 lines cross are called “Vertical angles”
So,
∠2 and ∠3 are vertical angles
∠4 and ∠5 are adjacent angles
∠1 and ∠8 are vertical angles
∠2 and ∠7 are vertical angles
Hence, from the above,
We can conclude that ∠4 and ∠5 angle-pair do not belong with the other three

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, think of each segment in the diagram as part of a line. All the angles are right angles. Which line(s) or plane(s) contain point B and appear to fit the description?
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 15
Question 3.
line(s) parallel to Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 16.
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 3

Question 4.
line(s) PerPendicular to Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 17.
Answer:
We know that,
The lines that are a straight angle with the given line and are coplanar is called “Perpendicular lines”
So,
From the given figure,
We can conclude that the line that is perpendicular to \(\overline{C D}\) is: \(\overline{A D}\) and \(\overline{C B}\)

Question 5.
line(s) skew to Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 18
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 5

Question 6.
plane(s) parallel to plane CDH
Answer:
From the given figure,
We can observe that the plane parallel to plane CDH is: Plane BAE

In Exercises 7-10, Use the diagram.

Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 19

Question 7.
Name a pair of parallel lines.
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 7

Question 8.
Name a pair of perpendicular lines.
Answer:
We know that,
The lines that have an angle of 90° with each other are called “Perpendicular lines”
Hence,
From the figure,
We can conclude that \(\overline{N P}\) and \(\overline{P O}\) are perpendicular lines

Question 9.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 20
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 9

Question 10.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 21
Answer:
We know that,
The lines that have an angle of 90° with each other are called “Perpendicular lines”
Hence,
From the figure,
We can conclude that \(\overline{P R}\) and \(\overline{P O}\) are not perpendicular lines

In Exercises 11-14, identify all pairs of angles of the given type.

Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 22
Question 11.
corresponding
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 11

Question 12.
alternate interior
Answer:
We know that,
“Alternate Interior angles” are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal.
Hence,
From the given figure,
We can conclude that the alternate interior angles are: ∠4 and ∠5; ∠3 and ∠6

Question 13.
alternate exterior
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 13

Question 14.
consecutive interior
Answer:
We know that,
The pair of angles on one side of the transversal and inside the two lines are called the “Consecutive interior angles”.
Hence,
From the given figure,
We can conclude that the consecutive interior angles are: ∠3 and ∠5; ∠4 and ∠6

USING STRUCTURE
In Exercises 15-18, classify the angle pair as corresponding. alternate interior, alternate exterior, or consecutive interior angles.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 23

Question 15.
∠5 and ∠1
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 15

Question 16.
∠11 and ∠13
Answer:
We know that,
The pair of angles on one side of the transversal and inside the two lines are called the “Consecutive interior angles”.
Hence,
From the given figure,
We can conclude that ∠11 and ∠13 are the “Consecutive interior angles”

Question 17.
∠6 and ∠13
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 17

Question 18.
∠2 and ∠11
Answer:
We know that,
“Vertical Angles” are the angles opposite each other when two lines cross
Hence, from the above figure,
We can conclude that ∠2 and ∠11 are the “Vertical angles”

ERROR ANALYSIS
In Exercises 19 and 20. describe and correct the error in the conditional statement about lines.

Question 19.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 24
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 19

Question 20.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 25
Answer:
We know that,
The “Perpendicular Postulate” states that if there is a line and a point not on the line, then there is exactly one line through the point perpendicular to the given line.
Hence, from the above,
We can conclude that the given statement is not correct

Question 21.
MODELING WITH MATHEMATICS
Use the photo to decide whether the statement is true or false. Explain Your reasoning.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 26
a. The plane containing the floor of the treehouse is parallel to the ground.
b. The lines containing the railings of the staircase, such as Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 27, are skew to all lines in the plane containing the ground.
c. All the lines containing the balusters. such as Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 28, are perpendicular to the plane containing the floor of the treehouse.
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 21

Question 22.
THOUGHT-PROVOKING
If two lines are intersected by a third line, is the third line necessarily a transversal? Justify your answer with a diagram.
Answer:
No, the third line does not necessarily be a transversal

Explanation:
We know that,
“Parallel lines” do not intersect each other
“Perpendicular lines” intersect at each other at right angles
The third intersecting line can intersect at the same point that the two lines have intersected as shown below:

Hence, from the above,
We can conclude that the third line does not need to be a transversal

Question 23.
MATHEMATICAL CONNECTIONS
Two lines are cut by a transversal. Is it possible for all eight angles formed to have the same measure? Explain your reasoning.
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 23

Question 24.
HOW DO YOU SEE IT?
Think of each segment in the figure as part of a line.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 29
a. Which lines are parallel to Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 30?
Answer:
We know that,
The lines that do not intersect to each other and are coplanar are called “Parallel lines”
Hence, from the above figure,
We can conclude that the line parallel to \(\overline{N Q}\) is: \(\overline{M P}\)

b. Which lines intersect Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 30?
Answer:
We know that,
The lines that are coplanar and any two lines that have a common point are called “Intersecting lines”
Hence, from the above figure,
We can conclude that the lines that intersect \(\overline{N Q}\) are: \(\overline{N K}\), \(\overline{N M}\), and \(\overline{Q P}\)

c. Which lines are skew to Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 30?
Answer:
We know that,
The lines that do not intersect or not parallel and non-coplanar are called “Skew lines”
Hence, from the above figure,
We can conclude that \(\overline{K L}\), \(\overline{L M}\), and \(\overline{L S}\)

d. Should you have named all the lines on the cube in parts (a)-(c) except \(\overline{N Q}\)? Explain.
Answer:
No, we did not name all the lines on the cube in parts (a) – (c) except \(\overline{N Q}\)

In exercises 25-28. copy and complete the statement. List all possible correct answers.

Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 31

Question 25.
∠BCG and __________ are corresponding angles.
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 25

Question 26.
∠BCG and __________ are consecutive interior angles.
Answer:
We know that,
When two lines are cut by a transversal, the pair of angles on one side of the transversal and inside the two lines are called the “Consecutive interior angles”
Hence,
From the given figure,
We can conclude that the consecutive interior angles of ∠BCG are: ∠FCA and ∠BCA

Question 27.
∠FCJ and __________ are alternate interior angles.
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 27

Question 28.
∠FCA and __________ are alternate exterior angles.
Answer:
“Alternate exterior angles” are the pair of angles that lie on the outer side of the two parallel lines but on either side of the transversal line
Hence,
From the above figure,
We can conclude that ∠FCA and ∠JCB are alternate exterior angles

Question 29.
MAKING AN ARGUMENT
Your friend claims the uneven parallel bars in gymnastics are not really Parallel. She says one is higher than the other. so they cannot be on the same plane. Is she correct? Explain.

Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 29

Maintaining Mathematical Proficiency

Use the diagram to find the measure of all the angles.

Question 30.
m∠1 = 76°
Answer:
We know that,
The sum of the adjacent angles is: 180°
The vertical angles are congruent i.e., the angle measures of the vertical angles are equal
Now,
From the figure,
The adjacent angles are: ∠1 and ∠2; ∠2 and ∠3; ∠3 and ∠4; and ∠4 and ∠1
The vertical angles are: ∠1 and ∠3; ∠2 and ∠4
Now,
∠1 + ∠2 = 180°
∠2 = 180° – ∠1
= 180° – 76°
= 104°
So,
∠3 = 76° and ∠4 = 104°
Hence, from the above,
We can conclude that,
∠1 = 76°, ∠2 = 104°, ∠3 = 76°, and ∠4 = 104°

Question 31.
m∠2 = 159°
Answer:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 3.1 a 31

3.2 Parallel Lines and Transversals

Exploration 1

Exploring parallel Lines

Work with a partner: Use dynamic geometry software to draw two parallel lines. Draw a third line that intersects both parallel lines. Find the measures of the eight angles that are formed. What can you conclude?
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 34
Answer:
By using the dynamic geometry,
The representation of the given coordinate plane along with parallel lines is:

Hence, from the coordinate plane,
We can observe that,
∠3 = 53.7° and ∠4 = 53.7°
We know that,
The angle measures of the vertical angles are congruent
So,
∠1 = 53.7° and ∠5 = 53.7°
We know that,
In the parallel lines,
All the angle measures are equal
Hence, from the above,
We can conclude that
∠1 = ∠2 = ∠3 = ∠4 = ∠5 = ∠6 = ∠7 = 53.7°

Exploration 2

Writing conjectures

Work with a partner. Use the results of Exploration 1 to write conjectures about the following pairs of angles formed by two parallel lines and a transversal.
ATTENDING TO PRECISION
To be proficient in math, you need to communicate precisely with others.
a. corresponding angles
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 35
Answer:
We know that,
When two lines are crossed by another line (which is called the Transversal), the angles in matching corners are called “Corresponding angles”
Hence, from the given figure,
We can conclude that
The corresponding angles are: ∠ and ∠5; ∠4 and ∠8

b. alternate interior angles
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 36
Answer:
We know that,
“Alternate Interior Angles” are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal.
Hence, from the above figure,
We can conclude that
The alternate interior angles are: ∠3 and ∠5; ∠2 and ∠8

c. alternate exterior angles
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 37
Answer:
We know that,
“Alternate exterior angles” are the pair of angles that lie on the outer side of the two parallel lines but on either side of the transversal line.
Hence, from the above figure,
We can conclude that
The alternate exterior angles are: ∠1 and ∠7; ∠6 and ∠4

d. consecutive interior angles
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 38
Answer:
We know that,
When two lines are cut by a transversal, the pair of angles on one side of the transversal and inside the two lines are called the “Consecutive interior angles”
Hence, from the above figure,
We can conclude that
The consecutive interior angles are: ∠2 and ∠5; ∠3 and ∠8

Communicate Your Answer

Question 3.
When two parallel lines are cut by a transversal, which of the resulting pairs of angles are congruent?
Answer:
If two parallel lines are cut by a transversal, then the pairs of “Corresponding angles” are congruent.
If two parallel lines are cut by a transversal, then the pairs of “Alternate interior angles” are congruent.
If two parallel lines are cut by a transversal, then the pairs of “Alternate exterior angles” are congruent.

Question 4.
In Exploration 2. m∠1 = 80°. Find the other angle measures.
Answer:
It is given that
In Exploration 2,
∠1 = 80°
We know that,
Exploration 2 comes from Exploration 1
From Exploration 1,
We can say that all the angle measures are equal in Exploration 1
Hence, from the above,
We can conclude that
∠1 = ∠2 = ∠3 = ∠4 = ∠5 = ∠6 = ∠7 = ∠8 = 80°

Lesson 3.2 Parallel Lines and Transversals

Monitoring Progress

Use the diagram
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 39

Question 1.
Given m∠1 = 105°, find m∠4, m∠5, and m∠8. Tell which theorem you use in each case.
Answer:
It is given that ∠1 = 105°
Now,
We have to find ∠4, ∠5, and ∠8
Now,
To find ∠4:
Verticle angle theorem:
Vertical Angles Theorem states that vertical angles, angles that are opposite each other and formed by two intersecting straight lines, are congruent
So,
∠1 = ∠4
Hence,
∠4 = 105°

To find ∠5:
Alternate Interior angles theorem:
The Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal, the resulting alternate interior angles are congruent
So,
∠4 = ∠5
Hence,
∠5 = 105°

To find ∠8:
Verticle angle theorem:
Vertical Angles Theorem states that vertical angles, angles that are opposite each other and formed by two intersecting straight lines, are congruent
So,
∠5 = ∠8
Hence,
∠8 = 105°

Question 2.
Given m∠3 = 68° and m∠8 = (2x + 4)°, what is the value of x? Show your steps.
Answer:
It is given that
∠3 = 68° and ∠8 = (2x + 4)°
From the given figure,
We can observe that ∠3 and ∠8 are consecutive exterior angles.
So,
According to the consecutive exterior angles theorem,
If parallel lines are cut by a transversal line, then consecutive exterior angles are supplementary.
So,
∠3 + ∠8 = 180°
68° + (2x + 4)° = 180°
2x + 72° = 180°
2x° = 180° – 72°
2x° = 108°
x = \(\frac{108}{2}\)
x = 54°
Hence, from the above,
We can conclude that the value of x is: 54°

Question 3.
In the proof in Example 4, if you use the third statement before the second statement. could you still prove the theorem? Explain.
Answer:
In Example 4, the given theorem is “Alternate interior angle theorem”
If you even interchange the second and third statements, you could still prove the theorem as the second line before interchange is not necessary
Hence, from the above,
We can conclude that if you use the third statement before the second statement, you could still prove the theorem

Question 4.
WHAT IF?
In Example 5. yellow light leaves a drop at an angle of m∠2 = 41°. What is m∠1? How do you know?
Answer:
In Example 5,
If we observe ∠1 and ∠2, then they are alternate interior angles
Now,
According to Alternate interior angle theorem,
∠1 = ∠2
It is given that,
∠2 = 41°
So,
∠1 = 41°
Hence, from the above,
We can conclude that
∠1 = 41°

Exercise 3.2 Parallel Lines and Transversals

Vocabulary and Core Concept Check

Question 1.
WRITING
How are the Alternate Interior Angles Theorem (Theorem 3.2) and the Alternate Exterior
Angles Theorem (Theorem 3.3) alike? How are they different?
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 1

Question 2.
WHICH ONE did DOESN’T BELONG?
Which pair of angle measures does not belong with the other three? Explain.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 40
m∠1 and m∠3
m∠2 and m∠4
m∠2 and m∠3
m∠1 and m∠5
Answer:
The given figure is:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 40
From the given figure,
∠1 and ∠3 are the vertical angles
∠2 and ∠4 are the alternate interior angles
∠2 and ∠3 are the consecutive interior angles
∠1 and ∠5 are the alternate exterior angles
So,
From the above,
We can observe that all the angles except ∠1 and ∠3 are the interior and exterior angles
Hence, from the above,
We can conclude that ∠1 and ∠3 pair does not belong with the other three

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, find m∠1 and m∠2. Tell which theorem you use in each case.

Question 3.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 41
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 3

Question 4.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 42
Answer:
The given figure is:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 42
From the given figure,
We can observe that,
∠1 = ∠2 (By using the Vertical Angles theorem)
∠2 = 150° (By using the Alternate exterior angles theorem)
Hence, from the above,
We can conclude that
∠1 = ∠2 = 150°

Question 5.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 43
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 5

Question 6.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 44
Answer:
The given figure is:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 44
From the given figure,
We can observe that,
∠1 + ∠2 = 180° (By using the consecutive interior angles theorem)
∠2 = 140° (By using the Vertical angles theorem)
So,
∠1 = 180° – 140°
∠1 = 40°
Hence, from the above,
We can conclude that
∠1 = 40° and ∠2 = 140°

In Exercises 7-10. find the value of x. Show your steps.

Question 7.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 45
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 7

Question 8.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 46
Answer:
The given figure is:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 46
From the given figure,
We can observe that
72° + (7x + 24)° = 180° (By using the Consecutive interior angles theory)
(7x + 24)° = 180° – 72°
(7x + 24)° = 108°
7x° = 108° – 24°
7x° = 84°
x° = \(\frac{84}{7}\)
x° = 12°
Hence, from the above,
We can conclude that the value of x is: 12°

Question 9.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 47
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 9

Question 10.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 48
Answer:
The given figure is:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 48
From the given figure,
We can observe that
(8x + 6)° = 118° (By using the Vertical Angles theorem)
8x° = 118° – 6°
8x° = 112°
x° = \(\frac{112}{8}\)
x° = 14°
Hence, from the above,
We can conclude that the value of x is: 14°

In Exercises 11 and 12. find m∠1, m∠2, and m∠3. Explain our reasoning.

Question 11.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 49
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 11

Question 12.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 50
Answer:
The given figure is:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 50
From the given figure,
We can observe that
∠3 + 133° = 180° (By using the Consecutive Interior angles theorem)
∠3 = 180° – 133°
∠3 = 47°
Now,
We can observe that
∠2 + ∠3 = 180°
∠2 = 180° – ∠3
∠2 = 180° – 47°
∠2 = 133°
Now,
We can observe that
∠1 = ∠2
Hence, from the above,
We can conclude that
∠1 = ∠2 = 133° and ∠3 = 47°

Question 13.
ERROR ANALYSIS
Describe and correct the error in the students reasoning
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 51
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 13

Question 14.
HOW DO YOU SEE IT?
Use the diagram
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 52
a. Name two pairs of congruent angles when \(\overline{A D}\) and \(\overline{B C}\) are parallel? Explain your reasoning?
Answer:
Let the congruent angle be ∠P
So,
From the figure,
We can observe that the pair of angle when \(\overline{A D}\) and \(\overline{B C}\) are parallel is: ∠APB and ∠DPB

b. Name two pairs of supplementary angles when \(\overline{A B}\) and \(\overline{D C}\) are parallel. Explain your reasoning.
Answer:
From the given figure,
The two pairs of supplementary angles when \(\overline{A B}\) and \(\overline{D C}\) are parallel is: ∠ACD and ∠BDC

PROVING A THEOREM
In Exercises 15 and 16, prove the theorem.

Question 15.
Alternate Exterior Angles Theorem (Thm. 3.3)
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 15

Question 16.
Consecutive Interior Angles Theorem (Thm. 3.4)
Answer:
Statement of consecutive Interior angles theorem:
If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles formed are supplementary
Proof:

Question 17.
PROBLEM-SOLVING
A group of campers ties up their food between two parallel trees, as shown. The rope is pulled taut. forming a straight line. Find m∠2. Explain our reasoning.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 53
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 17

Question 18.
DRAWING CONCLUSIONS
You are designing a box like the one shown.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 54
a. The measure of ∠1 is 70°. Find m∠2 and m∠3.
b. Explain why ∠ABC is a straight angle.
c. If m∠1 is 60°, will ∠ABC still he a straight angle? Will the opening of the box be more steep or less steep? Explain.
Answer:

Question 19.
CRITICAL THINKING
Is it possible for consecutive interior angles to be congruent? Explain.
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 19

Question 20.
THOUGHT-PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry, is it possible that a transversal intersects two parallel lines? Explain your reasoning.
Answer:
We know that,
According to Euclidean geometry,
For a parallel line, there will be no intersecting point
But, In spherical geometry, even though there is some resemblance between circles and lines, there is no possibility to form parallel lines as the lines will intersect at least at 1 point on the circle which is called a tangent
Hence, from the above,
We can conclude that it is not possible that a transversal intersects two parallel lines

MATHEMATICAL CONNECTIONS
In Exercises 21 and 22, write and solve a system of linear equations to find the values of x and y.

Question 21.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 55
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 21

Question 22.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 56
Answer:
The given figure is:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 56
From the given figure,
We can observe that the given pairs of angles are consecutive interior angles
So,
2y° + 4x° = 180°
(2x + 12)° + (y + 6)° = 180°
2x° + y° + 18° = 180°
2x° + y° = 180° – 18°
2x° + y° = 162°———(1)
4x° + 2y° = 180°——–(2)
Solve eq. (1) and eq. (2) to get the values of x and y
2x° = 18°
x° = \(\frac{18}{2}\)
x° = 9°
Now,
y° = 162° – 2 (9°)
y° = 162° – 18°
y° = 144°
Hence, from the above,
We can conclude that the values of x and y are: 9° and 14° respectively

Question 23.
MAKING AN ARGUMENT
During a game of pool. your friend claims to be able to make the shot Shown in the diagram by hitting the cue ball so that m∠1 = 25°. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 57
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 23

Question 24.
REASONING
In diagram. ∠4 ≅∠5 and \(\overline{S E}\) bisects ∠RSF. Find m∠1. Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 58
Answer:
It is given that ∠4 ≅∠5 and \(\overline{S E}\) bisects ∠RSF
So,
∠FSE = ∠ESR
From ΔESR,
We know that,
The sum of the angle measures of a triangle is: 180°
So,
∠3 + ∠4 + ∠5 = 180°
So,
∠3 = 60° (Since ∠4 ≅ ∠5 and the triangle is not a right triangle)
From the given figure,
We can observe that,
∠1 = ∠3 (By using the Corresponding angles theorem)
So,
∠1 = 60°
Hence, from the above,
We can conclude that ∠1 = 60°

Maintaining Mathematical Proficiency

Write the converse of the conditional statement. Decide whether it is true or false.

Question 25.
If two angles are vertical angles. then they are congruent.
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 25

Question 26.
If you go to the zoo, then you will see a tiger.
Answer:
The given statement is:
If you go to the zoo, then you will see a tiger
The converse of the given statement is:
If you will see a tiger, then you go to the zoo———-> False

Question 27.
If two angles form a linear pair. then they are supplementary.
Answer:
Big Ideas Math Answers Geometry Chapter 3 Parallel and Perpendicular Lines 3.2 a 27

Question 28.
If it is warm outside, then we will go to the park.
Answer:
The given statement is:
If it is warm outside, then we will go to the park
The converse of the given statement is:
If you will go to the park, then it is warm outside —————-> False

3.3 Proofs with Parallel Lines

Exploration 1

Exploring Converses

Work with a partner: Write the converse of each conditional statement. Draw a diagram to represent the converse. Determine whether the converse is true. Justify your conclusion.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.

a. Corresponding Angles Theorem (Theorem 3.1): If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 59
Converse:
If the pairs of corresponding angles are

congruent, then the two parallel lines are

cut by a transversal.

Answer:
The Converse of the Corresponding Angles Theorem:
The Converse of the Corresponding Angles Theorem says that if two lines and a transversal form congruent corresponding angles, then the lines are parallel.
So,
When we compare the converses we obtained from the given statement and the actual converse,
We can conclude that both converses are the same
Hence,
We can conclude that the converse we obtained from the given statement is true
Now,
Proof of Converse of Corresponding Angles Theorem:
Consider the 2 lines L1 and L2 intersected by a transversal line L3 creating 2 corresponding angles 1 and 2 which are congruent
We want to prove L1 and L2 are parallel and we will prove this by using “Proof of Contradiction”
Now,
According to Contradiction,
Assume L1 is not parallel to L2
Then, according to the parallel line axiom, there is a different line than L2 that passes through the intersection point of L2 and L3 (point A in the drawing), which is parallel to L1.
Let’s draw that line, and call it P. Let’s also call the angle formed by the traversal line and this new line angle 3, and we see that if we add some other angle, call it angle 4, to it, it will be the same as angle 2.
Now,
P || L1
So,
∠1 ≅ ∠3,
as corresponding angles formed by a transversal of parallel lines, and so,
m∠1=m∠3
This contradicts what was given,  that angles 1 and 2 are congruent.
This contradiction means our assumption (“L1 is not parallel to L2) is false, and so L1 must be parallel to L2.
The representation of the Converse of Corresponding Angles Theorem is:

b. Alternate Interior Angles Theorem (Theorem 3.2): If two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 60
Converse:
If the pairs of alternate interior angles are

congruent, then the two parallel lines are

cut by a transversal.

Answer:
The converse of the Alternate Interior angles Theorem:
The “Converse of the Alternate Interior Angles Theorem” states that if two lines are cut by a transversal and the alternate interior angles are congruent, then the lines are parallel
So,
When we compare the actual converse and the converse according to the given statement,
we can conclude that the converse we obtained from the given statement is false

c. Alternate Exterior Angles Theorem (Theorem 3.3): If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 61
Converse:
If the pairs of alternate exterior angles

are congruent, then the two parallel

lines are cut by a transversal. 

Answer:
The Converse of the alternate exterior angles Theorem:
The “Converse of the Alternate Exterior Angles Theorem” states that if alternate exterior angles of two lines crossed by a transversal are congruent, then the two lines are parallel.
So,
When we compare the converses we obtained from the given statement and the actual converse,
We can conclude that both converses are the same
Hence,
We can conclude that the converse we obtained from the given statement is true
Now,
Proof of Alternate exterior angles Theorem:
Given: ∠1 ≅ ∠2
Prove: l || m
The flow proof for the Converse of Alternate exterior angles Theorem is:

The representation of the Converse of the Exterior angles Theorem is:

d. Consecutive Interior Angles Theorem (Theorem 3.4): If two parallel lines are cut by a transversal. then the pairs of consecutive interior angles are supplementary.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 62
Converse:
If the pairs of consecutive interior angles

are supplementary, then the two parallel lines

are cut by a transversal

Answer:
The Converse of the Consecutive Interior angles Theorem:
The “Converse of the consecutive Interior angles Theorem” states that if the consecutive interior angles on the same side of a transversal line intersecting two lines are supplementary, then the two lines are parallel.
So,
When we compare the converses we obtained from the given statement and the actual converse,
We can conclude that both converses are the same
Hence,
We can conclude that the converse we obtained from the given statement is true
Now,
Proof of the Converse of the Consecutive Interior angles Theorem:
a.  m∠5 + m∠4 = 180° //From the given statement
b.  m∠1 + m∠4 = 180° // Linear pair of angles are supplementary
c. m∠5=m∠1 // (1), (2), transitive property of equality
d.  AB||CD // Converse of the Corresponding Angles Theorem
The representation of the Converse of the Consecutive Interior angles Theorem is:

Communicate Your Answer

Question 2.
For which of the theorems involving parallel lines and transversals is the converse true?
Answer:
From Exploration 1,
The theorems involving parallel lines and transversals that the converse is true are:
a. Corresponding Angles Theorem
b. Alternate Exterior angles Theorem
c. Consecutive Interior angles Theorem

Question 3.
In Exploration 1, explain how you would prove any of the theorems that you found to be true.
Answer:
For the proofs of the theorems that you found to be true, refer to Exploration 1

Lesson 3.3 Proofs with Parallel Lines

Monitoring Progress

Question 1.
Is there enough information in the diagram to conclude that m || n? Explain.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 63
Answer:
Yes, there is enough information in the diagram to conclude m || n.

Explanation:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 63
From the given figure,
We can observe that the given angles are the consecutive exterior angles
Now,
We have to prove that m || n
So,
We will use “Converse of Consecutive Exterior angles Theorem” to prove m || n
Proof of the Converse of the Consecutive Exterior angles Theorem:
a.  m∠1 + m∠8 = 180° //From the given statement
b.  m∠1 + m∠4 = 180° // Linear pair of angles are supplementary
c. m∠5=m∠1 // (1), (2), transitive property of equality
d.  AB||CD // Converse of the Corresponding Angles Theorem
The representation of the Converse of the Consecutive Interior angles Theorem is:

Question 2.
Explain why the Corresponding Angles Converse is the converse of the Corresponding Angles Theorem (Theorem 3.1).
Answer:
Corresponding Angles Theorem:
The “Corresponding Angles Postulate” states that, when two parallel lines are cut by a transversal, the resulting corresponding angles are congruent
Converse:
When the corresponding angles are congruent, the two parallel lines are cut by a transversal
Now,
The Converse of Corresponding Angles Theorem:
If the corresponding angles formed are congruent, then two lines l and m are cut by a transversal.
So,
When we observe the Converse of the Corresponding Angles Theorem we obtained and the actual definition, both are the same
Hence, from the above,
We can conclude that the Corresponding Angles Converse is the converse of the Corresponding Angles Theorem

Question 3.
If you use the diagram below to prove the Alternate Exterior Angles Converse. what Given and Prove statements would you use?
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 64
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 64
It is given that the given angles are the alternate exterior angles
Now,
Alternate Exterior angle Theorem:
If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent
The Converse of the Alternate Exterior Angles Theorem:
The “Converse of the Alternate Exterior Angles Theorem” states that if alternate exterior angles of two lines crossed by a transversal are congruent, then the two lines are parallel.
Hence, from the above,
We can conclude that
For the Converse of the alternate exterior angles Theorem,
The given statement is: ∠1 ≅ 8
To prove: l || k

Question 4.
Copy and complete the following paragraph proof of the Alternate Interior Angles Converse using the diagram in Example 2.

It is given that ∠4 ≅∠5. By the _______ . ∠1 ≅ ∠4. Then by the Transitive Property of Congruence (Theorem 2.2), _______ . So, by the _______ , g || h.
Answer:
The completed proof of the Alternate Interior Angles Converse using the diagram in Example 2 is:
It is given that ∠4 ≅∠5.
By the Vertical Angles Congruence Theorem (Theorem 2.6). ∠1 ≅ ∠4.
Then by the Transitive Property of Congruence (Theorem 2.2), ∠1 ≅∠5.
So, by the Corresponding Angles Converse, g || h.

Question 5.
Each step is parallel to the step immediately above it. The bottom step is parallel to the ground. Explain why the top step is parallel t0 the ground.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 65
Answer:
From the given figure,
We can observe that not any step is intersecting at each other
In the same way, when we observe the floor from any step,
We can say that they are also parallel
Hence, from the above,
We can conclude that the  top step is also parallel to the ground since they do not intersect each other at any point

Question 6.
In the diagram below. p || q and q || r. Find m∠8. Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 66
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 66
From the figure,
We can observe that the given angles are the consecutive exterior angles
We know that,
According to the Consecutive Exterior angles Theorem,
∠8 + 115° = 180°
∠8 = 180° – 115°
∠8 = 65°
Hence, from the above,
We can conclude that
∠8 = 65°

Exercise 3.3 Proofs with Parallel Lines

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Two lines are cut by a transversal. Which angle pairs must be congruent for the lines to be parallel?
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 1

Question 2.
WRITING
Use the theorems from Section 3.2 and the converses of those theorems in this section to write three biconditional statements about parallel lines and transversals.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3-8. find the value of x that makes m || n. Explain your reasoning.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 67
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 68
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 68
From the given figure,
We can observe that the given angles are the corresponding angles
Now,
According to Corresponding Angles Theorem,
(2x + 15)° = 135°
2x° = 135° – 15°
2x° = 120°
x° = \(\frac{120}{2}\)
x° = 60°
Hence, from the above,
We can conclude that the value of x is: 60°

Question 5.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 69
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 70
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 70
From the given figure,
We can observe that the given angles are the corresponding angles
Now,
According to Corresponding Angles Theorem,
(180 – x)° = x°
180° = x°  + x°
2x° = 180°
x° = \(\frac{180}{2}\)
x° = 90°
Hence, from the above,
We can conclude that the value of x is: 90°

Question 7.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 71
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 7

Question 8.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 72
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 72
From the given figure,
We can observe that the given angles are the corresponding angles
Now,
According to Corresponding Angles Theorem,
(2x + 20)° = 3x°
20° = 3x°  – 2x°
x° = 20°
Hence, from the above,
We can conclude that the value of x is: 20°

In Exercises 9 and 10, use a compass and straightedge to construct a line through point P that is parallel to line m.

Question 9.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 73
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 9

Question 10.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 74
Answer:
Let A and B be two points on line m.
Draw \(\overline{A P}\) and construct an angle ∠1 on n at P so that ∠PAB and ∠1 are corresponding angles
Hence,
The representation of the complete figure is:

PROVING A THEOREM
In Exercises 11 and 12. prove the theorem.
Question 11.
Alternate Exterior Angles Converse (Theorem 3.7)
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 11

Question 12.
Consecutive Interior Angles Converse (Theorem 3.8)
Answer:
Proof of the Converse of the Consecutive Interior angles Theorem:
Given: m∠5 + m∠4 = 180°
Prove: AB || CD

Now,
a.  m∠5 + m∠4 = 180° //From the given statement
b.  m∠1 + m∠4 = 180° // Linear pair of angles are supplementary
c. m∠5=m∠1 // (1), (2), transitive property of equality
d.  AB||CD // Converse of the Corresponding Angles Theorem

In Exercises 13-18. decide whether there is enough information to prove that m || n. If so, state the theorem you would use.

Question 13.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 75
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 13

Question 14.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 76
Answer:
Yes, there is enough information to prove m || n
The theorem we can use to prove that m || n is: Alternate Exterior angles Converse theorem

Question 15.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 77
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 15

Question 16.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 78
Answer:
No, there is no enough information to prove m || n

Question 17.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 79
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 17

Question 18.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 80
Answer:
Yes, there is enough information to prove m || n
The theorem we can use to prove that m || n is: Alternate Exterior angles Converse theorem

ERROR ANALYSIS
In Exercises 19 and 20, describe and correct the error in the reasoning.

Question 19.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 81
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 19

Question 20.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 82
Answer:
The given figure shows that angles 1 and 2 are Consecutive Interior angles
It also shows that a and b are cut by a transversal and they have the same length
So,
From the converse of the Consecutive Interior angles Theorem,
We can conclude that a || b

In Exercises 21-24. are Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 83 and Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 84 parallel? Explain your reasoning.

Question 21.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 85
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 21

Question 22.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 86
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 86
From the given figure,
We can observe that
The sum of the given angle measures is: 180°
From the given figure,
We can observe that the given angles are consecutive exterior angles
So,
From the Consecutive Exterior angles Converse,
We can conclude that AC || DF

Question 23.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 87
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 23

Question 24.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 88
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 88
From the given figure,
We can observe that the sum of the angle measures of all the pairs i.e., (115 + 65)°, (115 + 65)°, and (65 + 65)° is not 180°
Since,
The sum of the angle measures are not supplementary, according to the Consecutive Exterior Angles Converse,
AC is not parallel to DF

Question 25.
ANALYZING RELATIONSHIPS
The map shows part of Denser, Colorado, Use the markings on the map. Are the numbered streets parallel to one another? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 89
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 25

Question 26.
ANALYZING RELATIONSHIPS
Each rung of the ladder is parallel to the rung directly above it. Explain why the top rung is parallel to the bottom rung.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 90
Answer:
When we observe the ladder,
The rungs are not intersecting at any point i.e., they have different points
We know that,
The parallel lines do not have any intersecting points
Hence, from the above,
We can conclude that the top rung is parallel to the bottom rung

Question 27.
MODELING WITH MATHEMATICS
The diagram of the control bar of the kite shows the angles formed between the Control bar and the kite lines. How do you know that n is parallel to m?
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 91
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 27

Question 28.
REASONING
Use the diagram. Which rays are parallel? Which rays are not parallel? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 92
Answer:

Question 29.
ATTENDING TO PRECISION
Use the diagram. Which theorems allow you to conclude that m || n? Select all that apply. Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 93
(A) Corresponding Angles Converse (Thm 3.5)
(B) Alternate Interior Angles Converse (Thm 3.6)
(C) Alternate Exterior Angles Converse (Thm 3.7)
(D) Consecutive Interior Angles Converse (Thm 3.8)
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 29

Question 30.
MODELING WITH MATHEMATICS
One way to build stairs is to attach triangular blocks to angled support, as shown. The sides of the angled support are parallel. If the support makes a 32° angle with the floor, what must m∠1 so the top of the step will be parallel to the floor? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 94
Answer:
It is given that the sides of the angled support are parallel and the support makes a 32° angle with the floor
So,
To make the top of the step where ∠1 is present to be parallel to the floor, the angles must be “Alternate Interior angles”
We know that,
The “Alternate Interior angles” are congruent
So,
∠1 = 32°
Hence, from the above,
We can conclude that
∠1 = 32°

Question 31.
ABSTRACT REASONING
In the diagram, how many angles must be given to determine whether j || k? Give four examples that would allow you to conclude that j || k using the theorems from this lesson.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 95
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 31

Question 32.
THOUGHT-PROVOKING
Draw a diagram of at least two lines cut by at least one transversal. Mark your diagram so that it cannot be proven that any lines are parallel. Then explain how your diagram would need to change in order to prove that lines are parallel.
Answer:
The diagram that represents the figure that it can not be proven that any lines are parallel is:

From the above,
The diagram can be changed by the transformation of transversals into parallel lines and a parallel line into transversal
Hence,
The diagram that represents the figure that it can be proven that the lines are parallel is:

PROOF
In Exercises 33-36, write a proof.

Question 33.
Given m∠1 = 115°, m∠2 = 65°
Prove m||n
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 96
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 33

Question 34.
Given ∠1 and ∠3 are supplementary.
Prove m||n
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 97
Answer:
Given: ∠1 and ∠3 are supplementary
Prove: m || n
Hence,

Question 35.
Given ∠1 ≅ ∠2, ∠3 ≅ ∠4
Prove \(\overline{A B} \| \overline{C D}\)
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 98
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 35

Question 36.
Given a||b, ∠2 ≅ ∠3
Prove c||d
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 99
Answer:
Given: a || b, ∠2 ≅ ∠3
Prove: c || d
Hence,

Question 37.
MAKING AN ARGUMENT
Your classmate decided that Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 100 based on the diagram. Is your classmate correct? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 101
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 37

Question 38.
HOW DO YOU SEE IT?
Are the markings on the diagram enough to conclude that any lines are parallel? If so. which ones? If not, what other information is needed?
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 102
Answer:
The given diagram is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 102
From the given diagram,
We can observe that
∠1 and ∠4; ∠2 and ∠3 are the pairs of corresponding angles
We know that,
According to the Converse of the Corresponding angles Theorem,
If the corresponding angles are congruent, then the two lines that cut by a transversal are parallel lines
Hence,
We can conclude that p and q; r and s are the pairs of parallel lines

Question 39.
PROVING A THEOREM
Use these steps to prove the Transitive Property of Parallel Lines Theorem
a. Cops the diagram with the Transitive Property of Parallel Lines Theorem on page 141.
b. Write the Given and Prove statements.
c. Use the properties of angles formed by parallel lines cut by a transversal to prove the theorem.
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 39

Question 40.
MATHEMATICAL CONNECTIONS
Use the diagram
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 103
a. Find the value of x that makes p || q.
Answer:
From the given figure,
We can observe that when p || q,
The angles are: (2x + 2)° and (x + 56)°
We can observe that the given angles are corresponding angles
Hence,
(2x + 2)° = (x + 56)°
2x – x = 56° – 2°
x° = 54°
Hence, from the above,
We can conclude that the value of x when p || q is: 54°

b. Find the value of y that makes r || s.
Answer:
We can observe that when r || s,
The angles are (y + 7)° and (3y – 17)°
We can observe that the given angles are corresponding angles
Hence,
(y + 7)° = (3y – 17)°
y – 3y = -17° – 7°
-2y° = -24°
y = \(\frac{24}{2}\)
y = 12°
Hence, from the above,
We can conclude that the value of y when r || s is: 12°

c. Can r be parallel to s and can p, be parallel to q at the same time? Explain your reasoning.
Answer:
No, p ||q and r ||s will not be possible at the same time because when p || q, r, and s can act as transversal and when r || s, p, and q can act as transversal

Maintaining Mathematical Proficiency
Use the Distance Formula to find the distance between the two points.

Question 41.
(1, 3) and (- 2, 9)
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 41

Question 42.
(- 3, 7) and (8, – 6)
Answer:
The given points are:
(-3, 7), and (8, -6)
Compare the given points with (x1, y1), and (x2, y2)
We know that,
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
d = \(\sqrt{(8 + 3)² + (7 + 6)²}\)
d = \(\sqrt{(11)² + (13)²}\)
d = \(\sqrt{290}\)
d = 17.02
Hence, from the above,
We can conclude that the distance between the given 2 points is: 17.02

Question 43.
(5, – 4) and (0, 8)
Answer:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 3.3 a 43

Question 44.
(13, 1) and (9, – 4)
Answer:
The given points are:
(13, 1), and (9, -4)
Compare the given points with (x1, y1), and (x2, y2)
We know that,
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
d = \(\sqrt{(13 – 9)² + (1 + 4)²}\)
d = \(\sqrt{(4)² + (5)²}\)
d = \(\sqrt{41}\)
d = 6.40
Hence, from the above,
We can conclude that the distance between the given 2 points is: 6.40

3.1 – 3.3 Study Skills: Analyzing Your Errors

Mathematical Practices

Question 1.
Draw the portion of the diagram that you used to answer Exercise 26 on page 130.
Answer:
The portion of the diagram that you used to answer Exercise 26 on page 130 is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 31

Question 2.
In Exercise 40 on page 144. explain how you started solving the problem and why you started that way.
Answer:
In Exercise 40 on page 144,
You started solving the problem by considering the 2 lines parallel and two lines as transversals
So,
If p and q are the parallel lines, then r and s are the transversals
If r and s are the parallel lines, then p and q are the transversals

3.1 – 3.3 Quiz

Think of each segment in the diagram as part of a line. Which lines(s) or plane(s) contain point G and appear to fit the description?

Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 104

Question 1.
line(s) parallel to Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 105.
Answer:
The line parallel to \(\overline{E F}\) is: \(\overline{D H}\)

Question 2.
line(s) perpendicular to Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 105.
Answer:
The lines perpendicular to \(\overline{E F}\) are: \(\overline{F B}\) and \(\overline{F G}\)

Question 3.
line(s) skew to Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 105.
Answer:
The lines skew to \(\overline{E F}\) are: \(\overline{C D}\), \(\overline{C G}\), and \(\overline{A E}\)

Question 4.
plane(s) parallel to plane ADE
Answer:
The plane parallel to plane ADE is: Plane GCB

Identify all pairs of angles of the given type.

Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 106

Question 5.
consecutive interior
Answer:
We know that,
When two lines are cut by a transversal, the pair of angles on one side of the transversal and inside the two lines are called the consecutive interior angles.
Hence, from the given figure,
We can conclude that the consecutive interior angles are: 3 and 5; 4 and 6

Question 6.
alternate interior
Answer:
We know that,
Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal.
Hence, from the given figure,
We can conclude that the alternate interior angles are: 3 and 6; 4 and 5

Question 7.
corresponding
Answer:
We know that,
When two lines are crossed by another line (which is called the Transversal), the angles in matching corners are called corresponding angles.
Hence, from the given figure,
We can conclude that the corresponding angles are: 1 and 5; 3 and 7; 2 and 4; 6 and 8

Question 8.
alternate exterior
Answer:
Alternate exterior angles are the pair of angles that lie on the outer side of the two parallel lines but on either side of the transversal line
Hence, from the above figure,
We can conclude that the alternate exterior angles are: 1 and 8; 7 and 2

Find m∠1 and m∠2. Tell which theorem you use in each case.

Question 9.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 107
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 107
From the given figure,
By using the linear pair theorem,
∠1 + 138° = 180°
∠1 = 180° – 138°
∠1 = 42°
Now,
By using the Alternate Exterior Angles Theorem,
∠1 = ∠2
Hence, from the above,
We can conclude that
∠1 = ∠2 = 42°

Question 10.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 108
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 108
From the given figure,
We can observe that
By using the Vertical Angles Theorem,
∠2 = 123°
Now,
By using the vertical Angles Theorem,
∠1 = ∠2
Hence, from the above,
We can conclude that
∠1 = ∠2 = 123°

Question 11.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 109
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 109
From the given figure,
By using the linear pair theorem,
∠1 + 57° = 180°
∠1 = 180° – 57°
∠1 = 123°
Now,
By using the consecutive interior angles theorem,
∠1 + ∠2 = 180°
∠2 = 180° – 123°
∠2 = 57°
Hence, from the above,
We can conclude that
∠1 = 123° and ∠2 = 57°

Decide whether there is enough information to prove that m || n. If so, state the theorem you would use.

Question 12.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 110
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 110
We know that,
By using the “Consecutive Interior angles Converse”,
If the angle measure of the angles is a supplementary angle, then the lines cut by a transversal are parallel
Now,
69° + 111° = 180°
Hence, from the above,
We can conclude that m || n by using the Consecutive Interior angles Theorem

Question 13.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 111
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 111
We know that,
By using the Corresponding Angles Theorem,
If the corresponding angles are congruent, then the lines cut by a transversal are parallel
Hence, from the above,
We can conclude that m || n by using the Corresponding Angles Theorem

Question 14.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 112
Answer:
The given figure is:
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 112
From the given figure,
It is given that l || m and l || n,
So,
We know that,
By using the parallel lines property,
If a || b and b || c, then a || c
Hence, from the above,
We can conclude that m || n

Question 15.
Cellular phones use bars like the ones shown to indicate how much signal strength a phone receives from the nearest service tower. Each bar is parallel to the bar directly next to it.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 113
a. Explain why the tallest bar is parallel to the shortest bar.
Answer:
From the given bars,
We can observe that there is no intersection between any bars
If we represent the bars in the coordinate plane, we can observe that the number of intersection points between any bar is: 0
We know that,
The number of intersection points for parallel lines is: 0
Hence, from the above,
We can conclude that the tallest bar is parallel to the shortest bar

b. Imagine that the left side of each bar extends infinitely as a line.
If m∠1 = 58°, then what is m∠2?
Answer:
From the given figure,
We can observe that ∠1 and ∠2 are the consecutive interior angles
We know that,
The sum of the angle measure between 2 consecutive interior angles is: 180°
So,
∠1 + ∠ 2 = 180°
It is given that ∠1 = 58°
So,
∠2 = 180° – 58°
∠2 = 122°
Hence, from the above,
We can conclude that
∠2 = 122°

Question 16.
The diagram shows lines formed on a tennis court.
Big Ideas Math Geometry Answer Key Chapter 3 Parallel and Perpendicular Lines 114
a. Identify two pairs of parallel lines so that each pair is in a different plane.
Answer:
From the given figure,
We can observe that there are a total of 5 lines.
Hence,
The two pairs of parallel lines so that each pair is in a different plane are: q and p; k and m

b. Identify two pairs of perpendicular lines.
Answer:
Fro the given figure,
We can observe that there are 2 perpendicular lines
Hence, from the above,
The two pairs of perpendicular lines are l and n

c. Identify two pairs of skew line
Answer:
From the given figure,
We can observe that there are 2 pairs of skew lines
Hence,
The 2 pair of skew lines are: q and p; l and m

d. Prove that ∠1 ≅ ∠2.
Answer:
From the given figure,
We can observe that ∠1 and ∠2 are the alternate exterior angles
We know that,
According to the Alternate Exterior angles Theorem,
If the line cut by a transversal is parallel, then the corresponding angles are congruent
Hence,
According to the above theorem,
We can conclude that ∠1 ≅ ∠2

3.4 Proofs with Perpendicular Lines

Exploration 1

Writing Conjectures

Work with a partner: Fold a piece of pair in half twice. Label points on the two creases. as shown.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 115
a. Write a conjecture about \(\overline{A B}\) and \(\overline{C D}\). Justify your conjecture.
Answer:
The conjecture about \(\overline{A B}\) and \(\overline{c D}\) is:
If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular.

b. Write a conjecture about \(\overline{A O}\) and \(\overline{O B}\) Justify your conjecture.
Answer:
The conjecture about \(\overline{A O}\) and \(\overline{O B}\) is:
In a plane, if two lines are perpendicular to the same line, then they are parallel to each other.

Exploration 2

Exploring a segment Bisector

Work with a partner: Fold and crease a piece of paper. as shown. Label the ends of the crease as A and B.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 116
a. Fold the paper again so that point A coincides with point B. Crease the paper on that fold.
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 116

b. Unfold the paper and examine the four angles formed by the two creases. What can you conclude about the four angles?
Answer:
When we unfold the paper and examine the four angles formed by the two creases, we can conclude that the four angles formed are the right angles i.e., 90°

Exploration 3

Writing a conjecture

Work with a partner.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 117
a. Draw \(\overline{A B}\), as shown.
b. Draw an arc with center A on each side of AB. Using the same compass selling, draw an arc with center B on each side \(\overline{A B}\). Label the intersections of arcs C and D.
c. Draw \(\overline{C D}\). Label its intersection with \(\overline{A B}\) as O. Write a conjecture about the resulting diagram. Justify your conjecture.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.
Answer:
The resultant diagram is:

From the above diagram,
We can conclude that
The angles formed at all the intersection points are: 90°
The lengths of the line segments are equal i.e., AO = OB and CO = OD

Communicate Your Answer

Question 4.
What conjectures can you make about perpendicular lines?
Answer:
The conjectures about perpendicular lines are:
a. If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular.
b. In a plane, if a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other line. c. In a plane, if two lines are perpendicular to the same line, then they are parallel to each other.

Question 5.
In Exploration 3. find AO and OB when AB = 4 units.
Answer:
From Exploration 2,
We can observe that,
AB = AO + OB
AO = OB
It is given that
AB = 4 units
Hence,
The values of AO and OB are: 2 units

Lesson 3.4 Proofs with Perpendicular Lines

Monitoring Progress

Question 1.
Find the distance from point E to Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 118
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 119
Answer:
The given figure is:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 119
It is given that E is ⊥ to \(\overline{F H}\)
So,
To find the distance between E and \(\overline{F H}\), we need to find the distance between E and G i.e., EG
Now,
From the coordinate plane,
E (-4, -3), G (1, 2)
Compare the given points with
E (x1, y1), G (x2, y2)
So,
EG = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
EG = \(\sqrt{(1 + 4)² + (2 + 3)²}\)
EG = \(\sqrt{(5)² + (5)²}\)
EG = \(\sqrt{50}\)
EG = 7.07
Hence, from the above,
We can conclude that the distance from point E to \(\overline{F H}\) is: 7.07

Question 2.
Prove the Perpendicular Transversal Theorem using the diagram in Example 2 and the Alternate Exterior Angles Theorem (Theorem 3.3).
Answer:
Perpendicular transversal theorem:
In a plane, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also

Proof:
Given: k || l, t ⊥ k
Prove: t ⊥ l

Alternate Exterior Angles Theorem:
The Alternate Exterior Angles Theorem states that, when two parallel lines are cut by a transversal, the resulting alternate exterior angles are congruent

Proof:
Given: k || l
Prove: ∠1 ≅ ∠7 and ∠4 ≅ ∠6
Since k || l, by the Corresponding Angles Postulate,
∠1 ≅ ∠5
Also, by the Vertical Angles Theorem,
∠5 ≅ ∠7
Then, by the Transitive Property of Congruence,
∠1 ≅ ∠7
You can prove that 4 and 6 are congruent using the same method.

Use the lines marked in the photo.

Question 3.
Is b || a? Explain your reasoning.
Answer:
From the given figure,
There is not any intersection between a and b
Hence, from the above,
We can conclude that b || a

Question 4.
Is b ⊥ c? Explain your reasoning.
Answer:
From the given figure,
We can observe that the angle between b and c is 90°
Hence, from the above,
We can conclude that b is perpendicular to c

Exercise 3.4 Proofs with Perpendicular Lines

Vocabulary and core Concept Check

Question 1.
COMPLETE THE SENTENCE
The perpendicular bisector of a segment is the line that passes through the _______________ of the segment at a _______________ angle.
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 120
Find the distance from point X to Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 121
Answer:
The given figure is:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 120
To find the distance from point X to \(\overline{W Z}\),
We have to find the distance between X and Y i.e., XY
Now,
The given points are:
X (-3, 3), Y (3, 1)
Compare the given points with
(x1, y1), (x2, y2)
Now,
XY = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
XY = \(\sqrt{(3 + 3)² + (3 – 1)²}\)
XY = \(\sqrt{(6)² + (2)²}\)
XY = 6.32
Hence, from the above,
We can conclude that the distance from point X to \(\overline{W Z}\) is: 6.32

Find XZ
Answer:
The given points are:
X (-3, 3), Z (4, 4)
Compare the given points with
(x1, y1), (x2, y2)
Now,
XZ = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
XZ = \(\sqrt{(4 + 3)² + (3 – 4)²}\)
XZ = \(\sqrt{(7)² + (1)²}\)
XZ = 7.07
Hence, from the above,
We can conclude that the value of XZ is: 7.07

Find the length of \(\overline{X Y}\)
Answer:
The given points are:
X (-3, 3), Y (3, 1)
Compare the given points with
(x1, y1), (x2, y2)
Now,
XY = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
XY = \(\sqrt{(3 + 3)² + (3 – 1)²}\)
XY = \(\sqrt{(6)² + (2)²}\)
XY = 6.32
Hence, from the above,
We can conclude that the value of XY is: 6.32

Find the distance from line l to point X.
Answer:
To find the distance from line l to point X,
We have to find the distance between X and Y i.e., XY
Now,
The given points are:
X (-3, 3), Y (3, 1)
Compare the given points with
(x1, y1), (x2, y2)
Now,
XY = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
XY = \(\sqrt{(3 + 3)² + (3 – 1)²}\)
XY = \(\sqrt{(6)² + (2)²}\)
XY = 6.32
Hence, from the above,
We can conclude that the distance from line l to point X is: 6.32

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4. find the distance from point A to Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 122.

Question 3.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 123
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 3

Question 4.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 124
Answer:
To find the distance from point A to \(\overline{X Z}\),
We have to find the distance between A and Y i.e., AY
Now,
The given points are:
X (3, 3), Y (2, -1.5)
Compare the given points with
(x1, y1), (x2, y2)
Now,
XY = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
XY = \(\sqrt{(3 + 1.5)² + (3 – 2)²}\)
XY = \(\sqrt{(4.5)² + (1)²}\)
XY = 4.60
Hence, from the above,
We can conclude that the distance from point A to \(\overline{X Z}\) is: 4.60

CONSTRUCTION
In Exercises 5-8, trace line m and point P. Then use a compass and straightedge to construct a line perpendicular to line m through point P.

Question 5.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 125
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 5

Question 6.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 126
Answer:
The given figure is:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 126
Now,
Using P as the center, draw two arcs intersecting with line m.
Label the intersections as points X and Y.
Using X and Y as centers and an appropriate radius, draw arcs that intersect.
Label the intersection as Z. Draw \(\overline{P Z}\)

Question 7.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 127
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 7

Question 8.
Math
Answer:
The given figure is:
Math
Now,
Using P as the center and any radius, draw arcs intersecting m and label those intersections as X and Y.
Using X as the center, open the compass so that it is greater than half of XP and draw an arc.
Using Y as the center and retaining the same compass setting, draw an arc that intersects with the first
Label the point of intersection as Z. Draw \(\overline{P Z}\)

CONSTRUCTION
In Exercises 9 and 10, trace \(\overline{A B}\). Then use a compass and straightedge to construct the perpendicular bisector of \(\overline{A B}\)

Question 9.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 129
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 9

Question 10.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 130
Answer:
The given figure is:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 130
Now,
Using a compass setting greater than half of AB, draw two arcs using A and B as centers
Connect the points of intersection of the arcs with a straight line

ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in the statement about the diagram.
Question 11.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 131
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 11

Question 12.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 132
Answer:
We know that,
According to the Perpendicular Transversal theorem,
The distance from the perpendicular to the line is given as the distance between the point and the non-perpendicular line
So,
From the given figure,
The distance from point C to AB is the distance between point C and A i.e., AC
Hence, from the above,
We can conclude that the distance from point C to AB is: 12 cm

PROVING A THEOREM 
In Exercises 13 and 14, prove the theorem.
Question 13.
Linear Pair Perpendicular Theorem (Thm. 3. 10)
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 13

Question 14.
Lines Perpendicular to a Transversal Theorem (Thm. 3.12)
Answer:
In a plane, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also

Proof:
Given: k || l, t ⊥ k
Prove: t ⊥ l

PROOF
In Exercises 15 and 16, use the diagram to write a proof of the statement.

Question 15.
If two intersecting lines are perpendicular. then they intersect to form four right angles.
Given a ⊥ b
Prove ∠1, ∠2, ∠3, and ∠4 are right angles.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 133
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 15.1
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 15.2

Question 16.
If two sides of two adjacent acute angles are perpendicular, then the angles are complementary.
Given \(\overrightarrow{B A}\) ⊥\(\vec{B}\)C
Prove ∠1 and ∠2 are complementary
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 134
Answer:

In Exercises 17-22, determine which lines, if any, must be parallel. Explain your reasoning.

Question 17.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 135
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 17

Question 18.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 136
Answer:
The given figure is:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 136
From the given figure,
We can observe that a is perpendicular to both the lines b and c
Hence, from the above,
We can conclude that
By using the Perpendicular transversal theorem,
a is both perpendicular to b and c and b is parallel to c

Question 19.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 139
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 19

Question 20.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 140
Answer:
The given figure is;
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 140
From the given figure,
We can observe that
a is perpendicular to d and b is perpendicular to c
For parallel lines, we can’t say anything
Hence, from the above,
By using the Perpendicular transversal theorem,
a is perpendicular to d and b isperpendicular to c

Question 21.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 141
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 21

Question 22.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 142
Answer:
The given figure is:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 142
From the given figure,
We can observe that
w ⊥ v and w⊥ y
So,
We can say that w and v are parallel lines by “Perpendicular Transversal Theorem”
We can observe that
z ⊥ x and w ⊥ z
So,
We can say that w and x are parallel lines by “Perpendicular Transversal theorem”

Question 23.
USING STRUCTURE
Find all the unknown angle measures in the diagram. Justify your answer for cacti angle measure.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 143
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 23

Question 24.
MAKING AN ARGUMENT
Your friend claims that because you can find the distance from a point to a line, you should be able to find the distance between any two lines. Is your friend correct? Explain your reasoning.
Answer:
No, your friend is not correct

Explanation:
It is given that your friend claims that because you can find the distance from a point to a line, you should be able to find the distance between any two lines
Now,
This can be proven by following the below steps:

Hence, from the above,
We can conclude that your friend is not correct

Question 25.
MATHEMATICAL CONNECTIONS
Find the value of x when a ⊥ b and b || c.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 144
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 25

Question 26.
HOW DO YOU SEE IT?
You are trying to cross a stream from point A. Which point should you jump to in order to jump the shortest distance? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 145
Answer:
From the given figure,
We can observe that
Point A is perpendicular to Point C
We know that,
According to Perpendicular Transversal Theorem,
The distance between the perpendicular points is the shortest
Hence, from the above,
We can conclude that in order to jump the shortest distance, you have to jump to point C from point A

Question 27.
ATTENDING TO PRECISION
In which of the following diagrams is \(\overline{A C}\) || \(\overline{B D}\) and \(\overline{A C}\) ⊥ \(\overline{C D}\)? Select all that apply.
(A) Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 146
(B) Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 147
(C) Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 148
(D) Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 149
(E) Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 150
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 27

Question 28.
THOUGHT-PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry. how many right angles are formed by two perpendicular lines? Justify your answer.
Answer:
It is given that in spherical geometry, all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere.
We know that,
In Euclidean geometry, the two perpendicular lines form 4 right angles whereas, In spherical geometry, the two perpendicular lines form 8 right angles according to the “Parallel lines Postulate” in spherical geometry.
Hence, from the above,
We can conclude that 8 right angles are formed by two perpendicular lines in spherical geometry

Question 29.
CONSTRUCTION
Construct a square of side length AB
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 151
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 29

Question 30.
ANALYZING RELATIONSHIPS
The painted line segments that brain the path of a crosswalk are usually perpendicular to the crosswalk. Sketch what the segments in the photo would look like if they were perpendicular to the crosswalk. Which type of line segment requires less paint? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 152
Answer:

Question 31.
ABSTRACT REASONING
Two lines, a and b, are perpendicular to line c. Line d is parallel to line c. The distance between lines a and b is x meters. The distance between lines c and d is y meters. What shape is formed by the intersections of the four lines?
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 31

Question 32.
MATHEMATICAL CONNECTIONS
Find the distance between the lines with the equations y = \(\frac{3}{2}\) + 4 and – 3x + 2y = – 1.
Answer:
The given equations are:
y = \(\frac{3}{2}\) + 4 and -3x + 2y = -1
y = \(\frac{3}{2}\) + 4 and y = \(\frac{3}{2}\)x – \(\frac{1}{2}\)
Compare the given equations with
y = mx + c
So,
m1 = m2 = \(\frac{3}{2}\)
c1 = 4
c2= –\(\frac{1}{2}\)
We know that,
The distance between the two parallel lines is:
d = | c1 – c2 |
So,
The distance between the given 2 parallel lines = | c1 – c2 |
= | 4 + \(\frac{1}{2}\) |
= \(\frac{9}{2}\)
Hence, from the above,
We can conclude that the distance between the given lines is: \(\frac{7}{2}\)

Question 33.
WRITING
Describe how you would find the distance from a point to a plane. Can you find the distance from a line to a plane? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 33

Maintaining Mathematical Proficiency

Simplify the ratio.

Question 34.
\(\frac{6-(-4)}{8-3}\)
Answer:
The given expression is:
\(\frac{6 – (-4)}{8 – 3}\)
= \(\frac{6 + 4}{8 – 3}\)
= \(\frac{10}{5}\)
= 2
Hence,f rom the above,
We can conclude that the value of the given expression is: 2

Question 35.
\(\frac{3-5}{4-1}\)
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 35

Question 36.
\(\frac{8-(-3)}{7-(-2)}\)
Answer:
The given expression is:
\(\frac{8 – (-3)}{7 – (-2)}\)
= \(\frac{8 + 3}{7 + 2}\)
= \(\frac{11}{9}\)
Hence,f rom the above,
We can conclude that the value of the given expression is: \(\frac{11}{9}\)

Question 37.
\(\frac{13-4}{2-(-1)}\)
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 37

Identify the slope and the y-intercept of the line.

Question 38.
y = 3x + 9
Answer:
The given equation is:
y = 3x + 9 ——-(1)
Compare the above equation with
y = mx + c
Where,
m is the slope
c is the y-intercept
By comparing eq. (1) with the y = mx + c,
m = 3 and c = 9
Hence, from the above,
We can conclude that
The slope is: 3
The y-intercept is: 9

Question 39.
y = –\(\frac{1}{2}\)x + 7
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 39

Question 40.
y = \(\frac{1}{6}\)x – 8
Answer:
The given equation is:
y = \(\frac{1}{6}\)x – 8
Compare the above equation with
y = mx + c
So,
m = \(\frac{1}{6}\) and c = -8
Hence, from the above,
We can conclude that
The slope is: \(\frac{1}{6}\)
The y-intercept is: -8

Question 41.
y = – 8x – 6
Answer:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 3.4 a 41

3.5 Equations of Parallel and Perpendicular Lines

Exploration 1

Writing Equations of Parallel and Perpendicular Lines

Work with a partner: Write an equation of the line that is parallel or perpendicular to the given line and passes through the given point. Use a graphing calculator to verify your answer. What is the relationship between the slopes?
a.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 153
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 153
We know that,
From the given figure,
We can observe that the given lines are parallel lines
Now,
The equation for another line is:
y = \(\frac{3}{2}\)x + c
Substitute (0, 2) in the above equation
So,
2 = 0 + c
c = 2
So,
The equation for another parallel line is:
y = \(\frac{3}{2}\)x + 2
When we compare the given equation with the obtained equation,
We can observe that the slopes are the same and the y-intercepts are different
We know that,
The lines that have the same slope and different y-intercepts are “Parallel lines”
Hence, from the above,
We can conclude that the parallel lines are:
y = \(\frac{3}{2}\)x – 1
y = \(\frac{3}{2}\)x + 2

b.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 154
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 154
We know that,
From the given figure,
We can observe that the given lines are perpendicular lines
The equation for another line is:
y = \(\frac{3}{2}\)x + c
We know that,
The slope of perpendicular lines is: -1
So,
m1m2 = -1
\(\frac{3}{2}\) . m2 = -1
So,
m2 = –\(\frac{2}{3}\)
So,
y = –\(\frac{2}{3}\)x + c
Substitute (0, 1) in the above equation
So,
1 = 0 + c
c = 1
So,
The equation for another perpendicular line is:
y = –\(\frac{2}{3}\)x + 1
When we compare the given equation with the obtained equation,
We can observe that the product of the slopes are -1 and the y-intercepts are different
We know that,
The lines that have the slopes product -1 and different y-intercepts are “Perpendicular lines”
Hence, from the above,
We can conclude that the perpendicular lines are:
y = \(\frac{3}{2}\)x – 1
y = –\(\frac{2}{3}\)x + 1

c.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 155
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 155
We know that,
From the given figure,
We can observe that the given lines are parallel lines
Now,
The equation for another line is:
y = \(\frac{1}{2}\)x + c
Substitute (2, -2) in the above equation
So,
-2 = \(\frac{1}{2}\) (2) + c
-2 = 1 + c
c = 2 – 1
c = -3
So,
The equation for another parallel line is:
y = \(\frac{1}{2}\)x – 3
When we compare the given equation with the obtained equation,
We can observe that the slopes are the same and the y-intercepts are different
We know that,
The lines that have the same slope and different y-intercepts are “Parallel lines”
Hence, from the above,
We can conclude that the parallel lines are:
y = \(\frac{1}{2}\)x + 2
y = \(\frac{1}{2}\)x – 3

d.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 156
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 156
We know that,
From the given figure,
We can observe that the given lines are perpendicular lines
The equation for another line is:
y = \(\frac{1}{2}\)x + c
We know that,
The slope of perpendicular lines is: -1
So,
m1m2 = -1
\(\frac{1}{2}\) . m2 = -1
So,
m2 = -2
So,
y = -2x + c
Substitute (2, -3) in the above equation
So,
-3 = -2 (2) + c
-3 = -4 + c
c = 4 – 3
c = 1
So,
The equation for another perpendicular line is:
y = -2x + 1
When we compare the given equation with the obtained equation,
We can observe that the product of the slopes are -1 and the y-intercepts are different
We know that,
The lines that have the slopes product -1 and different y-intercepts are “Perpendicular lines”
Hence, from the above,
We can conclude that the perpendicular lines are:
y = \(\frac{1}{2}\)x + 2
y = -2x + 1

e.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 157
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 157
We know that,
From the given figure,
We can observe that the given lines are parallel lines
Now,
The equation for another line is:
y = -2x + c
Substitute (0, -2) in the above equation
So,
-2 = 0 + c
c = -2
So,
The equation for another parallel line is:
y = -2x – 2
When we compare the given equation with the obtained equation,
We can observe that the slopes are the same and the y-intercepts are different
We know that,
The lines that have the same slope and different y-intercepts are “Parallel lines”
Hence, from the above,
We can conclude that the parallel lines are:
y = -2x + 2
y = -2x – 2

f.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 158
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 158
We know that,
From the given figure,
We can observe that the given lines are perpendicular lines
The equation for another line is:
y = -2x + c
We know that,
The slope of perpendicular lines is: -1
So,
m1m2 = -1
-2 . m2 = -1
So,
m2 = \(\frac{1}{2}\)
So,
y = \(\frac{1}{2}\)x + c
Substitute (4, 0) in the above equation
So,
0 = \(\frac{1}{2}\) (4) + c
0 = 2 + c
c = 0 – 2
c = -2
So,
The equation for another perpendicular line is:
y = \(\frac{1}{2}\)x – 2
When we compare the given equation with the obtained equation,
We can observe that the product of the slopes are -1 and the y-intercepts are different
We know that,
The lines that have the slopes product -1 and different y-intercepts are “Perpendicular lines”
Hence, from the above,
We can conclude that the perpendicular lines are:
y = \(\frac{1}{2}\)x – 2
y = -2x + 2

Exploration 2

Writing Equations of Parallel and Perpendicular Lines

Work with a partner: Write the equations of the parallel or perpendicular lines. Use a graphing calculator to verify your answers.

a.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 159
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 159
From the given graph,
We can observe that
The given lines are the parallel lines
Now,
The coordinates of the line of the first equation are: (-1.5, 0), and (0, 3)
The coordinates of the line of the second equation are: (1, 0), and (0, -2)
Compare the given coordinates with
A (x1, y1), and B (x2, y2)
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
Now,
The slope of the line of the first equation is:
m = \(\frac{3 – 0}{0 + 1.5}\)
m = \(\frac{3}{1.5}\)
m = 2
Now,
We know that,
The standard linear equation is:
y = mx + c
So,
y = 2x + c
We know that,
For parallel lines,
The slopes are the same but the y-intercepts are different
Hence,
The given parallel line equations are:
y = 2x + c1
y = 2x + c2

b.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 160
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 160
From the given figure,
We can observe that
The given lines are perpendicular lines
So,
The coordinates of the line of the first equation are: (0, -3), and (-1.5, 0)
The coordinates of the line of the second equation are: (-4, 0), and (0, 2)
Compare the given coordinates with
A (x1, y1), and B (x2, y2)
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
Now,
The slope of the line of the first equation is:
m = \(\frac{0 + 3}{0 – 1.5}\)
m = \(\frac{3}{-1.5}\)
m = \(\frac{-30}{15}\)
m = -2
Now,
We know that,
The standard linear equation is:
y = mx + c
So,
y = -2x + c
We know that,
For perpediclar lines,
The product of the slopes is -1 and the y-intercepts are different
So,
m1 × m2 = -1
-2 × m2 = -1
m2 = \(\frac{1}{2}\)
Hence,
The given perpendicular line equations are:
y = -2x + c1
y = \(\frac{1}{2}\)x + c2

Communicate Your Answer

Question 3.
How can you write an equation of a line that is parallel or perpendicular to a given line and passes through a given point?
MODELING WITH MATHEMATICS
To be proficient in math, you need to analyze relationships mathematically to draw conclusions.
Answer:
We know that,
The standard form of a linear equation is:
y = mx + c
Now,
For parallel lines,
We know that,
The slopes are the same and the y-intercepts are different
So,
To find the y-intercept of the equation that is parallel to the given equation, substitute the given point and find the value of c
Now,
For perpendicular lines,
we know that,
The product of the slopes is -1
So,
The slope of the equation that is perpendicular to the given equation is: –\(\frac{1}{m}\)
To find the y-intercept of the equation that is perpendicular to the given equation, substitute the given point and find the value of c

Question 4.
Write an equation of the line that is (a) parallel and (b) perpendicular to the line y = 3x + 2 and passes through the point (1, -2).
Answer:
The given equation is:
y = 3x + 2
The given point is: (1, -2)
a) Parallel line equation:
We know that,
The slope of the parallel equations are the same
So,
The slope of the equation that is parallel t the given equation is: 3
Now,
The equation that is parallel to the given equation is:
y = 3x + c
Substitute (1, -2) in the above equation
So,
-2 = 3 (1) + c
-2 – 3 = c
c = -5
Hence,
The equation of the line that is parallel to the given equation is:
y = 3x – 5
b) Perpendicular line equation:
We know that,
The product of the slope of the perpendicular equations is: -1
So,
m1 m2 = -1
3m2 = -1
So,
m2 = –\(\frac{1}{3}\)
So,
The slope of the equation that is parallel t the given equation is: –\(\frac{1}{3}\)
Now,
The equation that is perpendicular to the given equation is:
y = –\(\frac{1}{3}\)x + c
Substitute (1, -2) in the above equation
So,
-2 = –\(\frac{1}{3}\) (-2) + c
-2 – \(\frac{2}{3}\) = c
c = –\(\frac{8}{3}\)
Hence,
The equation of the line that is perpendicular to the given equation is:
y = –\(\frac{1}{3}\)x –\(\frac{8}{3}\)

Lesson 3.5 Equations of Parallel and Perpendicular Lines

Monitoring Progress

Find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio.

Question 1.
A(1, 3), B(8, 4); 4 to 1
Answer:
The given coordinates are: A (1, 3), and B (8, 4)
Compare the given points with
A (x1, y1), and B (x2, y2)
It is given that
AP : PB = 4 : 1
So,
We have to divide AB into 5 parts
Now,
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of AB = \(\frac{4 – 3}{8 – 1}\)
Slope of AB = \(\frac{1}{7}\)
Now,
To find the coordinates of P, add slope to AP and PB
So,
P = (4 + (4 / 5) × 7, 1 + (4 / 5) × 1)
P = (22.4, 1.8)
Hence, from he above,
We can conclude that
The coordinates of P are (22.4, 1.8)

Question 2.
A(- 2, 1), B(4, 5); 3 to 7
Answer:
The given coordinates are: A (-2, 1), and B (4, 5)
Compare the given points with
A (x1, y1), and B (x2, y2)
It is given that
AP : PB = 3 : 7
So,
We have to divide AB into 10 parts
Now,
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of AB = \(\frac{5 – 1}{4 + 2}\)
Slope of AB = \(\frac{4}{6}\)
Slope of AB = \(\frac{2}{3}\)
Now,
To find the coordinates of P, add slope to AP and PB
So,
P = (3 + (\(\frac{3}{10}\) × 3), 7 + (\(\frac{3}{10}\) × 2))
P = (3.9, 7.6)
Hence, from the above,
We can conclude that
The coordinates of P are (3.9, 7.6)

Question 3.
Determine which of the lines are parallel and which of the lines are perpendicular.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 161
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 161
From the given figure,
The coordinates of line a are: (0, 2), and (-2, -2)
The coordinates of line b are: (2, 3), and (0, -1)
The coordinates of line c are: (4, 2), and (3, -1)
The coordinates of line d are: (-3, 0), and (0, -1)
Now,
Compare the given coordinates with (x1, y1), and (x2, y2)
So,
The slope of line a (m) = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{-2 – 2}{-2 – 0}\)
= \(\frac{-4}{-2}\)
= 2

The slope of line b (m) = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{-1 – 3}{0 – 2}\)
= \(\frac{-4}{-2}\)
= 2

The slope of line c (m) = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{-1 – 2}{3 – 4}\)
= \(\frac{-3}{-1}\)
= 3

The slope of line d (m) = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{-1 – 0}{0 + 3}\)
= \(\frac{-1}{3}\)
= –\(\frac{1}{3}\)
We know that,
The parallel lines have the same slopes
The perpendicular lines have the product of slopes equal to -1
Hence, from the above,
We can conclude that
Linea and Line b are parallel lines
Line c and Line d are perpendicular lines

Question 4.
Write an equation of the line that passes through the point (1, 5) and is
(a) parallel to the line y = 3x – 5 and
Answer:
The given equation is:
y = 3x – 5
The given point is: (1, 5)
We know that,
The parallel lines have the same slope
Compare the given equation with
y = mx + c
So,
The slope of the parallel line that passes through (1, 5) is: 3
So,
The equation of the parallel line that passes through (1, 5) is
y = 3x + c
To find the value of c, substitute (1, 5) in the above equation
So,
5 = 3 (1) + c
c = 5 – 3
c = 2
Hence,
The equation of the parallel line that passes through (1, 5) is:
y = 3x + 2

(b) perpendicular to the line y = 3x – 5.
Answer:
The given equation is:
y = 3x – 5
The given point is: (1, 5)
We know that,
The product of the slopes of the perpendicular lines is equal to -1
Now,
Compare the given equation with
y = mx + c
So,
m = 3
Now,
The slope of the perpendicular line that passes through (1, 5) is:
m1m2 = -1
3m2 = -1
m2 = –\(\frac{1}{3}\)
Now,
The equation of the perpendicular line that passes through (1, 5) is:
y = –\(\frac{1}{3}\)x + c
To find the value of c, substitute (1, 5) in the above equation
So,
5 = –\(\frac{1}{3}\) + c
c = 5 + \(\frac{1}{3}\)
c = \(\frac{16}{3}\)
Hence,
The equation of the perpendicular line that passes through (1, 5) is:
y = –\(\frac{1}{3}\)x + \(\frac{16}{3}\)

Question 5.
How do you know that the lines x = 4 and y = 2 are perpendiculars?
Answer:
The given lines are:
x = 4 and y = 2
We know that,
The line x = 4 is a vertical line that has the right angle i.e., 90°
The line y = 4 is a horizontal line that have the straight angle i.e., 0°
So,
The angle at the intersection of the 2 lines = 90° – 0° = 90°
Hence, from the above,
We can conclude that the lines x = 4 and y = 2 are perpendicular lines

Question 6.
Find the distance from the point (6, 4) to the line y = x + 4.
Answer:
The given equation is:
y = x + 4
Compare the given equation with
ax + by + c = 0
So,
x – y + 4 = 0
So,
a = 1, and b = -1
The given point is: (6, 4)
We know that,
The distance from the point (x, y) to the line ax + by + c = 0 is:
d = | ax + by + c| /\(\sqrt{a² + b²}\)
d = | x – y + 4 | / \(\sqrt{1² + (-1)²}\)
d = | x – y + 4 | / \(\sqrt{2}\)}
Substitute (6, 4) in the above equation
So,
d = | 6 – 4 + 4 |/ \(\sqrt{2}\)}
d = 3√2
Hence, from the above,
We can conclude that the distance from the given point to the given line is: 3√2

Question 7.
Find the distance from the point (- 1, 6) to the line y = – 2x.
Answer:
The given equation is:
y = -2x
Compare the given equation with
ax + by + c = 0
So,
2x + y = 0
So,
a = 2, and b = 1
The given point is: (-1, 6)
We know that,
The distance from the point (x, y) to the line ax + by + c = 0 is:
d = | ax + by + c| /\(\sqrt{a² + b²}\)
d = | 2x + y | / \(\sqrt{2² + (1)²}\)
d = | 2x + y | / \(\sqrt{5}\)}
Substitute (-1, 6) in the above equation
So,
d = | -2 + 6 |/ \(\sqrt{5}\)
d = \(\frac{4}{√5}\)
Hence, from the above,
We can conclude that the distance from the given point to the given line is: \(\frac{4}{√5}\)

Exercise 3.5 Equations of Parallel and Perpendicular Lines

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A _________ line segment AB is a segment that represents moving from point A to point B.
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 1

Question 2.
WRITING
How are the slopes of perpendicular lines related?
Answer:
We know that,
The “Perpendicular lines” are lines that intersect at right angles.
If you multiply the slopes of two perpendicular lines in the plane, you get −1 i.e., the slopes of perpendicular lines are opposite reciprocals

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio.

Question 3.
A(8, 0), B(3, – 2); 1 to 4
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 3

Question 4.
A(- 2, – 4), B(6, 1); 3 to 2
Answer:
The given coordinates are: A (-2, -4), and B (6, 1)
Compare the given points with
A (x1, y1), and B (x2, y2)
It is given that
AP : PB = 3 : 2
So,
We have to divide AB into 5 parts
Now,
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of AB = \(\frac{1 + 4}{6 + 2}\)
Slope of AB = \(\frac{5}{8}\)
Now,
To find the coordinates of P, add slope to AP and PB
So,
P = (3 + (3 / 5) × 8, 2 + (3 / 5) × 5)
P = (7.8, 5)
Hence, from he above,
We can conclude that
The coordinates of P are (7.8, 5)

Question 5.
A(1, 6), B(- 2, – 3); 5 to 1
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 5

Question 6.
A(- 3, 2), B(5, – 4); 2 to 6
Answer:
The given coordinates are: A (-3, 2), and B (5, -4)
Compare the given points with
A (x1, y1), and B (x2, y2)
It is given that
AP : PB = 2 : 6
So,
We have to divide AB into 8 parts
Now,
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of AB = \(\frac{-4 – 2}{5 + 3}\)
Slope of AB = \(\frac{-6}{8}\)
Now,
To find the coordinates of P, add slope to AP and PB
So,
P = (2 + (2 / 8) × 8, 6 + (2 / 8) × (-6))
P = (4, 4.5)
Hence, from the above,
We can conclude that
The coordinates of P are (4, 4.5)

In Exercises 7 and 8, determine which of the lines are parallel and which of the lines are perpendicular.

Question 7.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 162
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 7

Question 8.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 163
Answer:

The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 163
From the given figure,
The coordinates of line a are: (2, 2), and (-2, 3)
The coordinates of line b are: (3, -2), and (-3, 0)
The coordinates of line c are: (2, 4), and (0, -2)
The coordinates of line d are: (0, 6), and (-2, 0)
Now,
Compare the given coordinates with (x1, y1), and (x2, y2)
So,
The slope of line a (m) = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{3 – 2}{-2 – 2}\)
= \(\frac{1}{-4}\)
= –\(\frac{1}{4}\)

The slope of line b (m) = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{0 + 2}{-3 – 3}\)
= \(\frac{2}{-6}\)
= –\(\frac{1}{3}\)

The slope of line c (m) = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{-4 – 2}{0 – 2}\)
= \(\frac{-6}{-2}\)
= 3

The slope of line d (m) = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{6 – 0}{0 + 2}\)
= \(\frac{6}{2}\)
= 3
We know that,
The parallel lines have the same slopes
The perpendicular lines have the product of slopes equal to -1
Hence, from the above,
We can conclude that
Line c and Line d are parallel lines
Line b and Line c are perpendicular lines

In Exercises 9 – 12, tell whether the lines through the given points are parallel, perpendicular, or neither. justify your answer.

Question 9.
Line 1: (1, 0), (7, 4)
Line 2: (7, 0), (3, 6)
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 9

Question 10.
Line 1: (- 3, 1), (- 7, – 2)
Line 2: (2, – 1), (8, 4)
Answer:
The coordinates of line 1 are: (-3, 1), (-7, -2)
The coordinates of line 2 are: (2, -1), (8, 4)
Compare the given coordinates with
(x1, y1), (x2, y2)
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of line 1 = \(\frac{-2 – 1}{-7 + 3}\)
= \(\frac{-3}{-4}\)
= \(\frac{3}{4}\)
Slope of line 2 = \(\frac{4 + 1}{8 – 2}\)
= \(\frac{5}{6}\)
Hence,
By comparing the slopes,
We can conclude that the given lines are neither parallel nor perpendicular

Question 11.
Line 1: (- 9, 3), (- 5, 7)
Line 2: (- 11, 6), (- 7, 2)
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 11

Question 12.
Line 1: (10, 5), (- 8, 9)
Line 2: (2, – 4), (11, – 6)
Answer:
The coordinates of line 1 are: (10, 5), (-8, 9)
The coordinates of line 2 are: (2, -4), (11, -6)
Compare the given coordinates with
(x1, y1), (x2, y2)
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of line 1 = \(\frac{9 – 5}{-8 – 10}\)
= \(\frac{4}{-18}\)
= –\(\frac{2}{9}\)
Slope of line 2 = \(\frac{4 – 6}{11 – 2}\)
= \(\frac{-2}{9}\)
= –\(\frac{2}{9}\)
Hence,
By comparing the slopes,
We can conclude that the given lines are parallel

In Exercises 13 – 16. write an equation of the line passing through point P that ¡s parallel to the given line. Graph the equations of the lines to check that they are parallel.

Question 13.
P(0, – 1), y = – 2x + 3
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 13

Question 14.
P(3, 8), y = \(\frac{1}{5}\)(x + 4)
Answer:
The given equation is:
y = \(\frac{1}{5}\) (x + 4)
The given point is: P (3, 8)
So,
y = \(\frac{1}{5}\)x + \(\frac{4}{5}\)
Now,
We know that,
The slopes are equal fot the parallel lines
So,
The equation that is parallel to the given equation is:
y = \(\frac{1}{5}\)x + c
Substitute P (3, 8) in the above equation to find the value of c
So,
8 = \(\frac{1}{5}\) (3) + c
c = 8 – \(\frac{3}{5}\)
c = \(\frac{37}{5}\)
Hence,
The parallel line equation that is parallel to the given equation is:
y = \(\frac{1}{5}\)x + \(\frac{37}{5}\)
The representation of the parallel lines in the coordinate plane is:

Question 15.
P(- 2, 6), x = – 5
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 15

Question 16.
P(4, 0), – x + 2y = 12
Answer:
The given equation is:
-x + 2y = 12
The given point is: P (4, 0)
So,
y = \(\frac{1}{2}\)x + 6
Now,
We know that,
The slopes are equal fot the parallel lines
So,
The equation that is parallel to the given equation is:
y = \(\frac{1}{2}\)x + c
Substitute P (4, 0) in the above equation to find the value of c
So,
0 = \(\frac{1}{2}\) (4) + c
c = 2 – 0
c = 2
Hence,
The parallel line equation that is parallel to the given equation is:
y = \(\frac{1}{2}\)x + 2
The representation of the parallel lines in the coordinate plane is:

In Exercises 17 – 20. write an equation of the line passing through point P that is perpendicular to the given line. Graph the equations of the lines to check that they are perpendicular.

Question 17.
P(0, 0), y = – 9x – 1
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 17

Question 18.
P(4, – 6)y = – 3
Answer:
The given equation is:
y = -3
The given point is: P (4, -6)
We know that,
The line that is perpendicular to y=n is:
x = n
So,
The line that is perpendicular to the given equation is:
x = n
Substitute P (4, -6) in the above equation
So,
x = 4
Hence, from the above,
We can conclude that
The equation that is perpendicular to y = -3 is:
x = 4
The representation of the perpendicular lines in the coordinate plane is:

Question 19.
P(2, 3), y – 4 = – 2(x + 3)
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 19.1
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 19.2

Question 20.
P(- 8, 0), 3x – 5y = 6
Answer:
The given equation is:
3x – 5y = 6
The given point is: P (-8, 0)
Now,
5y = 3x – 6
y = \(\frac{3}{5}\)x – \(\frac{6}{5}\)
We know that,
The product of the slopes of perpendicular lines is equal to -1
So,
m = –\(\frac{5}{3}\)
So,
The equation that is perpendicular to the given line equation is:
y = –\(\frac{5}{3}\)x + c
Substitute P(-8, 0) in the above equation
0 = –\(\frac{5}{3}\) ( -8) + c
c = \(\frac{40}{3}\)
Hence,
The equation that is perpendicular to the given equation is:
y = –\(\frac{5}{3}\)x + \(\frac{40}{3}\)
The representation of the perpendicular lines in the coordinate plane is:

In Exercises 21 – 24, find the distance from point A to the given line.

Question 21.
A(- 1, 7), y = 3x
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 21.1
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 21.2

Question 22.
A(- 9, – 3), y = x – 6
Answer:
The given equation is:
y = x – 6 ——-(1)
The given point is: A (-9, -3)
Compare the given equation with
y = mx + c
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1m2 = -1
1 (m2) = -3
m2 = -1
So,
The equation that is perpendicular to the given line equation is:
y = -x + c
Substitute A (-9, -3) in the above equation to find the value of c
So,
-3 = 9 + c
c = -9 – 3
c = -12
Hence,
The equation that is perpendicular to the given line equation is:
y = -x – 12 ——(2)
Now,
We have to find the point of intersection
So,
Eq. (1) = Eq. (2)
x – 6 = -x – 12
x + x = -12 + 6
2x = -6
x = \(\frac{-6}{2}\)
x = -3
So,
y = x – 6
y = -3 – 6
y = -9
So,
The point of intersection = (-3, -9)
Now,
The points are: (-9, -3), (-3, -9)
So,
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
= \(\sqrt{(9 – 3)² + (9 – 3)²}\)
= \(\sqrt{(6)² + (6)²}\)
= 8.48
Hence, from the above,
We can conclude that the distance from point A to the given line is: 8.48

Question 23.
A(15, – 21), 5x + 2y = 4
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 23.1
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 23.2

Question 24.
A(- \(\frac{1}{4}\), 5), – x + 2y = 14
Answer:
The given equation is:
-x + 2y = 14
y = \(\frac{1}{2}\)x + 7 ——-(1)
The given point is: A (-\(\frac{1}{4}\), 5)
Compare the given equation with
y = mx + c
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1m2 = -1
\(\frac{1}{2}\) (m2) = -1
m2 = -2
So,
The equation that is perpendicular to the given line equation is:
y = -2x + c
Substitute A (-\(\frac{1}{4}\), 5) in the above equation to find the value of c
So,
5 = -2 (-\(\frac{1}{4}\)) + c
c = 5 – \(\frac{1}{2}\)
c = \(\frac{9}{2}\)
Hence,
The equation that is perpendicular to the given line equation is:
y = -2x + \(\frac{9}{2}\) ——(2)
Now,
We have to find the point of intersection
So,
Eq. (1) = Eq. (2)
\(\frac{1}{2}\)x + 7 = -2x + \(\frac{9}{2}\)
\(\frac{1}{2}\)x + 2x = -7 + 9/2
\(\frac{5}{2}\)x = –\(\frac{5}{2}\)
x = -1
So,
y = -2x + 3
y = -2 (-1) + \(\frac{9}{2}\)
y = \(\frac{13}{2}\)
So,
The point of intersection = (-1, \(\frac{13}{2}\))
Now,
The points are: (-\(\frac{1}{4}\), 5), (-1, \(\frac{13}{2}\))
So,
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
= \(\sqrt{(3 / 2)² + (3 / 4)²}\)
= 1.67
Hence, from the above,
We can conclude that the distance from point A to the given line is: 1.67

Question 25.
ERROR ANALYSIS
Describe and correct the error in determining whether the lines are parallel. perpendicular, or neither.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 164
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 25

Question 26.
ERROR ANALYSIS
Describe and correct the error in writing an equation of the line that passes through the point (3, 4) and is parallel to the line y = 2x + 1.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 165
Answer:
The given equation of the line is:
y = 2x + 1
The given point is: (3, 4)
We know that,
The slopes of the parallel lines are the same
Now,
Compare the given equation with
y = mx + c
So,
m = 2
So,
The slope of the line that is aprallle to the given line equation is:
m = 2
So,
The equation of the line that is parallel to the given equation is:
y = 2x + c
To find the value of c,
Substitute (3, 4) in the above equation
So,
4 = 2 (3) + c
4 – 6 = c
c = -2
Hence, from the above,
We can conclude that the equation of the line that is parallel to the given line is:
y = 2x – 2

In Exercises 27-30. find the midpoint of \(\overline{P Q}\). Then write
an equation of the line that passes through the midpoint and is perpendicular to \(\overline{P Q}\). This line is called the perpendicular bisector.

Question 27.
P( – 4, 3), Q(4, – 1)
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 27

Question 28.
P(- 5, – 5), Q(3, 3)
Answer:
The given points are: P (-5, -5), Q (3, 3)
Compare the given points with (x1, y1), (x2, y2)
We know that,
The midpoint of PQ = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\))
= (\(\frac{-5 + 3}{2}\), \(\frac{-5 + 3}{2}\))
= (\(\frac{-2}{2}\), \(\frac{-2}{2}\))
= (-1, -1)
Now,
The slope of PQ = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{3 + 5}{3 + 5}\)
= \(\frac{8}{8}\)
= 1
Now,
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1m2 = -1
So,
m2 = -1
We know that,
The equation of the perpendicular line that passes through the midpoint of PQ is:
y = -x + c
To find the value of c,
Substitute (-1, -1) in the above equation
So,
-1 = -1 + c
c = 0
Hence, from the above,
We can conclude that the equation of the line that is perpendicular bisector is:
y = -x

Question 29.
P(0, 2), Q(6, – 2)
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 29

Question 30.
P(- 7, 0), Q(1, 8)
Answer:
The given points are: P (-7, 0), Q (1, 8)
Compare the given points with (x1, y1), (x2, y2)
We know that,
The midpoint of PQ = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\))
= (\(\frac{8 + 0}{2}\), \(\frac{-7 + 1}{2}\))
= (\(\frac{8}{2}\), \(\frac{-6}{2}\))
= (4, -3)
Now,
The slope of PQ = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{8 – 0}{1 + 7}\)
= \(\frac{8}{8}\)
= 1
Now,
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1m2 = -1
So,
m2 = -1
We know that,
The equation of the perpendicular line that passes through the midpoint of PQ is:
y = -x + c
To find the value of c,
Substitute (4, -3) in the above equation
So,
-3 = -4 + c
c = -3 + 4
c = 1
Hence, from the above,
We can conclude that the equation of the line that is perpendicular bisector is:
y = -x + 1

Question 31.
MODELING WITH MATHEMATICS
Your school lies directly between your house and the movie theater. The distance from your house to the school is one-fourth of the distance from the school to the movie theater. What point on the graph represents your school?
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 166
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 31

Question 32.
REASONING
Is quadrilateral QRST a parallelogram? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 167
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 167
From the above figure,
The coordinates of the quadrilateral QRST is:
Q (2, 6), R (6, 4), S (5, 1), and T (1, 3)
Compare the given points with (x1, y1), and (x2, y2)
Now,
We know that,
If both pairs of opposite sides of a quadrilateral are parallel, then it is a parallelogram
So,
If the slopes of the opposite sides of the quadrilateral are equal, then it is called as “Parallelogram”
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of QR = \(\frac{4 – 6}{6 – 2}\)
Slope of QR = \(\frac{-2}{4}\)
Slope of QR = –\(\frac{1}{2}\)

Slope of RS = \(\frac{1 – 4}{5 – 6}\)
Slope of RS = \(\frac{-3}{-1}\)
Slope of RS = 3

Slope of ST = \(\frac{3 – 1}{1 – 5}\)
Slope of ST = \(\frac{2}{-4}\)
Slope of ST = –\(\frac{1}{2}\)

Slope of TQ = \(\frac{3 – 6}{1 – 2}\)
Slope of TQ = \(\frac{-3}{-1}\)
Slope of TQ = 3
Now,
From the slopes,
We can observe that the slopes of the opposite sides are equal i.e., the opposite sides are parallel
Hence, from the above,
We can conclude that the quadrilateral QRST is a parallelogram

Question 33.

REASONING
A triangle has vertices L(0, 6), M(5, 8). and N(4, – 1), Is the triangle a right triangle? Explain ‘your reasoning.
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 33

Question 34.
MODELING WITH MATHEMATICS
A new road is being constructed parallel to the train tracks through points V. An equation of the line representing the train tracks is y = 2x. Find an equation of the line representing the new road.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 168
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 168
It is given that a new road is being constructed parallel to the train tracks through points V. An equation of the line representing the train tracks is y = 2x.
So,
From the figure,
V = (-2, 3)
We know that,
The slopes of the parallel lines are the same
So,
By comparing the given equation with
y = mx + c
We get,
m = 2
So,
The equation of the line that is parallel to the line that represents the train tracks is:
y = 2x + c
Now,
To find the value of c,
Substitute (-2, 3) in the above equation
So,
3 = 2 (-2) + x
3 + 4 = c
c = 7
Hence, from the above,
We can conclude that the equation of the line that is parallel to the line representing railway tracks is:
y = 2x + 7

Question 35.
MODELING WITH MATHEMATICS
A bike path is being constructed perpendicular to Washington Boulevard through point P(2, 2). An equation of the line representing Washington Boulevard is y = –\(\frac{2}{3}\)x. Find an equation of the line representing the bike path.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 169
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 35

Question 36.
PROBLEM-SOLVING
A gazebo is being built near a nature trail. An equation of the line representing the nature trail is y = \(\frac{1}{3}\)x – 4. Each unit in the coordinate plane corresponds to 10 feet. Approximately how far is the gazebo from the nature trail?
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 170
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 170
It is given that a gazebo is being built near a nature trail. An equation of the line representing the nature trail is y = \(\frac{1}{3}\)x – 4. Each unit in the coordinate plane corresponds to 10 feet
So,
It can be observed that
1 unit either in the x-plane or y-plane = 10 feet
So,
y = \(\frac{1}{3}\)x – 4
y = \(\frac{1}{3}\) (10) – 4
y = \(\frac{10 – 12}{3}\)
y = –\(\frac{2}{3}\)
We know that,
The distance won’t be in negative value,
y = \(\frac{2}{3}\)
y = 0.66 feet
Hence, from the above,
We can conclude that the distance of the gazebo from the nature trail is: 0.66 feet

Question 37.
CRITICAL THINKING
The slope of line l is greater than 0 and less than 1. Write an inequality for the slope of a line perpendicular to l. Explain your reasoning.
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 37

Question 38.
HOW DO YOU SEE IT?
Determine whether quadrilateral JKLM is a square. Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 171
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 171
From the given figure,
The coordinates of a quadrilateral are:
J (0 0), K (0, n), L (n, n), M (n, 0)
Compare the given points with (x1, y1), and (x2, y2)
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
So,
Slope of JK = \(\frac{n – 0}{0 – 0}\)
= Undefined
Slope of KL = \(\frac{n – n}{n – 0}\)
= 0
Slope of LM = \(\frac{0 – n}{n – n}\)
= Undefined
Slope of MJ = \(\frac{0 – 0}{n – 0}\)
= 0
We know that,
For a square,
The slopes of perpendicular lines are undefined and 0 respectively
Hence, from the above,
We can conclude that quadrilateral JKLM is a square

Question 39.
CRITICAL THINKING
Suppose point P divides the directed line segment XY So that the ratio 0f XP to PY is 3 to 5. Describe the point that divides the directed line segment YX so that the ratio of YP Lo PX is 5 to 3.
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 39

Question 40.
MAKING AN ARGUMENT
Your classmate claims that no two nonvertical parallel lines can have the same y-intercept. Is your classmate correct? Explain.
Answer:
Yes, your classmate is correct

Explanation:
It is given that your classmate claims that no two nonvertical parallel lines can have the same y-intercept
We know that,
The parallel lines have the same slope but have different y-intercepts and do not intersect
The intersecting lines intersect each other and have different slopes and have the same y-intercept
Hence, from the above,
We can conclude that the claim of your classmate is correct

Question 41.
MATHEMATICAL CONNECTIONS
Solve each system of equations algebraically. Make a conjecture about what the solution(s) can tell you about whether the lines intersect. are parallel, or are the same line.
a. y = 4x + 9
4x – y = 1
b. 3y + 4x = 16
2x – y = 18
c. y = – 5x + 6
10x + 2y = 12
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 41.1
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 41.2

Question 42.
THOUGHT-PROVOKING
Find a formula for the distance from the point (x0, Y0) to the line ax + by = 0. Verify your formula using a point and a line.
Answer:





MATHEMATICAL CONNECTIONS
In Exercises 43 and 44, find a value for k based on the given description.

Question 43.
The line through (- 1, k) and (- 7, – 2) is parallel to the line y = x + 1.
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 43

Question 44.
The line through (k, 2) and (7, 0) is perpendicular to the line y = x – \(\frac{28}{5}\).
Answer:
The given points are: (k, 2), and (7, 0)
The given line that is perpendicular to the given points is:
y = x – \(\frac{28}{5}\)
Now,
Compare the given points with (x1, y1), and (x2, y2)
So,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{0 – 2}{7 – k}\)
m = \(\frac{-2}{7 – k}\)
Now,
The slope that is perpendicular to the given line is:
m = -1 [ Since we know that m1m2 = -1]
So,
-1 = \(\frac{-2}{7 – k}\)
k – 7 = -2
k = -2 + 7
k = 5
Hence, from the above,
We can conclude that the value of k is: 5

Question 45.
ABSTRACT REASONING
Make a conjecture about how to find the coordinates of a point that lies beyond point B along \(\vec{A}\)B. Use an example to support your conjecture.
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 45

Question 46.
PROBLEM-SOLVING
What is the distance between the lines y = 2x and y = 2x + 5? Verify your answer.
Answer:
The given lines are:
y = 2x and y = 2x + 5
Compare the given equations with
y = mx + c
Now,
To find the distance between the two lines, we have to find the intersection point of the line
So,
The intersection of the line is the y-intercept
So,
From y = 2x + 5,
The intersection point is: (0, 5)
Now,
For the intersection point of y = 2x,
Find the perpendicular line of y = 2x and find the intersection point of the two lines
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
2m2 = -1
m2 = –\(\frac{1}{2}\)
So,
The perpendicular equation of y = 2x is:
y = –\(\frac{1}{2}\)x + c
To find the value of c in the above equation, substitue (0, 5) in the above equation
So,
5 = c
c = 5
So,
The perpendicular line equation of y = 2x is:
y = –\(\frac{1}{2}\)x + 5
Now,
For the intersection point,
2x = –\(\frac{1}{2}\)x + 5
2x + \(\frac{1}{2}\)x = 5
\(\frac{5}{2}\)x = 5
x = 2
So,
y = 2x
= 2 (2)
= 4
So,
The intersection point of y = 2x is: (2, 4)
Now,
The points are: (0, 5), and (2, 4)
Compare the given points with (x1, y1), and (x2, y2)
So,
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
= \(\sqrt{(4 – 5)² + (2 – 0)²}\)
= \(\sqrt{1 + 4}\)
= 2.23
Hence, from the above,
We can conclude that the distance between the lines y = 2x and y = 2x + 5 is: 2.23

PROVING A THEOREM
In Exercises 47 and 48, use the slopes of lines to write a paragraph proof of the theorem.

Question 47.
Lines Perpendicular to a Transversal Theorem (Theorem 3.12): In a plane. if two lines are perpendicular to the same line. then they are parallel to each other.
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 47

Question 48.
Transitive Property of Parallel Lines Theorem (Theorem 3.9),/+: If two lines are parallel to the same line, then they are parallel to each other.
Answer:
The given statement is:
If two lines are parallel to the same line, then they are parallel to each other
Proof:
Let the two parallel lines be E and F and the plane they lie be plane x
Let the two parallel lines that are parallel to the same line be G
Hence,
According to the Transitive Property of parallel lines,
If line E is parallel to line F and line F is parallel to line G, then line E is parallel to line G.

Question 49.
PROOF
Prove the statement: If two lines are vertical. then they are parallel.
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 49

Question 50.
PROOF
Prove the statement: If two lines are horizontal, then they are parallel.
Answer:
The given statement is:
If two lines are horizontal, then they are parallel
Proof:
If two lines x and y are horizontal lines and they are cut by a vertical transversal z, then
x ⊥ z and y ⊥ z
Hence,
x || y is proved by the Lines parallel to Transversal Theorem

Question 51.
PROOF
Prove that horizontal lines are perpendicular to vertical lines.
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 51

Maintaining Mathematical Proficiency

Plot the point in a coordinate plane.

Question 52.
A(3, 6)
Answer:
The given point is: A(3, 6)
The representation of the given point in the coordinate plane is:

Question 53.
B(0, – 4)
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 53

Question 54.
C(5, 0)
Answer:
The given point is: C (5, 0)
The representation of the given point in the coordinate plane is:

Question 55.
D( – 1, – 2)
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 55

Copy and complete the table.

Question 56.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 172
Answer:
The given table is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 172
From the above table,
The given equation is:
y = x + 9
Hence,
The completed table is:

Question 57.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 173
Answer:
Big Ideas Math Geometry Solutions Chapter 3 Parallel and Perpendicular Lines 3.5 a 57

3.4 – 3.5 Performance Task: Navajo Rugs

Mathematical Practices

Question 1.
Compare the effectiveness of the argument in Exercise 24 on page 153 with the argument “You can find the distance between any two parallel lines” What flaw(s) exist in the argument(s)? Does either argument use correct reasoning? Explain.
Answer:
From the argument in Exercise 24 on page 153,
We can say that
The claim of your friend is not correct
We know that,
If we want to find the distance from the point to a given line, we need the perpendicular distance of a point and a line
Hence, from the above,
We can conclude that we can not find the distance between any two parallel lines if a point and a line is given to find the distance

Question 2.
Look back at your construction of a square in Exercise 29 on page 154. How would your
construction change if you were to construct a rectangle?
Answer:
From the construction of a square in Exercise 29 on page 154,
We can observe that the length of all the line segments are equal
Now,
If you were to construct a rectangle,
We have to keep the lengths of the length of the rectangles the same and the widths of the rectangle also the same

Question 3.
In Exercise 31 on page 161, a classmate tells you that our answer is incorrect because you should have divided the segment into four congruent pieces. Respond to your classmates’ argument by justifying your original answer.
Answer:
In Exercise 31 on page 161, from the coordinate plane,
We can observe that we divided the total distance into the four congruent segments or pieces
Hence, from the above,
We can conclude that the argument of your friend that the answer is incorrect is not correct

Parallel and Perpendicular Lines Chapter Review

3.1 Pairs of Lines and Angles

Think of each segment in the figure as part of a line. Which line(s) or plane(s) appear to fit the description?
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 174
Question 1.
line(s) perpendicular to Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 175
Answer:
We know that,
The lines that are at 90° are “Perpendicular lines”
Hence,
From the above figure,
The lines perpendicular to \(\overline{Q R}\) are: \(\overline{R M}\) and \(\overline{Q L}\)

Question 2.
line(s) parallel to Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 175
Answer:
We know that,
The lines that do not have any intersection points are called “Parallel lines”
Hence,
From the above figure,
The line parallel to \(\overline{Q R}\) is: \(\overline {L M}\)

Question 3.
line(s) skew to Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 175
Answer:
We know that,
The lines that do not intersect and are not parallel and are not coplanar are “Skew lines”
Hence,
From the above figure,
The lines skew to \(\overline{Q R}\) are: \(\overline{J N}\), \(\overline{J K}\), \(\overline{K L}\), and \(\overline{L M}\)

Question 4.
plane(s) parallel to plane LMQ
Answer:
From the given figure,
We can conclude that the plane parallel to plane LMQ is: Plane JKL

3.2 Parallel Lines and Transversals

Find the values of x and y.

Question 5.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 176
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 176
From the given figure,
We can observe that x and 35° are the corresponding angles
We know that,
By using the Corresponding Angles Theorem,
x = 35°
Now,
We can observe that 35° and y are the consecutive interior angles
So,
35° + y = 180°
y = 180° – 35°
y = 145°
Hence, from the above,
We can conclude that
x° = 35° and y° = 145°

Question 6.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 177
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 177
From the given figure,
We can observe that 48° and y are the consecutive interior angles and y and (5x – 17)° are the corresponding angles
So,
By using the Consecutive Interior Angles Theorem,
48° + y° = 180°
y° = 180° – 48°
y° = 132°
Now,
By using the corresponding angles theorem,
y° = (5x – 17)°
132° = (5x – 17)°
5x = 132° + 17°
5x = 149°
x = \(\frac{149}{5}\)
x° = 29.8°
Hence, from the above,
We can conclude that
x° = 29.8° and y° = 132°

Question 7.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 178
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 178
From the above figure,
We can observe that
2x° and 2y° are the alternate exterior angles
2y° and 58° are the alternate interior angles
So,
2x° = 2y° = 58°
So,
x° = y° =29°
Hence, from the above,
We can conclude that
x° = y° = 29°

Question 8.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 179
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 179
From the given figure,
We can observe that
(5y – 21)° and 116° are the corresponding angles
So,
(5y – 21)° = 116°
5y° = 116° + 21°
5y° = 137°
y° = \(\frac{137}{5}\)
y° = 27.4°
Now,
We can observe that
(5y – 21)° ad (6x + 32)° are the alternate interior angles
So,
(5y – 21)° = (6x + 32)°
5 (28)° – 21° = (6x + 32)°
140 – 21 – 32 = 6x°
6x° = 140° – 53°
6x = 87°
x = \(\frac{87}{6}\)
x° = 14.5°
Hence, from the above,
We can conclude that
x° = 14.5° and y° = 27.4°

3.3 Proofs with Parallel Lines

Find the value of x that makes m || n.

Question 9.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 180
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 180
We know that,
m || n is true only when x and 73° are the consecutive interior angles according to the “Converse of Consecutive Interior angles Theorem”
Now,
It is given that m || n
So,
x + 73° = 180°
x = 180° – 73°
x = 107°
Hence, from the above,
We can conclude that the value of x is: 107°

Question 10.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 181
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 181
We know that,
m || n is true only when 147° and (x + 14)° are the corresponding angles by using the “Converse of the Alternate Exterior Angles Theorem”
Now,
It is given that m || n
So,
(x + 14)°= 147°
x° = 147° – 14°
x° = 133°
Hence, from the above,
We can conclude that the value of x is: 133°

Question 11.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 182
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 182
m || n is true only when 3x° and (2x + 20)° are the corresponding angles by using the “Converse of the Corresponding Angles Theorem”
Now,
It is given that m || n
So,
(2x + 20)°= 3x°
3x° – 2x° = 20°
x° = 20°
Hence, from the above,
We can conclude that the value of x is: 20°

Question 12.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 183
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 183
We know that,
m || n is true only when (7x – 11)° and (4x + 58)° are the alternate interior angles by the “Convesre of the Consecutive Interior Angles Theorem”
Now,
It is given that m || n
So,
(7x – 11)° = (4x + 58)°
7x° – 4x° = 58° + 11°
3x° = 69°
x° = \(\frac{69}{3}\)
x° = 23°
Hence, from the above,
We can conclude that the value of x is: 23°

3.4 Proofs with Perpendicular Lines

Determine which lines, if any, must be parallel. Explain your reasoning.

Question 13.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 184
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 184
From the given figure,
We can observe that
x ⊥ z and y ⊥ z
We know that,
According to the Perpendicular Transversal Theorem,
In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other line also.
Hence, from the above,
We can conclude that x and y are parallel lines

Question 14.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 185
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 185
From the given figure,
We can observe that
w ⊥ y and z ⊥ x
We can also observe that w and z is not both ⊥ to x and y
We know that,
According to the Perpendicular Transversal Theorem,
In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other line also.
Hence, from the above,
We can conclude that there are not any parallel lines in the given figure

Question 15.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 186
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 186
From the given figure,
We can observe that
m ⊥ a, n ⊥ a, l ⊥ b, and n ⊥ b
We know that,
According to the Perpendicular Transversal Theorem,
In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other line also.
Hence, from the above,
We can conclude that m and n are parallel lines

Question 16.
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 187
Answer:
The given figure is:
Big Ideas Math Answer Key Geometry Chapter 3 Parallel and Perpendicular Lines 187
From the given figure,
We can observe that
a ⊥ n, b ⊥ n, and c ⊥ m
We know that,
According to the Perpendicular Transversal Theorem,
In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other line also.
Hence, from the above,
We can conclude that there are not any parallel lines in the given figure

3.5 Equations of Parallel and Perpendicular Lines

Write an equation of the line passing through the given point that is parallel to the given line.

Question 17.
A(3, – 4),y = – x + 8
Answer:
The given equation is:
y = -x + 8
The given point is: A (3, -4)
So,
y = -x + 8
Now,
We know that,
The slopes are equal fot the parallel lines
So,
The equation that is parallel to the given equation is:
y = -x + c
Substitute A (3, -4) in the above equation to find the value of c
So,
-4 = -3 + c
c = -4 + 3
c = -1
Hence,
The parallel line equation that is parallel to the given equation is:
y = -x – 1

Question 18.
A(- 6, 5), y = \(\frac{1}{2}\)x – 7
Answer:
The given equation is:
y = \(\frac{1}{2}\)x – 7
The given point is: A (-6, 5)
So,
y = \(\frac{1}{2}\)x – 7
Now,
We know that,
The slopes are equal fot the parallel lines
So,
The equation that is parallel to the given equation is:
y = \(\frac{1}{2}\)x + c
Substitute A (-6, 5) in the above equation to find the value of c
So,
5 = \(\frac{1}{2}\) (-6) + c
c = 5 + 3
c = 8
Hence,
The parallel line equation that is parallel to the given equation is:
y = \(\frac{1}{2}\)x + 8

Question 19.
A(2, 0), y = 3x – 5
Answer:
The given equation is:
y = 3x – 5
The given point is: A (2, 0)
So,
y = 3x – 5
Now,
We know that,
The slopes are equal for the parallel lines
So,
The equation that is parallel to the given equation is:
y = 3x + c
Substitute A (2, 0) in the above equation to find the value of c
So,
0 = 3 (2) + c
c = 6 – 0
c = 6
Hence,
The parallel line equation that is parallel to the given equation is:
y = 3x – 6

Question 20.
A(3, – 1), y = \(\frac{1}{3}\)x + 10
Answer:
The given point is: A (3, -1)
The given equation is:,
y = \(\frac{1}{3}\)x + 10
Now,
We know that,
The slopes are equal fot the parallel lines
So,
The equation that is parallel to the given equation is:
y = \(\frac{1}{3}\)x + c
Substitute A (3, -1) in the above equation to find the value of c
So,
-1 = \(\frac{1}{3}\) (3) + c
c = -1 – 1
c = -2
Hence,
The parallel line equation that is parallel to the given equation is:
y = \(\frac{1}{3}\)x – 2

Write an equation of the line passing through the given point that is perpendicular to the given line.

Question 21.
A(6, – 1), y = – 2x + 8
Answer:
The given equation is:
y = -2x + 8
The given point is:A (6, -1)
Now,
y = -2x + 8
We know that,
The product of the slopes of perpendicular lines is equal to -1
So,
m = \(\frac{1}{2}\)
So,
The equation that is perpendicular to the given line equation is:
y = \(\frac{1}{2}\)x + c
Substitute A (6, -1) in the above equation
-1 = \(\frac{1}{2}\) ( 6) + c
c = -1 – 3
c = -4
Hence,
The equation that is perpendicular to the given equation is:
y = \(\frac{1}{2}\)x – 4

Question 22.
A(0, 3), y = – \(\frac{1}{2}\)x – 6
Answer:
The given equation is:
y = –\(\frac{1}{2}\)x – 6
The given point is: A (0, 3)
Now,
y = \(\frac{1}{2}\)x – 6
We know that,
The product of the slopes of perpendicular lines is equal to -1
So,
m = 2
So,
The equation that is perpendicular to the given line equation is:
y = 2x + c
Substitute A (0, 3) in the above equation
3 = 2 ( 0) + c
c = 3
Hence,
The equation that is perpendicular to the given equation is:
y = 2x + 3

Question 23.
A(8, 2),y = 4x – 7
Answer:
The given equation is:
y = 4x – 7
The given point is: A (8, 2)
Now,
y = 4x – 7
We know that,
The product of the slopes of perpendicular lines is equal to -1
So,
m = –\(\frac{1}{4}\)
So,
The equation that is perpendicular to the given line equation is:
y = –\(\frac{1}{4}\)x + c
Substitute A (8, 2) in the above equation
2 = –\(\frac{1}{4}\) (8) + c
c = 2 + 2
c = 4
Hence,
The equation that is perpendicular to the given equation is:
y = –\(\frac{1}{4}\)x + 4

Question 24.
A(-1, 5), y = \(\frac{1}{7}\)x + 4
Answer:
The given equation is:
y = \(\frac{1}{7}\)x + 4
The given point is: A (-1, 5)
Now,
y = \(\frac{1}{7}\)x + 4
We know that,
The product of the slopes of perpendicular lines is equal to -1
So,
m = -7
So,
the equation that is perpendicular to the given line equation is:
y = -7x + c
Substitute A (-1, 5) in the above equation
5 = -7 ( -1) + c
c = 5 – 7
c = -2
Hence,
The equation that is perpendicular to the given equation is:
y = -7x – 2

Find the distance front point A to the given line.

Question 25.
A(2, – 1), y = – x + 4
Answer:
The given equation is:
y = -x + 4 ——-(1)
The given point is: A (2, -1)
Compare the given equation with
y = mx + c
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1m2 = -1
(-1) (m2) = -1
m2 = 1
So,
The equation that is perpendicular to the given line equation is:
y = x + c
Substitute A (2, -1) in the above equation to find the value of c
So,
-1 = 2 + c
c = -1 – 2
c = -3
Hence,
The equation that is perpendicular to the given line equation is:
y = x – 3 ——(2)
Now,
We have to find the point of intersection
So,
Eq. (1) = Eq. (2)
-x + 4 = x – 3
-x – x = -3 – 4
2x = 7
x = \(\frac{7}{2}\)
So,
y = x – 3
y = \(\frac{7}{2}\) – 3
y = \(\frac{1}{2}\)
So,
The point of intersection = (\(\frac{7}{2}\), \(\frac{1}{2}\))
Now,
The points are: (2, -1), (\(\frac{7}{2}\), \(\frac{1}{2}\))
So,
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
= \(\sqrt{(3 / 2)² + (3 / 2)²}\)
= 2.12
Hence, from the above,
We can conclude that the distance from point A to the given line is: 2.12

Question 26.
A(- 2, 3), y = \(\frac{1}{2}\)x + 1
Answer:
The given equation is:
y = \(\frac{1}{2}\)x + 1 ——-(1)
The given point is: A (-2, 3)
Compare the given equation with
y = mx + c
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1m2 = -1
(\(\frac{1}{2}\)) (m2) = -1
m2 = -2
So,
The equation that is perpendicular to the given line equation is:
y = -2x + c
Substitute A (-2, 3) in the above equation to find the value of c
So,
3 = -2 (-2) + c
c = 3 – 4
c = -1
Hence,
The equation that is perpendicular to the given line equation is:
y = -2x – 1 ——(2)
Now,
We have to find the point of intersection
So,
Eq. (1) = Eq. (2)
\(\frac{1}{2}\)x + 1 = -2x – 1
\(\frac{5}{2}\)x = 2
x = \(\frac{4}{5}\)
So,
y = -2x – 1
y = –\(\frac{8}{5}\) – 1
y = –\(\frac{13}{5}\)
So,
The point of intersection = (\(\frac{4}{5}\), –\(\frac{13}{5}\))
Now,
The points are: (-2, 3), (\(\frac{4}{5}\), –\(\frac{13}{5}\))
So,
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
= \(\sqrt{31.36 + 7.84}\)
= 6.26
Hence, from the above,
We can conclude that the distance from point A to the given line is: 6.26

Parallel and Perpendicular Lines Test

Find the values of x and y. State which theorem(s) you used.

Question 1.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 188
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 188
From the given figure,
We can observe that
x and 61° are the vertical angles
61° and y° are the alternate interior angles
We know that,
According to the “Alternate Interior Angles theorem”, the alternate interior angles are congruent
According to the “Vertical Angles Theorem”, the vertical angles are congruent
Hence,
x° = y° = 61°

Question 2.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 189
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 189
From the given figure,
We can observe that
8x° and 96° are the alternate interior angles
(11y + 19)° and 96° are the corresponding angles
We know that,
According to the “Alternate Interior Angles Theorem”, the alternate interior angles are congruent
According to the “Corresponding Angles Theorem”, the corresponding angles are congruent
So,
8x° = 96°
x° = \(\frac{96}{8}\)
x° = 12°
Now,
(11y + 19)° = 96°
11y° = 96° – 19°
11y° = 77°
y° = \(\frac{77}{11}\)
y° = 7°
Hence, from the above,
We can conclude that
x° = 12° and y° = 7°

Question 3.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 190
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 190
From the given figure,
We can observe that
42° and 6(2y – 3)° are the consecutive interior angles
42° and (8x + 2)° are the vertical angles
We know that,
According to the “Consecutive Interior Angles Theorem”, the sum of the consecutive interior angles is 180°
According to the “Vertical Angles Theorem”, the vertical angles are congruent
So,
42° + 6 (2y – 3)° = 180°
6 (2y°) –  6(3)° = 180° – 42°
12y° – 18° = 138°
12y° = 138° + 18°
12y° = 156°
y° = \(\frac{156}{12}\)
y° = 13°
Now,
42° = (8x + 2)°
8x° = 42° – 2°
x° = 40°
x° = \(\frac{40}{8}\)
x° = 5°
Hence, from the above,
We can conclude that
x° = 5° and y° = 13°

Find the distance from point A to the given line.

Question 4.
A(3, 4), y = – x
Answer:
The given equation is:
y = -x ——-(1)
The given point is: A (3, 4)
Compare the given equation with
y = mx + c
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1m2 = -1
(-1) (m2) = -1
m2 = 1
So,
The equation that is perpendicular to the given line equation is:
x = c
Substitute A (3, 4) in the above equation to find the value of c
So,
c = 3
Hence,
The equation that is perpendicular to the given line equation is:
x = 3 ——(2)
Now,
We have to find the point of intersection
So,
Eq. (1) = Eq. (2)
-x = x – 3
-x – x = -3
2x = 3
x = \(\frac{3}{2}\)
So,
y = -x
y = –\(\frac{3}{2}\)
So,
The point of intersection = (\(\frac{3}{2}\), –\(\frac{3}{2}\))
Now,
The points are: (3, 4), (\(\frac{3}{2}\), –\(\frac{3}{2}\))
So,
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
= \(\sqrt{30.25 + 2.25}\)
= 5.70
Hence, from the above,
We can conclude that the distance from point A to the given line is: 5.70

Question 5.
A(- 3, 7), y = \(\frac{1}{3}\)x – 2
Answer:
The given equation is:
y = \(\frac{1}{3}\)x – 2 ——-(1)
The given point is: A (-3, 7)
Compare the given equation with
y = mx + c
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1m2 = -1
(\(\frac{1}{3}\)) (m2) = -1
m2 = -3
So,
The equation that is perpendicular to the given line equation is:
y = -3x + c
Substitute A (-3, 7) in the above equation to find the value of c
So,
7 = -3 (-3) + c
c = 7 – 9
c = -2
Hence,
The equation that is perpendicular to the given line equation is:
y = -3x – 2 ——(2)
Now,
We have to find the point of intersection
So,
Eq. (1) = Eq. (2)
\(\frac{1}{3}\)x – 2 = -3x – 2
\(\frac{1}{3}\)x + 3x = -2 + 2
x = 0
So,
y = -3x – 2
y = -3 (0) – 2
y = -2
So,
The point of intersection = (0, -2)
Now,
The points are: (-3, 7), (0, -2)
So,
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
= \(\sqrt{(-2 – 7)² + (0 + 3)²}\)
= 9.48
Hence, from the above,
We can conclude that the distance from point A to the given line is: 9.48

Find the value of x that makes m || n.

Question 6.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 191
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 191
From the given figure,
We can observe that
x° and 97° are the corresponding angles
We know that,
According to the “Converse of the Corresponding Angles Theorem”, m || n is true only when the corresponding angles are congruent
It is given that m || n
So,
x° = 97°
Hence, from the above,
We can conclude that
x° = 97°

Question 7.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 192
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 192
From the given figure,
We can observe that
8x° and (4x + 24)° are the alternate exterior angles
We know that,
According to the “Converse of the Alternate Exterior Angles Theorem”, m || n is true only when the alternate exterior angles are congruent
It is given that m || n
So,
8x° = (4x + 24)°
8x° – 4x° = 24°
4x° = 24°
x° = \(\frac{24}{4}\)
x° = 6°
Hence, from the above,
We can conclude that
x° = 6°

Question 8.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 193
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 193
From the given figure,
We can observe that
(11x + 33)° and (6x – 6)° are the interior angles
We know that,
According to the “Converse of the Interior Angles Theory”, m || n is true only when the sum of the interior angles are supplementary
It is given that m || n
So,
(11x + 33)°+(6x – 6)° = 180°
17x° + 27° = 180°
17x° = 180° – 27°
x° = –\(\frac{153}{17}\)
x° = 9°
Hence, from the above,
We can conclude that
x° = 9°

Write an equation of the line that passes through the given point and is
(a) parallel to and
(b) perpendicular to the given line.

Question 9.
(- 5, 2), y = 2x – 3
Answer:
The given equation is:
y= 2x – 3
The given point is: (-5, 2)
a) Parallel to the given line:
We know that,
The slopes of the parallel lines are the same
Compare the given equation with
y = mx + c
So,
The equation of the line that is parallel to the given line is:
y = 2x + c
To find the value of c,
Substitute (-5, 2) in the above equation
So,
2 = 2 (-5) + c
2 + 10 = c
c = 12
Hence,
The equation of the line that is parallel to the given line equation is:
y = 2x + 12
b) Perpendicular to the given line:
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1 m2 = -1
2m2 = -1
m2 = –\(\frac{1}{2}\)
So,
The equation of the line that is perpendicular to the given line equation is:
y = –\(\frac{1}{2}\)x + c
To find the value of c,
Substitute (-5, 2) in the given equation
2 = –\(\frac{1}{2}\) (-5) + c
c = –\(\frac{1}{2}\)
Hence,
The equation of the line that is perpendicular to the given line equation is:
y = –\(\frac{1}{2}\)x – \(\frac{1}{2}\)

Question 10.
(- 1, – 9), y = – \(\frac{1}{3}\)x + 4
Answer:
The given equation is:
y= –\(\frac{1}{3}\)x + 4
The given point is: (-1, -9)
a) Parallel to the given line:
We know that,
The slopes of the parallel lines are the same
Compare the given equation with
y = mx + c
So,
The equation of the line that is parallel to the given line is:
y = –\(\frac{1}{3}\)x + c
To find the value of c,
Substitute (-1, -9) in the above equation
So,
-9 = –\(\frac{1}{3}\) (-1) + c
c = \(\frac{26}{3}\)
Hence,
The equation of the line that is parallel to the given line equation is:
y = –\(\frac{1}{3}\)x + \(\frac{26}{3}\)
b) Perpendicular to the given line:
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1 m2 = -1
–\(\frac{1}{3}\)m2 = -1
m2 = 3
So,
The equation of the line that is perpendicular to the given line equation is:
y = 3x + c
To find the value of c,
Substitute (-1, -9) in the given equation
-9 = 3 (-1) + c
c = -6
Hence,
The equation of the line that is perpendicular to the given line equation is:
y = 3x – 6

Question 11.
A student says. “Because j ⊥ K, j ⊥ l’ What missing information is the student assuming from the diagram? Which theorem is the student trying to use?
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 194
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 194
It is given that a student claimed that j ⊥ K, j ⊥ l
We know that,
According to the “Perpendicular Transversal Theorem”,
In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other line also.
So,
From the above definition,
The missing information the student assuming from the diagram is:
The line l is also perpendicular to the line j
Hence, from the above,
We can conclude that the theorem student trying to use is the “Perpendicular Transversal Theorem”

Question 12.
You and your family are visiting some attractions while on vacation. You and your mom visit the shopping mall while your dad and your sister visit the aquarium. You decide to meet at the intersection of lines q and p. Each unit in the coordinate plane corresponds to 50 yards.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 195
Answer:
You and your family are visiting some attractions while on vacation. You and your mom visit the shopping mall while your dad and your sister visit the aquarium. You decide to meet at the intersection of lines q and p. Each unit in the coordinate plane corresponds to 50 yards.

a. Find an equation of line q.
Answer:
From the figure,
The coordinates of line q are:
(50, 500), (200, 50)
Compare the given points with
(x1, y1), (x2, y2)
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{50 – 500}{200 – 50}\)
= \(\frac{-450}{150}\)
= –\(\frac{45}{15}\)
= -3
Now,
We know that,
The standard form of the equation is:
y – y1 = m (x – x1)
So,
y – 500 = -3 (x -50)
y – 500 = -3x + 150
y = -3x + 150 + 500
y = -3x + 650
Hence, from the above,
We can conclude that
The equation of line q is:
y = -3x + 650

b. Find an equation of line p.
Answer:
From the figure,
The coordinates of line p are:
(50, 175), (500, 325)
Compare the given points with
(x1, y1), (x2, y2)
We know that,
Slope (m) = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{325 –  175}{500 – 50}\)
= \(\frac{15}{45}\)
= \(\frac{1}{3}\)
Now,
We know that,
The standard form of the equation is:
y – y1 = m (x – x1)
So,
y – 175 = \(\frac{1}{3}\) (x -50)
3 (y – 175) = x – 50
3y – 525 = x – 50
3y = x – 50 + 525
3y = x + 475
y = \(\frac{1}{3}\)x + \(\frac{475}{3}\)
Hence, from the above,
We can conclude that
The equation of line p is:
y = \(\frac{1}{3}\)x + \(\frac{475}{3}\)

c. What are the coordinates of the meeting point?
Answer:
From the given figure,
We can conclue that
The coordinates of the meeting point are: (150. 200)

d. What is the distance from the meeting point to the subway?
Answer:
From the given figure,
The coordinates of the meeting point are: (150, 200)
The coordinates of the subway are: (500, 300)
Now,
The distance between the meeting point and the subway is:
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
d = \(\sqrt{(300 – 200)² + (500 – 150)²}\)
d = 364.5 yards
Hence, from the above,
We can conclude that the distance between the meeting point and the subway is: 364.5 yards

Question 13.
Identify an example on the puzzle cube of each description. Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 196
a. a pair of skew lines
Answer:
We know that,
The “Skew lines” are the lines that are non-intersecting, non-parallel and non-coplanar
Hence,
From the given figure,
We can conclude that the pair of skew lines are:
\(\overline{A B}\) and \(\overline{G H}\)

b. a pair of perpendicular lines
Answer:
We know that,
The “Perpendicular lines” are the lines that are intersected at the right angles
Hence,
From the given figure,
We can conclude that the pair of perpendicular lines are:
\(\overline{I J}\) and \(\overline{C D}\)

c. a pair of paralIeI lines
Answer:
We know that,
The “parallel lines” are the lines that do not have any intersection point
Hence,
From the given figure,
We can conclude that the pair of parallel lines are:
\(\overline{C D}\) and \(\overline{E F}\)

d. a pair of congruent corresponding angles
Answer:
From the given figure,
We can conclude that
∠1 and ∠3 are the corresponding angles

e. a pair of congruent alternate interior angles
Answer:
From the given figure,
We can conclude that
∠2 and ∠3 are the congruent alternate interior angles

Parallel and Perpendicular Lines Cumulative Assessment

Question 1.
Use the steps in the construction to explain how you know that\(\overline{C D}\) is the perpendicular bisector of \(\overline{A B}\).
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 198
Answer:
Step 1:
Draw a line segment of any length and name that line segment as AB
Step 2:
Draw an arc by using a compass with above half of the length of AB by taking the center at A above AB
Step 3:
Draw another arc by using a compass with above half of the length of AB by taking the center at B above AB
Step 4:
Repeat steps 3 and 4 below AB
Step 5:
Draw a line segment CD by joining the arcs above and below AB
Step 6:
Measure the lengths of the midpoint of AB i.e., AD and DB.
By measuring their lengths, we can prove that CD is the perpendicular bisector of AB

Question 2.
The equation of a line is x + 2y = 10.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 199
a. Use the numbers and symbols to create the equation of a line in slope-intercept form
that passes through the point (4, – 5) and is parallel to the given line.
Answer:
The given line equation is:
x + 2y = 10
The given point is: (4, -5)
Now,
The given equation in the slope-intercept form is:
y = –\(\frac{1}{2}\)x + 5
We know that,
The slopes of the parallel lines are the same
So,
The equation of the line that is parallel to the given line equation is:
y = –\(\frac{1}{2}\)x + c
To find the value of c,
Substitute (4, -5) in the above equation
So,
-5 = –\(\frac{1}{2}\) (4) + c
c = -5 + 2
c = -3
Hence, from the above,
We can conclude that the line that is parallel to the given line equation is:
y = –\(\frac{1}{2}\)x – 3

b. Use the numbers and symbols to create the equation of a line in slope-intercept form
that passes through the point (2, – 1) and is perpendicular to the given line.
Answer:
The given line equation is:
x + 2y = 10
The given point is: (4, -5)
Now,
The given equation in the slope-intercept form is:
y = –\(\frac{1}{2}\)x + 5
We know that,
The product of the slopes of the perpendicular lines is equal to -1
So,
m1m2 = -1
–\(\frac{1}{2}\) (m2) = -1
m2 = 2
So,
The equation of the line that is perpendicular to the given line equation is:
y = 2x + c
To find the value of c,
Substitute (4, -5) in the above equation
So,
-5 = 2 (4) + c
-5 – 8 = c
c = -13
Hence, from the above,
We can conclude that a line equation that is perpendicular to the given line equation is:
y = 2x – 13

Question 3.
Classify each pair of angles whose measurements are given.
a.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 200
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 200
From the given figure,
We can conclude that 44° and 136° are the adjacent angles

b.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 201
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 201
From the given figure,
We can conclude that 18° and 23° are the adjacent angles

c.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 202
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 202
From the given figure,
We can conclude that 75° and 75° are alternate interior angles

d.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 203
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 203
From the given figure,
We can conclude that 42° and 48° are the vertical angles

Question 4.
Your school is installing new turf on the football held. A coordinate plane has been superimposed on a diagram of the football field where 1 unit = 20 feet.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 204
a. What is the length of the field?
Answer:
It is given that a coordinate plane has been superimposed on a diagram of the football field where 1 unit is 20 feet.
So,
From the given figure,
The length of the field = | 20 – 340 |
= 320 feet
Hence, from the above,
We can conclude that the length of the field is: 320 feet

b. What is the perimeter of the field?
Answer:
From the figure,
The width of the field is: 140 feet
From the figure,
We can observe that the figure is in the form of a rectangle
We know that,
The perimeter of the field = 2 ( Length + Width)
= 2 (320 + 140)
= 2 (460)
= 920 feet
Hence, from the above,
We can conclude that the perimeter of the field is: 920 feet

c. Turf costs $2.69 per square foot. Your school has a $1,50,000 budget. Does the school have enough money to purchase new turf for the entire field?
Answer:
We know that,
The area of the field = Length × Width
So,
The area of the field = 320 × 140
= 44,800 square feet
it is given that the turf costs $2.69 per square foot
So,
The total cost of the turf = 44,800 × 2.69
= $1,20,512
It is given that your school has a budget of $1,50,000 but we only need $1,20,512
Hence, from the above,
We can conclude that the school have enough money to purchase new turf for the entire field

Question 5.
Enter a statement or reason in each blank to complete the two-column proof.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 205
Given ∠1 ≅∠3
Prove ∠2 ≅∠4

Answer:
The given table is:

Hence,
The completed table is:

Question 6.
Your friend claims that lines m and n are parallel. Do you support your friend’s claim? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 206
Answer:
Yes, I support my friend’s claim

Explanation:
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 206
From the given figure,
We can observe that 141° and 39° are the consecutive interior angles
We know that,
According to the consecutive Interior Angles Theorem,
If the sum of the angles of the consecutive interior angles is 180°, then the two lines that are cut by a transversal are parallel
Hence, from the above,
We can conclude that the claim of your friend can be supported

Question 7.
Which of the following is true when Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 207 are skew?
(A) Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 207 are parallel.
(B) Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 207 intersect
(C) Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 207 are perpendicular
(D) A, B, and C are noncollinear.
Answer:
We know that,
The “Skew lines” are the lines that are not parallel, non-intersect, and non-coplanar
Hene, from the given options,
We can conclude that option D) is correct because parallel and perpendicular lines have to be lie in the same plane

Question 8.
Select the angle that makes the statement true.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 208
∠1    ∠2    ∠3    ∠4    ∠5     ∠6     ∠7     ∠8
a. ∠4 ≅ ________ b the Alternate Interior Angles Theorem (Thm. 3.2).
Answer:
From the given figure,
We can conclude that
By using the Alternate interior angles Theorem,
∠4 ≅ ∠5

b. ∠2 ≅ ________ by the Corresponding Angles Theorem (Thm. 3. 1)
Answer:
From the given figure,
We can conclude that
By using the Corresponding angles Theorem,
∠2 ≅ ∠6

c. ∠1 ≅ ________ by the Alternate Exterior Angles Theorem (Thm. 3.3).
Answer:
From the given figure,
We can conclude that
By using the Alternate exterior angles Theorem,
∠1 ≅ ∠8

d. m∠6 + m ________ = 180° by the Consecutive Interior Angles Theorem (Thm. 3.4).
Answer:
From the given figure,
We can conclude that
By using the Consecutive interior angles Theorem,
∠6 + ∠4 = 180°

Question 9.
You and your friend walk to school together every day. You meet at the halfway point between your houses first and then walk to school. Each unit in the coordinate plane corresponds to 50 yards.
Big Ideas Math Geometry Answers Chapter 3 Parallel and Perpendicular Lines 209
Answer:
It is given that you and your friend walk to school together every day. You meet at the halfway point between your houses first and then walk to school. Each unit in the coordinate plane corresponds to 50 yards.
a. What are the coordinates of the midpoint of the line segment joining the two houses?
Answer:
From the given figure,
We can conclude that the midpoint of the line segment joining the two houses is:
M = (150, 250)

b. What is the distance that the two of you walk together?
Answer:
From the given figure,
We can observe that
The coordinates of the school = (400, 300)
The coordinates of the midpoint of the line segment joining the two houses = (150, 250)
It is given that the two friends walk together from the midpoint of the houses to the school
Now,
We know that,
The distance that the two of you walk together is:
d = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)
= \(\sqrt{(250 – 300)² + (150 – 400)²}\)
= \(\sqrt{2500 + 62,500}\)
= 255 yards
Hence, from the above,
We can conclude that the distance that the two of the friends walk together is: 255 yards

Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes

Big Ideas Math Answers Grade 2 Chapter 15

Big Ideas Math Book Grade 2 Chapter 15 Identify and Partition Shapes Answer Key is available here. Students can download BIM Grade 2 Chapter 15 Identify and Partition Shapes Solutions PDF for free of cost. With the help of this Big Ideas Math Answers Grade 2 Ch 15, you can finish homework or assignments within time. This free BIM Book 2nd Grade 15th Chapter Identify and Partition Shapes Answer Key contains solutions for all the questions from the textbook.

Big Ideas Math Book 2nd Grade Answer Key Chapter 15 Identify and Partition Shapes

Every student must have Big Ideas Math 2nd Grade 15th Chapter Identify and Partition Shapes Answer Key to improve their math skills. By referring solution key, no one feels solving problems is difficult. Students have to practice all the topics included in Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes to get better marks in the exam.

The different lessons of Big Ideas Math Book Grade 2 Chapter 15 Identify and Partition Shapes Answer Key are Describe Two-Dimensional Shapes, Identify Angles of Polygons, Draw Polygons, Identify and Draw Cubes, Compose Rectangles, Identify Two, Three, or Four Equal Shares, Partition Shapes into Equal Shares, and Analyze Equal Shares of the Same Shape. Click on the quick link attached here to get the questions and answers for all those topics.

Vocabulary

Lesson: 1 Describe Two-Dimensional Shapes

Lesson: 2 Identify Angles of Polygons

Lesson: 3 Draw Polygons

Lesson: 4 Identify and Draw Cubes

Lesson: 5 Compose Rectangles

Lesson: 6 Identify Two, Three, or Four Equal Shares

Lesson: 7 Partition Shapes into Equal Shares

Lesson: 8 Analyze Equal Shares of the Same Shape

Chapter – 15: Identify and Partition Shapes

Identify and Partition Shapes Vocabulary

Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 1
Organize It
Use the review words to complete the graphic organizer.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 2

Define It
Use your vocabulary cards to identify the word. Find the word in the word search
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 3

Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 4
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 5

Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 6
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 7

Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 8
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 9

Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 10
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes v 11

Lesson 15.1 Describe Two-Dimensional Shapes

Explore and Grow

Create a shape with 3 sides on your geoboard. Draw your shape. Did everyone in your class make the same shape?
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 1
Circle the word that makes the sentence true.
_______ are shapes with 3 sides.
Circles Squares Triangles
Answer: Triangles

Show and Grow

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 2
______ sides
_______ vertices
Shape : _________
Answer: 4 sides
4 vertices
Shape: Irregular Quadrilateral

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 3
______ sides
_______ vertices
Shape : _________
Answer: 5 sides
5 vertices
Shape: Irregular Pentagon

Apply and Grow: Practice

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 4
______ sides
_______ vertices
Shape : _________
Answer: 6 sides
6 vertices
Shape: Irregular Hexagon

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 5
______ sides
_______ vertices
Shape : _________
Answer: 4 sides
4 vertices
Shape: Irregular Quadrilateral

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 6
______ sides
_______ vertices
Shape : _________
Answer: 8 sides
8 vertices
Shape: Octagon

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 7
______ sides
_______ vertices
Shape : _________
Answer: 3 sides
3 vertices
Shape: Scalene Triangle

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 8
______ sides
_______ vertices
Shape : _________
Answer: 5 sides
5 vertices
Shape: Concave Pentagon

Question 8.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 9
______ sides
_______ vertices
Shape : _________
Answer: 4 sides
4 vertices
Shape: Concave Quadrilateral

Question 9.
Writing
How are a pentagon and an octagon different?
____________________
_____________________
Answer: Pentagon has 5 sides and Octagon has 8 sides.

Think and Grow: Modeling Real Life

Draw a pentagon to make a house. Draw 2 quadrilaterals to make windows and 1 quadrilateral to make a door. Draw an octagon to make a chimney.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 10
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1-10-Answer

Show and Grow

Question 10.
Draw a pentagon to make a fish. Draw 4 triangles to make the fins. Draw a hexagon to make an eye.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 11
Answer:

Question 11.
You draw 5 quadrilaterals. How many sides and vertices do you draw in all?
______ sides _______ vertices
Answer: 20 sides 20 vertices

Question 12.
DIG DEEPER!
You draw an octagon and two pentagons. How many sides and vertices do you draw in all?
______ sides ______ vertices
Answer: 18 sides 18 vertices

Describe Two-Dimensional Shapes Homework & Practice 15.1

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 12
______ sides
_______ vertices
Shape : _________
Answer: 6 sides
6 vertices
Shape: Irregular hexagon

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 13
______ sides
_______ vertices
Shape : _________
Answer: 5 sides
5 vertices
Shape: Irregular Pentagon

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 14
______ sides
_______ vertices
Shape : _________
Answer: 8 sides
8 vertices
Shape: Irregular Octagon

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 15
______ sides
_______ vertices
Shape : _________
Answer: 3 sides
3 vertices
Shape: Right angle Triangle

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 16
______ sides
_______ vertices
Shape : _________
Answer: 4 sides
4 vertices
Shape: Quadrilateral

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 17
______ sides
_______ vertices
Shape : _________
Answer: 5 sides
5 vertices
Shape: Irregular Pentagon

Question 7.
Precision
Describe the shape in 3 ways.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 18
Answer: A 5 sided shape called Pentagon
It has 2 parallel sides
All sides are not equal.

Question 8.
Modeling Real Life
Draw a hexagon to make a dog’s body. Draw quadrilaterals for the head and tail. Draw two triangles for the ears.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.1 19
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1-19-Answer

Question 9.
DIG DEEPER!
You draw a triangle and two hexagons. How many sides and vertices do you draw in all?
______ sides _______ vertices
Answer: 15 sides 15 vertices

Review & Refresh

Question 10.
You are building a 34-foot fence. You build 15 feet on Saturday and 13 feet on Sunday. How many feet are left to build?
______ feet
Answer: 6 feet are left to build
Explanation: Total 34 foot fence
15 feet on Saturday
13 feet on Sunday
15+13= 28 built
34-28= 6 feet left to build

Lesson 15.2 Identify Angles of Polygons

Explore and Grow

Color the triangle blue. Color the quadrilateral red. Color the pentagon green. Color the hexagon orange.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 1
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-1-Answer

Which shape is not colored? How is it different from the other shapes?
________________________
________________________
________________________
Answer: Circle is not colored. Circle has no straight lines.

Show and Grow

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 2
_____ angles
How many right angles? ______
Shape: ______
Answer: 6 angles
2 right angles
Shape: Irregular hexagon

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 3
______ angles
How many right angles? ______
Shape: _______
Answer: 3 angles
1 right angles
Shape: Right angle triangle

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 4
_____ angles
How many right angles? ______
Shape: ______
Answer: 4 angles
4 right angles
Shape: Square

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 5
______ angles
How many right angles? ______
Shape : _______
Answer: 4 angles
No right angles
Shape: Trapezium

Apply and Grow: Practice

Question 5.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 6
_____ angles
How many right angles? ______
Shape: ______
Answer: 3 angles
1 right angle
Shape: Right angle triangle

Question 6.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 7
_____ angles
How many right angles? ______
Shape: ______
Answer: 5 angles
2 right angles
Shape: irregular pentagon

Question 7.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 8
_____ angles
How many right angles? ______
Shape: ______
Answer: 8 angles
No right angle
Shape: Irregular Octagon

Question 8.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 9
_____ angles
How many right angles? ______
Shape: ______
Answer: 4 angles
2 right angles
Shape: Irregular Quadrilateral

Question 9.
Draw and name a polygon with 6 angles.
________
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.2-18-Answer

Question 10.
Draw and name a polygon with 2 right angles
_________
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-10-Answer

Question 11.
Writing
Can you draw a polygon with 4 sides and 5 angles? Explain.
__________________
____________________
Answer: No, A polygon with 4 sides and 5 angles cannot be drawn as the number of sides and angles are always equal.

Think and Grow: Modeling Real Life

You are designing a road sign. The new sign must be a pentagon with only 2 right angles. Which signs might be yours?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 10
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-10-Answer

Show and Grow

Question 12.
You are making a sign for your lemonade stand. Your sign must be a quadrilateral with 4 right angles. Which signs might be yours?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 11
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-11-Answer

Question 13.
You draw 3 pentagons. How many angles do you draw in all?
______ angles
Answer: 15 angles

Question 14.
DIG DEEPER!
You draw a quadrilateral and three triangles. Your friend draws an octagon and a hexagon. Who draws more angles in all? How many more?
You Friend ______ more angles
Answer: I draw 13 angles in all. My friend draws 14 angles in all. Friend draws more angles in all with 1 more angle.

Identify Angles of Polygons Homework & Practice 15.2

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 12
______ angles
How many right angles? ______
Shape: ______
Answer: 6 angles
2 right angles
Shape: Irregular hexagon

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 13
______ angles
How many right angles? _____
Shape: ______
Answer: 4 angles
4 right angles
Shape: Rhombus

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 14
______ angles
How many right angles? ______
Shape: ______
Answer: 8 angles
No right angles
Shape: Irregular octagon

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 15
_____ angles
How many right angles? ______
Shape: _______
Answer: 5 angles
1 right angle
Shape: Irregular pentagon

Question 5.
Draw and name a polygon with 4 sides and 1 right angle.
_______
Answer: Right Quadrilateral
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.2-16-Answer

Question 6.
Draw and name a polygon with 6 angles.
________
Answer: Hexagon

Question 7.

DIG DEEPER!
Draw two polygons that have 9 angles in all.
Answer: Triangle and Hexagon

Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.2-19-Answer

Question 8.
Modeling Real Life
You are designing a company logo. Your logo must be a hexagon with 2 right angles. Which logos might be yours?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 16
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.2-20-Answer

Question 9.
DIG DEEPER!
You draw an octagon and two triangles. Your friend draws two quadrilaterals and a pentagon. Who draws more angles in all? How many more?
You Friend _______ more angles
Answer: You draw more angles. 1 angle more
You draw 14 angle in total, your friend draw 13 angle in total.

Review & Refresh

Draw to show the time.
Question 10.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 17
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-17-Answer

Question 11.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 18
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-18-Answer

Question 12.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.2 19
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-19-Answer

Lesson 15.3 Draw Polygons

Explore and Grow

Compare the shapes.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.3 1
Answer: Sqaure and right Quadrilateral

How are the shapes the same? How are they different?
_____________________
_____________________
_____________________
Answer: Both the shapes have 4 sides and 4 angles
They are different as length of the shapes are not equal.

Show and Grow

Question 1.
Draw a polygon with 6 sides. Two of the sides are the same length.
_______ angles
Polygon: ________
Answer: 6 angles
Polygon: Irregular Hexagon
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-117-Answer

Question 2.
Draw a polygon with 5 angles. One of the angles is a right angle.
______ sides
Polygon: ________
Answer: 5 sides
Polygon: Irregular Pentagon
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-118-Answer

Apply and Grow: Practice

Question 3.
Draw a polygon with 3 angles. One of the angles is a right angle.
______ sides
Polygon: _______
Answer: 3 sides
Right angle Triangle
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-119-Answer

Question 4.
Draw a polygon with 1 more side than a triangle. No sides are equal.
________ sides
Polygon: ______
Answer: 4 sides
Polygon: Irregular Quadrilateral
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-10-Answer

Question 5.
Draw a polygon with 4 fewer angles than an octagon. All sides are equal. All angles are right angles.
______ sides
Polygon: _______
Answer: 4 sides
Polygon: Square
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-chp-12

Question 6.
Draw a polygon with 4 sides. Two pairs of sides are the same length.
______ angles
Polygon: ______
Answer: 4 angles
Polygon: Rectangle
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-21-Answer

Question 7.
Precision
Which is not a polygon with only 4 angles?
square rectangle rhombus
trapezoid pentagon quadrilateral
Answer: Pentagon
Explanation: Pentagon has 5 angles

Think and Grow: Modeling Real Life

You have 9 straws. You use all the straws to create two polygons. Draw two polygons you can create. Write the names of the polygons.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.3 2
Polygon 1: ______ Polygon 2: _______
Answer: Polygon 1: Square   Polygon 2: Pentagon

Show and Grow

Question 8.
You have 7 clay balls and some toothpicks. You create two polygons using the clay balls as vertices and the toothpicks as sides. Draw two polygons you can create. Write the names of the polygons.
Polygon 1: ______ Polygon 2: _______
Answer: Polygon 1: Triangle Polygon 2: Rhombus

Question 9.
DIG DEEPER!
You draw two different polygons. One of the polygons is a pentagon. You draw 11 sides in all. Draw a possible shape for your other polygon. Write the name of the polygon.
Polygon: _______
Answer: Polygon: Hexagon

Draw Polygons Homework & Practice 15.3

Question 1.
Draw a polygon with 4 angles. There are no right angles. No sides are equal.
______ sides
Polygon: _______
Answer: 4 sides
Polygon: Trapezoid
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-120-Answer

Question 2.
Draw a polygon with 8 sides. Two of the angles are right angles.
_____ angles
Polygon: ______
Answer: 8 angles
Polygon: Irregular octagon
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-120-Answer

Question 3.
Draw a polygon with 2 more angles than a quadrilateral. Two of the angles are right angles.
______ sides
Polygon: _______
Answer: 6 sides
Polygon: Irregular hexagon
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-117-Answer

Question 4.
Draw a polygon with 3 fewer sides than an octagon.
_____ angles
Polygon: _____
Answer: 5 angles
Polygon: Pentagon
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-30b-Answer

Question 5.
Patterns
Draw 3 shapes. The first shape is a quadrilateral. The number of angles in each shape increases by two.
Name the third shape. ________
Answer: Octagon
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-29-Answer

Question 6.
Modeling Real Life
You have 9 apples and some sticks. You create two polygons using the apples for vertices and the sticks for sides. Draw two polygons you can create. Write the names of the polygons.
Polygon 1: ______ Polygon 2: ________
Answer: Polygon 1: Sqaure     Polygon 2: Pentagon

Question 7.
DIG DEEPER!
You draw two different polygons. One of the polygons is an octagon. You draw 14 sides in all. Draw a possible shape for your other polygon. Write the name of the polygon.
Polygon: ________
Answer: Polygon: Hexagon

Review & Refresh

Question 8.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.3 3
How many more students chose math than science? _____
Answer: 5 Students

Lesson 15.4 Identify and Draw Cubes

Explore and Grow

Draw an X on the shapes with curved surfaces. Circle the remaining shape with flat surfaces that are all the same.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 1
Name the shape you circled.
Answer: Cube, Cuboid
Big-Ideas-Math-Solutions-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.4-1-Answer

Show and Grow

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 2
_____ faces
______ vertices
______ edges
Is it a cube? Yes No
Answer: 6 faces
8 vertices
12 edges
Yes, it is a cube

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 3
____ faces
_____ vertices
_______ edges
Is it a cube? Yes No
Answer: 6 faces
8 vertices
12 edges
No, it is not a Cube

Question 3.
Use the dot paper to draw a cube.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 4
Answer:

Apply and Grow: Practice

Question 4.
Which shapes are cubes?
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 5
Answer: Big-Ideas-Math-Solutions-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.4-5-Answer

Question 5.
Use the dot paper to draw a cube.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 6
Answer: Big-Ideas-Math-Solutions-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.4-4-Answer

Question 6.
How many faces do two cubes have in all?
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 7
______ faces
Answer: 12 faces

Question 7.
Structure
Which two-dimensional shape makes up a cube? Name the shape.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 8
A cube is made up of ______.
Answer: A cube is made up of Sqaure.

Think and Grow: Modeling Real Life

You make a ballot box for a school election. Your box is in the shape of a cube. Each face of the ballot box is a different color. How many colors do you use?
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 9
_____ colors
Answer: 6 colors
Explanation: Cube has 6 faces, So 6 different colors will be used.

Show and Grow

Question 8.
You construct a cube. You use clay balls for the vertices and straws for the edges. How many clay balls do you make? How many straws do you use?
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 10
______ clay balls _____ straws
Answer: 8 clay balls 12 straws

Question 9.
The faces of the number cube are numbered, starting with 1. Draw and label all the faces of the number cube.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 11
Answer: Big-Ideas-Math-Solutions-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.4-7-Answer

Question 10.
DIG DEEPER!
You have 48 toothpicks and 32 grapes. You use all of the materials to make cubes using the toothpicks as edges and the grapes as vertices. How many cubes do you make?
_____ cubes
Answer: 4 cubes

Identify and Draw Cubes Homework & Practice 15.4

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 12
_____ faces
______ vertices
______ edges
Is it a cube? Yes No
Answer: 6 faces
8 vertices
12 edges
No, it is not a cube.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 13
_____ faces
______ vertices
______ edges
Is it a cube? Yes No
Answer: 6 faces
8 vertices
12 edges
Yes, it is a cube

Question 3.
Use the dot paper to draw a cube.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 14
Answer: Big-Ideas-Math-Solutions-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.4-6-Answer

Question 4.
How many vertices do two cubes have in all?
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 15
_____ vertices
Answer: 16 vertices

Question 5.
YOU BE THE TEACHER
Newton says the cube has 3 faces. Is he correct? Explain.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 16
___________________
____________________
Answer: Cube has 6 faces

Question 6.
Modeling Real Life
You construct a cube. You use marshmallows for the vertices and pretzel rods for the edges. How many marshmallows do you use? How many pretzel rods do you use?
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 17
_______ marshmallows ________ pretzel rods
Answer: 8 marshmallows 12 pretzel rods

Question 7.
DIG DEEPER!
You have 24 cotton balls and 36 toothpicks. You use all of the materials to make cubes using the cotton balls as vertices and the toothpicks as edges. How many cubes do you make?
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.4 18
______ cubes
Answer: 3 cubes

Review & Refresh

Question 8.
43 − 5 = ____
Answer: 38
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-122-Answer

Question 9.
62 − 6 = _____
Answer: 56
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-123-Answer

Question 10.
______ = 41 − 4
Answer: 37
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-124-Answer

Question 11.
______ = 44 − 7
Answer: 37
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.2-125-Answer

Lesson 15.5 Compose Rectangles

Explore and Grow

How many square tiles do you need to cover the rectangle?
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.5 1
_____ squares
Answer: 28 squares
Write an equation to match your model.
Answer: l × b
7 × 4 = 28

Show and Grow

Question 1.
Use square tiles to cover the rectangle. Draw to show your work.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.5 2
Complete the statements.
Add by rows:
_____ + _____ + = ______
Add by columns: _____ + _____ = ______
Total square tiles: _______
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.5-2
Add by rows:
3 + 3 + 3 + 3 + 3 = 15
Add by columns:
5 + 5 + 5 = 15

Apply and Grow: Practice

Question 2.
Use square tiles to cover the rectangle. Draw to show your work.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.5 3
Complete the statements.
Add by rows: _____ + ____ + _____ = ______
Add by columns: ____ + _____ + _____ + ____ + _____ = ______
Total square tiles: ______
Answer:
5 + 5 + 5 = 15
3 + 3 + 3 + 3 + 3 = 15

Question 3.
Precision
Divide the rectangle into 6 equal parts.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.5 4
Answer: Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.5-4-Answer

Think and Grow: Modeling Real Life

You use foam mats to cover the entire floor of a square room. You fit 4 mats across one side of the room. How many rows and columns of mats will you have?
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.5 5
_____ rows _____ columns
How many foam mats do you use to cover the entire floor?
Addition equation:
______ foam mats
Answer: 4 rows 4 columns
16 foam mats

Show and Grow

Question 4.
You use square tiles to cover the floor of a square room. You fit 5 tiles across one side of the room. How many rows and columns of tiles will you have?
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.5 6
_____ rows _____ columns
DIG DEEPER!
How many tiles do you use to cover the entire floor?
_____ tiles
Answer: 5 rows 5 columns
25 tiles

Compose Rectangles Homework & Practice 15.5

Question 1.
Use square tiles to cover the rectangle. Draw to show your work.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.5 7
Complete the statements.
Add by rows: _____ = _____
Total square tiles: ______
Add by columns:
_____ + _____ + _____ + _____ + ______ + _____ = _____
Answer:

1 = 1
1 + 1 + 1 + 1 + 1 + 1 = 6

Question 2.Writing
Newton wants to cover the rectangle with square tiles. Explain what he is doing wrong.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.5 8
__________________
____________________
Answer: The Square titles are not arranged properly in the rectangle.

Question 3.
Modeling Real Life
You use square glass tiles to make a square mosaic picture. You fit 6 tiles across one side of the picture. How many rows and columns of tiles will you have?
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.5 9
______ rows _______ columns

Answer: 6 rows 6 columns

DIG DEEPER!
How many glass tiles do you use to cover the entire picture?
______ glass tiles
Answer: 36 glass tiles

Review & Refresh

Question 4.
Circle a.m. or p.m.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 15.5 10
Answer: Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.5-10-Answer

Lesson 15.6 Identify Two, Three, or Four Equal Shares

Explore and Grow

Sort the Equal Share Cards.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 1

Show and Grow

Circle the shape that shows halves.
Question 1.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 2
Answer: Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-2-Answer

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 3
Answer: Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-3-Answer

Circle the shape that shows thirds.
Question 3.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 4
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-4-Answer

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 5
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-5-Answer

Circle the shape that shows fourths.
Question 5.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 6
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-6-Answer

Question 6.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 7
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-6-Answer

Apply and Grow: Practice

Question 7.
Which shapes show halves?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 8
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-8-Answer

Question 8.
Which shapes show thirds?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 9
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-9-Answer

Question 9.
Which shapes show fourths?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 10
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-10-Answer

Question 10.
Color a third of the shape.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 11
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-11-Answer

Question 11.
Color half of the shape.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 12
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-12-Answer

Question 12.
Color a fourth of the shape.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 13
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-13-Answer

Question 13.
YOU BE THE TEACHER
Newton says the shape shows fourths. Is he correct? Explain.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 14
________________________
_________________________
Answer: No The shape doesnot shows fourths. The square is not divided into equal parts.

Think and Grow: Modeling Real Life

You want to play 3 games in the pool. Each game needs an equal share of the pool. Show how you could divide the pool.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 15
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-15-Answer

Show and Grow

Question 14.
2 friends are making crafts. Each friend wants an equal share of the table. Show how the friends could divide the table.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 16
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-16-Answer

Question 15.
You and 3 friends want to share the piece of watermelon. Show how to cut the piece of watermelon so you and your friends each get an equal share.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 17
How did you know how many equal shares to cut?
__________________
__________________
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-17-Answer

Identify Two, Three, or Four Equal Shares Homework & Practice 15.6

Question 1.
Which shapes show halves?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 18
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-18-Answer

Question 2.
Which shapes show thirds?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 19
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-19-Answer

Question 3.
Which shapes show fourths?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 20
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-20-Answer

Question 4.
Color a third of the shape.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 21
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-21-Answer

Question 5.
Color half of the shape.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 22
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-22-Answer

Question 6.
Color a fourth of the shape.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 23
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-23-Answer

Question 7.
Patterns
Draw what comes next.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 24
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-24-Answer

Question 8.
Modeling Real Life
Newton and Descartes share a bedroom. Show how they could divide their room into equal shares.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 25
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-25-Answer

Question 9.
Modeling Real Life
You and 2 friends are making a poster. Each friend wants an equal share of the poster. Show how the friends could divide the poster.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes 15.6 26
How did you know how many equal shares you need?
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-15.6-2.6-Answer

Review & Refresh

Question 10.
A pen is 16 centimeters tall. A pen holder is 11 centimeters tall. How much taller is the pen than the pen holder?
________ centimeters
Answer: 5 centimeters
Explanation: 16-11 = 5

Lesson 15.7 Partition Shapes into Equal Shares

Explore and Grow

Which pattern blocks can you use to model equal shares of the hexagon?
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 1
Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.7-1-Answer

Show and Grow

Draw lines to show equal shares. Complete the sentences.
Question 1.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 2
Each share is a _______ of the whole.
The whole is _______.
Answer:
Each share is a part of the whole.
The whole is 4.

Question 2.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 3
Each share is a _______ of the whole.
The whole is ______.
Answer:
Each share is a half of the whole.
The whole is 2.

Apply and Grow: Practice

Question 3.
Draw lines to show equal shares. Complete the sentences.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 4
Each share is a ______ of the whole.
The whole is ________.
Answer:
Each share is a one third of the whole.
The whole is 1/3.

Draw lines to show equal shares. Which word describes the parts?
Question 4.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 5
Answer: Fourths

Question 5.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 6
Answer: Thirds

Question 6.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 7
Answer: Halves

Question 7.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 8
Answer: Thirds

Question 8.
Precision
Draw lines to show thirds. Color 3 thirds.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 9
Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.7-9

Think and Grow: Modeling Real Life

You have 2 towels that are the same size. Half of one towel is green. A fourth of the other towel is yellow. Is the green share or the yellow share larger? Explain.
Draw to show:
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 10
Which share is larger? Green share Yellow share
________________________
________________________
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-127-Answer
Green share
Explanation: First towel is divided into halves and other towel is divided into fourths, So halves is larger than fourths.

Show and Grow

Question 9.
You have 2 rugs that are the same size. A fourth of one rug is red. A third of the other rug is blue. Is the red share or the blue share smaller? Explain.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 11
________________________
________________________
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-124-Answer
The red share is smaller as it is divided into 4 parts.

Partition Shapes into Equal Shares Homework & Practice 15.7

Draw lines to show equal shares. Complete the sentences.
Question 1.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 12
Each share is a ______ of the whole.
The whole is ______.
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-123-AnswerAnswer:

Question 2.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 13
Each share is a _______ of the whole.
The whole is _____.
Answer:  Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-122-Answer

Draw lines to show equal shares. Complete the sentences.
Question 3.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 14
Each share is a ______ of the whole.
The whole is _______.
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-121-Answer

Question 4.
Logic
Complete the sentences.
1 whole is ______ halves.
1 whole is ______ thirds.
1 whole is ______ fourths.
Answer: 1 whole is 2 halves.
1 whole is 3 thirds.
1 whole is 4 fourths.

Question 5.
Structure
Does the shape show thirds? Explain.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 15
Answer: No, The circle is not divided into 3 equal parts.

Question 6.
Modeling Real Life
You have 2 blankets that are the same size. A third of one blanket is yellow. A half of the other blanket is purple. Is the yellow share or the purple share smaller? Explain.
Big Ideas Math Answers Grade 2 Chapter 15 Identify and Partition Shapes 15.7 16
_____________________
_____________________
Answer: The Yellow Share is smaller as the blanket is divided into 3 parts whereas, the purple share is divided into 2 parts.

Review & Refresh

Question 7.
150 + 610 = ______
Answer: 760
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-125-Answer

Question 8.
553 + 250 = ______
Answer: 803
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-126-Answer

Lesson 15.8 Analyze Equal Shares of the Same Shape

Explore and Grow

Color the squares that show equal shares.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 1
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-111-Answer

How are the squares you colored the same? How are they different?
________________________
________________________
________________________
Answer: No, the squares colored are not same because all are differently divided.

Show and Grow

Question 1.
Draw lines to show halves in two different ways. Color one-half of each circle.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 2
Think: How are the halves of each circle the same? How are they different?
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-112-Answer
The Circle are divided into 2 parts. The colored part are not same because all are differently divided.

Question 2.
Draw lines to show thirds in two different ways. Color one-third of each rectangle.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 3
Think: How are the thirds of each rectangle the same? How are they different?
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-12-Answer
The Rectangle are divided into 3 parts. The colored part are not same because all are differently divided.

Apply and Grow: Practice

Question 3.
Draw lines to show fourths two different ways. Color one-fourth of each circle.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 4
Answer:

Question 4.
Draw lines to show thirds two different ways. Color one-third of each square.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 5
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-110-Answer

Question 5.
Draw lines to show halves two different ways. Color one-half of each rectangle.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 6
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.1.2-09-Answer

Question 6.
DIG DEEPER!
Explain how you know each color is a fourth of the whole square.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 7
________________________
________________________
_________________________
Answer:

Think and Grow: Modeling Real Life

You and your friend each cut a sandwich into fourths different ways. The sandwiches are the same size. Show how you and your friend can cut the sandwiches.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 8
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-12-Answer

Show and Grow

Question 7.
You, Newton, and Descartes each cut a granola bar into halves different ways. The granola bars are the same size. Show how you, Newton, and Descartes can cut the granola bars.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 9
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-13-Answer

Question 8.
DIG DEEPER!
There are 2 pizzas that are the same size. 6 friends each want an equal share of the pizzas. Should the pizzas be cut into halves, thirds, or fourths? Explain.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 10
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-10-Answer
The Pizzas should be cut into thirds as there are 2 pizzas, if one pizza is cut into 3 equal parts and other pizza as 3 equal parts then there is total 6 pieces for 6 friends.

Analyze Equal Shares of the Same Shape Homework & Practice 15.8

Question 1.
Draw lines to show fourths two different ways. Color one-fourth of each rectangle.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 11
Think: How are the fourths of each rectangle the same? How are they different?
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-9-Answer

Question 2.
Draw lines to show thirds two different ways. Color one-third of each circle.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 12
Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.8-12

Question 3.
Reasoning
Descartes says there are only two ways to divide a rectangle into 3 equal shares. Is he correct? Explain.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 13
Answer:Yes. Descartes is correct. the rectangle can be divided equally horizontally and vertically.

Question 4.
Modeling Real Life
You and your friend each cut a loaf of bread into thirds different ways. The loaves of bread are the same size. Show how you and your friend can cut the loaves of bread.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 14
Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.8-14

Question 5.
DIG DEEPER!
There are 2 quesadillas that are the same size. 8 friends each want an equal share of the quesadillas. Should the quesadillas be cut into halves, thirds, or fourths? Explain.
Big Ideas Math Solutions Grade 2 Chapter 15 Identify and Partition Shapes 15.8 15
Answer: Fourths
Explanation: There are 2 quesadillas. It can be cut into fourths So, it will be 8 parts in total and given to 8 friends equally.

Review & Refresh

Question 6.
841 − 603 = _____
Answer: 238

Question 7.
439 − 210 = _______
Answer: 229

Identify and Partition Shapes Performance Task

You paint a square suncatcher in art class.
Question 1.
You paint each shape a different color. Color to show how you paint the sun catcher.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 1
a. Each triangle is red.
b. Each octagon is orange.
c. Each pentagon with more than 1 right angle is yellow.
d. The rest of the pentagons are green.
e. Each shape with 6 angles is blue.
f. Each quadrilateral with all right angles is purple.
g. The rest of the quadrilaterals are pink.
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-1-Answer

Question 2.
The length of one side of the suncatcher is 12 inches. What are the lengths of the other sides?
______ inches
Answer: 12 inches

Question 3.
Your friend paints a rectangular suncatcher.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 2
a. What share of your friend’s design is blue? _______
b. What share of your friend’s design is orange or yellow? ________
Answer: a. second
b. third or fourth

Identify and Partition Shapes Activity

Three In a Row: Equal Shares
To Play: Players take turns. On your turn, spin the spinner. Cover a square that matches your spin. Continue playing until a player gets three in a row.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes 3

Identify and Partition Shapes Chapter Practice

15.1 Describe Two-Dimensional Shapes

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 1
______ sides
_____ vertices
Shape: ______
Answer: 6 sides
6 vertices
Shape: Irregular hexagon

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 2
______ sides
_____ vertices
Shape: ______
Answer: 5 sides
5 vertices
Shape: Irregular pentagon

Question 3.
Modeling Real Life
You draw three quadrilaterals and an octagon. How many sides and vertices do you draw in all?
______ sides ______ vertices
Answer: 20 sides 20 vertices

15.2 Identify Angles of Polygons

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 4
______ angles
How many right angles? ______
Shape: ______
Answer: 5 angles
2 right angles
Shape: Irregular pentagon

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 5
______ angles
How many right angles? ______
Shape: ______
Answer: 6 angles
2 right angles
Shape: Irregular hexagon

Question 6.
Draw and name a shape with 2 right angles.
________
Answer: Pentagon
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-11-Answer

15.3 Draw Polygons

Question 7.
Draw a polygon with 4 angles. All sides are equal length.
______ sides
Polygon: _______
Answer: 4 sides
Polygon: Square
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-chp-12

Question 8.
Draw a polygon with 5 sides. Two of the angles are right angles.
______ angles
Polygon: ________
Answer: 5 angles
Polygon: Pentagon
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-11-Answer

15.4 Identify and Draw Cubes

Question 9.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 9
______ faces
______ vertices
______ edges
Is it a cube? Yes No
Answer: 6 faces
8 vertices
12 edges
Yes it is cube.

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 10
______ faces
______ vertices
______ edges
Is it a cube? Yes No
Answer: 6 faces
8 vertices
12 edges
No it is not cube.

Question 11.
Use the dot paper to draw a cube.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 11
Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-15-Identify-and-Partition-Shapes-15.4-6-Answer

15.5 Compose Rectangles

Question 12.
Use square tiles to cover the rectangle. Draw to show your work.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 12
Complete the statements.
Add by rows:
_____ + _____ = ______
Add by columns:
_____ + _____ = _______
Total square tiles: _______
Answer:
Add by rows:
2 + 2 = 4
Add by columns:
2 + 2 = 4
Total square tiles: 4

15.6 Identify Two, Three, or Four Equal Shares

Question 13.
Which shapes show halves?
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 13
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-chp-13-Answer

Question 14.
Which shapes show thirds?
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 14
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-chp-14-Answer

Question 15.
Which shapes show fourths?
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 15
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-chp-15-Answer

15.7 Partition Shapes into Equal Shares

Draw lines to show equal parts. Complete the sentences.
Question 16.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 16
Each share is a ______ of the whole.
The whole is ______.
Answer:
Each share is a fraction or part of the whole.
The whole is 3.

Question 17.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 17
Each share is a ______ of the whole.
The whole is _____.
Answer:
Each share is a part of the whole.
The whole is 2.

Question 18.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 18
Each share is a ______ of the whole.
The whole is _____.

Answer:
Each share is a part of the whole.
The whole is 4.

15.8 Analyze Equal Shares of the Same Shape

Question 19.
Modeling Real Life
There are 3 bagels that are the same size. 6 friends each want an equal share of the bagels. Should the bagels be cut into halves, thirds, or fourths? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 15 Identify and Partition Shapes chp 19
__________________
___________________
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-6-Answer
Explanation: 3 bagels cut into Halves will be 6 bagels.

Identify and Partition Shapes Cumulative Practice

Question 1.
Your bed is 39 inches long. Your comforter is 66 inches long. How much longer is the comforter than the bed?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 1
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-8-Answer

Question 2.
Find each difference.
700 – 465 = _____
910 – 186 = _______
302 – 176 = ______
Answer: 700 – 465 = 235
910 – 186 = 724
302 – 176 = 126

Question 3.
A dog park is 48 yards long. Your dog enters the park and runs 29 yards. You run 13 yards. How far is your dog from the other end of the park?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 3
Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-15-Identify-and-Partition-Shapes-7-Answer
Explanation: 29+13=42
48-42=6 yards

Question 4.
Which shapes show thirds?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 4
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-cp-4-Answer

Question 5.
The girls’ soccer team raises $237. The boys’ soccer team raises $113 more. How much money do both teams raise in all?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 5
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-cp-5-Answer
Explanation: 237+113=350

Question 6.
Complete the bar graph.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 6
What mascot got the most votes? _______
How many more votes did Warrior get than Tiger? ______
Answer: Knight
3 more votes Warrior get than Tiger.

Question 7.
Count on to find the total value.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 7
Answer: Total value: 91

Question 8.
Which expressions have a difference of 34?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 8
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-cp-8-Answer

Question 9.
Complete the line plot. Then choose all of the statements that are true.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 9
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-cp-9-Answer

Question 10.
Which clock shows 10:40?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 10
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-cp-10-Answer

Question 11.
The phone is about 12 centimeters long. What is the best estimate for the length of the tablet?
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 11
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-15-Identify-and-Partition-Shapes-cp-11-Answer

Question 12.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 12
______ sides
_______ vertices
Shape: _____
Answer: 6 sides
6 vertices
Shape: Irregular hexagon

Question 13.
Big Ideas Math Answers 2nd Grade Chapter 15 Identify and Partition Shapes cp 13
______ sides
_______ vertices
Shape: _______
Answer: 5 sides
5 vertices
Shape: pentagon

Conclusion:

I hope the details given here regarding Big Ideas Math Grade 2 Chapter 15 Identify and Partition Shapes Answer Key PDF is useful for you to prepare well for the exams. Stay in touch with our site to get the answer key of Big Ideas Math all grades and remaining chapters of grade 2.

Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies

Big Ideas Math Answers Grade 2 Chapter 5

Big Ideas Math Book Grade 2 Chapter 5 Subtraction to 100 Strategies Answer Key is helpful for the students who are willing to be perfect in Math skills. At this Big Ideas Math 2nd Grade 5th Chapter Subtraction to 100 Strategies, you can find the solutions for all the questions in a simple manner. Practice various topics covered in this chapter and prepare well for the exam. Download Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies PDF for free of cost.

Big Ideas Math Book 2nd Grade Answer Key Chapter 5 Subtraction to 100 Strategies

Students who are facing difficulties in solving Big Ideas Math Book Grade 2 Chapter 5 Subtraction to 100 Strategies problems have reached the correct place. This BIM Book 2nd Grade 5th Chapter Subtraction to 100 Strategies Answer Key gives the most accurate answers to all the problems you have related to this chapter. It has different methods of solving every question in a simple way to perform in their exams.

This Subtraction to 100 Strategies consists of Subtract Tens Using a Number Line, Subtract Tens and Ones Using a Number Line, Use Addition to Subtract, Decompose to Subtract, Decompose to Subtract Tens and Ones, Use Compensation to Subtract, Practice Subtraction Strategies and Problem Solving: Subtraction. It is also helpful for the students to have a real-life calculation go very smoothly and neat defining their quick responses to daily life tasks.

Lesson: 1 Subtract Tens Using a Number Line

Lesson: 2 Subtract Tens and Ones Using a Number Line

Lesson: 3 Use Addition to Subtract

Lesson: 4 Decompose to Subtract

Lesson: 5 Decompose to Subtract Tens and Ones

Lesson: 6 Use Compensation to Subtract

Lesson: 7 Practice Subtraction Strategies

Lesson: 8 Problem Solving: Subtraction

Chapter: 5 – Subtraction to 100 Strategies

Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 0.1

Organize It

Use the review words to complete the graphic organizer.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 0.2

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-0.2

Define It
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 0.3

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-0.3

Lesson 5.1 Subtract Tens Using a Number Line

Color to show how you can use the hundred chart to solve.

Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 1

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-1

Show and Grow

Question 1.
70 – 50 = __

Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 3

Answer : 20

Explanation: 70-50=20 first you subtract since difference is often a synonym for subtraction.

Question 2.
33 – 20 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 3

Answer : 13

Explanation : 33-20=13 first you subtract since difference is often a synonym for subtraction.

Apply and Grow: Practice

Question 3.
60 – 40 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 4

Answer : 20

Explanation : 60 – 40 = 20 first you subtract since difference is often a synonym for subtraction.

Question 4.
71 – 20 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 5

Answer : 51

Explanation : 71 –  20 = 51 first you substract since difference is often a synonym for substraction.

Question 5.
46 – 30 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 6

Answer :  16

Explanation : 46 – 30 = 16 first you subtract since difference is often a synonym for substraction.

Question 6.
YOU BE THE TEACHER
Your friend shows 79 − 40 on a number line. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 7

Answer : True

Think and Grow: Modeling Real Life

You have a 74-piece set of magnetic tiles. You use 60 of them to make buildings. How many pieces are left?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 8
Subtraction equation:
Model:
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 9

Answer : 74 – 60 = 14

Explanation : 14 pieces are left.

Show and Grow

Question 7.
A clown has 62 balloons. She uses 40 of them to make balloon animals. How many balloons are left?

Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 10

Answer : 62 – 40 = 12

Explanation:
Given,
A clown has 62 balloons.
She uses 40 of them to make balloon animals.
Subtract 40 from 62 you get 12.
12 balloons are left.

Question 8.
DIG DEEPER!
There are 35 people at a park. 20 of them leave. Then 10 more arrive at the park. How many people are at the park now?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 11

Answer: 25 people

Explanation:
Given,
There are 35 people at a park. 20 of them leave.
35 – 20 = 15
Then 10 more arrive at the park.
15 + 10 = 25 people
Therefore 25 people are at the park now.

Subtract Tens Using a Number Line Homework & Practice 5.1

Question 1.
90 – 50 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 12

Answer: 40

Explanation:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-12

Question 2.
84 – 60 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 13

Answer: 24
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-13

Question 3.
22 – 10 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 14

Answer: 12
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-13

Question 4.
54 – 50 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 15

Answer: 4
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-13

Question 5.
Number Sense
Write the equation shown by the number line.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 16

Answer: 100 – 60 = 40
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-16

Question 6.
Modeling Real Life
There are 52 cards in a deck. You pass out 20 of them. How many cards are left in the deck?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 17

Answer:
Given,
There are 52 cards in a deck. You pass out 20 of them.
52 – 20 = 32
32 cards are left in the deck.

Review & Refresh

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 18

Answer: 51

Question 8.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 19

Answer: 93

Question 9.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 20

Answer: 54

Lesson 5.2 Subtract Tens and Ones Using a Number Line

Explore and Grow

Color to sh

ow how you can use the hundred chart to solve.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 21

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-21

Show and Grow

Question 1.
80 – 34 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 22

Answer: 46
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-22

Question 2.
56 – 23 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 23

Answer: 33
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-23

Apply and Grow: Practice

Question 3.
74 – 51 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 24

Answer: 23

Question 4.
86 – 44 = ___

Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 25

Answer: 42

Question 5.
97 – 61 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 26

Answer: 36

Question 6.
46 – 15 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 27

Answer: 31

Question 7.
69 – 35 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 28

Answer: 34

Question 8.
38 – 22 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 29

Answer: 16

Question 9.
Reasoning
Complete the number line and the equation.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 30

Answer: 47 – 30 = 17
17 – 2 = 15

Think and Grow: Modeling Real Life

You have 85 baseball cards and 54 football cards. How many more baseball cards do you have?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 31
Subtraction equation:
Model:
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 32

Answer:
Given,
You have 85 baseball cards and 54 football cards.
85 – 54 = 31
Thus there are 31 more baseball cards.

Show and Grow

Question 10.
A carnival has 17 rides and 48 games. How many more games are there?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 33

Answer:
Given,
A carnival has 17 rides and 48 games.
48 – 17 = 31 games
Therefore there are 31 more games.

Question 11.
DIG DEEPER!
There are 63 people in a theater, 21 people in the lobby, and 10 people in the parking lot. How many more people are in the theater than in both the lobby and the parking lot?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 34

Answer:
Given,
There are 63 people in a theater, 21 people in the lobby, and 10 people in the parking lot.
21 + 10 = 31
63 – 31 = 32
Thus there are 32 people in the theater than in both the lobby and the parking lot.

Subtract Tens and Ones Using a Number Line Homework & Practice 5.2

Question 1.
95 – 40 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 35

Answer: 55

Question 2.
29 – 12 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 36

Answer: 17

Question 3.
58 – 14 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 37

Answer: 44

Question 4.
77 – 31 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 38

Answer: 46

Question 5.
86 – 26 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 39

Answer: 60

Question 6.
70 – 18 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 40

Answer: 52

Question 7.
Structure
Use the number lines to show 84 −62 in two ways.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 41

Answer: 22

Question 8.
Modeling Real Life
Your classroom has 26 desks and 38 chairs. How many more chairs are there?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 42

Answer:
Given,
Your classroom has 26 desks and 38 chairs.
38 – 26 = 12 chairs
Thus there are 12 more chairs.

Question 9.
Modeling Real Life
A grocery store orders 54 pineapples and 22 watermelons. How many more pineapples are there?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 43

Answer:
Given,
A grocery store orders 54 pineapples and 22 watermelons.
54 – 22 = 32 pineapples
Thus there are 32 more pineapples.

Review & Refresh

Is the number even or odd?

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 44

Answer: Even
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-44
8 is divisible by 2 so 8 is an even number.

Question 11.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 45

Answer: Odd
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-45

Lesson 5.3 Use Addition to Subtract

Explore and Grow

Show how you can use a number line to solve.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 46
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 47

Answer:
29 + x = 52
x = 52 – 29
x = 23
52 – 29 = 23

Show and Grow

Add to find the difference

Question 1.
43 – 15 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 48

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-48
43 – 15 = 28

Question 2.
76 – 59 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 49

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-49
76 – 59 = 17

Apply and Grow: Practice

Add to find the difference.

Question 3.
56 – 27 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 50

Answer: 29
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-50

Question 4.
21 – 13 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 51

Answer: 8
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-51

Question 5.
72 – 57 = ___
Big Ideas Math Answers 2nd Grade Chapter 5 Subtraction to 100 Strategies 52

Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-5-Subtraction-to-100-Strategies-52

Question 6.
33 – 15 = ___
Big Ideas Math Answers 2nd Grade Chapter 5 Subtraction to 100 Strategies 53

Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-5-Subtraction-to-100-Strategies-53

Question 7.
45 – 36 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 54

Answer: 9
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-54

Question 8.
61 – 46 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 55

Answer: 15
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-55

Question 9.
Structure
Use addition and the number lines to show 64 − 35 in two ways.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 56

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-55

Think and Grow: Modeling Real Life

A ship has a crew of 52 pirates. Some of them leave. There are 27 left. How many pirates got off the ship?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 57
Model:
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 58

Answer:
A ship has a crew of 52 pirates. Some of them leave. There are 27 left.
52 – 27 = 25
Thus 25 pirates got off the ship.

Show and Grow

Question 10.
A pumpkin patch has 85 pumpkins. Some of them are picked. There are 48 left. How many pumpkins were picked?
Big Ideas Math Answers 2nd Grade Chapter 5 Subtraction to 100 Strategies 59

Answer:
Given,
A pumpkin patch has 85 pumpkins. Some of them are picked. There are 48 left.
85 – 48 = 37 pumpkins
Thus 37 pumpkins were picked.

Question 11.
DIG DEEPER!
There are 96 treats in a bowl. Newton takes 15 treats. Descartes takes some treats. There are 68 treats left. How many treats did Descartes take?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 60

Answer:
Given,
There are 96 treats in a bowl. Newton takes 15 treats. Descartes takes some treats. There are 68 treats left.
96 – 15 = 81 treats
81 – 68 = 13 treats
Thus Descartes take 13 treats.

Use Addition to Subtract Homework & Practice 5.3

Add to find the difference.

Question 1.
92 – 67 = __
Big Ideas Math Answers 2nd Grade Chapter 5 Subtraction to 100 Strategies 61

Answer: 25
Big-Ideas-Math-Answers-2nd-Grade-Chapter-5-Subtraction-to-100-Strategies-61

Question 2.
43 – 24 = __
Big Ideas Math Answers 2nd Grade Chapter 5 Subtraction to 100 Strategies 62

Answer: 19
Big-Ideas-Math-Answers-2nd-Grade-Chapter-5-Subtraction-to-100-Strategies-62

Question 3.
71 – 42 = ___
Big Ideas Math Answers 2nd Grade Chapter 5 Subtraction to 100 Strategies 63

Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-5-Subtraction-to-100-Strategies-63

Question 4.
63 – 58 = __
Big Ideas Math Answers 2nd Grade Chapter 5 Subtraction to 100 Strategies 64

Answer: 5
Big-Ideas-Math-Answers-2nd-Grade-Chapter-5-Subtraction-to-100-Strategies-64

Question 5.
55 – 19 = __
Big Ideas Math Answers 2nd Grade Chapter 5 Subtraction to 100 Strategies 65

Answer: 36
Big-Ideas-Math-Answers-2nd-Grade-Chapter-5-Subtraction-to-100-Strategies-65

Question 6.
86 – 29 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 66

Answer: 57
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-66

Question 7.
YOU BE THE TEACHER
Descartes adds to find 35 − 12. Is he correct? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 67

Answer: Yes he is correct

Question 8.
Modeling Real Life
A farmer has 96 cornstalks. Some of them are sold. There are 38 left. How many cornstalks were sold?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 68

Answer:
Given,
A farmer has 96 cornstalks. Some of them are sold. There are 38 left.
96 – 38 = 58
58 cornstalks were sold

Question 9.
DIG DEEPER!
There are 56 bouncy balls in a pack. Your friend takes some. You take 23. There are 23 bouncy balls left. How many did your friend take?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 69

Answer:
Given,
There are 56 bouncy balls in a pack. Your friend takes some.
You take 23. There are 23 bouncy balls left.
56 – 23 = 33
33 – 23 = 10
Your friend take 10 bouncy balls.

Review & Refresh

Draw to show the time

Question 10.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 70

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-70

Question 11.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 71

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-71

Lesson 5.4 Decompose to Subtract

Explore and Grow

Color to show how you can get to a decade number by subtracting.

Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 72How can the equation above help you find 36 − 9?
______________________________
______________________________
______________________________

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-72

Show and Grow

Break apart the number being subtracted. Then find the difference. Use the number line to help.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 73

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 74

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-74

Question 2.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 75

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-75

Question 3.
75 – 7 = __

Answer: 68

Question 4.
82 – 6 = __

Answer: 76

Apply and Grow: Practice

Break apart the number being subtracted. Then find the difference.

Question 5.
47 – 8 = __

Answer: 39

Question 6.
56 – 9 = __

Answer: 47

Question 7.
43 – 5 = __

Answer: 38

Question 8.
62 – 6 = __

Answer: 56

Question 9.
__ = 41 – 4

Answer: 37

Question 10.
__ = 44 – 7

Answer: 37

Question 11.
Reasoning
Which equation is shown by the number line?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 76

Answer:
50 + 5 = 55
Mark 50 on the line and add 5 from 5 we get 55
55 – 8 = 47
First mark 55 on the line and count 8 before 55 we get 47
47 + 3 = 50
First, mark on 47 and add 3 from 47 we get 50 as an answer.

Think and Grow: Modeling Real Life

Your friend has 45 comic books. You have 8 fewer. How many comic books do you have?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 77
Subtraction equation:
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 78

Answer:
Given,
Your friend has 45 comic books. You have 8 fewer.
45 – 8 = 37 comic books

Show and Grow

Question 12.
Your friend can do 33 tricks on a yo-yo. You can do 9 fewer. How many tricks can you do?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 79

Answer:
Given,
Your friend can do 33 tricks on a yo-yo. You can do 9 fewer.
33 – 9 = 24

Question 13.
Descartes’s walk was 8 minutes longer than Newton’s. Descartes’s walk was 56 minutes. How long was Newton’s walk?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 80

Answer:
Given,
Descartes’s walk was 8 minutes longer than Newton’s.
Descartes’s walk was 56 minutes.
56 – 8 = 48 minutes
Thus Newton’s walk was 48 minutes.

Decompose to Subtract Homework & Practice 5.4

Break apart the number being subtracted. Then find the difference. Use the number line to help
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 81

Question 1.
95 – 6 = __
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 82

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-82

Question 2.
86 – 8 = __
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 83

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-83

Question 3.
89 – 9 = __

Answer: 80

Question 4.
82 – 7 = __

Answer: 75

Question 5.
__ = 83 – 5

Answer: 78

Question 6.
__ = 98 – 9

Answer: 89

Question 7.
Number Sense
Which way would you break apart 9 to find 25 – 9 ? Explain. Then find the difference.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 84

Answer: 16

Question 8.
Modeling Real Life
You build a train track with 32 pieces. You remove 6 pieces. How many pieces does the train track have now?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 85

Answer:
Given,
You build a train track with 32 pieces. You remove 6 pieces.
32 – 6 = 26 pieces

Question 9.
Modeling Real Life
You read 42 pages on Friday. You read 8 fewer pages on Saturday. How many pages do you read on Saturday?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 86

Answer:
Given,
You read 42 pages on Friday.
You read 8 fewer pages on Saturday.
42 – 8 = 34 pages

Review & Refresh

Question 10.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 87

Answer: 3 rows of 5
5 + 5 + 5 = 15

Lesson 5.5 Decompose to Subtract Tens and Ones

Explore and Grow

Color to show how you can break apart 16 to find the difference.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 88

Explain your strategy
______________________________
______________________________
______________________________

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-88

Show and Grow

Break apart the number being subtracted. Then find the difference.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 89

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-89

Question 2.
63 – 26 = __
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 90

Answer: 37
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-90

Apply and Grow: Practice

Break apart the number being subtracted. Then find the difference.

Question 3.
32 – 13 = __
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 91

Answer: 19
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-91

Question 4.
46 – 17 = __
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 92

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-92

Question 5.
93 – 45 = __

Answer: 48
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-92.1

Question 6.
71 – 24 = __

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-92.2

Question 7.
Structure
Can you use the equations to find 87 − 29?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 93

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-93

Think and Grow: Modeling Real Life

How many more pizzas do you sell than Newton?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 94
How many fewer pizzas does Newton sell than Descartes?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 95

Answer:
Number of pizzas Descartes sold = 57
Number of pizzas Newton sold = 38
57 – 38 = 19 fewer pizzas

Show and Grow

Question 8.
How many more tickets does Newton sell than Descartes?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 96

Answer:
85 – 47 = 38 more tickets
Thus Newton sell 38 more tickets than Descartes.

Question 9.
DIG DEEPER!
Your friend picks 49 apples. 11 apples are green and 14 apples are red. The rest are yellow. How many apples are yellow?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 97

Answer:
Given,
Your friend picks 49 apples. 11 apples are green and 14 apples are red. The rest are yellow.
11 + 14 = 25
49 – 25 = 24 yellow apples

Decompose to Subtract Tens and Ones Homework & Practice 5.5

Break apart the number being subtracted. Then find the difference.

Question 1.
45 – 16 = __
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 98

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-98

Question 2.
52 – 27 = __
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 99

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-99

Question 3.
84 – 55 = __

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-100

Question 4.
76 – 29 = __

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-101

Question 5.
__ = 23 – 14

Answer: 9

Question 6.
__ = 68 – 49

Answer: 19

Question 7.
Reasoning
Complete the number line and the equation.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 100

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-100 (2)

Question 8.
Modeling Real Life
How many more cups does Descartes sell than Newton?

Question 9.
DIG DEEPER!
In Exercise 8, 100 cups are sold for the day. Newton sold the rest. How many cups did Newton sell?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 101

Review & Refresh

Question 10.
47 + 32 = ___

Answer: 79

Question 11.
74 + 15 = __

Answer: 89

Lesson 5.6 Use Compensation to Subtract

Explore and Grow

Use mental math to find each difference.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 102

How did you use mental math and 41 − 20 to find each difference?
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 103

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-102

Show and Grow

Use compensation to subtract.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 104

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-104

Question 2.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 105

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-105

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 106

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-106

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 107

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-107

Apply and Grow: Practice

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 108

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-108

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 109

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-109

Question 7.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 110

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-110

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 111

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-111

Question 9.
35 – 7 = __

Answer: 28

Question 10.
53 – 37 = __

Answer: 16

Question 11.
Reasoning
Match the expressions that have the same difference.
Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies 112

Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-5-Subtraction-to-100-Strategies-112

Think and Grow: Modeling Real Life

You blow up 57 balloons for a carnival game. Some of them pop. There are 29 left. How many balloons popped?
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 113
Subtraction equation:
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 114

Answer:
Given,
You blow up 57 balloons for a carnival game. Some of them pop.
There are 29 left.
57 – 29 = 28
28 balloons are popped.

Show and Grow

Question 12.
There are 66 cars in a parking lot. Some of them leave. There are 31 left. How many cars leave the parking lot?
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 115

Answer:
Given,
There are 66 cars in a parking lot. Some of them leave. There are 31 left.
66 – 31 = 35
35 cars leave the parking lot.

Question 13.
You have 78 pictures on your tablet. You take 4 more pictures. Then you delete 17. How many pictures are on your tablet now?
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 116

Answer:
Given,
You have 78 pictures on your tablet. You take 4 more pictures. Then you delete 17.
78 + 4 = 82
84 – 17 = 67
67 pictures are on your tablet now.

Question 14.
You pick 59 strawberries. You eat 5 of them. Then you give 22 to your friend. How many strawberries do you have left?
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 117

Answer:
Given,
You pick 59 strawberries. You eat 5 of them. Then you give 22 to your friend.
59 – 5 = 54
54 – 22 = 32
Thus 32 strawberries are left.

Use Compensation to Subtract Homework & Practice 5.6

Use compensation to subtract.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 118

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-118

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 119

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-119

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 120

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-120

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 121

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-121

Question 5.
27 – 6 = __

Answer: 21

Question 6.
67 – 14 = __

Answer: 53

Use compensation to subtract.

Question 7.
77 – 52 = __

Answer: 25

Question 8.
69 – 44 = ___

Answer: 25

Question 9.
Reasoning
Use the numbers to complete the problem.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 122

Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-5-Subtraction-to-100-Strategies-122

Question 10.
Modeling Real Life
There are 36 boats on a lake. Some of them leave. There are 21 boats left. How many boats leave the lake?
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 123

Answer:
Given,
There are 36 boats on a lake. Some of them leave. There are 21 boats left.
36 – 21 = 15
15 boats leave the lake.

Review & Refresh

Question 11.
Circle groups of 3. Write a repeated addition equation to match.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 124

Answer: 4 groups of 3
3 + 3 + 3 + 3 =12

Lesson 5.7 Practice Subtraction Strategies

Explore and Grow

Use any strategy to find the difference.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 125

Compare your strategy to your partner’s strategy. How are your strategies alike? How are they different?
_______________________________
_______________________________
_______________________________

Answer: 42
Use mental math strategy to find the difference of the two numbers 76 and 34
76 – 34 = 42

Show and Grow

Question 1.
81 – 50 = ___

Answer: 31

Question 2.
94 – 8 = __

Answer: 86

Question 3.
58 – 49 = __

Answer: 9

Question 4.
77 – 35 = ___

Answer: 42

Apply and Grow: Practice

Question 5.
97 – 71 = __

Answer: 26

Question 6.
68 – 9 = __

Answer: 59

Question 7.
52 – 28 = ___

Answer: 24

Question 8.
83 – 60 = ___

Answer: 23

Question 9.
__ = 75 – 11

Answer: 64

Question 10.
___ = 46 – 35

Answer: 11

Question 11.
YOU BE THE TEACHER
Your friend uses compensation to subtract. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 126

Answer:
35 – 30 = 5
Yes, your friend is correct.

Question 12.
A store has 38 hats. 17 of them are sold. How many hats are left?
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 127

Answer:
Given,
A store has 38 hats. 17 of them are sold.
38 – 17 = 21
21 hats are left.

Think and Grow: Modeling Real Life

You have 41 toys. You put some away. There are 22 toys left. How many toys did you put away?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 128
Subtraction equation:
Big Ideas Math Solutions Grade 2 Chapter 5 Subtraction to 100 Strategies 129

Answer:
Given,
You have 41 toys. You put some away.
There are 22 toys left.
41 – 22 = 19 toys
Thus you put away 19 toys.

Show and Grow

Question 13.
A teacher has 62 prizes. He gives some away. There are 26 left. How many prizes did the teacher give away?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 130

Answer:
Given,
A teacher has 62 prizes. He gives some away. There are 26 left.
62 – 26 = 36
Thus the teacher gives away 36 prizes.

Question 14.
A roller-skating rink rents 28 pairs of skates. There are 52 pairs left. How many pairs of skates were there to start?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 131

Answer:
Given,
A roller-skating rink rents 28 pairs of skates. There are 52 pairs left.
52 + 28 = 80
80 pairs of skates were there to start.

Question 15.
A baker has 16 loaves of french bread and 31 loaves of wheat bread. She sells 18 loaves. How many loaves does the baker have left?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 132

Answer:
Given,
A baker has 16 loaves of french bread and 31 loaves of wheat bread. She sells 18 loaves.
16 + 31 = 47 loaves
47 – 18 = 29 loaves
The baker has left 29 loaves.

Practice Subtraction Strategies Homework & Practice 5.7

Question 1.
50 – 20 = __

Answer:
By using the mental math we can subtract the two numbers 50 and 20
50 – 20 = 30

Question 2.
62 – 30 = ___

Answer:
By using the mental math we can subtract the two numbers 62 and 30
62 – 30 = 32

Question 3.
88 – 64 = ___

Answer:
By using the mental math we can subtract the two numbers 88 and 64
88 – 64 = 24

Question 4.
42 – 17 = __

Answer:
By using the mental math we can subtract the two numbers 42 and 17
42 – 17 = 25

Question 5.
__ = 97 – 56

Answer:
By using the mental math we can subtract the two numbers 97 and 56
97 – 56 = 41

Question 6.
___ = 71 – 18

Answer:
By using the mental math we can subtract the two numbers 71 and 18
71 – 18 = 53

Question 7.
___ = 63 – 25

Answer:
By using the mental math we can subtract the two numbers 63 and 25
63 – 25 = 38

Question 8.
__ = 21 – 6

Answer:
By using the mental math we can subtract the two numbers 21 and 6
21 – 6 = 15

Question 9.
Number Sense
Which expressions have a difference of 24?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 133

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-133

Question 10.
Modeling Real Life
27 dogs were adopted from a shelter. There are 14 left. How many dogs were there to start?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 134

Answer:
Given,
27 dogs were adopted from a shelter. There are 14 left.
27 – 14 = 13

Question 11.
Modeling Real Life
86 hot dogs were sold at a baseball game. There are 14 left. How many hot dogs were there to start?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 135

Answer:
Given,
86 hot dogs were sold at a baseball game. There are 14 left.
86 – 14 = 72 hot dogs

Review and Refresh

Question 12.
Circle the shapes that show halves.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 136

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-136
The above two shapes are divided into two equal parts.
Thus the second figure and fourth figure are halves.

Lesson 5.8 Problem Solving: Subtraction

Explore and Grow

Model the story.

Newton collects 58 flowers. He gives 27 away. How many flowers does Newton have left?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 137

Answer:
Given,
Newton collects 58 flowers. He gives 27 away.
58 – 27 = 31
Thus 31 flowers are left.

Newton gives 11 more flowers away. How many flowers does Newton have now?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 138

Answer:
Given,
31 – 11 = 20
Newton has 20 flowers now.

Show and Grow

Question 1.
There are 60 kids at a summer camp. 26 are swimming. 15 are playing soccer. The rest are hiking. How many kids are hiking?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 139

Answer:
Total kids =  60
Swimming = 26
Playing soccer  = 15
Hiking = 60 – 26 – 15  = 19 .

Apply and Grow: Practice

Question 2.
Your class recycles 72 cans. You collected 18 of them. Your friend collected 9. The other students collected the rest. How many cans did the other students collect?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 140

Answer:
Given,
Your class recycles 72 cans. You collected 18 of them.
72 – 18 = 54
Your friend collected 9. The other students collected the rest
54 – 9 = 45
Therefore the other students collect 45 cans.

Question 3.
98 people visit the library in a week. 34 visit on Monday. 14 visit on Tuesday. How many people visit the library the rest of the week?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 141

Answer:
Given,
98 people visit the library in a week. 34 visit on Monday. 14 visit on Tuesday.
34 + 14 = 48
98 – 48 = 50
50 people visit the library the rest of the week.

Question 4.
You have 61 stickers. You give 24 stickers to your friend. Then you get 6 more stickers. How many stickers do you have now?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 142

Answer:
Given,
You have 61 stickers. You give 24 stickers to your friend.
61 – 24 = 37 stickers
Then you get 6 more stickers.
37 + 6 = 43 stickers
Thus you have 43 stickers now.

Think and Grow: Modeling Real Life

You collect 47 pine cones. Your friend collects 21 fewer than you. How many pine cones do you and your friend collect in all?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 143
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 144

Answer:
Given,
You collect 47 pine cones. Your friend collects 21 fewer than you.
47 – 21 = 26
Now we have to find how many pine cones do you and your friend collect in all.
47 + 26 = 73 pine cones

Show and Grow

Question 5.
A garbage truck collects trash from 68 trash cans. Another truck collects trash from 39 fewer cans. How many cans of trash do the trucks collect from in all?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 145

Explain the steps you used to solve the problem.
_________________________________
_________________________________
_________________________________

Answer:
Given,
A garbage truck collects trash from 68 trash cans. Another truck collects trash from 39 fewer cans
68 – 39 = 29 trash cans are collected by another cans.
68 + 29 = 97 trash cans

Problem Solving: Subtraction Homework & Practice 5.8

Question 1.
There are 39 people in a pool. 8 are floating on rafts. 11 are playing a game. The rest are swimming laps. How many people are swimming laps?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 146

Answer:
Given,
There are 39 people in a pool. 8 are floating on rafts. 11 are playing a game. The rest are swimming laps.
39 – 19 = 20
Thus 20 people are swimming laps.

Question 2.
Newton collects 24 rocks. He gives 13 to Descartes. Then Newton collects 8 more rocks. How many rocks does Newton have now?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 147

Answer:
Given,
Newton collects 24 rocks. He gives 13 to Descartes.
24 – 13 = 11
Then Newton collects 8 more rocks
11 + 8 = 19
Newton has 19 rocks now.

Question 3.
Modeling Real Life
You use 76 craft sticks. Your friend uses 62 fewer than you. How many craft sticks do you and your friend use in all?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 148

Answer:
Given,
You use 76 craft sticks. Your friend uses 62 fewer than you.
76 – 62 = 14 craft sticks
your friend uses 14 craft sticks
76 + 14 = 90 craft sticks

Question 4.
Modeling Real Life
You put 52 photos in an album. Your friend puts in 17 fewer than you. How many photos do you and your friend put in the album in all?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 149

Answer:
Given,
You put 52 photos in an album. Your friend puts in 17 fewer than you.
52 – 17 = 35 photos
52 + 35 = 87 photos
Thus 87 photos you and your friend put in the album in all.

Review & Refresh

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 150

Answer: A hexagon is a polygon that contains 6 straight sides and 6 vertices.

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 151

Answer: A triangle is a polygon that consists of 3 straight sides and 3 vertices.

Subtraction to 100 Strategies Performance Task

A science class uses an incubator to hatch chicken eggs.

Question 1.
The temperature of the incubator must be 99°F. The current temperature is 25°F less than the correct temperature. What is the current temperature?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 152

Answer:
Given,
The temperature of the incubator must be 99°F. The current temperature is 25°F less than the correct temperature.
99°F – 25°F = 74°F
The current temperature is 74°F.

Question 2.
a. The school has 4 incubators. Each one has 8 eggs. Each egg must be rotated 3 times every day. How many total rotations must be made to all of the eggs in all 4 incubators?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 153

Answer:
Given,
The school has 4 incubators.
Each one has 8 eggs. Each egg must be rotated 3 times every day.
8 × 4 = 32 eggs in all 4 incubators.
32 × 4 = 128 rotations

b. Two of your friends each complete 25 rotations. How many rotations are left?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 154

Answer:
Two of your friends each complete 25 rotations.
128 – 25 = 103 rotations are left

Question 3.
a. The first egg hatches at 11:00. The second egg hatches a half hour late. What time does the second egg hatch?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 155

Answer:
Given,
The first egg hatches at 11:00. The second egg hatches a half hour late.
11:00 + 00:30 = 11:30

b. The third egg hatches an hour after the second egg. What time does the third egg hatch?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 155.1

Answer:
Given,
The third egg hatches an hour after the second egg.
11:30 + 1:00 = 12:30

Subtraction to 100 Strategies Activity

Three in a Row: Subtraction

To Play: Players take turns. On your turn, spin both spinners. Subtract the numbers, and cover the difference on the game board. Continue until someone gets three in a row.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 156

Subtraction to 100 Strategies Chapter Practice 5

5.1 Subtract Tens and Ones Using a Number Line

Question 1.
90 – 40 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 157

Answer: 50
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-157

Question 2.
66 – 20 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 158

Answer: 46
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-157

5.2 Subtract Tens and Ones Using a Number Line

Question 3.
78 – 32 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 159

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-159

Question 4.
49 – 16 = ___
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 160

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-159

Question 5.
Modeling Real Life
You have 54 stickers and 21 key chains. How many more stickers do you have than key chains?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 161

Answer:
Given,
You have 54 stickers and 21 key chains.
54 – 21 = 33 stickers
Thus you have 33 stickers than key chains.

5.3 Use Addition to Subtract

Add to find the difference.

Question 6.
44 – 28 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 162

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-161

Question 7.
84 – 56 = __
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 163

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-162

5.4 Decompose to Subtract

Break apart the number being subtracted. Then find the difference.

Question 8.
44 − 5 = __

Answer: 39
Big Ideas Math Book 2nd Grade Answer Key Chapter 5 Subtraction to 100 Strategies img_164
Break the number into two parts and find the difference of two numbers.

Question 10.
67 − 9 = __

Answer: 58
Big Ideas Math Book 2nd Grade Answer Key Chapter 5 Subtraction to 100 Strategies img_165
Break the number into two parts and find the difference of two numbers.

Question 11.
32 − 4 = ___

Answer: 28
Big Ideas Math Book 2nd Grade Answer Key Chapter 5 Subtraction to 100 Strategies img_166
Break the number into two parts and find the difference of two numbers.

Question 12.
Number Sense
Which way would you break apart 8 to find 56 − 8? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 164

Answer: Both are correct to use break apart strategy to find the difference between 56 and 8.

5.5 Decompose to Subtract Tens and Ones

Break apart the number being subtracted. Then find the difference.

Question 13.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 165

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-165

Question 14.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 166

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-166

5.6 Use Compensation to Subtract

Use compensation to Subtract

Question 15.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 167

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-167

Question 16.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 168

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-168

Question 17.
Reasoning
Match the expressions that have the same difference.
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 169

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-5-Subtraction-to-100-Strategies-169

5.7 Practice Subtraction Strategies

Question 18.
58 − 36 = ___

Answer: 22

Question 19.
67 − 52 = ___

Answer: 15

5.8 Problem Solving: Subtraction

Question 20.
There are 31 papers to pass out. You pass out 12 papers. Then you pass out 8 more. How many papers do you have left?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 170

Answer:
Given,
There are 31 papers to pass out. You pass out 12 papers.
31 – 12 = 19 papers
Then you pass out 8 more.
19 + 8 = 27 papers are left.

Question 21.
47 students are on the playground. 23 fewer students are playing soccer than on the playground. How many students are on the playground and playing soccer in all?
Big Ideas Math Answer Key Grade 2 Chapter 5 Subtraction to 100 Strategies 171

Answer:
Given,
47 students are on the playground.
23 fewer students are playing soccer than on the playground.
47 – 23 = 24 students
47 + 24 = 71 students

Final Words:

I hope the data mentioned here about Big Ideas Math Answers Grade 2 Chapter 5 Subtraction to 100 Strategies are useful to score better marks in the exam. Students are advised to prepare well by referring Big Ideas Math Book Grade 2 Chapter 5 Subtraction to 100 Strategies Answer Key. Bookmark our site https://bigideasmathanswers.com/ to get the updates on other solution keys of Big Ideas Math Grade 2 Chapters.

Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions

Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions

Make your preparation perfect with the help of the Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions PDF. Every student can become a math expert if they prepare with the Big Ideas Grade 5 Chapter 2 Numerical Expressions Math Answers. The concepts are explained clearly with real-time examples. Therefore, students can easily refer to all the topics and understand them with the help of examples. Check out the problems and also take the practice test to test your knowledge. If you feel lag in any concept, then prepare that topic more and get a grip on all topics.

Big Ideas 5th Grade Chapter 2 Numerical Expressions Math Book Answer Key

Follow the step-by-step procedure to learn the complete concept deeply. We have given an explanation for every problem below. Therefore, students also check out the explanation if they feel confused while solving the problems. Quickly, refer to available Big Ideas Numerical Expressions 5th Grade Chapter 2 Math Book Answer Key and finish your preparation as early as possible. Click on the links provided below and prepare every individual topic on this page. Also, in this chapter, we were given different topics on Numerical Expressions, Order of operations, Number properties, Evaluating numerical expressions, and practice questions.

Lesson 1 Number Properties

Lesson 2 Order of Operations

Lesson 3 Write Numerical Expressions

Lesson 4 Evaluate Expressions with Grouping Symbols

Performance Task

Lesson 2.1 Number Properties

Explore and Grow

Use all four numbers on the game card below to write an expression that has a value of 24. You can use any number of the four operations: addition, subtraction, multiplication, and division.
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 1

Answer:
We will use addition, multiplication, and division to get the value as 24.

Explanation:
In the game card, we can see that the four numbers are 2, 3, 9, 1. So to get the value as 24 we will add 9 + 3 which is 12 and then we will multiply the value 12 with 2  which is 12 × 2 = 24 and then we will divide the value with 1 which is 24 ÷ 1= 24. So here the operations used are addition, multiplication, and division to get the value 24.

Reasoning
Write another expression that has a value of 24 using the game card above.
Answer:
We will use addition and multiplication to get the value as 24.

Explanation:
In the game card, we can see that the four numbers are 2, 3, 9, 1. So to get the value as 24 we will add 9 + 3 which is 12 and then we will multiply the value 12 with 2  which is 12 × 2 = 24. So to use another expression we will multiply the value with 1 which is 24 × 1= 24. So here the operations used are addition and multiplication to get the value 24.

Think and Grow: Use Number Properties

Key Idea
Here are several number properties.
Commutative Properties: Changing the order of addends or factors does not change the sum or product.
3 + 5 = 5 + 3
3 × 5 = 5 × 3
Associative Properties: Changing the grouping of addends or factors does not change the sum or product.
(2 + 4) + 1 = 2 + (4 + 1)
(2 × 4) × 1 = 2 × (4 × 1)
Addition Property of Zero: The sum of any number and 0 is that number.
8 + 0 = 8
Multiplication Properties of Zero and One:
The product of any number and 0 is 0. 5 × 0 = 0
The product of any number and 1 is that number. 7 × 1 = 7
Distributive Property: Multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products.
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 2
Example
Complete the equation. Identify the property shown.
32 + 29 + 8 = 29 + ______ + 8
_______ Property of Addition
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 3

Show and Grow

Question 1.
Complete the equation. Identify the property shown.
9 × 15 = 15 × ______
Answer:
The equation is 9 × 15 = 15 × 9. And the property which is used is the Commutative property.

Explanation:
The equation is 9 × 15 = 15 × 9. And the property which is used is the Commutative property.
Commutative Properties means changing the order of addends or factors does not change the sum or product. So the equation is 9 × 15 = 15  × 9.

Question 2.
Use the Distributive Property to find 8 × 49.
Answer:
The equation is (8 × 40) + (8 × 9) and the property used is Distributive property.

Explanation:
Given that 8 × 49, by distributive property the equation will be
= 8 × (40 + 9)
= (8 × 40) + (8 × 9)
So, Distributive Property is a property that multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products.

Apply and Grow: Practice

Complete the equation. Identify the property shown.
Question 3.
1 × _____ = 17
Answer:
The equation is 1 × 17= 17 and the property used is Multiplication Properties of One.

Explanation:
The equation is 1 × 17 = 17.
The property used to complete this equation is Multiplication Properties of One which means the product of any number and 1 is that number. So the equation will be 1 × 17 = 17.

Question 4.
248 + 0 = ______
Answer:
The equation is 248 + 0= 248 and the property used is the Addition Property of Zero.

Explanation:
The equation is 248 + 0= 248.
The property used is the Addition Property of Zero which means the sum of any number and 0 is that number. So the equation will be 248 + 0= 248.

Question 5.
23 + 145 + 7 = 23 + 7 + _____
Answer:
The equation is 23 + 145 + 7= 23 + 7 + 145. And the property which is used is the Commutative property.

Explanation:
The equation is 23 + 145 + 7= 23 + 7 + 145. And the property which is used is the Commutative property.
Commutative Properties means changing the order of addends or factors does not change the sum or product. So the equation is 23 + 145 + 7= 23 + 7 + 145.

Question 6.
3 × (10 + 2) = (3 × 10) + (3 × ______)
Answer:
The equation is 3 × (10 + 2)= (3 × 10) + (3 × 2) and the property used is the Distributive Property.

Explanation:
The equation is 3 × (10 + 2)= (3 × 10) + (3 × 2)
The Distributive Property is a property in which Multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. So the equation is 3 × (10 + 2)= (3 × 10) + (3 × 2).

Use the Distributive Property to find the product.
Question 7.
5 × 97
Answer:
The equation is
5 × 97 = 5 × (90 + 7)
= (5 × 90) + (5 × 7) and the property used is Distributive property.

Explanation:
Given that 5 × 97, by distributive property the equation will be
= 5 × (90 + 7)
= (5 × 90) + (5 × 7)
So, Distributive Property is a property that multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. So the equation is
5 × 97 = 5 × (90 + 7)
= (5 × 90) + (5 × 7).

Question 8.
83 × 7
Answer:
The equation is
83 × 7 = 7 × (80 + 3)
= (7 × 80) + (7 × 3) and the property used is Distributive property.

Explanation:
Given that 7 × 83, by distributive property the equation will be
= 7 × (80 + 3)
= (7× 80) + (7 × 3)
So, Distributive Property is a property that multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. So the equation is
7 × 83 = 7 × (80 + 3)
= (7 × 80) + (7 × 3)

Use a property to find the sum or product. Identify the property you used.
Question 9.
4 + (6 + 27)
Answer:
The sum of 4 + (6 + 27) is 37 and the property which is used is the Commutative property.

Explanation:
The sum of 4 + (6 + 27) is 37 and the property which is used is the Commutative property.
Commutative Properties means changing the order of addends or factors does not change the sum or product. So the sum is 37.

Question 10.
5 × 49 × 0
Answer:
The product of 5 × 49 × 0 is 0 and the property used is Multiplication Properties of Zero.

Explanation:
The product of 5 × 49 × 0 is 0 and the property used is Multiplication Properties of Zero.
Multiplication Properties of Zero is the product of any number and 0 is 0. So 5 × 49 × 0 is 0.

Question 11.
11 + 16 + 89
Answer:
The sum of 11 + 16 + 89 is 116 and the property which is used is the Commutative property.

Explanation:
The sum of 11 + 16 + 89 is 116 and the property which is used is the Commutative property.
Commutative Properties means changing the order of addends or factors does not change the sum or product. So the sum is 116.

Question 12.
YOU BE THE TEACHER
Your friend uses the Distributive Property to find 4 × 46. Is your friend correct? Explain.
4 × 46 = 4 × (50 – 4)
= (4 × 50) – (4 × 4)
= 200 – 16
= 184
Answer:
Yes, my friend is correct. He uses distributive property correctly.

Explanation:
Yes, my friend is correct. By distributive property, the equation will be
4 × 46 = 4 × (50 – 4)
= (4 × 50) – (4 × 4)
= 200 – 16
= 184.
So, Distributive Property is a property that multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. So the equation is

Question 13.
Can you use the Associative Property with subtraction? Explain. Use an example to justify your answer.
Answer:
No, we cannot use the Associative Property with subtraction.

Explanation:
No, we cannot use the Associative Property with subtraction. As Associative property means changing the grouping of addends or factors does not change the sum or product. So Associative property can only be used for addition or multiplication. And the subtraction doesn’t have the associative property because for example if we take 20 and10 and subtract the first two numbers 20 minus 10 then the result will be 10. So changing the way of associating the numbers in subtraction can change the result. So the subtraction doesn’t have the associative property.

Think and Grow: Modeling Real Life

Example
There are three types of animals at a shelter. How many animals are there in all?
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 4
Find the number of each type of animal at the shelter.
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 5
Add the numbers of dogs, cats, and rabbits.
16 + 22 + 4 = ____ + ____ + ____ Commutative Property of Addition
= _____ + _____ Add.
= _____ Add.
There are _____ animals in all.

Answer:
The total number of animals in the shelter is 42 animals.
Add the numbers of dogs, cats, and rabbits.
16 + 22 + 4 = 4 + 22 + 16 Commutative Property of Addition
= 26 + 16 Add.
= 42 Add.
There are 42 animals in all.

Explanation:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-2-Numerical-Expressions-2.1-5
Given that each paw is equal to four pets. So the dog has four paws which means 4 × 4 = 16. The total number of dogs is 16 dogs. And cats have five paws and a half paw so the total number of cats is 5 × 4= 20 as one paw is equal to four so the half paw is two animals. So the total number of cats is 20 + 2= 22 cats. And rabbit has one paw so the total number of rabbits is 4 rabbits. So to find the total number of animals is we will add all the animals and add the numbers of dogs, cats, and rabbits. So
16 + 22 + 4 = 4 + 22 + 16 Commutative Property of Addition
= 26 + 16 Add.
= 42 Add.
There are 42 animals in all.

Show and Grow

Question 14.
You play three different video games in an evening. How many minutes do you play video games in all?
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 6
Answer:
The total number of minutes the video game is playing is 66 minutes.

Explanation:
Given that each clock is six minutes. And given that dancing has three clocks and a half clock. So the total number of minutes in playing dancing is 3 × 6= 18 minutes, as each clock is 6 minutes, so the half clock will be 3 minutes. So the total number of minutes in playing dancing is 18 + 3= 21 minutes. And given that puzzle has two clocks and a half clock, so the total number of minutes in playing puzzle is 1 × 6= 6 as each clock is 6 minutes, so the half clock will be 3 minutes. So the total number of minutes in playing dancing is 6 + 3= 9 minutes. And given that racing has six clocks. So the total number of minutes in playing racing is 6 × 6= 36 minutes. So the total number of minutes in playing dancing is 18 + 3= 21 minutes. So the total number of minutes the video game is playing is
21 minutes + 9 minutes + 36 minutes which is 66 minutes.

Question 15.
DIG DEEPER!
Tickets for a school play are sold out. The auditorium has 4 sections. Each section has 25 rows with 15 seats in each row. Each ticket costs $2. How much money is raised in ticket sales?
Answer:
The total amount raised in the ticket sale is $3,000.

Explanation:
As there are four sections in the auditorium and each section has 25 rows with 15 seats in each row and each ticket costs $2. So first we need to find the total number of seats. For that, we should multiple numbers of rows and the seats which are 25 × 15= 375 seats. And there are four sections, so 4 × 375= 1,500. So there are 1500 total seats. And each ticket costs $2, for 1500 seats it will be 1500 × 2= 3,000. So the total amount raised in the ticket sale is $3,000.

Number Properties Homework & Practice 2.1

Complete the equation. Identify the property shown.
Question 1.
687 × ____ = 0
Answer:
The equation is 687 × 0 = 0 and the property used is Multiplication Properties of Zero.

Explanation:
The equation is 687 × 0 = 0 and the property used is Multiplication Properties of Zero.
As Multiplication Properties of Zero is the product of any number and 0 is 0. So 687 × 0 is 0.

Question 2.
15 + 13 = 13 + _____
Answer:
The equation is 15 + 13 = 13 + 15 and the property which is used is the Commutative property.

Explanation:
The equation is 15 + 13 = 13 + 15 and the property which is used is the Commutative property.
As the Commutative Properties means changing the order of addends or factors does not change the sum or product. So the equation will be 15 + 13 = 13 + 15.

Question 3.
4 × 7 × 25 = 4 × _____ × 7
Answer:
The equation is 4 × 7 × 25 = 4 × 25 × 7 and the property used is Associative Property.

Explanation:
The equation is 4 × 7 × 25 = 4 × 25 × 7 and the property used is Associative Property.
As Associative Property means changing the grouping of addends or factors does not change the sum or product.
4 × 7 × 25 = 4 × 25 × 7

Question 4.
6 × (20 – 3) = (6 × 20) – (6 × _____)
Answer:
The equation is 6 × (20 – 3) = (6 × 20) – (6 × 3) and the property used is Distributive property.

Explanation:
The equation is 6 × (20 – 3) = (6 × 20) – (6 × 3) and the property used is Distributive property.
Distributive Property is a property of Multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products.

Use the Distributive Property to find the product.
Question 5.
4 × 78
Answer:
The product is
4 × 78 = 4 × (70 + 8)
= (4 × 70) + (4 × 8)
= 280 + 32
= 312
and the property used is the Distributive Property.

Explanation:
The product is
4 × 78 = 4 × (70 + 8)
= (4 × 70) + (4 × 8)
= 280 + 32
= 312
The Distributive Property is a property in which Multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. So the product of 4 × 78 is
= 4 × (70 + 8)
= (4 × 70) + (4 × 8)
= 280 + 32
= 312.

Question 6.
23 × 8
Answer:
The product is
23 × 8 = 8 × (20 + 3)
= (8 × 20) + (8 × 3)
= 160 + 24
= 184
and the property used is the Distributive Property.

Explanation:
The product is
23 × 8 = 8 × (20 + 3)
= (8 × 20) + (8 × 3)
= 160 + 24
= 184
The Distributive Property is a property in which Multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. So the product of 23 × 8 is
= 8 × (20 + 3)
= (8 × 20) + (8 × 3)
= 160 + 24
= 184.

Use a property to find the sum or product. Identify the property you used.
Question 7.
6 × 43
Answer:
The product is
6 × 43 = 6 × (40 + 3)
= (6 × 40) + (6 × 3)
= 240 + 18
= 258.
and the property used is the Distributive Property.

Explanation:
The product is
6 × 43 = 6 × (40 + 3)
= (6 × 40) + (6 × 3)
= 240 + 18
= 258.
The Distributive Property is a property in which Multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. So the product of 6 × 43 is
= 6 × (40 + 3)
= (6 × 40) + (6 × 3)
= 240 + 18
= 258.

Question 8.
339 + 0 + 54
Answer:
The equation is 23 + 145 + 7= 23 + 7 + 145. And the property which is used is the Commutative property.

Explanation:
The sum is 339 + 0 + 54 = 393. And the property which is used is the Commutative property.
Commutative Properties means changing the order of addends or factors does not change the sum or product. So the equation is 339 + 0 + 54 = 54 + 339 + 0.

Question 9.
25 × 8 × 2
Answer:
The equation is 25 × 8 × 2 = 400. And the property which is used is the Commutative property.

Explanation:
The product is 25 × 8 × 2 = 400. And the property which is used is the Commutative property.
Commutative Properties means changing the order of addends or factors does not change the sum or product. So the equation is 25 × 8 × 2 = 25 × 8 × 2.

Question 10.
Number Sense
To find 29 + (11 + 16), your friend adds 29 and 11. Then he adds 16 to the sum. Which property did he use and why?
Answer:
The property which is used is the Commutative property. And the equation can be written as
29 + 11 + 16 = 29 + 11 + 16.

Explanation:
Given equation is 29 + (11 + 16) and my friend adds 29 and 11 after that he adds 16 to find the sum. And the property which is used is the Commutative property. And Commutative Properties means changing the order of addends or factors does not change the sum or product. So the equation can be written as
29 + 11 + 16 = 29 + 11 + 16.

Question 11.
Writing
Explain how using properties can help you mentally find answers to problems.
Answer:
Using the properties like Commutative property, Distributive property, Associative Properties, Addition Property of Zero, Multiplication Properties of Zero & One helps us mentally in finding answers to problems by this process and we can solve the problems easily by these properties and everyone can identify this property. With these properties, we can write the equivalent expressions which help us to solve mental problems.

Question 12.
Number Sense
Newton uses two properties to rewrite the expression (4 × 23) × 25. Identify the properties he uses. Why would Newton use these properties?
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 7
Answer:
The property Newton used is the Commutative property.

Explanation:
The property Newton used is the Commutative property. Newton uses Commutative Property because by Commutative property changing the order of addends or factors does not change the sum or product. So the equation can be written as 4 × 23 × 25. = 25 × 4 × 23.

Question 13.
Modeling Real Life
Before performing an experiment, students are asked to predict which substance will melt an ice cube the fastest. How many students make a prediction? Identify the property you used.
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 8
Answer:
The total number of students who have participated in the prediction is 76 students and the property used is commutative property.

Explanation:
Given that each emoji is 8 students, so the salt has two emojis that mean 2 × 8= 16 students. So the students who have predicted the salt are 16 students. And the sugar has three emojis and a half emoji that means 3 × 8= 24 students and a half emoji means, as each emoji is 8 students, so half emoji is 4 students and the total number of students is 24 + 4= 28 students. So the students who have predicted the sugar are 28 students. And the water has four emojis that means 4 × 8= 32 students. So the students who have predicted the water are 32 students.
The total number of students who have participated in the prediction is
16 students + 28 students + 32 students = 76 students. So the total number of students who have participated in the prediction is 76 students. And the property used is commutative property. And Commutative Properties means changing the order of addends or factors does not change the sum or product. So the equation can be written as
16 + 28 + 32= 32 + 16 +28.

Question 14.
Modeling Real Life
An apartment building has 35 floors with 12 apartments on each floor. There are 300 apartments that have 2 bedrooms. The rest of the apartments have 1 bedroom. How many 1-bedroom apartments are in the building? How can you use the Distributive Property to help solve this problem mentally?
Answer:

Review & Refresh

Find the sum. Check whether your answer is reasonable.
Question 15.
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 9
Answer:
By adding 8,968 and 4,683 we will get the sum of 13,651.

Explanation:
By adding 8,968 and 4,683 we will get the sum of 13,651. To check whether the answer is reasonable we will subtract 4,683 with 13,651 the difference will be 8,968. So the answer is reasonable.

Question 16.
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 10
Answer:
By adding 75,310 and 8,596 we will get the sum of 83,906.

Explanation:
By adding 75,310 and 8,596 we will get the sum of 83,906. To check whether the answer is reasonable we will subtract8,596 with 83,906 the difference will be 75,310. So the answer is reasonable.

Question 17.
Big Ideas Math Answer Key Grade 5 Chapter 2 Numerical Expressions 2.1 11
Answer:
By adding 90,583 and 19,877 we will get the sum of 110,460.

Explanation:
By adding 90,583 and 19,877 we will get the sum of 110,460. To check whether the answer is reasonable we will subtract 19,877 with 110,460 the difference will be 90,583. So the answer is reasonable.

Lesson 2.2 Order of Operations

Explore and Grow

Two students were asked to find the value of the expression below and they got different answers. Only one student has the correct answer. The students did not make any mistakes in their calculations. How did they get different answers?
24 + 16 ÷ 4 – 2
Answer:
The other student did not use the BODMAS rule so the other student calculation is not correct.

Explanation:
To find the value of the expression 24 + 16 ÷ 4 – 2 first we will divide 16 by 4 then the result will be 4. Now we will add result 4 with the number 24 which is 24 + 4= 28. And now we will subtract the number 2 with 28 which is
28 – 2= 26. This expression was done by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. And this was done by one student. Now the other student has done the same expression in a different way. The expression is 24 + 16 ÷ 4 – 2, so the other student solves the expression by first adding 24 and 16 which is 24 + 16= 40, and then we will divide 40 by 4, so the result will be 10. And now we will subtract 2 with 10 and the result will be 10 – 2= 8. So the other student’s answer will be 8 which is not correct because the other student did not use the BODMAS rule so the other student’s calculation is not correct.

Structure
Why is it important to have rules when finding values of expressions that contain more than one operation?
Answer:
It is important to have the rules when finding the values of expressions that contain more than one operation because the rules tell that the right order in which we will solve different parts of a math problem. And the order of operations is very important because it guarantees that we can all read and solve the problem in the same way.

Think and Grow: Use Order of Operations

Key Idea
A numerical expression is an expression that contains numbers and operations. When you evaluate a numerical expression, you find the value of the expression.
When evaluating a numerical expression, use a set of rules called the order of operations. These rules tell you the order in which to perform the operations.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 1
Order of Operations
1. Perform operations in parentheses
2. Multiply and divide from left to right
3. Add and subtract from left to right

Example
Evaluate 19 – 18 ÷ 6.
Using the order of operations, divide first. Then subtract.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 2
So, 19 – 18 ÷ 6 = _______.

Example
Evaluate 30 ÷ (3 + 7) × 2.
Using the order of operations, perform the addition in the parentheses first. Then multiply and divide from left to right.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 3
So, 30 ÷ (3 + 7) × 2 = ______.

Show and Grow

Evaluate the expression.
Question 1.
24 + 4 ÷ 2
Answer:
The value of the expression 24 + 4 ÷ 2 is 26.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right.  And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So first we will solve the division part of the expression which is 4 ÷ 2 = 2 and the result will be 2. Now we solve the addition part, we will add the result with 24 which is 24 + 2 = 26. So the value of the expression is 26.

Question 2.
12 + (10 – 3) × 8
Answer:
The value of the expression 12 + (10 – 3) × 8 is 68.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So first we will solve the parentheses part which is (10 – 3)= 7 and the result will be 7. Now we will solve the multiplication part which is 7 × 8= 56 and the result will be 56. And now we will solve the addition part which is 12 + 56= 68. So the value of the expression is 68.

Apply and Grow: Practice

Evaluate the expression.
Question 3.
25 + (10 ÷ 5)
Answer:
The value of the expression is 27.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So first we will solve the parentheses part which is (10 ÷ 5)= 2 and the result will be 2. Now we will add the result with 25 which is 25 + 2= 27. So the value of the expression is 27.

Question 4.
10 + 10 + 7
Answer:
The sum of the expression 10 + 10 +7 is 27.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. As there are no other expressions rather than addition we will perform addition to the given expression 10 + 10 + 7 and the sum of the expression
10 + 10 +7 is 27.

Question 5.
48 ÷ (8 – 2)
Answer:
The value of the expression 48 ÷ (8 – 2) is 8.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition and Subtraction. So first we will solve the parentheses which is (8 – 2)= 6. Now we will perform division operation for 48 by 6 and the result will be 8. So the value of the expression 48 ÷ (8 – 2) is 8.

Question 6.
(45 + 25) ÷ 10
Answer:
The value of the expression (45 + 25) ÷ 10 is 7.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition and Subtraction. So first will solve the parentheses part which is (45 + 25)= 70 and now we will perform division 70 ÷ 10 and the result will be 7. So the value of the expression (45 + 25) ÷ 10 is 7.

Question 7.
63 – 54 ÷ 9
Answer:
The value of the expression 63 – 54 ÷ 9 is 57.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition and Subtraction. Now first we will solve the division part which is 54 ÷ 9= 6 and the result is 6. Now we will subtract the result with 63 which is 63 – 6= 57. So the value of the expression 63 – 54 ÷ 9 is 57.

Question 8.
\(\left(\frac{1}{2}+\frac{1}{2}\right) \times 5\)
Answer:
The value of the expression \(\left(\frac{1}{2}+\frac{1}{2}\right) \times 5\) is 5.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the parentheses part which is \(\left(\frac{1}{2}+\frac{1}{2}\right) so the value will be 1. And now we will perform the multiplication part which is 1 × 5= 5. So the value of the expression [latex]\left(\frac{1}{2}+\frac{1}{2}\right) \times 5\) is 5.

Question 9.
15 + (16 – 6) × 1
Answer:
The value of the expression 15 + (16 – 6) × 1 is 25.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will perform the parentheses part which is (16 – 6)= 10 and the result will be 6. Now we will perform the multiplication part which is 10 × 1= 10 and the result will be 10. And now we will perform the addition which is 15 + 10 then the result will be 25. So the value of the expression 15 + (16 – 6) × 1 is 25.

Question 10.
(18 + 23 + 22) ÷ 9
Answer:
The value of the expression (18 + 23 + 22) ÷ 9 is 7.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will perform the parentheses part which is (18 + 23 + 22)= 63 and the result will be 63. Now we will perform division which is
63 ÷ 9= 7. So the value of the expression (18 + 23 + 22) ÷ 9 is 7.

Question 11.
80 – 6 × 5 × 2
Answer:
The value of the expression 80 – 6 × 5 × 2 will be 20.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the multiplication part which is 6 × 5 × 2= 60 and the result will be 60. Now we will solve the subtraction part which is 80 – 60= 20. So the value of the expression 80 – 6 × 5 × 2 will be 20.

Insert parentheses to make the statement true.
Question 12.
6 + 2 × 7 = 56
Answer:
The expression after inserting parentheses is (6 +2) × 7.

Explanation:
Given the expression is 6 + 2 × 7. So we will insert the parentheses for 6 + 2, then the expression will be
(6 +2) × 7. So we will solve the expression using the BODMAS rule, now first we will perform the parentheses part which is (6 +2)= 8. And now we will perform the multiplication part which is 8 × 7= 56. So the statement is true by inserting the parentheses for 6 + 2.

Question 13.
18 – 6 ÷ 2 + 5 = 11
Answer:
The expression after inserting parentheses is (18 – 6) ÷ 2 + 5.

Explanation:
Given the expression is 18 – 6 ÷ 2 + 5. So we will insert the parentheses for 18 – 6, then the expression will be
(18 – 6) ÷ 2 + 5. So we will solve the expression using the BODMAS rule, now first we will perform the parentheses part which is (18 – 6)= 12. And now we will perform division which is 12 ÷ 2= 6. Now we will perform the addition part which is 6 + 5= 11. So the statement is true by inserting the parentheses for 18 – 6.

Question 14.
10 + 2 × 4 – 1 = 16
Answer:
The statement is true by inserting parentheses for 4 – 1 and the expression is 10 + 2 ×(4 – 1).

Explanation:
Given the expression is 10 + 2 × 4 – 1. So we will insert the parentheses for 4 – 1, then the expression will be
10 + 2 ×(4 – 1). So we will solve the expression using the BODMAS rule, now first we will perform the parentheses part which is (4 – 1)= 3. And now we will perform the multiplication part which is 2 × 3= 6. And now we will add the result 6 with 10 which is 10 + 6= 16. So the statement is true by inserting parentheses for 4 – 1.

Question 15.
YOU BE THE TEACHER
Is Newton correct? Explain.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 4
Answer:
No, Newton is not correct.

Explanation:
No, Newton is not correct. Because Newton did not use the BODMAS rule to solve the expression. By using the BODMAS rule first Newton should solve the division part and then he should solve the addition part.

Question 16.
Writing
Describe how you can evaluate 9 × (40 + 5) two different ways.
Answer:
The expression 9 × (40 + 5) can be solved by the BODMAS way and the other way to solve the expression is by Distributive property.

Explanation:
Given the expression is 9 × (40 + 5), so to solve the expression we will use the BODMAS rule. So by the BODMAS rule first we will solve the parentheses part which is (40 + 5)= 45 and then we will solve the multiplication part which s 9 × 45= 405. This process is one way to solve the given expression. Another way to solve the expression is by Distributive property and the equation can be solved as
9 × (40 + 5) = (9 × 40) + (9 + 5)
= 360 + 451
= 405.
and this property is called as Distributive property which means Distributive Property is a property of Multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products.

Think and Grow: Modeling Real Life

Example
A robotics team orders 14 shirts. The order has a $9 shipping fee. Use the expression 14 × 12 + 9 to find how much the team spends on the order.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 5
Evaluate 14 × 12 + 9 using the order of operations.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 6
The team spends _____ on the order.

Answer:
The team spends $177 on the order.

Explanation:
As robotic team orders, 14 shirts and the order has a $9 shipping fee and the given expression is 14 × 12 + 9. We can solve this expression by the BODMAS rule in which BODMAS stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the multiplication part which is stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the multiplication part which is 14 × 12= 168 and then we will solve the addition part which is 168 + 9= 177. So the team spends $177 on the order.
Big-Ideas-Math-Answers-5th-Grade-Chapter-2-Numerical-Expressions-2.2-6

Show and Grow

Question 17.
A parking garage has 7 floors with 109 spaces on each floor. There are 486 spaces being used. Use the expression 7 × 109 – 486 to find how many spaces are not being used.
Answer:
The number of spaces that are not being used is 1,367.

Explanation:
As a parking garage has 7 floors with 109 spaces on each floor and there are 486 spaces being used, so given expression is 7 × 109 – 486. So the given expression first we will solve the multiplication part by the BODMAS rule in which BODMAS stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So now first we will solve the multiplication part which is 7 × 109= 1,853 and then we will solve the subtraction part which is
1853 – 486= 1,367. So the number of spaces that are not being used is 1,367.

Question 18.
At a fair, you ride the swinging ship 3 times and the bumper cars 5 times. Use the expression (3 × 2) + (5 × 4) to find how many tickets you use in all.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 7
Answer:
The number of tickets I have used is 26 tickets.

Explanation:
As a fair has a swinging ship and a bumper car and I rode the swinging ship 3 times and the bumper cars 5 times, so the given expression is (3 × 2) + (5 × 4). By the BODMAS rule first, we will solve the parentheses part and in which BODMAS stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So now first we will solve the parentheses part which is (3 × 2) + (5 × 4)= 6 + 20 and then we will solve the addition part which is 6 + 20= 26. So the number of tickets I have used is 26 tickets.

Question 19.
You download 128 songs and divide them into4 equal-sized playlists. You delete 1 playlist. Then you download 56 more songs. Use the expression 128 – (128 ÷ 4) + 56 to find how many songs you have now.
Answer:
The number of songs we had is 40 songs.

Explanation:
As I have downloaded 128 songs and divide them into 4 equal-sized playlists and then deleted 1 playlist. Then I have downloaded 56 more songs, so the given expression is 128 – (128 ÷ 4) + 56. By the BODMAS rule first we will solve the parentheses part and in which BODMAS stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So now first we will solve the parentheses part which is (128 ÷ 4)= 32 and now we will solve the addition part which is 32 + 56= 88, and now we will solve the subtraction part which is 128 – 88= 40. So the number of songs we have is 40 songs.

Question 20.
DIG DEEPER!
A politician buys 3 boxes of campaign buttons. There are 60 buttons in each box. He divides the buttons into4 equal groups. How many buttons are in each group?
Answer:
There will be 45 buttons in each group.

Explanation:
As a politician buys 3 boxes of campaign buttons and there are 60 buttons in each box. So the total number of buttons is 3 × 60= 180 buttons and then he divides the buttons into four equal groups. So the number of buttons in each group is 180 ÷ 4= 45. So there will be 45 buttons in each group.

Order of Operations Homework & Practice 2.2

Evaluate the expression.
Question 1.
(8 – 2) × 4
Answer:
The value of the expression (8 – 2) × 4 will be 24.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the parentheses part which is (8 – 2)= 6 and the result will be 6. Now we will solve the multiplication part which is 6 × 4= 24. So the value of the expression (8 – 2) × 4 will be 24.

Question 2.
7 + (6 ÷ 3)
Answer:
The value of the expression 7 + (6 ÷ 3) will be 10.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the parentheses part which is (6 ÷ 3)= 3 and the result will be 3. Now we will solve the addition part which is 7 + 3= 10. So the value of the expression 7 + (6 ÷ 3) will be 10.

Question 3.
5 × (6 + 2)
Answer:
The value of the expression 5 × (6 + 2) will be 40.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the parentheses part which is (6 + 2)= 8 and the result will be 8. Now we will solve the multiplication part which is 5 × 8= 40. So the value of the expression 5 × (6 + 2) will be 40.

Question 4.
21 + 42 ÷ 7
Answer:
The value of the expression 21 + 42 ÷ 7 will be 27.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the Division part which is 42 ÷ 7= 6 and the result will be 6. Now we will solve the addition part which is 21 + 6= 27. So the value of the expression 21 + 42 ÷ 7 will be 27.

Question 5.
\(\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)\) × 2
Answer:
The value of the expression \(\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)\) × 2 will be 3.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve parentheses part which is \(\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)\)= 6/4 and the result will be 3/2. Now we will solve the multiplication part which is 3/2 × 2= 3. So the value of the expression
\(\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)\) × 2 will be 3.

Question 6.
86 – 9 × 6
Answer:
The value of the expression 86 – 9 × 6 will be 32.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the multiplication part which is 9 × 6= 54 and the result will be 54. Now we will solve the subtraction part which is 86 – 54= 32. So the value of the expression 86 – 9 × 6 will be 32.

Question 7.
56 – 22 ÷ 2
Answer:
The value of the expression 56 – 22 ÷ 2 will be 45.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the division part which is 22 ÷ 2= 11 and the result will be 11. Now we will solve the subtraction part which is 56 – 11= 45. So the value of the expression 56 – 22 ÷ 2 will be 45.

Question 8.
27 + 11 × 4
Answer:
The value of the expression 27 + 11 × 4 will be 71.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the multiplication part which is 11 × 4= 44 and the result will be 44. Now we will solve the addition part which is 27 + 44= 71. So the value of the expression 27 + 11 × 4 will be 71.

Question 9.
21 – (12 + 6) ÷ 9
Answer:
The value of the expression 21 – (12 + 6) ÷ 9 will be 19.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the parentheses part which is (12 + 6)= 18 and the result will be 18. Now we will solve the division part which is 18 ÷ 9= 2 and now we will solve the subtraction part which is 21 – 2= 19. So the value of the expression 21 – (12 + 6) ÷ 9 will be 19.

Evaluate the expression.
Question 10.
9 × (5 + 15) – 42
Answer:
The value of the expression 9 × (5 + 15) – 42 will be 138.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the parentheses part which is (5 + 15)= 20 and the result will be 20. Now we will solve the multiplication part which is 9 × 20= 180 and now we will solve the subtraction part which is 180 – 42= 138. So the value of the expression 9 × (5 + 15) – 42 will be 138.

Question 11.
14 + 56 ÷ 7 – 6
Answer:
The value of the expression 14 + 56 ÷ 7 – 6 will be 16.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the division part which is 56 ÷ 7= 8 and the result will be 8. Now we will solve the addition part which is 14 + 8= 22 and now we will solve the subtraction part which is 22- 6= 16. So the value of the expression 14 + 56 ÷ 7 – 6 will be 16.

Question 12.
(106 + 350 + 244) ÷ 10
Answer:
The value of the expression (106 + 350 + 244) ÷ 10 will be 70.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will solve the parentheses part which is (106 + 350 + 244) and the result will be 700. Now we will solve the division part which is 700 ÷ 10= 70. So the value of the expression (106 + 350 + 244) ÷ 10 will be 70.

Insert parentheses to make the statement true.
Question 13.
36 – 9 ÷ 9 = 3
Answer:
The statement is true by inserting parentheses for 36 – 9.

Explanation:
Given the expression is 36 – 9 ÷ 9 = 3. So we will insert the parentheses for 36 – 9, then the expression will be
(36 – 9) ÷ 9. So we will solve the expression using the BODMAS rule, now first we will perform the parentheses part which is (36 – 9)= 27. And now we will perform the division part which is 27 ÷ 9= 3. So the statement is true by inserting parentheses for 36 – 9.

Question 14.
12 + 8 ÷ 4 + 1 = 6
Answer:
The statement is true by inserting parentheses for 12 + 8.

Explanation:
Given the expression is 12 + 8 ÷ 4 + 1 = 6. So we will insert the parentheses for 12 + 8, then the expression will be
(12 + 8) ÷ 4 + 1. So we will solve the expression using the BODMAS rule, now first we will perform the parentheses part which is (12 + 8)= 20. And now we will perform the division part which is 20 ÷ 4= 5, now we will solve the addition part which is 5 + 1=6. So the statement is true by inserting parentheses for 12 + 8.

Question 15.
10 + 4 × 12 – 6 = 34
Answer:
The statement is true by inserting parentheses for 12 – 6.

Explanation:
Given the expression is 10 + 4 × 12 – 6 = 34. So we will insert the parentheses for 12 + 8, then the expression will be 10 + 4 × (12 – 6). So we will solve the expression using the BODMAS rule, now first we will perform the parentheses part which is (12 – 6)= 6. And now we will perform the multiplication part which is 4 × 6= 24, now we will solve the addition part which is 10 + 24=34. So the statement is true by inserting parentheses for 12 – 6.

Question 16.
YOU BE THE TEACHER
Your friend says that because of the order of operations, the expressions are equivalent. Is your friend correct? Explain.
10 – (5 × 2) + 7    10 – 5 × 2 + 7
Answer:
Yes, my friend is correct.

Explanation:
Yes, my friend is correct. The expressions are equivalent because of the order of operations. By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So if we solve the expression without parentheses we will get the same result. So my friend is correct.

Question 17.
Number Sense
Which expressions have a value of 9?
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 8
Answer:
On solving all the expressions the expression 5 + 40 ÷ 5 – 4, 5 × (10 – 4) – 21 has got the value of 9.

Explanation:
To check which expression has a value of 9, we will follow the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So the first expression is
2 + 1 × 3= 2 + 3, here first we will solve the multiplication part by the order of operations and then we will solve the addition part.
on solving we will get the result as 5.
So the first expression didn’t get the value as 9.
The second expression is
3 × 4 – 1= 12 – 1, here first we will solve the multiplication part by the order of operations and then we will solve the addition part.
on solving we will get the result as 11.
So the second expression didn’t get the value as 9.
The third expression is
5 + 40 ÷ 5 – 4 = 5 + 8 – 4
= 13 – 4, here first we will solve the division part by the order of operations then we will solve the addition part, and then the subtraction part.
On solving we will get the result as 9.
So the expression has the value 9.
The fourth expression is
40 – 30 ÷ 10 + 8 = 40 – 3 +8
= 40 – 11, here first we will solve the division part by the order of operations then we will solve the addition part, and then the subtraction part.
On solving we will get the result as 29.
So the expression didn’t get the value as 9.
The fifth expression is
18 ÷ (8+0+18) = 18 ÷ (26), here first we will solve the parentheses part by the order of operations then we will solve the division part. By seeing the solution we can say that the expression didn’t get the value 9.
The sixth expression is
5 × (10 – 4) – 21 = 5 × (6) – 21
= 30 – 21, here first we will solve the parentheses part by order of operations then we will solve the multiplication part, and then we will solve the subtraction part.
On solving we will get the result as 9.
So the expression has the value 9.
On solving all the expressions the expression 5 + 40 ÷ 5 – 4, 5 × (10 – 4) – 21 has got the value of 9.

Question 18.
Modeling Real Life
Fifth graders at a school write a paper about a historical person for a contest. There are 5 classes of 25 students and 1 class of 28 students participating in the contest. Use the expression 5 × 25 + 28 to find how many students participate in the contest.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 9
Answer:
The number of students who participated in the contest is 153.

Explanation:
As there are 5 classes of 25 students and 1 class of 28 students participating in the contest and the given expression is 5 × 25 + 28. So, here first we will solve the multiplication part by the order of operations then we will solve the addition part. So
5 × 25 + 28= 125 + 28
On solving the expression we will get the result as 153. So the number of students who participated in the contest is 153.

Review & Refresh

Divide. Then check your answer.
Question 19.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 10
Answer:
On dividing 832 ÷ 8 we will get the result as 104.

Explanation:
On dividing 832 ÷ 8 we will get the result as 104. To check the answer we will multiply the result 104 and 8 which is 104 × 8= 832.

Question 20.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 11
Answer:
On dividing 215 ÷ 7 we will get the result as 30.71.

Explanation:
On dividing 215 ÷ 7 we will get the result as 30.71. To check the answer we will multiply the result 30.71 and 7 which is 30.71 × 7= 214.97.

Question 21.
Big Ideas Math Answers 5th Grade Chapter 2 Numerical Expressions 2.2 12
Answer:
On dividing 5,078 ÷ 5 we will get the result as 1015.6.

Explanation:
On dividing 5,078 ÷ 5 we will get the result as 1015.6. To check the answer we will multiply the result 1015.6 and 5 which is 1015.6 × 7= 5,078.

Lesson 2.3 Write Numerical Expressions

Explore and Grow

Write a real-life problem that can be represented by one of the expressions below. Switch papers with your partner. Which expression represents your partner’s problem? Explain.
7 × (8 + 5)
(7 × 8) + 5
Answer:
The expression represented by the partner is (7 × 8) + 5.

Explanation:
Given that the expressions are 7 × (8 + 5) and (7 × 8) + 5. So the solution for my problem is
7 × (8 + 5)= 7 × (12), here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. On solving we will get the result as 84. While checking my friend’s paper he solved the expression (7 × 8) + 5= 56 +5, here first we will solve the parentheses part by the order of operations then we will solve the addition part. On solving we will get the result as 61. AS the parentheses are different in the expression so the answer differs.

Make Sense of Problems
How can parentheses change the meaning of an expression? Explain.
Answer:
These parentheses in the expression represent that the expression or the equation should be solved first before any other calculation has done. The path between the parentheses represents as one number.

Think and Grow: Write an Interpret Expressions

Example
Write the words as an expression.
Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions 2.3 1
Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions 2.3 2
10 ÷ _______
The numerical expression is ______.

Answer:
The numerical expression is 10 ÷ (9 – 4).

Explanation:
The numerical expression using parenthesis is 10 ÷ (9 – 4). So, here first we will solve the parentheses part by the order of operations then we will solve the division part. So
10 ÷ (9 – 4) = 10 ÷ 5
on solving we will get the result as 5.

Example
Write the words as an expression. Then interpret the expression.
Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions 2.3 3
The numerical expression is _____.
The value of the expression is _______ times the sum 54 + 97.

Answer:
The numerical expression is (54 + 97) × 2.
The value of the expression is 2 times the sum of 54 + 97.

Explanation:
The numerical expression using parenthesis is (54 + 97) × 2. So, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. So
(54 + 97) × 2 = 151 × 2
on solving we will get the result as 302.
The numerical expression is (54 + 97) × 2.
The value of the expression is 2 times the sum of 54 + 97.

Show and Grow

Write the words as an expression
Question 1.
Multiply 8 and 5, then divide by 4.
Answer:
The numerical expression using parenthesis is (8 × 5) ÷ 4.

Explanation:
The numerical expression using parenthesis is (8 × 5) ÷ 4. So, here first we will solve the parentheses part by the order of operations then we will solve the division part. So
(8 × 5) ÷ 4= 40 ÷ 4
on solving we will get the result as 10.

Question 2.
Multiply the sum of 14 and 18 by 4.
Answer:
The numerical expression using parenthesis is (14 + 18) × 4.

Explanation:
The numerical expression using parenthesis is (14 + 18) × 4. So, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. So
(14 + 18) × 4= 32 × 4
on solving we will get the result as 128.

Write the words as an expression. Then interpret the expression.
Question 3.
Multiply 3 by the sum of 12 and 3.
Answer:
The numerical expression using parenthesis is 3 × (12 + 3).

Explanation:
The numerical expression using parenthesis is 3 × (12 + 3). So, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. So
3 × (12 + 3)= 3 × 15
on solving we will get the result as 45.

Question 4.
Subtract 30 from 50, then divide by 10.
Answer:
The numerical expression using parenthesis is (50 – 30) ÷ 10.

Explanation:
The numerical expression using parenthesis is (50 – 30) ÷ 10. So, here first we will solve the parentheses part by the order of operations then we will solve the division part. So
(50 – 30) ÷ 10= 20 ÷ 10
on solving we will get the result as 2.

Apply and Grow: Practice

Write the words as an expression. Then interpret the expression.
Question 5.
Add 238 and 12, then multiply by 3.
Answer:
The numerical expression using parenthesis is (238 + 12) × 3.

Explanation:
The numerical expression using parenthesis is (238 + 12) × 3. So, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. So
(238 + 12) × 3= 250 × 3
on solving we will get the result as 750.

Question 6.
Subtract 15 from 60, then divide by 9.
Answer:
The numerical expression using parenthesis is (60 – 15) ÷ 9.

Explanation:
The numerical expression using parenthesis is (60 – 15) ÷ 9. So, here first we will solve the parentheses part by the order of operations then we will solve the division part. So
(60 – 15) ÷ 9= 45 ÷ 9
on solving we will get the result as 5.

Write the words as an expression. Then evaluate the expression.
Question 7.
Multiply 5 by the difference of 25 and 20.
Answer:
The numerical expression is 5 × (25 – 20) and on evaluating we will get the value as 25.

Explanation:
The numerical expression using parenthesis is 5 × (25 – 20). So, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. So
5 × (25 – 20)= 5 × 5
on solving we will get the result as 25.

Question 8.
Add the product of 60 and 4 to the product of 5 and 4.
Answer:
The numerical expression is (60 × 4) + (5 × 4) and on evaluating we will get the value as 260.

Explanation:
The numerical expression using parenthesis is (60 × 4) + (5 × 4). So, here first we will solve the parentheses part by the order of operations then we will solve the addition part. So
(60 × 4) + (5 × 4)= 240 + 20
on solving we will get the result as 260.

Write the expression in words.
Question 9.
13 + (4 × 6)
Answer:
Add 13 to the product of 4 and 6.

Explanation:
To expand the given expression 13 + (4 × 6) in words we will be written as
Add 13 to the product of 4 and 6.

Question 10.
(4 + 8) ÷ 2
Answer:
Add 4 and 8 and then divide by 2.

Explanation:
To expand the given expression 13 + (4 × 6) in words we will be written as
add 4 and 8 and then divide by 2.

Question 11.
Newton has $20. He spends $4 on lunch and $13 at the store. Write an expression to represent the situation.
Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions 2.3 4
Answer:
The expression can be represented as
$20 – ($13 + $4)= $20 – ($17)

Explanation:
As Newton has $20 and he spends $4 on lunch and $13 at the store. So the expression can be represented as
$20 – ($13 + $4)= $20 – ($17)
here first we will solve the parentheses part by the order of operations then we will solve the subtraction part. On solving we will get $3.

Question 12.
Writing
Explain how you know which operation to use when writing words as an expression.
Answer:
The operation to use when writing words as expressions are
The sum is used as addition and the difference is used as subtraction and the product is used as the product and the quotient is used as the division.

Question 13.
Number Sense
Write the words as an expression. Then use a property of addition to write an equivalent expression.
Add 9 to the sum of 21 and 6.
Answer:
The numerical expression of the given expression is 9 + (21 + 6) and the property used to solve the expression is Distributive property.

Explanation:
Given that the expression in words is Add 9 to the sum of 21 and 6. The numerical expression of the given expression is 9 + (21 + 6). The property we can use to solve the expression is distributive property. By Distributive Property is a property in which Multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. So by the distributive property, the expression will be solved as
9 + (21 + 6)= 9 + 27, here first we will solve the parentheses part by the order of operations then we will solve the addition part. On solving we will get the result as 36.

Think and Grow: Modeling Real Life

Example
On your turn of a word game, you draw the card shown and create the word MATH. Your word score is the sum of the points of the letters you use. How many points do you earn on your turn?
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 2.3 5
Write an expression.
Think: Add _____, _____, _____, and _____, then multiply by _____. (____ + ____ + _____ + _____) × _____
Interpret the expression.
You earn ____ times as many points as the value of your word score.

Answer:
Add 3 + 1 + 1 + 4, then multiply by 3. (3 + 1 + 1 + 4) × 3
You earn 3 times as many points as the value of your word score.

Explanation:
Given that the word M has a score of 3 and the word A has a score of 1 and the word T has a score of 1 and the word H has a score of 4. As word score is the sum of the points of the letters you use so we will add the total score which is 3 + 1 + 1 + 4= 9 and now we will multiply by 3 to triple the score. So the expression will be
(3 + 1 + 1 + 4) × 3= (9) × 3, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. On solving we will get the result as 27.

Evaluate the expression.
(3 + 1 + 1 + 4) × 3 = ______ × _____
= ____
So, you earn _____ points on your turn.

Answer:
The points that I have earned in my turn is 27.

Explanation:
Given the expression is (3 + 1 + 1 + 4) × 3= (9) × 3, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. On solving we will get the result as 27. So the points that I have earned in my turn are 27.

Show and Grow

Question 14.
You make 2 batches of fruit salad. How many cups of fruit do you use in all?
Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions 2.3 6
Answer:
The number of cups of fruit does we use in all is 14 cups.

Explanation:
As there are two cups of strawberries, 1 1/2 cups blueberries, 2 1/2 cups grapes,1 cup pineapple and there are two batches of fruit salad. So the expression will be 2 × ( 2 + 1 1/2 + 2 1/2 + 1).
2 × ( 2 + 1 1/2 + 2 1/2 + 1)= 2 × ((4 + 3 + 5 + 2)/2)
= 2 × (14/2) here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. On solving we will get the result as 14. So the number of cups of fruit do we use in all is 14 cups.

Question 15.
A customer places an online order for a video game that costs $49 and 3 books that each cost $12. Shipping costs $6. What is the total cost of the order?
Answer:
The numerical expression will be 49 + (3 × 12) + 6 on solving the result will be 91.

Explanation:
As the customer places an online order for a video game that costs $49 and 3 books and each book costs $12 and the shipping cost is $6. So the numerical expression will be 49 + (3 × 12) + 6.
49 + (3 × 12) + 6= 49 + 36 +6, here first we will solve the parentheses part by the order of operations then we will solve the addition part. On solving we will get the result as 91.

Question 16.
DIG DEEPER!
Eight students use2 sets of 52 cards to play a game. The cards are divided equally among the players. How many cards does each player get?
Answer:
The numerical expression is (2 × 52) ÷ 8 on solving the result will be 13.

Explanation:
As there are eight students and uses two sets of 52 cards to play a game, so the cards are equally divided among the players. So the numerical expression will be (2 × 52) ÷ 8.
(2 × 52) ÷ 8= 104 ÷ 8, here first we will solve the parentheses part by the order of operations then we will solve the division part. On solving we will get the result as 13.

Question 17.
DIG DEEPER!
How much more does it cost to rent 2 adult bikes for 4 hours than a tandem bike for 4 hours? Explain.
Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions 2.3 7
Answer:
It costs four times more than the tandem bike.

Explanation:
To rent a two adult bike for four hours then a tandem bike for four hours.
So the numerical expression will be (2 × 4) – 4.
(2 × 4) – 4= 8 – 4, here first we will solve the parentheses part by the order of operations then we will solve the subtraction part. On solving we will get the result as 4.

Write Numerical Expressions Homework & Practice 2.3

Write the words as an expression.
Question 1.
Add 10 to the quotient of 72 and 8.
Answer:
The numerical expression is  (72 ÷ 8) + 10 and on evaluating we will get the value as 19.

Explanation:
The numerical expression using parenthesis is (72 ÷ 8) + 10. So, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. So
(72 ÷ 8) + 10= 9 + 10
on solving we will get the result as 19.

Question 2.
Subtract 55 from the sum of 124 and 56.
Answer:
The numerical expression is  (124 + 56) – 10 and on evaluating we will get the value as 170.

Explanation:
The numerical expression using parenthesis is (124 + 56) – 10. So, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. So
(124 + 56) – 10= 180 – 10
on solving we will get the result as 170.

Write the words as an expression. Then interpret the expression.
Question 3.
Add 14 and 13, then divide by 3.
Answer:
The numerical expression is  (14 + 13) ÷ 3 and on evaluating we will get the value as 9.

Explanation:
The numerical expression using parenthesis is (14 + 13) ÷ 3. So, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. So
(14 + 13) ÷ 3= 27 ÷ 3
on solving we will get the result as 9.

Question 4.
Subtract 29 from 39, then multiply by \(\frac{1}{2}\).
Answer:
The numerical expression is  (39 – 29) × 1/2 and on evaluating we will get the value as 5.

Explanation:
The numerical expression using parenthesis is (39 – 29) × 1/2. So, here first we will solve the parentheses part by the order of operations then we will solve the multiplication part. So
(39 – 29) × 1/2= 10 × 1/2
on solving we will get the result as 5.

Write the words as an expression. Then evaluate the expression.
Question 5.
Subtract the product of 5 and 8 from the product of 10 and 9.
Answer:
The numerical expression is (10 × 9) – (5 × 8) and on evaluating we will get the value as 50.

Explanation:
The numerical expression using parenthesis is (10 × 9) – (5 × 8). So, here first we will solve the parentheses part by the order of operations then we will solve the subtraction part. So
(10 × 9) – (5 × 8)= 90 – 40
on solving we will get the result as 50.

Question 6.
Add 15 to the quotient of 60 and 3.
Answer:
The numerical expression is 15 + (60 ÷ 3) and on evaluating we will get the value as 35.

Explanation:
The numerical expression using parenthesis is 15 + (60 ÷ 3). So, here first we will solve the parentheses part by the order of operations then we will solve the addition part. So
15 + (60 ÷ 3)= 15 + (20)
on solving we will get the result as 35.

Write the expression in words.
Question 7.
35 – (14 + 17)
Answer:
The expression in words is Subtract the sum of 14 and 17 with 35.

Explanation:
To expand the given expression 35 – (14 + 17) in words we will be written as
Subtract the sum of 14 and 17 with 35.

Question 8.
(30 – 20) × 10
Answer:
The expression in words is Subtract 30 and 20 and then multiply by 10.

Explanation:
To expand the given expression (30 – 20) × 10 in words we will be written as
Subtract 30 and 20 and then multiply by 10.

Question 9.
Descartes has $9. He works 5 hours and earns $8 each hour. Write an expression to represent the situation.
Answer:
The expression will be (5 × 8)+ 9 and on solving the expression we will get the result as 49.

Explanation:
As Descartes has $9 and he works 5 hours and earns $8 each hour, so the expression will be (5 × 8)+ 9.
(5 × 8)+ 9= 40 + 9, here first we will solve the parentheses part by the order of operations then we will solve the addition part. On solving the expression we will get the result as 49.

Question 10.
Open-Ended
Write a real-life problem that can be represented by the phrase “5 more than the sum of 15 and 7.”
Answer:
The numerical expression is (15 + 7) +5.

Explanation:
To represent the phrase 5 more than the sum of 15 and 7 in a numerical expression is (15 + 7) +5. As given that the 5 more than the sum of 15 and 7so we will add the number 5 to the sum of 15 and 7.

Question 11.
DIG DEEPER!
Write two different expressions that each represent the combined area of the rectangles. Then evaluate the expressions.
Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions 2.3 8
Answer:

Explanation:

Question 12.
Modeling Real Life
A music teacher replaces the strings on 3 violins, 2 violas, 4 cellos, and 1 bass. There are 4 strings on each of the instruments. How many strings does the teacher replace?
Answer:
The total number of strings replaced by the teacher is 37.

Explanation:
As music teacher replaces the strings on 3 violins, 2 violas, 4 cellos, and 1 bass and there are 4 strings on each of the instruments. So the teacher replaces 3 × 4= 12 violins and 2 × 4= 8 violas and 4 × 4= 16 cellos and 1 × 4= 1 bass. So the total number of strings replaced by the teacher is 12 + 8 + 16 + 1= 37.

Question 13.
DIG DEEPER!
A customer buys 2 shirts that cost $10 each and a pair of jeans that costs $14. What is the customer’s total after using the coupon?
Big Ideas Math Answers Grade 5 Chapter 2 Numerical Expressions 2.3 9
Answer:
The customer’s total after using the coupon is $24.

Explanation:
As the customer buys 2 shirts that cost $10 each which is 2 × 10= 20 and a pair of jeans that costs $14 which is
2 × 14= 28. So the total purchase is 20 + 28= 48 and the coupon is 1/2 off on the entire purchase. This means
48 × 1/2= 24. So the customer’s total after using the coupon is $24.

Review & Refresh

Subtract.
Question 14.
9\(\frac{3}{4}\) – 5\(\frac{1}{4}\) = _____
Answer:
On subtracting 9\(\frac{3}{4}\) – 5\(\frac{1}{4}\) we will get the result as
4\(\frac{2}{4}\).

Explanation:
To subtract 9\(\frac{3}{4}\) – 5\(\frac{1}{4}\) first we need to convert mixed fraction into improper fraction. So the improper fraction will be \(\frac{39}{4}\) – \(\frac{21}{4}\). As the denominator is equal which is 4 so on subtracting \(\frac{39}{4}\) – \(\frac{21}{4}\) we will get the result as \(\frac{18}{4}\). So the mixed fraction of the result is 4\(\frac{2}{4}\).

Question 15.
6\(\frac{1}{3}\) – 3\(\frac{1}{3}\) = ______
Answer:
On subtracting 6\(\frac{1}{3}\) – 3\(\frac{1}{3}\) we will get the result as
\(\frac{3}\).

Explanation:
To subtract 6\(\frac{1}{3}\) – 3\(\frac{1}{3}\) first we need to convert mixed fraction into improper fraction. So the improper fraction will be \(\frac{19}{3}\) – \(\frac{10}{3}\). As the denominator is equal which is 3 so on subtracting \(\frac{19}{3}\) – \(\frac{10}{3}\) we will get the result as \(\frac{9}{3}\) which is \(\frac{3}\).

Question 16.
4\(\frac{7}{12}\) – 1\(\frac{11}{12}\) = ______
Answer:
On subtracting 4\(\frac{7}{12}\) – 1\(\frac{11}{12}\) we will get the result as
\(\frac{8}{3}\).

Explanation:
To subtract 4\(\frac{7}{12}\) – 1\(\frac{11}{12}\) first we need to convert mixed fraction into improper fraction. So the improper fraction will be \(\frac{55}{12}\) – \(\frac{23}{12}\). As the denominator is equal which is 12 so on subtracting \(\frac{55}{12}\) – \(\frac{23}{12}\) we will get the result as \(\frac{32}{12}\) which is \(\frac{8}{3}\).

Lesson 2.4 Evaluate Expressions with Grouping Symbols

Explore and Grow

Write the words as an expression. How is the expression different from the expressions you wrote in previous lessons?

Multiply the sum of 4 and 5 by the difference of 8 and 7.
Answer:
The numerical expression is ( 4 + 5 ) × (8 – 7 ) and on evaluating we will get the result as 9.

Explanation:
Given, Multiply the sum of 4 and 5 by the difference of 8 and 7.
the sum of 4 and 5 is ( 4 + 5 )
the difference between 8 and 7 is ( 8 – 7 )
And the Multiplication of these two is  ( 4 + 5 ) × (8 – 7 ) = 9 × 1 = 9 .
Here we used simple mathematic rules for the expression other than using any properties and methods. And here we have only one pair of grouping symbols.

Precision
How can you evaluate an expression that has more than one pair of parentheses?
Answer: The expressions having more than one parenthesis can be evaluated by performing grouping symbols and in the next step, multiply and divide from left to right, Add and subtract from left to write if there are any,
parentheses( ), brackets [ ], braces { } are very helpful in grouping the symbols and to evaluate the numerical expression by performing inside operations parentheses, inside brackets, or inside braces.

Think and Grow: Evaluate with Grouping Symbols

Key Idea
Parentheses ( ), brackets [ ], and braces { } are called grouping symbols. You can write and evaluate numerical expressions that have more than one pair of grouping symbols.
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 2.4 1
Order of Operations (with Grouping Symbols)
1. Perform operations in grouping symbols.
2. Multiply and divide from left to right.
3. Add and subtract from left to right.
Example
Evaluate (25 – 5) ÷ (3 + 1).
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 2.4 2
So, (25 – 5) ÷ (3 + 1) = ______ .

Example
Evaluate [60 – (3 + 7)] × 5.
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 2.4 3

Show and Grow

Evaluate the expression.
Question 1.
(18 – 12) × (8 ÷ 4)
Answer:
On evaluating we will get the result as 12

Explanation:
Given that the numerical expression is (18 – 12) × (8 ÷ 4)
Firstly perform the operations within the parentheses,
Then we have,(8 ÷ 4) = 2,
(18 – 12) = 6, and now we will perform the multiplication part.
To evaluate the expression, multiply the resulting output from the inside parentheses operations, then 6 × 2 = 12
So , (18 – 12) × (8 ÷ 4) = 12

Question 2.
35 – [6 × (1 + 4)]
Answer:
On evaluating we will get the result as 5.

Explanation:
Given that the numerical expression is 35 – [6 × (1 + 4)]
Firstly perform the operations within the parentheses,
we have , (1 + 4) = 5 ,
Now performing operation inside brackets we get, [6 × 5] = 30,
and now we will  perform the subtraction part
Next, apply the resulted output in the expression then we have, 35 – 30 = 5.
So,  35 – [6 × (1 + 4)] = 5.

Apply and Grow: Practice

Evaluate the expression.
Question 3.
(60 – 12) ÷ (3 + 5)
Answer:
On evaluating we will get the result as 6

Explanation:
Given that the numerical expression is  (60 – 12) ÷ (3 + 5)
Firstly perform the operations within the parentheses,
we have , (60 – 12) = 48 ,
(3 + 5) = 8, now we will perform the subtraction part
Next, Divide the resulted outputs, we get 48 ÷ 8 = 6.
So , (60 – 12) ÷ (3 + 5) = 6

Question 4.
95 – (26 + 14) × (4 ÷ 2)
Answer:
On evaluating we will get the result as 15

Explanation:
Given that the numerical expression is 95 – (26 + 14) × (4 ÷ 2)
Firstly perform the operations within the parentheses,
we have (26 + 14) = 40 ,
(4 ÷ 2) = 2, now we will perform the multiplication part and then we will perform the subtraction part.
Now we have the given expression as 95 – ( 40 × 2 ), And calculating the inside parentheses we get ( 40 × 2 )= 80
Then 95 – 80 = 15
So , 95 – (26 + 14) × (4 ÷ 2) = 15 .

Question 5.
(8 + 2) × (13 + 7 + 5)
Answer:
On evaluating we will get the result as 250

Explanation:
Given that the numerical expression is (8 + 2) × (13 + 7 + 5)
Firstly perform the operations within the parentheses,
we have (8 + 2)  = 10,
(13 + 7 + 5) = 20, and now we will perform the multiplication part.
To evaluate the given expression apply the resulted output as 10 × 25= 250
So, (8 + 2) × (13 + 7 + 5) = 250.

Question 6.
2 × [(96 – 72) ÷ 8]
Answer:
On evaluating the numerical expression the result is 72.

Explanation:
Given that the numerical expression is 2 × [(96 – 72) ÷ 8]
Firstly perform the operations within the parentheses,
we have (96 – 72)  = 24,
[24 ÷ 8]= 3, and now we will perform the multiplication part.
To evaluate the given expression apply the resulted output as 24 × 3= 72
So, 2 × [(96 – 72) ÷ 8]= 72.

Question 7.
[(4 – 9) – (30 ÷ 6)] × 4
Answer:
On evaluating the expression we will get the result as 40.

Explanation:
Given that the numerical expression is [(4 – 9) – (30 ÷ 6)] × 4
Firstly perform the operations within the parentheses,
we have (4 – 9)  = (-5),
(30 ÷ 6) =5 which is [-5 -5]= 10 and now we will perform the multiplication part.
To evaluate the given expression apply the resulted output as 10 × 4= 40
So, [(4 – 9) – (30 ÷ 6)] × 4 = 40.

Question 8.
36 + {[(7 + 5) ÷ 6 × 9] + 24}
Answer:

Write the words as an expression. Then evaluate the expression.
Question 9.
Divide the sum of 35 and 28 by the sum of 3 and 4.
Answer:
The numerical expression is (35 + 28) ÷ (3 + 4) and on evaluating we will get the value as 9.

Explanation:
The numerical expression using parenthesis is (35 + 28) ÷ (3 + 4). So, here first we will solve the parentheses part by the order of operations then we will solve the division part. So
(35 + 28) ÷ (3 + 4)= 63 ÷ 7
on solving we will get the result as 9.

Question 10.
Add 11 to the product of 4 and 6, then divide by 5.
Answer:
The numerical expression is [(4 × 6) + 11] ÷ 5 and on evaluating we will get the value as 9.

Explanation:
The numerical expression using parenthesis is [(4 × 6) + 11] ÷ 5. So, here first we will solve the parentheses part by the order of operations then we will solve the addition part after that we will solve the division part. So
(35 + 28) ÷ (3 + 4)= 63 ÷ 7
on solving we will get the result as 9.

Question 11.
Reasoning
Will the value of {3 + [(38 – 10) – 1]} ÷ 3 change when the braces are removed? Explain.
Answer: Yes, the value will be changed if we remove the braces.

Explanation:
Yes, the value will be changed if we remove the braces because if the expression with braces means first we will solve the part of the braces and then we will solve by the order of operations. So by that, the value of the expression will be changed when the braces are removed.

Question 12.
DIG DEEPER!
Use each symbol once to make the number sentence true.
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 2.4 4
Answer:
The numerical expression will be (8 + 2) × (20 ÷ 4) – 11.

Explanation:
The numerical expression will be (8 + 2) × (20 ÷ 4) – 11. Let’s solve the expression to check the answer.
(8 + 2) × (20 ÷ 4) – 11= (10) × (5) – 11
= 50 – 11, here first we will solve the parentheses part by the order of operations then we will solve the addition part after that we will solve the subtraction part. On solving we will get the result as 39.

Think and Grow: Modeling Real Life

Example
A woman walks 2 miles from her house to the lake, walks 3 laps on the sidewalk around the lake, then walks2 miles back to her house. She does this each day for 3 days. How many miles does the woman walk in all?
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 2.4 5
Think: How many laps does she walk around the lake? How far does she walk to and from her house?
Write an expression.
[____ + (_____ × _____) + _____] × _____
Evaluate the expression.
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 2.4 6
So, the woman walks ______ miles in all.

Show and Grow

Question 13.
You collect 12 coins and 4 gems, complete 2 missions, and fail to complete 1 mission. How many points do you earn?
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 2.4 7
Answer:
The total number of points earned is 175.

Explanation:
The number of coins collected is 12, so the number of points earned for one coin is 5 so for 12 coins it will be
12 × 5= 60 points and the number of gems collected is 4, so the number of points earned for each gem is 10 so for 4 gems it will be 4 × 10= 40 points. The completed missions is 2, so the number of points earned for completing one mission is 50 so for 2 missions it will be 2 × 50= 100 points and the missions failed is 1, so the number of points lost for failing one mission is -25 so for 1 failed mission it will be 1× -25= -25 points. So the total number of points earned is (60 + 40 + 100) – 25= 200 -25, here first we will solve the parentheses part by the order of operations then we will solve the addition part after that we will solve the subtraction part. On solving we will get the result as 175. So the total number of points earned is 175.

Question 14.
DIG DEEPER!
A book has 200 pages. You read 10 pages twice each day for 1 week. What fraction of the book, in tenths, do you have left to read?
Answer:
Th fraction in tenths is 3/10.

Explanation:
As the book has 200 pages and 10 pages were read twice a day for a week as a week means seven days and twice a day means 2 × 7= 14. So the number of pages were read is 14 × 10= 140 pages. And the number of pages left is 200 – 140= 60. So to represent this in a fraction it will be 60/200 which is 3/10.

Evaluate Expressions with Grouping Symbols Homework & Practice 2.4

Evaluate the expression.
Question 1.
15 ÷ (5 × 1) – (2 × 1)
Answer:
On evaluating the given expression result will be 5.

Explanation:
Given the numerical expression is 15 ÷ (5 × 1) – (2 × 1). By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will perform the parentheses part which is (5 × 1) – (2 × 1)= 5 – 2= 3 and the result will be 3. Now we will perform division which is
15 ÷ 3= 5. So the value of the expression 15 ÷ (5 × 1) – (2 × 1) is 5.

Question 2.
(7 + 9) ÷ (2 + 4 + 2)
Answer:
On evaluating the given expression we will get the result as 2.

Explanation:
Given the numerical expression is (7 + 9) ÷ (2 + 4 + 2). By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will perform the parentheses part which is (7 + 9) ÷ (2 + 4 + 2)= 16 ÷ 8 and the result will be 16 ÷ 8. Now we will perform division which is
16 ÷ 8= 2. So the value of the expression (7 + 9) ÷ (2 + 4 + 2) is 2.

Question 3.
(1 + 5) × (2 + 3)
Answer:
On evaluating the given expression result will be 25.

Explanation:
Given the numerical expression is (1 + 5) × (2 + 3). By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will perform the parentheses part which is (1 + 5) × (2 + 3)= 5 × 5 and the result will be 5 × 5. Now we will perform the multiplication part which is 5 × 5= 25. So the value of the expression
(1 + 5) × (2 + 3) is 25.

Question 4.
[(16 – 14) + (9 × 4)] ÷ 2
Answer:
On evaluating the given expression result will be 9.

Explanation:
Given the numerical expression is [(16 – 14) + (9 × 4)] ÷ 2. By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will perform the parentheses part which is (16 – 14) + (9 × 4)= 2 + 36 and the result will be 38. Now we will perform the division part which is 38 ÷ 2= 19. So the value of the expression
[(16 – 14) + (9 × 4)] ÷ 2 is 19.

Question 5.
70 ÷ [(463 – 443) ÷ 2]
Answer:
On evaluating the given expression result will be 7.

Explanation:
Given the numerical expression is 70 ÷ [(463 – 443) ÷ 2]. By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will perform the parentheses part which is (463 – 443)= 20 and the result will be 20. Now we will perform the division part which is 20 ÷ 2= 10. And there is another division part which is 70 ÷ 10= 70. So the value of the expression 70 ÷ [(463 – 443) ÷ 2] is 7.

Question 6.
9 × {[14 ÷ 2 + (4 – 1)] – 8}
Answer:
On evaluating the given expression result will be 18.

Explanation:
Given the numerical expression is 9 × {[14 ÷ 2 + (4 – 1)] – 8}. By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Now first we will perform the parentheses part which is (4 – 1)= 3 and the result will be 3. Now we will perform the other parentheses part which is [14 ÷ 2 + (3)] = 7 + 3 and the result will be 10. Now we will perform the other parentheses part which is {10 – 8}= 2. Now we will perform the multiplication part which is 9 × 2= 18. So the value of the expression 9 × {[14 ÷ 2 + (4 – 1)] – 8} is 18.

Write the words as an expression. Then evaluate the expression.
Question 7.
Multiply the sum of 4 and 5 by the difference of 14 and 8.
Answer:
The numerical expression is (4 + 5) × (14 – 8) on evaluating the numerical expression we will get the result as 54.

Explanation:
Given that Multiply the sum of 4 and 5 by the difference of 14 and 8, so to represent this in an expression it will be (4 + 5) × (14 – 8). Now we will solve the numerical expression which is (4 + 5) × (14 – 8) = 9 × 6
Now first we have solved the parentheses part which is (4 + 5) × (14 – 8) and the result will be 9 × 6. And then we will perform the multiplication part which is 9 × 6= 54. So on evaluating the numerical expression we will get the result as 54.

Question 8.
Add 23, 26, and 17, then divide by 7.
Answer:
The numerical expression is (23 +26 +17) ÷ 7 on evaluating the numerical expression we will get the result as 9.4.

Explanation:
Given that Add 23, 26, and 17, then divide by 7, so to represent this in an expression it will be (23 +26 +17) ÷ 7. Now we will solve the numerical expression which is (23 +26 +17) ÷ 7= 66 ÷ 7
Now first we have solved the parentheses part which is (23 +26 +17) and the result will be 66. And then we will perform the division part which is 66 ÷ 7= 9.4. So on evaluating the numerical expression we will get the result as 9.4.

Question 9.
Writing
Explain how to evaluate the expression.
8 × [(36 – 33) × 2]
Answer:
On evaluating the numerical expression we will get the result as 48.

Explanation:
Given that the numerical expression is 8 × [(36 – 33) × 2]= 8 × [3 × 2]
Now first we have solved the parentheses part which is (36 – 33) and the result will be 3 and then we will perform the other parentheses part which is [3 × 2]= 6. And then we will perform the multiplication part which is
8 × 6= 48. So on evaluating the numerical expression we will get the result as 48.

Question 10.
Open-Ended
Write and evaluate two equivalent numerical expressions that show the Distributive Property.
Answer:
The two numerical equations are 5 × (90 + 7) and 83 × 7.

Explanation:
Let the equation be 5 × 97, by distributive property the equation will be
= 5 × (90 + 7)
= (5 × 90) + (5 × 7)
So, Distributive Property is a property that multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. So the equation is
5 × 97 = 5 × (90 + 7)
= (5 × 90) + (5 × 7).

Let the equation is 83 × 7, by distributive property the equation will be
= 7 × (80 + 3)
= (7× 80) + (7 × 3)
So, Distributive Property is a property that multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. So the equation is
7 × 83 = 7 × (80 + 3)
= (7 × 80) + (7 × 3)

Question 11.
Modeling Real Life
Your friend has 3 California postcards, 2 Hawaii postcards, and 4 New York postcards. He gives 1 postcard away, then divides the rest equally among 4 pages of a scrapbook. How many postcards are on each page?
Answer:

Question 12.
Modeling Real Life
A nutritionist recommends that fifth graders should eat about 305 grams of fruit each day.You eat the apple shown. How many more grams of fruit should you eat today?
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 2.4 8
Answer:

Review & Refresh

Find the product. Check whether your answer is reasonable.
Question 13.
639 × 5 = _____
Answer:
The product of 639 × 5= 3195.

Explanation:
The product of 639 × 5 is 3,195. So to check the answer is reasonable we will divide the result by 5 which is
3195 ÷ 5= 639. So the answer is reasonable.

Question 14.
7 × 1,926 = _____
Answer:
The product of 7 × 1,926 is  13,482.

Explanation:
The product of 7 × 1,926 is 13,482. So to check the answer is reasonable we will divide the result by 7 which is
13,482 ÷ 7= 1,926. So the answer is reasonable.

Question 15.
507 × 3 = ______
Answer:
The product of 507 × 3 is 1,521.

Explanation:
The product of 507 × 3 is 1,521. So to check the answer is reasonable we will divide the result by 3 which is
1,521 ÷ 3= 507. So the answer is reasonable.

Numerical Expressions Performance Task

Atoms are the basic building blocks of matter. When two or more atoms bond together, they form a molecule or a compound.
• The chemical formula for a water molecule is H2O because it has 2 atoms of hydrogen (H) and 1 atom of oxygen (O).
• The chemical formula for the compound sodium chloride, also known as table salt, is NaCl because it has 1 atom of sodium (Na) and 1 atom of chlorine (Cl) in a repeating pattern.
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 1
Question 1.
How many atoms are in 52 molecules of water?
Answer:

Question 2.
You and your friend want to make models for a science fair.
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 2
a. Your friend wants to make a model of sodium chloride. The model has the same number of green and orange atoms. There are 4 × 4 × 4 atoms in all. How many orange atoms do you need?
b. You make a model of a salt-water mixture. You use 14 atoms of sodium, 14 atoms of chlorine, and 10 molecules of H2O. Write an expression that represents the number of atoms in your salt-water model. Use the Distributive Property to rewrite your expression. Find the number of atoms in your model.
c. You and your friend buy 3 boxes of foam balls to represent atoms. There are 45 foam balls in each box. How many foam balls are left after you and your friend make the models above?
Answer:

Numerical Expressions Activity

Expression Boss
Directions:
1. Each player rolls three dice to complete the numerical expression. Players can arrange the numbers however they choose.
2. Each player evaluates the numerical expression.
3. Players compare their values. The player with the greater value earns one point.
4. If the values are equal, each player earns one point.
5. The player with the most points at the end of the game wins!
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions 3
Answer:

Numerical Expressions Chapter Practice

2.1 Number Properties

Complete the equation. Identify the property shown.
Question 1.
56 × _____ = 0
Answer:
The product of 56 × 0 is 0 and the property used is Multiplication Properties of Zero.

Explanation:
The product of 56 × 0 is 0 and the property used is Multiplication Properties of Zero.
As Multiplication Properties of Zero is the product of any number and 0 is 0. So 56 × 0 is 0.

Question 2.
8 + (242 + 32) = (8 + 242) + _____
Answer:
The sum of 8 + (242 + 32) = (8 + 242) + 32 and the property which is used is the Commutative property. And the sum of the given expression is 282.

Explanation:
Given the expression is 8 + (242 + 32) = (8 + 242) + 32 and the property which is used is the Commutative property. Commutative Properties means changing the order of addends or factors does not change the sum or product. And the sum of the given expression is 282.

Use the Distributive Property to find the product.
Question 3.
8 × 64
Answer:
The equation is 8 × 64= 8 × (60 + 4)
= (8 × 60) + (8 × 4) and the property used is Distributive property.
And the product of the given expression is 512.

Explanation:
Given that 8 × 64, by distributive property the equation will be
= 8 × (60 + 4)
= (8 × 60) + (8 × 4)
So, Distributive Property is a property that multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. And the product of the given expression is 512.

Question 4.
57 × 9
Answer:
The equation is 57 × 9= 9 × (50 + 7)
= (9 × 50) + (9 × 7) and the property used is Distributive property.
And the product of the given expression is 513.

Explanation:
Given that 57 × 9, by distributive property the equation will be
= 9 × (50 + 7)
= (9 × 50) + (9 × 7)
So, Distributive Property is a property that multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. And the product of the given expression is 513.

Use a property to find the sum or product. Identify the property you used.
Question 5.
3 × 92
Answer:
The equation is 3 × 92= 3 × (90 + 2)
= (3 × 90) + (3 × 2) and the property used is Distributive property.
And the product of the given expression is 276.

Explanation:
Given that 3 × 92, by distributive property the equation will be
= 3 × (90 + 2)
= (3 × 90) + (3 × 2)
So, Distributive Property is a property that multiplying a sum (or difference) by a number is the same as multiplying each number in the sum (or difference) by the number and adding (or subtracting) the products. And the product of the given expression is 276.

Question 6.
41 × 6 × 0
Answer:
The product of 41 × 6 × 0 is 0 and the property used is Multiplication Properties of Zero.

Explanation:
The product of 41 × 6 × 0 is 0 and the property used is Multiplication Properties of Zero.
As Multiplication Properties of Zero is the product of any number and 0 is 0. So 41 × 6 × 0 is 0.

Question 7.
13 + 24 + 57
Answer:
The sum of 13 + 24 + 57 = 24 + 57 + 13 and the property which is used is the Commutative property. And the sum of the given expression is 94.

Explanation:
Given the expression is 13 + 24 + 57 = 24 + 57 + 13 and the property which is used is the Commutative property. Commutative Properties means changing the order of addends or factors does not change the sum or product. And the sum of the given expression is 94.

Question 8.
Modeling Real Life
The graph shows the number Favorite Apple of votes each apple received. How many people were surveyed? Identify the property you used.
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions chp 8
Answer:
The total number of people surveyed is 46. And the property which can be used is the Commutative property

Explanation:
As given that each apple image is equal to four votes, so to know how many people are surveyed let’s first find the count of the apples and then the count of the votes. So Fuji has four apple images and a half apple image, so the total number of votes for Fuji is 4×4= 16 and a half apple image means two votes. So the total number of votes is 16 + 2= 18 votes. So the total number of votes for Fuji is 18 votes. And Granny Smith has three apple images, so the total number of votes for Granny Smith is 3×4= 12. And Honeycrisp has four apple images, so the total number of votes for Granny Smith is 4×4= 16. So the total number of people surveyed is 18 + 12 + 16= 46.
And the property which can be used is the Commutative property, as the Commutative Properties means changing the order of addends or factors does not change the sum or product. So the expression can be written as 18 + 12 + 16= 16 + 18 +12.

2.2 Order of Operations

Evaluate the expression.
Question 9.
18 × (9 – 3) ÷ 2
Answer:
The value of the expression 18 × (9 – 3) ÷ 2 is 54.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So first we will solve the parentheses part which is (9 – 3)= 6 and the result will be 6. Now we will solve the division part which is 6 ÷ 2= 3 and the result will be 3. And now we will solve the multiplication part which is 18 × 3= 54. So the value of the expression is 54.

Question 10.
50 – 18 ÷ 3 × 7
Answer:
The value of the expression 50 – 18 ÷ 3 × 7 is 8.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So first we will solve the division part which is 18 ÷ 3= 6 and the result will be 6. Now we will solve the multiplication part which is 6 × 7= 42 and the result will be 42. And now we will solve the subtraction part which is 50 – 42= 8. So the value of the expression is 8.

Question 11.
(36 + 14) × 4 ÷ 5
Answer:
The value of the expression (36 + 14) × 4 ÷ 5 is 40.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So first we will solve the parentheses part which is (36 + 14)×4= 50 × 4  and the result will be 200. Now we will solve the division part which is
200 ÷ 5 = 40 and the result will be 40. So the value of the expression is 40.

Question 12.
Number Sense
Which expressions have a value of 6?
Big Ideas Math Solutions Grade 5 Chapter 2 Numerical Expressions chp 12
Answer:

2.3 Write Numerical Expressions

Write the words as an expression. Then interpret the expression.
Question 13.
Multiply 8 by the difference of 54 and 49.
Answer:
The numerical expression is (54 – 49) × 8 on evaluating the numerical expression we will get the result as 40.

Explanation:
Given that Multiply 8 by the difference of 54 and 49, so to represent this in an expression it will be (54 – 49) × 8. Now we will solve the numerical expression which is (54 – 49) × 8 = 5 × 8
Now first we have solved the parentheses part which is (54 – 49) and the result will be 5. And then we will perform the multiplication part which is 5 × 8= 40. So on evaluating the numerical expression we will get the result as 40.

Question 14.
Subtract 58 from 94, then divide by4.
Answer:
The numerical expression is (94 – 58) ÷ 4 on evaluating the numerical expression we will get the result as 9.

Explanation:
Given that Subtract 58 from 94, then divide by4, so to represent this in an expression it will be (94 – 58) ÷ 4. Now we will solve the numerical expression which is (94 – 58) ÷ 4= 36 ÷ 4
Now first we have solved the parentheses part which is (94 – 58) and the result will be 36. And then we will perform the division part which is 36 ÷ 4= 9. So on evaluating the numerical expression we will get the result as 9.

Question 15.
Your friend buys a bag of 24 party favors and a bag of 16 party favors. She shares them equally among 5 friends. Write an expression to represent the problem.
Answer:

Write the expression in words.
Question 16.
45 + (7 × 8)
Answer:
Add 45 to the product of 7 and 8

Explanation:
Given that the numerical expression is 45 + (7 × 8), so in words, it will be represented as add 45 to the product of 7 and 8

Question 17.
(18 ÷ 2) – 5
Answer:
Divide 18 by 2 and then subtract 5.

Explanation:
Given that the numerical expression is (18 ÷ 2) – 5 so in words, it will be represented as divide 18 by 2 and then subtract 5.

2.4 Evaluate Expressions with Grouping Symbols

Evaluate the expression.
Question 18.
(28 – 14) × (42 ÷ 6)
Answer:
The value of the expression (28 – 14) × (42 ÷ 6) is 98.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So first we will solve the parentheses part which is (28 – 14) × (42 ÷ 6)= 14 × 7. Now we will solve the multiplication part which is 14 × 7= 98 and the result will be 98. So the value of the expression is 98.

Question 19.
6 × [(21 – 9) ÷ 3]
Answer:
The value of the expression 6 × [(21 – 9) ÷ 3] is 24.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So first we will solve the parentheses part which is(21 – 9)= 12. Now we will solve the other parentheses part which is [12 ÷ 3]= 4 and the result will be 4. Now we will solve the multiplication part which is 6 × 4= 24. So the value of the expression is 24.

Question 20.
[(2 × 2) + (10 ÷ 5)] × 4
Answer:
The value of the expression [(2 × 2) + (10 ÷ 5)] × 4 is 32.

Explanation:
By the order of operations, we will perform the addition in the parentheses first and then multiply and divide from left to right. And then we will add and subtract from left to right. So this is performed by the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. So first we will solve the parentheses part which is (2 × 2) + (10 ÷ 5)= 4 + 2 and the result will be 8. Now we will solve the multiplication part which is
8 × 4= 32. So the value of the expression is 32.

Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem

Big Ideas Math Answers Grade 8 Chapter 9

Redefine your true self using the Big Ideas Math Answers Grade 8 Ch 9 Real Numbers and the Pythagorean Theorem. Shed the inhibitions that are holding you back and begin your practice using the BIM Grade 8 Chapter 9 Solutions and clear the exam with better grades. Big Ideas Math Grade 8 Answers Chapter 9 Real Numbers and the Pythagorean Theorem is given as per the latest syllabus guidelines. Enhance your math skills by making the most out of the BIM Book Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem Solutions.

Big Ideas Math Book 8th Grade Answer Key Chapter 9 Real Numbers and the Pythagorean Theorem

Access the Topicwise Big Ideas Math Grade 8 Chapter 9 Real Numbers and Pythagorean Theorem Answers provided below via quick links. Help your child with the homework help he might need right after class and master the concepts. Kickstart your preparation taking the help of the BIM 8th Grade Ch 9 Real Numbers and Pythagorean Theorem Answer Key and identify the areas of need and allot time accordingly.

Performance Task

Lesson: 1 Finding Square Roots

Lesson: 2 The Pythagorean Theorem

Lesson: 3 Finding Cube Roots

Lesson: 4 Rational Numbers

Lesson: 5 Irrational Numbers

Lesson: 6 The Converse of the Pythagorean Theorem

Chapter: 9 – Real Numbers and the Pythagorean Theorem

Real Numbers and the Pythagorean Theorem STEAM Video/Performance Task

STEAM Video

Metronome Design
A metronome is a device that ticks at a constant rate. A metronome includes a pendulum, which swings back and forth in a precise time called a period. Why do musicians use metronomes?
Watch the STEAM Video “Metronome Design.” Then answer the following questions. The equation T = 0.2√L relates the period T (in seconds) and the length L (in centimeters) of a pendulum, where √L is the square root of L.
1. The table shows the square roots of several values of L. Use the pattern to find the values of √36 , √49 , and √64 .
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 1
2. What is the period of a pendulum that is 100 centimeters long? Justify your answer.

Answer:
1. √36 = 6
√49 = 7
√64 = 8
2. √100 = 10

Explanation:
In the above given table the square roots of 1,2, 3,4,5 are given
square root of 36 = 6
6 x 6 = 36
square root of  49 = 7
7 x 7 =49
square root of  64 = 8
8 x 8 =64
2. given that period of the pendulum is 100 centimeters.
square root of 100 =10
10 x 10 = 100

Performance Task

Identify and Correct the Error!
After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given the lengths of several pendulums.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 2
You will be asked to identify and correct errors in calculations of periods. Why is it important to pay attention to units when substituting values into a formula?

Answer:
0.438 sec, 1.219 sec, 1.612sec.

Explanation:
1 feet = 0.3048 seconds
so in the question given that the length of 1.44= 1.44 x 0.3048 sec
0.438 sec
4 x 0.3048 = 1.219 sec
5.29 x 0.3048 = 1.612 sec.

Real Numbers and the Pythagorean Theorem Getting Ready for Chapter 9

Chapter Exploration

When you multiply a number by itself, you square the number.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 3
To“undo” this, take the square root of the number.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 4
1. Work with a partner. Find the radius of each circle.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 5

Answer:
a. 6 in
b. 1 yd
c. 0.5 ft
d. (3/9) m
e. 0.7 cm
f. 1.2 in

Explanation:
a. The radius of circle = (√A/π)
where A = area of a circle, π = 3.14
given that area = 36 π in²
radius = (√ 36 π / π )
both in the numerator and denominator π  get canclled.
radius = ( √ 36)
square root of 36 is 6 x 6
√ 36 = √ 6²
where square and square root get cancelled.
radius = 6 in
b. The radius of circle = (√A/π)
where A = area of a circle, π = 3.14
given that area = π yd²
radius = (√ π / π )
both in the numerator and denominator π  get canceled.
radius = ( √ 1)
the square root of 1 is 1 x 1
√ 1= √ 1²
where square and square root get canceled.
radius =1 yd
c. The radius of circle = (√A/π)
where A = area of a circle, π = 3.14
given that area = 0.25 π ft
radius = (√ 0.25π / π )
both in the numerator and denominator π  get canceled.
radius = ( √ 0.25)
the square root of 0.25 is 0.5 x 0.5
√ 0.25 = √ 0.5²
where square and square root get canceled.
radius = 0.5 ft
c. The radius of circle = (√A/π)
where A = area of a circle, π = 3.14
given that area = 0.25 π ft
radius = (√ 0.25π / π )
both in the numerator and denominator π  get canceled.
radius = ( √ 0.25)
the square root of 0.25 is 0.5 x 0.5
√ 0.25 = √ 0.5²
where square and square root get canceled.
radius = 0.5 ft
d. The radius of circle = (√A/π)
where A = area of a circle, π = 3.14
given that area = (9/16) π m²
radius = (√ (9/16)π / π )
both in the numerator and denominator π  get canceled.
radius = ( √ (9/16)
the square root of (9/16) is (3/4) x (3/4)
√ (9/16) = √ (3/4)²
where square and square root get canceled.
radius = (3/4) m
e. The radius of circle = (√A/π)
where A = area of a circle, π = 3.14
given that area = (0.49) πcm²
radius = (√ (0.49)π / π )
both in the numerator and denominator π  get canceled.
radius = ( √ (0.49)
the square root of (0.49) is (0.7) x (0.7)
√ (0.49) = √ (0.7)²
where square and square root get canceled.
radius = (0.7) cm
f. The radius of circle = (√A/π)
where A = area of a circle, π = 3.14
given that area = (1.44) π in²
radius = (√ (1.44)π / π )
both in the numerator and denominator π  get canceled.
radius = ( √ (1.44)
the square root of (1.44) is (1.2) x (1.2)
√ (1.44) = √ (1.2)²
where square and square root get canceled.
radius = (1.2) in

2. WRITING GUIDELINES Work with a partner. Explain how you can find the radius and diameter of a circular object when you are given its area. Justify your answer using an example that is different from those in Exercise 1.

Answer:
The radius of circle = 8 cm
The diameter of circle = 16 cm

Explanation:
for example, we have to consider the area of a circle is  64π cm² given.
we have to find the radius and diameter of the circle.
The radius of circle = (√A/π)
where A = area of a circle, π = 3.14
the radius of circle = (√64π/π)
both in the numerator and denominator π  get canceled.
radius = ( √ (64)
the square root of (64) is (8) x (8)
√ (64) = √ (8)²
where square and square root get canceled.
radius = 8 cm
The radius of circle = (2√A/π)
where A = area of a circle, π = 3.14
the radius of circle = (2√64π/π)
both in the numerator and denominator π  get canceled.
diameter = ( 2√ (64)
the square root of (64) is (8) x (8)
√ (64) = 2√ (8)²
where square and square root get canceled.
diameter = 2 x 8 cm
diameter = 16 cm
Vocabulary
The following vocabulary terms are defined in this chapter. Think about what the terms might mean and record your thoughts.
square root
cube root
irrational number
perfect square
perfect cube

Answer:
square root = a number that produces a specified quantity when multiplied by itself.
cube root = the number which produces a given number when cubed.
irrational number = an irrational number is a number that cannot be expressed as a fraction for any integers.
perfect square = perfect square is another term for square number.
perfect cube = a perfect cube is an integer that is equal to some other integer raised to the third power
square root  of 36

Explanation:
36 = 6 x 6
√ 36 = √ 6 ²
where square and square root get canceled.
36 = 6
cube root of 343
343 = 7 x 7 x 7
3√ 343 = √ 7  ³
where cube and cube root get canceled.
343 = 7
irrational number
Example : √ 2, √ 3, √ 5, √ 11, √ 12, π (pi) are  all irrational.
perfect square
9 = 3 x 3
perfect cube
125 = 5 x 5 x 5

Lesson 9.1 Finding Square Roots

EXPLORATION 1

Finding Side Lengths
Work with a partner. Find the side length of each square. Explain your method.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 1

Answer:
a. 9yd
b. 18 cm
c. 19 mi
d. 15 mi
e. 1.7 in
f. 0.2 ft

Explanation:
Area of a square = a ²
where a = side
Given that A = 81yd ²
A = s x s
A = 9 x 9
a = 9
b.Given that A =324cm ²
A = s x s
A = 18 x 18
a = 18 cm
c. Given that A =361 mi ²
A = s x s
A = 19 x 19
a = 19 mi
d. Given that A =225 mi²
A = s x s
A = 15 x 15
a = 15 mi
e. Given that A =2.89in ²
A = s x s
A = 1.7 x 1.7
a = 1.7 in
f. Given that A =(4/9) ft ²
A = s x s
A = (2/3) x (2/3)
a = (2/3) ft

EXPLORATION 2

Finding Solutions of Equations
Work with a partner. Use mental math to solve each equation. How many solutions are there for each equation? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 2

Answer:
x = +0, -0
x = +1, -1
x = +2, -2
x = +3, -3
x = +4, -4
x = +5, -5

Explanation:
x² = 0, x = +0, -0
x² = 1, x =+1, -1
x² = 4, x = +2, -2
x² = 9, x = +3, -3
x² = 16, x = +4, -4
x² = 25, x = +5, -5

Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 3

Try It

Find the two square roots of the number.
Question 1.
36

Answer:
6 , -6

Explanation:
square root of 36 = 6x 6
6 and -6

Question 2.
100

Answer:
10 , – 10

Explanation:
square root of 100 = 10 x 10
10 and – 10

Question 3.
121

Answer:
11,  – 11

Explanation:
square root of 121 = 11 x 11
11 and – 11

Find the square root(s).
Question 4.
\(\sqrt{4}\)

Answer:
2

Explanation:
square root of 4 = 2 x 2
2

Question 5.
– \(\sqrt{81}\)

Answer:
9

Explanation:
square root of 81 = 9 x 9
9

Question 6.
± \(\sqrt{64}\)

Answer:
8

Explanation:
square root of 64 = 8 x 8
8

Find the square root(s).
Question 7.
– \(\sqrt{\frac{1}{100}}\)

Answer:
– 0.1

Explanation:
square root of (1/100) = (1/10 x 10)
(1/10) = 0.1
– 0.1

Question 8.
± \(\sqrt{\frac{4}{25}}\)

Answer:
± 0.4

Explanation:
square root of (4/25) = 0.16
0.16 = 0.4 x 0.4

Question 9.
\(\sqrt{12.25}\)

Answer:
3.5

Explanation:
square root of (12.25) = 3.5
12.25 = 3.5 x 3.5

Evaluate the expression.
Question 10.
12 – 3\(\sqrt{25}\)

Answer:
-3

Explanation:
square root of 25 = 5
12 – 3 (5)
12 – 15
-3

Question 11.
\(\sqrt{\frac{28}{7}}\) + 2.4

Answer:
4.4

Explanation:
(28/7) = 4
square root of 4 = 2
2 + 2.4
4.4

Question 12.
15 – (√4)2

Answer:
11

Explanation:
square root of 16  = 4
15 – 4
11

Solve the equation.
Question 13.
k2 = 169

Answer:
k = 13

Explanation:
square root of 169 = 13 x 13
k = 13

Question 14.
7n2 = 175

Answer:
n = 5

Explanation:
n² = (175/7)
n = 25
square root of 25  = 5 x 5
n = 5

Question 15.
190 = 4b2 – 6

Answer:
b = 7

Explanation:
190 = 4b2 – 6
4b² = 190 + 6
4b²=196
b² = (196/4)
b² = 49
b = 7

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

FINDING SQUARE ROOTS Find the square root(s).
Question 16.
\(\sqrt{256}\)

Answer:
16

Explanation:
square root of 256  = 16 x 16
16

Question 17.
– \(\sqrt{\frac{1}{9}}\)

Answer:
– (1/3)

Explanation:
square root of (1/9)  = (1/3)
square root of 9 = 3 x 3
given in – ve symbol
– (1/3)

Question 18.
± \(\sqrt{1.44}\)

Answer:
± 1.2

Explanation:
square root of 1.44 = 1.2
square root of 1.44 = 1.2 x 1.2
given in ± symbol
± 1.2

EVALUATING EXPRESSIONS Evaluate the expression.
Question 19.
\(\sqrt{\frac{81}{9}}\) – 7

Answer:
– 4

Explanation:
square root of 81 = 9 x 9
square root of  9 = 3x   3
(9/3) = 3
3 – 7 = -4

Question 20.
– 1 – \(\sqrt{121}\)

Answer:
-12

Explanation:
square root of 121  = 11 x 11
– 1 – 11
– 12

Question 21.
5 + (\(\sqrt{2}\))2

Answer:
7

Explanation:
square root of 4   = 2 x 2
5 + 2
7

SOLVING EQUATIONS Solve the equation.
Question 22.
2r2 = 162

Answer:
r = 9

Explanation:
2r2 = 162
r ² = (162/2)
r² = 81
square root of 81   = 9 x 9
r = 9

Question 23.
d2 + 5 = 41

Answer:
d = 6

Explanation:
d2 + 5 = 41
d ² = 41 – 5
d ²  = 36
square root of 36   = 6 x 6
d = 6

Question 24.
– 42 = 7b2 – 385

Answer:
b = – 7

Explanation:
– 42 = 7b2 – 385
7b ²  = – 385 + 42
7b ²  = – 343
b ²  = (- 343/7)
b ²  = – 49
square root of  – 49 = 7 x 7
b = – 7

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 25.
Your distance d (in miles) from the horizon can be approximated by d = 1.22√h, where h is your eye level (in feet above ground level). What is your eye level when you are 9.76 miles from the horizon?

Answer:
The eye level when you are 9.76 miles from the horizon = 11.907 ft

Explanation:
Given that the horizon can be approximated by 1.22 feet
so 1.22 x 11.907
11.907 ft
the eye level when you are 9.76 miles from the horizon = 11.907 ft.

Question 26.
DIG DEEPER!
The speed s (in meters per second) of a tsunami can be modeled by the function s = √9.8d, where d is the water depth (in meters).
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 4
a.What is the speed of the tsunami when the water depth is 500 meters?
b. What happens to the speed of the tsunami as the depth decreases? Explain.

Answer:
a. The speed of tsunami  = 4,900 sq meters.
b. if the speed of the tsunami as depth decreases it has low flow.

Explanation:
a. Given that the speed of the tsunami = √9.8d
√9.8 x 500 where d = 500 given
4,900 sq meters
b. if the speed of the tsunami as depth decreases the flow of the water is also less.

Finding Square Roots Homework & Practice 9.1

Review & Refresh

Evaluate the expression. Write your answer in scientific notation.
Question 1.
(4.3 × 103) + (2.4 × 103)

Answer
6.7 x 10 ³

Explanation:
(4.3 × 103) + (2.4 × 103)
(4.3 + 2.4) x 10³
6.7 x 10 ³

Question 2.
(1.5 × 10-2) – (3.5 × 10-3)

Answer:
2 × 10-2

Explanation:
(1.5 × 10-2) – (3.5 × 10-3)
(1.5 – 3.5) x 10-2
2 × 10-2

Question 3.
9 × (7 × 10-2)

Answer:
6.3 x  10 ¹

Explanation:
9 × (7 × 10-2)
(9 x 10) x 10-2
6.3 x  10 ¹

Question 4.
(6.6 × 10-5) ÷ (1.1 × 104)

Answer:
6 x 10-5

Explanation:
(6.6 × 10-5) ÷ (1.1 × 104)
(6.6/ 1.1) x 10-5
6 x 10-5

Make a scatter plot of the data. Identify any outliers, gaps, or clusters.
Question 5.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 5

Answer:

Explanation:
outliers = (8,33) and (9,36)
gap = 3.5 meters to 8 meters.
fillers=2 meters to 3 meters.

Question 6.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 6

Answer:

Explanation:
outliers = (1,4.05) and (1, 4.00)
gap = 0.5 gallons to 1 gallon
fillers =1meter to 1.5 meters.

Concepts, Skills, &Problem Solving

FINDING SIDE LENGTHS Find the side length s of the square. (See Exploration 1, p. 373.)
Question 7.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 7

Answer:
S = 21 cm

Explanation:
Area = 441 cm²
Area of square = a ²
where a = side
square root of 441 = 21 x 21
side length = 21

Question 8.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 8

Answer:
S = 1.3 km

Explanation:
Area = 1.69 km²
Area of square = a ²
where a = side
square root of 1.69 = 1.3 x 1.3
side length = 1.3km

Question 9.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 9

Answer:
S = (6/7) yd

Explanation:
Area = (36/49)²
Area of square = a ²
where a = side
square root of (36/49) = (6/7) x (6/7)
side length =(6/7) yd

FINDING SQUARE ROOTS OF A PERFECT SQUARE Find the two square roots of the number.
Question 10.
9

Answer:
3, – 3

Explanation:
square root of 9 = 3 x 3
3
given that two square roots
3, -3

Question 11.
64

Answer:
8, – 8

Explanation:
square root of 64 = 8 x 8
8
given that two square roots
8, -8

Question 12.
4

Answer:
2, -2

Explanation:
square root of 4 = 2 x 2
2
given that two square roots
2, -2

Question 13.
144

Answer:
12, -12

Explanation:
square root of 144 = 12 x 12
12
given that two square roots
12, -12

FINDING SQUARE ROOTS Find the square root(s).
Question 14.
√625

Answer:
25

Explanation:
√625
√25² therefore square and square root get canceled.
the square root of 625 is 25

Question 15.
± √196

Answer:
± 14

Explanation:
± √196
√14² therefore square and square root get canceled.
the square root of 196 is 14

Question 16.
– √1600

Answer:
– 40

Explanation:
– √1600
√40² therefore square and square root get canceled.
the square root of 1600 is 40

Question 17.
± √2500

Answer:
± 50

Explanation:
± √2500
√50² therefore square and square root get canceled.
the square root of 2500 is 50

Question 18.
\(\sqrt{\frac{1}{16}}\)

Answer:
(1/4)

Explanation:
√(1/16)
√(1/4)² therefore square and square root get canceled.
the square root of (1/16) is (1/4)

Question 19.
\(\sqrt{\frac{49}{576}}\)

Answer:
(7/24)

Explanation:
√(49/576)
√(7/24)² therefore square and square root get canceled.
the square root of (49/576) is (7/24)

Question 20.
± \(\sqrt{\frac{1}{961}}\)

Answer:
±(1/31)

Explanation:
√(1/961)
√(1/31)² therefore square and square root get canceled.
the square root of (1/961) is (1/31)

Question 21.
– \(\sqrt{\frac{9}{100}}\)

Answer:
– (3/10)

Explanation:
√(9/100)
√(3/10)² therefore square and square root get canceled.
the square root of (9/100) is (3/10)

Question 22.
± √4.84

Answer:
± 2.2

Explanation:
√4.84
√2.2² therefore square and square root get canceled.
the square root of 4.84 is 2.2

Question 23.
√7.29

Answer:
2.7

Explanation:
√7.29
√27² therefore square and square root get canceled.
the square root of 7.29 is 2.7

Question 24.
– √361

Answer:
– 19

Explanation:
-√361
√19² therefore square and square root get canceled.
the square root of  361 is 19

Question 25.
– √2.25

Answer:
– 1.5

Explanation:
– √2.25
√15² therefore square and square root get canceled.
the square root of  2.25 is 1.5

Question 26.
YOU BE THE TEACHER
Your friend finds ± \(\sqrt{\frac{1}{4}}\). Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 10
Answer:
± (1/2)

Explanation:
±√(1/4)
√(1/2)² therefore square and square root get canceled.
the square root of  (1/4) is (1/2)

Question 27.
MODELING REAL LIFE
The area of a square patch of fabric is 2.25 square inches. What is the side length of the patch?

Answer:
1.5 inch

Explanation:
Area of a square patch of fabric is 2.25
area of a square is s²
area = 1.5 x 1.5
side = 1.5 inch

Question 28.
CRITICAL THINKING
There are two square roots of 25. Why is there only one answer for the radius of the button?
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 11

Answer:
The radius of the button = 5 mm

Explanation:
Given that the area of a circle = πr²
25π = πr²
where π get canceled on both sides.
25 = r²
r = 5

NUMBER SENSE Copy and complete the statement with <, >, or =.
Question 29.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 12

Answer:
√81 >  8

Explanation:
square root  of 81 = 9 x 9
9 is greater then  8

Question 30.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 13

Answer:
0.5  = √0.25

Explanation:
square root  of 0.25 = 0.5 x 0.5
0.5 is equal to 0.5

Question 31.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 14

Answer:
(3/2)  < √(25/4)

Explanation:
square root  of (25/4) = (5/2) x (5/2)
(3/2) is less than (5/2)
(3/2) = 1.5
(5/2) = 3.5

EVALUATING EXPRESSIONS Evaluate the expression.
Question 32.
(√9)2 + 5

Answer:
14

Explanation:
square and square root get canceled.
9 + 5
14

Question 33.
28 – (√144)2

Answer:
-116

Explanation:
square and square root get canceled.
28 – 144
-116

Question 34.
3√16 – 5

Answer:
7

Explanation:
the square root of 16 is 4
16 = 4 x 4
3 x 4 – 5
12 – 5
7

Question 35.
10 – 4\(\sqrt{\frac{1}{16}}\)

Answer:
9

Explanation:
the square root of 16 is 4
16 = 4 x 4
10 – 4 (1/4)
10 – 4 (0.25)
10 – 1
9

Question 36.
√6.76 + 5.4

Answer:
12.56

Explanation:
square and square root get canceled
6.76 + 5.4
12.16

Question 37.
8√8.41 + 1.8

Answer:
25

Explanation:
square and square root get canceled
8 x 8.41 + 1.8
67.28 + 1.8
69.08

Question 38.
2(\(\sqrt{\frac{80}{5}}\) – 5)

Answer:
3

Explanation:
the square root of 16 is 4 x 4
(80/5) = 16
2 x 4 -5
8 – 5
3

Question 39.
4(\(\sqrt{\frac{147}{3}}\) + 3

Answer:
40

Explanation:
the square root of 49 is 7 x 7
(147/3) = 49
4 x 7 + 3
37 + 3
40

Question 40.
NUMBER SENSE
Without calculating, describe how the value of \(\sqrt{\frac{1}{a}}\) changes as a increases. Assume a > 0.
Answer:
0.5

Explanation:
Assuming a > 0
a = 4
the square root of 4 is 2
4 = 2 x 2
(1/2) = 0.5

SOLVING EQUATIONS Solve the equation.
Question 41.
x2 = 100

Answer:
x = 10 , x = -10

Explanation:
x2 = 100
x² = 10²
when powers are equal exponents must be equal.
x = 10

Question 42.
42 = d2 – 22

Answer:
d = 8 , d = 8

Explanation:
42 = d2 – 22
d² = 42 + 22
d² = 64
d² = 8²
when powers are equal exponents must be equal.
d = 8

Question 43.
4z2 = 144

Answer:
z = 6, z = -6

Explanation:
4z2 = 144
z² = (144/4)
z² = 36
z² = 6²
when powers are equal exponents must be equal.
z = 6

Question 44.
\(\sqrt{\frac{136}{9}}\) = \(\frac{1}{3}\) m2 – 10

Answer:
m = 4.8335, m = -4.8335

Explanation:
(136/9) = 45.33
(1/3) = 0.333
0.3333 m2 – 10
m2  = 10 – 0.33
m2 = 9.67
m = 4.835

Question 45.
0.25r2 = 49

Answer:
r = 14, r = -14

Explanation:
0.25r2 = 49
r² = (49/0.25)
r² = 196
r² = 14²
when powers are equal exponents must be equal.
r = 14

Question 46.
3h2 = h2 + 18

Answer:
h = 3, h = -3

Explanation:
3h2 = h2 + 18
3h² – h²  = 18
2h²  = 18
h²  = (18/2)
h²  = 9
h²  = 3²
when powers are equal exponents must be equal.
h = 3

Question 47.
YOU BE THE TEACHER
Your friend solves the equation 9x2 = 36. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 15

Answer:
Yes, my friend is correct.

Explanation:
9x2 = 36
x² = (36/9)
x² = 4
x² = 2²
when powers are equal exponents must be equal.
x =2

Question 48.
PROBLEM SOLVING
The period of a pendulum is the time the pendulum takes to complete one back-and-forth T swing. The period T (in seconds) can be modeled by the function T = 1.1√L, where L is the length (in feet) of the pendulum. Estimate the length of a pendulum with a period of 1.65 seconds.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 16

Answer:
The length of the pendulum with a period of 1.65 seconds = 0.50292 sec

Explanation:
Given that the period (in seconds) is 1.1
1.65 x 0.3048
1 feet = 0.3048 sec
0.50292

Question 49.
MODELING REAL LIFE
The area of a sail is 40\(\frac{1}{2}\) square feet. The base and the height of the sail are equal. What is the height of the sail?
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 17

Answer:
The height of the sail = 9 ft

Explanation:
Area of triangle = (b h/2)
area = 40 x (1/2)
40 x 0.5 = 20
h = 9 , b = 4
(9 x 4/2) = (36/2)
18 = area

Question 50.
REASONING
Is the product of two perfect squares always a perfect square? Explain your reasoning.

Answer:
Yes, the product of two perfect squares always a perfect square.

Explanation:
for example, we take 9
the square root of 9 = 3 x 3
3 is also a perfect square.

Question 51.
PROBLEM SOLVING
The kinetic energy (in joules) of a falling apple is represented byK = \(\frac{v^{2}}{2}\), where v is the speed of the apple (in meters per second). How fast is the apple traveling when the kinetic energy is 32 joules?

Answer:
The apple traveling when the kinetic energy is 32 joules = 8 meter per second.

Explanation:
Given that the apple is  represented by k
k = 2 x 2
k = 4
also given that 32 joules.
32 / 4 = 8 meter per sec.

Question 52.
PRECISION
The areas of the two watch faces have a ratio of 16:25.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 18
a. What is the ratio of the radius of the smaller watch face to the radius of the larger watch face?
b. What is the radius of the larger watch face?

Answer:
a. The ratio of the radius of the smaller watch face to the radius of the larger watch face = 4: 5
b. the radius of the larger watch face = 25

Explanation:
a. square root of (16/25) = (4/5)
ratio = 4: 5
b. the radius of the larger watch = 25

Question 53.
PROBLEM SOLVING
The cost (in dollars) of making a square window with a side length of n inches is represented by C = \(\frac{n^{2}}{2}\) + 175. A window costs $355. What is the side length (in feet) of the window? Justify your answer.

Answer:
The side length of the window = 2.5  feet

Explanation:
Given that the window costs = 355$
the side length of n inches is represented by c
the side length of the window  355 =  n x n /5+ 175
n = 30
(30/12)
2.5 ft
Question 54.
DIG DEEPER!
Albert Einstein’s most famous equation is E = mc2, where E is the energy of an object (in joules), is the mass of the object (in kilograms), and c is the speed of light (in meters per second). A hydrogen atom has 15.066 × 10-7 joule of energy and a mass of 1.674 × 10-27 kilogram. What is the speed of light? Write your answer in scientific notation.

Answer:
The speed of the light =3 x 10²⁰ meters per second.

Explanation:
Given that mass of the object = 1.674 x  10-27
energy of the object = 15.066 x 10-7
Albert Einsteins equation = E = mc2
where m = mass , e = energy, c = speed
c² = (E/M)
c² = (15.066/1.674)
c² = 9 x 10²⁰
c² = 3²
c = 3 x 10²⁰

Question 55.
GEOMETRY
The area of the triangle is represented by the formula A = \(\sqrt{s(s-21)(s-17)(s-10)}\), where s is equal to half the perimeter. What is the height of the triangle?
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.1 19
Answer:
The height of the triangle = 8 cm

Explanation:
Area of the triangle = 2 (A/b)
where b = base A = area
area = 2 x 4
area = 8 cubic cm

Question 56.
WRITING
Can you find the square root of a negative number? Explain.

Answer:
Yes, we can find the square root of a negative number.

Explanation:
The square root of – 144 is – 12 x 12
the square root of – ve number is also a + ve.

REASONING Without solving, determine the number of solutions of the equation.
Question 57.
x2 = 1

Answer:
2 solutions
x = 1, x = -1

Explanation:
x² = 1²
when powers are equal exponents must be equal.
x = 1

Question 58.
b2 = – \(\sqrt{\frac{1}{9}}\)

Answer:
b = -0.16, b = 0.16

Explanation:
b2 = – \(\sqrt{\frac{1}{9}}\)
square root of (1/9) = (1/3)
b² = – (1/3)
b² = – 0.33
b = 0.16

Question 59.
z = √-144

Answer:
z = 0

Explanation:
√-144 = √(12 x 12 )²
√-144 = 12
square and square root get cancelled.
z = -12

Lesson 9.2 The Pythagorean Theorem

Pythagoras was a Greek mathematician and philosopher who proved one of the most famous rules in mathematics. In mathematics, a rule is called a theorem. So, the rule that Pythagoras proved is called the Pythagorean Theorem.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 1

EXPLORATION 1

Discovering the Pythagorean Theorem
Work with a partner.

  • On grid paper,draw a right triangle with one horizontal side and one vertical side.
  • Label the lengths of the two shorter sides a and b. Label the length of the longest side c.
  • Draw three squares that each share a side with your triangle. Label the areas of the squares a2, b2, and c2.
  • Cut out each square. Then make eight copies of the right triangle and cut them out.

Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 2
a. Arrange the figures to show how a2 and b2 relate to c2. Use an equation to represent this relationship.
b. Estimate the side length of your triangle. Then use the relationship in part(a) to find c. Compare the values.

Answer:
a. a² + b² = c²
b. 9 + 16 = 25

Explanation:
a. a² + b² = c²
b. we are estimating a = 3 , b = 4 , c =5
a² + b² = c²
3² + 4² = 5²
9 + 16 = 25

Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 3

Try It

Find the length of the hypotenuse of the triangle.
Question 1.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 4

Answer:
Length = -6, 6

Explanation:
a² + b² = c²
15 – 8 = 6

Question 2.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 5

Answer:
Length = 0.7

Explanation:
a² + b² = c²
0.4² + 0.3² =0.7²
length = 0.7

Find the missing length of the triangle.
Question 3.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 6

Answer:
Length of the triangle = 8 , -8

Explanation:
a² + b² = c²
16² -8 =8²
length = 8yd

Question 4.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 7

Answer:
Length of the triangle = 0.8 m

Explanation:
a² + b² = c²
0.8² + 9.6² =10.4²
length = 0.8 m

Find x
Question 5.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 8

Answer:
x = 10 yd

Explanation:
a² + b² = x²
9² + 1² =10²
length = 10 yd

Question 6.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 9

Answer:
x = 15.4 ft

Explanation:
a² + b² = x²
6.6² + 8.8² =15.4²
length = 15.4 ft

Find the distance between the points.
Question 7.
(3, 6) and (7, 9)

Answer:
3, 2

Explanation:
(3,6) = (6- 3) = 3
(7 , 9) = (9 – 7) = 2

Question 8.
(- 3, – 4) and (2, 8)

Answer:
-7 , 6

Explanation:
(-3 ,-4) = (-4- -3) = -7
(2 , 8) = (8 – 2) = 6

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

FINDING A MISSING LENGTH Find x.
Question 9.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 10

Answer:
The missing length = 1.6 ft

Explanation:
a² + x² = c²
2.4² + 1.6² =4²
length = 1. 6 ft

Question 10.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 11

Answer:
The missing length = 2.8 cm

Explanation:
a² + x² = c²
1.6² + 1.2² =2.8²
length = 2.8 cm

Question 11.
FINDING A DISTANCE
Find the distance between (- 5, 2) and (7, – 7).

Answer:
– 3 , 0

Explanation:
(-5 ,2) = (-4 -2) = -3
(7 , -7) = (7 – 7) = 0
-3 , 0

Question 12.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 12

Answer:
The hypotenuse = c
longest side = c
leg = c
opposite to right angle = c

Explanation:
By using Pythagoras theorem
a² + x² = c²
The hypotenuse = c
longest side = c
leg = c
opposite to right angle = c

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 13.
A zoo keeper knows that an escaped red panda is hiding somewhere in the triangular region shown. What is the area (in square miles) that the zoo keeper needs to search? Explain.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 13

Answer:
The area that zoo keeper needs to search = 9.8 sq m

Explanation:
By using Pythagoras theorem
a² + b² = c²
2.4 + 7.4 = 9.8 sq m
c² = a² + b²

Question 14.
DIG DEEPER!
Objects detected by radar are plotted in a coordinate plane where each unit represents 1 mile. The point (0, 0) represents the location of a shipyard. A cargo ship is traveling at a constant speed and in a constant direction parallel to the coastline. At9 A.M., the radar shows the cargo ship at(0, 15). At 10 A.M., the radar shows the cargo ship at(16, 15). How far is the cargo ship from the shipyard at 4 P.M.? Explain.

Answer:
The cargo ship from the shipyard at 4 p.m = (16, 0)

Explanation:
Given that cargo ship at (0, 15) 9 a.m
cargo ship at (16, 15) 10 a.m
(16 – 0) , (15 – 15)
(16, 0) at 4 p.m

The Pythagorean Theorem Homework & Practice 9.2

Review & Refresh

Solve the equation.
Question 1.
7z2 = 252

Answer:
z = 6

Explanation:
7z2 = 252
z² = (252/7)
z² = 36
z² = 6²
when powers are equal exponents must be equal.
z = 6

Question 2.
0.75 q2 = 108

Answer:
q = 12

Explanation:
0.75 q2 = 108
q² = (108/0.75)
q² = 144
q² = 12²
when powers are equal exponents must be equal.
q = 12

Question 3.
\(\sqrt{\frac{1000}{10}}\) = n2 – 54

Answer:
n = 8

Explanation:
(1000/10) = 100
square root of 100 = 10 x 10
10 = n² – 54
n² = 10 + 54
n² = 64
n² = 8²
when powers are equal exponents must be equal.
n = 8

Question 4.
What is the solution of the system of linear equations y = 4x + 1 and 2x + y = 13?
A. (1, 5)
B. (5, 3)
C. (2, 9)
D. (9, 2)

Answer:
option c is correct.

Explanation:
y = 4x + 1, 2x + y = 13
if we consider x = 2 and y = 9 the values are equal.
9 = 4(2) + 1 , 2 (2) + 9 = 13
9 = 8 + 1, 4 + 9 = 13
9 = 9 , 13 = 13.

Concepts, Skills, &Problem Solving

USING GRID PAPER Find c. (See Exploration 1, p. 381.)
Question 5.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 14

Answer:
The value of c = 13

Explanation:
given that grid paper
a = 6 ( 6 boxes touching.)
b = 6 ( 8 boxes touching.)
c = 13 (13 boxes touching.)

Question 6.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 15

Answer:
The value of c = 15

Explanation:
given that grid paper
a = 5 ( 5 boxes touching.)
b = 12 ( 12 boxes touching.)
c = 15 (15 boxes touching.)

FINDING A MISSING LENGTH Find the missing length of the triangle.
Question 7.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 16
Answer:
The missing length = 41km

Explanation:
By using Pythagoras theorem
a² + b² = c²
20² + 21² =41²
length = 41 km

Question 8.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 17

Answer:
The missing length = 16.8 sq ft

Explanation:
By using Pythagoras theorem
a² + b² = c²
7.2² + 9.6² =16.8²
length = 16.8 sq ft

Question 9.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 18

Answer:
The missing length = 5.0 sq in

Explanation:
By using Pythagoras theorem
a² + b² = c²
5.0² + 5.6² =10.6²
length = 10.6 sq in

Question 10.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 19

Answer:
The missing length = 6  mm

Explanation:
By using Pythagoras theorem
a² + b² = c²
9² + 6² =15²
length = 6 mm

Question 11.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 20

Answer:
The missing length = 16  cm

Explanation:
By using Pythagoras theorem
a² + b² = c²
10² + 16² =26²
length = 16 cm

Question 12.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 21

Answer:
The missing length = 5.0 yd

Explanation:
By using Pythagoras theorem
a² + b² = c²
5.0² + 1.1² = 6.1²
length = 5.0 yd

Question 13.
YOU BE THE TEACHER
Your friend finds the missing length of the triangle. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 22

Answer:
No my friend is not correct.

Explanation:
By using Pythagoras theorem
a² + b² = c²
7² + 18² = 25²
length = 18 ft

FINDING LENGTHS OF THREE-DIMENSIONAL FIGURES Find x.
Question 14.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 23

Answer:
The missing length = 14 ft

Explanation:
By using Pythagoras theorem
a² + b² = c²
10² + 14² = 24²
length = 14 ft

Question 15.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 24

Answer:
The missing length = 50 in

Explanation:
By using Pythagoras theorem
a² + b² = c²
14² + 48² = 50²
length = 50 in

Question 16.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 25

Answer:
The missing length = 0.6 m

Explanation:
By using Pythagoras theorem
a² + b² = c²
0.6² + 0.9² = 1.5²
length = 0.6 m

Question 17.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 26

Answer:
The missing length = 6 cm

Explanation:
By using Pythagoras theorem
a² + b² = c²
4.5² + 2.5² = 6²
length = 6 cm

FINDING DISTANCES IN THE COORDINATE PLANE Find the distance between the points.
Question 18.
(0, 0), (9, 12)

Answer:
(9,12)

Explanation:
(0, 0), (9, 12)
(9-0) = 9
(12 – 0) = 12

Question 19.
(1, 2), (- 3, 5)

Answer:
5 units
Explanation:
(1, 2), (-3, 5)
(-3-1) = -4
(5- 2) = 5

Question 20.
(—18, 9), (22, 0)

Answer:
(4,9)

Explanation:
(—18, 9), (22, 0)
(22-18) = 4
(9- 0) = 9

Question 21.
(- 7, – 2), (13, – 23)

Answer:
29 units

Explanation:
(- 7, – 2), (13, – 23)
(13-7) = 6
(-23- 2) = -25

Question 22.
(15, – 17), (- 20, – 5)

Answer:
(-35, -22)

Explanation:
(15, – 17), (- 20, – 5)
(-20-15) = -35
(-5- 17) = -22

Question 23.
(- 13, – 3.5), (17, 2)

Answer:
(30, 18.5)

Explanation:
(- 13, – 3.5), (17, 2)
(17+13) = 30
(15+ 3.5) = 18.5

FINDING A MISSING LENGTH Find x.
Question 24.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 27

Answer:
The missing length = 8 cm

Explanation:
By using Pythagoras theorem
a² + b² = c²
8² + 12² = 20²
length = 8 cm

Question 25.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 28

Answer:
The missing length = 37 mm

Explanation:
By using Pythagoras theorem
a² + b² = c²
13² + 22² = 37²
length = 37 mm

Question 26.
MODELING REAL LIFE
The figure shows the location of a golf ball after a tee shot. How many feet from the hole is the ball?
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 29

Answer:
The ball from the hole = 540 feet

Explanation:
Given that the hole and  ball is
180 x 3
540 feet
1 yd = 3 ft

Question 27.
MODELING REAL LIFE
A tennis player asks the referee a question. The sound of the player’s voice travels 30 feet. Can the referee hear the question? Explain.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 30

Answer:
Yes, the referee hears the voice.

Explanation:
In the question given that the sound of the player’s voice travels 30 feet.
in the figure given that 24 feet.
so the referee can hear the voice.

Question 28.
PROBLEM SOLVING
You are cutting a rectangular piece of fabric in half along a diagonal. The fabric measures 28 inches wide and 1\(\frac{1}{4}\) yards long. What is the length (in inches) of the diagonal?

Answer:
The length of the diagonal = 28.25 in

Explanation:
Given that the fabric wides = 28 inches.
(1/4) = 0.25 in
The length of the diagonal  = 28 x 0.25
28.25 in

Question 29.
PROJECT
Measure the length, width, and height of a rectangular room. Use the Pythagorean Theorem to find the distance from B to C and the distance from A to B.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 31

Answer:
a² + b²  = c²

Explanation:
By using Pythagoras theorem
a² + b² = c²
length + width = height

Question 30.
STRUCTURE
The legs of a right triangle have lengths of 28 meters and 21 meters. The hypotenuse has a length of 5x meters. What is the value of x?

Answer:
The value of x = 9.8 meters.

Explanation:
Given that the right triangle has length  = 28 meters and 21 meters.
the hypotenuse = 5x
28 + 21 = 5x
49 = 5x
x = (49/5)
x = 9.8 meters.

Question 31.
PRECISION
You and a friend stand back-to-back. You run 20 feet forward, then 15 feet to your right. At the same time, your friend runs 16 feet forward, then 12 feet to her right. She stops and hits you with a snowball.
a. Draw the situation in a coordinate plane.
b. How far does your friend throw the snowball?
Answer:

Question 32.
MODELING REAL LIFE
The coordinate plane shows dig sites for archaeological research. Each unit on the grid represents 1 square foot. What is the distance from Site A to Site C?

Answer:
7 square feet.

Explanation:
Given that each grid represents 1 sq feet.
in the below figure shows that the distance.
distance from site A to site C is 7 sq ft.
the boxes 7 are touching the edges of the site A to B

Question 33.
PRECISION
A box has a length of 30 inches, a width of 40 inches, and a height of 120 inches. Can a cylindrical rod with a length of 342.9 centimeters fit in the box? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 32

Answer:
No, the cylindrical rod with a length of 135 cm will fit in the box.

Explanation:
Given that the box has a length = 30 in
width = 40 in
height = 120 in
area = l x w x h
area = 30 x 40 x 120
area = 1,44,000 cm

Question 34.
MODELING REAL LIFE
A green roof is like a traditional roof but covered with plants. Plants used for a green roof cost $0.75 per square foot. The roof at the right is 40 feet long. How much does it cost to cover both sides of the roof? Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 33

Answer:
The cost to cover both sides of the roof = 60 sq ft

Explanation:
Given that the plants used for a green roof cost = $0.75 sq ft.
the roof at the right = 40 ft long
the roof at both sides = 40 + 40 =80
1 sq feet = 0.75 $
80 x $0.75 = 60 sq ft

Question 35.
CRITICAL THINKING
A triangle has coordinates A(2, 1), B(2, 4), and C(5, 1). Write an expression for the length of \(\overline{B C}\). Use a calculator to find the length of \(\overline{B C}\) to the nearest hundredth.

Answer:
4.24 units

Question 36.
DIG DEEPER!
Write an equation for the distance d between the points (x1, y1) and (x2, y2). Explain how you found the equation.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.2 34

Answer:
x1 + y1 = x2 + y2

Explanation:
Given that the points (x1, y1) and (x2 , y2)
the equation is
x1 + y1 = x2 + y2

Lesson 9.3 Finding Cube Roots

EXPLORATION 1

Finding Edge Lengths
Work with a partner. Find the edge length of each cube. Explain your method.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 1

Answer:
a. 2 cm
b. 3 ft
c. 5m
d. 7 in
e. 0.1 cm
f. 0.5 yd

Explanation:
a. volume = 8 = 2 x 2 x 2
b. 27 = 3 x 3 x 3
c. 125 = 5 x 5 x 5
d. 343 = 7 x 7 x 7
e. 0.001 = 0.1 x 0.1 x 0.1
f. (1/8) = 0.5 x 0.5 x 0.5

EXPLORATION 2

Finding Solutions of Equations
Work with a partner. Use mental math to solve each equation. How many solutions are there for each equation? Explain your reasoning.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 2

Answer:
x = -3
x = -2
x = -1
x = 1
x = 2
x = 3

Explanation:
x³ = -27
x³ = -3³
when powers are equal exponents must be equal.
x = -3
x³ = -8
x³ = -2³
when powers are equal exponents must be equal.
x = -2
x³ = -1
x³ = -1³
when powers are equal exponents must be equal.
x = -1
x³ = 27
x³ = 3³
when powers are equal exponents must be equal.
x = 3
x³ = 8
x³ = 2³
when powers are equal exponents must be equal
x = 2
x³ = 1
x³ =1³
when powers are equal exponents must be equal
x = 1

Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 3

Try It

Find the cube root.
Question 1.
\(\sqrt [ 3 ]{ 1 } \)

Answer:
1

Explanation:
cube root of 1 is  1
3√1 = 1

Question 2.
\(\sqrt [ 3 ]{ -343 } \)

Answer:
-7

Explanation:
cube root of -343 is  -7
3√ -343= 7 x 7 x 7
(-343)³ = 7³
when powers are equal exponents must be equal.
-343 = 7

Question 3.
\(\sqrt[3]{-\frac{27}{1000}}\)

Answer:
(-3/10)

Explanation:
cube root of -27 is  -3
3√ -27= 3 x 3 x 3
(-27)³ = 3³
when powers are equal exponents must be equal.
-27 = 3
cube root of -1000 is  -10
3√ -1000= 10 x 10 x 10
(-1000)³ = 10³
when powers are equal exponents must be equal.
-1000= 10

Evaluate the expression.

Question 4.
18 – 4 \(\sqrt [ 3 ]{ 8 } \)

Answer:
10

Explanation:
cube root of 8 is  2
3√ 8 = 2 x 2 x 2
(8)³ = 2³
when powers are equal exponents must be equal.
8= 2
18 – 4 (2)
18 – 8
10

Question 5.
(\(\sqrt[3]{-64})^{3}\)) + 43

Answer:
44.333

Explanation:
cube root of -64 is  -4
3√ -64 = 4 x 4 x 4
(-64)³ = 4³
when powers are equal exponents must be equal.
-64= 4
1.333 + 43
44.333

Question 6.
5\(\sqrt [ 3 ]{ 512 } \) – 19

Answer:
21

Explanation:
cube root of 512 is  8
3√512 = 8 x 8 x 8
(512)³ = 8³
when powers are equal exponents must be equal.
512= 8
5 x 8 -19
40 – 19
21

Solve the equation.
Question 7.
z3 = – 1000

Answer:
z = -10

Explanation:
cube root of -1000 is  -10
3√-1000 = 10 x 10 x 10
z3  = 10³
when powers are equal exponents must be equal.
z= -10

Question 8.
3b3 = 1029

Answer:
b = 7

Explanation:
3b3 = 1029
b³ = (1029/3)
b³ = 343
cube root of 343 is  7
3√343 = 7 x 7 x 7
b³  = 7³
when powers are equal exponents must be equal.
b = 7

Question 9.
33 = – \(\frac{1}{5}\) m3 + 8

Answer:
m³ = 25.8

Explanation:
33 = – 0.2 m³  + 8
m³  = 33+0.2 -8
m³  = 25.8

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

FINDING CUBE ROOTS Find the cube root.
Question 10.
\(\sqrt [ 3 ]{ 64 } \)

Answer:
4

Explanation:
cube root of 64 is  4
3√64 = 4 x 4 x 4
64³ = 4³
when powers are equal exponents must be equal.
64 = 4

Question 11.
\(\sqrt [ 3 ]{ -216 } \)

Answer:
-6

Explanation:
cube root of -216 is  -6
3√-216 = 6 x 6 x 6
216³ = 6³
when powers are equal exponents must be equal.
216 = 6

Question 12.
\(\sqrt[3]{-\frac{343}{100}}\)

Answer:
– 7

Explanation:
cube root of  -343 is  -7
3√-343 = 7 x 7 x 7
-343³ = 7³
when powers are equal exponents must be equal.
-343 = 7

EVALUATING EXPRESSIONS Evaluate the expression.
Question 13.
\((\sqrt[3]{-27})^{3}\) + 61

Answer:
52

Explanation:
cube root of  -27 is  -3
3√-27 = 3 x 3 x 3
-27³ = 3³
when powers are equal exponents must be equal.
-27 = -3
-3 x 3 + 61
-9 + 61
52

Question 14.
15 + 3\(\sqrt[3]{125}\)

Answer:
30

Explanation:
cube root of  125 is  5
3√125 = 5 x 5 x 5
125³ = 5³
when powers are equal exponents must be equal.
125 = 5
15 + 3 (5)
15 + 15
30

Question 15.
z\(\sqrt[3]{-729}\) – 5

Answer:
-14

Explanation:
cube root of  -729 is  -9
3√-729 = 9 x 9 x 9
729³ = 9³
when powers are equal exponents must be equal.
729 = 9
-9 -5 = -14

SOLVING EQUATIONS Solve the equation.
Question 16.
d3 = 512

Answer:
d = 8

Explanation:
cube root of  512 is  8
3√512 = 8 x 8 x 8
512³ = 8³
when powers are equal exponents must be equal.
d³ = 8³
d = 8

Question 17.
w3 – 12 = – 76

Answer:
w = -4

Explanation:
w3 – 12 = – 76
w³ = -76 + 12
w³ = -64
cube root of  -64 is  -4
3√-64 = 4 x 4 x 4
w³ = 4³
when powers are equal exponents must be equal.
w = 4

Question 18.
– \(\frac{1}{3}\)m3 + 13 = 4

Answer:
-8.67

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 19.
You have 275 square inches of wrapping paper. Do you have enough wrapping paper to wrap the gift box shown? Explain.

Answer:
Yes i have enough wrapping paper to wrap the gift box.

Explanation:
Given that 275 sq in of wrapping paper.
6.5 x 6.5 x 6.5 = 274.6
275 sq in

Question 20.
A cube-shaped end table has a volume of 16,000 cubic centimeters. Does the end table fit in the corner shown? Justify your answer.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 4
Answer:
No the end table does not fit in the corner.

Explanation:
volume = 343
343³ = 7³
when powers are equal exponents must be equal.
343 = 7

Question 21.
DIG DEEPER!
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 5
The relationship between the volumes and the lengths of two cereal boxes is represented by
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 6
Box A has a volume of 192 cubic inches and a length of 8 inches. Box B has a volume of 375 cubic inches. What is the length of Box B? Justify your answer.

Answer:
The length of the box B = 1.6 inches.

Explanation:
(v of A/ v of B ) = ( l of A / l of B) ³
B³ = (192/375) x 8
b³ = 0.512 x 8
b³ = 4.096
b ³ = 1.6³
when powers are equal exponents must be equal.
b = 1.6

Finding Cube Roots Homework & Practice 9.3

Review & Refresh

Find the missing length of the triangle.
Question 1

Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 7

Answer:
The missing length = 40 m

Explanation:
By using Pythagoras theorem
a² + b² = c²
24² + 32² = 40²
576 + 1024 = 1600
1600 = 1600
length = 56m

Question 2.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 8

Answer:
The missing length = 10 ft

Explanation:
By using Pythagoras theorem
a² + b² = c²
7.8² + 10² = 17.8²
length = 10 ft

Question 3.
Which linear function is shown by the table?
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 9
Answer:
option C is correct.

Explanation:
y = 3x + 1
x = 1, y = 4
4 = 3(1) + 1
y= 4

Concepts, Skills, &Problem Solving

FINDING EDGE LENGTHS Find the edge length s of the cube. (See Exploration 1, p. 389.)
Question 4.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 10

Answer:
The edge length of the cube = 6

Explanation:
cube root of  216 is  6
3√216 = 6 x 6 x 6
216³ = 6³
when powers are equal exponents must be equal.
s³ = 6³
s = 6

Question 5.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 11

Answer:
The edge length of the cube = (1/3) ft

Explanation:
cube root of  (1/27) is  (1/3)
3√(1/27) = (1/3)x (1/3)  x (1/3)
(1/27)³ = (1/3)³
when powers are equal exponents must be equal.
s³ = (1/3)³
s = (1/3)

Question 6.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 12

Answer:
The edge length of the cube = 0.4

Explanation:
cube root of  0.064 is  0.4
3√0.064 = 0.4 x 0.4  x 0.4
0.064³ = 0.4³
when powers are equal exponents must be equal.
s³ = 0.4³
s = 0.4

FINDING CUBE ROOTS Find the cube root.
Question 7.
\(\sqrt [ 3 ]{ 729 } \)

Answer:
9

Explanation:
cube root of 729 is 9
3√729 = 9 x 9 x 9
729³ = 9³
when powers are equal exponents must be equal.
729 = 9

Question 8.
\(\sqrt [ 3 ]{ -125 } \)

Answer:
-5

Explanation:
cube root of -125 is -5
3√-125 = 5 x 5 x 5
-125³ = -5³
when powers are equal exponents must be equal.
125 = -5

Question 9.
\(\sqrt [ 3 ]{ -1000} \)

Answer:
-10

Explanation:
cube root of -1000 is -10
3√-1000 = 10 x 10 x 10
-1000³ = -10³
when powers are equal exponents must be equal.
-1000 = -10

Question 10.
\(\sqrt [ 3 ]{ 1728 } \)

Answer:
12

Explanation:
cube root of 1728 is 12
3√1728 = 12 x 12 x 12
1728³ = 12³
when powers are equal exponents must be equal.
1728 = 12

Question 11.
\(\sqrt[3]{-\frac{1}{512}}\)

Answer:
(-1/6)

Explanation:
cube root of – 512 is -(1/6)
3√-512 = (1/6) x (1/6) x (1/6)
-512³ = -(1/6)³
when powers are equal exponents must be equal.
-512 = (-1/6)

Question 12.
\(\sqrt[3]{\frac{343}{64}}\)

Answer:
(7/4)

Explanation:
cube root of (343/64) is (7/4)
3√343 = (7/4) x (7/4) x (7/4)
343³ = (7/4)³
when powers are equal exponents must be equal.
343 = (7/4)

EVALUATING EXPRESSIONS Evaluate the expression.
Question 13.

18 – (\(\sqrt [ 3 ]{ 27 }\))3

Answer:
-9

Explanation:
3√27³ = 27
where cube and cube root get cancelled.
18 – 27
-9

Question 14.
\(\left(\sqrt[3]{-\frac{1}{8}}\right)^{3}\) + 3\(\frac{3}{4}\)

Answer:
1.75

Explanation:
cube root of (1/18) is -(1/2)
3√(1/18) = (1/2) x (1/2) x (1/2)
(1/18)³ = (1/2)³
when powers are equal exponents must be equal.
(1/18) =- (1/2)
-(1/2) + 3(3/4)
– 0.5 + 3 (0.75)
-0.5 + 2.25
1.75

Question 15.
5\(\sqrt[3]{729}\) – 24

Answer:
21

Explanation:
cube root of (729) is (9)
3√729 = (9) x (9) x (9)
729³ = (9)³
when powers are equal exponents must be equal.
729 = 9
5 x 9 -24
45 – 24
21

Question 16.
\(\frac{1}{4}-2 \sqrt[3]{\frac{1}{216}}\)

Answer:
-11.75

Explanation:
cube root of (216) is (6)
3√216 = (6) x (6) x (6)
216³ = (6)³
when powers are equal exponents must be equal.
216 = 6
(1/4) -2 (6)
(1/4) – 12
0.25 – 12
-11.75

Question 17.
54 + \(\sqrt[3]{-4096}\)

Answer:
38

Explanation:
cube root of (-4096) is (-16)
3√-4096 = (-16) x (-16) x (-16)
-4096³ = (-16)³
when powers are equal exponents must be equal.
4096 = -16
54 +(-16)
54 – 16
38

Question 18.
4\(\sqrt[3]{8000}\) – 6

Answer:
166

Explanation:
cube root of (8000) is (43)
3√8000 = (43) x (43) x (43)
8000³ = (43)³
when powers are equal exponents must be equal.
8000 = 43
4 x 43 -6
172 – 6
166

EVALUATING EXPRESSIONS Evaluate the expression for the given value of the variable.
Question 19.
\(\sqrt[3]{\frac{n}{4}}+\frac{n}{10}\), n = 500

Answer:
55

Explanation:
(500/4) = 125
(500/10) = 50
cube root of (125) is (5)
3√125 = (5) x (5) x (5)
125³ = (5)³
when powers are equal exponents must be equal.
125 = 5
5 + 50
55

Question 20.
\(\sqrt[3]{6 w}\) – w, w = 288

Answer:
276

Explanation:
w = 288 given
6 x 288 = 1728

Question 21.
2d + \(\sqrt[3]{-45 d}\), d = 75

Answer:
165

Explanation:
d = 75 given
-45 x 75 = -3375
cube root of (-3375) is (-15)
3√-3375 = (-15) x (15) x (15)
(-3375)³ = (-15)³
when powers are equal exponents must be equal.
3375 = 15
2x 75 + 15
150 +15
165

SOLVING EQUATIONS Solve the equation.
Question 22.
x3 = 8

Answer:
x = 2

Explanation:
x³= 8
x³ = 2³
when powers are equal exponents must be equal.
x = 2

Question 23.
t3 = – 343

Answer:
t = – 7

Explanation:
t³ = -343
t³ = -7³
when powers are equal exponents must be equal.
t = -7

Question 24.
– 75 = y3 + 50

Answer:
y = -5

Explanation:
– 75 = y3 + 50
y³ = -75 – 50
y³ = -125

y³ = -5³
when powers are equal exponents must be equal.
y = -5

Question 25.
– \(\frac{1}{2}\)z3 = – 108

Answer:
z = 6

Explanation:
-(1/2)z³ = -108
(1/2) = 0.5
-0.5z³ = -108
z³ = (-108 /-0.5)
z³ = 216
z³ = 6³
when powers are equal exponents must be equal.
z = 6

Question 26.
2h3 – 11 = 43

Answer:
h = 17

Explanation:
2h3 – 11 = 43
2h³ = 43 + 11
2h³ = 54
h³ = (54/2)
h³ = 17
h = 17

Question 27.
– 600 = \(\frac{2}{5}\)k3 + 750

Answer:
k = -15

Explanation:
(2/5) = 0.4
-600 = 0.4 k³ + 750
0.4 k³ = -600 -750
0.4k³ = -1350
k³ = -(1350/0.4)
k³ = -3375
k³ = -15³
when powers are equal exponents must be equal.
k = -15

Question 28.
MODELING REAL LIFE
The volume of a cube-shaped compost bin is 27 cubic feet. What is the edge length of the compost bin?

Answer:
The edge length of the compost bin = 3 feet.

Explanation:
Given that the volume of a cube-shaped compost bin = 27 cubic ft.
volume of cube = s x sx s
volume = 3 x 3 x3
volume = 27

Question 29.
MODELING REAL LIFE
The volume of a cube of ice for an ice sculpture is 64,000 cubic inches.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 13
a. What is the edge length of the cube of ice?
b. What is the surface area of the cube of ice?

Answer:
a. The edge length of the cube of ice = 40 inches.
b. The surface area of the cube of ice = 9600 sq in

Explanation:
a. The edge length of the cube of ice = 40 in
volume = s x s x s
v = 40 x 40 x 40
v = 40 in
b. The surface area of cube  = 6a²
area = 6 x 40 x 40
where a = edge
area = 6 x 1600
area = 9600 sq in

Question 30.
NUMBER SENSE
There are three numbers that are their own cube roots. What are the numbers?

Answer:
-1, 1 , 0

Explanation:
cube root of -1 = -1
cube root of 1 = 1
cube root of 0 = 0
The 3 numbers have their own cube roots.

REASONING Copy and complete the statement with <, >, or =.
Question 31.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 14

Answer:
-(1/4) <   3√(-8/125)

Explanation:
(1/4) = -0.25
(8/125) = (-2/5)
(-2/5) = 0.4

Question 32.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 15

Answer:
3√ 0.001 >   0.01

Explanation:
3√ 0.001 = 0.1
0.1 > 0.01

Question 33.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 16

Answer:
3√64 <    √64

Explanation:
3√64  = 4 x 4 x4
3√64  = 4
√64 = 8 x 8
√64 = 8
4  <  8

Question 34.
DIG DEEPER!
You bake a dessert in the baking pan shown. You cut the dessert into cube-shaped pieces of equal size. Each piece has a volume of 8 cubic inches. How many pieces do you get from one pan? Justify your answer.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 17

Answer:
24 pieces.

Explanation:
Given that volume = 8 cubic inches
s = 2 in
Area = 64 sq in given
area = 6 x a²
area = 6 x 2 x2
area = 6 x 4
area = 24 sq in

Question 35.
LOGIC
Determine whether each statement is true for square roots. Then determine whether each statement is true for cube roots. Explain your reasoning.
a. You cannot find the square root of a negative number.
b. Every positive number has a positive square root and a negative square root.

Answer:
a.Yes
b. yes

Explanation:
a. we can find the square root of a negative number.
for example -2
the square root of -2 = -4
b. yes, every positive number has a positive square root and a negative square root.
for example = 4
-2 and 2 are the square roots of 4

Question 36.
GEOMETRY
The pyramid has a volume of 972 cubic inches. What are the dimensions of the pyramid?

Answer:
The dimensions of the pyramid = 14 in

Explanation:
Given that volume of pyramid = 972 cubic in
volume = ( l w h/3)
volume = (14 x 14 x 14/3)
v = 972 cubic in

Question 37.
REASONING
The ratio 125 : x is equivalent to the ratio x2 :125. What is the value of x?
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.3 18

Answer:
The value of x = 25

Explanation:
Given that the ratio = 125 : x
x ² : 125  = x = 25
when powers are equal exponents must be equal.
x = 25

CRITICAL THINKING Solve the equation.
Question 38.
(3x + 4)3 = 2197

Answer:
x = 6

Explanation:
(3x + 4)3 = 2197
9x³+ 64 = 2197
9x³ = 2197 – 64
9 x³ = 2133
x³ = (2133/9)
x³ =  237
x = 6

Question 39.
(8x3 – 9)3 = 5832

Answer:
x³ = 11.86
x = 3/2

Explanation:
512 x³ – 243 = 5832
512 x³ = 5832 + 243
512 x³ = 6075
x³ = 6075/512
x³ = 11.86

Question 40.
((5x – 16)3 – 4)3 = 216,000

Answer:
x = 56

Explanation:
125 x³ – 4096 – 64 = 216000
125 x³ – 4096 = 216000 + 64
125 x³ = 216064 + 4096
125 x³ = 220160
x³ = 1761.28
x = 56

Lesson 9.4 Rational Numbers

EXPLORATION 1

Work with a partner.
a. Complete the table.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 1
b. For each row of the table, use the two equations and what you know about solving systems of equations to write a third equation that does not involve a repeating decimal. Then solve the equation. What does your solution represent?
c. Write each repeating decimal below as a fraction. How is your procedure similar to parts (a) and (b)? How is it different?
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 2
d. Explain how to write a repeating decimal with n repeating digits as a fraction.

Answer:
a. x = 0.666 , 10 x = 10 x 0.666 = 6.666
x = 0.111, 10 x = 10 x0.111 = 1.111
x = 0.2444, 10 x = 10 x 0.24444 = 2.4444
c. x = (12/100)
x = (0.45/100)
x = (0.27/100)
x = (0.940/1000)

Explanation:
Given that x = 0.333
10 x = 10 x 0.333 = 3.3333
x = 0.666
10 x = 10 x 0.666 = 6.6666
x = 0.111
10 x = 10 x 0.111 = 1.111
x = 0.244
10 x = 10 x 0.244 = 2.444

Try It

Write the fraction or mixed number as a decimal.
Question 1.
\(\frac{3}{15}\)

Answer:
0.2

Explanation:
(3/15) = 0.2

Question 2.
–\(\frac{2}{9}\)

Answer:
-0.222

Explanation:
(-2/9) = -0.222

Question 3.
4\(\frac{3}{8}\)

Answer:
1.5

Explanation:
(3/8) = 0.375
4 x 0.375 = 1.5

Question 4.
2\(\frac{6}{11}\)

Answer:
1.09

Explanation:
(6/11) = 0.5454
2 x 0.5454 = 1.0909

Write the decimal as a fraction or a mixed number.
Question 5.
0.888 …..

Answer:
(111/125)

Explanation:
0.888
(888/1000)
(444/500)
(222/250)
(111/125)

Question 6.
\(2.0 \overline{6}\)

Answer:
(33/100)

Explanation:
(2.0 /6)
0.333
(33/100)

Question 7.
\(0 . \overline{64}\)

Answer:
0

Explanation:
(0/64)
64

Question 8.
\(-4 . \overline{50}\)

Answer:
-(8/100)

Explanation:
-4 /50
– 0.08
-8/100

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
VOCABULARY
How can you identify a rational number?

Answer:
A rational number is a number that can be written as a ratio, which means it can be written as a fraction, in which both the numerator and denominator are whole numbers.

Explanation:
8
(8/1)
the number 8 is a rational number because the number written as a fraction.

WRITING FRACTIONS OR MIXED NUMBERS AS DECIMALS Write the fraction or mixed number as a decimal.
Question 10.
\(\frac{9}{50}\)

Answer:
0.18

Explanation:
(9/50)
(9/50) = 0.18

Question 11.
– \(\frac{7}{18}\)

Answer:
-0.388

Explanation:
-(7/18)
-(7/18) = -0.388

Question 12.
3\(\frac{4}{9}\)

Answer:
1.333

Explanation:
(4/9)
4/9 = 0.444
3 x 0.444
1.333

Question 13.
– 12 \(\frac{1}{6}\)

Answer:
– 1.9992

Explanation:
(1/6)
(1/6) = 0.1666
0.1666 x -12
-1.9992

WRITING A REPEATING DECIMAL AS A FRACTION Write the repeating decimal as a fraction or a mixed number.
Question 14.
\(-1 . \overline{7}\)

Answer:
(-1/7)

Explanation:
-1: 7
-(1/7)

Question 15.
\(0 . \overline{2}\)

Answer:
(0/2)

Explanation:
0 : 2
(0/2)

Question 16.
\(8 . \overline{93}\)

Answer:
(8 /93)

Explanation:
8 : 93
(8/93)

Question 17.
\(-6 . \overline{235}\)

Answer:
-(6/235)

Explanation:
-6 : 235
-(6 /235)

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.
A fun house mirror distorts the image it reflects. Objects reflected in the mirror appear \(1 . \overline{3}\) times taller. When a five-foot-tall person looks in the mirror, how tall does he appear?
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 3

Answer:
He appears in the mirror  = 5 feet.

Explanation:
1 x 3 = 3 given that
1 x 5 = 5

Question 19.
An exchange rate represents the value of one currency relative to another. Your friend visits a country that uses a local currency with an exchange rate of \(1 . \overline{265}\) units of the local currency to $1. If a bank charges $2 to change currency, how many units of the local currency does your friend receive when she gives the bank $200?

Answer:
Friend receive local currency = 1,06,000 units.

Explanation:
Given that local currency of $1 = 265 units.
if a bank charges to $ 2
265 + 265 = 530
the local currency of the friend receive when she gives the bank $ 200
200 x 530
1,06,000 units.

Question 20.
DIG DEEPER!
A low fuel warning appears when a particular car has \(0.014 \overline{6}\) 6 of a tank of gas remaining. The car holds 18.5 gallons of gas and can travel 36 miles for each gallon used. How many miles can the car travel after the low fuel warning appears?

Answer:
9.324 gallons.

Explanation:
(0.014/6) = 0.0023
0.0023 x 6
0.014
Given that car holds 18.5 gallons of gas can travel 36 miles for each gallon.
36 x 0.014
0.504
0.504 x 18.5 gallons
9.324 gallons.

Rational Numbers Homework & Practice 9.4

Review & Refresh

Evaluate the expression.
Question 1.
2 + \(\sqrt [ 3 ]{ 27 } \)

Answer:
5

Explanation:
cube root of (27) is (3)
3√27 = (3) x (3) x (3)
27³ = (3)³
when powers are equal exponents must be equal.
27 = 3
2 + 3
5

Question 2.
1 – \(\sqrt [ 3 ]{ 8 } \)

Answer:
-1

Explanation:
cube root of (8) is (2)
3√8 = (2) x (2) x (2)
8³ = (2)³
when powers are equal exponents must be equal.
8 = 2
1- 2
-1

Question 3.
7\(\sqrt [ 3 ]{ 125 } \) – 12

Answer:
23

Explanation:
cube root of (125) is (5)
3√125 = (5) x (5) x (5)
125³ = (5)³
when powers are equal exponents must be equal.
125 = 5
7 x 5 -12
35 -12
23

Find the measures of the interior angles of the triangle.
Question 4.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 4

Answer:
x = 75⁰

Explanation:
Right angle triangle = 90⁰
x = 90⁰ – 15⁰
given that 75⁰
x = 75⁰

Question 5.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 5

Answer:
x = 80⁰

Explanation:
x + 40⁰ = 60⁰
x = 60⁰ +40⁰ +80
x = 80⁰

Question 6.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 6
Answer:

Explanation:
given that (x – 5)⁰ = 140⁰
x – 5⁰ = 140⁰
x = 140⁰ + 5⁰
x = 145⁰
2x = 145⁰ x 2
2x = 290⁰
x = 290⁰/2
x = 145⁰

Concepts, Skills, &Problem Solving

WRITING REPEATING DECIMALS AS FRACTIONS Write the repeating decimal as a fraction. (See Exploration 1, p. 395.)
Question 7.
0.777 . . .

Answer:
(7/9)

Explanation:
0.7777. . .
(7/9)

Question 8.
0.858585 . . .

Answer:
(85/99)

Explanation:
0.858585 . . .
(85/99)

Question 9.
0.232323 . . .

Answer:
(23/99)

Explanation:
0.232323 . . .
(23/99)
it is a repeating decimal

WRITING FRACTIONS OR MIXED NUMBERS AS DECIMALS Write the fraction or mixed number as a decimal.
Question 10.
–\(\frac{3}{20}\)

Answer:
-0.15

Explanation:
(3/20) = 0.15

Question 11.
9\(\frac{1}{12}\)

Answer:
9.083

Explanation:
(1/12) = 0.833
0.833 x 9 = 9.083

Question 12.
\(\frac{5}{36}\)

Answer:
0.1388

Explanation:
(5/36) = 0.1388

Question 13.
6\(\frac{1}{40}\)

Answer:
6.025

Explanation:
(1/40) = 0.025
6 x 0.025
6.025

Question 14.
\(\frac{11}{75}\)

Answer:
0.1466

Explanation:
(11/75)
0.1466

Question 15.
– 2\(\frac{7}{18}\)

Answer:
-2.38
Explanation:
(17/18)
0.3888
-2 x 0.3888
-2.38

Question 16.
PRECISION
Your hair is \(\frac{5}{16}\) inch long. Write this length as a decimal.

Answer:
0.3125 in

Explanation:
(5/16)
0.3125

WRITING A REPEATING DECIMAL AS A FRACTION Write the repeating decimal as a fraction or a mixed number.
Question 17.
\(-0 . \overline{5}\)

Answer:
-5/9

Explanation:
-5/9

Question 18.
\(4 . \overline{1}\)

Answer:
4

Explanation:
4 x 1
4

Question 19.
\(-0.3 \overline{56}\)

Answer:
-353/990
Explanation:
-0.3 x 56
-26.7
-(353/990)

Question 20.
\(6.0 \overline{89}\)

Answer:
(674/1000)

Explanation:
(6/89)
0.0674
(674/1000)

Question 21.
\(0.18 \overline{72}\)

Answer:
103/550
Explanation:
(0.18/72)
103/550

Question 22.
\(11 . \overline{510}\)

Answer:
(2156/10000)

Explanation:
(11/510)
0.02156
(2156/10000)

Question 23.
STRUCTURE
A forecast cone defines the probable path of a tropical cyclone. The probability that the center of a particular tropical cyclone remains within the forecast cone is \(0 . \overline{8}\). Write this probability as a fraction.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 7

Answer:
8 : 9

Explanation:
Given that cone is 8 x 9
they said to write it in the fraction.

Question 24.
STRUCTURE
Describe how to write a decimal with 12 repeating digitsas a fraction.

Answer:
12: 0

Explanation:
given that 12 is a decimal
and they said to write in decimal.
12 : 0

Question 25.
STRUCTURE
An approximation for the value of π is \(\frac{22}{7}\). Write this number as a repeating decimal.

Answer:
3.14 2857

Explanation:
π = (22/7)
(22/7) = 3.14
3.142857

Question 26.
MODELING REAL LIFE
The density of iodine is about \(6.28 \overline{1}\) times the density of acetone. The density of acetone is about 785 kilograms per cubic meter. What is the density of iodine? Write your answer as a repeating decimal.

Answer:
4929.8. . .

Explanation:
The density of iodine = 6.28 given
the density of acetone is 785 kg for cubic meter.
6.28 x 785
4,929.8

Question 27.
MODELING REAL LIFE
A disinfectant manufacturer suggests that its product kills \(99.9 \overline{8}\) % of germs. Write this percent as a repeating decimal and then as a fraction. How many germs would survive when the disinfectant is applied to an object with 18,000 germs?
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 8

Answer:
Fraction = (8999/9000)
2 germs

Explanation:
Given that the product kills (99.9/8)%
8999/9000
2 germs
Question 28.
MODELING REAL LIFE
You and your friend are making pear tarts for a bake sale. Your recipe uses \(\frac{7}{6}\) times the weight of the diced pears used in your friend’s recipe. Your friend’s recipe calls for 0.3 pound of diced pears. How many pounds of pears should you buy to have enough for both recipes?
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 9

Answer:
0.35 pounds of pears should buy to have enough for both recipes.

Explanation:
Given that my recipe uses (7/6) times the weight of the diced pears used in friends recipe.
my friends recipie calls for 0.3 pounds of diced pears.
(7/6) x 0.3
1.16666 x 0.3
0.35 pounds

Question 29.
PROBLEM SOLVING
The table shows the principal and interest earned per year for each of three savings accounts with simple annual interest. Which account has the greatest interest rate? Justify your answer.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 10

Answer:
Account c has the greatest interest rate.

Explanation:
Simple interest = (px r x t /100)
where p = principal amount, rate of interest, and time = t
Account A = $ 4/$90 =0.04 = 4.4 %
Account B = $5.50/$120.00 = 0.04583 = 4.583%
Account C = $4.80/$100.00 = 0.048 = 4.8%

Question 30.
DIG DEEPER!
The probability that an athlete makes a half-court basketball shot is 22 times the probability that the athlete makes a three-quarter-court shot. The probability that the athlete makes three-quarter-court shot is \(0.00 \overline{9}\). What is the probability that the athlete makes a half-court shot? Write your answer as a percent.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.4 11

Answer:
1.98 %

Explanation:
Given that the athlete makes three-quarter-court shot is 0.00 x 9
basketball shot is 22 times the probability that the athlete makes 3 -quarter-court shot.
0.00 x 9
0
22 x 9
198 writing it as decimal
1.98%
NUMBER SENSE Determine whether the numbers are equal. Justify your answer.
Question 31.
\(\frac{9}{22}\) and \(0.4 \overline{09}\)

Answer:
The numbers are not equal.

Explanation:
(9/22) = 0.4090
(0.4/9)  = 0.0444

Question 32.
\(\frac{1}{999}\) and 0

Answer:
The numbers are not equal.

Explanation:
(1/999) = 0.001
0 = 0.001

Question 33.
\(\frac{135}{90}\) and 1.5

Answer:
The numbers are not equal.

Explanation:
(35/90) = 1.5
1.5 = 1.5

ADDING AND SUBTRACTING RATIONAL NUMBERS Add or subtract.
Question 34.
\(0.4 \overline{09}+0.6 \overline{81}\)

Answer:
0.45180

Explanation:
(0.4/9) = .04444
(0.6/81) = 0.0074
0.0074 +0.444

Question 35.
\(-0 . \overline{63}+\frac{5}{99}\)

Answer:
-58/99

Explanation:
(5/99) = 0.0505
63 + 0.0505

Question 36.
\(\frac{11}{6}-0 . \overline{27}\)

Answer:
-25.1666

Explanation:
(11/6) = 1.833333
27
1.83333  – 27
-25.1666

Question 37.
\(0 . \overline{03}-0 . \overline{04}\)

Answer:
-1/99

Question 38.
STRUCTURE
Write a repeating decimal that is between \(\frac{9}{7}\) and \(\frac{10}{7}\). Justify your answer.

Answer:
1.3546

Explanation:
(9/7)
1.2857
(10/7)
1.4285
The number between 1.2857 and 1.4285 is 1.3546

Lesson 9.5 Irrational Numbers

EXPLORATION 1

Approximating Square Roots
Work with a partner. Use the square shown.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 1
a. Find the exact length x of the diagonal. Is this number a number a rational number or an irrational number? Explain.
b. The value of x is between which two whole numbers? Explain your reasoning.
c. Use the diagram below to approximate the length of the diagonal to the nearest tenth.Explain your method.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 2
d. Which of the following is the closest approximation of the length of the diagonal? Justify your answer using inverse operations.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 3
Answer:
a. The length of x = √2
a number is an irrational number.
d. 1.412 is the closest approximation of the length of the diagonal.

Try It

Classify the real number.
Question 1.
0.121221222 ……

Answer:
Irrational number.

Explanation:
Irrational number : the decimal form neither terminates nor repeats.
so the given number is a irrational number.

Question 2.
– √196
Answer:
The number is real number

Explanation:
the real number are the numbers which includes both rational and irrational numbers.

Question 3.
\(\sqrt [ 3 ]{ 2 } \)

Answer:
The number is real number

Explanation:
cube root of 3 is 2
where cube and cube root get cancelled.

Approximate the number to the nearest (a) integer and (b) tenth.
Question 4.
√8

Answer:
Integer = 3
tenth = 3 .0

Explanation:
square root of 8 = 2 x 2 x2
2 x2 = 4
3 is nearest integer to the 2

Question 5.
– √13

Answer:
-14

Question 6.
– √24
Answer:
13

Question 7.
√20
Answer:
11

Which number is greater? Explain.
Question 8.
√8, π

Answer:
√8 is greater.

Explanation:
√8 = 8
π = 3.14
so 8 is greater.

Question 9.
\(\sqrt [ 3 ]{ 65 }\), √26

Answer:
√26 is greater.

Explanation:
cube root of 65 = 4 x 4 x4
square root of 26 = 26
26 is greater.

Question 10.
– √2, – \(\sqrt [ 3 ]{ 10 }\)
Answer:
10 is greater.

Explanation:
cube root of 10 = 10
square root of -2 = -2
soo 10 is greater.

Approximate the distance between the points to the nearest tenth.
Question 11.
(- 3, – 1) and (- 2, – 2)
Answer:

Question 12.
(1, – 1) and (5, 4)
Answer:

Question 13.
(5, 4) and (9, 8)
Answer:

Question 14.
(- 7, 10) and (3, – 5)
Answer:

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 15.
VOCABULARY
How are rational numbers and irrational numbers different?

Answer:
rational number: a rational number is a number such as -3/7 that can be expressed as the quotient or fractionp/q of two integers, a numerator p, and non zero denominator q. Every integer is a rational number.
irrational number: irrational numbers are all the real numbers that are not rational numbers.

Explanation:
rational number: a rational number is a number such as -3/7 that can be expressed as the quotient or fractionp/q of two integers, a numerator p, and non zero denominator q. Every integer is a rational number.
irrational number: irrational numbers are all the real numbers that are not rational numbers.

CLASSIFYING REAL NUMBERS Classify the real number.
Question 16.
\(\frac{48}{16}\)
Answer:
Real number

Explanation:
(48/16) = 3
3 is a real number.

Question 17.
– √76
Answer:
Irrational numbers

Explanation:
76 is irrational number
irrational numbers are real numbers.

Question 18.
\(\sqrt [ 3 ]{ -216 }\)

Answer:
-6 is a rational number.

Explanation:
cube root of 216 = -6

APPROXIMATING AN IRRATIONAL NUMBER Approximate the number to the nearest (a) integer and (b) tenth.
Question 19.
√51
Answer:
26

Question 20.
– √87
Answer:
44

Question 21.
\(\sqrt [ 3 ]{ 60 }\)
Answer:
5

Question 22.
WHICH ONE DOESN’T BELONG?
Which number does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 4

Answer:
-3.3 is different from the three.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 23.
The equation 3600b2 = hw represents the relationship among the body surface area b (in square meters), height h (in centimeters), and weight w (in kilograms) of a person. To the nearest tenth, approximate the body surface area of a person who is 168 centimeters tall and weighs 60 kilograms.

Answer:
2.8 cm

Explanation:
Given that the surface area of a person = 168 cm
w = 60
3600 b² = 168 x 60
3600 b² = 10080
b² = (10080/3600)
b² = 2.8

Question 24.
Which plane is closer to the base of the airport tower? Justify your answer.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 5

Answer:
Plane B is closer to the base of the airport tower.

Irrational Numbers Homework & Practice 9.5

Review & Refresh

Write the repeating decimal as a fraction or a mixed number.
Question 1.
\(0 . \overline{4}\)
Answer:

Question 2.
\(1.0 \overline{3}\)
Answer:

Question 3.
\(0 . \overline{75}\)

Answer:

Question 4.
\(2 . \overline{36}\)
Answer:

Simplify the expression. Write your answer as a power.
Question 5.
(54)2

Answer:
5 ⁸

Explanation:
(54)2
5 ⁸ = 5 x 5 x 5 x 5 x5 x5 x5
5⁸

Question 6.
(- 9)4 . (- 9)7

Answer:
(-9)¹¹

Explanation:
(-9)⁴ . (-9)⁷
when exponents are equal powers must be added.
(-9)⁴ +⁷
(-9)¹¹

Question 7.
a8 . a

Answer:
a⁹

Explanation:
(a)⁸ . (a)¹
when exponents are equal powers must be added.
(a)⁸ +¹
(a)⁹

Question 8.
(y3)6

Answer:
y ⁶

Explanation:
cube root of y = y x y x y
y ⁶

Concepts, Skills, &Problem Solving

APPROXIMATING SQUARE ROOTS Find the exact length x of the diagonal of the square or rectangle. The value of x is between which two whole numbers? (See Exploration 1, p. 401.)
Question 9.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 6

Answer:
x = 18

Explanation:
Given that the sides of a square = s x s
square =2 3 x 3
s = 18

Question 10.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 7

Answer:
x = 24

Explanation:
Given that the sides of a rectangle  = 2 (l + b)
rectangle = 2( 8 + 4)
rectangle = 2(12)
rectangle = 24

CLASSIFYING REAL NUMBERS Classify the real number.
Question 11.
0
Answer:
whole
integer
rational

Question 12.
\(\sqrt [ 3 ]{ 343 }\)

Answer:
7

Explanation:
cube root of 343 = 7 x 7 x 7
7

Question 13.
\(\frac{π}{6}\)

Answer:
18.84
irrational number.

Question 14.
– √81

Answer:
-9

Explanation:
square root of -81 = -9 x -9

Question 15.
– 1.125

Answer:
rational number

Question 16.
\(\frac{52}{13}\)

Answer:
676

Explanation:
(52/13)
676

Question 17.
\(\sqrt [ 3 ]{ -49 }\)

Answer:
-7
irrational

Question 18.
√15

Answer:
15

Question 19.
YOU BE THE TEACHER
Your friend classifies √144. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 8

Answer:
No

Explanation:
144 is a perfect square.
square root of 144 is rational number.

Question 20.
MODELING REAL LIFE
You cut a photograph into a right triangle for a scrapbook. The lengths of the legs of the triangle are 4 inches and 6 inches. Is the length of the hypotenuse a rational number? Explain.

Answer:
√52

Explanation:
By using Pythagoras theorem
a² + b² = c²
4² + 6² = c²
16 + 36 = c²
52 = c²
c =√52

Question 21.
REASONING
Place each number in the correct area of the Venn diagram.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 9
a. the last digit of your phone number
b. the square root of any prime number
c. the quotient of the circumference of a circle and its diameter

Answer:
b. irrational number.
c. irrational number.

APPROXIMATING AN IRRATIONAL NUMBER Approximate the number to the nearest (a) integer and (b) tenth.
Question 22.
√46

Answer:
a. 46
b.√36

Explanation:
square root of 46 = 46
nearest tenth = 36
46 – 10 = 36

Question 23.
– √105

Answer:
a.-10
b.-10 .2

Explanation:
square root of -105 = -10
nearest tenth = -10.2
Question 24.
\(\sqrt [ 3 ]{ -12 }\)

Answer:
a.³√-12
b..³√-2

Explanation:
square root of -12= -12
nearest tenth = -2
12 – 10 = 2

Question 25.
\(\sqrt [ 3 ]{ 310 }\)

Answer:
a. 7
b. 6.8

Explanation:
cube root of 310 = 7
nearest tenth = 7
6.8

Question 26.
\(\sqrt{\frac{27}{4}}\)

Answer:
6.75
6.65

Explanation:
(27/4)
6.75
6.75 – 10 = 6.65

Question 27.
\(-\sqrt{\frac{335}{2}}\)

Answer:
-13
12.9

Explanation:
(335/2)
-13
-13 – 10
-12.9

COMPARING IRRATIONAL NUMBERS Which number is greater? Explain.
Question 28.
√125, √135

Answer:
√135

Explanation:
135 is greater than 125

Question 29.
√22, \(\sqrt [ 3 ]{ 34 }\)

Answer:
34

Explanation:
34 is greater then 22

Question 30.
– \(\sqrt [ 3 ]{ 100 }\), – √42

Answer:
-42

Explanation:
-42 is greater than -100

Question 31.
√5, π

Answer:
√5

Explanation:
5 is greater than 3.14

Question 32.
\(\sqrt [ 3 ]{ 130 }\), √28

Answer:
130

Explanation:
130 is greater than 28

Question 33.
– √38, \(\sqrt [ 3 ]{ -250 }\)

Answer:
-250

Explanation:
250 is greater than 38

USING TOOLS Use the graphing calculator screen to determine whether the statement is true or false.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 10
Question 34.
To the nearest tenth, √10 = 3.1.

Answer:
3.16227766

Question 35.
The value of √14 is between 3.74 and 3.75.

Answer:
True

Explanation:
Given that the √14 = 3.74165

Question 36.
√10 lies between 3.1 and 3.16 on a number line.

Answer:
True

Explanation:
Given that the √10 = 3.1622

USING THE PYTHAGOREAN THEOREM Approximate the distance between the points to the nearest tenth.
Question 37.
(1, 2), (7, 6)

Answer:
7.2 units.

Question 38.
(2, 4), (7, 2)
Answer:

Question 39.
(- 1, – 3), (1, 3)

Answer:
6.3 units.

Question 40.
(- 6, – 7), (0, 0)
Answer:

Question 41.
(- 1, 1), (7, 4)

Answer:
8.5 units.

Explanation:
7 + 1 = 8
4 + 1 = 5
8.5 units

Question 42.
(- 6, 5), (- 4, – 6)
Answer:

Question 43.
MODELING REAL LIFE
The locations of several sites in a forest are shown in the coordinate plane. Approximate each distance to the nearest tenth.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 11
a. How far is the cabin from the peak?
b. How far is the fire tower from the lake?
c. How far is the lake from the peak?
d. You are standing at(- 5, – 6). How far are you from the lake?

Answer:
a. 5.8 km
b. 8.6 km
c. 7.1 km
d. 11.7 km

Question 44.
WRITING
Explain how to continue the method in Example 2 to approximate √71 to the nearest hundredth.

Answer:
29

Question 45.
MODELING REAL LIFE
The area of a four square court is 66 square feet. Approximate the side length s of the four square court to the nearest whole number.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 12

Answer:
The side length = 8 ft

Explanation:
The area of a four square court is 66 square feet given.
the side lengths = 8 x 8
s = 8 ft

Question 46.
MODELING REAL LIFE
A checker board is 8 squares long and 8 squares wide. The area of each square is 14 square centimeters. Approximate the perimeter (in centimeters) of the checker board to the nearest tenth.

Answer:
The perimeter of the checker board = 384

Explanation:
Given that the checker board is 8 squares long and 8 squares wide.
perimeter = 6 x side x side
perimeter = 6 x 8 x 8
p= 6 x 64
p = 384  cm

Question 47.
GEOMETRY
The cube has a volume of 340 cubic inches. Approximate the length d of the diagonal to the nearest whole number. Justify your answer.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 13

Answer:
The length of the diagonal = 12 in

Explanation:
Given that the volume of cube = 340 cubic inches.
v = s x s
v =  7 x 7
d = 7 x 7 + 10 x 10 = 149
12 in
Question 48.
CRITICAL THINKING
Ona number line, π is between 3 and 4.
a. Use this information to draw a number line and shade a region that represents the location of π2. Explain your reasoning.
b. Repeat part(a) using the fact that π is between 3.1 and 3.
c. Repeat part(a) using the fact that π is between 3.14 and 3.15.
Answer:

NUMBER SENSE Approximate the square root to the nearest tenth.
Question 49.
√0.39

Answer:
0.6

Explanation:
√0.39²
where square and square root get cancelled.
0.39  – 10
0.6

Question 50.
√1.19

Answer:
1.09

Explanation:
√1.19²
where square and square root get cancelled.
1.19  – 10
1.09

Question 51.
√1.52

Answer:
1.2

Explanation:
√1.52²
where square and square root get cancelled.
1.52  – 10
1.2

Question 52.
STRUCTURE
Is \(\sqrt{\frac{1}{4}}\) a rational number? Is \(\sqrt{\frac{3}{16}}\) a rational number? Explain.
Answer:

Question 53.
MODELING REAL LIFE
The equation s2 = 54r represents the relationship between the speed s (in meters per second) of a roller-coaster car and the radius r (in meters) of the loop. Approximate the speed of a roller-coaster car going around the loop shown to the nearest tenth.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 14

Answer:
s = 30.1 meters per second.

Explanation:
s2 = 54r
where r = 16.764 m
s² = 54 x 16.764 m
s² = 905.256
s =30.1

Question 54.
OPEN-ENDED
Find two numbers a and b that satisfy the diagram.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 15

Answer:
√a = 9.3
√b = 9.8

Explanation:
Given that a = 9
b = 10
√a = 9.3
√b = 9.8
the distance between the two points is 1 inch

Question 55.
DIG DEEPER!
The equation d3 = t3 represents the relationship between the mean distance d (in astronomical units) of a planet from the Sun and the t time (in years) it takes the planet to orbit the Sun.
a. Jupiter takes about 11.9 years to orbit the Sun. Approximate the mean distance of Jupiter from the Sun to the nearest tenth.
b. The mean distance of Saturn from the Sun is about 9.5 astronomical units. Approximate the time it takes Saturn to orbit the Sun to the nearest tenth.

Answer:
a. d = 5.2 astronomical units
b. 29.3 years

Explanation:
Given that the
a. Jupiter takes about 11.9 years to orbit the sun.
b. the mean distance of Saturn from the sun is about 9.5 astronomical units.
Question 56.

MODELING REAL LIFE
The equation h = – 16t2 + 26 represents the height h (in feet) of a water balloon t seconds after it is dropped. Approximate the time it takes the water balloon to reach the ground to the nearest tenth. Justify your answer.
Big Ideas Math Answer Key Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem 9.5 16

Answer:
– 28 feet.

Explanation:
Given that the equation h = – 16t2 + 26
we assuming t = 2
h = -16 x 4 + 26
h = -64 + 26
h = -38
nearest tenth = 38 – 10
– 28 feet

Question 57.
NUMBER SENSE
Determine whether the statement is sometimes, always, or never true. Explain your reasoning.
a. A rational number multiplied by a rational number is rational.
b. A rational number multiplied by an irrational number is rational.
c. An irrational number multiplied by an irrational number is rational.

Answer:
a. always.
b. sometimes.
c. sometimes.

Explanation:
a. if a/b and c/d are rational numbers, then ac and bd are integers and bd not equal to zero. so ac/bd is rational.
b. for example, π . 0 = 0 is rational, but 2. √3 is irrational.
c. for example, √2 . π  is irrational, but π  . 1/ π  is rational.

Lesson 9.6 The Converse of the Pythagorean Theorem

The converse of a statement switches the hypothesis and the conclusion.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 1

EXPLORATION 1

Analyzing the Converse of a Statement
Work with a partner.
a.Write the converse of each statement. Then determine whether each statement and its converse are true or false.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 2

  • If I live in California,then I live in the United States.
  • If my heart is beating, then I am alive.
  • If one figure is a translation of another figure,then the figures are congruent.

b. Write your own statement whose converse is true. en write your own statement whose converse is false.

EXPLORATION 2

The Converse of the Pythagorean Theorem
Work with a partner.
a.Write the converse of the Pythagorean Theorem. Do you think the converse is true or false?
b. Consider △DEFwith side lengths a, b, and c such that a2 + b2 = c2. Also consider △JKL with leg lengths a and b, where the measure of ∠K is 90°. Use the two triangles and the Pythagorean Theorem to show that the converse of the Pythagorean Theorem is true.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 3

Try It

Tell whether the triangle with the given side lengths is a right triangle.
Question 1.
28 in., 21 in., 20 in.

Answer:
The given sides does not form a right triangle.
Explanation:
20² + 21² = 28²
400  + 441 = 784
841 = 784
the given sides does not form a right triangle.

Question 2.
1.25 mm, 1 mm, 0.75 mm

Answer:
The given sides form a right triangle.
Explanation:
1² + 0.75² = 1.25²
1 + 0.5625= 1.5623
1.5625 = 784
the given sides  form a right triangle.

Tell whether the points form a right triangle.
Question 4.
D (- 4, 0), E (- 2, 3), F (1, 0)

Answer:
Yes the given points form a right triangle.

Question 5.
J (4, 1), K (1, – 3), L (- 3, 0)

Answer:
Yes the given points form a right triangle.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
WRITING
Explain the converse of the Pythagorean Theorem.

Answer:
a² + b² ≠ c²

Explanation:
a² + b² = c²
where a, b and c are side lengths of a triangle.
converse of pythagoras theorem
a² + b² ≠ c²

Question 6.
IDENTIFYING A RIGHT TRIANGLE
Is a triangle with side lengths of 2 millimeters, 2.5 millimeters, and 3 millimeters a right triangle?

Answer:
No the side lengths does not form a right triangle.

Explanation:
2 ² + 2.5² = 3²
4  + 6.25 = 9
10.25 = 9

Question 7.
IDENTIFYING A RIGHT TRIANGLE
Do the points (- 1, 1), (- 3, 5), and (0, 8) form a right triangle?

Answer:
No the given points does not form a right triangle.

Question 8.
WHICH ONE DOESN’T BELONG?
Which set of numbers does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 4

Answer:
3, 6, 8 set does not belong with the other 3 sets.

Explanation:
3 ² + 6² = 8²
9  + 36 = 64
45 = 64
the given sides does not form a right triangle.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
You practice archery as shown. Determine whether the arrow is perpendicular to the vertical support. Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 5

Answer:
Yes the archery is perpendicular to vertical support.

Explanation:
2 times of 7 is 14.
so archery is perpendicular to vertical support.

Question 10.
DIG DEEPER!
Three fire hydrants in a neighborhood are represented on a map. The coordinates of the fire hydrants are (0, 0), (2, 5), and (7, y). The fire hydrants are arranged in a right triangle, where y is a natural number less than 10. Find y.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 6

Answer:
y = 5

Explanation:
Given that the coordinates of the fire hydrants are (0, 0), (2, 5), and (7, y).
y can meet the y-axis on 5
y = 5
7 -2 = 5

The Converse of the Pythagorean Theorem Homework & Practice 9.6

Review & Refresh

Approximate the number to the nearest (a) integer and (b) tenth.
Question 1.
√31

Answer:
a. 6
b. 5.6

Explanation:
square root of 31 = 31
√31 ² = 6
where square and square root get cancelled.
nearest tenth = it is decreased by 10
5.6

Question 2.
– √7

Answer:
a. 7
b. -3

Explanation:
square root of 7 = 7
√7 ² = 7
where square and square root get cancelled.
nearest tenth = it is decreased by 10
7 -10
-3

Question 3.
\(\sqrt [ 3 ]{ 25 }\)

Answer:
a. 3
b. 2.9

Explanation:
cube root of 25 = 25
3√25 = 3
where square and square root get cancelled.
nearest tenth = it is decreased by 10
2.9

The figures are similar. Find x.
Question 4.
The ratio of the perimeters is 2 : 5.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 7

Answer:
x = 2

Explanation:
Given that the ratio of perimeters = 2 : 5
x = 2
5 x 2 = 10

Question 5.
The ratio of the perimeters is 4 : 3.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 8

Answer:
x = 4.5

Explanation:
Given that the ratio of perimeters = 4 : 3
x = 4.5
3 x 2 = 4.5

Concepts, Skills, &Problem Solving

ANALYZING THE CONVERSE OF A STATEMENT Write the converse of the statement. Then determine whether the statement and its converse are true or false. Explain. (See Exploration 1, p. 409.)
Question 6.
If a is an odd number, then a2 is odd.

Answer:
False.
if a is not an odd number, then square of a is not odd.

Explanation:
The given statement is true.
converse of a given statement is false.

Question 7.
If ABCD is a square, then ABCD is a parallelogram.

Answer:
True.
If ABCD is  not a square, then ABCD is not a parallelogram.

Explanation:
The given statement is false.
converse of a given statement is true.
IDENTIFYING A RIGHT TRIANGLE Tell whether the triangle with the given side lengths is a right triangle.
Question 8.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 9

Answer:
Yes the given side lengths is a right triangle.

Explanation:
8 ² + 15² = 17²
64  + 225 = 289
289 = 625

Question 9.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 10

Answer:
Yes the given side lengths is a right triangle.

Explanation:
27 ² + 36² = 45²
729  + 1296 = 2025
2025 = 625

Question 10.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 11

Answer:c

Question 11.

14 mm, 19 mm, 23 mm

Answer:
No the given side lengths is not a right triangle.

Explanation:
14 ² + 19² = 23²
196  + 361  = 529
557 = 625

Question 12.
\(\frac{9}{10}\) mi, 1 \(\frac{1}{5}\) mi, 1 \(\frac{1}{2}\) mi

Answer:

Question 13.
1.4 m, 4.8 m, 5 m

Answer:
Yes the given side lengths is  a right triangle.

Explanation:
1.4 ² + 4.8² = 5²
1.96  + 23.04  = 25
25 = 25

Question 14.
MODELING REAL LIFE
A post-and-beam frame for a shed is shown in the diagram. Does the brace form a right triangle with the post and beam? Explain.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 12

Answer:
Yes the given side lengths is a right triangle.

Explanation:
15 ² + 20² = 25²
225  + 400 = 625
625 = 625

Question 15.
MODELING REAL LIFE
A traffic sign has side lengths of 12.6 inches, 12.6 inches, and 12.6 inches. Is the sign a right triangle? Explain.

Answer:
No the given side lengths is not a right triangle.

Explanation:
12.6 ² + 12.6² = 12.6²
158.76 + 158.76 = 158.76
317.52 = 158.76

IDENTIFYING A RIGHT TRIANGLE Tell whether a triangle with the given side lengths is a right triangle.
Question 16.
√63, 9, 12

Answer:
Yes the given side lengths is a right triangle.

Explanation:
√63 ² + 9² = 12²
63 + 81 = 144
144 = 144

Question 17.
4, √15, 6

Answer:
No the given side lengths is not a right triangle.

Explanation:
√15 ² + 4² = 6²
15 + 16 = 36
31 = 36

Question 18.
√18, √24, √42

Answer:
Yes the given side lengths is a right triangle.

Explanation:
√18 ² + √24² = √42²
18 +24 = 42
42= 42

Question 19.
YOU BE THE TEACHER
Your friend determines whether a triangle with side lengths of 3, √58, and 7 is a right triangle. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 13

Answer:
No my friend is  not correct.

Explanation:
By using Pythagoras theorem
a² + b² = c²
13² + 58² = 7²
9 + 58  is not equal to 49

IDENTIFYING A RIGHT TRIANGLE Tell whether the points form a right triangle.
Question 20.
(0, 0), (0, 5), (2, 0)

Answer:
Yes the points  forms a right triangle

Explanation:
the above given points are all positive

Question 21.

(0, 8), (2, 2), (11, 6)

Answer:
No the points not forms a right triangle

Explanation:
the above given points are all positive.

Question 22.
(- 1, 0), (5, 0), (2, – 3)

Answer:
No the points not forms a right triangle

Explanation:
the above given points are all positive and negative.

Question 23.
(- 1, – 2), (2, 6), (4, – 1)

Answer:
No the points does not forms a right triangle

Explanation:
the above given points are all positive

Question 24.

(- 8, 6), (7, 9), (0, – 13)

Answer:
Yes the points  forms a right triangle

Explanation:
the above given points are all positive and negative

Question 25.
(0.5, 1.5), (7.5, 5.5), (9.5, 0.5)

Answer:
No  the points does not  forms a right triangle

Explanation:
the above given points are all positive and negative

Question 26.
LOGIC
The equation a2 + b2 = c2 is not true for a particular triangle with side lengths of a, b, and c. What can you conclude about the type of triangle?

Answer:
If the triangle is right triangle the above condition satisfies.

Explanation:
The equation a2 + b2 = c2 is a pythagoras theroem.
it satisfies for only right triangle.
where a,b, c are side lengths.
Question 27.
MODELING REAL LIFE
You spend the day looking for hidden containers in a wooded area using a Global Positioning System (GPS). You park your bike on the side of the road, and then locate Container 1 and Container 2 before going back to your bike. Does your path form a right triangle? Explain. Each grid line represents 10 yards.
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 14

Answer:
No path does not forms a right triangle.

Explanation:
(10, 50) , (20, -20) and (80, -10)
the given points form a right triangle.

Question 28.
DIG DEEPER!
The locations of a fishing boat, buoy, and kayak are represented by the points (0, 0), (16, 12), and (10, – 5). Each unit represents 1 nautical mile
Big Ideas Math Answers 8th Grade Chapter 9 Real Numbers and the Pythagorean Theorem 9.6 15
a. Do the boat, kayak, and buoy form a right triangle?
b. The boattravels at8 nautical miles per hour. How long does the boat take to reach the buoy if the boat travels directly toward it?

Answer:
a. yes the boat, kayak, buoy form a right triangle.
b. the boat travels at 8 nautical miles per hr= 80 miles per hr

Explanation:
a.yes the boat, kayak, buoy form a right triangle.
b. the boat travels at 8 nautical miles per hr.
8 x 10
80 miles per hr

Question 29.
The vertices of a quadrilateral are (1, 2), (5, 4), (6, 2), and (2, 0). Use the converse of the Pythagorean Theorem to determine whether the quadrilateral is a rectangle.
Answer:
Yes the given points forms a rectangle.

Real Numbers and the Pythagorean Theorem Connecting Concepts

Using the Problem-Solving Plan
Question 1.
The scale drawing of a baseball field has a scale factor of \(\frac{1}{270}\). Approximate the distance from home plate to second base on the actual baseball field to the nearest tenth.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cc 1
Understand the problem.
You know several measurements and the scale factor in a scale drawing of a baseball field. You are asked to approximate the distance from home plate to second base on the actual baseball field.

Make a plan.
The distance from home plate to second base is the hypotenuse of a right triangle. Approximate the distance in the scale drawing to the nearest tenth. Then use the scale factor to approximate the distance on the actual field.

Solve and check.
Use the plan to solve the problem. Then check your solution.

Answer:
0.048 sq in

Explanation:
Given that the scale factor of (1/270)
(1/270) = 0.003
from home plate to second base
4 x 4
16 x 0.003
0.048

Question 2.
You cut the wood cube shown into two identical triangular prisms. Approximate the surface area of each triangular prism to the nearest tenth. Justify your answer.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cc 2
Answer:
The surface area of triangular prism = 64 cm

Explanation:
the volume of the cube
v = s x s x
v = 8 x 8 x 8
v = 64 x 8
v = 512

Question 3.
Complete the mapping diagram representing the relationship between the lengths of the hypotenuse and the legs of an isosceles right triangle. Is the relationship linear? Explain.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cc 3

Answer:
a√2

Performance Task

Identify and Correct the Error!
At the beginning of this chapter, you watched a STEAM Video called “Metronome Design.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cc 4

Answer:

Real Numbers and the Pythagorean Theorem Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 1

Graphic Organizers
You can use a Four Square to organize information about a concept. Each of the four squares can be a category, such as definition, vocabulary, example, non-example, words, algebra, table, numbers, visual, graph, or equation. Here is an example of a Four Square for Pythagorean Theorem.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 2

Choose and complete a graphic organizer to help you study the concept.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 3
1. square roots
2. cube roots
3. rational numbers
4. irrational numbers
5. real numbers
6. converse of the Pythagorean Theorem

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 4

9.1 Finding Square Roots (pp. 373–380)
Learning Target: Understand the concept of a square root of a number.

Find the square root(s).
Question 1.
√1

Answer:
1

Explanation:
square root of 1 = 1 x 1

Question 2.
– \(\sqrt{\frac{9}{25}}\)

Answer:
-(3/5)

Explanation:
square root of 9 = 3x 3
square root of 25 = 5 x 5

Question 3.
± √1.69

Answer:
± 1.3

Explanation:
square root of 1.69 = 1.3 x 1.3

Evaluate the expression.
Question 4.
15 – 4 √36

Answer:
-9

Explanation:
15 – 4 √36
square root of 36 = 6 x 6
15 – 4 x 6
15 – 24
-9

Question 5.
\(\sqrt{\frac{54}{6}}+\frac{2}{3}\)

Answer:
9.66

Explanation:
(54/6) = 9
(2/3) = 0.66
9 + 0.66
9.66

Question 6.
\((\sqrt{9})^{2}\) – 12

Answer:
-7.5

Explanation:
(9/2) = 4.5
4.5 – 12
-7.5

Question 7.
The total area of a checker board is 256 square inches. What is the side length (in inches) of one of the small squares?
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 7

Answer:
The side length = 16 in

Explanation:
Given that total area = 256 sq in
area of square = s x s
area = 16 x 16
side = 16 in

9.2 The Pythagorean Theorem (pp. 381–388)
Learning Target: Understand the Pythagorean Theorem.

Find the missing length of the triangle.
Question 8.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 8

Answer:
The missing length = 47 in

Explanation:
By using Pythagoras theorem
a² + b² = c²
12² + 35² = 47²
length = 47 in

Question 9.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 9

Answer:
The missing length = 0.2 cm

Explanation:
By using Pythagoras theorem
a² + b² = c²
0.3² + 0.2² = 0.5²
length = 0.2 cm

Question 10.
You lean a 13-foot ladder on a house so the bottom of the ladder is 5 feet from the house. From the top of the ladder, you can safely reach another 4 feet higher. Can you reach a window that is located 13 feet above the ground? Explain.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 10

Answer:
No we cannot reach a window.

Explanation:
Given that we lean a 13-foot ladder on a house
so that the bottom of the ladder is 5 feetfrom the house.
from the top of ladder, we can reach safely another 4 feet higher.
5 + 4
we can reach 9 feet.

Question 11.
Find the distance between (- 6, 8) and (10, – 4).
Answer:

9.3 FindingCube Roots (pp. 389–394)
Learning Target: Understand the concept of a cube root of a number.

Find the cube root.
Question 12.
\(\sqrt [ 3 ]{ -2197 }\)

Answer:
-13

Explanation:
cube root of (-2197) is (-13)
3√-2197= (-13) x (-13) x (-13)
-2197³ = (-13)³
when powers are equal exponents must be equal.
-2197 = -13

Question 13.
\(\sqrt[3]{\frac{64}{343}}\)

Answer:-
(4/7)

Explanation:
cube root of (64/343) is (4/7)
3√(64/343)= (4/7) x (4/7) x (4/7)
(64/343)³ = (4/7)³
when powers are equal exponents must be equal.
(64/343)= (4/7)

Question 14.
\(\sqrt[3]{-\frac{8}{27}}\)

Answer:
-2/3

Explanation:
cube root of (8/27) is (2/3)
3√(8/27)= (2/3) x (2/3) x (2/3)
(8/27)³ = (2/3)³
when powers are equal exponents must be equal.
(8/27)= (2/3)

Question 15.
Evaluate the expression 25 + 2\(\sqrt [ 3 ]{ -64 }\).
Answer:
17

Explanation:
cube root of (-64) is (-4)
3√(64)= (4) x (4) x (4)
(64)³ = (4)³
when powers are equal exponents must be equal.
(64)= 4
25 + 2(-4)
25 -8
17

Question 16.
Solve the equation – 55 = \(\frac{1}{4}\)x3 + 73.
Answer:
x =8

Explanation:
-55 = 0.25 x ³ + 73
0.25 x³ = -55 -73
0.25 x³ = -128
x³ = -(128/0.25)
x³ = 512
x = 8

Question 17.
You are shipping a puzzle cube to your friend using the cube-shaped box shown. What is the difference between the height of the puzzle cube and the top of the box when you place the cube in the box?
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 17

Answer:
The height of the puzzle cube = 6 cm

Explanation:
Given that the volume of the cube-shaped box = 216 cubic cm
volume of cube = s x s x s
v = 6 x 6 x 6
v = 36 x 6
v = 216
the height of puzzle cube= 6 cm

9.4 Rational Numbers (pp. 395–400)
Learning Target: Convert between different forms of rational numbers.

Write the fraction or mixed number as a decimal.
Question 18.
– 2\(\frac{5}{6}\)

Answer:
-1.66

Explanation:
(5/6) = 0.83
-1.66

Question 19.
\(\frac{27}{80}\)

Answer:
0.33

Explanation:
(27/80)
0.3375

Question 20.
3\(\frac{8}{9}\)

Answer:
2.66

Explanation:
(8/9)  = 0.888
3 x 8/9
3 x 0.88
2.66

Question 21.
Write \(1 . \overline{36}\) as a mixed number.
Answer:

Question 22.
The gas mileage of a hybrid car is \(3.0 \overline{3}\) times the gas mileage of a regular car. The regular car averages 24 miles per gallon. Find the gas mileage of the hybrid car.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 22
Answer:
The gas mileage of the hybrid car is 72 miles

Explanation:
Given that the mileage of the hybrid car is 3
the regular car averages 24 miles per gallon.
the gas mileage of the hybrid car is
24 x 3
72 miles

Question 23.
Your friend’s cat weighs 0.8 3 times the weight of your cat. Your friend’s cat weighs 10 pounds. How much more does your cat weigh than your friend’s cat?

Answer:
2 pounds

Explanation:
Given that the my friend’s cat weighs = 0.8
3 times the weight of my cat.
my friend cat weighs = 10 pounds
10 x 0.2 = 2 pounds.

Question 23.
An apple dessert recipe makes \(2 . \overline{3}\) pounds of dessert and serves 6 people. What is the serving size (in pounds)?

Answer:
The serving size is 18 pounds.

Explanation:
given that an apple dessert recipe makes 3 pounds.
the serving size is
6 x 3
18 pounds.

9.5 Irrational Numbers (pp. 401–408)
Learning Target: Understand the concept of irrational numbers.

Classify the real number.
Question 24.
\(0.81 \overline{5}\)

Answer:
real number

Explanation:
given expression is a real number.

Question 25.
√101
Answer:
no

Explanation:
square root of 101 is real algebraic number
so it is not a real number.

Question 26.
√4
Answer:
no

Explanation:
square root of 4 is real algebraic number
so it is not a real number.

Approximate the number to the nearest (a) integer and (b) tenth.
Question 27.
√14

Answer:
a. 3
b. 4

Explanation:
3 is a integer for square root of 14
4 is the nearest tenth for 14

Question 28.
√90

Answer:
89
80

Explanation:
89 is a integer for square root of 90
80 is the nearest tenth for 90

Question 29.
√175

Answer:
174
165

Explanation:
174 is a integer for square root of 175
165 is the nearest tenth for 175

Question 31.
Which is greater, √48 or \(\sqrt [ 3 ]{ 127 }\) ? Explain.

Answer:
cube root of 127 is greater.

Explanation:
cube root of 127 = 127
cube root of 48 = 48

Question 32.
Approximate the distance between (- 2, – 5) and (3, 5) to the nearest tenth.
Answer:

Question 33.
The equation d = \(\frac{v^{2}}{15.68}\)represents the relationship between the distance d (in meters) needed to stop a vehicle and the velocity v (in meters per second) of the vehicle. Approximate the velocity of the vehicle when it takes 40 meters to stop.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 33

Answer:
The velocity of vehicle  = 5.102 meters

Explanation:
given that the equation d = (2/15.68)
(2/15.68 = 0.12755
the velocity of the vehicle when it takes 40 meters
0.12755 x 40
5.102 meters.

9.6 The Converse of the Pythagorean Theorem (pp. 409–414)
Learning Target: Understand the converse of the Pythagorean Theorem.

Tell whether the triangle with the given side lengths is a right triangle.
Question 34.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 34

Answer:
yes.

Explanation:
the given triangle is a right triangle.

Question 35.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 35

Answer:
no

Explanation:
the given measures not forming a right triangle.

Question 36.
Tell whether the points A (1, – 1), B (3, – 4), and C (4, 1) form a right triangle.

Answer:
no

Explanation:
given that the points
A(1, -1) B(3, -4) and C(4, 1)
the given measures does not form the right triangle.

Question 37.
You want to make a wooden border around a flower bed in the shape of a right triangle. You have three pieces of wood that measure 3.5 meters, 1.2 meters, and 3.9 meters. Do these pieces of wood form a right triangle? If not, explain how you can cut the longest piece of wood to make a right triangle.
Big Ideas Math Answers Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cr 37

Answer:
Yes these pieces of wood form a right triangle.

Explanation:
Given that the length of wooden pieces have measures of 3.5 m, 1.2 m, 3.9 m
so it can cut the wooden border in the shape of right triangle.

Real Numbers and the Pythagorean Theorem Practice Test

Question 1.
Find − √1600 .

Answer:
-40

Explanation:
square root of 1600 = 40 x 40
√1600 = 40²
when powers are equal exponents must be equal.
40

Question 2.
Find \(\sqrt[3]{-\frac{729}{64}}\).

Answer:
(-9/4)

Explanation:
cube root of -729 = 9
cube root of -64 = 4
-(9/4)

Evaluate the expression.
Question 3.
12 + 8√16

Answer:
80

Explanation:
12 + 8√16
square root of 16 is 4
12 + 8 x 4
20 x 4
80

Question 4.
\((\sqrt[3]{-125})^{3}+75\)

Answer:
60

Explanation:
cube root of -125 is -5
-5 x 3 + 75
-15 + 75
60
Question 5.
Find the missing length of the triangle.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem pt 5

Answer:
The missing length = 2 in

Explanation:
By using Pythagoras theorem
a² + b² = c²
2² + 24² = 26²
length = 2 in

Classify the real number.
Question 6.
16 π

Answer:

Explanation:
square root of 16 = 4 x 4
square root of 16 = 4

Question 7.
– √49
Answer:
7

Explanation:
square root of 49 = 7 x 7
square root of 49 = 7

Question 8.
Approximate √83 to the nearest (a) integer and (b) tenth.
Answer:
83

Question 9.
Write \(1 . \overline{24}\) as a mixed number.

Answer:
23

Question 10.
Tell whether the triangle is a right triangle.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem pt 10

Answer:
The given triangle is not a right triangle.

Explanation:
The above given triangle is not a right triangle,
right triangle = 90 degrees.

Approximate the distance between the points to the nearest tenth, if necessary.
Question 11.
(- 2, 3), (6, 9)

Answer:
(4, 6)

Explanation:
(- 2, 3), (6, 9)
(6-2, 9-3)
(4, 6)

Question 12.
(0, – 5), (4, 1)
Answer:

Question 13.
How high is the hand of the superhero balloon above the ground?
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem pt 13

Answer:
The hand of  superhero ballon above the ground = 78 cubic feet.

Explanation:
The height of the ground = 6 ft
height from the man hand = 11ft
the distance from super hero to hand of the man is
6 + 11 + 61
17 + 61
78 cubic feet

Question 14.
The area of a circular pool cover is 314 square feet. Write and solve an equation to find the diameter of the pool cover. Use 3.14 for π.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem pt 14

Answer:
46,735.76

Explanation:
Area  of the circle = π x r x r
Area = 3.14 x 122 x 122
Area = 46,735.76 ft

Question 15.
Five roads form two triangles. What is the value of x? Justify your answer.
Answer:

Real Numbers and the Pythagorean Theorem Cumulative Practice

Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cp 1
Question 1.
The period T of a pendulum is the time (in seconds) it takes the pendulum to swing back and forth once. The period can be found using the formula T = 1.1 √L, where is the length (in feet) of the pendulum. A pendulum has a length of 4 feet. What is the period of the pendulum?
A. 2.2 sec
B. 3.1 sec
C. 4.4 sec
D. 5.1 sec

Answer:
option C is correct.

Explanation:
given that the pendulum has a length of 4 feet.
T = 1.1 l
where t = time and l = length
T = 4.4 sec

Question 2.
Whatis the value of y = 5 – 2x when x = – 3?
F. – 1
G. 1
H. 4
I. 11

Answer:
option F is correct.

Explanation:
given that x = -3
y = 5 -2 x
y = 5- 2 (-3)
y = 5 -6
y = -1

Question 3.
Which graph represents the linear equation 3x + 2y = 12?
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cp 3

Answer:
option B is correct.

Explanation:
In the above 4 graphs option B is correct.
3x + 2y = 12
x = 2 and y= 3
3 x 2 + 2 x 3
6 + 6 = 12

Question 4.
Which expression is equivalent to \(\frac{(-3)^{12}}{(-3)^{3}}\)?
F. (- 3)4
G. (- 3)9
H. 09
I. 19

Answer:
option F is correct.

Explanation:
given that -3 x -3 x -3 x -3
(-3)⁴

Question 5.
A football field is 40 yards wide and 120 yards long. Approximate the distance between opposite corners of the football field to the nearest tenth. Show your work and explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cp 5

Answer:
The distance between opposite corners of the football field is 80 yards

Explanation:
Given that the football field is 40 yards and 120 yards long.
the distance between opposite corners = 120 – 40
120 – 40 = 80 yards

Question 6.
A computer consultant charges $50 plus $40 for each hour she works. The consultant charged $650 for one job. This can be represented by the equation below, where h represents the number of hours worked.
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cp 6
40h + 50 = 650
How many hours did the consultant work?

Answer:
The consultant work = 7 hours.

Explanation:
Given that the consultant charges $ 50 + $ 40
for 1 hour the consultant charges $ 90
The consultant charged = 650$
seven hours worked = 650
9 x 7 = 63

Question 7.
Which triangle is not a right triangle?
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cp 7

Answer:
option D is not a right triangle.

Explanation:
Given that the three triangles are right triangle.
right triangle = 90 degree.

Question 8.
What is the distance between (- 3,- 1) and (- 1, – 5)?
F. \(\sqrt{12}\)
G. \(\sqrt{20}\)
H. \(\sqrt{40}\)
I. \(\sqrt{52}\)

Answer:
option F is correct.

Explanation:
given that the distance is (-3, -1) and (-1, -5)
(-3, -1) and (-1, -5)
(-4 , -6)

Question 9.
An airplane flies from City 1 at (0, 0) to City 2 at (33, 56) and then to City 3 at (23, 32). Whatis the total number of miles it flies? Each unit represents 1 mile.
Answer:
The total number of miles it flies = 56 miles

Explanation:
given that city 1 at (0, 0)
city 2 at (33, 56)
city 3 at (23, 32)
33 + 23
56

Question 10.
The national debt of Country A is $398,038,013,519. The national debt of Country B is $2,137,348,918. Approximately how many times greater is the debt of Country A than the debt of Country B?
A. 2 times greater
B. 20 times greater
C. 133 times greater
D. 200 times greater

Answer:
option B is correct.

Explanation:
country A = $398,038,013,519
country B = $2,137,348,918.

Question 11.
What is the solution of the system?
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cp 11
F.(2, 4)
G. (3, 0)
H.(4, 2)
I. (5, 0)

Answer:
option H is correct.

Explanation:
In the given graph
the line x-axis meets on 4 and y-axis meet on 2
(4, 2)

Question 12.
In the diagram, lines ℓ and m are parallel. Which angle has the same measure as ∠1?
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cp 12
A. ∠2
B. ∠5
C. ∠7
D. ∠8

Answer:
option D

Explanation:
given that l and m are parallel lines.
˂ 1 is equal to ˂8

Question 12.
Which graph represents the linear equation y = – 2x – 2?
Big Ideas Math Solutions Grade 8 Chapter 9 Real Numbers and the Pythagorean Theorem cp 13

Answer:
option F is correct.

Explanation:
graph F represents the linear equation.
y = -2x -2

Conclusion

Hope the information shared regarding the Big Ideas Math Answers Grade 8 Ch 9 Real Numbers and Pythagorean Theorem has shed some light on you. For any further clarifications on the same do leave us your suggestions and we will look into them and respond on the same.

Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume

Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume

A perfect preparation will come to your hand when you have the best source with you. Finding the best source will help you to save your time while preparing and also get a grip on the concept. The only best source to learn all the Grade 5 Chapter 13 Understand Volume is our Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume. Why every student love to use our Big Ideas Math Answers 5th Grade Answer Key Chapter 13 Understand Volume? Do you want to know why? Because we have provided from basic concepts to difficult concepts.

Big Ideas Math Book 5th Grade Chapter 13 Understand Volume Answer Key

We are providing different methods to solve problems, practice tests, quiz, objective question and answers, ad a detailed explanation for problems, etc. Get a free pdf of Big Ideas Math Answer Key Grade 5th. Good practice will help you to score the best marks in the exam. Therefore, without wasting your time, quickly start your preparation now. We have given quick links along with various problems. Try to practice all problems and become a pro in your math subject.

Lesson: 1 Understand the Concept of Volume

Lesson: 2 Find Volumes of Right Rectangular Prisms

Lesson: 3 Apply the Volume Formula

Lesson: 4 Find Unknown Dimensions

Lesson: 5 Find Volumes of Composite Figures

Chapter: 13 – Understand Volume

Lesson 13.1 Understand the Concept of Volume

Explore and Grow

Use centimeter cubes to make each figure.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 1

Which figure takes up the most space? Which figure takes up the least space? How do you know?

Answer:
Figure-3 occupies more space.

Explanation:
figure-1
volume = length x width x height
where l = length, w = width and h = height
volume = 9 x 9 x 1
In the above-given figure l=9 ft, w =9 ft and h = 1ft
volume = 9 x 9 x 1
volume = 81x 1
volume = 81 cubic ft
figure-2
volume = length x width x height
where l = length, w = width and h = height
volume = 3 x 3 x 2
In the above-given figure l=3 ft, w =3 ft and h = 2 ft
volume = 3 x 3 x 2
volume = 9 x 2
volume = 18 cubic ft
figure-3
volume = length x width x height
where l = length, w = width and h = height
volume = 7 x 3 x 5
In the above-given figure l=7 ft, w =3 ft and h = 5 ft
volume = 7 x 3 x 5
volume = 21 x 5
volume = 105 cubic ft
figure-4
volume = length x width x height
where l = length, w = width and h = height
volume = 3 x 3 x 1
In the above-given figure l=3 ft, w =3 ft and h = 1 ft
volume = 3 x 3 x 1
volume = 9 x 1
volume = 9 cubic ft

Structure
Can two different figures take up the same amount of space? Explain.

Answer:
Yes.

Explanation:
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 3
In example  1 :if we consider l= 3 cm and w = 2 cm and h = 2cm
v = l x w x h
volume = length x width x height
v = 3 x 2 x 2
v = 6 x 2
v = 12 cubic cm
In example 2 : if we consider l= 2 cm and w = 3 cm and h = 2cm.
v = l x w x h
volume = length x width x height
v = 2 x 3 x 2
v = 6 x 2
v = 12  cubic cm.
both the volumes are same.

Think and Grow: Use Unit Cubes to Find Volume

Key Idea
Volume is a measure of the amount of space that a solid figure occupies. The volume of a unit cube is 1 cubic unit. You can count unit cubes to find the volume of a solid figure.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 2
Example
Find the volume of the figure.
The figure is made of __ unit cubes.
So, the volume of the figure is _1_ cubic units.

Answer:
The volume of the figure = 12 cubic feet.
The figure is made up of 7 unit cubes.
So, the volume of the figure is 1 cubic units.

Explanation:
The volume of the figure = l x w x h
where l = length, w = width and h = height.
given that l = 4 , w= 3 and h = 1
v = 4 x 3 x 1
v = 12 x 1
v = 12 cubic feet.
The figure is made up of 7 unit cubes.
So, the volume of the figure is 1 cubic units.

Example
Find the volume of the figure.
Each unit cube has an edge length of ___
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 3
So, each unit cube has a volume of ___.
The figure is made of ___ unit cubes.
So, the volume of the figure is ___.

Answer:
Each cube has an edge length of 1 unit
So, each cube has a volume of 1 cubic unit.
The figure is made of 12 unit cubes.
So, the volume of the figure is 12 cubic cm

Explanation:
Each cube has an edge length of 1 unit
So, each cube has a volume of 1 cubic unit.
v = l x w x h
v = 1 x 1 x 1
v = 1 cubic unit
volume of figure = l x w x h
v = 3 x 2 x 2
where l = 3 cm , w = 2 cm and h = 2 cm given.
v = 3 x 2 x 2
v = 6 x 2
v = 12 cubic cm.
Show and Grow

Find the volume of the figure.

Question 1.
Volume = 6 cubic units
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 4

Answer:
The volume of the cube = 16 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 4 x 1 x 4
In the above-given figure l=4 ft, w =1 ft and h = 4ft
volume = 4 x 1 x 4
volume = 4 x 4
volume = 16 cubic ft

Question 2.
Volume = ___
Answer :- 16
Explanation:-  width*length*height(w*l*h)
= 2*4*2
=16
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 5                   

 

 

 

Answer:
The volume of cube = 16 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 4 x 2 x 2
In the above-given figure l=4 ft, w = 2 ft and h = 2ft
volume = 4 x 2 x 2
volume = 8 x 12
volume = 16 cubic ft

Apply and  Grow: Practice

Find the volume of the figure.

Question 3.
Volume = ___
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 6

Answer:
The volume of the figure = 9 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 3 x 1 x 3
In the above-given figure l=3 ft, w = 1 ft and h = 3ft
volume = 3 x 3
volume = 9 cubic ft

Question 4.
Volume = ___
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 7

Answer:
The volume of the figure = 18 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 3 x 3 x 2
In the above-given figure l=3 ft, w = 3 ft and h = 2ft
volume = 3 x 3 x 2
volume = 9 x 2
volume = 18 cubic feet

Question 5.
Volume = ____
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 8

Answer:
The volume of the figure = 9 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 3 x 1 x 3
In the above-given figure l=3 ft, w = 1 ft and h = 3 ft
volume = 3 x 1 x 3
volume = 3 x 3
volume = 9 cubic feet

Question 6.
Volume = ____
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 9

Answer:
The volume of the figure = 40 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 4 x 2 x 5
In the above-given figure l=4 ft, w = 2 ft and h = 5 ft
volume = 4 x 2 x 5
volume = 8 x 5
volume = 40 cubic feet

Question 7.
Cube-shaped shipping boxes, like the one shown, are packed into a van. They form a solid without any gaps or overlaps that is 3 feet long, 2 feet wide, and 3 feet tall. What is the volume of the solid formed?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 10

Answer:
The volume of the solid formed = 40 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 3 x 2 x 3
In the above-given figure l=3 ft, w = 2 ft and h = 3 ft
volume = 3 x 2 x 3
volume = 6 x 3
volume = 18 cubic feet

Question 8.
YOU BE THE TEACHER
Your friend says the volume of the figure is 12 cubic centimeters. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 11

Answer:
No my friend is not correct.
The volume of the figure = 24 cubic cm.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 4 x 2 x 3
In the above-given figure l=4 ft, w = 2 ft and h = 3 ft
volume = 4 x 2 x 3
volume = 8 x 3
volume = 24 cubic feet

Question 9.
DIG DEEPER!
What is the volume of the figure formed by the missing cubes?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 12

Answer:
The volume of the figure formed by the missing cubes = 32 cubic inches.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 4 x 4 x 2
In the above-given figure l=4 ft, w = 4 ft and h = 2 ft
volume = 4 x 4 x 2
volume = 16 x 2
volume = 32 cubic feet

Think and Grow: Modeling Real Life

Example
The puzzle is made of unit cubes. A box can hold 20 of the puzzles without any gaps or overlaps. What is the volume of the box?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 13
Find the volume of one puzzle.
The puzzle is made of unit cubes that each have an edge length of ___.
So, each unit cube has a volume of ___.
One puzzle is made of ___ unit cubes.
So, the volume of one puzzle is ___.

Answer:
The puzzle is made of unit cubes that each have an edge length of  3cm.
so, each unit cube has a volume of 1 cm.
one puzzle is made up of  3 cm
so, the volume of one puzzle is 27 cubic cm.

Explanation:
In the above-given figure,
length = 3 cm, width = 3 cm, height = 3cm
volume of the cube = l x w x h
where l = length, w = width and h = height
volume = 3 x 3 x 3
volume = 9 x 3
volume = 27 cubic cm

Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 14
Multiply the volume of one puzzle by the number of puzzles
___ × ___ = ____
So, the volume of the box is ___.

Answer:
The volume of the box =

Show and Grow

Question 10.
A container holds 8 of the snowglobe boxes shown, without any gaps or overlaps. What is the volume of the container?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 15

Answer:
The volume of the container = 36 cubic in.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 4 x 3 x 3
In the above-given figure l=4 in, w = 3 in and h = 3 in
volume = 4 x 3 x 3
volume = 13 x 3
volume = 36 cubic in

Question 11.
The shelving unit is made of six small compartments, one medium compartment, and one large compartment. Each small compartment has a length, a width, and a height of 1 foot. What is the volume of the shelving unit?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 16

Answer:
The volume of the shelving unit = 4 cubic feet.

Explanation:
In the above-given question the shelving unit is made of  six small compartments, one medium compartment, and one large compartment.
volume = length x width x height
where l = length, w = width and h = height
volume = 4 x 1 x 1
In the above-given figure l=4 ft, w = 1 ft and h = 1 ft
volume = 4 x 1 x 1
volume = 4 x 1
volume = 4 cubic feet.

Understand the Concept of Volume Homework & Practice 13.1

Find the volume of the figure.

Question 1.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 17

Answer:
The volume of the figure = 15 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 5 x 1 x 3
In the above-given figure l=5 ft, w = 1 ft and h = 3 ft
volume = 5 x 1 x 3
volume = 5 x 3
volume = 15 cubic feet.

Question 2.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 18

Answer:
The volume of the figure = 27 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 3 x 3 x 3
In the above-given figure l=3 ft, w = 3 ft and h = 3 ft
volume = 3 x 3 x 3
volume = 9 x 3
volume = 27 cubic feet.

Question 3.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 19

Answer:
The volume of the figure = 10 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 2 x 1 x 5
In the above-given figure l=2 ft, w = 1 ft and h = 5 ft
volume = 2 x 1 x 5
volume = 2 x 5
volume = 10 cubic feet.

Question 4.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 20

Answer:
The volume of the figure = 8 cubic cm.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 2 x 2 x 2
In the above-given figure l=2 cm, w = 2 cm and h = 2 cm
volume = 2 x 2 x 2
volume = 4 x 2
volume = 8 cubic cm.

Question 5.
Cube-shaped boxes, like the one shown, are placed together for a display at a store. They form a solid without any gaps or overlaps that is 5 feet long, 4 feet wide, and 1 foot tall. What is the volume of the solid formed?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 21

Answer:
The volume of the solid formed = 20 cubic ft.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 5 x 4 x 1
In the above-given figure l= 5 ft , w = 4 ft and h = 1 ft
volume = 5 x 4 x 1
volume = 20  x 1
volume = 20 cubic ft.

Question 6.
YOU BE THE TEACHER
Your friend says the figures have the same volume. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 22

Answer:
Yes, my friend is correct.

Explanation:
In the 1st given figure l= 4 cm, w = 1cm h = 1 cm.
volume = l x w x h
volume = 4 x 1 x 1
volume = 4 x 1
volume = 4 cubic cm
In the 2nd given figure l=  2 in , w = 1 in h = 2 in .
volume = l x w x h
volume = 2 x 1 x 2
volume = 2 x 2
volume = 4 cubic in

Question 7.
DIG DEEPER!
Newton uses 32 centimeter cubes to build a solid. The base of the solid is shown. What is the height of the solid?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 23

Answer:
The height of the solid = 24 cm

Explanation:
In the above-given figure, Newton uses 32-centimeter cubes to build a solid.
The number of cubes filled in the solid = 8 cm cubes.
The remaining cubes used to fill the solid is
32 – 8
24
The height of the solid = 24 cm

Question 8.
Modeling Real Life
A container holds 27 of the spring toy boxes shown, without any gaps or overlaps. What is the volume of the container?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 23.1

Answer:
The volume of the container = 8 cubic in.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 2 x 2 x 2
In the above-given figure l= 2 in , w = 2 in  and h = 2 in.
volume = 2 x 2 x 2
volume = 4  x 2
volume = 8 cubic in.

Question 9.
Modeling Real Life
A store stacks boxes in an organized display. Each shaded box has a length, a width, and a height of 1 foot. What is the volume of the display?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 24

Answer:
The volume of the display = 28 cubic ft.

Explanation:
In the above-given figure,
the shaded box has a length, width, and height of 1 foot.
volume = length x width x height
where l = length, w = width and h = height
volume = 7 x 2 x 2
In the above-given figure l= 7 ft , w = 2 ft  and h = 2 ft.
volume = 7 x 2 x 2
volume = 14  x 2
volume = 28 cubic ft.

Review & Refresh

Divide.

Question 10.
3 ÷ 5 = ___

Answer:
Divisor = 3
dividend = 5
Quotient = 1.6

Explanation:

Question 11.
1 ÷ 4 = ___

Answer:
Divisor = 1
dividend = 4
Quotient = 4

Explanation:

Question 12.
7 ÷ 8 = ___

Answer:
Divisor = 7
dividend = 8
Quotient = 4

Explanation:

Lesson 13.2 Find Volumes of Right Rectangular Prisms

Explore and Grow

You can stack identical layers of unit cubes to build a rectangular prism.
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 25
Use centimeter cubes to make one layer of a rectangular prism. Then stack identical layers and complete the table.
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 26

Answer:
The volume of one layer = 9 cubic cm
volume of 2 layers = 216 cubic cm
volume of 3 layers = 729 cubic cm
volume of 4 layers = 1,728 cubic cm
the volume of prism for 1 layer = 9 cubic cm
the volume of prism for 2 layers = 72 cubic cm
the volume of prism for 3 layers = 729 cubic cm
volume of prism for  4 layers = 1,728 cubic cm

Explanation:
Given that,
the number of layers are 1, 2, 3, 4.
volume of 1 layer = l x w x h
l = 3, w = 1 , h = 3
volume = 3 x 1 x 3
v = 9 cubic cm
volume of 2nd layer = l x w x h
l = 6, w = 6 , h = 6
volume = 6 x 6 x 6
v = 9 cubic cm
volume of 3rd layer = l x w x h
l = 9, w = 9 , h = 9
volume = 9 x 9 x 9
v = 729 cubic cm
volume of 4th layer = l x w x h
l = 12, w = 12 , h = 12
volume = 12 x 12 x 12
v = 1,728 cubic cm
volume of rectangular prism = l x w x h
where l= length, w = width and h= height
volume of 1 layer = l x w x h
l = 3, w = 1 , h = 3
volume = 3 x 1 x 3
v = 9 cubic cm
volume of 2nd layer = l x w x h
l = 6, w = 6 , h = 6
volume = 6 x 6 x 6
v = 9 cubic cm
volume of 3rd layer = l x w x h
l = 9, w = 9 , h = 9
volume = 9 x 9 x 9
v = 729 cubic cm
volume of 4th layer = l x w x h
l = 12, w = 12 , h = 12
volume = 12 x 12 x 12
volume = 1,728 cubic cm

Precision
Explain the relationship among the values in each row of the table.

Think and Grow: Find Volumes of Right Rectangular Prisms

Key Idea
A right rectangular prism is a solid figure with six rectangular faces.
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 27
To find the volume of a right rectangular prism, multiply the number of unit cubes that cover the base by the number of layers of unit cubes.

Example
Find the volume of the rectangular prism.
Each unit cube has an edge length of ___
So, each unit cube has a volume of ___.

Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 28

Answer:
The volume of the rectangular prism = 24 cubic in.
Each unit cube has an edge length  = 1 cm
so, each unit cube has a volume = 1 cubic cm

Explanation:
The volume of the rectangular prism = l x w x h
where l = length, w = width and h = height
In the above figure, given that l = 4 in, w = 3 in, and h = 2 in.
v= 4 x 3 x 2
v = 12 x 2
v = 24 cubic in.
each cube has an edge length of 1 cm
in the above-figure shown that each cube has a length of 1 cm.
each cube has a volume of 1 cubic cm.
v = 1 x 1 x 1
v = 1 x 1
v = 1 cubic cm

Find the number of unit cubes in a base layer. Then multiply by the number of layers to find the volume.
A base layer is made of ___ × ___ = ___ unit cubes.
The prism is made of _____ layers of unit cubes.
So, the prism is made of ___ × ___ = ___ unit cubes.
The volume of the prism is ___.

Answer:
A base layer is made up of 4 x 1 = 4 cm cubes.
The prism is made up of 2 layers of unit cubes.
So, the prism is made up of  8 x 1 = 8 unit cubes.
The volume of the prism = 24 cubic in

Explanation:
In the above-given figure,
the base layer is made up of 4 cubes and it has 1 unit
4 x 1 = 4 unit cubes.
In the above-given figure,
the prism is made up of 2 layers of unit cubes.
so the prism is made up of 8 unit cubes.
8 x 1 = 8 unit cubes.
volume of the prism = length x width x height
v = l x w x h
v = 4 x 3 x 2
v = 24 cubic in

Show and Grow

Find the volume of the rectangular prism.

Question 1.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 29

Answer:
The volume of rectangular prism = 20 cubic  cm.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given figure, l= 5 cm , w = 2 cm , h = 2 cm.
volume = l x w x h
v = 5 x 2 x 2
v = 10 x 2
v = 20 cubic cm.

Question 2.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 30

Answer:
The volume of rectangular prism = 18 cubic  ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given figure, l= 2 ft  , w = 3 ft  , h = 3 ft .
volume = l x w x h
v = 2 x 3 x 3
v = 6 x 3
v = 18 cubic ft .

Apply and Grow: Practice

Find the volume of the rectangular prism.

Question 3.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 31

Answer:
The volume of rectangular prism = 60 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given figure, l= 4 ft  , w = 3 ft  , h = 5 ft .
volume = l x w x h
v = 4 x 3 x 5
v = 12 x 5
v = 60 cubic ft .

Question 4.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 32

Answer:
The volume of rectangular prism = 36 cubic cm.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given figure, l= 6cm  , w = 2 cm  , h = 3 cm.
volume = l x w x h
v = 6 x 2 x 3
v = 12 x 3
v = 36 cubic cm .

Question 5.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 33

Answer:
The volume of rectangular prism = 108 cubic in.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given figure, l= 9 in  , w = 3 in  , h = 4 in.
volume = l x w x h
v = 9 x 3 x 4
v = 27 x 4
v = 108  cubic in .

Question 6.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 34

Answer:
The volume of rectangular prism = 80 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given figure, l= 5 ft  , w = 8 ft  , h = 2 ft.
volume = l x w x h
v = 5 x 8 x 2
v = 40 x 2
v = 80  cubic ft .

Question 7.
A rectangular prism is made of unit cubes, like the one shown. The prism is 6 centimeters long, 3 centimeters wide, and 4 centimeters tall. What is the volume of the prism?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 35

Answer:
The volume of rectangular prism = 72 cubic cm.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given question, l= 6 cm  , w = 3 cm  , h = 4 cm.
volume = l x w x h
v = 6 x 3 x 4
v = 18 x 4
v = 72  cubic cm .

Question 8.
Structure
What happens to the volume of a rectangular prism when you double its height? Justify your answer by giving an example.

Answer:
The volume of the rectangular prism is increased.

Explanation:
For example, we are assuming l = 3 in , w = 2 in and h = 2 in.
volume of the rectangular prism = l x w x h
v = 3 x 2 x 2
In the above given question height is doubled.
h = 4
v = 3 x 2 x 4
v = 6 x 4
v = 24 cubic in

Question 9.
DIG DEEPER!
How many unit cubes can fit inside the rectangular prism? Explain.

Answer:
80 unit cubes can fit inside the rectangular prism.

Explanation:
For example, we are assuming l = 5 , w = 8 and h = 2 ft .
v = l x w x h
v = 5 x 8 x 2
v = 40  x 2
v = 80 cubiic feets.

Think and Grow: Modeling Real Life

Example
A package containing video games has a volume of 150 cubic inches. The diagram shows the number of packages that can fit in a shipping box. Estimate the volume of the shipping box.
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 36
Find the number of packages in a base layer of the shipping box. Then multiply by the number of layers to find the total number of packages.
A base layer of the shipping box can hold 2 × 3 = 6 packages. There are 4 layers. So, the shipping box holds 4 × 6 = 24 packages.
Multiply the volume of 1 package by the number of packages that fit in the shipping box.
24 × __ = __ cubic inches
So, the volume of the shipping box is about __ cubic inches.

Answer:
a. The number of packages in a base layer of the shipping box = 6 packages.
number of layers = 4 and total number of packages = 6
6 x 4 = 24
The number of packages that fit in the shipping box = 6 boxes
1 x  6 = 6
volume of 1 package = 1 cube.

Explanation:
a. The number of packages in a base layer of the shipping box = 6 packages.
6 x 4 = 24
number of layers = 4 and total number of packages = 6
6 x 4 = 24
The number of packages that fit in the shipping box = 6 boxes
1 x  6 = 6
volume of 1 package = 1 cube.

Show and Grow

Question 10.
A book has a volume of 91 cubic inches. The diagram shows the number of books that can fit in a container.
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 37
Estimate the volume of the container.

Answer:
The volume of the container = 819 in

Explanation:
In the above-given figure,
A book has a volume of 91 cubic inches.
9 books can fit in the container .
91 x 9 = 819 in.

Question 11.
DIG DEEPER!
Your teacher stores 12 of the crates in a supply closet. The closet is 3 feet long, 2 feet wide, and 7 feet tall. How much extra space is in the closet?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 38

Answer:
The extra space in the closet = 43 cubic ft.

Explanation:
volume of closet = l x w x h
where l = length, w = width and h = height.
l = 3 ft , w = 2 ft and h = 7 ft.
v = 3 x 2 x 7
v = 6 x 7
v = 42 cubic feet
In the above given figure,
l = 1 ft , w = 1 ft, and h = 1 ft
v = 1 x 1 x 1
v = 1 cubic cm
The extra space = 42 + 1
43 cubic ft

Find Volumes of Right Rectangular Prisms Homework & Practice 13.2

Find the volume of the rectangular prism.

Question 1.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 39

Answer:
The volume of the right rectangular prism = 30 cubic cm.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given figure, l= 5 cm  , w = 2 cm  , h = 3 cm.
volume = l x w x h
v = 5 x 2 x 3
v = 10 x 3
v = 30  cubic cm.

Question 2.
Volume = ___
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 40

Answer:
The volume of the right rectangular prism = 120 cubic in.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given figure, l= 4 in  , w = 3 in  , h = 3 in.
volume = l x w x h
v = 4 x 3 x 3
v = 40 x 3
v = 120 cubic in

Question 3.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 41

Answer:
The volume of the right rectangular prism = 50 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given figure, l= 5 ft  , w = 5 ft , h = 2 ft.
volume = l x w x h
v = 5 x 5 x 2
v = 25 x 2
v = 50 cubic ft.

Question 4.
Volume = ___
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 42

Answer:
The volume of the right rectangular prism = 120 cubic cm.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given figure, l=10 cm  , w = 3 cm , h = 4cm.
volume = l x w x h
v = 10 x 3 x 4
v = 30 x 4
v = 120 cubic cm.

Question 5.
A rectangular prism is made of unit cubes, like the one shown. The prism is 7 inches long, 2 inches wide, and 4 inches tall. What is the volume of the prism?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 43

Answer:
The volume of the right rectangular prism = 120 cubic in.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
In the above-given question , l=7 in  , w = 2 in , h = 4 in.
volume = l x w x h
v = 7 x 2 x 4
v = 14 x 4
v = 176 cubic in.

Question 6.
Reasoning
A cube has a volume of 27 cubic inches. What is the edge length of the cube?

Answer:
The edge length of the cube = 3 in

Explanation:
In the above-given question,
the volume of cube = 27 cubic inches.
v = l x w x h
v = 3 x 3 x 3
The edge length of the cube = 3 in

Question 7.
Reasoning
You, Newton, and Descartes each build the rectangular prism shown. You stack all three prisms on top of each other. What is the volume of the new prism? How many times greater is the volume of the new prism than the volume of your original prism?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 44

Answer:
The volume of the new prism = 24 cubic cm

Explanation:
In the above-given figure,
l = 4 cm , w = 1 cm and h = 2 cm given.
v = l x w x h
v = 4 x 1 x 2
v = 4 x 2
v = 8 cubic cm
I stack all 3 prisms on top of each other.
The volume of the new prism = 12 x 1 x 2
v = 12 x 2
v = 24 cubic cm.

Question 8.
Modeling Real Life
A shoe box has a volume of 336 cubic inches. The diagram shows the number of shoe boxes that can fit in a shipping box. Estimate the volume of the shipping box.
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 45

Answer:
The volume of the shipping box = 3,024 cubic in.

Explanation:
Given that a shoebox has a volume of 336 cubic inches.
In the shipping box, we can fit 9 shoe boxes.
volume of the shipping box = 336 x 9
volume = 3,024 cubic in

Question 9.
Modeling Real Life
The wooden pallet can hold a maximum of 5 layers of boxes.A worker places 48 boxes, like the one shown, on to the pallet with no gaps. How many more boxes can fit?
Big Ideas Math Solutions Grade 5 Chapter 13 Understand Volume 46

Answer:
The more boxes can fit  = 3 boxes.

Explanation:
Given that the wooden pallet can hold a maximum of 5 layers of boxes.
A worker places 48 boxes
The more boxes can fit = 5 x 9
5 x 9 = 45 cubic ft
48 – 45 = 3 boxes.

Review & Refresh

Use a common denominator to write an equivalent fraction for each fraction.

Question 10.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 47

Answer:
(18/27) , (15/27)

Explanation:
Factors of 3 = 1, 3
factrs of 9 =  1, 3 , 9
from those the common are 1, 3.
1 x 3 = 3
3 x 3 = 9
2 x 9 = 18
5 x 3 = 15
(18/27) , (15/27)

Question 11.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 48

Answer:
(7/28) , (12/28)

Explanation:
Factors of 4 = 1, 4 and 2.
factrs of 7 =  1, 7
from those the common are 1
7 x 4 = 28
7 x 1 = 7
3 x 4 = 12
(7/28) , (12/28)

Lesson 13.3 Apply the Volume Formula

Explore and Grow

Work with a partner. Use centimeter cubes to create several different rectangular prisms, each with a volume of 12 cubic centimeters. Record your results in the table.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 49
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 50
What do you notice about the dimensions of each prism?

Answer:
The dimensions of each prism = 2cm, 3 cm ,2cm.
length = 2 width = 3 and height = 2.
Base length, base width, and height
2, 3, 2.
3, 2, 2.
2, 2, 3.
4, 3, 1.

Explanation:
In the above-given question,
volume of each cube = 12 cubic cm.
we can assume length = 2cm width = 3cm and height = 2cm.
otherwise we can assume length = 3 cm width = 2 cm and height = 2cm.
length = 2 cm width = 2 cm and height = 3 cm.
length = 4 cm width = 3 cm and height = 1 cm.

Structure
How can you find the volume of a rectangular prism without using unit cubes?

Answer:
we can find the volume of a rectangular prism by using cm cubes.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 4 in  , w = 3 in  , h = 3 in.
volume = l x w x h
v = 4 x 3 x 3
v = 40 x 3
v = 120 cubic in

Think and Grow: Use a Formula to Find Volume

Key Idea
You can use the length, width, and height of a rectangular prism to find its volume.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 51
Volume of a Rectangular Prism

Example
Find the volume of the rectangular prism.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 52

Answer:
The volume of rectangular prism = 160 cubic feet.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 4 in  , w = 3 in  , h = 3 in.
volume = l x w x h
v = 4 x 3 x 3
v = 40 x 3
v = 120 cubic in

Show and Grow

Find the volume of the rectangular prism.

Question 1.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 55

Answer:
The Volume of rectangular prism = 125 cubic cm.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 5 cm  , w = 5 cm  , h = 5 cm.
volume = l x w x h
v = 5 x 5 x 5
v = 25 x 5
v = 125 cubic cm.

Question 2.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 56

Answer:
The Volume of rectangular prism = 540 cubic in.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 6 in  , w = 9 in  , h = 10 in.
volume = l x w x h
v = 6 x 9 x 10
v = 54 x 10
v = 540 cubic in.

Apply and Grow: Practice

Find the volume of the rectangular prism.

Question 3.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 57

Answer:
The Volume of rectangular prism = 60 cubic in.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 5 in  , w = 2 in  , h = 5 in.
volume = l x w x h
v = 5 x 2 x 6
v = 10 x 6
v = 60 cubic in.

Question 4.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 58

Answer:
The Volume of rectangular prism = 108 cubic cm.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 9 cm  , w = 4 cm  , h = 3 cm .
volume = l x w x h
v = 9 x 4 x 3
v = 36 x 3
v = 108 cubic cm.

Question 5.
Newton’s dog house is a rectangular prism. The house is 5 feet long, 4 feet wide, and 5 feet tall. What is the volume of the house?

Answer:
The Volume of house = 100 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 5 ft  , w = 4 ft  , h = 5 ft .
volume = l x w x h
v = 5 x 4 x 5
v = 20 x 5
v = 100 cubic ft.

Question 6.
A hole dug for an in-ground swimming pool is a rectangular prism. The hole is 20 feet long, 10 feet wide, and 6 feet tall. What is the volume of the hole?

Answer:
The Volume of the hole = 1200 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 20 ft  , w = 10 ft  , h = 6 ft .
volume = l x w x h
v = 20 x 10 x 6
v = 200 x 6
v = 1200 cubic ft.

Question 7.
Structure
Compare the dimensions and volumes of the three rectangular prisms. What do you notice?
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 59

Answer:
The dimensions and volumes of the 3 rectangular prisms are equal volumes.

Explanation:
volume of the 1st rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 4 in  , w = 6 in  , h = 3 in .
volume = l x w x h
v = 4 x 6 x 3
v = 24 x 3
v = 72 cubic in.
volume of the 3rd rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 6 in  , w = 3 in  , h = 4 in .
volume = l x w x h
v = 6 x 3 x 4
v = 18 x 4
v = 72 cubic in.
volume of the 2nd rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 3 in  , w = 4 in  , h = 6 in .
volume = l x w x h
v = 3 x 4 x 6
v = 12 x 6
v = 72 cubic in.

Question 8.
DIG DEEPER!
What is the volume of Newton’s rectangular prism?
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 60

Answer:
volume of the Newtons rectangular prism = 400 cubic cm

Explanation:
volume of the Newtons rectangular prism = length x width x height
where l = length, w= width and h = height
given, l= 2 cm , w = 6 in  , h = 3 in .
height = 5times of length = 5 x 2 = 10 cm
width = 2 times of height = 10 x 2 = 20 cm
volume = l x w x h
v = 2 x 20 x 10
v = 40 x 10
v = 400 cubic cm.

Think and Grow: Modeling Real Life

Example
The dump truck bed is a rectangular prism. The area of the base is 8 square yards.The height is 2 yards. The driver needs to transport 18 cubic yards of gravel. Can he transport all of the gravel at once?
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 61
Use a formula to find the volume of the dump truck bed.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 62
The volume of the dump truck bed is ___.
Compare the volume of the bed to the amount of gravel.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 63
So, the driver __ transport all of the gravel at once.

Answer:
The volume of dump truck bed = 16 sq yards.
the volume of bed to the amount of gravel = 16: 18
so, the driver cannot transport all of the gravel at once.

Explanation:
The volume of dump truck bed  = base x height
v = b x h
where base = 8 yds and height = 2 yards.
v = 8 x 2
v = 16 sq yds.
volume of bed to the amount of gravel is 18 yards given.
16: 18
so, the driver cannot transport all of the gravel at once.

Show and Grow

Question 9.
The aquarium is a rectangular prism. The area of the base is 288 square inches. Can the aquarium hold 4,500 cubic inches of water? Explain.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 64

Answer:
Yes, the aquarium can hold 4,500 cubic inches of water.

Explanation:
the volume of rectangular prism = base x height
area = 288 sq inches given . height = 16 in
volume = 288 x 16
v = 4608 cubic in.
The aquarium can hold 4,500 cubic in of water.

Question 10.
DIG DEEPER!
A principal orders two types of lockers for the school. Which locker has a greater volume? How much greater?
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 65

Answer:
Locker B has greater volume than Locker A.

Explanation:
volume of the Locker A = length x width x height
where l = length, w= width and h = height
Assume, l= 12 in  , w = 15 in  , h = 6 in .
volume = l x w x h
v = 12 x 15 x 6
v = 180 x 6
v = 1080 cubic in.
volume of the Locker B = length x width x height
where l = length, w= width and h = height
Assume, l= 15 in  , w = 15 in  , h = 5 in .
volume = l x w x h
v = 15 x 15 x 5
v = 225 x 5
v = 1125 cubic in.

Apply the Volume Formula Homework & Practice 13.3

Find the volume of the rectangular prism.

Question 1.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 66

Answer:
The Volume of the rectangular prism = 64 cubic in.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 4 in , w = 4 in , h = 4 in .
volume = l x w x h
v = 4 x 4 x 4
v = 16 x 4
v = 64 cubic in.

Question 2.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 67

Answer:
The Volume of the rectangular prism = 120 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 8 ft , w = 5 ft , h = 3 ft .
volume = l x w x h
v = 8 x 5 x 3
v = 40 x 3
v = 120 cubic ft.

Question 3.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 68

Answer:
The Volume of the rectangular prism = 56 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 4 ft , w = 2ft , h = 7 ft .
volume = l x w x h
v = 4 x 2 x 7
v = 8 x 7
v = 56 cubic ft.

Question 4.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 69

Answer:
The Volume of the rectangular prism = 120 cubic cm.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 6 cm , w = 5 cm , h = 4 cm .
volume = l x w x h
v = 6 x 5 x 4
v = 30 x 4
v = 120 cubic cm.

Question 5.
A toy chest is a rectangular prism. The chest is 4 feet long, 2 feet wide, and 2 feet tall. What is the volume of the chest?

Answer:
The Volume of the chest = 16 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 4 ft , w = 2 ft , h = 2 ft .
volume = l x w x h
v = 4 x 2 x 2
v = 8 x 2
v = 16 cubic ft.

Question 6.
YOU BE THE TEACHER
Your friend uses the formula V = B × h and finds V = 10 × 9 = 90 cubic inches. Is your friend correct? Explain.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 70

Answer:
No, my friend is not correct.

Explanation:
volume = base x height
v = 9 x 5
where base = 9 in, height = 5 in given.
v = 45 cubic in.

Question 7.
Number Sense
Find the volume of the rectangular prism. Then double the dimensions of the prism and find the new volume. How does the new volume compare to the original volume?
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 71

Answer:
The new Volume of the prism= 3600 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 4 in , w = 3 in , h = 5 in .
volume = l x w x h
v = 4 x 3 x 5
v = 12  x 5
v =  60 cubic in.
Given that the dimensions are doubled.
4 x 4 = 16 , 3 x 3 = 9, 5 x 5 = 25
volume of the new rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l= 16 in , w = 9 in , h = 25 in .
volume = l x w x h
v = 16 x 9 x 25
v = 144  x 25
v =  3600 cubic in.

Question 8.
Modeling Real Life
A sandbox is a rectangular prism. The area of the base is 3,600 square inches. The height is 11 inches. You add 38,000 cubic inches of sand to the box. Do you fill the sandbox to the top? Explain.

Answer:
Yes, we can fill the sandbox to the top.

Explanation:
volume = base x height
v = 3600 x 11
v = 39,600 cubic in.
if  we add 38,000 cubic inches of sand to the box
39,600 + 38,000
77,600  inches.

Question 9.
DIG DEEPER!
Each piece of Descartes’s clubhouse is a cube. What is the volume of the clubhouse? Explain.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 72

Answer:
The volume of the clubhouse = 6 cubic feet.

Explanation:
volume of the cube = length x width x height
where l = length, w= width and h = height
Assume, l= 3 ft , w = 2 ft , h = 1 ft .
volume = l x w x h
v = 3 x 2 x 1
v = 6  x 1
v =  6 cubic ft.

Review & Refresh

Find the quotient. Then check your answer.

Question 10.
3.38 ÷ 2.6 = ___

Answer:
Quotient = 1.3

Explanation:

Question 11.
6.12 ÷ 1.53 = __

Answer:
Quotient = 4.0

Explanation:

Question 12.
0.63 ÷ 0.9 = ___

Answer:
Quotient = 0.7

Explanation:

Lesson 13.4 Find Unknown Dimensions

Explore and Grow

For each row of the table, use centimeter cubes to create the rectangular prism described. Then complete the table.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 73
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 74

Construct Arguments
Explain how you determined the unknown dimension for each rectangular prism.

Answer:
volume = 8, length = 2cm, width = 2 cm, height = 2 cm.
volume = 15, length = 5cm, width = 3 cm, height = 1 cm.
volume = 24, length = 8cm, width = 2 cm, height = 4 cm.
volume = 36, length = 4cm, width = 3 cm, height = 3 cm.

Explanation:
Volume of rectangular prism = l x w x h
length = 2cm, width = 2cm given.
volume = 2 x 2 x 2
v = 8 cubic cm, h = 2 cm
Volume of rectangular prism = l x w x h
length = 5cm, height= 1cm given.
volume = 2 x 3 x 1
v = 15 cubic cm, w = 3 cm
Volume of rectangular prism = l x w x h
width = 2cm, height= 4cm given.
volume =3  x 2 x 4
v = 24 cubic cm, l = 3 cm
Volume of rectangular prism = l x w x h
length = 4cm, width= 3cm given.
volume =4 x 3 x 3
v = 36 cubic cm, h = 3 cm

Think and Grow: Find Unknown Dimensions

Example
The volume of the rectangular prism is 56 cubic feet. Find the height.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 75

Answer:
The height of the rectangular prism = cm

Explanation:
Given that volume of rectangular prism = 56 cubic feet.
volume = base x height
v = 56 given base = 8
56 = 8 x h
h = (56/8)
h = 7 cm

Example
The volume of the rectangular prism is 180 cubic centimeters. Find the length.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 76

Answer:
The length of prism = 9 cm

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
given , w = 5 cm , h = 4 cm .
volume = l x w x h
180 = l x 5 x 4
180 = l x 20
180 = 20 l
l = (180/20)
l = 9 cm.

Show and Grow

Find the unknown dimension of the rectangular prism.

Question 1.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 77

Answer:
The height of prism = 1 cm

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
given , l = 3 cm , w = 4 cm .
volume = l x w x h
12 = 3 x 4 x h
12 = l2 x h
12 = 12h
h= (20/20)
h = 1 cm.

Question 2.
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 78

Answer:
The width of prism = 1 cm

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
given , l = 3 cm , h = 9 cm .
volume = l x w x h
27 = 3 x w x 9
27 = 27 w
w= (27/27)
w = 1 cm.

Apply and Grow: Practice

Find the unknown dimension of the rectangular prism.

Question 3.
Volume = 400 cubic inches
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 79

Answer:
The height of the rectangular prism = 8 in

Explanation:
Given that volume of rectangular prism = 56 cubic feet.
volume = base x height
v = 400 given base = 50
400 = 50 x h
h = (400/50)
h = 8 in

Question 4.
Volume = 343 cubic centimeters
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 80

Answer:
The length of the rectangular prism = 8 in

Explanation:
Given that volume of rectangular prism = 56 cubic feet.
volume = base x height
v = 343 given base = 49 sq cm
343 = 49 x l
l = (343/49)
l = 8 in

Question 5.
Volume = 132 cubic feet
Big Ideas Math Answers Grade 5 Chapter 13 Understand Volume 81

Answer:
The height of the rectangular prism = 11 ft

Explanation:
Given that volume of rectangular prism = 132cubic feet.
volume = base x height
v = 132 given base = 12 sq ft
132 = 12 x h
h = (132/12)
h = 11 ft

Question 6.
Volume = 512 cubic inches
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 82

Answer:
The Width of the rectangular prism = 8 in.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l=16 in , v = 512 cubicin , h = 4 in.
volume = l x w x h
512 = 16 x w x 4
512 = w x 64
512 = 64 w
w = (512/64)
w = 8 in

Question 7.
Writing
Explain how to find an unknown dimension of a rectangular prism given its volume and the other two dimensions.

Answer:
The Width of the rectangular prism = 8 in.

Explanation:
We have to assume v = 512 cubic in, l= 16 and h= 4 in
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
Assume, l=16 in , v = 512 cubicin , h = 4 in.
volume = l x w x h
512 = 16 x w x 4
512 = w x 64
512 = 64 w
w = (512/64)
w = 8 in

Question 8.
DIG DEEPER!
A rectangular prism has a volume of 720 cubic centimeters. The height of the prism is 10 centimeters. The length is twice the width. What are the dimensions of the base?

Answer:
The dimensions of the base = 72 cm

Explanation:
Given that volume of rectangular prism = 720 cubic cm.
volume = base x height
v = 720 given h = 10
720 = b x 10
720 =10 b
b = (720/10)
b = 72 cm

Think and Grow: Modeling Real Life

Example
A traveler buys the painting shown. The volume of the suitcase is 1,820 cubic inches. Can the painting lie flat inside the suitcase?
Use a formula to find the length of the suitcase.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 83

Answer:
Yes, the painting lies flat inside the suitcase.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height.
given , v = 1,820 ,  h = 7 in, w = 13 in .
volume = l x w x h
1820 = l x 13x 7
1820 = l x 91
l = (1820/91)
l = 20 in
The area of the painting = l x w
painting is in the shape of a rectangle
area of rectangle = l x w
area = 14 x 11
area = 154

Show and Grow

Question 9.
The volume of the candle is 576 cubic centimeters. Will the candle fit completely inside the glass jar? Explain.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 85

Answer:
Yes, the candle will fit completely inside the glass jar.

Explanation:
Given that volume of candle = 576 cubic cm.
volume = base x height
v = 576 given b = 36 sq cm
576 = 36 x h
576= 36 h
h = (576/36)
h = 16 cm
Given that volume of glass jar = 576 cubic cm.
volume = base x height
given b = 64 sq cm h = 12 cm
v = 64 x 12
v= 768 cubic cm

Question 10.
DIG DEEPER!
The popcorn boxes have the same volume, but different dimensions. What is the width of Box B?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 86

Answer:
The Width of the Box B  = 3 in.

Explanation:
volume of the Box A = length x width x height
where l = length, w= width and h = height
given, l=5 in , w = 2 in , h = 9 in.
volume = l x w x h
v = 5 x 2 x 9
v = 10 x 9
v = 90 cubic in
volume of the Box B = length x width x height
where l = length, w= width and h = height
given, l=5 in , v = 90 cubic in , h = 6 in.
volume = l x w x h
90 = 5 x w x 6
90 = 30 x w
90 = 30 w
w = (90/30)
w = 3 in

Find Unknown Dimensions Homework & Practice 13.4

Question 1.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 87

Answer:
The unknown dimensions are l= 6in, w= 2 in,and h = 1 in.

Explanation:
given that area of the base = 12 sq in
area = l x w
12 = 6 x 2
v = l x w x h
12 = 6 x 2h
12 = 12 h
h = (12/12)
h = 1 in

Question 2.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 88

Answer:
length of the prism = 1 cm

Explanation:
Given that volume of prism = 30 cubic cm.
volume = length x base x height
given b = 6 cm h = 5 cm
30 =  l x 6 x 5
30= 30 l
l = (30/30)
l = 1 cm

Find the unknown dimension of the rectangular prism.

Question 3.
Volume = 480 cubic feet
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 89

Answer:
The height of the rectangular prism = 2 ft

Explanation:
Given that volume of rectangular prism = 56 cubic feet.
volume = base x height
v = 480 cubic feet given base = 240 sq ft
480 = 240 x h
480 = 240 h
h = (480/240)
h = (48/24)
h = 2 ft

Question 4.
Volume = 351 cubic inches
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 90

Answer:
The Width of the prism =7.3125 in

Explanation:
volume of the Box B = length x width x height
where l = length, w= width and h = height
given, l=3 in , v = 351 cubic in ,  h = 13 in.
volume = l x w x h
351 = 3 x w x 16
351 = 48 x w
351 = 48 w
w = (351/48)
w = 7.3125 in

Question 5.
Open-Ended
A rectangular prism has a volume of 144 cubic feet. The height of the prism is 6 feet. Give one possible pair of dimensions for the base.

Answer:
The base of the rectangular prism = 24 ft

Explanation:
Given that volume of rectangular prism = 144 cubic feet.
volume = base x height
v = 144 cubic feet given height = 6 ft
144 = b x 6
144 = 6b
b = (144/6)
b = 24 feet

Question 6.
DIG DEEPER!
The volume of the rectangular prism is 64 cubic centimeters. What is the value of y?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 91

Answer:
The value of y = 4 cm

Explanation:
Given that the volume of rectangular prism = 64 cubic in.
v = l x w x h
64 = lwh
64 = 4 x 4 x 4
64 = 16  x 4
64 = 64

Question 7.
Modeling Real Life
The volume of the gift is 900 cubic inches. Can the gift fit inside the shipping box? Explain.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 92

Answer:
Yes, the gift will fit  inside the shipping box.

Explanation:
Given that volume of gift = 900 cubic in.
volume = base x height
v = 900 given b = 100 sq in
900 = 100 x h
900= 100h
h = (900/100)
h = 9 in
Given that base of shipping box = 576 cubic cm.
volume = base x height
given b = 144 sq in h = 10 in
v = 144 x 10
v= 1440 cubic in.

Question 8.
DIG DEEPER!
You fill the entire fish tank with water. Then you dump out some of the water.The volume of the water in the tank decreases by 96 cubic inches. How deep is the water now?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 93

Answer:
The water in the tank  = 480 cubic in.

Explanation:
volume of the Box A = length x width x height
where l = length, w= width and h = height
given, l=8 in , w = 6 in , h = 12 in.
volume = l x w x h
v = 8 x 6 x 12
v = 48 x 12
v = 576 cubic in
given that volume of water in the tank is decreased by 96 cubic in.
v = 576 – 96
v = 480 cubic in.

Review & Refresh

Divide.

Question 9.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 94

Answer:
Divisor = 0.3
Dividend = 2
Remainder = 2
Quotient = 6.6

Explanation:
(1/3) = 0.33

Question 10.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 95

Answer:
Divisor = 0.1
Dividend = 6
Remainder = 0
Quotient = 60

Explanation:
(1/10) = 0.1

Question 11.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 96

Answer:
Divisor = 0.125
Dividend = 4
Remainder = 0
Quotient = 32

Explanation:
(1/10) = 0.125

Lesson 13.5 Find Volumes of Composite Figures

Explore and Grow

Use centimeter cubes to find the volume of the figure. Explain your method.
Answer:
volume of figure 1 = 60 cubic cm.
volume of figure 2 = 36 cubic cm.

Explanation:
volume of the figure 1( left side) = length x width x height
where l = length, w= width and h = height
given, l=6 cm , w = 2 cm , h = 5 cm.
volume = l x w x h
v = 6 x 2 x 5
v = 12 x 5
v = 60 cubic cm
volume of the figure 2 (right side) = length x width x height
where l = length, w= width and h = height
given, l=6 cm , v = 3 cm , h = 2 cm.
volume = l x w x h
v = 6 x 3 x 2
v = 18 x 2
v = 36 cubic cm.

Structure
Describe another way to find the volume of the figure

Think and Grow: Find the Volume of a Composite Figure

Key Idea

A composite figure is made of two or more composite figure solid figures. You can break apart a composite figure to find its volume.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 98

Example
Find the volume of the composite figure.

Break apart the figure into rectangular prisms. Find the dimensions of each prism.

Answer:
volume of figure 1 = 320 cubic ft.
volume of figure 2 = 1200 cubic ft.

Explanation:
volume of the figure 2( right side) = length x width x height
where l = length, w= width and h = height
given, l=10 ft , w = 4 ft , h = 3 ft.
volume = l x w x h
v = 10 x 4 x 3
v = 40 x 3
v = 1200 cubic ft.
volume of the figure 1 (left side) = length x width x height
where l = length, w= width and h = height
given, l=10 ft , w = 4 ft , h = 8 ft.
volume = l x w x h
v = 10 x 4 x 8
v = 40 x 8
v = 320 cubic feet.

Show and Grow

Question 1.
Find the volume of the composite figure.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 102

Answer:
volume of figure 1 = 60 cubic cm.
volume of figure 2 = 990 cubic cm.

Explanation:
volume of the figure 1( left side) = length x width x height
where l = length, w= width and h = height
given, l=10 cm , w = 2 cm , h = 3 cm.
volume = l x w x h
v = 10 x 2 x 3
v = 20 x 3
v = 60 cubic cm.
volume of the figure 2 (right side) = length x width x height
where l = length, w= width and h = height
given, l=10 cm , w = 11 cm , h = 9 cm.
volume = l x w x h
v = 10 x 11 x 9
v = 110 x 9
v = 990 cubic cm

Apply and Grow: Practice

Find the volume of the composite figure.

Question 2.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 103

Answer:
volume of figure 1 = 360 cubic in.
volume of figure 2 = 360 cubic in.

Explanation:
volume of the figure 1( left side) = length x width x height
where l = length, w= width and h = height
given, l=12 in , w = 3 in , h = 10 in.
volume = l x w x h
v = 12 x 3 x 10
v = 36 x 10
v = 360 cubic in.
volume of the figure 2 (right side) = length x width x height
where l = length, w= width and h = height
given, l=12 in , w = 5 in , h = 6 in.
volume = l x w x h
v = 12 x 5 x 6
v = 60 x 6
v = 360 cubic in .

Question 3.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 104

Answer:
volume of figure 1 = 720 cubic ft.
volume of figure 2 = 120 cubic ft.

Explanation:
volume of the figure 1( left side) = length x width x height
where l = length, w= width and h = height
given, l=15 ft , w = 6 ft , h = 8 ft.
volume = l x w x h
v = 15 x 6 x 8
v = 90 x 8
v = 720 cubic ft.
volume of the figure 2 (right side) = length x width x height
where l = length, w= width and h = height
given, l=10 ft , w = 4 ft , h = 3 ft.
volume = l x w x h
v = 10 x 4 x 3
v = 40 x 3
v = 120 cubic ft

Question 4.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 105

Answer:
volume of figure 1 = 720 cubic ft.
volume of figure 2 = 120 cubic ft.

Explanation:
volume of the figure 1( left side) = length x width x height
where l = length, w= width and h = height
given, l=20 cm , w = 5 cm , h = 1 cm.
volume = l x w x h
v = 20 x 5 x 1
v = 100 x 8
v = 720 cubic ft.
volume of the figure 2 (right side) = length x width x height
where l = length, w= width and h = height
given, l=10 ft , w = 4 ft , h = 3 ft.
volume = l x w x h
v = 10 x 4 x 3
v = 40 x 3
v = 120 cubic ft

Question 5.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 106

Answer:
volume of figure 1 = 132 cubic ft.
volume of figure 2 = 52 cubic ft.

Explanation:
volume of the figure 2( right side) = length x width x height
where l = length, w= width and h = height
given, l=9 ft , w = 3 ft , h = 2 ft.
volume = l x w x h
v = 9 x 3 x 2
v = 27 x 2
v = 52 cubic ft.
volume of the figure 1 (left side) = length x width x height
where l = length, w= width and h = height
given, l=11ft , w = 6 ft , h = 2 ft.
volume = l x w x h
v = 11 x 6 x 2
v = 66 x 2
v = 132 cubic ft.

Question 6.
Number Sense
Write one possible equation for the volume of the composite figure.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 107

Answer:
4 x + 6 y = 13
4 x + 2 y = 1

Question 7.
YOU BE THE TEACHER
Your friend finds the volume of the composite figure. Is your friend correct? Explain.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 108

Answer:
yes, my friend is correct.

Explanation:
volume = l x w x h
v = 12 x 2 x 3
v = 72 cubic ft
v = l x w x h
v = 5 x 2 x 5
v = 10 x 5
v = 50 cubic ft.
v = 72 + 50
v = 122 cubic ft.

Think and Grow: Modeling Real Life

Example
The depth of the swimming pool is 5 feet. How many cubic feet of water can the pool hold?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 109
Break apart the pool into rectangular prisms. Find the dimensions of each prism.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 110

Answer:
so, the cube can hold 1200 cubic feet of water.

Explanation:
volume of prism A = l x w x h
l = 12 , w = 5 , h = 20
v = 12 x 5 x 20
v = 60 x 20
v = 1200 cubic feet.
volume of prism B = l x w x h
l = 40 , w = 5 v = 1200
1200 = 40 x 5 x h
1200 = 200 x h
1200 = 200h
h = (1200/200)
h = 6 ft
volume of prism C = l x w x h
h = 6, w = 5 , v = 1200
1200 = l x 5 x 6
1200 = l x 30
1200 = 30 l
l = (1200/30)
l = 40 ft

Show and Grow

Question 8.
How many cubic centimeters of wood are needed to make the game piece?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 112

Answer:
The game needed to make the game piece = 90 cubic cm.

Explanation:

volume  of prism A= l x w x h
v = 9 x 2 x 3
v = 18 x 3
v = 54 cubic cm
volume of pridm B= l x w x h
v = 6 x 2 x 3
v = 12 x 3
v = 36 cubic cm.

Question 9.
DIG DEEPER!
What is the volume of the building surrounding the outdoor courtyard? Explain your method.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 113

Answer:
The building surrounding the outdoor courtyard = 1,008 cubic m

Explanation:
volume of building surrounding inside  = l x w x h
h = 6, w = 6 m , h = 14
v = 6 x 6 x 14
v = 36 x 14
v = 504 cubic m.
The outdoor courtyard is 2 times the inside courtyard.
v = 504 x 2
v = 1,008 cubic m

Find Volumes of Composite Figures Homework & Practice 13.5

Question 1.
Find the volume of the composite figure.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 114

Answer:
volume of figure 1 = 160 cubic ft.
volume of figure 2 = 12 cubic ft.

Explanation:
volume of the figure 1( left side) = length x width x height
where l = length, w= width and h = height
given, l=5 ft , w = 4 ft , h = 8 ft.
volume = l x w x h
v = 5 x 4 x 8
v = 20 x 8
v = 160 cubic ft.
volume of the figure 2 (right side) = length x width x height
where l = length, w= width and h = height
given, l= 3ft , w = 2 ft , h = 2 ft.
volume = l x w x h
v = 3 x 2 x 2
v = 6 x 2
v = 12 cubic ft.

Find the volume of the composite figure.

Question 2.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 115

Answer:
volume of figure 1 = 40 cubic in.
volume of figure 2 = 42 cubic in.

Explanation:
volume of the figure 1( left side) = length x width x height
where l = length, w= width and h = height
given, l=10 ft , w = 2 in , h = 2 in.
volume = l x w x h
v = 10 x 2 x 2
v = 20 x 2
v = 40 cubic in.
volume of the figure 2 (right side) = length x width x height
where l = length, w= width and h = height
given, l= 3in , w = 7 in , h = 2 in.
volume = l x w x h
v = 3 x 7 x 2
v = 21 x 2
v = 42 cubic in.

Question 3.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 116\

Answer:
volume of figure 1 = 25 cubic cm.
volume of figure 2 = 48 cubic cm.

Explanation:
volume of the figure 1( top ) = length x width x height
where l = length, w= width and h = height
given, l=5 cm , w = 1 cm , h = 5 cm.
volume = l x w x h
v = 5 x 1 x 5
v = 5 x 5
v = 25 cubic cm.
volume of the figure 2 (bottom ) = length x width x height
where l = length, w= width and h = height
given, l= 8 cm , w = 6 cm , h = 1 cm.
volume = l x w x h
v = 8 x 6 x 1
v = 48 x 1
v = 48 cubic cm.

Question 4.
Structure
Find the volume of the composite figure two different ways.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 117

Answer:
volume of figure 1 = 288 cubic ft.
volume of figure 2 = 60  cubic ft.

Explanation:
volume of the figure 1( bottom) = length x width x height
where l = length, w= width and h = height
given, l= 12 ft , w = 6 ft , h = 4 ft.
volume = l x w x h
v = 12 x 6 x 4
v = 72 x 4
v = 288 cubic ft.
volume of the figure 2 (top) = length x width x height
where l = length, w= width and h = height
given, l= 5 ft , w = 4 ft , h = 3 ft.
volume = l x w x h
v = 5 x 4 x 3
v = 20 x 3
v = 60 cubic ft.

Question 5.
Modeling Real Life
The height of the garden bed is 2 feet. How many cubic feet of soil are in the garden bed?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 118

Answer:
The soil in the garden bed= 48 cubic feet.

Explanation:
v = l x w x h
where l = 16 – 2 (5)
l = 16 -10 = 6 ft
w = 12 – 8
w = 4 ft
h = 2 ft given
v = 6 x 4 x 2
v = 24 x 2
v = 48 cubic ft

Question 6.
DIG DEEPER!
The dimensions of each step are the same. How many cubic inches of concrete does a worker need to make the staircase?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 119

Answer:
The concrete does a worker need to make the staircase = 10,368 cubic in.

Explanation:
volume of the figure = length x width x height
where l = length, w= width and h = height
given, l= 36 in , w = 9 in , h = 8 in.
volume = l x w x h
v = 36 x 9 x 8
v = 324 x 8
v = 2,592 cubic in.
In the above-given figure there are  4 steps.
v = 2592 x 4
v = 10,368 cubic in

Review & Refresh

Question 7.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 120

Answer: 4

Question 8.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 121

Answer: 3

Question 9.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 122

Answer: 21/8

Understand Volume Performance Task 13

Mechanical engineers follow guidelines when designing and installing elevators to make sure passengers are safe and the elevators are wheelchair accessible. Many elevators have heating and air conditioning units attached to them, so it is important to know the volume inside the elevator to keep the air a comfortable temperature.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 123

Question 1.
The inside of the elevator shown is in the shape of a rectangular prism.
a. What is the volume of the inside of the elevator?
b. Another elevator has the same height and width. The length of the inside is doubled. How does this affect the volume?
c. Five people ride with their luggage on the elevator. Each person weighs about 200 pounds. The total combined weight of their luggage is about 425 pounds. About how many more pounds can the elevator hold?

Answer:
a. The volume of the inside elevator = 240 cubic ft
b. The volume increases.
c.The more pounds the elevator can hold = 1,875 pounds.

Explanation:
a. The volume of the inside elevator = l x w x h
given that elevator is in the shape of a rectangular prism.
v = 5 x 6 x 8
where l = 5 , w = 6 and h = 8
v = 5 x 6 x 8
v = 30 x 8
v = 240 cubic ft.
b. given that if the length is doubled = 2 x 5
l = 10
v = 10 x 6 x 8
v = 60 x 8
v = 480 cubic ft
c. Five people ride with their luggage on the elevator. Each person weighs about 200 pounds. The total combined weight of their luggage is about 425 pounds.
425 + 200
625 pounds.
Maximum capacity = 2500 pounds
2500 – 625 = 1,875 pounds

Question 2.
The inside of all new elevators must be wheelchair accessible. The minimum requirements are shown. The inside volume of an older elevator is 144 cubic feet. The base is a square and the height is 9 feet.
a. Explain why the older elevator does not meet the wheelchair requirements.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 125
b. A new wheelchair-accessible elevator is added to the building. Its height is 9 feet. Find the minimum volume of the new elevator. How does this compare to the volume of the old elevator?

Answer:
a. The older elevator does not meet the wheelchair requirements because the elevator is added to the building.
b. The minimum volume of the new elevator = 30 cubic ft

Explanation:
a. The older elevator does not meet the wheelchair requirements because the elevator is added to the building.
b. The minimum volume of the new elevator = base x height
h = 6 x 2/3 = 20/3 = 6.66
6.66/2 = 3.33
v = 3.33 x 9
v = 30 cubic ft.

Understand Volume Activity

Directions:

  1. Players take turns rolling a die.
  2. On your turn, move your piece the number of spaces shown on the die.
  3. Find the volume of the rectangular prism.
  4. Cover the volume with a counter. Your turn is over.
  5. The first player to get three in a row, horizontally, vertically, or diagonally, wins!

Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 126

Understand Volume Chapter Practice

13.1 Understand the Concept of Volume

Find the volume of the figure.

Question 1.
Volume = ____
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 127

Answer:
The volume of the cube = 28 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 7 x 4 x 1
In the above-given figure l=7 ft, w =4 ft and h = 1ft
volume = 7 x 4 x 1
volume = 28 x 1
volume = 28 cubic ft

Question 2.
Volume = ____
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 128

Answer:
The volume of the cube = 9 cubic feet.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 3 x 1 x 3
In the above-given figure l=3 ft, w =1 ft and h = 3ft
volume = 3 x 1 x 3
volume = 3 x 3
volume = 9 cubic ft

Question 3.
Volume = ____
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 129

Answer:
The volume of the cube = 8 cubic in.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 2 x 1 x 4
In the above-given figure l=2 in, w =1 in and h = 4 in.
volume = 2 x 1 x 4
volume = 2 x 4
volume = 8 cubic in.

Question 4.
Volume = ____
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 130

Answer:
The volume of the cube = 18 cubic cm.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 3 x 3 x 2
In the above-given figure l= 3 cm, w =3cm and h = 2 cm.
volume = 3 x 3 x 2
volume = 9 x 2
volume = 18 cubic cm.

Question 5.
Volume = ____
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 131

Answer:
The volume of the cube = 16 cubic ft.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 4 x 1 x 4
In the above-given figure l= 4 ft, w = 1 ft and h = 4ft.
volume = 4 x 1 x 4
volume = 4 x 4
volume = 16 cubic ft.

Question 6.
Volume = ____
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 132

Answer:
The volume of the cube = 16 cubic ft.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 4 x 2 x 2
In the above-given figure l= 4 ft, w = 2 ft and h = 2ft.
volume = 4 x 2 x 2
volume = 8 x 2
volume = 16 cubic ft.

Question 7.
Modeling Real Life
A container holds 30 of the phone boxes shown, without any gaps or overlaps. What is the volume of the container?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 133

Answer:
The volume of the container = 36 cubic in.

Explanation:
volume = length x width x height
where l = length, w = width and h = height
volume = 6 x 3 x 2
In the above-given figure l= 6 in, w = 3 in and h = 2 in.
volume = 6 x 3 x 2
volume = 18 x 2
volume = 36 cubic in.

13.2 Find Volumes of Right Rectangular Prisms

Question 8.
Volume =

Question 9.
Volume =___
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 135

Answer:
The Volume of the rectangular prism = 84 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
given, l= 4 ft , w = 3 ft, h = 7 ft .
volume = l x w x h
v = 4 x 3 x 7
v = 12 x 7
v = 84 cubic ft.

Question 10.
A rectangular prism is made of unit cubes, like the one shown. The prism is 9 centimeters long, 3 centimeters wide, and 5 centimeters tall. What is the volume of the prism?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 136

Answer:
The Volume of the rectangular prism = 135 cubic cm.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
given, l= 9 cm , w = 3 cm, h = 5 cm.
volume = l x w x h
v = 9 x 3 x 5
v = 27 x 5
v = 135 cubic cm.

13.3 Apply the Volume Formula

Find the volume of the rectangular prism.

Question 11.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 137

Answer:
The Volume of the rectangular prism = 48 cubic ft.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
given, l= 6 ft , w = 2 ft, h = 4 ft .
volume = l x w x h
v = 6 x 2 x 4
v = 12 x 4
v = 48 cubic ft.

Question 12.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 138

Answer:
The Volume of the rectangular prism = 270 cubic cm.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
given, l= 9 cm , w = 6 cm, h = 5 cm .
volume = l x w x h
v = 9 x 6 x 5
v = 54 x 5
v = 270 cubic cm.

Question 13.
Descartes’s cat cave is a rectangular prism. The cave is 14 inches long, 13 inches wide, and 12 inches tall. What is the volume of the cave?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 139

Answer:
The Volume of the cave = 2,184 cubic in.

Explanation:
volume of the rectangular prism = length x width x height
where l = length, w= width and h = height
given, l= 14 in , w = 13 in, h = 12 in .
volume = l x w x h
v = 14 x 13 x 12
v = 182 x 12
v = 2,184 cubic in.

13.4 Find Unknown Dimensions

Find the unknown dimension of the rectangular prism.

Question 14.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 140

Answer:
The unknown dimensions are l= 9 ft, w= 3 ft,and h = 1 ft.

Explanation:
given that area of the base = 27 sq ft
area = l x w
27 = 9 x 3
v = l x w x h
27 = 9 x 3 x h
27 = 27 h
h = (27/27)
h = 1 ft

Question 15.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 141

Answer:
The unknown dimensions are l= 6 cm, w= 1 cm,and h = 8 cm.

Explanation:
given that area of the base = 48 sq cm.
area = l x w x h
48 = 6 x w x 8
v = l x w x 8
48 = 48 x w
48 = 48 w
w = (48/48)
w = 1 cm

Question 16.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 142

Answer:
The unknown dimensions are l=9 in , w= 9 in,and h = 1 in.

Explanation:
given that area of the base = 81 sq in.
area = l x w x h
81 = 9 x 9 x h
v = l x w x h
81 = 81 x h
81 = 81 h
h = (81/81)
h = 1 in

Question 17.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 143

Answer:
The unknown dimensions are l=1 ft , w= 6 ft ,and h = 3 ft.

Explanation:
given that area of the base = 18 sq ft.
area = l x w x h
18 = l x 6 x 3
v = l x w x h
18 = l x 6 x 3
18 = 18 l
l = (18/18)
h = 1 ft

Question 18.
Open-Ended
A rectangular prism has a volume of 128 cubic feet. The height of the prism is 8 feet. Give one possible set of dimensions for the base.

Answer:
The dimensions of the base = 16 sq ft.

Explanation:
Given that volume of rectangular prism = 128 cubic feet.
volume = base x height
v = 128 given height = 8 ft
128 = b x 8
b = (128/8)
b = 16 sq ft.

Question 19.
Modeling Real Life
You buy the drawing pad shown. The volume of the drawer is 800 cubic inches. Can the drawing pad lie flat inside the drawer?
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 144

Answer:
Yes, the  drawing pad will fit  inside the drawer.

Explanation:
Given that volume of drawer = 800 cubic in.
volume =length x  width x height
v = 800 given w = 10 inand h = 8in
800 = l x w x h
800= l x 10 x 8
800 = 80 l
l = (800/80)
l = 10 in
Given that base of  drawig pad = 10 in.
volume = base x height
given b = 10 in h = 12 in
v = 10 x 12
v= 120 cubic in.

13.5 Find Volumes of Composite Figures

Find the volume of the composite figure

Question 20.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 145

Answer:
volume of figure 1 = 300 cubic ft.
volume of figure 2 = 480 cubic ft.

Explanation:
volume of the figure 1( left side) = length x width x height
where l = length, w= width and h = height
given, l=15 ft , w = 10 ft , h = 2 ft.
volume = l x w x h
v = 15 x 10 x 2
v = 150 x 2
v = 300 cubic ft.
volume of the figure 2 (right side) = length x width x height
where l = length, w= width and h = height
given, l=6 ft , w = 10 ft , h = 8 ft.
volume = l x w x h
v = 6 x 10 x 8
v = 60 x 8
v = 480 cubic ft.

Question 21.
Big Ideas Math Answers 5th Grade Chapter 13 Understand Volume 146

Answer:
volume of figure 1 = 88 cubic cm.
volume of figure 2 = 12 cubic cm.

Explanation:
volume of the figure 1( top) = length x width x height
where l = length, w= width and h = height
given, l=11 cm , w = 8 cm , h = 1 cm.
volume = l x w x h
v = 11 x 8 x 1
v = 88 x 1
v = 88 cubic cm.
volume of the figure 2 (bottom) = length x width x height
where l = length, w= width and h = height
given, l=4 cm , w = 3 cm , h = 1 cm.
volume = l x w x h
v = 4 x 3 x 1
v = 12 x 1
v = 12 cubic cm.

Conclusion:

The question and answers seen in the above article are prepared as per the latest syllabus. So, the students of the 5th Class can make use of the links and kickstart your preparation. Clarify your doubts by posting the comments in the below comment box. Keep in touch with us to get the Answer Key for all Big Ideas Math Grade 5 Chapters.

Big Ideas Math Answers Grade 8 Chapter 1 Equations

Big Ideas Math Answers Grade 8 Chapter 1

Big Ideas Math Answers Grade 8 Ch 1 Equations provided for all the Topics aligned as per the Textbooks makes it easy for you to understand the concepts. Become a Pro in the Concepts of BIM 8th Grade Chapter 1 Equations by solving the questions from here. Step by Step Solutions provided for all the Questions in the Big Ideas Math  Grade 8 Answers Chapter 1 Equations help you understand the concepts easily. Download the BIM 8th Grade Chapter 1 Equations Solution Key available here via quick links and ace up your preparation.

Big Ideas Math Book 8th Grade Chapter 1 Equations

Answer Key

Refer to the Topicwise Big Ideas Math Grade 8 Ch 1 Equations Solutions whenever you have any doubt on the related concepts. The BIM Textbook 8th Grade Chapter 1 Answer Key has questions belonging to Lessons 1.1 to 1.4, Assessment Tests, Chapter Tests, Cumulative Assessments, etc. You will have the topics related to Solving Simple Equations, Multi-Step Equations, Equations having Variables on Both Sides, etc. Simply click on the respective topic you wish to prepare for and excel in the exams.

performance

Lesson: 1 Solving Simple Equations

Lesson: 2 Solving Multi-step Equations

Lesson: 3 Solving Equations with Variables on Both Sides

Lesson: 4 Rewriting Equations and Formulas

Chapter: 1 – Equations 

STEAM Video/performance

STEAM Video

A half marathon is a race that is 13.1 miles long. How can a runner develop a routine to help train for a half marathon?

Big Ideas Math Answers Grade 8 Chapter 1 Equations 1

Watch the STEAM Video “Training for a Half Marathon.” Then answer the following questions.

Question 1.
Alex and Enid are training for a half marathon. They run four days each week, as shown in the table. How far do they have to run on Saturday to average 4.75 miles per running day in Week Nine?
Big Ideas Math Answers Grade 8 Chapter 1 Equations 2

Answer:
They have to run 13.1 miles on Saturday in Week Nine.

Explanation:
The average on Saturday = 4.7 miles per running day
Sum of distance ran miles/number of weeks = 4.7
(7.0 + 7.0 + 8.5 + x)/4 = 4.7
(22.5 + x)/4 = 4.7
22.5 + x = 4.7 x 4
22.5 + x = 9.4
x = 9.4 – 22.5
x = -13.1
They have to run 13.1 miles on Saturday in Week Nine.

Question 2.
Assuming they meet their goal on Saturday in Week Nine, what is the average number of miles per running day over the 4 weeks in the table?

Answer:
The average number of miles per running day over the 4 weeks in the table is 2.875 miles per day.

Explanation:
Given the running details on week 4 is
Monday – 2.0, Wednesday – 2.1, Friday – 1.9, Saturday – 5.5
The average number of miles per running day of week 4 = (2.0 + 2.1 + 1.9 + 5.5)/4
= 11.5/4
= 2.875 miles per day.

Performance Task

Target Heart Rates

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given information about a person’s heart rate.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 3
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 4
You will be asked to find the range of a person’s target heart rate. What factors might affect the range of a person’s target heart rate?

Getting Ready for Chapter 1

Chapter Exploration

Question 1.
Work with a partner. Use algebra tiles to model and solve each equation.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 5
c. x – 4 = 1
d. x + 5 = -2
e. -7 = x + 4
f. x + 6 = 7
g. -5 + x = -3
h. -4 = x – 4

Answer:
a. x = -6
b. x = -1
c. x = 5
d. x = -7
e. x = -11
f. x = 1
g. x = 2
h. x = 0

Explanation:
a. x + 3 = -3
Subtract 3 from both sides
x + 3 – 3 = -3 – 3
x = -6
b. -3 = x – 2
Add 2 to each side
-3 + 2 = x – 2 + 2
-1 = x
c. x – 4 = 1
Add 4 to each side
x – 4 + 4 = 1 + 4
x = 5
d. x + 5 = -2
Subtract 5 from each side
x + 5 – 5 = -2 – 5
x = -7
e. -7 = x + 4
Subtract 4 from each side
-7 – 4 = x + 4 – 4
-11 = x
f. x + 6 = 7
Subtract 6 from each side
x + 6 – 6 = 7 – 6
x = 1
g. -5 + x = -3
Add 5 to each side
-5 + x + 5 = -3 + 5
x = 2
h. -4 = x – 4
Add 4 to each side
-4 + 4 = x – 4 + 4
0 = x

Question 2.
WRITE GUIDELINES
Work with a partner. Use your models in Exercise 1 to summarize the algebraic steps that you can use to solve an equation.

Answer:
Write the given equation as it is.
And add or subtract the same quantity on both sides of the equation that makes variable one side of the equation.
Simplify the equation to get the value variable.

Vocabulary

The following vocabulary term is defined in this chapter. Think about what the term might mean and record your thoughts.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 7

Lesson 1.1 Solving Simple Equations

EXPLORATION 1

Work with a partner.
a. You have used the following properties in a previous course. Explain the meaning of each property.

  • Addition Property of Equality
  • Subtraction Property of Equality
  • Multiplication Property of Equality
  • Division Property of Equality

b. Which property can you use to solve each of the equations modeled by the algebra tiles? Solve each equation and explain your method.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 8
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 9
c. Write an equation that can be solved using one property of equality. Exchange equations with another pair and find the solution.

1.1 Lesson

Try It

Solve the equation. Check your solution.

Question 1.
b + 2 = -5

Answer:
b = -7

Explanation:
Given equation is b + 2 = -5
Subtract 2 from each side
b + 2 – 2 = -5 – 2
b = -7
Putting b = -7 in b + 2 = -5
-7 + 2 = -5

Question 2.
-3 = k + 3

Answer:
k = -6

Explanation:
Given equation is -3 = k + 3
Subtract 3 from each side
-3 -3 = k + 3 – 3
-6 = k
Putting k = -6 in -3 = k + 3
-3 = -6 + 3

Question 3.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 10

Answer:
t = -1/2

Explanation:
Given equation is t – 1/4 = -3/4
Add 1/4 to each side
t – 1/4 + 1/4 = -3/4 + 1/4
t = (-3 + 1)/4
t = -2/4
t = -1/2
Putting t = -1/2 in t – 1/4 = -3/4
-1/2 – 1/4 = (-2 – 1)/4
= -3/4

Try It

Solve the equation. Check your solution.

Question 4.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 10.1

Answer:
y = -28

Explanation:
Given equation is y/4 = -7
Multiply each side by 4
y/4 x 4 = -7 x 4
y = -28
Putting y = -28 in y/4 = -7
-28/4 = -7

Question 5.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 11

Answer:
z = -9

Explanation:
Given equation is -2z/3 = 6
Multiply each side by 3
-2z/3 x 3 = 6 x 3
-2z = 18
Divide each side by -2
-2z/-2 = 18/-2
z = -9
Putting z = -9 in -2z/3 = 6
[-2(-9)]/3 = 18/3 = 6

Question 6.
0.09w = 1.8

Answer:
w = 20

Explanation:
Given equation is 0.09w = 1.8
Divide each side by 0.09
0.09w/0.09 = 1.8/0.09
w = 20
putting w = 20 in 0.09w = 1.8
0.09 x 20 = 1.8

Question 7.
6π = πx

Answer:
x = 6

Explanation:
Given equation is 6π = πx
Divide each side by π
6π/π = πx/π
6 = x
Putting x = 6 in 6π = πx
6π = π(6)

Try It

Solve the equation. Check your solution.

Question 8.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 12

Answer:
p = 13

Explanation:
Given equation is p – 8 ÷ 1/2 = -3
p – 16 = -3
Add 16 to each side
p – 16 + 16 = -3 + 16
p = 13
Putting p = 13 in p – 8 ÷ 1/2 = -3
13 – 8 ÷ 1/2 = -3
13 – 16 = -3

Question 9.
q + | -10 | = 2

Answer:
q = -8

Explanation:
q + | -10 | = 2
Mod of negative 10 is 10
q + 10 = 2
Subtract 10 from each side
q + 10 – 10 = 2 – 10
q = -8
Putting q = -8 in q + | -10 | = 2
-8 + | -10 | = -8 + 10 = 2

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

WRITING
Are the equations equivalent? Explain.

Question 10.
x + 3 = 4 and x = 1

Answer:
Yes, equations are equivalent.

Explanation:
Given equations are x + 3 = 4 and x = 1
x = 4 – 3 & x = 1
x = 1 & x = 1

Question 11.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 14

Answer:
No, equations are not equivalent.

Explanation:
Given equations are -y/5 = 2 and y = 10
-y = 2 x 5 & y = 10
-y = 10 & y = 10
y = -10 & y = 10

Question 12.
OPEN-ENDED
Write an equation that you can use the Division Property of Equality to solve.

Answer:
3x + 6 = 36

Explanation:
3x + 6 = 36
Subtract 6 from each side
3x + 6 – 6 = 36 – 6
3x = 30
Dvide each side by 3
3x/3 = 30/3
x = 10

SOLVING EQUATIONS
Solve the equation. Check your solution.

Question 13.
-5 = w – 3

Answer:
w = -2

Explanation:
-5 = w – 3
Add 3 to each side
-5 + 3 = w – 3 + 3
-2 = w
Putting w = -2 in -5 = w – 3
-5 = -2 – 3

Question 14.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 15

Answer:
n = -12

Explanation:
Given equation is -2/3 n = 8
multiply each side by 3
-2/3 n x 3 = 8 x 3
-2n = 24
Divide each side by -2
-2n/-2 = 24/-2
n = -12
Putting n = -12 in -2/3 n = 8
-2/3 (-12) = -2 x -4 = 8

Question 15.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 16

Answer:
p = 33

Explanation:
Given equation is p – 9 ÷ 1/3 = 6
p – 27 = 6
Add 27 to each side
p – 27 + 27 = 6 + 27
p = 33
Putting p = 33 in p – 9 ÷ 1/3 = 6
33 – 9 ÷ 1/3 = 33 – 27
= 6

Question 16.
q + | 3 | = -5

Answer:
q = -8

Explanation:
q + | 3 | = -5
q + 3 = -5
Subtract 3 from each side
q + 3 – 3 = -5 – 3
q = -8
Putting q = -8 in q + | 3 | = -5
-8 + | 3 | = -8 + 3
= -5

Question 17.
WHICH ONE DOESN’T BELONG?
Which equation does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 17

Answer:
x – 3 = 9 does not belong with the other three.

Explanation:
x – 2 = 4
x = 4 + 2
x = 6
x – 3 = 9
x = 9 + 3
x = 12
x – 5 = 1
x = 1 + 5
x = 6
x – 6 = 0
x = 6
So, x – 3 = 9 does not belong with the other three.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.
A shipwreck is 300 meters away from a diving station. An undersea explorer travels away from the station at a speed of 2 meters per second. The explorer is x meters away from the station and will reach the shipwreck in 100 seconds. What is the value of x?
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 18

Answer:
The value of x is 100.

Explanation:
Given that,
The distance from shipwreck to the diving station = 300 m
The distance formula is d = speed x time
The distance from the explorer to the shipwreck = (2 m/s) x (100s)
= 200 m
He is also x meters away from the diving station
So, the distance between the diving station and shipwreck is x + 20 m
x + 20 = 300
Subtract 200 from both sides
x + 200 – 200 = 300 – 200
x = 100

Question 19.
You conduct an inventory for a hardware store and count 40 rolls of duct tape. Your manager wants to keep 7 boxes of duct tape in stock. If each box holds 8 rolls of duct tape, how many boxes should you order? Justify your answer.

Answer:
I should order 2 boxes of duct tapes

Explanation:
Number of rolls of duct tape available = 40
Number of rolls of duct tape each box can hold = 8 rols
Number of boxes of duct tape the manager wants to keep in stock = 7
Let x represent the number of boxes of duct tape available
Then 8x = 40
Divide each side by 8
8x/8 = 40/8
x = 5
5 boxes of duct tapes are available in stock.
Number of boxes I order = Number of boxes in the stock – Number of boxes available = 7 – 5
= 2
So, I should order 2 boxes of duct tapes

Question 20.
DIG DEEPER!
Your fitness tracker overestimates the number of steps you take by 5%. The tracker indicates that you took 7350 steps today. Write and solve an equation to find the actual number of steps you took today.

Answer:
Your actual number of steps is 7000.

Explanation:
Let x be the actual number of steps
Since the fitness tracker overestimates, the sum of the number of actual steps and 5% of its equal to 7350
Then, the equation will be
x + 0.05x = 7350
1.05x = 7350
Divide each side by 1.05
1.05x/1.05 = 7350/1.05
x = 7000
So, your actual number of steps is 7000.

Solving Simple Equations Homework & Practice 1.1

Review & Refresh

Evaluate the expression.

Question 1.
(32 – 8) + 4

Answer:
(32 – 8) + 4 = 5

Explanation:
Given expression is (32 – 8) + 4
= (9 – 8) + 4
= 1 + 4
= 5

Question 2.
1 + 5 × 32

Answer:
1 + 5 × 3² = 41

Explanation:
Given expression is 1 + 5 × 3²
= 1 + 5 x 9
= 1 + 40
= 41

Question 3.
4 × 3 + 102

Answer:
4 × 3 + 10² = 112

Explanation:
Given expression is 4 × 3 + 10²
= 4 x 3 + 10 x 10
= 12 + 100
= 112

Identify the terms, coefficients, and constants in the expression.

Question 4.
11q + 2

Answer:
The constant = 2
Terms = p
Coefficients = 11

Explanation:
Given that,
11q + 2
The constant = 2
Terms = p
Coefficients = 11

Question 5.
h + 9 + g

Answer:
The constant = 9
Terms = h, g
Coefficients = 1, 1

Explanation:
Given that,
h + 9 + g
The constant = 9
Terms = h, g
Coefficients = 1, 1

Question 6.
6m2 + 7n

Answer:
The constant = 0
Terms = m², n
Coefficients = 6, 7

Explanation:
Given that,
6m² + 7n
The constant = 0
Terms = m², n
Coefficients = 6, 7

Write the phrase as an expression.

Question 7.
the quotient of 22 and a number a

Answer:
22/a

Explanation:
Given that,
the quotient of 22 and a number a
= 22/a

Question 8.
the difference of a number t and 9

Answer:
t – 9

Explanation:
Given that,
the difference of a number t and 9
t – 9

Concepts, Skills, &Problem Solving
USING PROPERTIES OF EQUALITY
Which property of equality can you use to solve the equation modeled by the algebratiles? Solve the equation and explain your method. (See Exploration 1, p. 3.)

Question 9.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 19

Answer:
x = 4

Explanation:
x – 5 = -1
Add 5 to each side
x – 5 + 5 = -1 + 5
x = 4

Question 10.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 20

Answer:
x = 3

Explanation:
x + x + x = + 1 +1 +1 +1 +1 +1 +1 +1 +1
3x = 9
Divide each side by 3
3x/3 = 9/3
x = 3

SOLVING EQUATIONS USING ADDITION OR SUBTRACTION
Solve the equation. Check your solution.

Question 11.
x + 12 = 7

Answer:
x = -5

Explanation:
Given equation is
x + 12 = 7
Subtract 12 from both sides
x + 12 – 12 = 7 – 12
x = -5
Putting x = -5 in x + 12 = 7
-5 + 12 = 7

Question 12.
g – 16 = 8

Answer:
g = 24

Explanation:
Given equation is
g – 16 = 8
Add 16 to each side
g – 16 + 16 = 8 + 16
g = 24
Putting g = 24 in g – 16 = 8
24 – 16 = 8

Question 13.
-9 + p = 12

Answer:
p = 21

Explanation:
Given equation is -9 + p = 12
Add 9 to each side
-9 + p + 9 = 12 + 9
p = 21
putting p = 21 in -9 + p = 12
-9 + 21 = 12

Question 14.
2.5 + y = -3.5

Answer:
y = -5.5

Explanation:
Given equation is
2.5 + y = -3.5
Subtract 2.5 from each side
2.5 + y -2.5 = -3.5 – 2.5
y = -5.5
putting y = -5.5 in 2.5 + y = -3.5
2.5 – 5.5 = -3.5

Question 15.
x – 8π = π

Answer:
x = 9π

Explanation:
Given equation is
x – 8π = π
Add 8π to each side
x – 8π + 8π = π + 8π
x = 9π
Putting x = 9π in x – 8π = π
9π – 8π = π

Question 16.
4π = w – 6π

Answer:
w = 10π

Explanation:
Given equation is
4π = w – 6π
add 6π to each side
4π + 6π= w – 6π + 6π
10π = w
putting w = 10π in 4π = w – 6π
4π = 10π – 6π

Question 17.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 21

Answer:
d = 2/3

Explanation:
Given equation is 5/6 = 1/6 + d
Subtract 1/6 from each side
5/6 -1/6 = 1/6 + d – 1/6
(5 – 1)/6 = d
d = 4/6
d = 2/3
Putting d = 2/3 in 5/6 = 1/6 + d
5/6 = 1/6 + 2/3
= (1 + 4)/6 = 5/6

Question 18.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 22

Answer:
r = -7/24

Explanation:
Given equation is 3/8 = r + 2/3
Subtract 2/3 from each side
3/8 – 2/3 = r + 2/3 – 2/3
(9 – 16)/24 = r
-7/24 = r
Putting r = -7/24 in 3/8 = r + 2/3
3/8 = -7/24 + 2/3
= (-7 + 16)/24 = 9/24
= 3/8

Question 19.
n – 1.4 = -6.3

Answer:
n = -4.9

Explanation:
Given equation is n – 1.4 = -6.3
Add 1.4 to each side
n – 1.4 + 1.4 = -6.3 + 1.4
n = -4.9
Putting n = -4.9 in n – 1.4 = -6.3
-4.9 – 1.4 = -6.3

Question 20.
MODELING REAL LIFE
A discounted concert ticket costs $14.50 less than the original price p. You pay $53 for a discounted ticket. Write and solve an equation to find the original price.

Answer:
53 = p − 14.5
The original price was $67.50.

Explanation:
Let c represent what you paid for the ticket, the discounted price.
We’re told that the discount price, which you pay, is $ 14.50 less than the original price, which will be represented by p.
Algebraically, this means that
c = p –  14.5
Furthermore, we’re told you pay $53, i.e c = 53
53 = p − 14.5
Now, solve for p
p = 53 + 14.5
p = 67.50
So, the original price was $67.50.

Question 21.
PROBLEM SOLVING
A game of bowling has ten frames. After five frames, your friend’s bowling score is 65 and your bowling score is 8 less than your friend’s score.
a. Write and solve an equation to find your score.
b. By the end of the game, your friend’s score doubles and your score increases by 80. Who wins the game? Explain.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 23

Answer:
a) x + 8= 65
My score is 57
b) I won the game.

Explanation:
a) Friends bowling score = 65
My score = 8 less than friends sore
Let us take x as my score
Equation is x + 8= 65
Subtrcat 8 from each side
x + 8 – 8 = 65 – 8
x = 57
My score is 57
b) Friend’s score is doubled = 65 x 2 = 130
My score is increased by 80 = 57 + 80 = 137
So, I won the game.

SOLVING EQUATIONS USING MULTIPLICATION OR DIVISION
Solve the equation. Check your solution.

Question 22.
7x = 35

Answer:
x = 5

Explanation:
Given equation is 7x = 35
divide each side by 7
7x/7 = 35/7
x = 5
putting x = 5 in 7x = 35
7(5) = 35

Question 23.
4 = -0.8n

Answer:
n = -5

Explanation:
Given equation is 4 = -0.8n
Divide each side by -0.8
4/-0.8 = -0.8n/-0.8
-5 = n
Putting n = -5 in 4 = -0.8n
4 = -0.8(-5)

Question 24.
6 = –\(\frac{w}{8}\)

Answer:
w = -48

Explanation:
Given equation is 6 = -w/8
Multiply each side by 8
6 x 8 = -w/8 x 8
48 = -w
Multiply each side by (-1)
48 x -1 = -w x -1
w = -48
putting w = -48 in 6 = -w/8
6 = -(-48)/8 = 48/8

Question 25.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 24

Answer:
m = 7.3π

Explanation:
Given equation is m/π = 7.3
multiply each side by π
m/π x π = 7.3 x π
m = 7.3π
Putting m = 7.3π in m/π = 7.3
7.3π/π = 7.3

Question 26.
-4.3g = 25.8

Answer:
g = -6

Explanation:
Given equation is -4.3g = 25.8
divide each side by -4.3
-4.3g/-4.3 = 25.8/-4.3
g = -6
putting g = -6 in -4.3g = 25.8
-4.3(-6) = 25.8

Question 27.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 25

Answer:
k = 5/3

Explanation:
Given equation is 3/2 = (9/10) k
Multiply each side by 10/9
3/2 x (10/9) = (9/10) k x (10/9)
5/3 = k
Putting k = 5/3 in 3/2 = (9/10) k
3/2 = (9/10) x (5/3)

Question 28.
-7.8x = -1.56

Answer:
x = 0.2

Explanation:
Given equation is -7.8x = -1.56
Divide each side by -7.8
-7.8x/-7.8 = -1.56/-7.8
x = 0.2
Putting x = 0.2 in -7.8x = -1.56
-7.8(0.2) = -1.56

Question 29.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 26

Answer:
p = -7/3

Explanation:
Given equation is -2 = (6/7)p
Multiply each side by (7/6)
-2 x (7/6) = (6/7)p x (7/6)
-7/3 = p
putting p = -7/3 in -2 = (6/7)p
-2 = (6/7) x (-7/3)

Question 30.
3πd = 12π

Answer:
d = 4π

Explanation:
Given equation is 3πd = 12π
Divide each side by 3π
3πd/3π = 12π/3π
d = 4π
Putting d = 4π in 3πd = 12π
3π x 4π = 12π

Question 31.
YOU BE THE TEACHER
Your friend solves the equation. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 27

Answer:
Wrong

Explanation:
-1.5 + k = 8.2
k = 8.2 + 1.5
k = 9.7

Question 32.
STRUCTURE
A gym teacher orders 42 tennis balls. The tennis balls come in packs of 3. Which of the following equations represents the number x of packs?

Answer:
3x = 42

Explanation:
A gym teacher orders 42 tennis balls
Each pack has 3 balls
To find the number of packages of tennis balls, take the total number of tennis balls and divide that number by the number of packages.
42/3 = 14
3x = 42

Question 33.
MODELING REAL LIFE
You clean a community park for 6.5 hours. You earn $42.25. How much do you earn per hour?

Answer:
The amount I earn per hour is $6.50

Explanation:
The amount I earned for cleaning a community park = $42.25
Time = 6.5 hours
Let the amount I earned per hour is x
Equation is 6.5x = 42.25
Divide each side by 6.5
6.5x/6.5 = 42.25/6.5
x = 6.5
The amount I earn per hour is $6.50

Question 34.
MODELING REAL LIFE
A rocket is scheduled to launch from a command center in 3.75 hours. What time is it now?
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 28

Answer:
The time is 7:35

Explanation:
A rocket is scheduled to launch at 11:20 AM from a command center in 3.75 hours.
11(20/60) – 3(3/4) = 11(1/3) – 3(3/4)
= 34/3 – 15/4
= (136 – 45)/12
= 91/12 = 7(7/12)
The time is 7:35

Question 35.
MODELING REAL LIFE
After earning interest, the balance of an account is $420. The new balance is \(\frac{7}{6}\) of the original balance. How much interest did it earn?

Answer:
You earned $60 as interest.

Explanation:
x/420 = 6/7
x = 420 x 6/7
x = 360 x 6
x = 360
The amountt of interest = new balance – original balance
= 420 – 360 = 60
So, you earned $60 as interest.

Question 36.
MODELING REAL LIFE
After a cleanup, algae covers 2 miles of a coastline. The length of the coastline1covered after the cleanup is of the previous length. How many miles of the coast did the algae previously cover?

Answer:
Miles of the coast the algae previously covered is 6 miles.

Explanation:
Number of miles the algae covers after cleanup = 2 miles
Let x as the miles algae previously covered
Equation is (1/3)x = 2
x = 2 x 3
x = 6
Miles of the coast the algae previously covered is 6 miles.

Question 37.
PROBLEM SOLVING
Cedar Point, an amusement park, has some of the tallest roller coasters in the United States. The Mantis is 165 feet shorter than the Millennium Force. What is the height of the Mantis?
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 29

Answer:
The height of Mantis is 145 feet.

Explanation:
Height of the Millennium Force = 310 Feet
The mantis is shorter than Millennium Force by 165 feet
Let us take x as the height of the mantis
x = 310 – 165
x = 145 feet
The height of Mantis is 145 feet.

SOLVING AN EQUATION
Solve the equation. Check your solution.

Question 38.
-3 = h + 8 ÷ 2

Answer:
h = -7

Explanation:
Given equation is -3 = h + 8 ÷ 2
-3 = h + 4
Subtract 4 from each side
-3 – 4 = h + 4 – 4
-7 = h
Putting h = -7 in -3 = h + 8 ÷ 2
-3 = -7 + 8 ÷ 2
= -7 + 4

Question 39.
12 = w – | -7 |

Answer:
w = 19

Explanation:
Given equation is 12 = w – | -7 |
12 = w – 7
Add 7 to each side
12 + 7 = w – 7 + 7
19 = w
Putting w = 19 in 12 = w – | -7 |
12 = 19 – | -7 | = 19 – 7

Question 40.
q + | 6.4 | = 9.6

Answer:
q = 3.2

Explanation:
Given equation is q + | 6.4 | = 9.6
q + 6.4 = 9.6
Subtract 6.4 from each side
q + 6.4 – 6.4 = 9.6 – 6.4
q = 3.2
Putting q = 3.2 in q + | 6.4 | = 9.6
3.2 + | 6.4 | = 3.2 + 6.4 = 9.6

Question 41.
d – 2.8 ÷ 0.2 = -14

Answer:
d = 0

Explanation:
Given equation is d – 2.8 ÷ 0.2 = -14
d – 14 = -14
Add 14 to each side
d – 14 + 14 = -14 + 14
d = 0
Put d = 0 in d – 2.8 ÷ 0.2 = -14
0 – 2.8 ÷ 0.2 = -14

Question 42.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 30

Answer:
x = -3/9

Explanation:
Given equation is 8/9 = x + (1/3) x 7
8/9 = x + 7/3
Subtract 7/3 from each side
8/9 -7/3 = x + 7/3 -7/3
(8 – 21)/9 = x
-13/9 = x
Put x = -3/9 in 8/9 = x + (1/3) x 7
8/9 = -13/9 + (1/3) x 7
= -13/9 + 7/3 = (-13 + 21)/3

Question 43.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 31

Answer:
p = -1/12

Explanation:
Given equation is p – 1/4 . 3 = -5/6
p – 3/4 = -5/6
Add 3/4 to each side
p – 3/4 + 3/4 = -5/6 + 3/4
p = (-10 + 9)/12
p = -1/12
put p = -1/12 in p – 1/4 . 3 = -5/6
-1/12 – 1/4 . 3 = -1/12 – 3/4
= (-1 – 9)/12 = -10/12
= -5/6

Question 44.
GEOMETRY
The volume V of the prism is 1122 cubic inches. Use the formula V = Bh to find the height h of the prism.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 32

Answer:
The height of the prism is 12 inches.

Explanation:
Prism volume = 1122 cubic inches
Base b = 93.5 sq in
Volume formula V = Bh
1122 = 93.5 x h
h = 1122/93.5
h = 12 inches

SOLVING AN EQUATION
Write and solve an equation to find the value of x.

Question 45.
The angles are complementary.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 33

Answer:
x = 45°

Explanation:
Complementary angles mean the sum of angles is 90 degrees.
x + 45 = 90
x = 90 – 45
x = 45°

Question 46.
The angles are supplementary.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 34

Answer:
x° = 130°

Explanation:
Two angles are called supplementary when their measures add up to 180 degrees.
So, x° + 50° = 180°
x° = 180° – 50°
x° = 130°

Question 47.
CRITICAL THINKING
Which of the operations +, −, ×, ÷, are inverses of each other? Explain.

Answer:
addition (+), Subtraction (-) are inverse to each other.
Multiplication (x), division (÷) are inverse to each other.
Because, when you move the quantity which is having + sign from one side to the other side of the equation, it automatically converts to – sign.
In the same way other operations also.

Question 48.
LOGIC
Without solving, determine whether the solution of -2x = -15 is greater than or less than -15. Explain.

Answer:
The solution is greater than -15

Explanation:
The solution is greater than -15
Here both sides of the equation have – sign. They are eliminated
When you divide 15 by 2 you will get obviously answer greater than -15.

Question 49.
OPEN-ENDED
Write a subtraction equation and a division equation so that each has a solution of 2. Justify your answer.

Answer:
The subtraction equation is x + 4 = 6
The division equation is 2x = 4

Explanation:
The subtraction equation is x + 4 = 6
x = 6 – 4
x = 2
The division equation is 2x = 4
x = 4/2
x = 2

Question 50.
MODELING REAL LIFE
Ants of a particular species can carry 50 times their body weight. It takes 32 ants of that species to carry the cherry shown. About how much does each ant weigh?
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 35

Answer:
The weight of each ant is 3 mg

Explanation:
Weight of the cherry = 4800 mg
Number of Ants that carry the cherry = 32
Ants of a particular species can carry 50 times their body weight.
Let w is the weight of each ant
w. 50 . 32 = 4800
1600w = 4800
w = 4800/1600
w = 3
The weight of each ant is 3 mg

Question 51.
REASONING
One-fourth of the girls and one-eighth of the boys in a grade retake their school pictures. The photographer retakes pictures for 16 girls and 7 boys. How many students are in the grade?
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 36

Answer:
Total number of students = 120

Explanation:
Let the total number of girls be g, hence according to the question only one-fourth of a total number of girls(g) have retaken their pictures which is 16.
(1/4)g = 16
g = 16 x 4
g= 64
According to the question, only one-eighth of the total number of boys(b) have retaken their pictures which is 7.
(1/8)b = 7
b = 7 x 8
b = 56
Total number of students = boys + girls
= 56 + 64 = 120

Question 52.
DIG DEEPER!
You use a crowdfunding website to raise money. The website keeps 5% of each donation. Five of your friends each donate the same amount. The total funding you receive is $47.50. How much does each friend donate?

Answer:
Each friend donate $10

Explanation:
Assume x as the amount each friend donate
the website keeps 5% of each donation
The total funding you receive is $47.50.
4.75x = 47.50
x = 47.50/4.75
x = 10
Each friend donate $10

Question 53.
CRITICAL THINKING
A neighbor pays you and two friends $90 to paint her garage. You divide the money three ways in the ratio 2 : 3 : 5.
a. How much does each person receive?
b. What is one possible reason the money is not divided evenly?

Answer:
a. They receive $18, &=$27, $45
b. The possible reason for unequal distribution is that the ratios are different for each of them.

Explanation:
a. let the common multiplier for each of the people be x
So, each one gets the value of 2x, 3x, 5x
As per the question, the total amount paid is$90
So, 2x + 3x + 5x = 90
10x = 90
x = 90/10
x = 9
Hence, person receives 9(2) = $18, 9(3) = $27, 9(5) = $45
b. The possible reason for unequal distribution is that the ratios are different for each of them.

Lesson 1.2 Solving Multi-step Equations

EXPLORATION 1
Work with a partner. Find each angle measure in each figure. Use equations to justify your answers.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 37

1.2 Lesson

Try It

Solve the equation. Check your solution.

Question 4.
-4n – 8n + 17 = 23

Answer:
n = -1/2

Explanation:
Given equation is -4n – 8n + 17 = 23
-12n + 17 = 23
-12n = 23 – 17
-12n = 6
n = -6/12
n = -1/2
Put n = -1/2 in -4n – 8n + 17 = 23
-4(-1/2) – 8(-1/2) + 17 = 2 + 4 + 17 = 23

Question 5.
10 = 3n + 20 – n

Answer:
n = -5

Explanation:
Given equation is 10 = 3n + 20 – n
10 = 2n + 20
10 – 20 = 2n
-10 = 2n
n = -10/2
n = -5
put n = -5 in 10 = 3n + 20 – n
10 = 3(-5) + 20 – (-5)
= -15 + 20 + 5 = -15 + 25

Try It

Solve the equation. Check your solution.

Question 6.
-3(x + 2) + 5x = -9

Answer:
x = -3/2

Explanation:
Given equation is -3(x + 2) + 5x = -9
-3x – 6 + 5x = -9
2x – 6 = -9
2x = -9 + 6
2x = -3
x = -3/2
Put x = -3/2 in -3(x + 2) + 5x = -9
-3((-3/2) + 2) + 5(-3/2) = 9/2 – 6 – 15/2
= (9 – 15 – 12)/2 = -18/2 = -9

Question 7.
5 + 1.5(2d – 1) = 0.5

Answer:
d = -1

Explanation:
Given equation is 5 + 1.5(2d – 1) = 0.5
5 + 3d – 1.5 = 0.5
3d + 3.5 = 05
3d = 0.5 – 3.5
3d = -3
d = -3/3
d = -1
Put d = -1 in 5 + 1.5(2d – 1) = 0.5
5 + 1.5(2(-1) – 1) = 5 + 1.5(-2 – 1)
= 5 + 1.5(-3) = 5 – 4.5 = 0.5

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

SOLVING AN EQUATION
Solve the equation. Check your solution.

Question 8.
-5x + 1 = 31

Answer:
x = -6

Explanation:
Given equation is -5x + 1 = 31
Subtract 1from each side
-5x + 1 -1 = 31 – 1
-5x = 30
Divide each side by -5
-5x/-5 = 30/-5
x = -6
Put x = -6 in -5x + 1 = 31
-5(-6) + 1 = 30 + 1 = 31

Question 9.
\(\frac{1}{3}\)x – 9 = -12

Answer:
x = -9

Explanation:
Given equation is (1/3)x – 9 = -12
(1/3)x = -12 + 9
(1/3)x = -3
x = -3 x 3
x = -9
Put x = -9 in (1/3)x – 9 = -12
(1/3)(-9) – 9 = -3 – 9
= -12

Question 10.
-n – 6n + 4 = 53

Answer:
n = -7

Explanation:
Given equation is -n – 6n + 4 = 53
-7n + 4 = 53
-7n = 53 – 4
-7n = 49
n = -49/7
n = -7
put n = -7 in -n – 6n + 4 = 53
-(-7) – 6(-7) + 4 = 7 + 42 + 4 = 53

Question 11.
14 = 6n + 6 – 2n

Answer:
n = 2

Explanation:
Given equation is 14 = 6n + 6 – 2n
14 = 4n + 6
14 – 6 = 4n
8 = 4n
8/4 = n
n = 2
Put n = 2 in 14 = 6n + 6 – 2n
14 = 6(2) + 6 – 2(2)
= 12 + 6 – 4 = 18 – 4

Question 12.
-8(x + 1) + 2x = -32

Answer:
x = 4

Explanation:
Given equation is -8(x + 1) + 2x = -32
-8x – 8 + 2x = -32
-6x – 8 = -32
-6x = -32 + 8
-6x = -24
x = -24/-6
x = 4
Put x = 4 in -8(x + 1) + 2x = -32
-8(4 + 1) + 2(4) = -8(5) + 8
= -40 + 8 = -32

Question 13.
3 + 4.5(2d – 3) = 7.5

Answer:
d = 2

Explanation:
Given equation is 3 + 4.5(2d – 3) = 7.5
3 + 9d – 13.5 = 7.5
9d – 10.5 = 7.5
9d = 7.5 + 10.5
9d = 18
d = 18/9
d = 2
Put d = 2 in 3 + 4.5(2d – 3) = 7.5
3 + 4.5(2(2) – 3) = 3 + 4.5(4 – 3)
= 3 + 4.5(1)
= 3 + 4.5 = 7.5

Question 14.
WRITING
Write the sentence as an equation, then solve.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 38

Answer:
2 + 3x = 17, x = 5

Explanation:
2 + 3x = 17
Subtract 2 from each side
2 + 3x – 2 = 17 – 2
3x = 15
Divide each side by 3
3x/3 = 15/3
x = 5

Question 15.
OPEN-ENDED
Explain how to solve the equation
2(4x – 11) + 9 = 19.

Answer:
x = 4

Explanation:
Given equation is 2(4x – 11) + 9 = 19.
Expand brackets
8x – 22 + 9 = 19
Simpify
8x – 13 = 19
Add 13 to each side
8x – 13 + 13 = 19 + 13
8x = 32
Divide each side by 8
8x/8 = 32/8
x = 4

Question 16.
CRITICAL THINKING
How can you solve 3(x + 2) = 9 without distributing the 3?

Answer:
x = 1

Explanation:
3(x + 2) = 9
Divide both sides by 3
3(x + 2)/3 = 9/3
x + 2 = 3
Subtract 2 from both sides
x + 2 -2 = 3 – 2
x = 1

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.
Find the number x of action figures that a small business needs to produce on Friday so that the mean number of action figures produced per day is 50.
Big Ideas Math Answer Key Grade 8 Chapter 1 Equations 39

Answer:
On friday the number of action figures that a sma business should produce is 53.

Explanation:
Mean = 50
(55 + 45 + 53 + 44 + x)/5 = 50
(197 + x)/5 = 50
197 + x = 50 x 5
197 + x = 250
x = 250 – 197
x = 53
On friday the number of action figures that a sma business should produce is 53.

Question 18.
DIG DEEPER!
A hard drive is 80% full and has 12,000 MB of free space. One minute of video uses 60 MB of storage. How many minutes of the video should be deleted so that the hard drive is 75% full?

Answer:
50 minutes video should be deleted so that the hard drive is 75% full

Explanation:
Let x be the capacity of the hard drive.
A hard drive is 80% full and has 12,000 MB of free space
0.8x + 12000 = x
12000 = x – 0.8x
0.2x = 12000
x = 12000/0.2
x = 60,000 MB
One minute of video uses 60 MB of storage.
Let t be the number of minutes of video that should be deleted
60t = 0.8x – 0.75x
60t = 0.05x
60t = 0.05 x 60000
60t = 3000
t = 3000/60
t = 50
50 minutes video should be deleted so that the hard drive is 75% full

Question 19.
A teacher spends $354 on costumes and microphones for six cast members in a play. Each cast member receives a costume that costs $38 and a microphone that costs $c. What did the teacher spend on each microphone? Justify your answer.

Answer:
The teacher spends $21 on each microphone

Explanation:
The amount the teacher spends on six costumes and microphone = $354
Each cast member receives a costume that costs $38 and a microphone that costs $c.
6(38 + c) = 354
38 + c = 354/6
38 + c = 59
c = 59 – 38
c = 21
The teacher spends $21 on each microphone

Solving Multi-step Equations Homework & Practice 1.2

Review & Refresh

Solve the equation.

Question 1.
y + 8 = 3

Answer:
y = -5

Explanation:
Given equation is y + 8 = 3
y + 8 – 8 = 3 – 8
y = -5

Question 2.
h – 1 = 7.2

Answer:
h = 8.2

Explanation:
Given equation is h – 1 = 7.2
Add 1 to each side
h – 1 + 1 = 7.2 + 1
h = 8.2

Question 3.
5 = -2n

Answer:
n = -5/2

Explanation:
Given equation is 5 = -2n
Divide each side by -2
5/-2 = -2n/-2
-5/2 = n

Question 4.
-3.3m = -1.1

Answer:
m = 1/3

Explanation:
Given equation is -3.3m = -1.1
Divide each side by -3.3
-3.3m/-3.3 = -1.1/-3.3
m = 1/3

Write the decimal as a fraction or mixed number in simplest form.

Question 5.
-0.2

Answer:
-0.2 = -1/5

Explanation:
Given decimal value is -0.2
Fraction form of the decimal = -2/10 = -1/5

Question 6.
3.82

Answer:
3.82 = 191/50

Explanation:
Given decimal value is 3.82
Fraction form of the decimal = 382/100 = 191/50

Question 7.
-0.454

Answer:
-0.454 = -217/500

Explanation:
Given decimal value is -0.454
Fraction form of the decimal = -454/1000 = -217/500

Question 8.
-0.125

Answer:
-0.125 = -1/8

Explanation:
Given decimal value is -0.125
Fraction form of the decimal = -125/1000 = -1/8

Concepts, Skills, &Problem Solving
FINDING ANGLE MEASURES
Find each angle measure in the figure. Use equations to justify your answers. (See Exploration 1, p. 11.)

Question 9.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 40

Answer:
Angles of a triangle are 20°, 24°, 136°.

Explanation:
The sum of angles in triangle = 180°
4y + 20 + (y – 10)= 180°
4y + 20 + y – 10= 180
5y + 10 = 180
5y = 180 – 10
5y = 170
y = 170/5
y = 34°
So, ranges are 4 x 34 = 136°, (34 – 10) = 24°, 20

Question 10.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 41

Answer:
The angles of the triangle are 37.5°, 75°, 67.5°.

Explanation:
The sum of angles in triangle = 180°
x + 2x + (x + 30) = 180
4x + 30 = 180
4x = 180 – 30
4x = 150
x = 150/4
x = 37.5°
The measure of each angle is 37.5°, 2 x 37.5 = 75°, (37.5 + 30) = 67.5°

SOLVING AN EQUATION
Solve the equation. Check your solution.

Question 11.
10x + 2 = 32

Answer:
x = 3

Explanation:
Given equation is 10x + 2 = 32
10x = 32 – 2
10x = 30
x = 30/10
x = 3
Put x = 3 in 10x + 2 = 32
10(3) + 2 = 30 + 2 = 32

Question 12.
19 – 4c = 17

Answer:
c = 1/2

Explanation:
Given equation is 19 – 4c = 17
Add 4c to each side
19 – 4c + 4c = 17 + 4c
19 = 17 + 4c
Subtract 17 from each side
19 – 17 = 17 + 4c – 17
2 = 4c
c = 2/4
c = 1/2
Put c = 1/2 in 19 – 4c = 17
19 – 4(1/2) = 19 – 2 = 17

Question 13.
5x + 2x + 4 = 18

Answer:
x = 2

Explanation:
Given equation is 5x + 2x + 4 = 18
7x + 4 = 18
7x = 18 – 4
7x = 14
x = 14/7
x = 2
Put x = 2 in 5x + 2x + 4 = 18
5(2) + 2(2) + 4 = 10 + 4 + 4 = 18

Question 14.
2 = -9n + 22 – n

Answer:
n = 2

Explanation:
Given equation is 2 = -9n + 22 – n
2 = -10n + 22
2 – 22 = -10n
-20 = -10n
n = -20/-10
n = 2
Put n = 2 in 2 = -9n + 22 – n
2 = -9(2) + 22 – 2
= -18 + 22 – 2 = -20 + 22

Question 15.
1.1x + 1.2x – 5.4 = -10

Answer:
x = -2

Explanation:
Given equation is 1.1x + 1.2x – 5.4 = -10
2.3x – 5.4 = -10
2.3x = -10 + 5.4
2.3x = -4.6
x = -4.6/2.3
x = -2
Put x = -2 in 1.1x + 1.2x – 5.4 = -10
1.1(-2) + 1.2(-2) – 5.4 = -2.2 – 2.4 – 5.4 = -10

Question 16.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 42

Answer:
h = -9

Explanation:
Given equation is (2/3)h – (1/3)h + 11 = 8
(2 – 1)/3 h + 11 = 8
(1/3)h + 11 = 8
(1/3)h = 8 – 11
(1/3)h = -3
h = -3 x 3
h = -9
Put h = -9 in (2/3)h – (1/3)h + 11 = 8
(2/3)(-9) – (1/3)(-9) + 11 = -6 + 3 + 11
= 14 – 6 = 8

Question 17.
6(5 – 8v) + 12 = -54

Answer:
v = 2

Explanation:
Given equation is 6(5 – 8v) + 12 = -54
30 – 48v + 12 = -54
42 – 48v = -54
42 + 54 = 48v
96 = 48v
v = 96/48
v = 2
Put v = 2 in 6(5 – 8v) + 12 = -54
6(5 – 8(2)) + 12 = 6(5 – 16) + 12
= 6(-11) + 12 = -66 + 12 = -54

Question 18.
21(2 – x) + 12x = 44

Answer:
x = -2/9

Explanation:
Given equation is 21(2 – x) + 12x = 44
42 – 21x + 12x = 44
42 – 9x = 44
42 – 44 = 9x
-2 = 9x
x = -2/9
Put x = -2/9 in 21(2 – x) + 12x = 44
21(2 – (-2/9)) + 12(-2/9) = 21(2 + 2/9) – 24/9
= 42 + 42/9 – 24/9
= 42 + (42 – 24)/9
= 42 + 18/9 = 42 + 2 = 44

Question 19.
8.5 = 6.5(2d – 3) + d

Answer:
d = 2

Explanation:
Given equation is 8.5 = 6.5(2d – 3) + d
8.5 = 13d – 19.5 + d
8.5 = 14d – 19.5
8.5 + 19.5 = 14d
28 = 14d
d = 28/14
d = 2
Put d = 2 in 8.5 = 6.5(2d – 3) + d
8.5 = 6.5(2(2) – 3) + 2
= 6.5(4 – 3) + 2
= 6.5 x 1 + 2 = 6.5 + 2

Question 20.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 43

Answer:
x = -6

Explanation:
Given equation is -1/4 (x + 2) + 5 = -x
-x/4 – 2.(1/4) + 5 = -x
-x/4 – 1/2 + 5 = -x
-1/2 + 5 = -x + x/4
(-1 + 10)/2 = (-4x + x)/4
9/2 = -3x/4
9/2 x (-4/3) = x
x = -6
Put x = -6 in -1/4 (x + 2) + 5 = -x
-1/4 (-6 + 2) + 5 = -(-6)
-1/4(-4) + 5 = 6
1 + 5 = 6

YOU BE THE TEACHER
Your friend solves the equation. Is your friend correct? Explain your reasoning.

Question 21.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 44

Answer:
Wrong

Explanation:
-2(7 – y) + 4 = -4
-14 + 2y + 4 = -4
-10 + 2y = -4
2y = -4 + 10
2y = 6
y = 3

Question 22.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 45

Answer:
Correct

Explanation:
3(y – 1) + 8 = 11
3y – 3 + 8 = 11
3y + 5 = 11
3y = 11 – 5
3y = 6
y = 2

Question 23.
STRUCTURE
The cost C (in dollars) of making watches is represented C = 15n + 85. How many watches are made when the cost is $385?

Answer:
20 watches can be manufactured.

Explanation:
C = 15n + 85
385 = 15n + 85
15n = 385 – 85
15n = 300
n = 300/15
n = 20
So, 20 watches can be manufactured.

Question 24.
MODELING REAL LIFE
The height of the house is 26 feet. What is the height x of each story?
Big Ideas Math Answers 8th Grade Chapter 1 Equations 48

Answer:
Height of each story is 10 ft.

Explanation:
The height of the house is 26 feet
x + x + 6 = 26
2x + 6 = 26
2x = 26 – 6
2x = 20
x = 20/2
x = 10 ft
Height of each story is 10 ft.

Question 25.
MODELING REAL LIFE
After the addition of an acid, a solution has a volume of 90 milliliters. The volume of the solution is 3 milliliters greater than 3 times the volume of the solution before the acid was added. What was the original volume of the solution?

Answer:
The original volume of the solution is 29 milliliters

Explanation:
Volume of the solution = 90 milliliters
Let us take x as the original volume of the solution
3x + 3 = 90
3x= 90 – 3
3x = 87
x = 87/3
x = 29
So, the original volume of the solution is 29 milliliters

Question 26.
PROBLEM SOLVING
A grocer prepares free samples of a salad to give out during the day. By lunchtime, the grocer has given out 5 fewer than half the total number of samples. How many samples did the grocer prepare if she gives out 50 samples before lunch?

Answer:
The grocer prepares 110 samples

Explanation:
Let x be the total number of samples the grocer prepares
Total number of samples a grocer given out = 50
(1/2) x – 5 = 50
(1/2)x = 50 + 5
1/2 x = 55
x = 55 x 2
x = 110
So, the grocer prepares 110 samples.

Question 27.
GEOMETRY
What is the length of the missing base of the trapezoid?
Big Ideas Math Answers 8th Grade Chapter 1 Equations 49

Answer:
The missing base is 6 in.

Explanation:
The trapezoid area formula is (a + b)/2 x height
Given trapezoid area = 21 sq. in
(8 + x)/2 x 3 = 21
(8 + x)/2 = 21/3
(8 + x)/2 = 7
(8 + x) = 7 x 2
8 + x = 14
x = 14 – 8
x = 6
The missing base is 6 in.

Question 28.
MODELING REAL LIFE
You order two servings of pancakes and a fruit cup. The cost of the fruit cup is $1.50. You leave a 15% tip. Your total bill is $11.50. How much does one serving of pancakes cost?

Answer:
The cost of one serving of pancake is $4.25

Explanation:
The total cost excluding the tip would be a sum of 1.5 for a fruit cup and twice of price of each pancake servings(b),
hence the price should be 1.5 + 2b.
The customer pays 15% tip that means 0.15 times the bill.
1.15(1.5 + 2b) = 11.5
1.5 + 2b = 11.5/1.15
1.5 + 2b = 10
2b = 10 – 1.5
2b = 8.5
b = 8.5/2
b = 4.25
The cost of one serving of pancake is $4.25

Question 29.
PROBLEM SOLVING
How many people must attend the third show so that the average attendance per show is 3000?
Big Ideas Math Answers 8th Grade Chapter 1 Equations 50

Answer:
3500 people must attend the third show so that the average attendance per show is 3000.

Explanation:
The average attendance = 3000 per show
(2580 + 2920 + x)/3 = 3000
5500 + x = 3000 x 3
5500 + x = 9000
x = 9000 – 5500
x = 3500
3500 people must attend the third show so that the average attendance per show is 3000.

Question 30.
DIG DEEPER!
Divers in a competition are scored by an international panel of judges. The highest and the lowest scores are dropped. The total of the remaining scores is multiplied by the degree of difficulty of the dive. This product is multiplied by 0.6 to determine the final score.
a. A diver’sfinal score is 77.7. What is the degree of difficulty of the dive?
Big Ideas Math Answers 8th Grade Chapter 1 Equations 51
Big Ideas Math Answers 8th Grade Chapter 1 Equations 52

b. CRITICAL THINKING
The degree of difficulty of a dive is 4.0. The diver’s final score is 97.2. Judges award half or whole points from 0 to 10. What scores could the judges have given the diver?

Answer:
a. The degree of difficulty of dive is 3.5
b. The scores are 7.5, 8.0, 8.0, 8.0, 8.0, 8.5, 9.0

Explanation:
a. Let x be the degree of difficulty
By eliminating the scores6.5 and 8.5, we ge
(7 + 7 + 7.5 + 7.5 + 8). x . 0.6 = 77.7
37 . 0.6x = 77.7
22.2x = 77.7
x = 77.7/22.2
x = 7/2
x = 3.5
The degree of difficulty of dive is 3.5
b. Let s be the sum of the scores given by 5 judges
s x 4 x 0.6 = 97.2
2.4s = 97.2
s = 97.2/2.4
s = 40.5
40.5/5 = 8.1
So, the average score is 8.1
The scores are 7.5, 8.0, 8.0, 8.0, 8.0, 8.5, 9.0

Lesson 1.3 Solving Equations with Variables on Both Sides

EXPLORATION 1
Finding Missing Measures in Figures
Work with a partner.
a. If possible, find the value of x so that the value of the perimeter (in feet) is equal to the value of the area (in square feet) for each figure. Use an equation to justify your answer.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 53
b. If possible, find the value of y so that the value of the surface area (in square inches) is equal to the value of the volume (in cubic inches) for each figure. Use an equation to justify your answer.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 54
c. How are the equations you used in parts (a) and (b) different from equations used in previous sections? Explain how to solve this type of equation.

1.3 Lesson

Try It

Solve the equation. Check your solution.

Question 1.
-3x = 2x + 20

Answer:
x = -4

Explanation:
Given equation is -3x = 2x + 20
-3x – 2x = 20
-5x = 20
x = -20/5
x = -4
Put x = -4 in -3x = 2x + 20
-3(-4) = 2(-4) + 20
12 = -8 + 20

Question 2.
2.5y + 6 = 4.5y – 1

Answer:
y = 3.5

Explanation:
Given equation is 2.5y + 6 = 4.5y – 1
6 + 1 = 4.5y – 2.5y
7 = 2y
y = 7/2
y = 3.5
Put y = 3.5 in 2.5y + 6 = 4.5y – 1
2.5(3.5) + 6 = 4.5(3.5) – 1
8.75 + 6 = 15.75 – 1
14.75 = 14.75

Try It

Solve the equation. Check your solution.

Question 3.
6(4 – z) =2z

Answer:
z = 3

Explanation:
Given equation is 6(4 – z) =2z
24 – 6z = 2z
24 = 2z + 6z
8z = 24
z = 24/8
z = 3
Put z = 3 in 6(4 – z) =2z
6(4 – 3) = 2(3)
6(1) = 6

Question 4.
5(w – 2) = -2(1.5w + 5)

Answer:
w = 0

Explanation:
Given equation is 5(w – 2) = -2(1.5w + 5)
5w – 10 = -3w – 10
5w + 3w = -10 + 10
8w = 0
w = 0
Put w = 0 in 5(w – 2) = -2(1.5w + 5)
5(0 – 2) = -2(1.5(0) + 5)
5(-2) = -2(5)
= -10 = -10

Try It

Solve the equation.

Question 5.
2x + 1 = 2x – 1

Answer:
The equation 2x + 1 = 2x – 1 is never true. So, it does not have a soution

Explanation:
Given equation is 2x + 1 = 2x – 1
2x + 1 = 2x – 1
2x – 2x = -1 – 1
0 = -2

Question 6.
6(5 – 2v) = -4(3v + 1)

Answer:
The equation 6(5 – 2v) = -4(3v + 1) is never true. So, it does not have a soution

Explanation:
Given equation is 6(5 – 2v) = -4(3v + 1)
30 – 12v = -12v – 4
30 – 12v + 12v = -4
30 – 0 = -4
30 = -4
The equation 6(5 – 2v) = -4(3v + 1) is never true. So, it does not have a soution

Try It

Solve the equation

Question 7.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 56

Answer:
The equation 1/2 (6t – 4) = 3t – 2 is always true. So, it has infinitely many solutions.

Explanation:
Given equation is 1/2 (6t – 4) = 3t – 2
3t – 2 = 3t – 2
The equation 1/2 (6t – 4) = 3t – 2 is always true. So, it has infinitely many solutions.

Question 8.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 57

Answer:
The equation 1/3 (2b + 9) = 2/3 (b + 9/2) is always true. So, it has infinitely many solutions.

Explanation:
Given equation is 1/3 (2b + 9) = 2/3 (b + 9/2)
2b + 9 = 2/3 (b + 9/2) x 3
2b + 9 = 2(b + 9/2)
2b + 9 = 2b + 9
The equation 1/3 (2b + 9) = 2/3 (b + 9/2)is always true. So, it has infinitely many solutions.

Try It

Question 9.
WHAT IF?
The diameter of the purple circle is 3x. What is the area of each circle?

Answer:
The areas of circles are 25π, 36π.

Explanation:
The diameter of the purple circle is 3x
Diameter = double the radius
3x = 2r
r = 3x/2
Two circles are identical
x + 2 = 3x/2
3x/2 – x = 2
x/2 = 2
x = 4
The radius of green circle is (1 + 4) = 5
The radius of purple circle is 3(4)/2 = 6
The area of each circle = πr²
= π (5)² = 25π
The area of second circle = π (6)² = 36π

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
OPEN-ENDED
Write an equation with variables on both sides that has a single solution of -1. Explain how to solve your equation.

Answer:
5x + 6= 6x + 7

Explanation:
Write an equation with variables on both sides that has a single solution of -1
5x + 6= 6x + 7
Subtract 5x from each side
5x + 6 – 5x = 6x + 7 – 5x
6 = x + 7
Subtract 7 from each side
6 – 7 = x+ 7 – 7
-1 = x

STRUCTURE
Without solving, determine whether the equation has one solution, no solution, or infinitely many solutions. Justify your answer.

Question 11.
3(x – 1) = -3

Answer:
The equation has only one solution

Explanation:
Given equation is 3(x – 1) = -3
(x – 1) = -1
x = -1 + 1
x = 0

Question 12.
6x + 6 = 6(x + 1)

Answer:
The equation has infinitely many solutions

Explanation:
Given equation is 6x + 6 = 6(x + 1)
The equation has infinitely many solutions
6x + 6 = 6x + 6

Question 13.
z + 1 = z + 6

Answer:
The equation has no solution

Explanation:
Given equation is z + 1 = z + 6
The equation has no solution
1 = 6

SOLVING AN EQUATION
Solve the equation. Check your solution, if possible.

Question 14.
-7x = x + 24

Answer:
x = -3

Explanation:
Given equation is -7x = x + 24
-7x – x = 24
-8x = 24
x = -24/8
x = -3
Put x = -3 in -7x = x + 24
-7(-3) = -3 + 24
= 21

Question 15.
8(3 – z) = 4z

Answer:
z = 2

Explanation:
Given equation is 8(3 – z) = 4z
24 – 8z = 4z
24 = 4z + 8z
24 =12z
z = 24/2
z = 2
Put z = 2 in 8(3 – z) = 4z
8(3 – 2) = 4(2)
8(1) = 8

Question 16.
2(t – 3) = 2t – 6

Answer:
The equation has infinitely many solutions.

Explanation:
Given equation is 2(t – 3) = 2t – 6
2t – 6 = 2t – 6
The equation has infinitely many solutions.

Question 17.
WRITING AND SOLVING AN EQUATION
The squares are identical. What is the area of each square?
Big Ideas Math Answers 8th Grade Chapter 1 Equations 58

Answer:
The areas of squares are 256 sq units, 256 sq units.

Explanation:
What is the area of each square?
So, 3x + 7 = 5x + 1
7 – 1 = 5x – 3x
6 = 2x
x = 6/2
x = 3
The side of first square = 3(3) + 7 = 9 + 7
= 16
First square area = side²
= 16² = 256 sq units
Second square side length = 5(3) + 1 = 15 + 1 = 16
Second sqaure area = 16² = 256 sq units

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.
Your cousin renews his apartment lease and pays a new monthly rent. His new rent is calculated by applying a discount of $50 to his original rent and then applying a 10% increase to the discounted amount. What was your cousin’s original monthly rent when his new rent is 5% greater?

Answer:
Cousin’s original monthly rent is $1100.

Explanation:
Let us take rent as x
Then, discounted rent = x – 50
The increase applied is 10%
So, final rent is (x – 50) + 10% = (x – 50)1.1
x + 5% = 1.05x
1.1(x – 50) = 1.05x
1.1x – 1.05x = 55
0.05x = 55
x = 55/0.05
x = 1100
Therefore, cousin’s original monthly rent is $1100.

Question 19.
DIG DEEPER!
You and your friend race on a trail that is 10 miles long. In each situation, does your friend pass you before the end of the trail? Justify your answer.
a. You have a four-mile head start and jog at 6 miles per hour. Your friend bikes at 8 miles per hour.
b. You have a five-mile head start and run at 7 miles per hour. Your friend bikes at 17 miles per hour.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 59

Answer:
a) Your friend will pass you after the end of the trial.
b) Your friend will pass you before the end of the trail.

Explanation:
a) Let x be the time after which your friend passes you
In x hours you jog the distance 6x, your friend bikes 8x
8x = 4 + 6x
8x – 6x = 4
2x = 4
x = 4/2
x = 2
So in 2 hours, your friend bikes 2. 8 = 16 miles
Therefore, he will pass you after the end of the trial because 16 > 10.
b) In x hours you jog the distance 7x, while your friend bikes 17x
17x = 5 + 7x
17x – 7x = 5
10x = 5
x = 5/10
x = 0.5
In 0.5 hours, your friend bikes 0.5 x 17 = 8.5 miles
So he will pass you before the end of the trail because 8.5 < 10.

Solving Equations with Variables on Both Sides Homework & practice 1.3

Review & Refresh

Solve the equation. Check your solution.

Question 1.
-9z + 2 = 11

Answer:
z = -1

Explanation:
Given equation is -9z + 2 = 11
-9z + 2 – 2 = 11 – 2
-9z = 9
z = -9/9
z = -1
Put z = -1 in -9z + 2 = 11
-9(-1) + 2 = 9 + 2 = 11

Question 2.
-3n – 4n – 17 = 25

Answer:
n = -6

Explanation:
Given equation is -3n – 4n – 17 = 25
-7n – 17 = 25
-7n = 25 + 17
-7n = 42
n = -42/7
n = -6
Put n = -6 in -3n – 4n – 17 = 25
-3(-6) – 4(-6) – 17 = 18 + 24 – 17
= 42 – 17 = 25

Question 3.
-2(x + 3) + 5x = -39

Answer:
x = -11

Explanation:
Given equation is -2(x + 3) + 5x = -39
-2x – 6 + 5x = -39
3x – 6 = -39
3x = -39 + 6
3x = -33
x = -33/3
x = -11
Put x = -11 in -2(x + 3) + 5x = -39
-2(-11 + 3) + 5(-11) = -2(-8) – 55
= 16 – 55 = 3-9

Question 4.
-15 + 7.5(2d – 1) = 7.5

Answer:
d = 2

Explanation:
Given equation is -15 + 7.5(2d – 1) = 7.5
-15 + 15d – 7.5 = 7.5
-22.5 + 15d = 7.5
15d = 7.5 + 22.5
15d = 30
d = 30/15
d = 2
Put d = 2 in -15 + 7.5(2d – 1) = 7.5
-15 + 7.5(2(2) – 1) = -15 + 7.5(4 – 1)
= -15 + 7.5(3) = -15 + 22.5 = 7.5

Find the volume of the solid.

Question 5.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 60

Answer:
Volume = 27 cubic cm

Explanation:
Length = 3 cm, width = 2 cm height = 4.5 cm
Solid volume formula is length x width x height
Volume V = 3 x 2 x 4.5
= 27 cubic cm

Question 6.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 61

Answer:
Volume is 15.75 cubic cm.

Explanation:
Given that,
b = 4.5, h = 2, l = 3.5
The volume of Triangular Prism = ½ × b × h × l
V = ½ × 4.5 × 2 × 3.5
= 4.5 x 1 x 3.5
= 15.75 cubic cm

Question 7.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 62

Answer:
Volume is 24 cubic in

Explanation:
Given that,
Area of pentagon = 18 in²
Height h = 4 in
The volume of the pentagonal pyramid = 1/3 x base area x height
= 1/3 x 18 x 4
= 6 x 4 = 24 cubic in

Concepts, Skills, &Problem Solving
FINDING MISSING MEASURES IN FIGURES
If possible, find the value of so that the value of the surface area (in square inches) is equal to the value of the volume (in cubic inches). Use an equation to justify your answer. (See Exploration 1, p. 17.)

Question 8.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 63

Answer:
2(14x + 33) = 33x, x = 66/5

Explanation:
Given that,
length l =x in, width w = 11 in, height h = 3 in
Surface area of rectangular prism = Volume of the rectangular prism
2(wl + hl + hw) = whl
2(11x + 3x + 33) = 11 . 3. x
2(14x + 33) = 33x
28x + 66 = 33x
66 = 33x – 28x
66 = 5x
x = 66/5

Question 9.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 64

Answer:
2(36 + 13x) = 36x, x = 7.2

Explanation:
Given that,
length l = 9 in, width w = 4 in, height h = x in
The surface area of rectangular prism = Volume of the rectangular prism
2(wl + hl + hw) = whl
2(36 + 9x + 4x) = 9 . 4 . x
2(36 + 13x) = 36x
72 + 26x = 36x
72 = 36x – 26x
10x = 72
x = 72/10
x = 7.2

Question 10.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 65

Answer:
2(9x + 18) = 18x, It does not have a solution

Explanation:
Given that,
length l = 6 in, width w = 3 in, height h = x in
The surface area of rectangular prism = Volume of the rectangular prism
2(wl + hl + hw) = whl
2(3x + 6x + 18) = 6 . 3 . x
2(9x + 18) = 18x
18x + 18 = 18x
18x – 18x = 18
0 = 18

SOLVING AN EQUATION
Solve the equation. Check your solution.

Question 11.
m – 4 = 2m

Answer:
m = -4

Explanation:
Given equation is m – 4 = 2m
2m – m = -4
m = -4
Put m = -4 in m – 4 = 2m
-4 – 4 = 2(-4)
-8 = -8

Question 12.
3k – 1 = 7k + 2

Answer:
k = -3/4

Explanation:
Given equation is 3k – 1 = 7k + 2
-1 – 2 = 7k – 3k
-3 = 4k
k = -3/4
Put k = -3/4 in 3k – 1 = 7k + 2
3(-3/4) – 1 = 7(-3/4) + 2
-9/4 – 1 = -21/4 + 2
(-9 – 4)/4 = (-21 + 8)/4
-13/4 = -13/4

Question 13.
6x = 5x + 22

Answer:
x = 22

Explanation:
Given equation is 6x = 5x + 22
6x – 5x = 22
x = 22
Put x = 22 in 6x = 5x + 22
6(22) = 5(22) + 22
132 = 110 + 22

Question 14.
-24 – 8p = 4p

Answer:
p = -2

Explanation:
Given equation is -24 – 8p = 4p
-24 = 4p + 8p
-24 = 12p
p = -24/12
p = -2
Put p = -2 in -24 – 8p = 4p
-24 – 8(-2) = 4(-2)
-24 + 16 = -8

Question 15.
12(2w – 3) = 6w

Answer:
w = 2

Explanation:
Given equation is 12(2w – 3) = 6w
24w – 36 = 6w
24w – 6w = 36
18w = 36
w = 36/18
w = 2
Put w = 2 in 12(2w – 3) = 6w
12(2w – 3) = 6w
12(2(2)- 3) = 6(2)
12(4 – 3) = 12
12 = 12

Question 16.
2(n – 3) = 4n + 1

Answer:
n = -7/2

Explanation:
Given equation is 2(n – 3) = 4n + 1
2n – 6 = 4n + 1
-6 – 1 = 4n – 2n
-7 = 2n
n = -7/2
Put n = -7/2 in 2(n – 3) = 4n + 1
2(-7/2 – 3) = 4(-7/2) + 1
2(-7 – 6)/2 = 2(-7) + 1
-13 = -14 + 1

Question 17.
2(4z – 1) = 3(z + 2)

Answer:
z = 8/5

Explanation:
Given equation is 2(4z – 1) = 3(z + 2)
8z – 2 = 3z + 6
8z – 3z = 6 +2
5z = 8
z = 8/5
Put z = 8/5 in 2(4z – 1) = 3(z + 2)
2(4(8/5) – 1) = 3(8/5 + 2)
64/5 – 2 = 24/5 + 6
(64 – 10)/5 = (24 + 30)/5
54/5 = 54/5

Question 18.
0.1x = 0.2(x + 2)

Answer:

Explanation:
Given equation is 0.1x = 0.2(x + 2)
0.1x = 0.2x + 0.4
0.2x – 0.1x = -0.4
0.1x = -0.4
x = -0.4/0.1
x = -4
Put x = -4 in 0.1x = 0.2(x + 2)
0.1(-4) = 0.2(-4 + 2)
-0.4 = 0.2(-2)

Question 19.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 66

Answer:
d = 14

Explanation:
Given equation is (1/6)d + 2/3 = 1/4 (d – 2)
(1/6)d + 2/3 = d/4 – 2/4
(1/6)d + 2/3 = d/4 – 1/2
d/6 – d/4 = -1/2 – 2/3
(4d – 6d)/24 = (-3 – 4)/6
-2d/24 = -7/6
d = -7/6 x (-24/2)
d = 14
Put d = 14 in (1/6)d + 2/3 = 1/4 (d – 2)
(1/6)14 + 2/3 = 1/4 (14 – 2)
7/3 + 2/3 = 1/4 (12)
9/3 = 3

Question 20.
YOU BE THE TEACHER
Your friend solves the equation shown. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 67

Answer:
Wrong

Explanation:
3x – 4 = 2x + 1
3x – 2x = 1 + 4
x = 5

Question 21.
MODELING REAL LIFE
Write and solve an equation to find the number of miles you must drive to have the same cost for each of the car rentals.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 68

Answer:
The number of miles I should drive to have the same cost for each of the car rentals is 40 miles

Explanation:
Let x represent the number of miles
20 + 0.5x = 30 + 0.25x
20 – 30 = 0.25x – 0.5x
-10 = -0.25x
x = 10/0.25
x = 40
The number of miles I should drive to have the same cost for each of the car rentals is 40 miles

SOLVING AN EQUATION
Solve the equation. Check your solution, if possible.

Question 22.
x + 6 = x

Answer:
The equation has no solution

Explanation:
Given equation is x + 6 = x
x – x = -6
0 = -6
The equation has no solution

Question 23.
3x – 1 = 1 – 3x

Answer:
x = 1/3

Explanation:
Given equation is 3x – 1 = 1 – 3x
3x + 3x = 1 + 1
6x = 2
x = 2/6
x = 1/3
Put x = 1/3 in 3x – 1 = 1 – 3x
3(1/3) – 1 = 1 – 3(1/3)
1 – 1 = 1 – 1

Question 24.
3x + 15 = 3(x + 15)

Answer:
The equation has no solution

Explanation:
Given equation is 3x + 15 = 3(x + 15)
3x + 15 = 3x + 45
3x – 3x = 45 – 15
0 = 30
The equation has no solution

Question 25.
4x – 9 = 3.5x – 9

Answer:
x = 0

Explanation:
Given equation is 4x – 9 = 3.5x – 9
4x – 3.5x = -9 + 9
-0.5x = 0
x = 0
Put x = 0 in 4x – 9 = 3.5x – 9
4(0) – 9 = 3.5(0) – 9
-9 = -9

Question 26.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 69

Answer:
The equation has infinitely many solutions

Explanation:
Given equation is 1/3 (9x + 3) = 3x + 1
3x + 1 = 3x + 1
3x – 3x = 1 – 1
0 = 0
The equation has infinitely many solutions

Question 27.
5x – 7 = 4x – 1

Answer:
x = 6

Explanation:
Given equation is 5x – 7 = 4x – 1
5x – 4x = 7 – 1
x = 6
Put x = 6 in 5x – 7 = 4x – 1
5(6) – 7 = 4(6) – 1
30 – 7 = 24 – 1
23 = 23

Question 28.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 70

Answer:
The equation has no solution

Explanation:
Given equation is ½ x + ½ x = x + 1
x = x + 1
x – x = 1
0 = 1
The equation has no solution

Question 29.
2x + 4 = -(-7x + 6)

Answer:
x = 2

Explanation:
Given equation is 2x + 4 = -(-7x + 6)
2x + 4 = 7x – 6
4 + 6 = 7x – 2x
10 = 5x
x = 10/5
x = 2
Put x = 2 in 2x + 4 = -(-7x + 6)
2(2) + 4 = -(-7(2) + 6)
4 + 4 = -(-14 + 6)
8 = -(-8)

Question 30.
5.5 – x = -4.5 – x

Answer:
The equation has no solution

Explanation:
Given equation is 5.5 – x = -4.5 – x
5.5 + 4.5 = -x + x
10 = 0

Question 31.
-3(2x – 3) = -6x + 9

Answer:
The equation has infinitely many solutions

Explanation:
Given equation is -3(2x – 3) = -6x + 9
-6x + 9 = -6x + 9
-6x + 6x = 9 – 9
The equation has infinitely many solutions

Question 32.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 71

Answer:
The equation has no solution

Explanation:
Given equation is 10x – 8/3 – 4x= 6x
6x – 8/3 = 6x
6x – 6x = 8/3
0 = 8/3
The equation has no solution

Question 33.
6(7x + 7) = 7(6x + 6)

Answer:
The equation has infinitely many solutions

Explanation:
Given equation is 6(7x + 7) = 7(6x + 6)
42x + 42 = 42x + 42
The equation has infinitely many solutions

Question 34.
YOU BE THE TEACHER
Your friend solves the equation shown. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 72

Answer:
The equation has infinitely many solutions

Explanation:
-4(2n – 3) = 12 – 8n
-8n + 12 = 12 – 8n
-8n + 8n = 12 – 12
0 = 0
The equation has infinitely many solutions

Question 35.
OPEN-ENDED
Write an equation with variables on both sides that has no solution. Explain why it has no solution.

Answer:
5(2x – 3) = 10x – 15

Explanation:
The equation is 5(2x – 3) = 10x – 15
10x – 15 = 10x – 15
10x – 10x = -15 + 15
0 = 0
For any values of x, the equation satisies.
So, The equation has infinitely many solutions

Question 36.
MODELING REAL LIFE
A cable television provider charges $75 for installation and $39.96 per month for a basic entertainment package.A satellite television provider others free installation and charges $13.32 per month for service for each television. Your neighbor subscribes to the cable provider the same month you subscribe to the satellite provider. After how many months is your neighbor’s total cost the same as your total cost when you own three televisions?

Answer:
Never my neighbor’s total cost, my total cost will be same.

Explanation:
Let x as the number of months
My satellite cost for 3 televisions is 13.32x * 3 = 39.96x
My friends cable cost = 75 + 39.96x
75 + 39.96x = 39.96x
75 = 39.96x – 39.96x
75 = 0
75 is never equal to zero.
So, never my neighbor’s total cost, my total cost will be same.

Question 37.
MODELING REAL LIFE
A pizza parlor makes 52 pizza crusts the first week of summer and 180 pizza crusts each subsequent week. A diner makes 26 pizza crusts the first week of summer and 90 pizza crusts each subsequent week. In how many weeks will the total number of pizza crusts made by the pizza parlor be twice the total number of pizza crusts made by the diner?
Big Ideas Math Answers 8th Grade Chapter 1 Equations 73

Answer:
The pizza parlor always makes twice the pizza crust than the dinner.

Explanation:
Let x is the number of weeks
Pizza parlor = 52 + 180x
Dinner = 26 + 90x
The number of pizza crusts made by the pizza parlor be twice the total number of pizza crusts made by the diner
52 + 180x = 2(26 + 90x
52 + 180x = 52 + 180x
52 – 52 = 180x – 180x
0 = 0
Infinite solutions.
So, the pizza parlor always makes twice the pizza crust than the dinner.

Question 38.
PRECISION
Is the triangle an equilateral triangle? Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 74

Answer:
The given triangle is not equilateral triangle

Explanation:
Equilateral triangle have all sides of equal length
2x + 5.2 = 3x + 1.2
2x – 3x = 1.2 – 5.2
-x = -4
x = 4
3x + 1.2 = 2x + 6.2
3x – 2x = 6.2 – 1.2
x = 5
So, the given triangle is not equilateral triangle

GEOMETRY
Find the perimeter of the regular polygon.

Question 39.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 75

Answer:
Perimeter = 3

Explanation:
Perimeter is the sum of all sides of the polygon
As it is a regular polygon all sides are equal
5 – 2x = -4x + 9
5 – 9 = -4x + 2x
-4 = -2x
x = 4/2
x = 2
One side = 5 – 2(2) = 5 – 4 = 1
Perimeter = 1 + 1 + 1 = 3

Question 40.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 76

Answer:
Perimeter = 7.5

Explanation:
Perimeter is the sum of all sides of the polygon
As it is a regular polygon all sides are equal
3(x – 1) = 5x – 6
3x – 3 = 5x – 6
5x – 3x = -3 + 6
2x = 3
x = 3/2 = 1.5
One side = 5(1.5) – 6
= 7.5 – 6 = 1.5
Perimeter = 5(1.5) = 7.5

Question 41.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 77

Answer:
Perimeter = 344/3

Explanation:
Perimeter is the sum of all sides of the polygon
As it is a regular polygon all sides are equal
4/3 x – 1/3 = x + 7
4/3 x – x = 7 + 1/3
x = (21 + 1)/3
x = 22/3
One side = 22/3 + 7
= (22 + 21)/3 = 43/3
Perimeter = 8(43/3)
= 344/3

Question 42.
PRECISION
The cost of mailing a DVD in an envelope using Company B is equal to the cost of mailing a DVD in a box using Company A. What is the weight of the DVD with its packing material? Round your answer to the nearest hundredth.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 78

Answer:
The weight of the DVD with its packing material is around 0.19 lb.

Explanation:
Let x is the weight of the DVD
Cost of mailing a DVD in a box using company A is 2.50x + 2.25
Cost of mailing a DVD in a box using company B 8.50x + 1.10
$8.50x + $1.10 = $2.50x + $2.25
8.50x – 2.50x = 2.25 – 1.10
6x = 1.15
x = 1.15/6
x = 0.19
The weight of the DVD with its packing material is around 0.19 lb.

Question 43.
WRITING
Would you solve the equation 0.25x + 7 = \(\frac{1}{3}\)x – 8 using fractions or decimals ? Explain.

Answer:
x = 180

Explanation:
0.25x + 7 = 1/3 x – 8
0.25x – 1/3 x = -8 – 7
25/100x – 1/3 x = -15
1/4 x – 1/3 x = -15
(3 – 4)/12 x = -15
-1/12 x = -15
x = -15 x -12
x = 180

Question 44.
NUMBER SENSE
The weight of an object is equal to \(\frac{3}{4}\) of its own weight plus \(\frac{3}{4}\) of a pound. How much does the object weigh? Explain.

Answer:
The object weight is 3 lb.

Explanation:
Let us take x as the object weight
x = 3/4 x + 3/4
4x = 3x + 3
4x – 3x = 3
x = 3
The object weight is 3 lb.

Question 45.
STRUCTURE
Fill in the blanks in three different ways to create an equation that has one solution, no solution, and infinitely many solutions.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 80

Answer:
7x + 3x + 10 = -2(10x + 10) has one solution.
7x + 3x + 10 = -2(-5x – 5) has infinite solutions
7x + 3x + 10 = -2(-5x + 1) has no solution.

Explanation:
Given that,
7x + 3x + 10 = -2(_ x + _)
One solution
7x + 3x + 10 = -2(10x + 10)
10x + 10 = -20x – 20
10x + 20x = -20 – 10
30x = -30
x = -30/3
x = -1
Infinite solutions
7x + 3x + 10 = -2(-5x – 5)
10x + 10 = 10x + 10
10x – 10x = 10 – 10
0 = 0
No solution
7x + 3x + 10 = -2(-5x + 1)
10x + 10 = 10x – 2
10 = -2

Question 46.
MODELING REAL LIFE
The volume of red blood cells in a blood sample is equal to the total volume of the sample minus the volume of plasma. What is the total volume x of blood drawn?
Big Ideas Math Answers 8th Grade Chapter 1 Equations 81

Answer:
The volume of blood is 5.95 mL.

Explanation:
The volume of red blood cells in a blood sample is equal to the total volume of the sample minus the volume of plasma.
45% = sample volume – 5.5
0.45 = sample volume – 5.5
sample volume = 045 + 5.5
= 5.95

Question 47.
PROBLEM SOLVING
One serving of oatmeal provides 16% of the fiber you need daily. You must get the remaining 21 grams of fiber from other sources. How many grams of fiber should you consume daily? Justify your answer.

Answer:
The total amount of fiber you need daily is 25 grams.

Explanation:
Let x is the total amount of fiber you needed daily.
Since one serving of oatmeal provides 16% of the fiber you need daily, then 0.16x grams is the amount of fiber you get from oatmeal
If you get another 21 grams of fiber from other sources, then 0.16x + 21 grams is the total amount of fiber you need daily
x and 0.16x + 21 both represent the total amount of fiber you need daily
So 0.16x + 21 = x
21 = x – 0.16x
21 = 0.84x
x = 21/0.84
x = 25
The total amount of fiber you need daily is 25 grams.

Question 48.
DIG DEEPER!
The floor of a six-foot-wide hallway is painted as shown, using equal amounts of white and black paint.
a. How long is the hallway?
b. Can this same hallway be painted with the same pattern, but using twice as much black paint as white paint? Explain.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 82

Answer:
a. The total length of the hallway = 9x + 4
b. Yes

Explanation:
a. The total length of the hallway = The length of white paints + lngth of black paints
= 5(x) + 4(x + 1)
= 5x + 4x + 4 = 9x + 4

Question 49.
PRECISION
Consider the equation c = ax – bx, where a, b, and c are whole numbers. Which of the following result in values a, b and c so that the original equation has exactly one solution? Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 83

Answer:
a ≠  b, c = 0

Explanation:
c = ax – bx,
Substitute a ≠  b, c = 0 values in above equation
c = ax – bx
0 = x(a – b)

Lesson 1.4 Rewriting Equations and Formulas

EXPLORATION

Work with a partner.
a. Write a formula for the height h of each figure. Explain your method.

  • A parallelogram with area A and base b
  • A rectangular prism with volume V, length l, and width w
  • A triangle with area A and base b

b. Write a formula for the length l of each figure. Explain your method.

  • A rectangle with perimeter P and width w
  • A rectangular prism with surface area S, width w, and height h.

c. Use your formulas in parts (a) and (b) to find the missing dimension of each figure.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 84

1.4 Lesson

Try It

Solve the equation for y.

Question 1.
5y – x = 10

Answer:
y = (10 + x)/5

Explanation:
Given equation is 5y – x = 10
Add x to both sides
5y – x + x = 10 + x
5y = 10 + x
Divide both sides by 5
5y/5 = (10 + x)/5
y = (10 + x)/5

Question 2.
4x – 4y = 1

Answer:
y = (4x – 1)/4

Explanation:
Given equation is 4x – 4y = 1
Subtract 4x from both sides
4x – 4y – 4x  = 1 – 4x
-4y = 1 – 4x
Divide both sides by -4
-4y/-4 = (1-4x)/-4
y = (4x – 1)/4

Question 3.
12 = 6x + 3y

Answer:
y = (12 – 6x)/3

Explanation:
Given equation is 12 = 6x + 3y
Subtract 6x from each side
12 – 6x = 6x + 3y – 6x
12 – 6x = 3y
Divide each side by 3
(12 – 6x)/3 = 3y/3
y = (12 – 6x)/3

Try It

Solve the formula for the red variable.

Question 4.
Area of rectangle: A = bh

Answer:
b = A/h

Explanation:
Given that,
Area of the rectangle A = bh
Divide each side by h
A/h = bh/h
A/h = b
Area/height = base

Question 5.
Simple interest: I = Prt

Answer:
P = I/rt

Explanation:
Given that,
Simple interest: I = Prt
Divide each side by rt
I/rt = Prt/rt
P = I/rt

Question 6.
Surface area of cylinder: S = 2πr2 + 2πrh

Answer:
h = (S – 2πr²)/2πr

Explanation:
Given that,
Surface area of cylinder: S = 2πr² + 2πrh
Subtract 2πr² from each side
S – 2πr² = 2πr² + 2πrh – 2πr²
S – 2πr² = 2πrh
Divide each side by 2πr
(S – 2πr²)/2πr = h

Try It

Question 7.
Solve the formula F = \(\frac{9}{5}\)C + 32 for C. Justify your answer.

Answer:
C = 5(F – 32)/9

Explanation:
Given that,
F = (9/5)C + 32
Subtract 32 from each side
F – 32 = (9/5)C + 32 – 32
F – 32 = (9/5)C
Multiply each side by (5/9)
(F – 32) x (5/9) = (9/5)C x (5/9
5(F – 32)/9 = C

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
REWRITING A Formula
The formula for the circumference of a circle is C = 2πr. Solve the formula for r.

Answer:
r = C/2π

Explanation:
Given that
The formula for the circumference of a circle is C = 2πr
Divide each side by 2π
C/ 2π = 2πr/2π
C/2π = r

Question 9.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 85

Answer:
2y – 4x = 6

Explanation:
1. 4x = 6 + 2y
4x – 6 = 2y
(4x – 6)/2 = y
y = [2(2x – 3)]/2
y = (2x – 3)
2. 6 = 4x – 2y
6 – 4x = -2y
y = (4x – 6)/2
y = (2x – 3)
3. 2y – 4x = -6
2y = -6 + 4x
y = (4x – 6)/2
y = 2x – 3
4. 2y – 4x = 6
2y = 6 + 4x
y = 3 + 2x

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
Room temperature is considered to be 70°F. The temperature outside is currently 23°C. Is this greater than or less than room temperature?

Answer:
The outside temperature is greater than room temperature

Explanation:
Room temperature is considered to be 70°F. The temperature outside is currently 23°C
F = (9/5) C + 32
Put C = 23°
F = (9/5)23 + 32
F = 41.1+ 32
F = 73.4°
C = 23° is greater than F = 70°
So the outside temperature is greater than room temperature

Question 11.
DIG DEEPER!
A bird flies at a top speed of 20,000 meters per hour. The bird flies 30,000 meters without stopping.
a. For how many hours did the bird fly if it flew at top speed?
b. In part(a), did you rewrite a formula to find the number of hours the bird flew, or did you use another approach? Explain.

Answer:
a) The bird flew 1.5 hours at Top speed
b) Yes, I rewrite a formula to find the number of hours the bird flew.

Explanation:
Speed of the bird = 20000 meters per hour
Distance travelled = 30000 meters
a) Distance = speed x time
d = s x t
t = d/s
t = 30000/20000
t = 1.5
The bird flew 1.5 hour at Top speed
b) Yes, I rewrite a formula to find the number of hours the bird flew.

Question 12.
A ball pit is in the shape of a cylinder with a lateral surface area of 245 square feet. The diameter of the ball pit is 312 inches. What is the height of the ball pit? Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 86

Answer:
The height of the ball pit is 3 feet.

Explanation:
Diameter of the ball pit = 312 inches = 26 feet
Radius of the ball pit = 26/2 = 13 feet
Lateral surface area = 245 sq. ft
Surface area = 2πrh
245 = 2 x 3.14 x 13 x h
245 = 81.64h
h = 245/81.64
h = 3 feet
The height of the ball pit is 3 feet

Rewriting Equations and Formulas Homework & Practice 1.4

Solve the equation. Check your solution, if possible.

Question 1.
-2x = x + 15

Answer:
x = -5

Explanation:
Given equation is -2x = x + 15
-2x – x = 15
-3x = 15
x = -15/3
x = -5
Put x = -5 in -2x = x + 15
-2(-5) = -5 + 15
10 = 10

Question 2.
4(z – 3) = 2z

Answer:
z = 6

Explanation:
Given equation is 4(z – 3) = 2z
4z – 12 = 2z
4z – 2z = 12
2z = 12
z = 12/2
z = 6
Put z = 6 in 4(z – 3) = 2z
4(6 – 3) = 2(6)
4(3) = 12

Question 3.
x – 8 = x – 1

Answer:
The equation has no solution

Explanation:
Given equation is x – 8 = x – 1
x – x = -1 + 8
0 = 7
The equation has no solution

Question 4.
5(4 + t) = 5t + 20

Answer:
The equation has infinitely many solutions

Explanation:
Given equation is 5(4 + t) = 5t + 20
20 + 5t = 5t + 20
20 + 5t – 20 = 5t
5t = 5t
The equation has infinitely many solutions

Find the unit rate.

Question 5.
60 miles in 5 hours

Answer:
Speed = 12 miles per 1 hour

Explanation:
60 miles in 5 hours
Speed = 60 miles in 5 hours
= 60/5 = 12 miles per 1 hour

Question 6.
$8.50 : 5 ounces

Answer:
$8.50 : 5 ounces = 1.7 per ounce

Explanation:
$8.50 : 5 ounces
Unit rate = $8.50/5
= 1.7 per ounce

Question 7.
9 pounds per 6 crates

Answer:
9 pounds per 6 crates = 1.5

Explanation:
9 pounds per 6 crates
Unit rate = 9/6
= 1.5

Concepts, Skills, &Problem Solving
REWRITING FORMULA
Solve the formula for the height of the figure. Then use the new formula to find the height. (See Exploration 1, p. 25.)

Question 8.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 87

Answer:
Height = 6mm

Explanation:
A = ½bh
36 = ½ x 12 x h
36 = 6h
h = 36/6
h = 6 mm

Question 9.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 88

Answer:
Height = 6 inch

Explanation:
Volume V = Base area x height
h = V/B
h = 36/6
h = 6 inch

IDENTIFYING LITERAL EQUATIONS
Is the equation a literal equation? Explain.

Question 10.
y = 4

Answer:
y = 4 is not a liter equation.

Explanation:
y = 4 is not a liter equation. A literal equation is an equation that has two or more variables.

Question 11.
t + 8y = 7

Answer:
t + 8y = 7 is a literal equation.

Explanation:
t + 8y = 7 is a literal equation because it has two variables t, and y.

Question 12.
z = 4x + 9y

Answer:
z = 4x + 9y is a literal equation.

Explanation:
z = 4x + 9y is a literal equation because it has 3 variables x, y , and z.

REWRITING AN EQUATION
Solve the equation for y.

Question 13.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 89

Answer:
y = (12 – x)/3

Explanation:
Given equation is (1/3)x + y = 4
Subtract each side by (1/3)x
(1/3)x + y – (1/3)x = 4 – (1/3)x
y = 4 – (1/3)x
y = (12 – x)/3

Question 14.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 90

Answer:
y = (7 – 3x)/5

Explanation:
Given equation is 3x + (1/5)y = 7
Subtract 3x from each side
3x + (1/5)y – 3x = 7 – 3x
(1/5)y = 7 – 3x
Multiply each side by 5
(1/5)y x 5 = (7 – 3x)/5
y = (7 – 3x)/5

Question 15.
6 = 4x + 9y

Answer:
y = (6 – 4x)/9

Explanation:
Given equation is 6 = 4x + 9y
Subtract 4x from each side
6 – 4x = 4x + 9y – 4x
6 – 4x = 9y
Divide each side by 9
(6 – 4x)/9 = 9y/9
y = (6 – 4x)/9

Question 16.
π = 7x – 2y

Answer:
y = (7x – π)/2

Explanation:
Given equation is π = 7x – 2y
Subtract 7x from each side
π – 7x = 7x – 2y – 7x
π – 7x = -2y
Divide eachside by -2
(π – 7x)/-2 = -2y/-2
(7x – π)/2 = y

Question 17.
4.2x – 1.4y = 2.1

Answer:
y = 3(2x – 1)/2

Explanation:
Given equation is 4.2x – 1.4y = 2.1
Subtract 4.2x from each side
4.2x – 1.4y – 4.2x = 2.1 – 4.2x
-1.4y = 2.1 – 4.2x
Divide each side by -1.4y
-1.4y/-1.4 = (2.1 – 4.2x)/-1.4
y = (4.2x – 2.1)/1.4
= 2.1(2x – 1)/1.4
y = 3(2x – 1)/2

Question 18.
6y – 1.5x = 8

Answer:
y = (8 – 1.5x)/6

Explanation:
Given equation is 6y – 1.5x = 8
Add 1.5x to each side
6y – 1.5x – 1.5x = 8 – 1.5x
6y = 8 – 1.5x
Divide each side by 6
6y/6 = (8 – 1.5x)/6
y = (8 – 1.5x)/6

Question 19.
YOU BE THE TEACHER
Your friend rewrites the equation 2x – y = 5. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 91

Answer:
Wrong

Explanation:
2x – y = 5
Subtract 2x from each side
2x – y – 2x = 5 – 2x
-y = -2x + 5
y = 5 – 2x

REWRITING A FORMULA
Solve the formula for the red variable.

Question 20.
d = rt

Answer:
t = d/r

Explanation:
Given that,
d = rt
t = d/r

Question 21.
e = mc2

Answer:
m = e/c²

Explanation:
Given that,
e = mc²
e/c² = m

Question 22.
R – C = P

Answer:
C = R – P

Explanation:
Given that,
R – C = P
-C = P – R
C = R – P

Question 23.
P = a + b + c

Answer:
a = P – b – c

Explanation:
Given that,
P = a + b + c
P – b – c = a
a = P – b – c

Question 24.
B = 3\(\frac{V}{h}\)

Answer:
V = Bh/3

Explanation:
Given that,
B = 3(V/h)
Multiply each side by h
Bh = 3(V/h) x h
Bh = 3V
Divide each side by 3
Bh/3 = 3V/3
V = Bh/3

Question 25.
D = \(\frac{m}{V}\)

Answer:
V = m/D

Explanation:
Given that,
D = m/V
Multiply each side by V
DV = (m/V)V
DV = m
Divide each side by D
DV/D = m/D
V = m/D

Question 26.
MODELING REAL LIFE
The formula K = C + 273.15 converts temperatures from degrees Celsius to Kelvin K.
a. Convert 200 degrees Celsius to Kelvin.
b. Solve the formula for C.
c. Convert 300 Kelvin to degrees Celsius.

Answer:
a. K = 473.15
b. C = K – 273.15
c. C = 26.85°

Explanation:
Given formula is K = C + 273.15
a) Convert 200 degrees Celsius to Kelvin.
K = C + 273.15
Put C = 200
K = 200 + 273.15
K = 473.15
b) Solve the formula for C
K = C + 273.15
C = K – 273.15
c) Convert 300 Kelvin to degree Celsius
C = 300 – 273.15
C = 26.85°

Question 27.
PROBLEM SOLVING
The formula for simple interest is I = Prt.
a. Solve the formula for t, when r is the simple interest per year.
b. Use the new formula to find the value of in the table.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 91.1

Answer:
a. t = I/Pr
b. t =1/20

Explanation:
The formula for simple interest is I = Prt.
a. Solve the formula for t, when r is the simple interest per year.
I = Prt
Divide each side by Pr
I/Pr = Prt/Pr
t = I/Pr
b. Given that,
I = $75, P = $500, r = 5%, t = ?
t = I/Pr
Put above values in the formula
t = 75/(500 x 5)
= 1/20

Question 28.
GEOMETRY
Use the triangle shown.
a. Write a formula for the perimeter P of the triangle.
b. Solve the formula for b.
c. Use the new formula to find b when a is 10 feet and c is 17 feet.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 92

Answer:
a. P = a + b + c
b. b = P – a – c
c. b = 15 feet

Explanation:
a. Write a formula for the perimeter P of the triangle.
We know that the perimeter is nothing but the sum of the outer edges of the triangle.
Perimeter = a + b + c
P = a + b + c
b. Solve the formula for b.
P = a + b + c
Subtract a, c from each side
P – a – c = a + b + c – a – c
P – a – c = b
c. Given that,
P = 42 feet, a = 10 ft, c = 17 ft
b = P – a – c
b = 42 – 10 – 17
b = 42 – 27
b = 15 feet

Question 29.
REASONING
The formula Big Ideas Math Answers 8th Grade Chapter 1 Equations 93 converts temperatures from degrees Fahrenheit F to Kelvin K.
a. Solve the formula for F.
b. The freezing point of helium is 0.95 Kelvin. What is this temperature in degrees Fahrenheit?
c. The temperature of dry ice is -78.5°C. Which is colder, dry ice or liquid nitrogen?

Answer:
a. F = 9/5 (K – 273.15) + 32
b. F = = -454.96
c. liquid nitrogen is colder than dry ice

Explanation:
K = 5/9 (F – 32) + 273.15
a. Solve the formula for F.
K = 5/9 (F – 32) + 273.15
Subtract 237.15 from each side
K – 273.15 = 5/9 (F – 32)
Multiply each side by 9
9(K – 273.15) = 5(F – 32)
Divide each side by 5
9(K – 273.15)/5 = 5(F – 32)/5
9/5 (K – 273.15) = F – 32
Subtract 32 from each side
9/5 (K – 273.15) + 32 = F – 32 + 32
9/5 (K – 273.15) + 32 = F
b. The freezing point of helium is 0.95 Kelvin. What is this temperature in degrees Fahrenheit?
F = 9/5 (K – 273.15) + 32
Put K = 0.95
F = 9/5 (0.95  – 273.15) + 32
F = 9/5 (-272.2) + 35
= 9(-54.44) + 35
= -489.96 + 35
= -454.96
c. The temperature of dry ice is -78.5°C. Which is colder, dry ice or liquid nitrogen?
liquid nitrogen is colder than dry ice

Question 30.
MODELING REAL LIFE
In which city is the water temperature higher?
Big Ideas Math Answers 8th Grade Chapter 1 Equations 94

Answer:
The water temperature is higher in Portland.

Explanation:
By observing the image we can say that in Portland the water temperature is higher.

Question 31.
GEOMETRY
The volume of a square pyramid with a height of 30 feet is 360 cubic feet. What are the side lengths of the base? Justify your answer.

Answer:
The base side length is 6 feet.

Explanation:
Given that,
height h = feet
Volume V = 360 cubic feet
Volume V = Bh/3
3V = Bh
3V/h = B
B = (3 x 360)/30
B = 36
Side² = 36
Side = √(36)
Side = 6 feet
The base side length is 6 feet.

Question 32.
DIG DEEPER!
The Navy Pier Ferris Wheel in Chicago has a circumference that is 56% of the circumference of the first Ferris wheel built in 1893.
a. What is the radius of the Navy Pier Ferris Wheel?
b. What was the radius of the first Ferris wheel?
c. The first Ferris wheel took 9 minutes to make a complete revolution. How fast was the wheel moving?
Big Ideas Math Answers 8th Grade Chapter 1 Equations 95

Answer:
a. The radius of the Navy Pier Ferris Wheel is 70 feet
b. The radius of the first Ferris wheel is 125 feet
c. The wheel is moving with 87.2 ft/min

Explanation:
a. Circle circumference = 2πr
439 = 2πr
r = 439/2π
r = 439/(2 x 3.14) = 439/6.28
r = 70 feet
The radius of the Navy Pier Ferris Wheel is 70 feet
b. New circumference = 56% of old circumference
439.6 = 0.56 x C
C = 439.6/0.56
C = 785
Old circumference = 2πr(old)
785 = 2πr(old)
r(old) = 785/2π
= 125 feet
The radius of the first Ferris wheel is 125 feet
c. Old circumference/ time
Cold / Time = 785/9
= 87.2 ft/min
The wheel is moving with 87.2 ft/min

Equations Connecting Concepts

Connecting Concepts

Problem-Solving Strategies
Using an appropriate strategy will help you make sense of problems as you study the mathematics in this course. You can use the following strategies to solve problems that you encounter.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 96

Using the Problem-Solving Plan

Question 1.
The battery life of a one-year-old cell phone is 75% of its original battery life. When the battery is charged to 50% of its capacity, it dies after 4\(\frac{1}{2}\) hours. Find the original battery life of the phone. Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 97

Understand the problem.
You know how long a cell phone battery lasts when it is charged to 50% of its capacity. You also know that the battery life of the phone is 75% of its original battery life. You are asked to find the original battery life of the phone.
Make a plan.
First, find the battery life of the one-year-old cell phone. Then use this information to write and solve an equation for the original battery life of the phone.
Solve and check.
Use the plan to solve the problem. Then check your solution.

Answer:
The original battery life of the phone is 12 hours.

Explanation:
Let x be the battery life of the one-year-old cell phone.
(100% – 50%)x = 4.5
(1 – 0.5)x = 4.5
0.5x = 4.5
x = 4.5/0.5
x = 9 hours
Let take y as the original battery life
One year old battery life = 75% of original battery life
9 = 0.75y
y = 9/0.75
y = 12 hours
75% of original battery life is 75% x 12 = 0.75 x 12 = 9
50% of the battery life of the one year old cell phone is 0.50 x 9 = 4.5
The original battery life of the phone is 12 hours.

Question 2.
The triangular prism shown has a volume of 132 cubic centimeters. Find the height of the prism. Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 98

Answer:
Height h = 5.5 cm

Explanation:
The triangular prism shown has a volume of 132 cubic centimeters.
a = 6 cm, b = 8 cm, h = ?
V = 0.5 x b x a x h
132 = 0.5 x 8 x 6 x h
132 = 24 x h
h = 132/24
h = 5.5
Height h = 5.5 cm

Performance Task

Target Heart Rates
At the beginning of this chapter, you watched a STEAM Video called “Training for a Half Marathon.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 99

Equations Chapter Review

Review Vocabulary

Write the definition and give an example of the vocabulary term.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 100

Graphic Organizers

You can use an Information Frame to help organize and remember a concept. Here is an example of an Information Frame for solving equations with variables on both sides.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 101
Choose and complete a graphic organizer to help you study the concept.

  1. solving simple equations using addition
  2. solving simple equations using subtraction
  3. solving simple equations using multiplication
  4. solving simple equations using division
  5. inverse operations
  6. literal equation

Big Ideas Math Answers 8th Grade Chapter 1 Equations 102

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 103

1.1 Solving Simple Equations (pp. 3–10)

Solve the equation. Check your solution.

Question 1.
y + 8 = -11

Answer:
y = -19

Explanation:
Given equation is y + 8 = -11
y + 8 – 8 = -11 – 8
y = -19
Put y = -19 in y + 8 = -11
-19 + 8 = -11

Question 2.
3.2 = -0.4n

Answer:
n = -8

Explanation:
Given equation is 3.2 = -0.4n
3.2/-0.4 = -0.4n/-0.4
-8 = n
Put n = -8 in 3.2 = -0.4n
3.2 = -0.4(-8)

Question 3.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 104

Answer:
t = 12π

Explanation:
Given equation is -t/4 = -3π
-t = -3π x 4
-t = -12π
t = 12π
Put t = 12π in -t/4 = -3π
-12π/4 = -3π

Question 4.
v – | 2.4 | = 5.7

Answer:
v = 8.1

Explanation:
Given equation is v – | 2.4 | = 5.7
v – 2.4 = 5.7
v = 5.7 + 2.4
v = 8.1
put v = 8.1 in v – | 2.4 | = 5.7
8.1 – | 2.4 | = 5.7
8.1 – 2.4 = 5.7

Question 5.
-6 = -2 + w

Answer:
w = -4

Explanation:
Given equation is -6 = -2 + w
-6 + 2 = w
-4 = w
Put w = -4 in -6 = -2 + w
-6 = -2 – 4

Question 6.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 105

Answer:
x = -1/4

Explanation:
Given equation is x – 2/3 = -11/12
x = -11/12 + 2/3
x = (-11 + 8)/12
x = -3/12
x = -1/4
Put x = -1/4 in x – 2/3 = -11/12
-1/4 – 2/3 = -11/12
(-3 – 8)/12 = -11/12

Question 7.
The boiling point of a liquid is the temperature at which the liquid becomes a gas. The boiling point of mercury is about \(\frac{41}{200}\) of the boiling point of lead. Write and solve an equation to find the boiling point of lead.
Big Ideas Math Answers 8th Grade Chapter 1 Equations 106

Answer:
41/200 x = 357
The boiling point of lead is 1741.46°C

Explanation:
The boiling point of mercury = 357°C
The boiling point of lead is x
41/200 x = 357
x = 357 x (200/41)
x = 1741.46
The boiling point of lead is 1741.46°C

Question 8.
Write an equation that you can use the Addition Property of Equality to solve.

Answer:
x – 6 = 8

Explanation:
x – 6 = 8
Add 6 to each side
x – 6 + 6 = 8 + 6
x = 14

Question 9.
To solve \(\frac{2}{5}\)x = 14, you multiply both sides of the equation by \(\frac{5}{2}\). Your friend divides both sides of the equation by \(\frac{2}{5}\). Who is correct? Explain.

Answer:
Multiply both sides of the equation by 5/2

Explanation:
2/5 x = 14
Multiply both sides of the equation by 5/2
2/5 x (5/2) = 14(5/2)
x = 7 x 5
x = 35

Question 10.
Write and solve an equation to find the value of x.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 107

Answer:
x = 45°

Explanation:
(x + 10)° = 55°
x + 10 – 10 = 55 – 10
x = 45°

Question 11.
The circumference C of a circle is 24π inches. Use the formula C = 2πr to find the radius r of the circle.

Answer:
The radius r of the circle is 12 inches.

Explanation:
The circumference C of a circle is 24π inches
C = 2πr
24π = 2πr
24π/2π = r
12 = r
The radius r of the circle is 12 inches

1.2 Solving Multi-Step Equations (pp. 11–16)

Solve the equation. Check your solution.

Question 12.
3n + 12 = 30

Answer:
n = 6

Explanation:
Given equation is 3n + 12 = 30
3n = 30 – 12
3n = 18
n = 18/3
n = 6
Put n = 6 in 3n + 12 = 30
3(6) + 12 = 18 + 12 = 30

Question 13.
2(3 – p) – 17 = 41

Answer:
p = -26

Explanation:
Given equation is 2(3 – p) – 17 = 41
6 – 2p – 17 = 41
-2p – 11 = 41
-2p = 41 + 11
-2p = 52
p = 52/-2
p = -26
Put p = -26 in 2(3 – p) – 17 = 41
2(3 – (-26)) – 17 = 2(3 + 26) – 17
= 2(29) – 17 = 58 – 17 = 41

Question 14.
-14x + 28 + 6x = -44

Answer:
x = 9

Explanation:
Given equation is -14x + 28 + 6x = -44
-8x + 28 = -44
-8x = -44 – 28
-8x = -72
x = 72/8
x = 9
put x = 9 in -14x + 28 + 6x = -44
-14(9) + 28 + 6(9) = -126 + 28 + 54
= -44

Question 15.
1.06(12.95 + x) = 31.27

Answer:
x = 16.55

Explanation:
Given equation is 1.06(12.95 + x) = 31.27
12.95 + x = 31.27/1.06
12.95 + x = 29.5
x = 29.5 – 12.95
x = 16.55
Put x = 16.55 in 1.06(12.95 + x) = 31.27
1.06(12.95 + 16.55) = 1.06(29.5) = 31.27

Question 16.
The sum of the angle measures of a quadrilateral x is 360°. Find the value of x. Then find the angle measures of the quadrilateral.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 108

Answer:
x = 120°

Explanation:
The sum of the angle measures of a quadrilateral is 360°
x° + x° + ½x° + ½x° = 360°
2x° + x° = 360°
3x° = 360°
x = 360/3
x = 120°

Question 17.
The equation P = 2.5m + 35 represents the price (in dollars) of a bracelet, where is the cost of the materials (in dollars). The price of a bracelet is $115. What is the cost of the materials?

Answer:
The cost of the materials is $32.

Explanation:
The equation P = 2.5m + 35 represents the price (in dollars) of a bracelet
The price of a bracelet is $115
$115 = 2.5m + 35
115 – 35 = 2.5m
80 = 2.5m
80/2.5 = m
32 = m
The cost of the materials is $32.

Question 18.
A 455-foot fence encloses a pasture. What is the length of each side of the pasture?
Big Ideas Math Answers Grade 8 Chapter 1 Equations 109

Answer:
The length of each side of the pasture is 50 ft, 150 ft, 180 ft, 75 ft.

Explanation:
A 455-foot fence encloses a pasture
x + 3x + 1.5x + 180 = 455
5.5x + 180 = 455
5.5x = 455 – 180
5.5x = 275
x = 275/5.5
x = 50
The length of each side of the pasture is 50 ft, 150 ft, 180 ft, 75 ft.

1.3 Solving Equations with Variables on Both Sides (pp. 17–24)

Solve the equation. Check your solution, if possible.

Question 19.
3(x – 4) = -2(4 – x)

Answer:
x = 4

Explanation:
Given equation is 3(x – 4) = -2(4 – x)
3x – 12 = -8 + 2x
3x – 2x = -8 + 12
x = 4
Put x = 4 in 3(x – 4) = -2(4 – x)
3(4 – 4) = -2(4 – 4)
3(0) = -2(0)
0 = 0

Question 20.
4 – 5k = -8 – 5k

Answer:
The equation has no solution

Explanation:
Given equation is 4 – 5k = -8 – 5k
4 + 8 = -5k + 5k
12 = 0
The equation has no solution

Question 21.
5m – 1 = 4m + 5

Answer:
m = 6

Explanation:
Given equation is 5m – 1 = 4m + 5
5m – 4m = 5 + 1
m = 6
Put m = 6 in 5m – 1 = 4m + 5
5(6) – 1 = 4(6) + 5
30 – 1 = 24 + 5
29 = 29

Question 22.
3(5p – 3) = 5(p – 1)

Answer:
p = 2/5

Explanation:
Given equation is 3(5p – 3) = 5(p – 1)
15p – 9 = 5p – 5
15p – 5p = -5 + 9
10p = 4
p = 4/10
p = 2/5
Put p = 2/5 in 3(5p – 3) = 5(p – 1)
3(5(2/5) – 3) = 5(2/5 – 1)
3(2 – 3) = 5(2 – 5)/5
3(-1) = -3

Question 23.
0.4n + 0.1 = 0.5(n + 4)

Answer:
n = -19

Explanation:
Given equation is 0.4n + 0.1 = 0.5(n + 4)
0.4n + 0.1 = 0.5n + 2
0.1 – 2 = 0.5n – 04n
-1.9 = 0.1n
-1.9/0.1 = n
n = -19
Put n = -19 in 0.4n + 0.1 = 0.5(n + 4)
0.4(-19) + 0.1 = 0.5(-19 + 4)
-7.6 + 0.1 = 0.5(-15)
-7.5 = -7.5

Question 24.
7t + 3 = 8 + 7t

Answer:
The equation has no solution

Explanation:
Given equation is 7t + 3 = 8 + 7t
7t – 7t = 8 – 3
0 = 5
The equation has no solution

Question 25.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 110

Answer:
The equation has no solution

Explanation:
Given equation is 1/5 (15b – 7) = 3b – 9
(15b – 7) = 5(3b – 9)
15b – 7 = 15b – 45
15b – 15b = -45 + 7
0 = -38
The equation has no solution

Question 26.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 111

Answer:
The equation has infinitely many solutions

Explanation:
Given equation is 1/6 (12z – 18) = 2z – 3
1/6 (6(2z – 3)) = 2z – 3
2z – 3 = 2z – 3
2z = 2z – 3 + 3
2z = 2z
The equation has infinitely many solutions

Question 27.
The side lengths of an isosceles triangle are (3x + 1) inches, (4x + 5) inches, and (2x + 7) inches. Find the perimeters of two possible triangles.

Answer:
The perimeters of two possible triangles are 11x + 11, 12x + 19.

Explanation:
The side lengths of an isosceles triangle are (3x + 1) inches, (4x + 5) inches, and (2x + 7) inches.
The sides of first triangle are (3x + 1), (4x + 5), (4x + 5)
Perimeter = sum of all sides
P = 3x + 1 + 4x + 5 + 4x + 5
= 11x + 11
The sides of the second triangle are (4x + 5), (2x + 7), (2x + 7)
Perimeter = (4x + 5) + (2x + 7) + (2x + 7)
= 12x + 19

Question 28.
A shuttle company charges $3.25 plus $0.55 per mile. A taxi company charges $2.50 plus $0.60 per mile. After how many miles will both companies charge the same amount?

Answer:
After 15 miles both company charge the same amount.

Explanation:
Assume x as the miles
3.25 + 0.55x = 2.50 + 0.60x
3.25 – 2.50 = 0.60x – 0.55x
0.75 = 0.05x
x = 0.75/0.05
x = 15
After 15 miles both company charge the same amount.

Question 29.
You begin the year with $25 in a savings account and $50 in a checking account. Each week you deposit $5 into the savings account and $10 into the checking account. In how many weeks is the amount in the checking account twice the amount in the savings account?

Answer:
Every week the amount in checking twice the amount in savings.

Explanation:
You begin the year with $25 in a savings account and $50 in a checking account.
50 + 10x
Savings is 25 + 5x
The amount in checking twice the amount in savings
50 + 10x = 2(25 + 5x)
50 + 10x = 50 + 10x
Every week the amount in checking twice the amount in savings.

Rewriting Equations and Formulas (pp. 25–30)

Question 30.
6y + x = 8

Answer:
y = (8 – x)/6

Explanation:
Given equation is 6y + x = 8
6y = 8 – x
y = (8 – x)/6

Question 31.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 112

Answer:
y = 30x – 45

Explanation:
Given equation is 10x – 1/3 y = 15
-1/3 y = 15 – 10x
y = (15 – 10x)-3
y = -45 + 30x
y = 30x – 45

Question 32.
20 = 5x + 10y

Answer:
y = (20 – 5x)/10

Explanation:
Given equation is 20 = 5x + 10y
20 – 5x = 10y
(20 – 5x)/10 = y

Question 33.
The formula Big Ideas Math Answers Grade 8 Chapter 1 Equations 113 converts a temperature from Kelvin K to Fahrenheit F.
a. Solve the formula for K.
b. Convert 240°F to Kelvin. Round your answer to the nearest hundredth.

Answer:
a. K = 5/9 (F – 32) + 273.15
b. K = 400

Explanation:
F = 9/5 (K – 273.15) + 32
a.
F – 32 = 9/5 (K – 273.15)
5(F – 32) = 9(K – 273.15)
5/9 (F – 32) = K – 273.15
5/9 (F – 32) + 273.15 = K
b.
K = 5/9 (F – 32) + 273.15
Put F = 240°F
K = 5/9 (240 – 32) + 273.15
= 5/9(208) + 273.15
= 115.5 + 273.15
= 388.7
Rounding 388 to nearest 100
So, K = 400

Question 34.
Use the trapezoid shown.
a. Write the formula for the area A of a trapezoid.
b. Solve the formula for h.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 113.1
c. Use the new formula to find the height of the trapezoid.

Answer:
a. A = (a + b) h/2
b. h = 2A/(a + b)
c. h = 6 cm

Explanation:
a.
Area of trapezoid formula is A = (a + b) h/2
b.
Area of trapezoid is A = (a + b) h/2
2A = (a + b)h
h = 2A/(a + b)
c.
A = 72 cm², a = 8 cm, b = 16 cm
h = (2 x 72)/(8 + 16)
= 144/24
h = 6 cm

Question 35.
The equation for a line in slope-intercept form is y = mx + b. Solve the equation for x.

Answer:
x = (y – b)/m

Explanation:
The equation for a line in slope-intercept form is y = mx + b
y = mx + b
y – b = mx
(y – b)/m = x

Question 36.
The formula for the volume of a cylinder is V = πr2h, where r is the radius of the circular base and is the height of the cylinder.
a. Solve the formula for h.
b. Use the new formula to find the height of the cylinder.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 114

Answer:
a. h = V/πr²
b. The height h = 3.84 in

Explanation:
The formula for the volume of a cylinder is V = πr²h
a. Solve the formula for h.
V = πr²h
Divide each side by πr²
V/πr² = πr²h/πr²
V/πr² = h
b.
Height h = V/πr²
Volume = 6π cubic inches
radius = 1.25 inch
h = V/πr²
= 6π / π(1.25)²
= 6/1.5625
= 3.84
The height h = 3.84 in

Equations Practice Test

Practice Test
Solve the equation. Check your solution, if possible.

Question 1.
4 + y = 9.5

Answer:
y = 5.5

Explanation:
Given equation is 4 + y = 9.5
y = 9.5 – 4
y = 5.5
Put y = 5.5 in 4 + y = 9.5
4 + 5.5 = 9.5

Question 2.
–\(\frac{x}{9}\) = -8

Answer:
x = 72

Explanation:
Given equation is -x/9 = -8
-x = -8 * 9
-x = -72
x = 72
Put x = 72 in -x/9 = -8
-72/9 = -8

Question 3.
z – \(\frac{2}{3}\) = \(\frac{1}{8}\)

Answer:
z = 19/24

Explanation:
Given equation is z – 2/3 = 1/8
z = 1/8 + 2/3
z = (3 + 16)/24
z = 19/24
Put z = 19/24 in z – 2/3 = 1/8
19/24 – 2/3 = (19 – 16)/24 = 3/24
= 1/8

Question 4.
15 = 9 – 3a

Answer:
a = -2

Explanation:
Given equation is 15 = 9 – 3a
15 – 9 = -3a
6 = -3a
6/-3 = a
-2 = a
put a = -2 in 15 = 9 – 3a
15 = 9 – 3(-2) = 9 + 6

Question 5.
4(b + 5) – 9 = -7

Answer:
b = -9/2

Explanation:
Given equation is 4(b + 5) – 9 = -7
4b + 20 – 9 = -7
4b + 11 = -7
4b = -7 – 11
4b = -18
b = -18/4
b = -9/2
Put b = -9/2 in 4(b + 5) – 9 = -7
4(-9/2 + 5) – 9 = 4(-9 + 10)/2 – 9
= 4/2 – 9 = 2 – 9 = -7

Question 6.
9j – 8 = 8 + 9j

Answer:
The equation has no solution

Explanation:
Given equation is 9j – 8 = 8 + 9j
9j = 8 + 9j + 8
9j – 9j = 16
0 = 16
The equation has no solution

Question 7.
3.8n – 13 = 1.4n + 5

Answer:
n = 7.5

Explanation:
Given equation is 3.8n – 13 = 1.4n + 5
3.8n – 1.4n = 5 + 13
2.4n = 18
n = 18/2.4
n = 7.5
Put n = 7.5 in 3.8n – 13 = 1.4n + 5
3.8(7.5) – 13 = 1.4(7.5) + 5
28.5 – 13 = 10.5 + 5
15.5 = 15.5

Question 8.
9(8d – 5) + 13 = 12d – 2

Answer:
d = 1/2

Explanation:
Given equation is 9(8d – 5) + 13 = 12d – 2
72d – 45 + 13 = 12d – 2
72d – 12d – 32 = -2
60d = -2 + 32
60d = 30
d = 30/60
d = 1/2
Put d = 1/2 in 9(8d – 5) + 13 = 12d – 2
9(8(1/2) – 5) + 13 = 12(1/2) – 2
9(4 – 5) + 13 = 6 – 2
9(-1) + 13 = 4
-9 + 13 = 4

Question 9.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 115

Answer:
t = -4

Explanation:
Given equation is 1/4 t + 4 = 3/4(t + 8)
1/4 t + 4 = 3/4t + 8(3/4)
1/4 t + 4 = 3/4t + 6
3/4 t – 1/4 t = 4 – 6
2/4 t = -2
1/2 t = -2
t = -2 * 2
t = -4
Put t = -4 in 1/4 t + 4 = 3/4(t + 8)
1/4 (-4) + 4 = 3/4(-4 + 8)
-1 + 4 = 3/4(4)
3 = 3

Question 10.
The sum of the angle measures of a triangle is 180°. Find the value of x. Then find the angle measures of the triangle.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 116

Answer:
x = 57.33°
The angles of traingle are 57.33°, 114.66°, 65.33°

Explanation:
The sum of the angle measures of a triangle is 180°.
x + 2x + x + 8 = 180
3x + 8 = 180
3x = 180 -8
3x = 172
x = 172/3
x = 57.33
The angles of traingle are 57.33, 114.66, 65.33

Question 11.
A formula for the perimeter of a rectangle P = 2l + 2w.
a. Solve the formula for w.
b. Use the new formula to find the width w (in meters) of a rectangle with a perimeter of 2 meters and a length of 40 centimeters.

Answer:
a. w = (P – 2l)/2
b. Width is 15.6 m

Explanation:
P = 2l + 2w
a.
P – 2l = 2w
(P – 2l)/2 = w
b.
P = 32 m = 3200 cm, l = 40 cm
w = (P – 2l)/2
w = (3200- 2(40))/2
= (3200 – 80)/2
= 3120/2
= 1560 cm
= 15.6 m
Width is 1560 cm

Question 12.
Solve 0.5 = 0.4y – 0.25 for y.

Answer:
y = 1.875

Explanation:
0.5 = 0.4y – 0.25
0.5 + 0.25 = 0.4y
0.75 = 0.4y
0.75/0.4 = y
1.875 = y

Question 13.
Your basketball team wins a game by 13 points. The opposing team scores 72 points. Explain how to find your team’s score.

Answer:
Your team’s score is 85 points

Explanation:
Your basketball team wins a game by 13 points. The opposing team scores 72 points.
Let your team score is x
x – 72 = 13
x = 13 + 72
x = 85
Your team’s score is 85 points

Question 14.
You are biking at a speed of 18 miles per hour. You are 3 miles behind your friend, who is biking at a speed of 12 miles per hour. Write and solve an equation to find the amount of time it takes for you to catch up to your friend.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 117

Answer:
It will take 1/2 hour for you to catch up with your friend.

Explanation:
Let t be the number of hours you and your friends ride bikes
Distance = rate x time and you are riding the bike at a rate of 18 miles per hour, then the distance you ride your bike is 18t miles
Your friend is riding the bike at 12 miles per hour, then your friend rides a distance of 12t miles.
You start 3 miles behind your friend so that distance you ride to catch up to your friend is 3 more than the distance your friend rides. Therefore,
the distance you ride = distance your friend rides + 3
18t = 12t + 3
18t – 12t = 3
6t = 3
t = 3/6
t = 1/2
It will take 1/2 hour for you to catch up with your friend.

Question 15.
Two scientists are measuring the temperatures of lava. One scientist records a temperature of 1725°F. The other scientist records a temperature of 950°C. Which is the greater temperature?

Answer:
The greater temperature is 950°C.

Explanation:
One scientist records a temperature of 1725°F. The other scientist records a temperature of 950°C.
C = 5/9 (F – 32)
Put F = 1725
C = 5/9 (1725 – 32)
= 5/9 (1693)
= 940.55
This is smaller than 950°C

Question 16.
Your profit for mowing lawns this week is $24. You are paid $8 per hour and you paid $40 for gas for the lawn mower. How many hours did you work this week?
Big Ideas Math Answers Grade 8 Chapter 1 Equations 118

Answer:
You work for 8 hours this week.

Explanation:
Let x be the number of hours you work this week
8x – 40 = 24
8x = 24 + 40
8x = 64
x = 64/8
x = 8
You work for 8 hours this week.

Equations Cumulative Practice

Question 1.
Which value of x makes the equation true?
Big Ideas Math Answers Grade 8 Chapter 1 Equations 119
A. 8
B. 28
C. 36
D. 128

Big Ideas Math Answers Grade 8 Chapter 1 Equations 120

Answer:
A. 8

Explanation:
4x = 32
Divide each side by 4
4x/4 = 32/4
x = 8

Question 2.
A taxi ride costs $3 plus $2 for each mile driven. You spend $39 on a taxi. This can be modeled by the equation 2m + 3 = 39, where m represents the number of miles driven. How long was your taxi ride?
F. 18 mi
G. 21 mi
H. 34 mi
I. 72 mi

Answer:
F. 18 mi

Explanation:
To know the miles you traveled,
we have to find the solution of the equation
2m + 3 = 39
2m = 39 – 3
2m = 36
m = 36/2
m = 18
So, you traveled 18 mi

Question 3.
Which of the following equations has exactly one solution?
Big Ideas Math Solutions Grade 8 Chapter 1 Equations 121

Answer:
None of the equation has exactly one solution.

Explanation:
A. 2/3 (x + 6) = 2/3 x + 4
2/3x + 2/3 (6) = 2/3 x + 4
2/3 x + 4 = 2/3 x + 4
B. 3/7 y + 13 = 13 – 3/7 y
3/7 y + 3/7y = 13 – 13
6/7y = 0
C. 4/5(n + 1/3) = 4/5 n + 1/3
4/5n + 4/15 = 4/5n + 1/3
4/5n – 4/5n = 1/3 – 4/15
0 = 1/15
D. 7/8(2t + 1/8) = 7/4t
7t/4 + 7/64 = 7t/4

Question 4.
The perimeter of the square is equal to the perimeter of the triangle. What are the side lengths of the square?
Big Ideas Math Answers Grade 8 Chapter 1 Equations 122
Big Ideas Math Answers Grade 8 Chapter 1 Equations 123

Answer:
The side length of the square is 12 units.

Explanation:
The perimeter of the square is equal to the perimeter of the triangle
4(3x + 3) = 2x + 4 + 7x – 2 + 7x – 2
12x + 12 = 16x
16x – 12x = 12
4x = 12
x = 12/4
x = 3
Side of square = 3x+ 3
= 3(3) + 3 = 9 + 3
= 12
The side length of the square is 12 units.

Question 5.
The formula d = rt relates distance, rate, and time. Solve the formula for t.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 124

Answer:
G. t = d/r

Explanation:
Given that,
d = rt
Divide both sides by r
d/r = rt/r
d/r = t

Question 6.
What is a possible first step to solve the equation 3x + 5 = 2(x + 7)?
A. Combine 3x and 5.
B. Multiply x by 2 and 7 by 2.
C. Subtract x from 3x.
D. Subtract 5 from 7.

Answer:
B. Multiply x by 2 and 7 by 2.

Explanation:
3x + 5 = 2(x + 7)
Multiply x by 2 and 7 by 2.
3x + 5 = 2x + 14
3x – 2x = 14 – 5
x = 9

Question 7.
You work as a sales representative. You earn $400 per week plus 5% of your total sales for the week.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 125
Part A
Last week, you had total sales of $5000. Find your total earnings. Show your work.
Part B
One week, you earned $1350. Let represent your total sales that week. Write an equation that you can use to find s.
Part C
Using your equation from Part B, find s. Show all steps clearly.

Answer:
Part A
$650
Part B
400 + 0.05s = 1350
Part C
s = 19,000

Explanation:
Earnings = 400 + 0.05s
a.
Earnings = 400 + 0.05s
Put s = 5000
400 + 0.05(5000) = 400 + 250 = 650
b.
400 + 0.05s = 1350
c.
400 + 0.05s = 1350
0.05s = 1350 – 400
0.05s = 950
s = 950/0.05
s = 19,000

Question 8.
In 10 years, your aunt will be 39 years old. Let m represent your aunt’s age today. Which equation can you use to find m?
F. m = 39 + 10
G. m – 10 = 39
H. m + 10 = 39
I. 10m = 39

Answer:
H. m + 10 = 39

Explanation:
In 10 years, your aunt will be 39 years old.
Let m represent your aunt’s age today.
Equation is
m = 39 – 10
m + 10 = 39

Question 9.
Which value of y makes the equation 3y + 8 = 7y + 11 true?
A. -4.75
B. -0.75
C. 0.75
D. 4.75

Answer:
C. 0.75

Explanation:
Given equation is 3y + 8 = 7y + 11
8 – 11 = 7y – 3y
-3 = -4y
y = 3/4

Question 10.
What is the value of x?
F. 23
G. 39
H. 58
I. 68
Big Ideas Math Answers Grade 8 Chapter 1 Equations 126

Answer:
F. 23

Explanation:
As shown in the figure
90 = 2x+ 12 + 32
90 = 2x + 44
90 – 44 = 2x
46 = 2x
46/2 = x
23 = x

Question 11.
You have already saved $35 for a new cell phone. You need $175 to buy the cell phone. You think you can save $10 per week. At this rate, how many more weeks will you need to save money before you can buy the new cell phone?
Big Ideas Math Answers Grade 8 Chapter 1 Equations 126.1

Answer:
14 more weeks are required to save money to buy the new cell phone

Explanation:
The equation for the total amount saved
35 + 10x = 175
10x = 175 – 35
10x = 140
x = 140/10
x = 14
14 more weeks are required to save money to buy the new cell phone

Question 12.
What is the greatest angle measure in the triangle?
Big Ideas Math Answers Grade 8 Chapter 1 Equations 127
A. 26°
B. 78°
C. 108°
D. 138°

Answer:
A. 26°

Explanation:
Sum of angle measures = 180°
3x + 2x + 50 = 180
5x + 50 = 180
5x = 180 – 50
5x = 130
x = 130/5
x = 26°

Question 13.
Which value x of makes the equation 6(x – 3) = 4x – 7 true?
F. -5.5
G. -2
H. 1.1
I. 5.5

Answer:
I. 5.5

Explanation:
6(x – 3) = 4x – 7
6x – 18 = 4x – 7
6x – 4x = -7 + 18
2x = 11
x = 11/2
x = 5.5

Question 14.
The drawing below shows equal weights on two sides of a balance scale.
Big Ideas Math Answers Grade 8 Chapter 1 Equations 128
What can you conclude from the drawing?
A. A mug weighs one-third as much as a trophy.
B. A mug weighs one-half as much as a trophy.
C. A mug weighs twice as much as a trophy.
D. A mug weighs three times as much as a trophy.

Answer:
A. A mug weighs one-third as much as a trophy.

Explanation:
The drawing shows 3mugs + 1trophy = 2trophy
3mugs = 2trophy – 1 trophy
3mugs = 1trophy
Mug = 1/3 trophy

Final Words:

Hope you are all satisfied with the information given in this article. Share this Big Ideas Math Answers Grade 8 Chapter 1 Equations pdf link with your besties and help them to overcome their difficulties in maths. Bookmark our site to get the Big Ideas Math Grade 8 Chapterwise Pdf links.

Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes

Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes

Learn the basics of Two-Dimensional Shapes with the help of our Big Ideas Math Grade K Answer Key for Chapter 11 Identify Two-Dimensional Shapes. So the students who are looking for Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes can Download the pdf form here. We provide step by step explanations with pictures for all the problems here. The solutions for all these questions are prepared by the math experts. Hence Download Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes pdf for free of cost.

Big Ideas Math Book Grade K Answer Key Chapter 11 Identify Two-Dimensional Shapes

Start answering all the questions in the exams by practicing from Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes. You can prove yourself in the tests by referring to Big Ideas Math Book Grade K Answer Key Chapter 11 Identify Two-Dimensional Shapes. Just tap on the topic you wish and learn various questions involved in it easily. We have listed all of them with examples and explained every problem clearly.

Vocabulary

Lesson: 1 Describe Two-Dimensional Shapes

Lesson: 2 Triangles

Lesson: 3 Rectangles

Lesson: 4 Squares

Lesson: 5 Hexagons and Circles

Lesson: 6 Join Two-Dimensional Shapes

Lesson: 7 Build Two-Dimensional Shapes

Chapter: 11 – Identify Two-Dimensional Shapes

Identify Two-Dimensional Shapes Vocabulary

Directions:
Classify the animals into the categories shown. Write the marks in the chart. Count the marks and write the numbers to tell how many animals are in each category. Draw a line through the number that is less than the other number.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes v 1

Answer:

Explanation:
In the above image their are 7 animals out of them 4 animals have spots and 3 animals do not have spots on their body.So, as 3 is lessthan 4 draw a line through 3.

Vocabulary Cards
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes v 3

Vocabulary Cards
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes v 4

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes v 5

Lesson 11.1 Describe Two-Dimensional Shapes

Explore and Grow

Directions:
Cut out the Two-Dimensional Shape Cards. Sort the cards into the categories shown.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 1

Answer:

Explanation:
In the above picture there is 1 shape that has curves and 4 shapes that have no curves.

Think and grow

Directions:

  • Color the shape that has a curve.
  • Color the shape that does not have any vertices.
  • Color the shape that has more than 4 vertices.
  • Color the shape that has only 3 sides.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 2
Answer:

Explanation:
In the above shapes, the shape that has curves and a shape that do not have any vertices is a circle. So, color the circle.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 3
Answer:

Explanation:
In the above shapes, the shape that has more than 4 vertices is a hexagon. So, color hexagon.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 4
Answer:

Explanation:
In the above shapes, the shape that has only 3 sides is a triangle. So, color triangle.

Apply and Grow: Practice

Directions:
1 Color the shapes that have only 4 sides. 2 Color the shapes that have only 3 vertices. 3 Color the shapes that have 6 vertices. 4 Color the shapes that have curves.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 5
Answer:

Explanation:
The shapes that have only 4 sides are square and rectangle. So, color square and rectangle.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 6
Answer:

Explanation:
The shape that has only 3 vertices is the triangle. So, color the triangles.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 7
Answer:

Explanation:
The shape that have 6 vertices is a hexagon. So, color the hexagons.

Question 4.
Answer:

Explanation:
The shape that has curves is a circle. So, color the circles.

Think and Grow: Modeling Real Life

Directions: Write the number that answers the question.

  • How many stickers in the picture have more than 4 vertices?
  • How many stickers in the picture have all straight sides?
  • How many stickers in the picture have both curves and straight sides?

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 9
Answer:

Explanation:
In the above picture, there are 2 stickers that have more than 4 vertices, 6 stickers that have all straight sides, and 4 stickers that have curves and straight sides.

Describe Two-Dimensional Shapes Homework & Practice 11.1

Directions:
1 Color the shape that has 6 sides. 2 Color the shape that has more than 3 straight sides.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 10
Answer:

Explanation:
The shape that has 6bsides is a hexagon. So, color the hexagon.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 11
Answer:

Explanation:
The shape that has more than 3 straight sides is a rectangle. So, color the rectangle.

Directions:
3 Color the shapes that have curves. 4 Color the shapes that have only 4 vertices. 5 How many stickers have less than 4 vertices? Write the number. How many stickers have more than 3 sides? Write the number.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 12
Answer:

Explanation:
The shape that has curves is a circle. So, color the circles.

Question 4.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 13
Answer:

explanation:
The shapes that have 4 vertices are rectangle and square. So, color the shapes rectangle and square.

Question 5.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.1 14
Answer:

Explanation:
In the above picture, there are 3 stickers that have less than 4 vertices, 5 stickers that have more than 3 sides.

Lesson 11.2 Triangles

Explore and Grow

Directions:
Cut out the Triangle or Not a Triangle Sort Cards. Sort the cards into the categories shown.

Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 1

Think and Grow

Directions:
Color any triangles. Tell why your answers are correct.

Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 2
Answer:

Explanation:
In the above shapes, the triangles are colored as triangles has 3 sides and 3 vertices.

Apply and Grow: Practice

Directions:
1 and 2 Color any triangles. Tell why your answers are correct. 3 Trace the shapes that are triangles. Write the number of sides and the number of vertices of a triangle.

Question 1.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 3
Answer:

Explanation:
In the above shapes, the triangle is colored as a triangle has 3 sides and 3 vertices.

Question 2.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 4
Answer:

Explanation:
In the above shapes, the triangles are colored as a triangle has 3 sides and 3 vertices.

Question 3.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 5
Answer:

Explanation:
In the above shapes, the shapes that have 3 sides and 3 vertices are triangles. So, trace the triangles.

Think and Grow: Modeling Real Life

Directions:
You use triangle-shaped flags to make two banners for a party. You use 10 flags in all.

  • Draw and color flags to make the banners.
  • Write an addition sentence to match your picture.

Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 6
Answer:

Explanation:
In the above picture we have to draw and color 5 triangle-shaped flags in 2 banners. So, there will be 2 banners with 5 flags each. Therefore the addition sentence is 5+5=10.

Triangles Homework & Practice 11.2

Directions:
1 – 2 Color any triangles. Tell why your answers are correct.

Question 1.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 7
Answer:

Explanation:
In the above shapes the triangle is colored as triangle has 3 sides and 3 vertices.

Question 2.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 8
Answer:

Explanation:
In the above shapes the triangle is colored as triangle has 3 sides and 3 vertices.

Question 3.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 9
Answer:

Explanation:
In the above shapes the triangles are colored as triangle has 3 sides and 3 vertices.
Directions:
4 Color any triangles. Tell why your answers are correct. 5 Trace the shapes that are triangles. Write the number of sides and the number of vertices of a triangle. 6 You use triangle-shaped flags to make two banners for a party. You use 8 flags in all. Draw and color flags to make the banners. Then write an addition sentence to match your picture.

Question 4.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 10
Answer:

Explanation:
In the above shapes the triangles are colored as triangle has 3 sides and 3 vertices.

Question 5.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 11
Answer:

Explanation:

In the above shapes the shapes that has 3 sides and 3 vertices are triangles. So, trace the triangles.

Question 6.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.2 12
Answer:

Explanation:
In the above picture we have to draw and color 4 triangle shaped flags in 2 banners. So, there will be 2 banners with 4 flags each. Therefore the addition sentence is 4+4=8.

Lesson 11.3 Rectangles

Explore and Grow

Directions:
Cut out the Rectangle or Not a Rectangle Sort Cards. Sort the cards into the categories shown.

Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 1
Answer:

Think and Grow

Directions:
Color any rectangles. Tell why your answers are correct.

Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 2
Answer:

Explanation:
In the above shapes the rectangles are colored as rectangles have 4 sides and 4 vertices.

Apply and Grow: Practice

Directions:
1 and 2 Color any rectangles. Tell why your answers are correct. 3 Trace the shapes that are rectangles. Write the number of sides and the number of vertices of a rectangle.

Question 1.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 3
Answer:

Explanation:
In the above shapes the rectangle is colored as rectangles have 4 sides and 4 vertices.

Question 2.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 4
Answer:

Explanation:
In the above shapes the rectangle is colored as rectangles have 4 sides and 4 vertices.

Question 3.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 5
Answer:

Explanation:
In the above shapes the rectangles are traced and rectangles have 4 sides and 4 vertices.

Think and Grow: Modeling Real Life

Directions:

  • Trace and color 4 rectangular posters on the wall.
  • You put 3 more posters on another wall. Draw and color 3 rectangular posters on the wall.
  • Write an addition sentence to tell how many posters are on the walls in all.

Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 6
Answer:

Explanation:
In the above picture 4 rectangular posters are traced and colored and 3 extra rectangular posters are drawn and colored. When we add all the posters ie 4+3 we have 7 rectangular posters in all.

Rectangles Homework & Practice 11.3

Directions:
1 – 3 Color any rectangles. Tell why your answers are correct.

Question 1.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 7
Answer:

Explanation:
In the above shapes the rectangle is colored as rectangles have 4 sides and 4 vertices.

Question 2.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 8
Answer:

Explanation:
In the above shapes the rectangle is colored as rectangles have 4 sides and 4 vertices.

Question 3.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 9
Answer:

Explanation:
In the above shapes the rectangle is colored as rectangles have 4 sides and 4 vertices.

Directions:
4 Color any rectangles. Tell why your answers are correct. 5 Trace the shapes that are rectangles. Write the number of sides and the number of vertices of a rectangle. 6 You make rectangle.

Question 4.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 10
Answer:

Explanation:
In the above shapes the rectangle is colored as rectangles have 4 sides and 4 vertices.

Question 5.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 11
Answer:

Explanation:
In the above shapes the rectangles are traced as rectangles have 4 sides and 4 vertices.

Question 6.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.3 12
Answer:

Explanation:
In the above picture we have to draw 1 large rectangular picture frame and 4 small rectangular picture frames. Therefore the addition sentence is 1+4=5. So, there are 5 rectangular picture frames in all.

Lesson 11.4 Squares

Explore and Grow

Directions:
Cut out the Square or Not a Square Sort Cards. Sort the cards into the categories shown.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 1
Answer:

Think and Grow

Directions:
Color any squares. Tell why your answers are correct.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 2
Answer:

Explanation:
In the above shapes the square are colored as the squares have 4 sides of equal length and 4 vertices.

Apply and Grow: Practice

Directions:
1 and 2 Color any squares. Tell why your answers are correct. 3 Trace the shapes that are squares. Write the number of sides and the number of vertices of a square.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 3
Answer:

Explanation:
In the above shapes the square are colored as the squares have 4 sides of equal length and 4 vertices.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 4
Answer:

Explanation:
In the above shapes the square are colored as the squares have 4 sides of equal length and 4 vertices.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 5
Answer:

Explanation:
In the above shapes the square are traced as the squares have 4 sides of equal length and 4 vertices.

Think and Grow: Modeling Real Life

Directions:

  • Use squares and rectangles to draw 6 windows and 1 door on the house. Color the house.
  • Write an addition sentence to tell how many squares and how many rectangles you draw in all.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 6
Answer:

Explanation:
In the house picture given above draw 6 square shaped windows and 1 rectangular door.Color all the shapes and the house. The addition sentence for shapes in all is 6+1=7.

Squares Homework & Practice 11.4

Directions:
1 – 3 Color any squares. Tell why your answers are correct.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 7
Answer:

Explanation:
In the above shapes the square is colored as the squares have 4 sides of equal length and 4 vertices.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 8
Answer:

Explanation:
In the above shapes the square is colored as the squares have 4 sides of equal length and 4 vertices.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 9
Answer:

Explanation:
In the above shapes the squares are colored as the squares have 4 sides of equal length and 4 vertices.

Directions:
4 Color any squares. Tell why your answers are correct. 5 Trace the shapes that are squares. Write the number of sides and the number of vertices of a square. 6 Use squares and rectangles to draw 4 windows and 1 door on the castle. Color the castle. Then write an addition sentence to tell how many squares and rectangles you draw in all.

Question 4.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 10
Answer:

Explanation:
In the above shapes the squares are colored as the squares have 4 sides of equal length and 4 vertices.

Question 5.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 11
Answer:

Explanation:
In the above shapes the squares are traced as the squares have 4 sides of equal length and 4 vertices.

Question 6.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.4 12
Answer:

Explanation:

In the castle given above draw 4 square shaped windows and 1 rectangular door.Color all the shapes and the castle. The addition sentence for shapes in all is 4+1=5.

Lesson 11.5 Hexagons and Circles

Explore and Grow

Directions:

  • Use your finger to trace around the yellow hexagon. Trace and color the shapes that are hexagons.
  • Use your finger to trace around the blue circle. Use a different color to trace and color the shapes that are circles.

Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 1
Answer:

Explanation:
In the above shapes the hexagons colored yellow and the circles are colored blue.

Think and Grow

Directions:
Color any hexagons red. Color any circles blue. Tell why your answers are correct.

Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 2
Answer:

Explanation:
In the above shapes the hexagons are colored red and the circles are colored blue as hexagons have 6 sides and 6 vertices and the circle has curves.

Apply and Grow: Practice

Directions:
1 and 2 Color any hexagons red. Color any circles blue. Tell why your answers are correct. 3 Trace the shapes that are hexagons. Write the number of sides and the number of vertices of a hexagon.

Question 1.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 3
Answer:

Explanation:
In the above shapes the hexagon is colored red and the circle is colored blue as hexagons have 6 sides and 6 vertices and the circle has curves.

Question 2.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 4
Answer:

Explanation:
In the above shapes the hexagon is colored red and the circle is colored blue as hexagons have 6 sides and 6 vertices and the circle has curves.

Question 3.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 5
Answer:

Explanation:
In the above shapes the hexagons are traced as hexagons have 6 sides and 6 vertices.

Think and Grow: Modeling Real Life

Directions: Follow the steps to complete the robot.

  • Draw 2 circles for the robot’s eyes, a triangle for the nose, and a rectangle for the mouth.
  • Draw a hexagon for each of the robot’s hands.
  • How many of each shape is used for the whole robot? Write the number next to each shape.
  • Color your robot.

Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 6
Answer:

Explanation:
In the Robot picture given above draw 2 circles for the robot’s eyes, a triangle for the nose, a rectangle for the mouth and a hexagon for each of the robot’s hands. Color all the shapes. So, there are 2 circles, 3 hexagons, 3 triangles and 7 rectangles in all.

Hexagons and Circles Homework & Practice 11.5

Directions:
1 – 3 Color any hexagons red. Color any circles blue. Tell why your answers are correct.

Question 1.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 7
Answer:

Explanation:
In the above shapes the hexagon is colored red and the circle is colored blue as hexagons have 6 sides and 6 vertices and the circle has curves.

Question 2.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 8
Answer:

Explanation:
In the above shapes the hexagons are colored red and the circle is colored blue as hexagons have 6 sides and 6 vertices and the circle has curves.

Question 3.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 9
Answer:

Explanation:
In the above shapes the hexagon is colored red and the circle is colored blue as hexagons have 6 sides and 6 vertices and the circle has curves.

Directions:
4 Color any hexagons red. Color any circles blue. Tell why your answers are correct. 5 Trace the shapes that are circles. Write the number of straight sides and the number of vertices of a circle. 6 Draw a hexagon at the end of the cat’s tail. Draw 2 circles for the eyes. How many of each shape is used for the cat? Write the numbers next to the shapes. Color your cat.

Question 4.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 10
Answer:

Explanation:
In the above shapes the hexagon is colored red and the circle is colored blue as hexagons have 6 sides and 6 vertices and the circle has curves.

Question 5.

Answer:

Explanation:
In the above shapes the hexagons are traced as the circle has curves.

Question 6.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 11.5 12
Answer:

Explanation:
in the cat picture given above draw 2 circles for eyes of cat, a hexagon at the end of the tail of the cat.Color all the shape.So, there are 5 circles, 2 hexagons, 5 triangles and 7 rectangles in all.

Lesson 11.6 Join Two-Dimensional Shapes

Explore and Grow

Directions:

  • Use 2 squares to make a rectangle. Trace your shape.
  • Add another square to make a larger rectangle. Trace your shape.
  • Add another square to make a larger square. Trace your shape.

Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.6 1
Answer:

Explanation:
In the above question the rectangle in first box is made by using 2 squares, the rectangle in second box is made by adding another square and the square in the third box is made by adding another square to it.Trace all the shapes.

Think and Grow

Directions:
Use the pattern block shown to form the shape. Count and write how many pattern blocks you use.

Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.6 2
Answer:

Explanation:
In the above shape 5 triangle pattern blocks are used to form the large triangle and 6 triangle pattern blocks are used to form a hexagon.

Apply and Grow: Practice

Directions:
1 and 2 Use the pattern blocks shown to form the shape. Count and write how many of each pattern block you use. 3 Draw a rectangle that can be formed by the 2 triangles shown.

Question 1.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.6 3
Answer:

Explanation:
In the above picture 3 triangle shape pattern blocks and 2 square shaped pattern blocks are used to form the given shape.

Question 2.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.6 4
Answer:

Explanation:
In the above picture 6 triangle shaped pattern blocks and 1 hexagon shaped pattern block are used to form the given shape.

Question 3.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.6 5
Answer:

Explanation:
The above picture the 2 triangle shaped pattern blocks are used to a rectangle.

Think and Grow: Modeling Real Life

Directions:
Use the pattern blocks shown to create the rocket ship. Count and write how many of each pattern block you use.

Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.6 6
Answer:

Explanation:
In the Rocket picture given above 3 triangle shaped pattern blocks, 2 square shaped pattern blocks and 1 hexagon pattern block are used to form the complex shape of a rocket.

Join Two-Dimensional Shapes Homework & Practice 11.6

Directions:
1 and 2 Use the pattern blocks shown to form the shape. Count and write how many of each pattern block you use.

Question 1.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.6 7
Answer:

Explanation:
In the above picture 4 square shaped pattern blocks are used to form the large square.

Question 2.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.6 8
Answer:

Explanation:
In the above picture 3 triangle shaped pattern blocks and 1 hexagon pattern blocks are used to form the large triangle shape .

Directions:
3 Draw a square that can be formed by the 2 triangles shown. 4 Use the pattern blocks shown to create the butterfly. Count and write how many of each pattern block you use.

Question 3.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.6 9
Answer:

Explanation:
In the above picture the 2 triangle shaped pattern blocks are used to form a square.

Question 4.
Big Ideas Math Solutions Grade K Chapter 11 Identify Two-Dimensional Shapes 11.6 10
Answer:

Explanation:
In the butterfly picture given above there are 6 triangle pattern blocks, 4 square pattern blocks and 1 hexagon pattern block. All these pattern blocks together form the shape of a butterfly.

Lesson 11.7 Build Two-Dimensional Shapes

Explore and Grow

Directions:
Use your materials to build one of the two-dimensional shapes shown. Circle the two-dimensional shape that you make.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 1

Think and Grow

Directions:

  • Use your materials to build a triangle. Draw your triangle or attach it to the page.
  • Use your materials to build a rectangle. Draw your shape or attach it to the page.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 2

Apply and Grow: Practice

Directions:
1 Use your materials to build a hexagon. Draw your hexagon or attach it to the page. 2 Use your materials to build a circle. Draw your circle or attach it to the page. 3 Use your materials to build a two-dimensional shape that has 4 vertices. Then build a different shape that has 4 vertices. Draw your shapes or attach them to the page.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 3
Answer:

Question 2.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 4
Answer:

Question 3.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 5
Answer:

Think and Grow: Modeling Real Life

Directions:

  • Use your materials to build the front of the house in the picture. Draw your shape or attach it to the page.
  • Circle the shapes that you use to make the front of the house.

Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 6
Answer:

Build Two-Dimensional Shapes Homework & Practice 11.7

Directions:
1 – 3 Use your materials to build the shape shown. Draw your shape or attach it to the page. 4 Use your materials to build a two-dimensional shape that has 6 vertices. Draw your shape or attach it to the page.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 7
Answer:

Question 2.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 8
Answer:

Question 3.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 9
Answer:

Question 4.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 10
Answer:

Directions:
5 Use your materials to build a two-dimensional shape that is not a rectangle. Then build a different two-dimensional shape that is not a rectangle. Draw your shapes or attach them to the page. 6 Use your materials to build the front of the sand castle tower in the picture. Draw your shape or attach it to the page. Circle the shapes that you use to make the front of the tower.

Question 5.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 11
Answer:

Question 6.
Big Ideas Math Answer Key Grade K Chapter 11 Identify Two-Dimensional Shapes 11.7 12
Answer:

Identify Two-Dimensional Shapes Performance Task

Directions: 1 Use the clues to draw each two-dimensional shape to make an animal.

  • The face is a shape that has 1 more than 5 sides.
  • The eyes are shapes that are curved and have no vertices.
  • The nose and ears are shapes that have 1 more than 2 vertices.
  • The mouth, the body, and the tail are shapes that have more than 3 sides, but less than 5 sides and have L-shaped vertices.
  • The 4 legs are shapes that have 1 less than 5 sides and have all equal side lengths.

2 Count and write how many of each shape you draw.

Question 1.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 1
Answer:

Explanation:
To form the above cat picture we need to draw 1 hexagon shape for face, 2 circles for eyes , 3 triangles for nose and ears,4 rectangles for body and tail of the cat and 2 squares for the legs.

Question 2.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 2
Answer:

Explanation:
In the above cat picture there are 2 circles, 2 squares, 1 hexagon, 4 rectangles and 3 triangles.

Identify Two-Dimensional Shapes Activity

Shape Flip and Find

Directions:
Place the Shape Flip and Find Cards face down in the boxes. Take turns flipping 2 cards. If your cards show the same shape, keep the cards. If your cards show different shapes, flip the cards back over. Repeat until all cards have been used.

Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes 3
Answer:

Identify Two-Dimensional Shapes Chapter Practice

Directions:
1 Color the shape that has only 3 vertices. 2 Color the shape that has a curve. 3 Color the shapes that have 6 sides. 4 Color the shapes that have 4 vertices.

11.1 Describe Two-Dimensional Shapes

Question 1.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 1
Answer:

Explanation:
In the above shapes color the triangle as the triangle has only 3 vertices.

Question 2.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 2
Answer:

Explanation:
In the above shapes color the circle as circle has a curve.

Question 3.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 3
Answer:

Explanation:
In the above shapes color the hexagons as the hexagons have 6 sides and 6 vertices.

Question 4.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 4
Answer:

Explanation:
In the above shapes color the square and the rectangle as the square and fectangle has 4 sides and 4 vertices.

Directions:
5 and 6 Color the triangle. Tell why your answer is correct. 7 and 8 Color any rectangles. Tell why your answers are correct.

11.2 Triangles

Question 5.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 5
Answer:

Explanation:
In the above shapes color the triangle as the triangle has 3 sides and 3 vertices.

Question 6.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 6
Answer:

Explanation:
In the above shapes color the triangle as the triangle has 3 sides and 3 vertices.

11.3 Rectangles

Question 7.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 7
Answer:

Explanation:
In the above shapes color the rectangle as the trectangle has 4 sides and 4 vertices.

Question 8.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 8
Answer:

Explanation:
In the above shapes color the rectangle as the rectangle has 4 sides and 4 vertices.

Directions:
9 Use squares and rectangles to draw 5 windows and 1 door on the bus. Color the bus. Then write an addition sentence to tell how many squares and rectangles you draw in all. 10 Trace the shapes that are hexagons. Write the number of sides and the number of vertices of a hexagon.

11.4 Squares

Question 9.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 9
Answer:

Explanation:
In the above bus picture given above draw 5 square shaped windows, 1 rectangular shaoed door and color all the shapes.Therefore the addition sentence of the shapes is 5+1=6.

11.5 Hexagons and Circles

Question 10.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 10
Answer:

Explanation:
In the above shapes trace the hexagon as the hexagon has 6 sides and 6 vertices.

Directions:
11 Use the pattern block shown to form the shape. Count and write how many pattern blocks you use. 12 Draw a larger triangle that can be formed by the 2 triangles shown. 13 Use your materials to build the shape. Draw your shape or attach it to the page. 14 Use your materials to build a two-dimensional shape that has 4 sides of the same length. Draw your shape or attach it to the page. 15 Use your materials to build a two-dimensional shape that is not a hexagon. Draw your shape or attach it to the page.

11.6 Join Two-Dimensional Shapes

Question 11.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 11
Answer:

Explanation:
In the above picture 4 triangle shaped pattern blocks are used to form the given shape.

Question 12.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 12
Answer:

Explanation:
The above given 2 triangle shaped pattern blocks are used to form a large triangle.

11.7 Build Two-Dimensional Shapes

Question 13.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 13
Answer:

Question 14.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 14
Answer:

Question 15.
Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes chp 15
Answer:

Conclusion:
Hope Big Ideas Math Answers Grade K Chapter 11 Identify Two-Dimensional Shapes pdf is beneficial for all the elementary school students. Make use of the above links and kickstart your preparation. Keep in touch with our site to get the solution key for all Big Ideas Math Grade K Chapters.

Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares

Big Ideas Math Answers Grade 1 Chapter 14

Students who are interested to learn Grade 1 concepts in depth can get a strong foundation by using Big Ideas Math Book Solution Key. People of highly subject expertise will prepare the solutions in a concise manner for easy grasping. Start answering all the questions given in exams with the help of Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares. Refer to our Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares to enhance your performance skills.

Big Ideas Math Book 1st Grade Answer Key Chapter 14 Equal Shares

In this chapter, we discuss Equal shares, Partition Shapes into Halves, Fourths. Apply Maths in your Real-Time and get the Tips & Tricks to Solve Various Problems using Big Ideas Math Book 1st Grade Answer Key Chapter 14 Equal Shares. We have provided shortcuts to solve all the Problems in Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares. Try to utilize the resource BIM Answers Grade 1 Ch 14 Equal Shares and make your preparation way more efficient.

Vocabulary

Lesson: 1 Equal Shares

Lesson: 2 Partition Shapes into Halves

Lesson: 3 Partition Shapes into Fourths

Performance Task

Equal Shares Vocabulary

Organize It

Review Words:
bar graph
picture graph

Use the review words to complete the graphic organizer.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 1
Answer:
In the above, given 2 figures the first figure indicates the bar graph and the second figure indicates the picture graph.
The number of students who like the subject math = 6
The number of students who like the subject science = 5

Explanation:
In the above, given 2 figures the first figure indicates the bar graph and the second figure indicates the picture graph.
The number of students who like the subject math = 6
The number of students who like the subject science = 5

Define It
Use your vocabulary cards to complete the puzzle.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 2
Answer:
In the above-given figures, the across indicates the horizontal lines.
down indicates the vertical lines.

Explanation:
In the above-given figures, the across indicates the horizontal lines.
down indicates the vertical lines.

Lesson 14.1 Equal Shares

Explore and Grow

Sort the Equal Shares Sort Cards.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 3
Answer:
The equal shares sort cards are equal, half, third, fourth.

Explanation:
The equal shares sort cards are equal, half, third, fourth.
half = divided into 2 parts.
third = divided into 3 parts.
fourth = divided into 4 parts.
Show and Grow

Circle the shape that shows equal shares.

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 4
Answer:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-14-Equal-Shares-4

Explanation:
In the above 2 given figures, the second figure is divided into equal halves
Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 5
Answer:

Explanation:
In the above 2 given figures, the second figure is divided into equal halves

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 6
Answer:

Explanation:
In the above 2 given figures, the first figure is divided into equal halves

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 7
Answer:

Explanation:
In the above 2 given figures, the first figure is divided into equal halves

Apply and Grow: Practice

Circle the shape that shows equal shares.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 8
Answer:

Explanation:
In the above 2 given figures, the first figure is divided into equal halves.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 9
Answer:

Explanation:
In the above 2 given figures, the second figure is divided into equal halves.

Question 7.
Answer:

Explanation:
In the above 2 given figures, the second figure is divided into equal halves.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 12
Answer:

Explanation:
In the above 2 given figures, the first and second figure is divided into equal halves.

Question 9.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 13
Answer:

Explanation:
In the above-given figure, the figure is divided into 2 equal halves.

Question 10.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 14
Answer:
The given figure is not divided into equal halves.

Explanation:
In the above-given figure, the figure is divided into 2 unequal halves.

Question 11.
YOU BE THE TEACHER
Newton says the shape shows equal shares. Is he correct? Explain.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 15
Answer:
No, he was not correct.

Explanation:
The shape of the trapezoid is not equal.

Think and Grow: Modeling Real Life

You and your friend each design a kite. Your kite has 2 equal shares. Your friend’s has 2 unequal shares. Draw to show the parts.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 16
Answer:

Explanation:
My kite has 2 equal shares and my friend kite has 2 unequal shares

Show and Grow

Question 12.
You and your friend each design a poster. Your poster has 4 unequal shares. Your friend’s has 4 equal shares. Draw to show the parts.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 17
Answer:

Explanation:
In the above given 2 figures my friend divided the rectangle into 4 equal parts.
I divided the rectangle into 4 unequal parts.

Equal Shares Practice 14.1

Circle the shape that shows equal shares.

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 18
Answer:
In the above 2 figures, the first figure shows the equal halves.

Explanation:
There are four boxes in each figure but in the first figure, the boxes are in equal size.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 19
Answer:
In the above 2 figures, the second figure shows the equal halves.

Explanation:
There are four two halves in each figure but in the first figure, the boxes are not in equal size.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 20
Answer:
In the above 2 figures, the second figure shows the equal halves.

Explanation:
There are four halves in 1st figure but in the second figure, the boxes are in equal size.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 21
Answer:
In the above 2 figures, the first figure shows the equal halves.

Explanation:
There are four halves in 1st figure but in the second figure, the boxes are not in equal size.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 22
Answer:
In the above figure, the figure shows the 4 equal halves.

Explanation:
There are four halves in the figure and, the boxes are in equal size.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 23
Answer:
The square is divided into two equal halves.

Explanation:
In the above figure the square is divided into 2 parts .
But it is divided in non equal halves.

Question 7.
MP Precision
Descartes makes a thank you card with 4 equal shares. Which cord does Descartes make?
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 24
Answer:
1st card makes the 4 equal shares.

Explanation:
In the first figure, the word Thank you so much! makes the 4 equal shares in the same size.
In the second figure, the word Thank you so much does not make equal shares.
so the first figure is correct.

Question 8.
Modeling Real Life
You and your friend each design a sticker. Your sticker has 2 unequal shares. Your friend’s has 2 equal shares. Draw to show the parts.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 25
Answer:

Explanation:
In the above heart shapes they said to draw that i has unequal shaped heart.
my friend has equal shaped heart.

Review & Refresh

Make quick sketches to find the sum.

Question 9.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 26
Answer:
57
Explanation:
In the ones place the given digits are 5 + 2 = 7
In the tens place the given digits are 3+2 = 5

Question 10.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 27
Answer:
76
Explanation:
In the ones place the given digits are 1 + 5 = 6
In the tens place the given digits are 6+ 1 = 7

Lesson 14.2 Partition Shapes into Halves

Explore and Grow

Build hexagons with the paffern blocks shown. Circle the hexagon that shows 2 equal shares.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 28
Answer:
In the above-given figures, the first figure has two equal halves.

Explanation:
In the above given 3 figures the first figure has two equal halves.
In the given 1st hexagon = 2 trapezoids
2nd hexagon = 3 parallelograms
3rd hexagon = 6 triangles

Show and Grow

Circle the shape that shows halves.

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 29
Answer:
In the above 2 figures, the ist figure shows the equal halves.

Explanation:
There are two boxes in each figure but in the first figure, the boxes are in equal size.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 30
Answer:
In the above 2 figures, the second figure shows the equal halves.

Explanation:
There are two boxes in each figure but in the second figure, the boxes are in equal size.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 31
Answer:
In the above 2 figures, the second figure shows the equal halves.

Explanation:
There are two boxes in each figure but in the second figure, the boxes are in equal size.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 32
Answer:
In the above 2 figures, the first figure shows the equal halves.

Explanation:
There are two boxes in each figure but in the first figure, the boxes are in equal size.

Apply and Grow: Practice

Circle the shapes that show halves.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 33
Answer:
In the above 4 figures, the first and the third figure shows the equal halves.

Explanation:
There are two halves in each figure but in the first and the 3rd figure, the boxes are in equal size.

Question 6.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 34
Answer:
In the above 4 figures, the second and the fourth figure shows the equal halves.

Explanation:
There are two halves in each figure but in the second and the fourth figure, the boxes are in equal size.

Color half of the shape.

Question 7.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 35
Answer:

Explanation:
In the above-given figure, half of the circle is filled.

Question 8.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 36
Answer:

Explanation:
In the above-given figure, half of the box is filled.

Question 9.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 37
Answer:

Explanation:
In the above-given figure, half of the square is filled.

Question 10.
MP Structure
Match each half with its whole.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 38
Answer:

Explanation:
The semicircle can fill the circle.
The rectangle can fill the rectangle.
The square can fill the rectangle.

Think and Grow: Modeling Real Life

Show three ways to cut the cheese in half.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 39
Answer:

Show and Grow

Question 11.
Show three ways to fold the rug in half.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 40
Answer:

Explanation:
In the above 3 figures we can divide the rug in many different ways.

Partition Shapes into Halves practice 14.2

Circle the shape that shows halves.

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 41
Answer:
In the above 2 figures, the first figure shows the equal halves.

Explanation:
There are two halves in each figure but in the first figure, the boxes are in equal size.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 42
Answer:
In the above 2 figures, the first figure shows the equal halves.

Explanation:
There are two halves in each figure but in the first figure, the boxes are in equal size.

Circle the shapes that show halves.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 43
Answer:
In the above 4 figures, the second and the third figure shows the equal halves.

Explanation:
There are two halves in each figure but in the second and the third figure, the boxes are in equal size.

Color half of the shape.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 44
Answer:
In the above 4 figures, the first and the third figure shows the equal halves.

Explanation:
There are two halves in each figure but in the first and the third figure, the boxes are in equal size.

Color half of the shape.

Color half of the shape.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 45
Answer:

Explanation:
In the above figure one part of the rectangle is filled.

Question 6.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 46
Answer:

Explanation:

In the above figure one part of the square is filled.

Question 7.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 47
Answer:

Explanation:
In the above figure one part of the circle is filled.

Question 8.
YOU BE THE TEACHER
Newton says there are only two ways to divide a rectangle into halves. Is he correct? Explain.

___________________________________________

___________________________________________
Answer:
No he was not correct.

Explanation:
rectangle can divided into 4 halves.

Question 9.
Modeling Real Life
Show three ways to fold the bandana in half.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 48
Answer:

Explanation:
In the above given the figures are divided into halves.

Review & Refresh

Question 10.
Circle the shapes that only have a curved surface.
Big Ideas Math Answers 1st Grade 1 Chapter 14 Equal Shares 49
Answer:
cone and cylinder have a curved surface.

Explanation:
In the above-given figures, cone and cylinder have the curved surface.

Lesson 14.3 Partition Shapes into Fourths

Explore and Grow

Sort the 2, 4, or Unequal Shares Sort Cards.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 50
Answer:

Show and Grow

Circle the shape that shows fourths.

Question 1.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 51
Answer:
In the above 2 figures, the first figure shows the equal halves.

Explanation:
There are four halves in each figure but in the first  figure, the boxes are in equal size.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 52
Answer:
In the above 2 figures, the second figure shows the equal halves.

Explanation:
There are four halves in each figure but in the second figure, the boxes are in equal size.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 53
Answer:
In the above 2 figures, the second figure shows the equal halves.

Explanation:
There are four halves in each figure but in the second  figure, the boxes are in equal size

Question 4.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 54
Answer:
In the above 2 figures, the first figure shows the equal halves.

Explanation:
There are four halves in each figure but in the figure, the boxes are in equal size

Apply and Grow: Practice

Circle the shapes that show fourths.

Question 5.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 55
Answer:
In the above 4 figures, the first, second, fourth figure shows fourths.

Explanation:
There are four halves in each figure but in the third  figure, the boxes are in halves.

Apply and Grow: Practice

Color a quarter of the shape.

Question 6.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 56
Answer:

Explanation:
In the above-given circle, one part of the circle is filled with color.

Question 7.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 57
Answer:

Explanation:
In the above-given Rectangle, one part of the circle is filled with color.

Question 8.
MP Precision
Draw more lines to show fourths.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 58
Answer:

Explanation:
In the above-given square, one part of the square is given, so we have to add  2 lines to show fourths.

Question 9.
DIG DEEPER!
You cut a circle into halves. Your friend cuts the same-sized circle into quarters. Who has the larger pieces? Think: How do you know?
You Friend
Answer:
I have the larger pieces.

Explanation:
My friend cuts the circle into quarts.
I cut the circle into halves.
I have the larger pieces.

Think and Grow: Modeling Real Life

You cut a pizza into quarters. Your friend eats 1 quarter. How many more friends could have a piece of pizza?
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 59

___________ friends
Answer:
3 friends

Explanation:
In the above-given pizza, the pizza is cut into quarters. My friend eats 1 quarter. so the remaining is 3 pieces.
3 friends have a piece of pizza

Show and Grow

Question 10.
You cut a granola bar into quarters. Your friend eats 2 quarters. How many more friends could have a piece of the granola bar?
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 60

___________ friends
Answer: 2 friends

Explanation:
In the above-given granola bar, the granola bar is cut into quarters. My friend eats 2 quarter. so the remaining is 2 quarters.
2 friends have a piece of a granola bar.

Partition Shapes into Fourths practice 14.3

Circle the shape that shows fourths.

Question 1.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 61
Answer:
In the above 2 figures, the first, second, figure fourth figure shows fourths.

Explanation:
There are four halves in each figure but in the second figure, the boxes did not have the same size.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 62
Answer:
In the above 2 figures, the first, second,  figure shows fourths.

Explanation:
There are four halves in each figure but in the second figure, the boxes have the same size.

Circle the shapes that show fourths.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 63
Answer:
In the above 4 figures, the first, third, and fourth figure shows fourths.

Explanation:
There are four fourths in each figure but in the second figure, the boxes are divided into 2 halves.
The remaining three figures are divided into quarts.

Circle the shapes that show fourths.

Color a quarter of the shape.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 64
Answer:

Explanation:
In the above-given figure, one part of the rectangle is filled with color.

Question 5.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 65
Answer:

Explanation:
In the above-given figure, one part of the circle is filled with color.

Question 6.
DIG DEEPER!
Which shape shows a fourth of a circle?
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 66
Answer:
The first figure shows the fourth of a circle.

Explanation:
In the above given 2 figures, the first figure shows the fourth of a circle.

Question 7.
MP Reasoning
Color half of the square. How many fourths did you color?
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 67
___________ fourths
Answer:
2 fourths

Explanation:

Question 8.
Modeling Real Life
You cut a slice of bread into quarters. Your friend eats 3 quarters. How many more friends could have a piece of bread?
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 68
___________ friend
Answer: 1 friend

Explanation:
We cut a slice of bread into quarters.
My friend eats 3  quarters so 1 friend could have a piece of bread.
Review & Refresh

Draw to show the time.

Question 9.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 69
Answer:

Explanation:
The time is 5:00

Question 10.
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 70
Answer:

Explanation:
The time is 7:30

Equal Shares Performance Task

Question 1.
You, your friend, and your cousin are having a picnic. Use the clues to match each person with a food item.

  • You bring an item that is cut into 4 unequal shares.
  • Your friend brings an item that is cut into halves.
  • Your cousin brings an item that is cut into quarters.
  • Answer:

Explanation:
I cut the item into 4 unequal shares
my friend cut the item into 2 equal shares
my cousin cut the item into 4 equal shares

Question 2.
a. You cut an apple into 2 equal shares. You cut each share in half. How many equal shares do you have now?
Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares 72

_____________ equal shares
Show how you know:
Answer:
4 equal shares we have now

Explanation:
We have to cut the apple into 2 equal shares.
Again we have to cut the 2 equal shares into 2 equal halves.

b. You give your friend a fourth of the apple. How many shares do you have left?

_____________ shares
Answer:
3 shares I have left

Explanation:
The apple is cut into 4 quarters.
I gave a friend a fourth of the apple
I have the 3 shares left

Equal Shares Chapter Practice

Equal Shares Homework & Practice 14.1

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 14 Equal Shares 73
___________ equal shares
Answer:
2 equal shares

Explanation:
In the above-given trapezoid, the figure is divided into 2 equal shares

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 14 Equal Shares 74
___________ equal shares
Answer:
4 equal shares

Explanation:
In the above-given parallelogram, the figure is divided into 4 equal shares

Question 3.
Modeling Real Life
Newton and Descartes each design a place mat. Newton’s has 4 equal shares. Descartes’s has 4 unequal shares. Draw to show the parts.
Big Ideas Math Solutions Grade 1 Chapter 14 Equal Shares 75
Answer:

Explanation:
In the above figure, the first figure is divided into 4 equal shares.
The second figure is divided into 4 unequal shares.

Partition Shapes into Halves Homework & Practice 14.2

Question 4.
Circle the shapes that show halves.
Big Ideas Math Solutions Grade 1 Chapter 14 Equal Shares 76
Answer:
In the above 4 figures, the first and fourth figure shows halves

Explanation:
There are two halves in each figure but in the second and fourth figure, the boxes are divided into 2 unequal halves.
The remaining three figures are divided into quarts.

Question 5.
MP Structure
Match each half with its whole.
Big Ideas Math Solutions Grade 1 Chapter 14 Equal Shares 77
Answer:

Explanation:
The triangle can fill in the square
The semicircle can fill in the circle
The square can fill in rectangle

Partition Shapes into Fourths Homework & Practice 14.3

Question 6.
Circle the shapes that show fourths.
Big Ideas Math Solutions Grade 1 Chapter 14 Equal Shares 78
Answer:
In the above 4 figures, the first, second, third and fourth figure shows fourths.

Explanation:
There are four fourths in each figure but in the 1st  and third figure, the boxes are divided into 4 equal fourths.
The remaining four figures are not equal divided into quarts.

Color a quarter of the shape.

Question 7.
Big Ideas Math Solutions Grade 1 Chapter 14 Equal Shares 79
Answer:

Explanation:
In the above-given figure, one part is filled with colour.

Question 8.
Big Ideas Math Solutions Grade 1 Chapter 14 Equal Shares 80
Answer:

Explanation:
In the above-given figure, one part is filled with color.

Equal Shares Cumulative Practice

Question 1.
Shade the circle next to the equation that tells how many fewer students chose manga comics than superhero comics.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 81
○ 5 + 8 + 2 = 15
○ 8 – 5 = 3
○ 8 – 2 = 10
○ 8 – 2 = 6
Answer: 8-5 = 3

Explanation:
3 fewer students chose manga comics than superhero comics.

Question 2.
Shade the circle next to the number that tells how many minutes are in a half hour.
○ 15
○ 30
○ 45
○ 60
Answer:
30 minutes

Explanation:
There are 60 minutes in an hour.
30 minutes in half hour

Question 3.
Shade the circle next to the shape that does not show fourths.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 82
Answer:
In the above 4 figures, the first, second,  and fourth figure shows fourths.

Explanation:
There are four fourths in each figure but in the 1st, second and fourth figure, the boxes are divided into 4 equal fourths.
The remaining third figure not divided into equal quarts.

Question 4.
Shade the circle next to the shape that has no straight sides.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 83
Answer:
the circle has no straight sides.

Explanation:
In the above, given four figures the second figure has no straight sides.

Question 5.
Shade the circle next to the difference.
17 – 8 = ________
○ 8
○ 9
○ 10
○ 12
Answer: 9

Explanation:
17-8=9

Question 6.
Draw lines to show halves.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 84
Answer:

Explanation:
In the above figure, the figures are divided into 2 halves

Question 7.
Write the time on the clock two ways.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 85
Answer:
8: 30

Explanation:
The time = 8 : 30
the hour’s hand is on 8, the minute’s hand is on 6
so the time = 8: 30

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 86
Answer:
vertices = 7
edges = 18
flat surfaces = 3

Explanation:
The vertices in the cube = 7
The edges in the cube= 18
The flat surfaces in the cube = 3

Question 9.
Shade the circles next to the choices that match the model.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 87
○ 7 ones
○ 2 tens + 5 ones
○ 70
○ 20 + 50
Answer:
2 tens  +  5 ones

Explanation:
In the above figure the there are 2 tens and 5 ones

Question 10.
Shade the circles next to the choices that show the shapes you can use to make a Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 88.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 89
Answer:
Trapezoid and parallelogram are used to make a hexagon

Explanation:
In the above 4 figures squares and triangles are not forming the hexagon.

Question 11.
Tell how many equal shares.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 90
Answer:
Triangle = 2 equal shares
square = 4 equal shares

Explanation:
In the above 2 figures the triangle = 2 equal shares
square = 4 equal shares

Question 12.
A group of students are at a park. 2 of them leave. There are 4 left. How many students were at the park to start?
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 91
__________ – __________ = __________ students
Answer:
There are 6 students at the park.

Explanation:
In the above question, given that 2 of them leave.
4 are left
6-2 = 4

Question 13.
Circle the shapes that make up the structure.
Big Ideas Math Answer Key Grade 1 Chapter 14 Equal Shares 92
Answer:
square and rectangle make the structure

Explanation:
In the above given 4 figures squares and rectangles are used to make the structure.
cylinder and cone are not used to make the structure.

Conclusion:
I hope the pdf regarding the Big Ideas Math Answers Grade 1 Chapter 14 Equal Shares is beneficial for you. Stay tuned to our page to get the latest updates regarding all Grade 1 Chapters in Big Ideas Math Answer Key.

Big Ideas Math Answers Grade K Chapter 2 Compare Numbers 0 to 5

Big Ideas Math Answers Grade K Chapter 2 Compare Numbers 0 to 5

Student’s who are in search of Big Ideas Math Answers Grade K Chapter 2 Compare Numbers 0 to 5? If our guess is correct, then you are on the right page. Learn the concepts in-depth with the help of Big Ideas Math Book K Grade Chapter 2 Compare Numbers 0 to 5 Solution Key. To excel in the exam, we suggest the elementary school students of the K Grade standard refer to the Big Ideas Math Answers Grade K Chapter 2 Compare Numbers 0 to 5. Hence Download Bigideas Math Kth Grade Answers Chapter 2 Compare Numbers 0 to 5 Pdf for free from here.

Big Ideas Math Book Grade K Answer Key Chapter 2 Compare Numbers 0 to 5

Make your learning more fun with the help of our Big Ideas Math Book K Grade Answer Key for Chapter 2 Compare Numbers 0 to 5. Get step-by-step explanations for all the questions by using the figures from BIM Grade 2 Answers Chapter 2 Compare Numbers 0 to 5. Just click on the below links and start practicing for the exams.

Vocabulary

Lesson: 1 Equal Groups

Lesson: 2 Greater Than

Lesson: 3 Less Than

Lesson: 4 Compare Groups to 5 by Counting

Lesson: 5 Compare Numbers to 5

Chapter 2 – Compare Numbers 0 to 5

Compare Numbers 0 to 5 Vocabulary

Big Ideas Math Answers Grade K Chapter 2 Compare Numbers 0 to 5 1
Big Ideas Math Answers Grade K Chapter 2 Compare Numbers 0 to 5 2

Answer:
The objects are:-
clouds = 3
toys = 5
slider = 1
cradle = 4

Explanation:
In the above-given figure,
given that count the objects. say the number. write the number.
clouds = 3
toys = 5
slider = 1
cradle = 4

Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2-1
Directions:
Count the objects. Say the number. Write the number.

Chapter 2 Vocabulary Cards

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 3
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 4

Lesson 2.1 Equal Groups

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 5

Answer:
The number of paintbrushes = 3
The number of paint jars = 4

Explanation:
In the above-given figure,
the number of paintbrushes = 3
the numbers of paint jars = 4
No there are no paintbrushes for each paint jars.
There are only 3 paintbrushes.
the paintbrushes = 4
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-1
Directions:
Use counters to show the number of paintbrushes. Use counters to show the number of paint jars. Is there a paint brush for each paint jar?

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 6

Answer:
In Group A the number of objects is equal.
In Group B the number of objects is not equal.

Explanation:
In the above-given figure,
In group A the number of red pencils is equal to the number of blue pencils.
In group B the number of green pencils is not equal to the number of purple pencils.

Directions:
Draw lines between the objects in each group. Are the numbers of objects equal? Circle the thumbs up for yes or the thumbs down for no.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 7

Answer:
The number of objects is not equal.

Explanation:
In the above-given figure,
the number of green pencils is not equal to the number of purple pencils.
so we have to circle down.

Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-2

Question 2.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 8

Answer:
The number of objects is equal.

Explanation:
In the above-given figure,
the number of blue balls is equal to the number of red balls.
so we have to circle up.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-3

Question 3.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 9

Answer:
The number of objects is not equal.

Explanation:
In the above-given figure,
the number of orange crayons is not equal to the number of pink crayons.
so we have to circle down.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-4

Question 4.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 10

Answer:
The number of objects is equal.

Explanation:
In the above-given figure,
the number of blue balls is equal to the number of red balls.
so we have to circle up.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-5

Directions:
1 – 4 Draw lines between the objects in each group. Are the numbers of objects equal? Circle the thumbs up for yes or the thumbs down for no.

Think and Grow: Modeling Real Life
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 11
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 12

Answer:
In group A the number of paint tubes and the number of paint spots is equal.

Explanation:
In the above-given figure,
the number of paint tubes and the number of paint spots are equal.
so we have to circle the thumb up.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-6

Answer:
In group B the number of paint tubes and the number of paint spots is not equal.

Explanation:
In the above-given figure,
the number of paint tubes and the number of paint spots is not equal.
so we have to circle the thumb down.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-7

Directions:

  • Draw paint spots so that the numbers of paint tubes and paint spots are equal. Draw lines between the objects in each group to show that you are correct.
  • Draw paint spots so that the numbers of paint tubes and paint spots are not equal. Draw lines between the objects in each group to show that you are correct.

Equal Groups Homework & Practice 2.1

Question 1.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 13

Answer:
The number of objects is equal.

Explanation:
In the above-given figure,
the number of blue bottles is equal to the number of yellow bottles.
so we have to circle up.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-8

Question 2.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 14

Answer:
The number of objects is not equal.

Explanation:
In the above-given figure,
the number of purple pencils is not equal to the number of red pencils.
so we have to circle down
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-9

Directions:
1 and 2 Draw lines between the counters in each group. Are the numbers of objects equal? Circle the thumbs up for yes or the thumbs down for no.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 15

Answer:
The number of objects is not equal.

Explanation:
In the above-given figure,
the number of orange bottles is not equal to the number of yellow bottles.
so we have to circle down.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-10

Question 4.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 16

Answer:
The number of objects is not equal.

Explanation:
In the above-given figure,
the number of blue brushes is not equal to the number of green brushes.
so we have to circle up.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-11

Question 5.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 17

Answer:
the number of erasers is equal to the number of pencils.

Explanation:
In the above-given question,
we have to draw the pencils for the erasers.
so the number of erasers is equal to the number of pencils.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.1-12
Directions:
3 and 4 Draw lines between the objects in each group. Are the numbers of objects equal? Circle the thumbs up for yes or the thumbs down for no. 5 Draw pencils so that the numbers of erasers and pencils are equal. Draw lines between the objects in each group to show that you are correct.

Lesson 2.2 Greater Than

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 18

Answer:
The number of suitcases is more than the number of stuffed animals.

Explanation:
In the above-given figure,
The number of suitcases is more than the number of stuffed animals.
suitcases are greater than the stuffed animals.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-1

Directions:
Use counters to show the number of suitcases. Use counters to show the number of stuffed animals. Are there more suitcases or more stuffed animals?

Think and Grow
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 19

Answer:
The number of cars is more than the number of airplanes.
so we have to circle the cars.

Explanation:
In the above-given 1st figure,
given that circle, the group is greater than the number of the other group.
The number of cars is more than the number of airplanes.
so we have to circle the cars.

Answer:
The number of mobiles is more than the number of watches.
so we have to circle the mobiles.

Explanation:
In the above-given 2nt figure,
given that circle, the group is greater than the number of the other group.
The number of mobiles is more than the number of watches.
so we have to circle the mobiles.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-2

Directions:
Draw lines between the objects in each group. Circle the group that is greater in number than the other group.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 20

Answer:
The number of pencils is more than the number of books.
so we have to circle the pencils.

Explanation:
In the above-given figure,
given that circle, the group is greater than the number of the other group.
The number of pencils is more than the number of books.
so we have to circle the pencils.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-3

Question 2.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 21

Answer:
The number of bottles is more than the number of apples.
so we have to circle the bottles.

Explanation:
In the above-given figure,
given that circle, the group is greater than the number of the other group.
The number of bottles is more than the number of apples.
so we have to circle the bottles.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-4

Question 3.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 22

Answer:
The number of shirts is more than the number of shorts.
so we have to circle the shirts.

Explanation:
In the above-given figure,
given that circle, the group is greater than the number of the other group.
The number of shirts is more than the number of shorts.
so we have to circle the shirts.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-5

Question 4.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 23

Answer:
The number of toothbrushes is more than the number of toothpaste.
so we have to circle the brushes.

Explanation:
In the above-given figure,
given that circle, the group is greater than the number of the other group.
The number of toothbrushes is more than the number of paste.
so we have to circle the brushes.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-6

Directions:
1 – 4 Draw lines between the objects in each group. Circle the group that is greater in number than the other group.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 24

Answer:
The number of photos is greater than the number of stickers.

Explanation:
In the above-given figure,
The number of photos is greater than the number of stickers.
given that draw lines between the object in each group.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-7

Directions:
Draw photos on the scrapbook so that the number of photos is greater than the number of stickers. Draw lines between the objects in each group to show that you are correct.

Greater Than Homework & Practice 2.2

Question 1.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 25

Answer:
The number of grey color coins is more than the number of yellow color coins.
so we have to circle the grey color coins.

Explanation:
In the above-given figure,
given that circle, the group is greater than the number of the other group.
The number of grey color coins is more than the number of yellow color coins.
so we have to circle the grey color coins.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-8

Question 2.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 26

Answer:
The number of spectacles is more than the number of toys.
so we have to circle the spectacles.

Explanation:
In the above-given figure,
given that circle, the group is greater than the number of the other group.
The number of spectacles is more than the number of toys.
so we have to circle the spectacles.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-9

Directions:
1 and 2 Draw lines between the counters in each group. Circle the group that is greater in number than the other group.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 27

Answer:
The number of mats is more than the number of spectacles.
so we have to circle the mats.

Explanation:
In the above-given figure,
given that circle, the group is greater than the number of the other group.
The number of mats is more than the number of spectacles.
so we have to circle the mats.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-110

Question 4.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 28

Answer:
The number of monitors is more than the number of pictures.
so we have to circle the monitors.

Explanation:
In the above-given figure,
given that circle, the group is greater than the number of the other group.
The number of monitors is more than the number of pictures.
so we have to circle the monitors.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-12

Question 5.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 29

Answer:
The number of photos is more than the number of stickers.
so we have to draw the lines between the objects.

Explanation:
In the above-given figure,
given that circle, the group is greater than the number of the other group.
The number of photos is more than the number of stickers.
so we have to draw the lines between the objects.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.2-13

Directions:
3 and 4 Draw lines between the objects in each group. Circle the group that is greater in number than the other group. 5 Draw photos on the scrapbook so that the number of photos is greater than the number of stickers. Draw lines between the objects in each group to show that you are correct.

Lesson 2.3 Less Than

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 30

Answer:
The number of holes is less than the number of rabbits.
There are fewer holes.

Explanation:
In the above-given figure,
The number of holes is less than the number of rabbits.
There are fewer holes.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-1

Directions:
Use counters to show the number of rabbits. Use counters to show the number of holes. Are there fewer rabbits or fewer holes?

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 31
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 32

Answer:
a.The number of pigeons is less than the number of clouds.
There are fewer pigeons.

Explanation:
In the above-given figure,
The number of pigeons is less than the number of clouds.
There are fewer pigeons.

Answer:
b.The number of aquariums is less than the number of fishes.
There are fewer aquariums.

Explanation:
In the above-given figure,
The number of aquariums is less than the number of fishes.
There are fewer aquariums.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-2

Directions:
Draw lines between the objects in each group. Draw a line through the group that is less in number than the other group.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 33

Answer:
The number of rooms is less than the number of horses.
There are fewer rooms.

Explanation:
In the above-given figure,
The number of rooms is less than the number of horses.
There are fewer rooms.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-3

Question 2.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 34

Answer:
The number of lands is less than the number of chicks.
There are fewer lands.

Explanation:
In the above-given figure,
The number of lands is less than the number of chicks.
There are fewer lands.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-4

Question 3.
Big Ideas Math Answers Grade K Chapter 2 Compare Numbers 0 to 5 35

Answer:
The number of spiders is less than the number of webs.
There are fewer webs.

Explanation:
In the above-given figure,
The number of spiders is less than the number of webs.
There are fewer webs.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-5

Question 4.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 36

Answer:
The number of rabbits is less than the number of borrows.
There are fewer borrows.

Explanation:
In the above-given figure,
The number of rabbits is less than the number of borrows.
There are fewer borrows.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-6

Directions:
1 – 4 Draw lines between the objects in each group. Draw a line through the group that is less in number than the other group.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 37

Answer:
The number of hummingbirds is less than the number of flowers.
There are fewer hummingbirds.

Explanation:
In the above-given figure,
The number of hummingbirds is less than the number of flowers.
There are fewer hummingbirds.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-7

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 38

Answer:
The number of flowers is less than the number of hummingbirds.
There are fewer flowers.

Explanation:
In the above-given figure,
The number of flowers is less than the number of hummingbirds.
There are fewer flowers.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-8

Directions

  • Draw flowers on the bush so that the number of hummingbirds is less than the number of flowers. Draw lines between the objects in each group to show that you are correct.
  • Draw flowers on the bush so that the number of owers is less than the number of hummingbirds. Draw lines between the objects in each group to show that you are correct.

Less Than Homework & Practice 2.3

Question 1.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 39

Answer:
The number of holes is less than the number of ants.
There are fewer holes.

Explanation:
In the above-given figure,
The number of holes is less than the number of ants.
There are fewer holes.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-9

Question 2.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 40

Answer:
The number of caterpillars is less than the number of leaves.
There are fewer caterpillars.

Explanation:
In the above-given figure,
The number of caterpillars is less than the number of leaves.
There are fewer caterpillars.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-10

Directions:
1 and 2 Draw lines between the objects in each group. Draw a line through the group that is less in number than the other group.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 41

Answer:
The number of grass is less than the number of foxes.
There is fewer grass.

Explanation:
In the above-given figure,
The number of grass is less than the number of foxes.
There is fewer grass.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-11

Question 4.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 42

Answer:
The number of snakes is less than the number of logs.
There are fewer logs.

Explanation:
In the above-given figure,
The number of snakes is less than the number of logs.
There are fewer logs.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-12

Question 5.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 43

Answer:
The number of berries is less than the number of bears.
There is one berry and there are four bears.

Explanation:
In the above-given figure,
The number of berries is less than the number of bears.
There are fewer berries.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.3-13

Directions:
3 and 4 Draw lines between the objects in each group. Draw a line through the group that is less in number than the other group. 5 Draw berries on the bush so that the number of berries is less than the number of bears. Draw lines between the objects in each group to show that you are correct.

Lesson 2.4 Compare Groups to 5 by Counting

Explore and Grow
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 44

Answer:
Group A has more red jerseys.
Group B has fewer yellow jerseys.

Explanation:
In  the above-given figure,
There are yellow and red jerseys.
the yellow jerseys are less than the red jerseys.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.4-1
Directions:
Use counters to show the number of yellow jerseys. Use counters to show the number of red jerseys. Tell which group has more and which group has less.

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 45

Answer:
The number of bats = 3
The number of balls = 1

Explanation:
In the above-given figure,
the shuttle bats = 3
balls = 1

Answer:
The number of bats = 4
The number of balls = 3

Explanation:
In the above-given figure,
the hockey bats = 4
balls = 3
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.4-2

Directions:
Count the objects in each group. Write each number.

  • Is the number of purple dots equal to the number of green dots? Circle the thumbs up for yes or the thumbs down for no.
  • Compare the numbers of tennis rackets and tennis balls. Circle the number that is greater than the other number.
  • Compare the numbers of hockey sticks and hockey pucks. Draw a line through the number that is less than the other number.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 47

Answer:
The number of objects is not equal.

Explanation:
In the above-given figure,
the number of balls is not equal to the number of nets.
so we have to circle down.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.4-3

Question 2.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 48

Answer:
The number of bubbles is greater than the number of stickers.

Explanation:
In the above-given figure,
The number of bubbles is greater than the number of stickers.
given that draw lines between the object in each group.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.4-4

Question 3.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 49

Answer:
The number of balls is less than the number of hands.
There are fewer balls.

Explanation:
In the above-given figure,
The number of balls is less than the number of hands.
There are fewer balls.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.4-5

Directions:
Count the objects in each group. Write each number. 1 Is the number of soccer balls equal to the number of soccer goals? Circle the thumbs up for yes or the thumbs down for no. 2 Circle the number that is greater than the other number. 3 Draw a line through the number that is less than the other number.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 50

Answer:
The number of golf balls is greater than the number of golf clubs.

Explanation:
In the above-given figure,
The number of balls is greater than the number of golf clubs.
given that draw lines between the object in each group.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.4-6

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 51

Answer:
The number of balls is less than the number of golf clubs.
There are fewer balls.

Explanation:
In the above-given figure,
The number of balls is less than the number of golf clubs.
There are fewer balls.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.4-7

Directions:

  • Draw golf balls on the ground so that the number of golf balls is greater than the number of golf clubs. Write the number of each object. Circle the number that is greater than the other number.
  • Draw golf balls on the ground so that the number of golf balls is less than the number of golf clubs. Write the number of each object. Draw a line through the number that is less than the other number.

Compare Groups to 5 by Counting Homework & Practice 2.4

Question 1.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 52

Answer:
No.

Explanation:
In the above-given figure,
the number of basketball hoops is not equal to the number of basketballs.
Given that circle the thumbs up for yes or the thumbs down for no.
so we have to circle down.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.4-8

Directions:
1 Count the objects in each group. Write each number. Is the number of basketball hoops equal to the number of basketballs? Circle the thumbs up for yes or the thumbs down for no.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 53

Answer:
The number of lamps is greater than the number of holes.

Explanation:
In the above-given figure,
The number of lamps is greater than the number of holes.
given that draw lines between the object in each group.

Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.4-9

Question 3.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 54

Answer:
The number of pom-poms is less than the number of megaphones.
There is one pom-pom and there are three megaphones.

Explanation:
In the above-given figure,
The number of pom-poms is less than the number of megaphones.
There are fewer pom-poms.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.4-10

Directions:
2 Count the objects in each group. Write each number. Circle the number that is greater than the other number. 3 Draw pom-poms on the floor in the picture so that the number of pom-poms is less than the number of megaphones. Write the number of each object. Draw a line through the number that is less than the other number.

Lesson 2.5 Compare Numbers to 5

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 55

Answer:
5 is greater than the other number 3.
3 is less than the other number 5.

Explanation:
Given that the which number is greater than the other number.
given numbers are 3 and 5.
5 is greater than the other number 3.
3 is less than the other number 5.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-1

Directions:
Which number is greater than the other number? Which number is less than the other number? Use counters to show how you know.

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 56

Answer:
Yes.
The numbers are equal.

Explanation:
Given that the numbers are equal.
yes, the numbers are equal.
so we have to circle up.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 56.1

Answer:
5 is greater than the 1, 3, 4.

Explanation:
In the above-given figure,
given that circle the number that is greater the other number.
5 is greater than the 1, 3, 4.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-2

Answer:
1 is less than the other number.

Explanation:
In the above-given figure,
given that draw, a line through the number is less than the other number.
1, 3, 4 is less than 5.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-3

Directions:
Compare the numbers.

  • Are the numbers equal? Circle the thumbs up for yes or the thumbs down for no. Draw to show how you know.
  • Circle the number that is greater than the other number. Draw to show how you know.
  • Draw a line through the number that is less than the other number. Draw to show how you know.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 57

Answer:
The numbers are not equal.

Explanation:
Given that the numbers are 3 and 2.
the numbers are not equal.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-4

Question 2.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 58

Answer:
2 is greater than the other number.

Explanation:
In the above-given figure,
the number  2 is greater than the other number 1.
given that the numbers are 1 and 2.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 59

Answer:
4 is less than 5.

Explanation:
In the above-given figure,
given that draw a line through the number that is less than the other number.
4 is less than 5.
the given numbers are 4 and 5.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-5

Directions:
Compare the numbers.
1 Are the numbers equal? Circle the thumbs up for yes or the thumbs down for no. Draw to show how you know. 2 Circle the number that is greater than the other number. Draw to show how you know. 3 Draw a line through the number that is less than the other number. Draw to show how you know.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 60

Answer:
a.The number equals to the number on the car is 4.

Explanation:
In the above-given figure,
the number which equals the number on the car is 4.
given that the number on the car is 4.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-6

Answer:
b.The number greater than the number on the car is 2.

Explanation:
In the above-given figure,
the number greater than the number on the car is 2.
given that the number on the car is 1.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-7

 

Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 61

Answer:
c.The number less than the number on the car is 4.

Explanation:
In the above-given figure,
the number less than the number on the car is 4.
given that the number on the car is 5.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-8

Directions:

  • Write the number that is equal to the number on the car. Draw to show how you know your number is correct.
  • Write a number that is greater than the number on the car. Draw to show how you know your number is correct.
  • Write a number that is less than the number on the car. Draw to show how you know your number is correct.

Compare Numbers to 5 Homework & Practice 2.5

Big Ideas Math Answers Grade K Chapter 2 Compare Numbers 0 to 5 61.1

Question 1.
Big Ideas Math Answer Key Grade K Chapter 2 Compare Numbers 0 to 5 62

Answer:
The number is equal.

Explanation:
In the above-given figure,
Given that the numbers are 1 and 1.
the numbers are equal.
so we have to circle up.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-9

Directions:
1 Compare the numbers. Are the numbers equal? Circle the thumbs up for yes or the thumbs down for no. Draw to show how you know.

Question 2.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 63

Answer:
The number are not equal.

Explanation:
In the above-given figure,
Given that the numbers are 4 and 2.
the numbers are not equal.
so we have to circle down.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-10

Question 3.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 64

Answer:
The number less than the number  5  is 2.

Explanation:
In the above-given figure,
the number less than the number 5 is 2.
given that the number are 2 and 5.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-11

Question 4.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 65

Answer:
c.The number greater than the number on the jersey is 4.

Explanation:
In the above-given figure,
the number greater than the number on the jersey is 4.
given that the number on the jersey is 3.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-12

Directions:
2 Compare the numbers. Are the numbers equal? Circle the thumbs up for yes or the thumbs down for no. Draw to show how you know. 3 Compare the numbers. Draw a line through the number that is less than the other number. Draw to show how you know. 4 Write a number that is greater than the number on the jersey. Draw to show how you know your number is correct.

Compare Numbers 0 to 5 Performance Task

Question 1.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 66

Answer:
The dots in each side of the domino = 5

Explanation:
In the above-given figure,
the dots on the left side = 1.
the dots on the right side = 4
the total number of dots on each side of the domino = 5
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-13

Question 2.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 67

Answer:
The number of 0’s is greater than the number of x’s.

Explanation:
In the above-given figure,
we have to draw the 0’s on the game board.
The number of 0’s is greater than the number of x’s.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-14

Question 3.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 68

Answer:
The numbers equal are 3.

Explanation:
In the above-given figure,
the numbers not equal are 4.
the numbers equal are 3.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-15

Directions:
1 Count the dots on each side of the domino. Write the 2 numbers. Draw a line through the number that is less than the other number. 2 Draw O’s on the game board so that the number of O’s is greater than the number of X’s. Write each number. Circle the number that is greater than the other number. 3 Write each number. Circle the numbers that are equal.

Compare Numbers 0 to 5 Activity

Toss and Compare

Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 69

Directions:
Take turns tossing a counter onto the board. If the counter lands on Newton or Descartes, choose any number from 0 to 5. Write the numbers on your Toss and Compare Numbers from 0 to 5 Recording Sheet. Circle the number that is greater than the other number. Circle both numbers if they are equal. Repeat this process until you fill your sheet.

Compare Numbers 0 to 5 Chapter Practice

2.1 Equal Groups

Question 1.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 70

Answer:
The number of objects is equal.

Explanation:
In the above-given figure,
the number of blue balls is equal to the number of green balls.
so we have to circle up.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-16

Question 2.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 71

Answer:
The number of objects is not equal.

Explanation:
In the above-given figure,
the number of ants is not equal to the number of ant holes.
so we have to circle down.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-17

Question 3.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 72

Answer:
The number of objects is not equal.

Explanation:
In the above-given figure,
the number of pencils not equal to the number of crayons.
so we have to circle down.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-18

Directions:
1 and 2 Draw lines between the objects in each group. Are the numbers of objects equal? Circle the thumbs up for yes or the thumbs down for no. 3 Draw pencils so that the numbers of crayons and pencils are equal. Draw lines between the objects in each group to show that you are correct.

2.2 Greater Than

Question 4.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 73

Answer:
The number of houses is greater than the number of dogs.

Explanation:
In the above-given figure,
The number of houses is greater than the number of dogs.
given that draw lines between the object in each group
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-19

2.3 Less Than

Question 5.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 74

Answer:
a.The number of logs is less than the number of dogs.
There are fewer logs.

Explanation:
In the above-given figure,
The number of logs is less than the number of dogs.
There are fewer logs.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-20

Directions:
Draw lines between the objects in each group.
4 Circle the group that is greater in number than the other group.
5 Draw a line through the group that is less in number than the other group

2.4 Compare Groups to 5 by Counting

Question 6.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 75

Answer:
The number of objects is not equal.

Explanation:
In the above-given figure,
the number of cycles is not equal to the number of helmets.
so we have to circle down.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-21

Question 7.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 76

The number of bottles is more than the number of balls.
so we have to circle the bottles.

Explanation:
In the above-given figure,
given that circle, the group is greater than the number of the other group.
The number of bottles is more than the number of balls.
so we have to circle the bottles.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-22

Question 8.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 77

Answer:
The number of tables is less than the number of golf.
There are fewer tables.

Explanation:
In the above-given figure,
The number of tables is less than the number of golf.
There are fewer tables.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-23

Directions:
Count the objects in each group. Write each number. 6 Is the number of helmets equal to the number of bikes? Circle the thumbs up for yes or the thumbs down for no. 7 Circle the number that is greater than the other number. 8 Draw a line through the number that is less than the other number.

2.5 Compare Numbers to 5

Question 9.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 78

Answer:
The number is equal.

Explanation:
In the above-given figure,
Given that the numbers are 5 and 5.
the numbers are equal.
so we have to circle up.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-24

Question 10.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 79

Answer:
The number greater than the number  3 is 4.

Explanation:
In the above-given figure,
the number greater than the number 3 is 4.
given that the number are 3 and 4.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-25

Question 11.
Big Ideas Math Solutions Grade K Chapter 2 Compare Numbers 0 to 5 80

Answer:
The number less than the number  4 is 2.

Explanation:
In the above-given figure,
the number less than the number 4 is 2.
given that the number are 2 and 4.
Big-Ideas-Math-Solutions-Grade-K-Chapter-2-Compare Numbers 0 to 5-2.5-26

Directions:
Compare the numbers.
9 Are the numbers equal? Circle the thumbs up for yes or the thumbs down for no. Draw to show how you know. 10 Circle the number that is greater than the other number. Draw to show how you know. 11 Draw a line through the number that is less than the other number. Draw to show how you know.

Conclusion:
All the Questions and Answers given on this page are as per the latest edition. Refer to the Big Ideas Math Answers Grade K Chapter 2 Compare Numbers 0 to 5 provided here and score maximum marks in the exams. We suggest the students make use of this platform to learn the concepts quickly and easily. Get a brief explanation for all the problems from our BIM Book Grade K Answer Key Chapter 2 Compare Numbers 0 to 5.

Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens

Big Ideas Math Answers Grade 1 Chapter 8

Big Ideas Math Book 1st Grade Answer Key Chapter 8 Add and Subtract Tens helps the students to overcome the difficulties in math. Learning in a fun way will make maths more interesting to students. You can see step-by-step explanations for all the questions by using the figures. You can easily understand the concepts deeply with the help of our Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens.

Big Ideas Math Book 1st Grade Answer Key Chapter 8 Add and Subtract Tens

In order to excel in the exam, we suggest the students of 1st standard refer to the Big Ideas Math Book 1st Grade Answer Key Chapter 8 Add and Subtract Tens. The topics shown in Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens are as per the latest syllabus. Hit the links and start solving the problems using easy methods.

Vocabulary

Lesson: 1 Mental Math: 10 More

Lesson: 2 Mental Math: 10 Less

Lesson: 3 Add Tens

Lesson: 4 Add Tens Using a Number Line

Lesson: 5 Subtract Tens

Lesson: 6 Subtract Tens Using a Number Line

Lesson: 7 Use Addition to Subtract Tens

Lesson: 8 Add Tens to a Number

Chapter: 8 Add and Subtract Tens

Add and Subtract Tens Vocabulary

Organize It

Review Words:
decade numbers
digits

Use the review words to complete the graphic organizer.

Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 1
The ____________ of 16 are 1 and 6.

Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-key-Chapter-8-Add-and-Subtract-Tens-Add-and-Subtract-Tens-Vocabulary

The digits of 16 Are 1 and 6.

Explanation:
10, 20, 30, 40 and so on are decade numbers.
Any of the numerals from 0 to 9, especially when forming part of a number are called as digits.

Define It

What am I?
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 2

Answer:

Big-Ideas-Math-Book-1st-Grade-Answer-key-Chapter-8-Add-and-Subtract-Tens-Define It

 

Explanation:
13 – 3 = 10= B                  4 + 4 = 8 = P
5 + 1 = 6 = O                   3 + 2 = 5 = L
5 + 4 = 9 = I                    12 – 8 = 4 = R
6 – 4 = 2 = M                   10 – 3 = U
1 + 2 = 3 = E
9 – 8 = 1 = N
Hence , YOU are open number line.

Lesson 8.1 Mental Math: 10 More

Explore and Grow

Find each sum. What do you notice?

13 + 10 = __________

39 + 10 = __________

52 + 10 = __________

Answer:
13 + 10 = 23
39 + 10 = 49
52 + 10 = 62

Explanation:
Addition is bringing two or more numbers (or things) together to make a new total.
The numbers to be added together are called the “Addends”:
1. Here, 13 and 10 are addends.
Now add both the numbers:
13 + 10 = 23.
2. Here, 39 and 10 are addends.
39 + 10 = 49.
3. Here, 52 and 10 are addends.
52 + 10 = 62.
Hence, its noticed that total number of groups are 10 in all the counters.

Show and Grow

Use mental math.

Question 1.
14 + 10 = __________
Answer:
14 + 10 = 24

Explanation:
Here, 14 and 10 are addends.
Now add both the numbers:
14 + 10 = 24.
Hence, Sum = 24.

Question 2.
46 + 10 = __________
Answer:
46 + 10 =56.

Explanation:
Here, 46 and 10 are addends.
Now add both the numbers:
46 + 10 = 56.
Hence, Sum = 56.

Question 3.
83 + 10 = __________
Answer:
83 + 10 = 93.

Explanation:
Here, 83 and 10 are addends.
Now add both the numbers:
83 + 10 = 93.
Hence, Sum = 93.

Question 4.
75 + 10 = __________
Answer:
75 + 10 = 85.

Explanation:
Here, 75 and 10 are addends.
Now add both the numbers:
75 + 10 = 85.
Hence, sum = 85.

Question 5.
21 + 10 = __________
Answer:
21 + 10 = 31.

Explanation:
Here, 21 and 10 are addends.
Now add both the numbers:
21 + 10 = 31.
Hence, Sum = 31.

Question 6.
60 + 10 = __________
Answer:
60 + 10 = 70.

Explanation:
Here, 60 and 10 are addends.
Now add both the numbers:
60 + 10 = 70.
Hence, Sum = 70.

Question 7.
10 + 89 = __________
Answer:
10 + 89 = 99.

Explanation:
Here, 10 and 89 are addends.
Now add both the numbers:
10 + 89 = 99.
Hence, Sum = 99.

Question 8.
10 + 68 = __________
Answer:
10 + 68 = 78.

Explanation:
Here, 10 and 68 are addends.
Now add both the numbers:
10 + 68 = 78.
Hence, Sum = 78.

Apply and Grow: Practice

Use Mental math.

Question 9.
16 + 10 = __________
Answer:
16 + 10 = 26.

Explanation:
Here, 16 and 10 are addends.
Now add both the numbers:
16 + 10 = 26.
Hence, Sum = 26.

Question 10.
63 + 10 = __________
Answer:
63 + 10 = 73.

Explanation:
Here, 63 and 10 are addends.
Now add both the numbers:
63 + 10 = 73.
Hence, Sum = 73.

Question 11.
8 + 10 = __________
Answer:
8 + 10 = 18.

Explanation:
Here, 8 and 10 are addends.
Now add both the numbers:
8 + 10 = 18.
Hence, Sum = 18.

Question 12.
44 + 10 = __________
Answer:
44 + 10 = 54.

Explanation:
Here, 44 and 10 are addends.
Now add both the numbers:
44 + 10 = 54.
Hence, Sum = 54.

Question 13.
19 + 10 = __________
Answer:
19 + 10 = 29.

Explanation:
Here, 19 and 10 are addends.
Now add both the numbers:
19 + 10 = 29.
Hence, Sum = 29.

Question 14.
59 + 10 = __________
Answer:
59 + 10 = 69.

Explanation:
Here, 59 and 10 are addends.
Now add both the numbers:
59 + 10 = 69.
Hence, Sum = 69.

Question 15.
10 + 22 = __________
Answer:
10 + 22 = 32.

Explanation:
Here, 10 and 22 are addends.
Now add both the numbers:
10 + 22 = 32.
Hence, Sum = 32.

Question 16.
10 + 50 = __________
Answer:
10 + 50 = 60.

Explanation:
Here, 10 and 50 are addends.
Now add both the numbers:
10 + 50 = 60.
Hence, Sum = 60.

Question 17.
10 + 71 = __________
Answer:
10 + 71 = 81.

Explanation:
Here, 10 and 71 are addends.
Now add both the numbers:
10 + 71 = 81.
Hence, Sum = 81.

Question 18.
10 + 38 = __________
Answer:
10 + 38 = 48.

Explanation:
Here, 10 and 38 are addends.
Now add both the numbers:
10 + 38 = 48.
Hence, Sum = 48.

Question 19.
55 + __________ = 65
Answer:
55 + _10__ = 65

Explanation:
Here, 55 and Missing number are addends.
Sum is given = 65
Let the missing number be X.
Now add the numbers:
55 + X= 65
X = 65 – 55
X = 10.
Hence, equation is  55 + 10 = 65.

Question 20.
87 + __________ = 97
Answer:
87 + __10__ = 97

Explanation:
Here, 87 and Missing number are addends.
Sum is given = 97
Let the missing number be X.
Now add the numbers:
87 + X = 97
X = 97 – 87
X = 10.
Hence, equation is 87 + 10 = 97.

Question 21.
DIG DEEPER!
Use each number once to complete the equations.
86       76       10         66
_________ + 10 = __________              ___________ + __________ = 96
Answer:
66 + 10 = 76
86 + 10 = 96

Explanation:
The numbers to be added together are called the “Addends”.
1.  _________ + 10 = __________
We know among two addends, one is 10.
2.     ___________ + __________ = 96
Given numbers 86,76,10,66.
Place every given number next to given addend 10:
86 + 10 = 96
76 + 10 = 86
10 + 10 = 20
66 + 10 =76.
Among this equations, for the first equation is  66 + 10 =76.
and the second  is 86 + 10 = 96 because the  it fulfills the given condition. (Use each number once to complete the equations.)

Think and Grow: Modeling Real Life

There are 33 students on a bus. 10 more get on. How many students are on the bus now?
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 3
Addition equation:

___________ students

Answer:
Total number of students on the bus now = 43.
Given:
Number of students in a bus = 33
Number of students more got into the bus = 10

Explanation:
Total number of students on the bus now = Number of students in a bus + Number of students more get into the bus = 33 + 10
= 43.
Hence, Total number of students on the bus now = 43.

Show and Grow

Question 22.
There are 61 tents at a campground. 10 more are put up. How many tents are at the campground now?
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 4
Addition equation:

____________ tents

Answer:
Number of tents at the campground now = 71.

Given:
Number of tents at a campground =61
Number of tents more put up at the campground= 10

Explanation:
Number of tents at the campground now = Number of tents at a campground – Number of tents more put up at the campground
= 61 + 10
= 71
Hence, Number of tents at the campground now = 71.

Mental Math: 10 More Practice 8.1

Use Mental math.

Question 1.
30 + 10 = __________
Answer:
30 + 10 = 40

Explanation:
Here, 30 and 10 are addends.
Now add both the numbers:
30 + 10 = 40
Hence, Sum = 40.

Question 2.
81 + 10 = __________
Answer:
81 + 10 = 91

Explanation:
Here, 81 and 10 are addends.
Now add both the numbers:
81 + 10 = 91.
Hence, Sum = 91.

Question 3.
6 + 10 = __________
Answer:
6 + 10 = 16

Explanation:
Here, 6 and 10 are addends.
Now add both the numbers:
6 + 10 = 16.
Hence, Sum = 16.

Question 4.
57 + 10 = __________
Answer:
57 + 10 = 67

Explanation:
Here, 57 and 10 are addends.
Now add both the numbers:
57 + 10 = 67.
Hence, Sum = 67.

Question 5.
48 + 10 = __________
Answer:
48 + 10 = 58.

Explanation:
Here, 48 and 10 are addends.
Now add both the numbers:
48 + 10 = 58.
Hence, Sum = 58.

Question 6.
26 + 10 = __________
Answer:
26 + 10 = 36.

Explanation:
Here, 26 and 10 are addends.
Now add both the numbers:
26 + 10 = 36.
Hence, Sum = 36.

Question 7.
10 + 43 = __________
Answer:
10 + 43 = 53.

Explanation:
Here, 10 and 43 are addends.
Now add both the numbers:
10 + 43 = 53.
Hence, Sum = 53.

Question 8.
10 + 65 = __________
Answer:
10 + 65 = 75.

Explanation:
Here, 10 and 65 are addends.
Now add both the numbers:
10 + 65 = 75.
Hence, Sum = 75.

Question 9.
10 + 82 = __________
Answer:
10 + 82 = 92.

Explanation:
Here, 10 and 82 are addends.
Now add both the numbers:
10 + 82 = 92.
Hence, Sum = 92.

Question 10.
10 + 79 = __________
Answer:
10 + 79 = 89.

Explanation:
Here, 10 and 79 are addends.
Now add both the numbers:
10 + 79 = 89.
Hence, Sum = 89.

Question 11.
22 + __________ = 32
Answer:
22 + 10 =32.

Explanation:
Here, 22 and Missing number are addends.
Let the missing number be X.
Sum is given =32
Now add the numbers:
22 + X = 32
X = 32 – 22
X = 10.
Hence, equation is 22 + 10 =32.

Question 12.
85 + __________ = 95
Answer:
85 + 10 = 95.

Explanation:
Here, 85 and Missing number are addends.
Let the missing number be X.
Sum is given = 95
Add the numbers.
85 + X = 95
X = 95 – 85
X = 10.
Hence, equation is 85 + 10 = 95.

Question 13.
64 + __________ = 74
Answer:
64 + 10 = 74.

Explanation:
Here, 64 and Missing number are addends.
Let the missing number be X.
Sum is given = 74.
Add the numbers.
64 + X = 74
X = 74 – 64
X = 10.
Hence, equation  is 64 + 10 = 74.

Question 14.
41 + __________ = 51
Answer:
41 + 10 = 51.

Here, 41 and Missing number are addends.
Let the missing number be X.
Sum is given = 51.
Add the numbers.
41 + X = 51
X = 51 – 41
X = 10.
Hence, equation is 41 + 10 = 51 .

Question 15.
DIG DEEPER!
Use each number once to complete the equations.
25        10       15       35
10 + __________ = _________              25 + __________ = ___________
Answer:
10 + 15 = 25
25 +10 = 35

Explanation:
The numbers to be added together are called the “Addends”.
1. 10 + __________ = _________
2. 25 + __________ = ___________
Given 10 and 25 as one of the addends among two addends in their respective addition.
Given numbers 25,10,15,35 are other addends.
Check out placing each addend next to 10.
Equation are ,
10 + 25 = 35
10 + 10 = 20
10 + 15 = 25
10 + 35 =45.
Among this equations, for the first equation is 10 + 15 = 25
and the second  is 25 + 10 = 35 because the  it fulfills the given condition .(Use each number once to complete the equations.)

Question 16.
Modeling Real Life
There are 142 teachers at a school. The school hires 10 more. How many teachers are there now?
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 5

_____________ teachers
Answer:
Number of teachers are there now = 152.

Given:
Number of teachers at a school = 142
Number of teachers the school hired more = 10

Explanation:
Number of teachers are there now = Number of teachers at a school + Number of teachers the school hired more
= 142 + 10
= 152.
Hence, Number of teachers are there now = 152.

Review & Refresh

Question 17.
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 6
10 less than 87 is _________ .
Answer:
87 – 10 = 77.

Explanation:
There are 87 mangoes in a  basket.
10 mangoes got spoiled. How many mangoes left??
Subtract the mangoes
87 – 10 = 77.
Hence, equation is 87 – 10 = 77.

Question 18.
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 7
1 less than 33 is __________ .
Answer:
33 – 1 = 32

Explanation:
There are 33 mangoes in a  basket.
1 mango got spoiled. How many mangoes left??
Subtract the mangoes
33 – 1 = 32.
Hence, equation is 33 – 1 = 32.

Lesson 8.2 Mental Math: 10 Less

Explore and Grow

Find each difference. What do you notice?

33 – 10 = __________

67 – 10 = __________

82 – 10 = __________
Answer:
33  -10 = 23
67 – 10 = 57
82 – 10 = 72

Explanation:
1. The number that is to be subtracted from is called Minuend.
The number that is to be subtracted is called Subtrahend:
The result of subtracting one number from another is called Difference:
In this case, 33 is minuend and 10 is subtrahend.
Now subtract this two:
=> 33  -10 = 23
Hence, difference is  23.
2. In this case, 67 is minuend and 10 is subtrahend.
Now subtract this two:
=> 67 – 10 = 57
Hence, difference is 57.
3. In this case, 82 is minuend and 10 is subtrahend.
Now subtract this two:
=> 82 – 10 = 72
Hence, difference is 72.

Show and Grow

Use mental math.

Question 1.
55 – 10 = __________
Answer:
55 – 10 = 45

Explanation:
In this case, 55 is minuend and 10 is subtrahend.
Now subtract this two:
55 – 10 = 45
Hence, Difference = 45.

Question 2.
21 – 10 = __________
Answer:
21 – 10 = 11

Explanation:
In this case, 21 is minuend and 10 is subtrahend.
Now subtract this two:
21 – 10 = 11
Hence, Difference = 11.

Question 3.
18 – 10 = __________
Answer:
18 – 10 =8.

Explanation:
In this case, 18 is minuend and 10 is subtrahend.
Now subtract this two:
18 – 10 =8.
Hence, Difference = 8.

Question 4.
74 – 10 = __________
Answer:
74 – 10 = 64.

Explanation:
In this case, 74 is minuend and 10 is subtrahend.
Now subtract this two:
74 – 10 = 64.
Hence, Difference = 64.

Question 5.
89 – 10 = __________
Answer:
89 – 10 = 79.

Explanation:
In this case, 89 is minuend and 10 is subtrahend.
Now subtract this two:
89 – 10 = 79.
Hence, Difference = 79.

Question 6.
72 – 10 = __________
Answer:
72 – 10 = 62.

Explanation:
In this case, 72 is minuend and 10 is subtrahend.
Now subtract this two:
72 – 10 = 62.
Hence, Difference = 62.

Apply and Grow: Practice

Use mental math

Question 7.
60 – 10 = __________
Answer:
60 – 10 = 50.

Explanation:
In this case, 60 is minuend and 10 is subtrahend.
Now subtract this two:
60 – 10 = 50.
Hence, Difference = 50.

Question 8.
45 – 10 = __________
Answer:
45 – 10 = 35

Explanation:
In this case, 45 is minuend and 10 is subtrahend.
Now subtract this two:
45 – 10 = 35
Hence, Difference = 35.

Question 9.
50 – 10 = __________
Answer:
50 – 10 = 40

Explanation:
In this case, 50 is minuend and 10 is subtrahend.
Now subtract this two:
50 – 10 = 40
Hence, Difference = 40.

Question 10.
34 – 10 = __________
Answer:
34 – 10 = 24

Explanation:
In this case, 34 is minuend and 10 is subtrahend.
Now subtract this two:
34 – 10 = 24
Hence, Difference = 24.

Question 11.
51 – 10 = __________
Answer:
51 – 10 = 41

Explanation:
In this case, 51 is minuend and 10 is subtrahend.
Now subtract this two:
51 – 10 = 41
Hence, Difference = 41.

Question 12.
86 – 10 = __________
Answer:
86 – 10 = 76.

Explanation:
In this case, 86 is minuend and 10 is subtrahend.
Now subtract this two:
86 – 10 = 76.
Hence, Difference = 76.

Question 13.
64 – 10 = __________
Answer:
64 – 10 = 54.

Explanation:
In this case, 64 is minuend and 10 is subtrahend.
Now subtract this two:
64 – 10 = 54.
Hence, Difference = 54.

Question 14.
97 – 10 = __________
Answer:
97 – 10 = 87.

Explanation:
In this case, 97 is minuend and 10 is subtrahend.
Now subtract this two:
97 – 10 = 87.
Hence, Difference = 87.

Question 15.
28 – 10 = __________
Answer:
28 – 10 = 18.

Explanation:
In this case, 28 is minuend and 10 is subtrahend.
Now subtract this two:
28 – 10 = 18.
Hence, Difference = 18.

Question 16.
73 – 10 = __________
Answer:
73 – 10 = 63.

Explanation:
In this case, 73 is minuend and 10 is subtrahend.
Now subtract this two:
73 – 10 = 63.
Hence, Difference = 63.

Question 17.
__________ – 10 = 22
Answer:
32 – 10 = 22.

Explanation:
In this case, Missing number is minuend and 10 is subtrahend.
Let missing number be X.
Sum is given = 22
Now subtract:
X – 10 = 80
X = 22 + 10
X = 32.
Hence, Equation = 32 – 10 = 22 .

Question 18.
__________ – 10 = 80
Answer:
90 – 10 = 80.

Explanation:
In this case, Missing number is minuend and 10 is subtrahend.
Let missing number be X.
Sum is given = 80.
Now subtract :
X – 10 = 80
X = 80 + 10
X = 90.
Hence, Equation = 90 – 10 = 80.

Question 19.
YOU BE THE TEACHER
Is Newton correct? Explain.
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 8
Answer:
94 – 10 = 84.

Explanation:
In this case, 94 is minuend and 10 is subtrahend.
Now subtract this two:
94 – 10 = 84.
Hence, Difference = 84.
Therefore, Newton is correct.

Think and Grow: Modeling Real Life

You want to ride all 47 rides at an amusement park. You ride 10 of them. How many rides are left?
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 9
Subtraction equation:

_________ rides.

Answer:
Number of rides left = 37.

Given:
Total number of rides you want to ride at an amusement park = 47
Number of rides you have taken out of them = 10

Explanation:
Number of rides left = Total number of rides you want to ride at an amusement park – Number of rides you have taken out of them
= 47 – 10
= 37.
Hence, Number of rides left = 37.

Show and Grow

Question 20.
You want to try all 65 flavors at a frozen yogurt shop. You try 10 of them. How many flavors are left?
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 10
Subtraction equation:

_________ flavors.

Answer:
Number of flavors left = 55.

Given:
Total number of flavors you want to try at a frozen yogurt shop = 65
Number of flavors you have tried out of them = 10

Explanation:
Number of flavors left = Total number of flavors you want to try at a frozen yogurt shop – Number of flavors you have tried out of them
= 65 – 10
= 55
Hence,  Number of flavors left = 55.

Mental Math: 10 Less Practice 8.2

Use mental math.

Question 1.
12 – 10 = __________
Answer:
12 – 10 = 2.

Explanation:
In this case, 12 is minuend and 10 is subtrahend.
Now subtract this two:
12 – 10 = 2.
Hence, Difference = 2.

Question 2.
49 – 10 = __________
Answer:
49 – 10 = 39.

Explanation:
In this case, 49 is minuend and 10 is subtrahend.
Now subtract this two:
49 – 10 = 39.
Hence, Difference = 39.

Question 3.
37 – 10 = __________
Answer:
37 – 10 = 27.

Explanation:
In this case, 37 is minuend and 10 is subtrahend.
Now subtract this two:
37 – 10 = 27.
Hence, Difference = 27.

Question 4.
26 – 10 = __________
Answer:
26 – 10 = 16.

Explanation:
In this case, 26 is minuend and 10 is subtrahend.
Now subtract this two:
26 – 10 = 16.
Hence, Difference = 16.

Question 5.
40 – 10 = __________
Answer:
40 – 10 = 30.

Explanation:
In this case, 40 is minuend and 10 is subtrahend.
Now subtract this two:
40 – 10 = 30.
Hence, Difference = 30.

Question 6.
62 – 10 = __________
Answer:
62 – 10 = 52.

Explanation:
In this case, 62 is minuend and 10 is subtrahend.
Now subtract this two:
62 – 10 = 52
Hence, Difference = 52.

Question 7.
88 – 10 = __________
Answer:
88 – 10 = 78.

Explanation:
In this case, 88 is minuend and 10 is subtrahend.
Now subtract this two:
88 – 10 = 78.
Hence, Difference =78.

Question 8.
91 – 10 = __________
Answer:
91 – 10 = 81.

Explanation:
In this case, 91 is minuend and 10 is subtrahend.
Now subtract this two:
91 – 10 = 81.
Hence, Difference = 81.

Question 9.
54 – 10 = __________
Answer:
54 – 10 = 44

Explanation:
In this case, 54 is minuend and 10 is subtrahend.
Now subtract this two:
54 – 10 = 44.
Hence, Difference = 44.

Question 10.
76 – 10 = __________
Answer:
76 – 10 = 66.

Explanation:
In this case, 76 is minuend and 10 is subtrahend.
Now subtract this two:
76 – 10 = 66.
Hence, Difference = 66.

Question 11.
__________ – 10 = 15
Answer:
25 – 10 = 15

Explanation:
In this case, Missing number  is minuend and 10 is subtrahend.
Let missing number be X.
Sum is given = 15.
Now subtract :
X – 10 = 15
X = 15 + 10
X = 25.
Hence, Equation is 25 – 10 = 15.

Question 12.
__________ – 10 = 77
Answer:
87 – 10 = 77

Explanation:
In this case, Missing number is minuend and 10 is subtrahend.
Let missing number be X.
Sum is given = 77.
Now subtract :
X – 10 = 77
X = 77 + 10
X = 87.
Hence, Equation is 87 – 10 = 77.

Question 13.
__________ – 10 = 34
Answer:
44 – 10 = 34

Explanation:
In this case, Missing number is minuend and 10 is subtrahend.
Let missing number be X.
Sum is given = 34.
Now subtract :
X – 10 = 34
X = 34 + 10
X = 44.
Hence, Equation is 44 – 10 = 34.

Question 14.
__________ – 10 = 53
Answer:
63 – 10 = 53.

Explanation:
In this case, Missing number is minuend and 10 is subtrahend.
Let missing number be X.
Sum is given = 53.
Now subtract :
X – 10 = 53
X = 53 + 10
X = 63.
Hence, Equation is 63 – 10 = 53.

Question 15.
YOU BE THE TEACHER
Is Descartes correct? Explain.
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 11
Answer:
40 – 10 = 30

Explanation:
In this case, 40 is minuend and 10 is subtrahend.
Now subtract this two:
40 – 10 = 30.
Difference = 30.
Therefore, Descartes answer 50 is incorrect because he did not subtract the numbers instead he added the number.

Question 16.
Modeling Real Life
There are 99 levels in a video game. You complete 10 of them. How many are left?
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 12

____________ levels

Answer:
Number of games left = 89.

Given:
Total number of levels in a video game = 99
Number of games I have completed out of them = 10

Explanation:
Number of games left = Total number of levels in a video game – Number of games I have completed out of them
= 99 – 10
= 89
Hence, Number of games left = 89.

Review & Refresh

Question 17.
A group of students are at the arcade. 4 of them leave. There are 5 left. How many students were there to start?

________ students

Answer:
Total number of students were there to start = 9.

Given:
Number of students leave = 4
Number of students left = 5

Explanation:
Total number of students were there to start = Number of students leave + Number of students left
= 4 + 5
= 9
Hence, total number of students were there to start = 9.

Lesson 8.3 Add Tens

Explore and Grow

Model each problem. How are the problems alike? How are they different?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 13

3 + 2 = _______

30 + 20 = _______
Answer:
3 + 2 = 5.
30 + 20 = 50.

Explanation:
A number that is added to another number is known as an addend.
1. In this case, 3 and 2 are the addends.
Sum is the result of adding two numbers.
3 + 2 = 5.
2. In this case, 30 and 20 are the addends.
Sum is the result of adding two numbers.
30 + 20 = 50.
Hence, the problem is alike in its addition process and different in its sum.

Show and Grow

Question 1.
40 + 50 = ?
____________ tens + ____________ tens = ____________ tens
So, 40 + 50 = ____________ .

Answer:
40 + 50 = 90.

Explanation:
Place value is the value of each digit in a number.
4 value in 40 = 4 tens
5 value in 50 = 5 tens
Add the two numbers:
=> 4 tens + 5 tens = 9 tens.
So, 40 + 50 = 90.

Question 2.
30 + 30 = ?
____________ tens + ____________ tens = ____________ tens
So, 30 + 30 = ____________ .

Answer:
30 + 30 = 60.

Explanation:
Place value is the value of each digit in a number.
3 value in 30 = 3 tens
3 value in 30 = 3 tens
Add the two numbers:
3 tens + 3 tens = 6 tens
So, 30 + 30 = 60.

Apply and Grow: Practice

Question 3.
20 + 40 = ?
____________ tens + ____________ tens = ____________ tens
So, 20 + 40 = ____________ .
Answer:
20 + 40 = 60

Explanation:
Place value is the value of each digit in a number.
2 value in 20 = 2 tens
4 value in 40 = 4 tens
Add the two numbers:
2 tens + 4 tens = 6 tens
So, 20 + 40 = 60.

Question 4.
50 + 30 = ?
____________ tens + ____________ tens = ____________ tens
So, 50 + 30 = ____________ .
Answer:
50 + 30 = 80.

Explanation:
Place value is the value of each digit in a number.
5 value in 50 = 5 tens
3 value in 30 = 3 tens
Add the two numbers:
5 tens + 3 tens = 8 tens
So, 50 + 30 = 80.

Question 5.
40 + 10 = _________
Answer:
40 + 10 = 50.

Explanation:
Place value is the value of each digit in a number.
4 value in 40 = 4 tens
1 value in 10 = 1 tens
Add the two numbers:
4 tens + 1 tens = 5 tens
So, 40 + 10 = 50.

Question 6.
70 + 20 = _________
Answer:
70 + 20 = 90.

Explanation:
Place value is the value of each digit in a number.
7 value in 70 = 7 tens
2 value in 20 = 2 tens
Add the two numbers:
7 tens + 2 tens = 9 tens
So, 70 + 20 = 90..

Question 7.
30 + 40 = _________
Answer:
30 + 40 = 70.

Explanation:
Place value is the value of each digit in a number.
3 value in 30 = 3 tens
4 value in 40 = 4 tens
Add the two numbers:
3 tens + 4 tens = 7 tens
So, 30 + 40 = 70.

Question 8.
20 + 60 = __________
Answer:
20 + 60 = 80.

Explanation:
Place value is the value of each digit in a number.
2 value in 20 = 2 tens
6 value in 60 = 6 tens
Add the two numbers:
2 tens + 6 tens = 8 tens
So, 20 + 60 = 80.

Question 9.
_________ + 10 = 40
Answer:
30 + 10 = 40.

Explanation:
Let the first addend be X.
Second addend given= 10
Sum is given = 40
Now add :
X + 10 = 40
X = 40 – 10
x = 30.
Hence, Equation is 30 + 10 = 40.

Question 10.
_________ + 60 = 90
Answer:
30 + 60 = 90.

Explanation:
Let the first addend be X.
Second addend given = 60
Sum is given = 90
Now add :
X + 60 = 90
X = 90 – 60
X = 30.
Hence, Equation is 30 + 60 = 90.

Question 11.
DIG DEEPER!
Which choices match the model?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 14
50                                       20 + 30
2 tens + 3 ones                  1 ten + 4 tens

Answer:
1 ten + 4 tens = 5 tens.

Explanation:
In the above picture, 5 lines are given.
Considered each line value as 10.
Here, two addends are given in each choice:
Now add this two :
1. 50
2. 20 + 30 = 50
3. 2 tens + 3 ones = 20 + 3=23
4. 1 ten + 4 tens = 5 tens
Hence, 4th choice matches the model because five lines of each value 10 = 50 not others as addition of two numbers it did not refer in the model.

Think and Grow: Modeling Real Life

One tray has 20 meatballs. Another tray has the same number of meatballs. How many meatballs are there in all?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 15
Model:

Addition equation:

__________ meatballs

Answer:
Total number of meatballs are there in all = 40.

Given:
Number of meatballs one tray has = 20
Number of meatballs same another tray has = 20
Explanation:
Total number of meatballs are there in all =  Number of meatballs one tray has – Number of meatballs same another tray has
= 20 + 20
= 40
Hence, Total number of meatballs are there in all = 40.

Show and Grow

Question 12.
One box has 40 bags of pretzels. Another box has the same number of bags. How many bags are there in all?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 16
Model:

Addition equation:

__________ bags

Answer:
Total number of bags are there in all = 80.

Given:
Number of bags of pretzels one box has = 40
Number of bags of pretzels another box has = 40

Explanation:
Total number of bags are there in all = Number of bags of pretzels one box has + Number of bags of pretzels another box has
= 40 + 40
= 80.
Hence, Total number of bags are there in all = 80.

Add Tens Practice 8.3

Question 1.
20 + 70 = ?
____________ tens + ____________ tens = ____________ tens
So, 20 + 70 = ____________ .
Answer:
2tens + 7tens = 9tens
20 + 70 = 90.

Explanation:
Here place value of 2 and 7 are considered.
Add the two numbers.
2tens + 7tens = 9tens
20 + 70 = 90.

Question 2.
50 + 30 = ?
____________ tens + ____________ tens = ____________ tens
So, 50 + 30 = ____________ .
Answer:
5tens + 3tens = 8tens
50 + 30 = 80.

Explanation:
Here place value of 2 and 7 are considered.
Add the two numbers.
5tens + 3tens = 8tens
50 + 30 = 80.

Question 3.
60 + 20 = ___________
Answer:
6tens + 2tens = 8tens.
60 + 20 = 80.

Explanation:
Here place value of 2 and 7 are considered.
Add the two numbers.
6tens + 2tens = 8tens.
60 + 20 = 80.

Question 4.
40 + 40 = ____________
Answer:
4tens + 4tens = 8tens.
40 + 40 = 80.

Explanation:
Here place value of 4 and 4 are considered.
Add the two numbers.
4tens + 4tens = 8tens.
40 + 40 = 80.

Question 5.
__________ + 40 = 50
Answer:
1 tens + 4 tens = 5 tens.
10 + 40 = 50.

Explanation:
Let the missing number be X.
Add the numbers now.
X + 40 = 50
X = 50 – 40
X = 10
1 tens + 4 tens = 5 tens.
10 + 40 = 50.

Question 6.
___________ + 30 = 60
Answer:
3 tens + 3 tens = 6 tens.
30 + 30 = 60.

Explanation:
Let the missing number be X.
Add the numbers now.
X + 30 = 60
X = 60 – 30
X = 30
3 tens + 3 tens = 6 tens.
30 + 30 = 60.

Question 7.
___________ + 20 = 70
Answer:
5 tens + 2 tens = 7 tens.
50 + 20 = 70.

Explanation:
Let the missing number be X.
Add the numbers now.
X + 20 = 70
X = 70 – 20
X = 50
5 tens + 2 tens = 7 tens.
50 + 20 = 70.

Question 8.
_____________ + 50 = 90
Answer:
5 tens + 4 tens = 90.
50 + 40 = 90.

Explanation:
Let the missing number be X.
Add the numbers now.
X + 50 = 90
X = 90 – 50
X = 40
5 tens + 4 tens = 90.
50 + 40 = 90.

Question 9.
DIG DEEPER!
Which choices match the model?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 17
3 tens + 4 ones                   5 tens + 2 tens
60                                        30 + 40
Answer:
3 tens + 4 tens = 7 tens
5 tens + 2 tens = 7 tens

Explanation:
1. 3 tens + 4 tens = 7 tens
2. 60
3. 5 tens + 2 tens = 7 tens
4. 30 + 40 = 70
Hence, 2nd and 3rd choice matches the model because five lines of each value 10  x 7 = 70 not others as addition of two numbers it did not refer in the model.

Question 10.
Modeling Real Life
One magic set has 30 pieces. Another set has the same number of pieces. How many pieces are there in all?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 18

_____________ pieces
Answer:
Total Number of  pieces are there in all = 60.

Explanation:
Number  of pieces magic sets has = 30
Number of  pieces another magic sets has same  = 30
Total Number of  pieces are there in all = Number  of pieces magic sets has + Number of  pieces another magic sets has same
= 30 + 30
= 60.
Hence, Total Number of  pieces are there in all = 60.

Review & Refresh

Question 11.
__________ = 6 + 5
Answer:
11 = 6 + 5

Explanation:
Let the sum of addends 6 and 5 be X.
add the two numbers.
X = 6 + 5
X = 11.
Hence, 11 = 6 + 5.

Question 12.
__________ = 3 + 17
Answer:
20 = 3 + 17

Explanation:
Let the sum of addends 3 and 17 be X.
Add the two numbers.
X = 3 + 17
X = 20.
Hence, 20 = 3 + 17.

Question 13.
__________ = 11 + 5
Answer:
16 = 11 + 5.

Explanation:
Let the sum of addends 11 and 5 be X.
Add the two numbers.
X = 11 + 5
X = 16.
Hence, 16 = 11 + 5.

Question 14.
__________ = 5 + 9
Answer:
14 = 5 + 9.

Explanation:
Let the sum of addends 5 and 9 be X.
Add the two numbers.
X = 5 + 9
X = 14.
Hence, 14 = 5 + 9.

Lesson 8.4 Add Tens Using a Number Line

Explore and Grow

Write the missing numbers. How do the hops help you solve?

30 + 20 = _________

Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 19

Answer:
30 + 20 = 50.

Explanation:
In math, a number line can be defined as a straight line with numbers placed at equal intervals or segments along its length.
Here, in the above picture the interval is calculated and added.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-
30 + 10 + 10
= 40 + 10
= 50 .
Hence, 30 + 20 = 50.

Show and Grow

Question 1.
50 + 40 = __________
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 20
Answer:
50 + 40 = 90.

Explanation:
On the number line, give the interval of 10 for 4 times  and add them to given number.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Show and Grow1
Hence, 50 + 40 = 90.

Question 2.
60 + 20 = _____________
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 21
Answer:
60 + 20 = 80

Explanation:
On the number line, give the interval of 10 for 2 times and add them to given number.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Show and Grow2
Hence, 60 + 20 = 80

Apply and Grow: Practice

Question 3.
40 + 20 = _________
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 22
Answer:
40 + 20 = 60.

Explanation:
On the number line, give the interval of 10 for 2 times and add them to given number.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Show and Grow3
Hence, 40 + 20 = 60.

Question 4.
50 + 30 = ___________
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 23
Answer:
50 + 30 = 80.

Explanation:
On the number line, give the interval of 10 for 3 times and add them to given number.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Show and Grow4
Hence, 50 + 30 = 80.

Question 5.
60 + 40 = __________
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 23
Answer:
60 + 40 = 100.

Explanation:
On the number line, give the interval of 10 for 4 times and add them to given number.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Show and Grow5

Hence, 60 + 40 = 100.

Question 6.
20 + 50 = ___________
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 23
Answer:
20 + 50 = 70.

Explanation:
On the number line, give the interval of 10 for 5 times and add them to given number.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Show and Grow6
Hence, 20 + 50 = 70.

Question 7.
MP Structure
Write an equation that matches the number line.
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 24
Answer:
Equation: 70 + 10 +10 = 90.

Explanation:
On the number line, give the interval of 10 for 2 times and add them to given number.

Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 24

Equation:
70 + 10 +10
= 80 + 10
= 90.
Hence, Equation: 70 + 10 +10 = 90.

Think and Grow: Modeling Real Life

You have 20 cans. You collect 20 more cans. Your friend collects 45 cans in all. Who collects more cans?
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 25
Model:
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 26
Addition equation:
Compare: ___________ ○ ___________
Who collects more cans?          You          Friend

Answer:
My friend collects  5 more cans than me .

Given:
Number of cans I have = 20
Number of cans I have collected more = 20

Explanation:
Total number of cans I have in all = Number of cans I have + Number of cans I have collected more
20 + 20 = 40.
Total number of cans my friend has = 45
Difference in cans collection = Total number of cans my friend has – Total number of cans I have in all
= 45 – 40
= 5

Comparison:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Think and Grow-Modeling Real Life

Hence, my friend collects  5 more cans than me .

Show and Grow

Question 8.
Your class makes 62 paper airplanes. Your friend’s class makes 30 small airplanes and 30 large airplanes. Whose class makes more airplanes?
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 27
Model:
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 28
Addition equation:
Compare: ___________ ○ ___________
Whose class makes more airplanes?
Your class                   Friend’s class

Answer:
Total number of paper airplanes I make are 2 more than my friends paper making airplanes.

Given:
Total number of paper airplanes I make = 62
Number of paper small airplanes my friend’s class makes = 30
Number of paper large airplanes my friend’s class makes  = 30

Explanation:
Total number of airplanes my friend’s class makes = Number of paper small airplanes my friend’s class makes + Number of paper large airplanes my friend’s class makes
= 30 + 30
= 60
Number of airplanes who makes more = Total number of paper airplanes class I makes – Total number of airplanes my friend’s class makes
=62 – 60
= 2

Comparison:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Think and Grow-Modeling Real Life-Show and Grow8
Hence, Total number of paper airplanes I make  2 more than my friends paper making airplanes.

Add Tens Using a Number Line Practice 8.4

Question 1.
60 + 30 = __________
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 29
Answer:
60 + 30 = 90.

Explanation:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Add-Tens-Using-a-Number-Line-Practice- 8.4-1
Equation:
60+ 10 +10 +10
=70 + 10 +10
= 80 + 10
=90.

Question 2.
20 + 20 = __________
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 30
Answer:
20 + 10+ 10 = 40.

Explanation:

Equation:
20 + 10+ 10
= 30 + 10
= 40.

Question 3.
30 + 70 = __________
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 30
Answer:
30 + 70 = 100.

Explanation:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Add-Tens-Using-a-Number-Line-Practice- 8.4-3
Equation:
30 + 10 + 10+ 10+ 10+ 10+ 10 +10
= 40 + 10+ 10+ 10+ 10+ 10 +10
= 50 +10+ 10+ 10+ 10 +10
= 60  +10+ 10+ 10 +10
= 70 + 10 + 10 +10
= 80 + 10 +10
= 90 + 10
= 100.

Question 4.
MP Structure
Write an equation that matches the number line.
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 31
Answer:
Equation:
10 + 40 = 50.

Explanation:
The number line shows the interval of 10 for 4 times.
10 + 10 + 10 + 10 +10
= 20 + 10 + 10 + 10
= 30 + 10 + 10
= 40 + 10
= 50.

Question 5.
Modeling Real Life
You make 3 snow bricks. Your friend makes 20 small snow bricks and 30 large snow bricks. Who makes more snow bricks?
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 32
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 33
Who makes more snow bricks?                You              Friend
Answer:
My friend has 47 more snow bricks than me.

Given:
Total number of snow bricks I have made = 3
Number of small snow bricks my friend has made = 20
Number of large snow bricks my friend has made = 30

Explanation:
Total number of small and large snow bricks my friend has made = Number of small snow bricks my friend has made + Number of large snow bricks my friend has made
= 20 + 30
= 50
Number of  snow bricks who makes more = Total number of small and large snow bricks my friend has made – total number of snow bricks I have made
= 50 – 3
= 47.

Comparison:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.4-Add-Tens-Using-a-Number-Line-Think and Grow-Modeling Real Life-Show and Grow5.
Hence, my friend has 47 more snow bricks than me.

Review & Refresh

Question 6.
Get to lo to subtract.
17 – 8 = ?
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 34
17 – __________ = 10
10 – __________ = __________
So, 17 – 8 = __________ .
Answer:
17 – 8 = 9.

Given:
1. 17 – ____ = 10
Let the missing subtrahend be X.
Difference is given = 10
Subtract the numbers.
17 – X = 10
17 = 10 + X
17 – 10 = X
7 = X
Equation: 17 – 7 = 10.

Lesson 8.5 Subtract Tens

Explore and Grow

Model each problem. How are the problems alike? How are they different?

Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 35

5 – 2 = ___________

50 – 20 = ___________
Answer:
5 – 2 = 3.
50 – 20 = 30.

Explanation:
Here, the minuend is 5 and the subtrahend is 2.
Now, Subtract the two numbers.
5 – 2 = 3.
Here, the minuend is 50 and the subtrahend is 20.
Now, Subtract the two numbers.
50 – 20 = 30.
Hence, the two equations are alike in the subtraction process yet different in the answers.

Show and Grow

Question 1.

70 – 30 = ?
____________ tens – ____________ tens = ____________ tens
So, 70 – 30 = ____________ .
Answer:
7 tens – 3 tens = 4 tens
70 – 30 = 40.

Explanation:
Place value is the value of each digit in a number.
7 value in 70 =7 tens
3 value in 30 = 3 tens
Subtract the two numbers:
7 tens – 3 tens = 4 tens
So, 70 – 30 = 40.

Question 2.
40 – 20 = __________
____________ tens – ____________ tens = ____________ tens
So, 40 – 20 = ____________ .
Answer:
4 tens – 2 tens = 2 tens.
40 – 20 = 20.

Explanation:
Place value is the value of each digit in a number.
4 value in 40 =4 tens
2 value in 20 = 2 tens
Subtract the two numbers:
4 tens – 2 tens = 2 tens.
So, 40 – 20 = 20.

Apply and Grow: Practice

Question 3.
90 – 30 = ?
____________ tens – ____________ tens = ____________ tens
So, 90 – 30 = ____________ .
Answer:
9 tens – 3 tens = 6 tens.
90 – 30 = 60.

Explanation:
Place value is the value of each digit in a number.
9 value in 90 =9 tens
3 value in 30 = 3 tens
Subtract the two numbers:
9 tens – 3 tens = 6 tens.
So, 90 – 30 = 60.

Question 4.
50 – 10 = ?
____________ tens – ____________ tens = ____________ tens
So, 50 – 10 = ____________ .
Answer:
5 tens – 1 tens = 4 tens.
50 – 10 = 40.

Explanation:
Place value is the value of each digit in a number.
5 value in 50 = 5 tens
1 value in 10 = 1 tens
Subtract the two numbers:
5 tens – 1 tens = 4 tens.
So, 50 – 10 = 40.

Question 5.
30 – 20 = __________
Answer:
30 – 20 = 10.

Explanation:
Place value is the value of each digit in a number.
3 value in 30 = 5 tens
2 value in 20 = 2 tens
Subtract the two numbers:
3 tens – 2 tens = 1 tens.
So, 30 – 20 = 10.

Question 6.
40 – 40 = __________
Answer:
40 – 40 = 0.

Explanation:
Place value is the value of each digit in a number.
4 value in 40 = 4 tens
4 value in 40 = 4 tens
Subtract the two numbers:
4 tens – 4 tens = 0 tens.
So, 40 – 40 = 0.

Question 7.
80 – 50 = __________
Answer:
80 – 50 = 30.

Explanation:
Place value is the value of each digit in a number.
8 value in 80 = 8 tens
5 value in 50 = 5 tens
Subtract the two numbers:
8 tens – 5 tens = 3 tens.
So, 80 – 50 = 30.

Question 8.
90 – 70 = ___________
Answer:
90 – 70 = 20.

Explanation:
Place value is the value of each digit in a number.
9 value in 90 = 9 tens
7 value in 70 = 7 tens
Subtract the two numbers:
9 tens – 7 tens = 2 tens.
So, 90 – 70 = 20.

Question 9.
20 – __________ = 10
Answer:
20 – 10 = 10.

Explanation:
Here, the minuend is 20.
Difference is 10.
Let the subtrahend be X.
Now, subtract the numbers.
20 – X = 10.
20 = 10 + X
20 – 10 = X
10 = X.
Hence, 20 – 10 = 10.

Question 10.
50 – __________ = 20
Answer:
50 – 30 = 20.

Explanation:
Here, the minuend is 50.
Difference is 20.
Let the subtrahend be X.
Now, subtract the numbers.
50 – X = 20.
50 = 20 + X
50 – 20 = X
30 = X.
Hence, 50 – 30 = 20.

Question 11.
DIG DEEPER!
Which choices match the model?
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 36
50 – 30                               80 – 30
5 tens – 3 tens                   8 tens – 3 tens
Answer:
8 tens – 3 tens = 5 tens.

Explanation:
In the given picture,  total there are eight lines.
Consider each line value is 10. Out of them 3 lines are removed.
=> 80 – 30 = 50.
2. 50 – 30 = 20
3. 5 tens – 3 tens = 2 tens.
4. 8 tens – 3 tens = 5 tens.
Hence, 4th equation is suitable for the model .

Think and Grow: Modeling Real Life

You have 80 math problems. You have 40 fewer spelling words. How many spelling words do you have?
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 37
Model:

Subtraction equation:

______________ spelling words

Answer:
Number of spelling words I have = 40.

Explanation:
Total number of math problems I have = 80
Number of fewer spelling words = 40

Subtraction equation:
Number of spelling words do you have = Total number of math problems I have – Number of fewer spelling words
= 80 – 40
= 40.
Hence, Number of spelling words I have = 40.

Show and Grow

Question 12.
There are 60 students in a play. A football team has 30 fewer students. How many students are on the football team?
Model:

Subtraction equation:

______________ students

Answer:
Number of students on the football team = 30.

Explanation:
Total number of students in a play = 60
Number of fewer students in a football team = 30

Subtraction equation:
Number of students on the football team = Total number of students in a play – Number of fewer students in a football team
= 60 – 30
= 30
Hence, Number of students on the football team = 30.

Subtract Tens Practice 8.5

Question 1.
70 – 50 = ?
____________ tens – ____________ tens = ____________ tens
So, 70 – 50 = ____________ .
Answer:
70 – 50 = 20.
7 tens – 5 tens = 2 tens.

Explanation:
Place value is the value of each digit in a number.
7 value in 70 = 7 tens
5 value in 50 = 5 tens
Subtract the two numbers:
7 tens – 5 tens = 2 tens.
So, 70 – 50 = 20.

Question 2.
60 – 20 = ?
____________ tens – ____________ tens = ____________ tens
So, 60 – 20 = ____________ .
Answer:
60 – 20 = 40.
6 tens – 2 tens = 4 tens.

Explanation:
Place value is the value of each digit in a number.
6 value in 60 = 6 tens
2 value in 20 = 2 tens
Subtract the two numbers:
6 tens – 2 tens = 4 tens.
So, 60 – 20 = 40.

Question 3.
60 – 60 = __________
Answer:
60 – 60 = 0.

Explanation:
Here, the minuend is 60 and subtrahend is 60.
Subtract the two numbers:
60 – 60 = 0.

Question 4.
30 – 10 = ____________
Answer:
30 – 10 = 20.

Explanation:
Here, the minuend is 30 and subtrahend is 10.
Subtract the two numbers:
30 – 10 = 20.

Question 5.
70 – __________ = 0
Answer:
70 – 70 =0.

Explanation:
Here, the minuend is 70.
Let the subtrahend be X
Difference = 0.
Subtract the two numbers:
70 – X = 0
70 = 0 + X
70 – 0 = X
70 = X
Equation is 70 – 70 =0.

Question 6.
50 – ___________ = 40
Answer:
50 – 10 =40.

Explanation:
Here, the minuend is 50.
Let the subtrahend be X
Difference = 40.
Subtract the two numbers:
50 – X = 40
50 = 40 + X
50 – 40 = X
10 = X
Equation is 50 – 10 =40.

Question 7.
40 – ____________ = 20
Answer:
40 – 20 = 20.

Explanation:
Here, the minuend is 40.
Let the subtrahend be X
Difference = 20.
Subtract the two numbers:
40 – X = 20
40 = 20 + X
40 – 20 = X
20 = X
Equation is 40 – 20 = 20.

Question 8.
90 – ____________ = 50
Answer:
90 – 40 =50.

Explanation:
Here, the minuend is 90.
Let the subtrahend be X
Difference = 50.
Subtract the two numbers:
90 – X = 50
90 = 50 + X
90 – 50 = X
40 = X
Equation is 90 – 40 =50.

Question 9.
DIG DEEPER!
Which choices match the model?
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 38
9o – 50                          40 – 50
4 tens – 5 ones              9 tens – 5 tens
Answer:
9 tens – 5 tens = 4 tens.

Explanation:
In the given picture,  total there are nine lines.
Consider each line value is 10. Out of them 5 lines are removed.
=> 9 tens – 5 tens = 4 tens
2. 90 – 50 = 40
3. 40 – 50 = -10
4. 4 tens – 5 ones = 40 + 5 = 45.
Hence, out of four choices, the first choice is correct for the model than other three equations.

Question 10.
Modeling Real Life
There are 40 chairs in the library. There are 30 fewer tables than chairs. How many tables are there?
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 39

___________ tables
Answer:
Number of tables are there = 10.

Explanation:
Total number of chairs in a library = 40
Number of fewer tables than chairs = 30
Number of tables are there = Total number of chairs in a library – Number of fewer tables than chairs
= 40 – 30
= 10.
Hence, Number of tables are there = 10.

Review & Refresh

Question 11.
11 – 7 = ____________
Answer:
11 – 7 = 4

Explanation:
Here, the minuend is 17 and subtrahend is 7.
Subtract the two numbers:
11 – 7 = 4
Hence, 11 – 7 = 4.

Question 12.
16 – 8 = ____________
Answer:
16 – 8 = 8.

Explanation:
Here, the minuend is 16 and subtrahend is 8.
Subtract the two numbers:
16 – 8 = 8.
Hence, 16 – 8 = 8.

Question 13.
15 – 8 = ____________
Answer:
15 – 8 = 7.

Explanation:
Here, the minuend is 15 and subtrahend is 8.
Subtract the two numbers:
15 – 8 = 7.
Hence, 15 – 8 = 7.

Question 14.
18 – 9 = ____________
Answer:
18 – 9 = 9.

Explanation:
Here, the minuend is 18 and subtrahend is 9.
Subtract the two numbers:
18 – 9 = 9.
Hence, 18 – 9 = 9.

Lesson 8.6 Subtract Tens Using a Number Line

Explore and Grow

Write the missing numbers. How do the hops help you solve?

40 – 20 = ___________

Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 40
Answer:
40 – 20 = 20.

Explanation:
In math, a number line can be defined as a straight line with numbers placed at equal intervals or segments along its length.
Here, we need to represent 20 subtracted from 40.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson-8.6-Subtract-Tens-Using-a-Number- Line1

Show and Grow

Question 1.
80 – 50 = __________
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 41
Answer:
80 – 50 = 30.

Explanation:
Here, we need to represent 50 subtracted from 80.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson-8.6-Subtract-Tens-Using-a-Number- Line2

Question 2.
70 – 30 = ___________
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 42
Answer:
70 – 30 = 40.

Explanation:
Here, we need to represent 30 subtracted from 70.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson-8.6-Subtract-Tens-Using-a-Number- Line-Show and Grow2

Apply and Grow: Practice

Question 3.
60 – 20 = __________
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 43
Answer:
60 – 20 = 40.

Explanation:
Here, we need to represent 20 subtracted from 60.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson-8.6-Subtract-Tens-Using-a-Number- Line-Apply-and- Grow-Practice3

Question 4.
40 – 30 = ____________
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 44
Answer:
40 – 30 = 10.

Explanation:
Here, we need to represent 30 subtracted from 40.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson-8.6-Subtract-Tens-Using-a-Number- Line-Apply-and-Grow-Practice4

Question 5.
90 – 40 = ____________
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 44
Answer:
90 – 40 = 50.

Explanation:
Here, we need to represent 40 subtracted from 90.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson-8.6-Subtract-Tens-Using-a-Number- Line-Apply-and-Grow-Practice5

Question 6.
90 – 70 = ____________
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 44
Answer:
90 – 70 = 20.

Explanation:
Here, we need to represent 70 subtracted from 90.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson-8.6-Subtract-Tens-Using-a-Number- Line-Apply-and-Grow-Practice6

Question 7.
MP Structure
Write the equation that matches the number line.
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 45
Answer:
80 – 40 = 40

Explanation:
80 – 10 -10 – 10 -10
= 70 -10 – 10 -10
= 60 – 10 -10
= 50 -10
= 40.

Think and Grow: Modeling Real Life

You have a bucket of 80 golf balls. You hit 60 of them. Your friend has 28 golf balls left. Who has more golf balls left?
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 46
Model:
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 47
Subtraction equation:
Compare: __________ ○ ___________
Who has more golf balls left?               You               Friend
Answer:
My friend has left 8 golf balls more than me.

Given:
Total number of golf balls I have = 80
Number of golf balls I hit out of them = 60
Explanation:
Number of gold balls I left = Total number of golf balls I have – Number of golf balls I hit out of them
= 80 – 60
= 20
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Think and Grow-Modeling-Real-Life
Number of golf balls my friend left = 28
Comparison:
Number of golf balls who left more = Number of golf balls my friend left –  Number of gold balls I left
= 28 – 20
= 8.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson-8.6-Subtract-Tens-Using-a-Number- Line-Think and Grow-Modeling-Real-Life
Hence, my friend has left 8 golf balls more than me.

Show and Grow

Question 8.
Pack A has 50 batteries. 40 of them have been used. Pack B has 15 batteries. Which pack has more bafferies left?
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 48
Model:
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 49
Subtraction equation:
Compare: __________ ○ ___________
Who has more golf balls left?             Pack A              Pack B
Answer:
5 batteries are left more in pack B than pack A.

Given:
Total number of batteries in pack A has= 50
Number of batteries used in pack A = 40
Total number of batteries in pack B has = 15
Explanation:
Number of batteries left in pack A = Total number of batteries in pack A has – Number of batteries used in pack A
= 50 – 40
= 10.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-8.6-Show-and-Grow-8.
Comparison:
Number of batteries which pack has left more = Total number of batteries in pack B has – Number of batteries left in pack A
= 15 – 10
= 5
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-8.6-Show-and-Grow8
Hence,5 batteries are left more in pack B than pack A.

Subtract Tens Using a Number Line Practice 8.6

Question 1.
50 – 30 = _____________
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 50
Answer:
50 – 30 = 20.

Explanation:
Here, we need to represent 30 subtracted from 50.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Subtract-Tens-Using-a-Number-Line-Practice-8.6.1

Question 2.
80 – 60 = ____________
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 51
Answer:
80 – 60 = 20.

Explanation:
Here, we need to represent 60 subtracted from 80.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Subtract-Tens-Using-a-Number-Line-Practice-8.6.2

Question 3.
90 – 20 = ____________
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 51
Answer:
90 – 20 = 70.

Explanation:
Here, we need to represent 20 subtracted from 90.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Subtract-Tens-Using-a-Number-Line-Practice-8.6.3

Question 4.
MP Structure
Write the equation that matches the number line.
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 52
Answer:
60 – 50 = 10.

Explanation:
Here we need to subtract 10 5times from 60.
60 – 10 -10 -10 -10 -10
= 50 -10 -10 -10 -10
=  40 – 10 -10 -10
= 30 – 10 -10
= 20 – 10
= 10.

Question 5.
Modeling Real Life
You have 80 raffle tickets and give away 30 of them. Your friend has 47 raffle tickets. Who has more raffle tickets?
Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 53

Big Ideas Math Answers 1st Grade 1 Chapter 8 Add and Subtract Tens 54

Who has more raffle tickets? You Friend

Answer:
I have 5 raffle tickets more than my friend.

Given:
Total number of raffle tickets = 80
Number of raffle tickets I gave away = 30
Explanation:
Number of raffle tickets I have now = Total number of raffle tickets – Number of raffle tickets I gave away
= 80 -30
= 50
Number of raffles tickets my friend has = 47

Comparison:
Number of raffle ticket who has more = Number of raffle tickets I have now – Number of raffles tickets my friend has
= 50 – 47
= 3
Hence, I have 5 raffle tickets more than my friend.

Review & Refresh

Question 6.
13 – 8 = ?
Think 8 + __________ = 13.
So, 13 – 8 = __________ .
Answer:
13 – 8 = 5.

Explanation:
Think 8 + ____5______ = 13.
So, 13 – 8 = ___5_______ .

Question 7.
15 – 7 = ?
Think 7 + _________ = 15
So, 15 – 7 = ____________ .
Answer:
15 – 7 = 8.

Explanation:
Think 7 + ____8_____ = 15
So, 15 – 7 = ______8______ .

Lesson 8.7 Use Addition to Subtract Tens

Explore and Grow

Complete each equation. What do you notice?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 55
Answer:
20 + 50 = 70
50 – 20 = 30.

Explanation:
Here, for number 20 we are adding 5 times 10.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.7-Use-Addition-to-Subtract-Tens

Here, we are subtracting 20 from number 50.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.7-Use-Addition-to-Subtract-Tens.

Show and Grow

Question 1.
90 – 70 = ?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 56
Think 70 + _________ = 90
So, 90 – 70 = _________ .
Answer:
90 – 70 = 20.

Explanation:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.7-Use-Addition-to-Subtract-Tens-Show and Grow1
Think 70 + ____20_____ = 90
So, 90 – 70 = ___20______ .

Question 2.
60 – 30 = ?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 57
Think 30 + _________ = 60
So, 60 – 30 = _________ .
Answer:
60 – 30 = 30.

Explanation:

Think 30 + __30___ = 60
So, 60 – 30 =____30___.

Apply and Grow: Practice

Question 3.
50 – 30 = ?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 58
Think 30 + _________ = 50
So, 50 – 30 = _________ .
Answer:
50 – 30 = 20.

Explanation:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.7-Use-Addition-to-Subtract-Tens-Apply-and-Grow-Practice3
Think 30 + __20_______ = 50
So, 50 – 30 = ___20______ .

Question 4.
70 – 20 = ?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 59
Think 20 + _________ = 70
So, 70 – 20 = _________ .
Answer:
70 – 20 = 50.

Explanation:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.7-Use-Addition-to-Subtract-Tens-Apply-and-Grow-Practice4

Think 20 + ____50_____ = 70
So, 70 – 20 = __50_______ .

Question 5.
90 – 50 = ?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 59
Think 50 + _________ = 90
So, 90 – 50 = _________ .
Answer:
90 – 50 = 40.

Explanation:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.7-Use-Addition-to-Subtract-Tens-Apply-and-Grow-Practice5
Think 50 + _____40____ = 90
So, 90 – 50 = _____40____ .

Question 6.
MP Structure
Match the related addition and subtraction equations.
60 + 10 = 70                      70 – 50 = 20
50 + 10 = 60                      70 – 60 = 10
50 + 20 = 70                      60 – 50 = 10
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson 8.7-Use-Addition-to-Subtract-Tens-MP Structure6
Explanation:
60 + 10 = 70 and 70 – 60 = 10 are related addition and subtraction equations because both deals with 60, 10 and 70.
50 + 10 = 60 and 60 – 50 = 10 are related addition and subtraction equations because both deals with 60, 10 and 50.
50 + 20 = 70 and 70 – 50 = 20 are related addition and subtraction equations because both deals with 70, 20 and 50.

Think and Grow: Modeling Real Life

A dentist has 40 toothbrushes. She gives away 20 of them. How many toothbrushes does she have left?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 60
Model:

Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 61
Subtraction equation:

______________ toothbrushes
Answer:
Number of tooth brushes left = 20.

Explanation:
Total number of tooth brushes a dentist has = 40
Number of tooth brushes gave away = 20
Number of tooth brushes left = Total number of tooth brushes a dentist has – Number of tooth brushes gave away
= 40 – 20
= 20
Hence, Number of tooth brushes left = 20.

Show and Grow

Question 7.
An art room has 70 bottles of glitter. 30 have been used. How many bottles are left?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 62
Model:

Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 61
Subtraction equation:

______________ bottles
Answer:
Number of bottles of glitter left = 40.

Explanation:
Total number bottles of glitter in an art room = 70
Number of  bottles of glitter used = 30
Number of bottles of glitter left = Total number bottles of glitter in an art room – Number of  bottles of glitter used
= 70 – 30
= 40.
Hence, Number of bottles of glitter left = 40.

Use Addition to Subtract Tens Practice 8.7

Question 1.
70 – 40 = ?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 63
Think 40 + __________ = 70
So, 70 – 40 = _________ .
Answer:
70 – 40 = 30.

Explanation:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Use-Addition-to-Subtract-Tens-Practice 8.7.1
Think 40 + ____30______ = 70
So, 70 – 40 = ___30______ .

Question 2.
90 – 30 = ?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 64
Think 60 + __________ = 90
So, 90 – 30 = _________ .
Answer:
90 – 30 = 60.

Explanation:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Use-Addition-to-Subtract-Tens-Practice 8.7.2

Think 60 + _____30_____ = 90
So, 90 – 30 = ____60_____ .

Question 3.
MP Structure
Match the related addition and subtraction equations.
30 + 10 = 40                  50 – 40 = 10
40 + 10 = 50                  50 – 30 = 20
30 + 20 = 50                  40 – 30 = 10
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Use-Addition-to-Subtract-Tens-Practice-MP Structure-3

Explanation:
30 + 10 = 40 and 40 – 30 = 10 are related addition and subtraction equations because both deals with 40, 10 and 30.
40 + 10 = 50 and 50 – 40 = 10 are related addition and subtraction equations because both deals with 40, 10 and 50.
30 + 20 = 50 and 50 – 30 = 20 are related addition and subtraction equations because both deals with 30, 20 and 50.

Question 4.
Modeling Real Life
Newton has 80 newspapers to deliver. He delivers 50 of them. How many newspapers does he have left?
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 65

Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 66

______________ newspapers
Answer:
Number of newspapers left = 30.

Explanation:
Total number of newspapers to deliver = 80
Number of newspapers delivered = 50
Number of newspapers left = Total number of newspapers to deliver – Number of newspapers delivered
= 80 – 50
= 30.
Hence, Number of newspapers left = 30.

Review & Refresh

Question 5.
Make quick sketches to compare the numbers.
Big Ideas Math Answers Grade 1 Chapter 8 Add and Subtract Tens 67
Answer:
43 > 34.
34 < 43.

Explanation:
43 is greater than 34.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Use-Addition-to-Subtract-Tens-Practice-5-
34 is lesser than 43.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Use-Addition-to-Subtract-Tens-Practice-5-

Lesson 8.8 Add Tens to a Number

Explore and Grow

Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 68

15 + 10 = ________               27 + 10 = __________
15 + 20 = ________               27 + 20 = __________
15 + 30 = ________               27 + 30 = __________
Answer:
15 + 10 = ___25_____               27 + 10 = ___37_______
15 + 20 = ___35_____               27 + 20 = ____47______
15 + 30 = _____45___               27 + 30 = ___57_______

Explanation:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson-8.8-Add-Tens-to-a-Number-Explore and Grow

Show and Grow

Question 1.
23 + 50 = ____________
Answer:
23 + 50 =  73.

Question 2.
6 + 70 = ____________
Answer:
6 + 70 = 76.

Apply and Grow: practice

Question 3.
27 + 40 = ____________
Answer:
27 + 40 = 67.

Question 4.
8 + 80 = ____________
Answer:
8 + 80 = 88.

Question 5.
60 + 35 = ____________
Answer:
60 + 35 =  95.

Question 6.
30 + 44 = ____________
Answer:
30 + 44 = 74.

Question 7.
____________ = 33 + 20
Answer:
53 = 33 + 20.

Question 8.
____________ = 70 + 22
Answer:
92= 70 + 22.

Question 9.
YOU BE THE TEACHER
Is Newton correct? Explain.
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 69
Answer:
36 +50 = 86.
Newton is correct.

Explanation:
36 +50 = 86.
Hence, Newton is correct because the addition of both the addends is correct.

Think and Grow: Modeling Real Life

You count 8 birds on your way to school. You count 40 more on your way home. Your friend counts 45 birds in all. Who counts more birds?
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 70
Model:

Addition equation:

Compare: ___________ ○ ____________

Who counts more birds?            You             Friend

Answer:
I have counted 3 birds more than my friend.

Explanation:
Number of birds I counted on the way to school = 8
Number of birds I counted more on your way home = 40
Total number of birds my friend counts in all = 45
Total number of birds I counted in all = Number of birds I counted on the way to school – Number of birds I counted more on your way home
= 8 + 40
= 48.

Comparison:
Number of birds who counted more = Total number of birds I counted in all – Total number of birds my friend counts in all
= 48 – 45
= 3
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-8.6-Show-and-Grow8--and-Grow-Real-Life

Hence, I have counted 3 birds more than my friend.

Show and Grow

Question 10.
You make 21 snowballs. Your friend makes 11 small snowballs and 20 large snowballs. Who makes more snowballs?
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 71
Model:

Addition equation:

Compare: ___________ ○ ____________

Who counts more birds?          You      Friend
Answer:
Number of snow balls who made more = 10.

Explanation:
Total number of snow balls I made = 21
Total number of small snow balls my friend made =11
Total number of large snow balls my friend made = 20
Total number of snow balls my friend made = Total number of small snow balls my friend made + Total number of large snow balls my friend made
= 11 + 20
= 31

Comparison:
Number of snow balls who made more = Total number of snow balls my friend made – Total number of snow balls I made
=  31 – 21
= 10.
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-8.6-Show and Grow10
Hence, the Number of snow balls who made more = 10.

Add Tens to a Number Practice 8.8

Question 1.
19 + 40 = __________
Answer:
19 + 40 = 59.

Question 2.
60 + 23 = __________
Answer:
60 + 23 = 83.

Question 3.
__________ = 5 + 90
Answer:
95 = 5 + 90.

Question 4.
__________ = 37 + 30
Answer:
67 = 37 + 30.

Question 5.
YOU BE THE TEACHER
Is Descartes correct? Explain.
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 72
Answer:
Descartes is incorrect.

Explanation:
46 + 20 = 66.
Hence, Descartes is incorrect because he did not do the addition correctly.

Question 6.
Modeling Real Life
You have 24 glow sticks and buy 40 more. Your friend has 66 glow sticks. Who has more glow sticks?
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 73

Who has more glow sticks?                 You             Friend

Answer:
My friend has 2 glow sticks more than me.

Explanation:
Number of glow sticks I have = 24
Number of glow sticks I have purchased more = 40
Total number of glow sticks I have = Number of glow sticks I have + Number of glow sticks I have purchased more
= 24 + 40
= 64
Number of  glow sticks my friend has = 66
Comparison:
Number of  glow sticks who has more = Number of  glow sticks my friend has – Total number of glow sticks I have
= 66 – 64
= 2
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-8.6-Show and Grow-Modeling Real Life6
Hence, my friend has 2 glow sticks more than me.

Review & Refresh

Question 7.
Circle the solid shapes that stack.
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 74

Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Lesson-8.8-Add-Tens-to-a-Number-Review & Refresh7

Explanation:
In Geometry, Solids are three-dimensional shapes because they have three dimensions such as length, breadth and height. The bodies which occupy space are called solids.
here in the above picture, cone , cylinder ,cube are solid shapes that stack not sphere shape ball.

Add and Subtract Tens Performance Task

Question 1.
The tables show the numbers of seconds 3 pinwheels and 3 fops spin.
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 75

a. How many more seconds does the yellow top spin than the red pinwheel?
__________ seconds
Answer:
Number of more seconds the yellow top spin than the red pinwheel taken = 50.

Explanation:
Number of seconds red pinwheel taken =  40
Number of seconds yellow pinwheel taken = 90
Number of more seconds the yellow top spin than the red pinwheel taken = Number of seconds red pinwheel taken – Number of seconds yellow pinwheel taken
= 90 – 40
= 50.
Hence, Number of more seconds the yellow top spin than the red pinwheel taken = 50.

b. The red pinwheel spins 30 fewer seconds than the red top. How long does the red top spin?
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 76
__________ seconds
Answer:
The number of seconds the red top taken = 70.

Explanation:
Number of seconds the red pinwheels spins = 40
Let the number of seconds the red top taken be X
The number of seconds the red top taken – 30 =  Number of seconds the red pinwheels spins
(The red pinwheel spins 30 fewer seconds than the red top)
=>  X – 30 = 40
=> X = 40 + 30
=> X = 70
Hence, The number of seconds the red top taken = 70.

c. The blue pinwheel and the blue top spin for 96 seconds in all. How long does the blue pinwheel spin?
Big Ideas Math Solutions Grade 1 Chapter 8 Add and Subtract Tens 77
__________ seconds
Answer:
Number of seconds blue  pinwheel spin taken = 60.

Explanation:
Number of seconds blue top spin taken = 36
Total number of seconds the blue pinwheel and the blue top spins = 96
Number of seconds blue pinwheel spin taken = Number of seconds blue top spin taken – Total number of seconds the blue pinwheel and the blue top spins
= 96 – 36
= 60.
Hence, Number of seconds blue pinwheel spin taken = 60.

d. Which pinwheel spins the longest?
Red                     Yellow                   Blue
Answer:
The yellow pinwheel spins the longest than other pinwheels.

Explanation:
Number of seconds blue pinwheel spin taken = 60
Number of seconds yellow pinwheel spin taken = 90
Number of seconds red pinwheel spin taken = 40
Among, the three pinwheels, the yellow pinwheel spins the longest than other pinwheels.

Add and Subtract Tens Chapter Practice

Mental Math: 10 More Homework & Practice 8.1

Use mental math.

Question 1.
58 + 10 = ___________
Answer:
58 + 10 = 68.

Question 2.
15 + 10 = ___________
Answer:
15 + 10 = 25.

Question 3.
29 + 10 = ___________
Answer:
29 + 10 = 39.

Question 4.
41 + 10 = ___________
Answer:
41 + 10 = 51.

Question 5.
10 + 7 = ___________
Answer:
10 + 7 =  17.

Question 6.
10 + 36 = ___________
Answer:
10 + 36 = 46.

Question 7.
84 + ___________ = 94
Answer:
84 + 10= 94

Explanation:
Let the missing number be X.
Add the addends.
84 + X = 94
X = 94 – 84
X = 10.

Question 8.
47 + ___________ = 57
Answer:
47 + 10 = 57.
Explanation:
Let the missing number be X.
add the addends.
47 + X = 57
X = 57 – 47
X =10.

Mental Math: 10 Less Homework & Practice 8.2

Use mental math.

Question 9.
24 – 10 = ___________
Answer:
24 – 10 = 14.

Question 10.
78 – 10 = ___________
Answer:
78 – 10 =  68.

Question 11.
31 – 10 = ___________
Answer:
31 – 10 = 21.

Question 12.
95 – 10 = ___________
Answer:
95 – 10 = 85.

Question 13.
___________ – 10 = 7
Answer:
17 – 10 = 7.

Explanation:
Let the Missing minuend be X.
Subtrahend = 10
Difference = 7
Subtract the numbers:
X – 10 = 7
X = 7 + 10
X = 17.
Hence, 17 – 10 = 7.

Question 14.
___________ – 10 = 43
Answer:
53 – 10 = 43.

Explanation:
Let the Missing minuend be X.
Subtrahend = 10
Difference = 43
Subtract the numbers:
X – 10 = 43
X = 43 + 10
X = 53.
Hence, 53 – 10 = 43.

Add Tens Homework & Practice 8.3

Question 15.
60 + 20 = ?
__________ tens + ___________ tens = ___________ tens
So, 60 + 20 = ___________ .
Answer:
60 + 20 = 80.

Explanation:
_______6___ tens + _______2____ tens = ______8_____ tens
So, 60 + 20 = _____80______ .

Question 16.
30 + 50 = ___________
Answer:
30 + 50 = 80.

Question 17.
___________ + 30 = 90
Answer:
60 + 30 = 90.

Explanation:
Let the missing addend be X.
sum = 90
Add the numbers
X + 30= 90
X = 90  – 30
X = 60.
Hence, 60 + 30 = 90.

Add Tens Using a Number Line Homework & Practice 8.4

Question 18.
50 + 40 = ___________
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 78
Answer:
50 + 40 = 90.

Explanation:

Question 19.
MP Structure
Write an equation that matches the number line.
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 79
Answer:
Equation:
20 + 30 =50.

Explanation:
20 + 10 +10 +10
= 30 +10 +10
= 40 + 10
= 50.

Subtract Tens Homework & Practice 8.5

Question 20.
90 – 40 = ?
__________ tens – ___________ tens = ___________ tens
So, 90 – 40 = ___________ .
Answer:
90 – 40 = 50.

Explanation:
___9__tens – __4_____ tens = ______5_____ tens
So, 90 – 40 = _50__________ .

Question 21.
70 – 40 = ___________
Answer:
70 – 40 = 20.

Question 22.
80 – __________ = 60
Answer:
80 – 20 = 60.

Explanation:
Let the missing number be X.
Add the addends.
80 – X = 60
80 = 60 + X
80 – 60 = X
20 = X.

Subtract Tens Using a Number Line Homework & Practice 8.6

Question 23.
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 80
60 – 40 = __________
Answer:
60 – 40 = 20.

Explanation:
Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Subtract-Tens-Using-a-Number- Line- Homework & Practice 8.6.23

Question 24.
MP Structure
Write the equation that matches the number line.
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 81
Answer:
Equation: 70 –  20 = 50.

Explanation:
Number line representation :
70 – 10 -10
= 60 -10
= 50.

Use Addition to Subtract Tens Homework & Practice 8.7

Question 25.
80 – 60 = ?
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 82
Think 60 + __________ = 80
So, 80 – 60 = __________ .
Answer:
80 – 60 = 20.

Explanation:

Big-Ideas-Math-Answers-Grade-1-Chapter 8-Add-and-Subtract-Tens-Use-Addition-to-Subtract-Tens-Homework- &- Practice 8.7-25
Think 60 + ____20______ = 80
So, 80 – 60 = _____20_____ .

Question 26.
Modeling Real Life
A mail carrier has 90 packages to deliver. She delivers 60 of them. How many packages are left?
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 83
Big Ideas Math Answer Key Grade 1 Chapter 8 Add and Subtract Tens 84

______________ packages
Answer:
Number of packages left = 30.

Explanation:
Number of packages are there for deliver = 90
number of packages she delivered = 60
Number of packages left = Number of packages are there for deliver – number of packages she delivered
= 90 – 60
= 30.
Hence, Number of packages left = 30.

Add Tens to a Number Homework & Practice 8.8

Question 27.
27 + 50 = ____________
Answer:
27 + 50 = 77.

Question 28.
_____________ = 80 + 12
Answer:
92 = 80 + 12.

Conclusion:

You can know how to use the number line in a simple method from our Big Ideas Math Book 1st Grade Answer Key Chapter 8 Add and Subtract Tens. Learn the fundamentals of maths like addition and subtraction from this page. Bookmark our page to get the Answer Key of all Big Ideas Math Grade 1 Chapters from 1 to 14.

Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes

Big Ideas Math Answers Grade 4 Chapter 14

Understand the concept of Two-Dimensional Shapes with the help of Big Ideas Math Grade 4 Answer Key Chapter 14. The lesson Identify Symmetry and Two-Dimensional Shapes involves line symmetry, symmetric shapes, Classify Quadrilaterals, Classify Triangles, and so on. We have provided a clear cut explanation for all the questions topic wise. Thus Download Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes pdf for free.

Big Ideas Math Book 4th Grade Answer Key Chapter 14 Identify Symmetry and Two-Dimensional Shapes

Quick learning is possible with the clear explanations of Grade 4 Chapter 14 Answer Key. So, instantly start your practicing now and strengthen the knowledge. Complete concepts are solved with simple steps included in the Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes. Tap the below-attached links and improve your math skills.

Lesson: 1 Line Symmetry

Lesson: 2 Draw Symmetric Shapes

Lesson: 3 Classify Triangles by Sides

Lesson: 4 Classify Triangles by Angles

Lesson: 5 Classify Quadrilaterals

Performance Task

Lesson 14.1 Line Symmetry

Explore and Grow

How many ways can you fold the Letter X so that the two parts match exactly? Draw models to support your Answer:
2 ways as shown in the below diagram.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 1

Structure
What other letters can you fold so that the two parts match exactly? Draw models to support your answer. Compare your models with a partner.
Answer:
The letter W can be folded. Both the letters formed after folding will be the same.

Think and Grow:

A shape has line symmetry when it can be folded on a line so that two parts match exactly. The fold line is called a line of symmetry.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 2
Example
Determine whether the shape has line symmetry.

The shape can be folded so that two parts match exactly.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 3
The shape has __4_ lines of symmetry.
So, the shape ___ line symmetry.

Example
Determine whether the shape has line symmetry.
The shape cannot be folded so that two parts match exactly.
The shape has ___ lines of symmetry.
So, the shape ___ line symmetry.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 4

Show and Grow

Determine whether the line is a line of symmetry.
We say there is symmetry when the exact reflection or mirror image of a line, shape or object gets created.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 5
Answer:
No, because it’s not dividing the figure into identical Halves.

Question 2.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 6
Answer:
Yes, divides it into identical halves.

Question 3.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 7
Answer:
Yes, divides it into identical halves.

Determine whether the shape has line symmetry. Draw each line of symmetry

Question 4.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 8
Answer:
Yes, divides it into identical halves.

Question 5.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 9
Answer:
Yes,divides it into identical halves.

Question 6.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 10
Answer:
No, As it cannot divide the shape into identical halves.

Apply and Grow: Practice

Determine whether the shape has line symmetry. Draw each line of symmetry.

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 11
Answer:
No, as it is a scalene triangle with different sides.

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 12
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 13
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 14
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 11.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 15
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 12.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 16
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 13.
Structure
Create a three-digit number that has a line of symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 17
Answer:

Question 14.
Writing
How can you tell when a line is not a line of symmetry?
Answer:
The line of symmetry can be defined as the axis or imaginary line that passes through the center of the shape or object and divides it into identical halves.
When the line of symmetry doesn’t divide the shape into identical halves then it is not a line of symmetry.

Question 15.q
DIG DEEPER!
Draw four lines of symmetry for the circle. Can you draw more lines of symmetry? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 18
Answer:
Yes.Circle has infinite lines of symmetry.
Since there are an infinite number of lines through the center, the circle has an infinite number of lines of symmetry. … This means that the parts of the circle on each side of the line  must  have the same area. So a line of symmetry divides the circle into two parts with equal area.

Think and Grow:Modeling Real Life

Example
Divide the archery target into multiple sections using each of its lines of symmetry. An archer gets 2 arrows in each section. How many arrows does the archer get on the target? Draw each line of symmetry on the target.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 19
The target has _4__ lines of symmetry.
Count the sections.
The lines of symmetry divide the target into _8__ sections.
The archer gets 2 arrows in each section, so multiply the number of sections by 2.
_8__ × _2_ = 16
So, the archer gets _16__ arrows on the target.
Answer:
16.

Show and Grow

Question 16.
Divide the soccer field into multiple sections using each of its lines of symmetry. At the start of the game, there are 3 players in each section. How many players are on the field?
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 20
Answer:
Number of players in each section = 3
Number of sections = 6.
Total number of players = Number of sections x Number of players in each section
= 3 X 6. = 18 players.

Question 17.
DIG DEEPER!
Your friend folds a rectangular sheet of paper in half horizontally, and then in half again vertically. She then draws the shape shown on the folded paper, cuts it out, and then unfolds it. How many lines of symmetry does your friend’s cutout have? Explain.
Answer:
The rectangle paper is folded into 4 small rectangle.
Then the rectangle is cut it out. it means 2 side edges cuts.
Two rectangles with adjacent are left.
lines of symmetry does your friend’s cutout = 2.

Line Symmetry Homework & Practice 14.1

Determine whether the shape has line symmetry.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 21
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 22
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 23
Answer:
Yes, as the line of symmetry divides it into identical halves.

Determine whether the line is a line of symmetry. Draw each line of symmetry.

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 24
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 25
Answer:
NO, as the line of symmetry cannot divide it into identical halves.

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 26
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 27
Answer:
No, as the line of symmetry cannot divide it into identical halves.

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 28
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 29
Answer:
Yes, as the line of symmetry divides it into identical halves.

Question 10.
Precision
Which figure correctly shows all the lines of symmetry of a square?
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 30
Answer:
Figure 4

Question 11.
YOU BE THE TEACHER
Your friend says the shape has exactly three lines of symmetry. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 31
Answer:
No
Explanation:
A regular hexagon with six equal sides has six lines of symmetry.
For all regular polygons, the number of lines of symmetry is equal to the number of sides.

Question 12.
Modeling Real Life
Divide the tennis court into multiple sections using each of its lines of symmetry. There are 4 players on the court with an equal number of players in each section. How many players are in each section?
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 32
Answer:
Number of players = 4.
Number of sections= 2.
Number of players in each section = 4 /2=2.

Question 13.
DIG DEEPER!
In art class, you are making a black and white art piece. Your teacher says it has to have exactly 1 line of diagonal symmetry. Shade the square below to show what the art piece could look like.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 33
Answer:

Review & Refresh

Question 14.
\(\frac{8}{12}\)
Answer:
8/12=2/3=0.66

Question 15.
\(\frac{5}{100}\)
Answer:
5/100=0.05

Lesson 14.2 Draw Symmetric Shapes

Explore and Grow

You want to create a mask that has at least 1 line of symmetry. Does either mask meet your requirement? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 34
Answer:

Complete the mask below so that it has 1 line of symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 35
Answer:

Reasoning
When given one half of a figure and a line of symmetry, how can you draw the other half of the figure? Explain.
Answer:
Line symmetry and mirror reflection are naturally related and linked to each other.Mirror image of the given image helps to draw the other half of the figure.

Think and Grow: Draw Symmetric Shapes

Example
One half of a symmetric shape is shown. Draw the rest of the shape.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 36

One way: Draw a line of Symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 37
Draw the other half of the shape on the opposite side of the line of symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 38

Another Way: Draw a different line of symmetry
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 39
Draw the other half of the shape on the opposite side of the line of symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 40
Answer:

Show and Grow

Question 1.
One half of a shape and a line of symmetry are shown. Draw the rest of the shape.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 400
Answer:

Question 2.
One half of a symmetric shape is shown. Draw the rest of the shape.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 41
Answer:

Apply and Grow: Practice

One half of a symmetric shape is shown. Draw the rest of the shape.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 42
Answer:

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 43
Answer:

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 44
Answer:

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 45
Answer:

Question 7.
Reasoning
Draw a shape that has exactly one line symmetry. Draw the line of symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 46
Answer:

Question 8.

YOU BE THE TEACHER
Your friend draws the other half of the shape on the opposite side of the line of symmetry. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 47
Answer:
Yes,The line of symmetry can be defined as the axis or imaginary line that passes through the center of the shape or object and divides it into identical halves.
As per above figure both are two identical halves.

Think and Grow: Modeling Real Life

Example
A rokkaku is a six-sided Japanese kite. One half of a symmetric rokkaku and a line of symmetry are shown. Draw the rest of the kite. Then draw each additional line of symmetry, if any.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 48
Draw the other half of the rokkaku on the opposite side of the line of symmetry.
The rokkaku can also be folded in half vertically so that two parts match exactly. So, draw a vertical line of symmetry.

Show and Grow

Question 9.
The Jamaican flag is symmetric. One half of the flag and a line of symmetry are shown. Draw the rest of the flag. Then draw each additional line of symmetry, if any.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 49
Answer:

Question 10.
DIG DEEPER!
A symmetric quilt is folded in half horizontally, and then in half again vertically. The folded quilt and fold lines are shown. Draw the unfolded quilt. Then draw each line of symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 50
Answer:

Draw Symmetric Shapes Homework & Practice 14.2

Example
One half of a symmetric shape is shown. Draw the rest of the shape.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 51

One Way: Draw a line symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 52
Draw the other half of the shape on the opposite side of the line of symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 53
Another Way: Draw a different line of symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 54
Drawthe other half of the shape on the opposite side of the line of symmetry.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 55

Question 1.
One half of a shape and a line of symmetry are shown. Draw the rest of the shape.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 56
Answer:

Question 2.
One half of a symmetric shape is shown. Draw the rest of the shape.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 57
Answer:

One half of a symmetric shape is shown. Draw the rest of the shape.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 58
Answer:

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 59
Answer:

Question 5.
Structure
Draw a shape that has 5 sides and exactly 1 line of symmetry.
Answer:

Question 6.
One fourth of a shape and two lines of symmetry are shown. Draw the rest of the shape.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 60
Answer:

Question 7.
Modeling Real Life
The flag of the Bahamas is symmetric. One half of the flag and a line of symmetry are shown. Draw the rest of the flag. Then draw each additional line of symmetry, if any.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 61
Answer:
No other line of symmetry is possible.

Question 8.
DIG DEEPER!
Snowflakes are symmetric. One half of a snowflake and a line of symmetry are shown. Draw the rest of the snowflake. Then draw each additional line of symmetry, if any.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 62
Answer:

Review & Refresh

Find the equivalent weight.

Question 9.
2T = lb
Answer:
1Ton =2,000 pounds
2Ton =4,000 pounds

Question 10.
15lb = ___oz
Answer:
1pound = 16 ounce
15lb = 240 oz

Lesson 4.3 Classify Triangles by Sides

Explore and Grow

Sort the triangles into two or more groups using their side lengths. Explain how you sorted the triangles.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 63
Answer:
Triangle A = Isosceles Triangle -An isosceles triangle has both two equal sides.
Triangle B = Scalene Triangle- no sides with same length
Triangle C -Isosceles Triangle -An isosceles triangle has both two equal sides.
Triangle D-Scalene Triangle- no sides with same length
Triangle E – Equilateral Triangle – All sides equal.
Triangle F – Equilateral Triangle – All sides equal.

Structure
Can you draw a triangle that does belong in any of your groups above? Explain.

Think and Grow: Classify Triangles by Sides

Triangles can be classified by their sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 64
Answer:

Example
Classify the triangle by its sides.
The triangle has _2_ sides with the same length.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 66
So, it is an ___ triangle.
Answer:
Isosceles Triangle.

Show and Grow

Classify the triangle by its sides.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 401

Answer:
Scalene Triangle-no sides with same length

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 402

Answer:
Equilateral Triangle – All sides equal.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 403

Answer:
Isosceles Triangle -An isosceles triangle has both two equal sides.

Question 4.
Use the triangular grid to draw any triangle. Classify the triangle by its sides
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 67
Answer:
It is Equilateral triangle with all sides equal.

Apply and Grow: Practice

Classify the triangle by its sides.

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 68
Answer:
Isosceles Triangle -An isosceles triangle has both two equal sides.

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 69
Answer:
Scalene Triangle-no sides with same length

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 70
Answer:
Equilateral Triangle – All sides equal.

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 71
Answer:
Scalene Triangle-no sides with same length

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 72
Answer:
Equilateral Triangle – All sides equal.

Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 73
Answer:
Isosceles Triangle -An isosceles triangle has both two equal sides.

Question 11.
Draw a triangle that has three lines of symmetry. Classify the triangle by its sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 74
Answer:
Equilateral Triangle – All sides equal.It has 3 lines of symmentry.

Question 12.
Draw a triangle that has no sides with the same length. Classify the triangle by its sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 75
Answer:

Question 13.
Reasoning
How many lines of symmetry does a scalene triangle have? Explain.
Answer:
A scalene triangle have is zero lines of symmetry
Explanation:
It has no equal sides and no equal angles.

Question 14.
DIG DEEPER!
Write an addition equation and a multiplication equation for the perimeter, P, of the triangle
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 76
Answer:
if a triangle has sides a, b, and c, then the perimeter of that triangle will be P = a + b + c.
As the above Triangle given is an Equilateral Triangle then perimeter = s +s+s=3s.

Think and Grow: Modeling Real Life

Example
Classify the yield sign by its sides. Then find the perimeter of the sign.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 77
The yield sign is a triangle that has 3 sides with the same length.
So, it is an _Equilateral__ triangle.
To find the perimeter of the yield sign, find the sum of all of its side lengths.
36 + 36 + 36 = _108__
So, the perimeter of the yield sign is _108__ inches.
Answer:
108 inches.

Show and Grow

Question 15.
Classify the sports pennant by its sides. Then find the perimeter of the pennant.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 78
Answer:
Perimeter of triangle = a + b+ c (Where a,b,c are 3 sides of triangle).
P=  30(5/10) +77(7/10) + 77(7/10)
P=(305+777+777)/10
P=1859/10=185.9cms.

Question 16.
DIG DEEPER!
The perimeter of the canoe sail is 29.75 feet. Find the unknown side length. Then classify the sail by its sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 79
Answer:
Perimeter of triangle = a + b+ c (Where a,b,c are 3 sides of triangle).
P=29.75feet
29.75=12+6.5+c
c=29.75-18.5=11.25 feet.

Question 17.
DIG DEEPER!
An emergency warning triangle is an equilateral triangle. It has a perimeter of 51 inches. What are the side lengths of the warning triangle?
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 410
Answer:
Perimeter of Equilateral Triangle= s +s+s=3s.(s=side of the triangle)
P = 51 inches.
51=3s.
s=51/3=17 inches.

Classify Triangles by Sides Homework & Practice 14.3

Classify the triangle by its sides.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 80
Answer:
Equilateral Triangle – All sides equal.

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 81
Answer:
Isosceles Triangle -An isosceles triangle has both two equal sides.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 82
Answer:
Scalene Triangle-no sides with same length

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 83
Answer:
Isosceles Triangle -An isosceles triangle has both two equal sides.

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 84
Answer:
Scalene Triangle-no sides with same length

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 85
Answer:
Equilateral Triangle – All sides equal.

Question 7.
Draw a triangle that has exactly 2 sides with the same length. Classify the triangle by its sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 86
Answer:
Isosceles Triangle -An isosceles triangle has both two equal sides.

Question 8.
Draw a triangle that has exactly 1 line of symmetry. Classify the triangle by its sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 87
Answer:
Isosceles Triangle -An isosceles triangle has two equal sides.

Question 9.
Draw a triangle that has no lines of symmetry. Classify the triangle by its sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 88
Answer:

A scalene triangle have is zero lines of symmetry
Explanation:
It has no equal sides and no equal angles.

Question 10.
Structure
Classify the triangle pattern block by its sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 89
Answer:
It is a Equilateral Triangle as it has 3 equal sides.

Question 11.
Reasoning
Can the triangle have 3 lines of symmetry? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 90
Answer:
The above given Triangle is the Isosceles triangle it have only one line of symmentry.
Isosceles Triangle -An isosceles triangle has both two equal sides.
Only an equilateral triangle has three lines of symmetry.

Question 12.
Reasoning
Your friend draws a triangle with side lengths of 25.3 centimeters, 15.2 centimeters, and 12.6 centimeters. Classify your friend’s triangle.
Answer:
Scalene Triangle have 3 different sides, none of the side is equal.

Question 13.
Modeling Real Life
Classify the roof truss by its sides. Then find the perimeter of the roof truss.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 91
Answer:
P=a+b+c where a,b and c are sides.
P=30+16+34=80ft.

Review & Refresh

Find the perimeter of the rectangle

Question 14.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 92
Answer:
Perimeter of a rectangle= 2(l+b) (where l=length and b=breadth of the rectangle).
Perimeter=2(32+20)=2(52)
Perimeter=104 yd.

Question 15.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 93
Answer:
Perimeter of a rectangle= 2(l+b) (where l=length and b=breadth of the rectangle).
Perimeter=2(36+45)=2(81)
Perimeter=162ft.

Lesson 14.4 Classify Triangles by Angles

Explore and Grow

Sort the triangles into two or more groups using their angle measures. Explain how you sorted the triangles.
Big Ideas Math Answers 4th Grade Chapter 14 Identify Symmetry and Two-Dimensional Shapes 94

Construct Arguments
Your friend saysa triangle that has three angles with the same measure also has three sides with the same length. Is your friend correct? Explain.

Think and Grow: Classify Triangles by Angles

Triangles can be classified by their angles.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 95
Example
Classify the triangle by its angles.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 96
The triangle has one _obtuse_ angle.
So, it is an _obtuse__ triangle.
Answer:
Obtuse Triangle

Example
Classify the triangle by its angles and its sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 97
The triangle has one _right_ angle.
and _2_ sides with the same length.
So, it is an _right angled_ ___ triangle.
Answer:
Right angled triangle.

Show and Grow

Question 1.
Classify the triangle by its angles.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 98
Answer:
Right-angled triangle-it have one right angle.

Question 2.
Classify the triangle by its angles and its sides
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 99
Answer:
Acute triangle.
Isosceles Triangle – 2 equal sides.

Apply and Grow: Practice

Classify the triangle by its angles.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 100
Answer:
Right angle triangle.

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 101
Answer:
Acute triangle

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 102
Answer:
Equiangular Triangle.

Classify the triangle by its angles and its sides.

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 103
Answer:
Acute triangle-angles
Isosceles triangle-sides

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 104
Answer:
Right angle triangle-angle
Scalene triangle-sides

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 105
Answer:
Obtuse triangle-angle
Scalene triangle-side

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 106
Answer:
Right angle triangle -angle
Isosceles triangle – side

Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 107
Answer:
Equiangular triangle-angles
Equilateral triangle – sides

Question 11.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 108
Answer:
Acute triangle-angle
Isosceles triangle-side

Question 12.
Precision
Draw an acute scalene triangle.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 109
Answer:

Question 13.
Can an obtuse triangle also be an equilateral triangle? Explain.
Answer:
An equilateral triangle can never be obtuse. Since an equilateral triangle has equal sides and angles, each angle measures 60°, which is acute.
Therefore, an equilateral angle can never be obtuse-angled.

Think and Grow: Modeling Real Life

Example
Classify the large triangle on the shuffleboard court by its angles. Verify your answer by finding the measure of each angle.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 110
The triangle does not have a right angle or an obtuse angle.
So, it is an _Acute__ triangle.
Verify: Use a protractor to find the measure of each angle.
The measure of ∠ABC is ___.
The measure of ∠BCA is ___.
The measure of ∠CAB is ___.
Because each angle measure is less than 90°, you can conclude that the triangle is a ___ triangle.
Answer:
Acute triangle
The measure of ∠CAB is 70°.
The measure of ∠BCA is 70°.
The measure of ∠ABC is 40°.


Show and Grow

Question 14.
Classify the triangle in the clothes hanger by its angles. Verify your answer by finding the measure of each angle.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 111
Answer:
Obtuse angle triangle.

Question 15.
DIG DEEPER!
In the picture, the tree is perpendicular to the ground. Draw the triangle formed by the tree, the wire, and the ground. Classify the triangle by its angles. Explain.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 112
Answer:
It is right angles triangle. As the angled formed here is 90°.

Classify Triangles by Angles Homework & Practice 14.4

Classify the triangle by its angles.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 113
Answer:
Right angle triangle

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 114
Answer:
Acute triangle

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 115
Answer:
Acute triangle

Classify the triangle by its angles and its sides

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 116
Answer:
Equiangular triangle-angles
Equilateral triangle – sides

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 117
Answer:
Acute triangle- angles
Scalene triangle-sides

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 118
Answer:
Obtuse triangle-angle
Isosceles triangle-side

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 119
Answer:
Right angle triangle-angle
Scalene triangle – side

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 120
Answer:
Equiangular triangle-angles
Equilateral triangle – sides

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 121
Answer:
Acute triangle – angles
Isosceles triangle – sides

Question 10.
Precision
Draw a right isosceles triangle.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 122
Answer:

Question 11.
Reasoning
Can a triangle have more than one right angle? Explain.
Answer:
No a triangle cannot have more than one right angle as if there are 2 right angles the their sum will be 180° and the third angle will exceed the sum.
Therefore the construction is not possible.

Question 12.
Which One Doesn’t Belong? Which triangle does not belong with the other three?
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 123
Answer:
Figure 2

Question 13.
DIG DEEPER!
Draw \(\overline{B D}\). Then classify each triangle by its angles and its sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 124
Answer:

△BAD and △BDC form Isosceles triangle and acute angle triangle.

Question 14.
Modeling Real Life
Classify the face of the bird house by its angles. Verify your answer by finding the measure of each angle.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 125
Answer:
It is a Right angled triangle. As the angle formed is 90°.

Question 15.
Modeling Real Life
A construction worker measures a triangle formed by the supports on a bridge. The angle measures are 120°, 36°, and 24°. The side lengths are 12.65 meters, 8.61 meters, and 5.91 meters. Classify the triangle by its angles and its sides.
Answer:
It is a Obtuse angle Triangle and Scalene Triangle (all different sides).

Review & Refresh

Add.

Question 16.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 126
Answer:
(14/6)+(27/6)=(14+27)/6
=41/6=6.833

Question 17.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 127
Answer:
=(51/5)+(44/5)=(51+44)/5
=95/5=19.

Question 18.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 128
Answer:
(35/10)+(63/10)+(6/10)
=(35+63+6)/10
=104/10
=10.4

Lesson 14.5 Classify Quadrilaterals

Explore and Grow

Sort the quadrilaterals into two or more groups. Explain how you sorted the quadrilaterals.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 129
Answer:
G and A are Trapezoids – one pair of parallel lines.
E and H are rhombus – all sides equal.
C is a Rectangle parallelogram – 2 pairs of parallel sides and 4 right angles.

Precision
You draw a quadrilateral with four right angles. Your friend draws a quadrilateral with two pairs of parallel sides. Did you and your friend draw the same type of quadrilateral? Explain.
Answer:
Figure can be drawn indifferent many ways. The figure can be Square , Rectangle as they too have with two pairs of parallel sides and four right angles.

Think and Grow: Classify Quadrilaterals

Quadrilaterals can be classified by their angles and sides.

A trapezoid is a quadrilateral that has exactly one pair of parallel sides.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 130

A parallelogram is a quadrilateral that has two pairs of parallel sides. Opposite sides have the same length.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 131

A rectangle is a parallelogram that has four right angles.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 132
A rhombus is a parallelogram that has four sides with the same length.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 133
A square is a parallelogram that has four right angles and four sides with the same length.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 134
Answer:
No square is not a rectangle because rectangle have only opposite sides equal. square cannot be rhombus because it does not have right angles.

Example
Classify the quadrilateral in as many ways as possible.
The quadrilateral has __ right angles,
___ pairs of parallel sides, and
___ pairs of opposite sides with the same length.
So, it is a ___ and a ___.
Answer:

Show and Grow

Classify the quadrilateral in as many ways as possible.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 135
Answer:
It is a Trapezoid
A trapezoid is a quadrilateral that has exactly one pair of parallel sides.

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 136
Answer:
It is a Rhombus
A rhombus is a parallelogram that has four sides with the same length.

Apply and Grow: Practice

Classify the quadrilateral in as many ways as possible.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 137
Answer:
It is a rectangle
A rectangle is a parallelogram that has four right angles.

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 138
Answer:
It is a trapezoid.
A trapezoid is a quadrilateral that has exactly one pair of parallel sides.

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 139
Answer:
It is a Square.
A square is a parallelogram that has four right angles and four sides with the same length.

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 140
Answer:
It is a parallelogram.
A parallelogram is a quadrilateral that has two pairs of parallel sides. Opposite sides have the same length.

Question 7.
The formula for the perimeter of a square with side length of s is P = 4 × s. Can you use this formula to find the perimeter of any other type of quadrilateral? If so, name the type of quadrilateral.
Answer:
It can be used to find the perimeter of a Rhombus.
A rhombus is a parallelogram that has four sides with the same length.

Question 8.
Structure
How many lines of symmetry can a parallelogram have?
Answer:
lines of symmetry can a parallelogram have is 2.

Question 9.
DIG DEEPER!
Write All, Some, or No for each statement
___ parallelograms are rectangles.
___ squares are rectangles.
___ trapezoids are parallelograms.
___ rectangles are squares.
___ trapezoids have four sides with the same length.
Answer:
_Some__ parallelograms are rectangles.
_All__ squares are rectangles.
__No_ trapezoids are parallelograms.
__No_ rectangles are squares.
_No__ trapezoids have four sides with the same length.

Think and Grow: Modeling Real Life

Example
Two artists each want to paint an identical half of a painting, like the one shown. Classify the shape of the painting in as many ways as possible. Then identify and show how many different ways the artists can divide the painting, if any.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 141
Classify the shape of the painting in as many ways as possible.
The painting has _4__ sides,
_2__ pairs of parallel sides,
_0__ right angles, and
_4__ sides with the same length.
So, it is a Rhombus.
Determine whether the painting has line symmetry. If so, find how many different ways the artists can divide the painting.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 142
The rhombus-shaped painting has __ lines of symmetry.
So, the artists can divide the painting __ different ways.
Answer:
The rhombus-shaped painting has _2_ lines of symmetry.
So, the artists can divide the painting _2_ different ways.

Show and Grow

Question 10.
You and your friend each want to share an identical half of the tabletop. Classify the shape of the tabletop in as many ways as possible. Then identify and show how many different ways you and your friend can divide the tabletop, if any.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 143
Answer:
The given shape is a Trapezoid.

Question 11.
A horse needs at least 4,046 square meters of pasture. Classify the shape of the pasture in as many ways as possible. Is the pasture large enough for the horse? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 144
Answer:
The side of the square = 64m
Area of the Square = side x side=64m x 64 m =4096sqmtrs.
Area required by horse = 4046 sqmtrs.
Therefore area of square is more than pasture of horse.

Classify Quadrilaterals Homework & Practice 14.5

Classify the quadrilateral in as many ways as possible.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 145
Answer:
It is a parallelogram
A parallelogram is a quadrilateral that has two pairs of parallel sides. Opposite sides have the same length.

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 146
Answer:
It is a Trapezoid.
A trapezoid is a quadrilateral that has exactly one pair of parallel sides.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 147
Answer:
It is a rhombus.
A rhombus is a parallelogram that has four sides with the same length.

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 148
Answer:
It is a Rectangle.
A rectangle is a parallelogram that has four right angles.

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 149
Answer:
It is a Trapezoid.
A trapezoid is a quadrilateral that has exactly one pair of parallel sides.

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 150
Answer:
It is a Square.
A square is a parallelogram that has four right angles and four sides with the same length.

Question 7.
Structure
Your friend sorts the shapes into two different groups. How do you think she sorted? Where does the shape below belong? Explain.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 151
Answer:
The above belong to the group 2 . The group 2 images have 2 pairs of sides equal.

Question 8.
Reasoning
All of the sides of an equilateral triangle have the same length. Is an equilateral triangle a rhombus? Explain.
Answer:
A Rhombus have all sides equal. 2 equilateral triangles form a rhombus.

Question 9.
DIG DEEPER!
Within the star, trace at least two different examples of trapezoids and parallelograms. Explain how you found each quadrilateral.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 152
Answer:

A parallelogram is a quadrilateral that has two pairs of parallel sides. Opposite sides have the same length. BJGD and CAIF are two parallelograms.
A trapezoid is a quadrilateral that has exactly one pair of parallel sides. LEGJ and KLFI are two trapezoids.

A Quadilateral can be defined as a closed, two-dimensional shape which has four straight sides.

Question 10.
Modeling Real Life
You and your friend each want to share an identical half of the whiteboard shown. Classify the shape of the whiteboard in as many ways as possible. Then identify and show how many different ways you and your friend can divide the whiteboard, if any.
Big Ideas Math Answers Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 153
Answer:
The image is divided into four identical parts with two lines of symmetry.

Review & Refresh

Question 11.
Write \(\frac{7}{10}\) as hundredths in fraction form and decimal form.
Answer:
7/10=0.7

Question 12.
Write \(\frac{50}{100}\) as tenths in fraction form and decimal form.
Answer:
50/100=0.5

Identify Symmetry and Two-Dimensional Shapes Performance Task 14

Roof trusses are wooden structures that support the roof of a building. They come in many different shapes and sizes.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 154
1. The truss has exactly one pair of parallel sides.
a. What polygon describes the shape of the truss?
Answer:   ABCD.
b. Does the truss have line symmetry? Explain.
Answer:
The line of symmetry divide the given truss into identical halves.

c. Name an acute angle, a right angle, and an obtuse angle.
Answer:
Right angle = ∠AED
Acute angle =∠DAE.
Obtuse angle =∠ABC.
d. Name a ray.
Answer: AD.

Question 2.
Most trusses are built in the shape of a triangle.
a. Classify each triangle by its sides. Identify any lines of symmetry.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 156
Answer:

b. Why do you think roof trusses come in different shapes and sizes?
Answer:
Roof trusses come in all shapes and sizes depending on your build project.

Question 3.
You see a truss in the shape of an obtuse isosceles triangle. The perimeter of the triangle is 36 feet. The length of the side opposite the obtuse angle is 16 feet. What are the lengths of the other two sides?
Answer:
An Isosceles triangle have 2 sides equal.
side of Isosceles triangle=16 feet.
Perimeter of isosceles triangle = s1+s2+s3. = s+s+16
36 feet =2s +16
2s=36-16
2s=20
s=10feet.
The other two sides lengths are 10 feet each.

Question 4.
Draw a triangular roof truss that has line symmetry. Include a few support lines inside the truss. Classify your triangle by its sides. What types of angles are on the inside of the truss?
Answer:

△ABF  = Isosceles triangle.
△BFC= Scalene Triangle.
△FEC = Isosceles Triangle.
△EDC=Scalene Triangle.

Identify Symmetry and Two-Dimensional Shapes Activity

Pyramid Climb and Slide

Directions:
1. Players take turns spinning the spinner.
2. On your turn, move your counter to the next triangle that best matches your spin.
3. If you land at the bottom of a ladder, climb to the top of the ladder. If you land at the top of a slide, slide down to the bottom of the slide.
4. The first player to reach the top of the pyramid wins!
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 157
Answer:

Identify Symmetry and Two-Dimensional Shapes Chapter Practice

14.1 Line Symmetry

Determine whether the line is a line of symmetry.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 158
Answer:
Yes it is a line of Symmetry .It divides the given shape into two identical halves.

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 159
Answer:
No, it is not a line of symmetry it doesnot divide the given shape into two identical halves.

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 160
Answer:
Yes it is a line of Symmetry .It divides the given shape into two identical halves.

Determine whether the shape has line symmetry. Draw each line of symmetry.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 161
Answer:
Yes it is a line of Symmetry .It divides the given shape into two identical halves.

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 162
Answer:
Yes it is a line of Symmetry .It divides the given shape into two identical halves.

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 163
Answer:
No, It doesnot divides the given shape into two identical halves.

Question 7.
Modeling Real Life
Divide the parking lot shown into multiple sections using each of its lines of symmetry. When there are 4 cars in each section, how many cars are in the parking lot?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 164
Answer:
Number of sections=4
Number of car parking slots in each section=4.
Total car parking slots in all sections = 4 x 4 =16 cars.

14.2 Draw Symmetric Shapes

Question 8.
One half of a shape and a line of symmetry are shown. Draw the rest of the shape.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 165
Answer:

Question 9.
One half of a symmetric shape is shown. Draw the rest of the shape.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 166
Answer:

14.3 Classify Triangles by Sides

Classify the triangles by its sides.

Question 10.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 167
Answer:
Scalene triangle

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 168
Answer:
Equilateral triangle

Question 12.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 169
Answer:
Isosceles triangle

14.4 Classify Triangles by Angles

Classify the triangle by its angles and its sides.

Question 13.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 170
Answer:
Equiangular -Angles
Equilateral-Sides

Question 14.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 171
Answer:
Obtuse triangle -Angle
Isosceles triangle- Sides

Question 15.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 172
Answer:
Right angled Triangle-Angle
Scalene Triangle-Sides.

Question 16.
Which One Doesn’t Belong?
Which triangle does not belong with the other three? Explain.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 173
Answer:
First triangle is obtuse angle triangle. where other 3 triangles are acute triangles.

14.5 Classify Quadrilaterals

Classify the quadrilateral in as many ways as possible

Question 17.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 174
Answer:
It is parallelogram.
A parallelogram is a quadrilateral that has two pairs of parallel sides. Opposite sides have the same length.

Question 18.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 175
Answer:
It is a Square.
A square is a parallelogram that has four right angles and four sides with the same length.

Identify Symmetry and Two-Dimensional Shapes Cumulative Practice 1-14

Question 1.
What is the difference of 10 and 4\(\frac{7}{8}\) ?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 176
Answer:
C – 5(1/8)

Question 2.
Which letters have at least one line of symmetry?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 177
Answer:
W,Z AND H

Question 3.
The table shows the amount of rain that falls in each city in 1 day. Which city has more rainfall than Detroit and less rainfall than Seattle?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 178
Answer:
Chicago – 0.72

Question 4.
Which shapes have four sides with the same length?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 179
Answer:
Rhombus and Square.

Question 5.
A movie starts at 2:55 P.M. and ends at 4:45 P.M. How long was the movie?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 180
Answer:
1 Hour 50 Mins

Question 6.
Which are equivalent to \(3 \times \frac{5}{6}\) ?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 181
Answer:
Expect 8/6 all the options are equal to \(3 \times \frac{5}{6}\) .

Question 7.
The rectangle has a perimeter of 48 centimeters. What is the width of the rectangle?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 182
Answer:
Perimeter of Rectangle = width + length.
48 = w + 15.
w=48-15
w=33cms.

Question 8.
Which two lines appear to be parallel?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 183
Answer:
Option A.

Question 9.
Your friend draws a triangle with side lengths of 10 centimeters,12 centimeters, and 10 centimeters. What type of triangle did your friend draw?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 184
Answer:
It is a Isosceles Triangle. It has 2 sides equal.

Question 10.
What is the quotient of 846 and 3?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 185
Answer:
282

Question 11.
Which rectangular dimensions have an area of 3,600 square miles?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 186
Answer:
72 miles long and 50 miles wide.

Question 12.
The angle measures of a triangle are 57°, 60°, and 63°. How can you classify the triangle?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 187
Answer:
An acute-angled triangle is a type of triangle in which all the three internal angles of the triangle are acute, that is, they measure less than 90°.

Question 13.
Newton is walking on Path X that continues straight ahead. He turns onto Path Y.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 188
Using a protractor, which statement is true about ∠FGH
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 189
Answer:
The angle is 45°, and is acute.

Question 14.
You jump 3\(\frac{1}{2}\) feet. How far do you jump in inches?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 190
Answer:
3(1/2)=7/2=3.5feet.

Question 15.
Which of the following shows \(\overrightarrow{D E}\) ?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 191
Answer:
Option DE.

Question 16.
A rectangular backyard has a perimeter of 280 feet. The length of the backyard is 75 feet. What is the area of the backyard?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 192
Answer:
The perimeter of the backyard =280 feet.
length of the backyard =75 feet
The breadth of the backyard = b feet.
280=75 + b
b=280-75
b=205.
Area of the backyard= l x b.
Area = 205 x 75.
Area=15375 sq feet.

Question 17.
Your friend makes a banner by taping the edges of 10 pieces of paper together. Each piece of paper is 28 centimeters long and 22 centimeters wide.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 193
Part A What is the length of the entire banner? Explain.
Answer:
28 centimeters long and 22 centimeters wide. As the tapping is done for the edges only.
Part B Your friend wants to place ribbon along the outside of the entire banner. How much ribbon will she need?
Answer:
Perimeter = length + breadth.
P= 28 + 22= 50 cms.

Identify Symmetry and Two-Dimensional Shapes Steam Performance Task 1-14

In computer graphic design, polygons are connected at the vertices and edges to create a two-dimensional image that appears three-dimensional. This is called polygon mesh.
1. The elephant graphic is made from different polygons. Do the polygons have the same number of sides? Why do you think that is?
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 194
2. Use the quadrilateral shown from the elephant graphic.
a. Exactly one pair of sides are parallel. Classify the quadrilateral.
Answer:
It is a Trapezoid.
b. Which two lines are parallel?
Answer:
BC ll AD.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 195
c. Which angles are acute?
Answer:
ADC.
d. The perimeter is 22 millimeters. Find the missing side lengths.
Answer:
Perimeter = 22 mm.
P= x +3 + x + 9 mm.
22= 2x + 12.
2x=10.
x=5 mm.

Question 3.
There are many different triangles in the elephant graphic. Use the side lengths and angle measures to classify two of the triangles. Trace your triangles in the elephant graphic
Answer:

Question 4.
In a polygon mesh, each polygon is shaded differently to make the object appear more three-dimensional. In the drawing, two polygons still need to be shaded. Which one should be darker, the triangle or the quadrilateral? Why?
Big Ideas Math Answer Key Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 600
Answer:
One should be darker is the triangle.A Triangle is a polygon with three edges and three vertices.

Question 5.
Triangle mesh is when a three-dimensional design is created using only triangles. The triangle mesh design below has line symmetry.
a. Use the line of symmetry to finish the design.
Big Ideas Math Solutions Grade 4 Chapter 14 Identify Symmetry and Two-Dimensional Shapes 196

b. Name two types of triangles you see in the design. Trace your triangles above. Explain.
Answer:
∆ DFE is Isosceles triangle.
∆ ABC is Scalene triangle.

Question 6.
Use the Internet or some other resource to find out more about polygon mesh and how it is used in graphic design. Write one thing that you learn.
Answer:
A polygon mesh is a collection of vertices, edges and faces that defines the shape of a polyhedral object.
In graphic design, it is used to make up a 3D object, 3D animated film, advertising, or video games.
We learn in making different animated pictures, videos.

Conclusion:

Enhance your performance by practicing the problems from Big Ideas Math Grade 4 Solution Key Chapter 14 Identify Symmetry and Two-Dimensional Shapes. Understand the concepts of symmetry and 2-D shapes and apply them to the real-time problems. Solving the problems of math is fun. Enjoy it by solving the questions given at the end of the chapter. So that you can know at which topic who were lagging. We are here to clarify your doubts. For any doubts, you can post the comments in the below section.