lesson 7 homework 2.7 answer key

school Campus Bookshelves
menu_book Bookshelves
perm_media Learning Objects
login Login
how_to_reg Request Instructor Account
hub Instructor Commons
Download Page (PDF)
Download Full Book (PDF)
Periodic Table
Physics Constants
Scientific Calculator
Reference & Cite
Tools expand_more
Readability

selected template will load here

This action is not available.

2.7E: Exercises for Section 2.7

Last updated
Save as PDF
Page ID 57426

1) Draw a number line, then plot the numbers $4,3,-4,7 / 8$, and $−8/3$ on your number line. Label each point with its value. Finally, list the numbers in order, from smallest to largest.

$Ans 2.6.1.png$

2) Draw a number line, then plot the numbers $5,3,-4,5 / 7$, and $−4/3$ on your number line. Label each point with its value. Finally, list the numbers in order, from smallest to largest.

3) Draw a number line, then plot the numbers $-5,5,4,2 / 3$, and $8/3$ on your number line. Label each point with its value. Finally, list the numbers in order, from smallest to largest.

$Ans 2.6.3.png$

4) Draw a number line, then plot the numbers $-3,-2,4,1 / 3$, and $5/2$ on your number line. Label each point with its value. Finally, list the numbers in order, from smallest to largest.

In Exercises 5-20, shade each of the following sets on a number line.

5) $\{x \, | \, x \geq-7\}$

$Ans 2.6.5.png$

6) $\{x \, | \, x \geq-1\}$

7) $\{x \, | \, x<2\}$

$Ans 2.6.7.png$

8) $\{x \, | \, x<-6\}$

9) $(-\infty, 2)$

$Ans 2.6.9.png$

10) $(-\infty,-9)$

11) $(6, \infty)$

$Ans 2.6.11.png$

12) $(5, \infty)$

13) $\{x \, | \, x>7\}$

$Ans 2.6.13.png$

14) $\{x \, | \, x>-8\}$

15) $[0, \infty)$

$Ans 2.6.15.png$

16) $[7, \infty)$

17) $\{x \, | \, x \leq-2\}$

$Ans 2.6.17.png$

18) $\{x \, | \, x \leq 7\}$

19) $(-\infty, 3]$

$Ans 2.6.19.png$

20) $(-\infty,-1]$

In Exercises 21-28, use set-builder notation to describe the shaded region on the given number line.

$Exercise 2.6.21.png$

$\{x \, | \, x \leq 9\}$

$Exercise 2.6.22.png$

$\{x \, | \, x<-8\}$

$Exercise 2.6.24.png$

$\{x \, | \, x>-2\}$

$Exercise 2.6.26.png$

$\{x \, | \, x \geq-3\}$

$Exercise 2.6.28.png$

In Exercises 29-36, use interval notation to describe the shaded region on the given number line.

$Exercise 2.6.29.png$

$(4, \infty)$

$Exercise 2.6.30.png$

$(-\infty,-2)$

$Exercise 2.6.32.png$

$(-\infty, 5]$

$Exercise 2.6.34.png$

$[1, \infty)$

$Exercise 2.6.36.png$

In Exercises 37-44, solve each of the given inequalities. Sketch the solution on a number line, then use set-builder and interval notation to describe your solution.

37) $x+10<19$

$(-\infty, 9)$

38) $x+17 \geq 7$

39) $4 x<8$

$(-\infty, 2)$

40) $16 x \geq-2$

41) $-2 x \leq-2$

42) $-18 x>-20$

43) $x-18>-10$

$(8, \infty)$

44) $x-8 \leq-18$

In Exercises 45-62, solve each of the given inequalities. Sketch the solution on a number line, then use set-builder and interval notation to describe your solution.

45) $-5 x-6 \geq 4-9 x$

$[5 / 2, \infty)$

46) $2 x-7 \geq-3-4 x$

47) $16 x-6 \leq 18$

$(-\infty, 3 / 2]$

48) $8 x-14 \leq-12$

49) $-14 x-6 \geq-10-4 x$

$(-\infty, 2 / 5]$

50) $-13 x-4 \geq-2-5 x$

51) $5 x+18<38$

$(-\infty, 4)$

52) $9 x+16<79$

53) $-16 x-5 \geq-11-6 x$

$(-\infty, 3 / 5]$

54) $-11 x-7 \geq-15-5 x$

55) $2 x-9 \geq 5-8 x$

$[7 / 5, \infty)$

56) $-3 x-6 \geq-2-9 x$

57) $-10 x-4 \leq 18$

$[-11 / 5, \infty)$

58) $-6 x-14 \leq 1$

59) $-12 x+4<-56$

$(5, \infty)$

60) $-18 x+6<-12$

61) $15 x+5<6 x+2$

$(-\infty,-1 / 3)$

62) $12 x+8<3 x+5$

In Exercises 63-76, solve each of the given inequalities. Sketch the solution on a number line, then use set-builder and interval notation describe your solution.

63) $\dfrac{3}{2} x>\dfrac{9}{8}$

$(3 / 4, \infty)$

64) $\dfrac{6}{7} x>\dfrac{3}{4}$

65) $x+\dfrac{3}{2}<\dfrac{9}{5}$

$(-\infty, 3 / 10)$

66) $x+\dfrac{1}{4}<-\dfrac{1}{5}$

67) $\dfrac{4}{7}-\dfrac{1}{6} x \leq \dfrac{4}{3} x-\dfrac{1}{2}$

$[5 / 7, \infty)$

68) $\dfrac{5}{3}-\dfrac{3}{4} x \leq \dfrac{7}{4} x-\dfrac{3}{5}$

69) $x-\dfrac{3}{8} \geq-\dfrac{9}{7}$

$[-51 / 56, \infty)$

70) $x-\dfrac{7}{2} \geq \dfrac{1}{5}$

71) $\dfrac{6}{5} x \leq-\dfrac{4}{7}$

$[10 / 21, \infty)$

72) $\dfrac{4}{3} x \leq \dfrac{2}{9}$

73) $-\dfrac{6}{5} x-\dfrac{7}{3} \leq \dfrac{5}{9}-\dfrac{2}{9} x$

$[-65 / 22, \infty)$

74) $-\dfrac{3}{7} x-\dfrac{1}{2} \leq \dfrac{3}{2}-\dfrac{2}{7} x$

75) $\dfrac{9}{7} x+\dfrac{9}{2}>\dfrac{1}{7} x+\dfrac{7}{2}$

$(-7 / 8, \infty)$

76) $\dfrac{5}{7} x+\dfrac{9}{2}>\dfrac{1}{3} x+\dfrac{5}{2}$

In Exercises 77-84, solve each of the given inequalities. Sketch the solution on a number line, then use set-builder and interval notation containing fractions in reduced form to describe your solution.

77) $-3.7 x-1.98 \leq 3.2$

$[-7 / 5, \infty)$

78) $-3.6 x-3.32 \leq 0.8$

79) $-3.4 x+3.5 \geq 0.9-2.2 x$

$(-\infty, 13 / 6]$

80) $-2.6 x+3.1 \geq-2.9-1.7 x$

81) $-1.3 x+2.9>-2.6-3.3 x$

$(-11 / 4, \infty)$

82) $2.5 x+2.1>1.4-3.8 x$

83) $2.2 x+1.9<-2.3$

$(-\infty,-21 / 11)$

84) $1.6 x+1.2<1.6$

Contributors and Attributions

David Arnold (Retired Professor (Mathematics) at College of the Redwoods )

Texas Go Math
Big Ideas Math
enVision Math
EngageNY Math
McGraw Hill My Math
180 Days of Math
Math in Focus Answer Key
Math Expressions Answer Key
Privacy Policy

$Go Math Answer Key$

Texas Go Math Grade 5 Lesson 2.7 Answer Key Adjust Quotients

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 2.7 Answer Key Adjust Quotients.

Unlock the Problem

A new music group makes 6,127 copies of its first CD. The group sells 75 copies of the CD at each of its shows. How many shows does it take the group to sell all of the CDs?

$Texas Go Math Grade 5 Lesson 2.7 Answer Key 1$

A. Is the estimate too high, too low, or correct? Answer: We know that, 75 × 9 = 675 So, 612 < 675 Hence, from the above, We can conclude that The estimate is too high

B. Adjust the number in the quotient if needed. Answer: We know that, 75 × 8 = 600 So, 612 > 600 Hence, from the above, We can conclude that The quotient can be adjusted to 8

STEP 2: Estimate the next digit in the quotient. Divide the ones. Estimate: 140 ÷ 70 = 2. Try 2 ones.

A. Is the estimate too high, too low, or correct? Answer: The remaining number in the given estimate is: 27 Now, 7 × 3 = 21 So, 21 is closer to 27 than 14 Hence, from the above, We can conclude that The estimate is too low

B. Adjust the number in the quotient if needed. Answer: We know that, 7 × 3 = 21 So, 27 > 21 Hence, from the above, We can conclude that The quotient can be adjusted to 3 So, It takes the group about 83 shows to sell all of the CDs.

Try This! When the difference is equal to or greater than the divisor, the estimate is too low.

Divide. 336 ÷ 48 Estimate. 300 ÷ 50 = 6

$lesson 7 homework 2.7 answer key$

Adjust the estimated digit in the quotient, if needed. Then divide. Try So, 336 ÷ 48 = 6

Math Talk Mathematical Processes

Explain why using the closest estimate could be useful in solving a division problem. Answer: You can determine if an answer in the division is reasonable by learning to estimate the answers. Round the divisor to the nearest 10 or 100, depending on how many digits are in the divisor. Compare your estimate to the exact answer to determine if the exact answer is reasonable

Share and Show

Adjust the estimated digit in the quotient, if needed. Then divide.

$Texas Go Math Grade 5 Lesson 2.7 Answer Key 3$

Problem Solving

H.O.T. Algebra Write the unknown number for each ☐.

Question 4. ☐ ÷ 33 = 11 ☐ = ________ Answer: Let the missing digit be: x So, The given expression is: x ÷ 33 = 11 Now, Multiply with 33 on both sides So, x = 11 × 33 x = 363 Hence, from the above, We can conclude that The missing digit in the given expression is: 363

Question 5. 1,092 ÷ 52 = ☐ ☐ = _________ Answer: Let the missing digit be: x So, The given expression is: 1,092 ÷ 52 = x So, x = 21 Hence, from the above, We can conclude that The missing digit in the given expression is: 21

Question 6. 429 ÷ ☐ = 33 ☐ = ___________ Answer: Let the missing digit be: x So, The given expression is: 429 ÷ x = 33 Now, Divide by 33 into both sides So, x = $\frac{429}{33}$ x = 13 Hence, from the above, We can conclude that The missing digit in the given expression is: 13

Go Math Grade 5 Lesson 2.7 Answer Key Question 7. How do you know whether an estimated quotient is too low or too high? Explain. Answer: If the multiplication of the divisor and quotient is greater than the dividend, then the estimate is too high If the multiplication of the divisor and quotient is less than the dividend, then the estimate is too low

Question 8. Multi-Step A banquet hall serves 2,394 pounds of turkey during a 3-week period. If the same amount is served each day, how many pounds of turkey does the banquet hall serve each day? (A) 50,274 pounds (B) 798 pounds (C) 342 pounds (D) 114 pounds

a. What do you need to find? Answer: It is given that A banquet hall serves 2,394 pounds of turkey during a 3-week period. If the same amount is served each day Hence, The information that you need to find is the number of pounds of turkey does the banquet hall serve each day

b. What information are you given? Answer: The given information is: A banquet hall serves 2,394 pounds of turkey during a 3-week period. If the same amount is served each day

c. What other information will you use? Answer: We know that, 1 week = 7 days So, 3 weeks = 3 × 7 = 21 days Hence, from the above, We can conclude that The other information you will use is the number of days present in a 3-week period

d. Find how many days there are in 3 weeks. There are ________ days in 3 weeks. Answer: We know that, 1 week = 7 days So, 3 weeks = 3 × 7 = 21 days Hence, from the above, We can conclude that There are 21 days in 3 weeks

$lesson 7 homework 2.7 answer key$

Question 10. The Box Sox company packs 12 pairs of socks in a box. How many boxes will the company need to pack 1,020 pairs of socks? Answer: It is given that The Box of Sox company packs 12 pairs of socks in a box So, According to the given information, The number of boxes will the company needed to pack 1,020 pairs of socks = $\frac{1,020}{12}$ = 85 boxes Hence, from the above, We can conclude that The number of boxes will the company needed to pack 1,020 pairs of socks is: 85 boxes

Daily Assessment Task

Fill in the bubble completely to show your answer.

$lesson 7 homework 2.7 answer key$

Texas Test Prep

$lesson 7 homework 2.7 answer key$

Texas Go Math Grade 5 Lesson 2.7 Homework and Practice Answer Key

$lesson 7 homework 2.7 answer key$

Write the unknown number for each ☐.

$lesson 7 homework 2.7 answer key$

Problem-Solving

$lesson 7 homework 2.7 answer key$

Lesson Check

$lesson 7 homework 2.7 answer key$

Answer Key 2.7

[latex]x=ky[/latex]
[latex]x=kyz[/latex]
[latex]x=\dfrac{k}{y}[/latex]
[latex]x=ky^2[/latex]
[latex]x=kzy[/latex]
[latex]x=\dfrac{k}{y^3}[/latex]
[latex]x=ky^2\sqrt{z}[/latex]
[latex]x=\dfrac{k}{y^6}[/latex]
[latex]x=\dfrac{ky^3}{\sqrt{z}}[/latex]
[latex]x=\dfrac{k}{y^2\sqrt{z}}[/latex]
[latex]x=\dfrac{kzy}{p^3}[/latex]
[latex]x=\dfrac{k}{y^3z^2}[/latex]
[latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} A&=&kB \\ \\ (15)&=&k(5) \\ \\ \dfrac{15}{5}&=&\dfrac{k(5)}{5} \\ \\ k&=&3 \end{array}[/latex]
[latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} P&=&kQR \\ \\ (12)&=&k(8)(3) \\ \\ \dfrac{12}{24}&=&\dfrac{k(8)(3)}{24} \\ \\ k&=&\dfrac{1}{2} \end{array}[/latex]
[latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} A&=&\dfrac{k}{B} \\ \\ (7)&=&\dfrac{k}{(4)} \\ \\ (4)7&=&\dfrac{k}{\cancel{4}}\cancel{(4)} \\ \\ k&=&28 \end{array}[/latex]
[latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} A&=&kB^2 \\ \\ (6)&=&k(3)^2 \\ \\ \dfrac{6}{9}&=&\dfrac{k(3)^2}{9} \\ \\ k&=&\dfrac{6}{9}\text{ or }\dfrac{2}{3} \end{array}[/latex]
[latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} C&=&kAB \\ \\ (24)&=&k(3)(2) \\ \\ \dfrac{24}{6}&=&\dfrac{k(3)(2)}{6} \\ \\ k&=&4 \end{array}[/latex]
[latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{k}{x^3} \\ \\ (54)&=&\dfrac{k}{(3)^3} \\ \\ 54&=&\dfrac{k}{27} \\ \\ 27\cdot 54&=&\dfrac{k}{\cancel{27}}\cdot \cancel{27} \\ \\ k&=&1458 \end{array}[/latex]
[latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} x&=&kY \\ \\ (12)&=&k(8) \\ \\ \dfrac{12}{8}&=&\dfrac{k(8)}{8} \\ \\ k&=&\dfrac{12}{8}\text{ or }\dfrac{3}{2} \end{array}[/latex]
[latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} A&=&kB^2\sqrt{C} \\ \\ (25)&=&k(5)^2\sqrt{(9)} \\ \\ 25&=&k(75) \\ \\ k&=&\dfrac{25}{75} \\ \\ k&=&\dfrac{1}{3} \end{array}[/latex]
[latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{kmn^2}{d} \\ \\ (10)&=&\dfrac{k(4)(5)^2}{(6)} \\ \\ k&=&\dfrac{\cancel{10}\cancel{5}\cdot \cancel{6}3}{\cancel{(4)}(5)^{\cancel{2}}} \\ \\ k&=&\dfrac{3}{5} \end{array}[/latex]
[latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} P&=&\dfrac{kT}{V} \\ \\ (10)&=&\dfrac{k(250)}{(400)} \\ \\ k&=&\dfrac{10(400)}{250} \\ \\ k&=&16 \end{array}[/latex]
[latex]I=kV[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ I&=&5 \text{ A} \\ V&=&15\text{ V} \\ k&=&\text{find} \\ \\ I&=&kV \\ 5\text{ A}&=&k(\text{15 V}) \\ \\ k&=&\dfrac{\text{5 A}}{\text{15 V}} \\ \\ k&=&\dfrac{1}{3}\text{ A/V} \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ I&=&\text{find} \\ k&=&\dfrac{1}{3} \\ \\ V&=&\text{25 V} \\ \\ I&=&kV \\ I&=&\left(\dfrac{1}{3}\right)(25) \\ \\ I&=&8\dfrac{1}{3}\text{ A} \end{array} \end{array}[/latex]
[latex]I=\dfrac{k}{R}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ I&=&\text{12 A} \\ k&=&\text{find} \\ R&=&240\Omega \\ \\ I&=&\dfrac{k}{R} \\ \\ \text{12 A}&=&\dfrac{k}{240\Omega} \\ \\ k&=&(\text{12 A})(240\Omega) \\ k&=&2880\text{ A}\Omega \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ I&=&\text{find} \\ k&=&2880 \\ R&=&540\Omega \\ \\ I&=&\dfrac{k}{R} \\ \\ I&=&\dfrac{2880\text{ A}\Omega}{540\Omega} \\ \\ I&=&5.\bar{3}\text{ A or }5\dfrac{1}{3} \end{array} \end{array}[/latex]
[latex]d_{\text{s}}=km[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ d_{\text{s}}&=&18\text{ cm} \\ k&=&\text{find} \\ m&=&3\text{ kg} \\ \\ 18\text{ cm}&=&k(3\text{ kg}) \\ \\ k&=&\dfrac{\text{18 cm}}{\text{3 kg}} \\ \\ k&=&\text{6 cm/kg} \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ d_{\text{s}}&=&\text{find} \\ k&=&\text{6 cm/kg} \\ m&=&\text{5 kg} \\ \\ d_{\text{s}}&=&(\text{6 cm/kg})(\text{5 kg}) \\ d_{\text{s}}&=&\text{30 cm} \end{array} \end{array}[/latex]
[latex]V=\dfrac{k}{P}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ P&=&32\text{ kg/cm}^2 \\ V&=&200\text{ cm}^3 \\ k&=&\text{find} \\ \\ 200\text{ cm}^3&=&\dfrac{k}{32\text{ kg/cm}^2} \\ \\ k&=&(200\text{ cm}^3)(32\text{ kg/cm}^2) \\ k&=&6400\text{ kg cm} \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ P&=&40 \\ V&=&\text{find} \\ k&=&6400 \\ \\ V&=&\dfrac{6400}{40} \\ \\ V&=&160\text{ cm}^3 \end{array} \end{array}[/latex]
[latex]c=kP[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ c&=&60,000 \\ k&=&\text{find} \\ P&=&250 \\ \\ 60,000&=&k(250) \\ \\ k&=&\dfrac{60,000}{250} \\ \\ k&=&240 \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} && \textbf{2nd Data} \\ c&=&\text{find} \\ k&=&240 \\ P&=&1,000,000 \\ \\ c&=&(240)(1,000,000) \\ c&=&240,000,000\text{ or 240 million} \end{array} \end{array}[/latex]
[latex]t=\dfrac{k}{b}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ t&=&5\text{ h} \\ k&=&\text{find} \\ b&=&7 \\ \\ 5\text{ h}&=&\dfrac{k}{7} \\ \\ k&=&\text{(5 h)}(7) \\ k&=&35 \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ t&=&\text{find} \\ k&=&35 \\ b&=&10 \\ \\ t&=&\dfrac{35}{10} \\ \\ t&=&3.5\text{ h} \end{array} \end{array}[/latex]
[latex]\lambda=\dfrac{k}{f}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ \lambda&=&250\text{ m} \\ k&=&\text{find} \\ f&=&1200\text{ kHz} \\ \\ 250&=&\dfrac{k}{1200} \\ \\ k&=&(250)(1200) \\ k&=&300,000 \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ \lambda&=&\text{find} \\ k&=&300,000 \\ f&=&60\text{ kHz} \\ \\ \lambda&=&\dfrac{300,000}{60} \\ \\ \lambda&=&5000\text{ m} \end{array} \end{array}[/latex]
[latex]w=km[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} && \textbf{1st Data} \\ w&=&64\text{ kg} \\ k&=&\text{find} \\ m&=&96\text{ kg} \\ \\ 64&=&k(96) \\ \\ k&=&\dfrac{64}{96} \\ \\ k&=&\dfrac{2}{3} \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ w&=&\text{find} \\ k&=&\dfrac{2}{3} \\ m&=&60\text{ kg} \\ \\ w&=&\left(\dfrac{2}{3}\right)(60\text{ kg}) \\ \\ w&=&40\text{ kg} \end{array} \end{array}[/latex]
[latex]t=\dfrac{d}{v}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ t&=&\text{5 h} \\ d&=&\text{find} \\ v&=&\text{80 km/h} \\ \\ \text{5 h}&=&\dfrac{d}{\text{80 km/h}} \\ \\ d&=&5(80) \\ d&=&\text{400 km} \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ t&=&\text{4.2 h} \\ d&=&\text{400 km} \\ v&=&\text{find} \\ \\ 4.2&=&\dfrac{400}{v} \\ \\ v&=&\dfrac{400}{4.2} \\ \\ v&=&95.24\text{ km/h} \end{array} \end{array}[/latex]
[latex]V=khr^2[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ V&=&33.5\text{ cm}^3 \\ k&=&\text{find} \\ h&=&\text{8 cm} \\ r&=&\text{2 cm} \\ \\ 33.5&=&k(8)(2)^2 \\ \\ k&=&\dfrac{33.5}{(8)(2)^2} \\ \\ k&=&1.046875 \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ V&=&\text{find} \\ k&=&1.046875 \\ h&=&\text{6 cm} \\ r&=&\text{4 cm} \\ \\ V&=&khr^2 \\ V&=&(1.046875)(6)(4)^2 \\ V&=&100.5\text{ cm}^3 \end{array} \end{array}[/latex]
[latex]F_{\text{e}}=\dfrac{kv^2}{r}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ F_{\text{e}}&=&100\text{ N} \\ k&=&\text{find} \\ v&=&10\text{ m/s} \\ r&=&\text{0.5 m} \\ \\ 100\text{ N}&=&\dfrac{k(10 \text{ m/s})^2}{\text{0.5 m}} \\ \\ k&=&\dfrac{(0.5)(100)}{(10)^2} \\ \\ k&=&0.5 \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ F_{\text{e}}&=&\text{find} \\ k&=&0.5 \\ v&=&25\text{ m/s} \\ r&=&1.0\text{ m} \\ \\ F_{\text{e}}&=&\dfrac{0.5(25)^2}{1.0} \\ \\ F_{\text{e}}&=&312.5\text{ N} \end{array} \end{array}[/latex]
[latex]L_{\text{max}}=\dfrac{kd^4}{h^2}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ L_{\text{max}}&=&64\text{ tonnes} \\ k&=&\text{find} \\ d&=&2.0\text{ m} \\ h&=&8.0\text{ m} \\ \\ 64&=&\dfrac{k(2)^4}{(8)^2} \\ \\ k&=&\dfrac{64(8)^2}{(2)^4} \\ \\ k&=&256 \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ L_{\text{max}}&=&\text{find} \\ k&=&256 \\ d&=&3.0\text{ m} \\ h&=&12.0\text{ m} \\ \\ L_{\text{max}}&=&\dfrac{(256)(3.0)^4}{(12.0)^2} \\ \\ L_{\text{max}}&=&144\text{ tonnes} \end{array} \end{array}[/latex]
[latex]V=\dfrac{kT}{P}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ V&=&225\text{ cc} \\ k&=&\text{find} \\ T&=&300\text{ K} \\ P&=&100\text{ N/cm}^2 \\ \\ V&=&\dfrac{kT}{P} \\ \\ 225&=&\dfrac{k(300)}{100} \\ \\ k&=&\dfrac{225(100)}{300} \\ \\ k&=&75 \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ V&=&\text{find} \\ k&=&75 \\ T&=&270 \\ P&=&150 \\ \\ V&=&\dfrac{75(270)}{150} \\ \\ V&=&135\text{ cc} \end{array} \end{array}[/latex]
[latex]R=\dfrac{kl}{d^2}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ R&=&20\Omega \\ k&=&\text{find} \\ l&=&5.0\text{ m} \\ d&=&0.25\text{ cm} \\ \\ R&=&\dfrac{kl}{d^2} \\ \\ 20\Omega&=&\dfrac{k(5.0\text{ m})}{\text{(0.25 cm)}^2} \\ \\ k&=&\dfrac{(20 \Omega)\text{(0.25 cm)}^2}{\text{5.0 m}} \\ \\ k&=&0.25 \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ R&=&\text{find} \\ k&=&0.25 \\ l&=&10.0\text{ m} \\ d&=&0.50\text{ cm} \\ \\ R&=&\dfrac{(0.25)\text{(10.0 m)}}{\text{(0.50 cm)}^2} \\ \\ R&=&10\Omega \end{array} \end{array}[/latex]
[latex]V=khd^2[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st Data} \\ V&=&377\text{ m}^3 \\ k&=&\text{find} \\ h&=&30\text{ m} \\ d&=&2.0\text{ m} \\ \\ 377\text{ m}^3&=&k(30)(2.0)^2 \\ \\ k&=&\dfrac{377}{(30)(2.0)^2} \\ \\ k&=&3.1416 \end{array} & \hspace{0.5in} \begin{array}[t]{rrl} &&\textbf{2nd Data} \\ V&=&225\text{ m}^3 \\ k&=&3.1416 \\ h&=&\text{find} \\ d&=&1.75\text{ m} \\ \\ 225&=&\pi h(1.75)^2 \\ \\ h&=&\dfrac{225}{\pi (1.75)^2} \\ \\ h&=&23.4\text{ m} \end{array} \end{array}[/latex]

Share This Book

Texas Go Math
Big Ideas Math
Engageny Math
McGraw Hill My Math
enVision Math
180 Days of Math
Math in Focus Answer Key
Math Expressions Answer Key
Privacy Policy

$CCSS Math Answers$

Eureka Math Grade 2 Module 7 Lesson 15 Answer Key

Engage ny eureka math 2nd grade module 7 lesson 15 answer key, eureka math grade 2 module 7 lesson 15 sprint answer key.

A. Adding and Subtracting by 2

$Eureka Math Grade 2 Module 7 Lesson 15 Sprint Answer Key 1$

Question 1. 0 + 2 = Answer: 0 + 2 = 2 Explanation: Adding any two numbers gives us Sum. Adding any number to 0 gives the sum the number itself.

Question 2. 2 + 2 = Answer: 2 + 2 = 4

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

Question 4. 6 + 2 = Answer: 6 + 2 = 8

Question 5. 8 + 2 = Answer: 8 + 2 = 10 Explanation: Adding 8 and 2 gives the sum 10. Here 1 is carried to the tens place and 0 is represented in the ones place.

Question 6. 10 + 2 = Answer: 10 + 2 = 12

Question 7. 12 + 2 = Answer: 12 + 2 = 14

Question 8. 14 + 2 = Answer: 14 + 2 = 16

Question 9. 16 + 2 = Answer: 16 +2 = 18

Question 10. 18 + 2 = Answer: 18 + 2 = 20

Question 11. 20 – 2 = Answer: 20 – 2 = 18

Question 12. 18 – 2 = Answer: 18 – 2 = 16

Question 13. 16 – 2 = Answer: 16 – 2 = 14

Question 14. 14 – 2 = Answer: 14 – 2 = 12

Question 15. 12 – 2 = Answer: 12 – 2 = 10

Question 16. 10 – 2 = Answer: 10 – 2 = 8 Explanation: Subtracting 10 and 2 gives the difference 8. Here 1 in tens place and 0 in ones place and subtracting 2 from ones place. Since 0 is less than 2 we borrow 1 from tens place 1 becomes 0 and 0 in the ones place becomes 10 and we subtract 2 from ones 10 we get the difference 8.

Question 17. 8 – 2 = Answer: 8 – 2 = 6

Question 18. 6 – 2 = Answer: 6 – 2 = 4

Question 19. 4 – 2 = Answer:

4 – 2 = 2

Question 20. 2 – 2 = Answer: 2 – 2 = 0

Question 21. 2 + 0 = Answer: 2 + 0 = 2

Question 22. 2 + 2 = Answer: 2 + 2 = 4

Question 23. 2 + 4 = Answer: 2 + 4 = 6

Question 24. 2 + 6 = Answer: 2 + 6 = 8

Question 25. 2 + 8 = Answer: 2 + 8 = 10

Question 26. 2 + 10 = Answer: 2 + 10 = 12

Question 27. 2 + 12 = Answer: 2 + 12 = 14

Question 28. 2 + 14 = Answer: 2 + 14 = 16

Question 29. 2 + 16 = Answer: 2 + 1 6 = 18

Question 30. 2 + 18 = Answer: 2 + 1 8 = 20

Question 31. 0 + 22 = Answer: 0 + 22 = 22

Question 32. 22 + 22 = Answer: 22 + 22 = 44

Question 33. 44 + 22 = Answer: 44 + 22 = 66

Question 34. 66 + 22 = Answer: 66 + 22 = 88

Question 35. 88 – 22 = Answer: 88 – 22 = 66

Question 36. 66 – 22 = Answer: 66 – 22 = 44

Question 37. 44 – 22 = Answer: 44 – 22 = 22

Question 38. 22 – 22 = Answer: 22 -22 = 0 Explanation: Subtracting any number by itself gives the difference zero.0

Question 39. 22 + 0 = Answer: 22 + 0 = 22 Explanation: Adding Zero to any number gives the sum the number itself.

Question 40. 22 + 22 = Answer: 22 + 22 = 44

Question 41. 22 + 44 = Answer: 22 + 44 = 66

Question 42. 66 + 22 = Answer: 66 + 22 = 88

Question 43. 888 – 222 = Answer: 888 – 222 = 666 Explanation: Here a 3 digit number is subtracted from a 3 digit number. First ones places are subtracted then tens place and then hundreds place. 8 – 6 = 2, 8 – 6 = 2 , 8 – 6 = 2. 222 is the difference.

Question 44. 666 – 222 = Answer: 666 – 222 = 444

B. Adding and Subtracting by 2

$Eureka Math Grade 2 Module 7 Lesson 15 Sprint Answer Key 2$

Question 1. 2 + 0 = Answer: 2 + 0 = 2

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

Question 4. 2 + 6 = Answer: 2 + 6 = 8

Question 5. 2 + 8 = Answer: 2 + 8 = 10

Question 6. 2 + 10 = Answer: 2 + 10 = 12

Question 7. 2 + 12 = Answer: 2 + 12= 14

Question 8. 2 + 14 = Answer: 2 + 14 = 16

Question 9. 2 + 16 = Answer: 2 + 16 = 18

Question 10. 2 + 18 = Answer: 2 + 18 = 20

Question 16. 10 – 2 = Answer: 10 – 2 = 8

Question 19. 4 – 2 = Answer: 4 – 2 = 2

Question 21. 0 + 2 = Answer: 0 + 2 = 2

Question 23. 4 + 2 = Answer: 4 + 2 = 6

Question 24. 6 + 2 = Answer: 6 + 2 = 8

Question 25. 8 + 2 = Answer: 8 + 2 = 10

Question 26. 10 + 2 = Answer: 10 + 2 = 12

Question 27. 12 + 2 = Answer: 12 + 2 = 14

Question 28. 14 + 2 = Answer: 14 + 2 = 16

Question 29. 16 + 2 = Answer: 16 + 2 = 18

Question 30. 18 + 2 = Answer: 18 + 2 = 20

Question 32. 22 + 22 = Answer: 22 +22 = 44

Question 33. 22 + 44 = Answer: 22 +44 = 66

Question 38. 22 – 22 = Answer: 22 -22 = 0

Question 39. 22 + 0 = Answer: 22 + 0 = 22

Question 40. 22 + 22 = Answer: 22 +22 = 44

Question 43. 666 – 222 = Answer: 666 – 222 = 444

Question 44. 888 – 222 = Answer: 888 – 222 = 666.

Eureka Math Grade 2 Module 7 Lesson 15 Problem Set Answer Key

Use your ruler to measure the length of the objects below in inches. Using your ruler, draw a line that is the same length as each object.

Question 1. a. A pencil is _____ inches. Answer: A pencil is _7_ inches.

$Engage-NY-Eureka-Math-2nd-Grade-Module-7-Lesson-15-Answer-Key-Eureka-Math-Grade-2-Module-7-Lesson-15-Problem-Set-Answer-Key-Question-1-b$

Question 2. a. An eraser is ______ inches. Answer: An eraser is _3__ inches.

$Engage-NY-Eureka-Math-2nd-Grade-Module-7-Lesson-15-Answer-Key-Eureka-Math-Grade-2-Module-7-Lesson-15-Problem-Set-Answer-Key-Question-2-b$

Question 3. a. A crayon is ______ inches. Answer: A crayon is __4____ inches.

$Engage-NY-Eureka-Math-2nd-Grade-Module-7-Lesson-15-Answer-Key-Eureka-Math-Grade-2-Module-7-Lesson-15-Problem-Set-Answer-Key-Question-3-b$

Question 4. a. A marker is _____ inches. Answer: A marker is _6____ inches.

$Engage-NY-Eureka-Math-2nd-Grade-Module-7-Lesson-15-Answer-Key-Eureka-Math-Grade-2-Module-7-Lesson-15-Problem-Set-Answer-Key-Question-4-b$

Question 5. a. What is the longest item that you measured? Answer: The longest item that i measured is Pencil which is 7 inches.

b. How long is the longest item? ______ inches Answer: The longest item is 7 inch long.

c. How long is the shortest item? _______ inches Answer: The shortest item is 3 inch long and it is an eraser.

$Engage-NY-Eureka-Math-2nd-Grade-Module-7-Lesson-15-Answer-Key-Eureka-Math-Grade-2-Module-7-Lesson-15-Problem-Set-Answer-Key-Question-5-d$

Question 6. Measure and label the length of each side of the triangle using your ruler.

$Eureka Math Grade 2 Module 7 Lesson 15 Problem Set Answer Key 3$

a. Which side is the shortest? 1. Side A 2. Side B 3. Side C Answer: Side A = 5 inches Side B = 4 inches Side C = 3 inches The shortest side is Side C with 3 inches.

b. What is the length of Side A? ______ inches Answer: The length of Side A = 5 inches.

c. What is the length of Sides C and B together? _____ inches Answer: The length of Side C and B together is 3 + 4 = 7 inches.

d. What is the difference between the shortest and longest sides? _____ inches Answer: The difference between the shortest and the longest side is 5 – 3 = 2 inches.

Question 7. Solve.

a. ______ inches = 1 foot Answer: 1 foot equal to 12 inches. _ 12 __ inches = 1 foot

b. 5 inches + ______ inches = 1 foot Answer: 1 foot = 12 inches 12 inches – 5 inches = 7 inches 5 inches + _ 7 __inches = 1 foot

c. ______ inches + 4 inches = 1 foot Answer: 1 foot = 12 inches 12 inches – 4 inches = 8 inches _ 8 __ inches + 4 inches = 1 foot

Eureka Math Grade 2 Module 7 Lesson 15 Exit Ticket Answer Key

$Eureka Math Grade 2 Module 7 Lesson 15 Exit Ticket Answer Key 4$

Eureka Math Grade 2 Module 7 Lesson 15 Homework Answer Key

Measure the length of each household object with your ruler, and then use your ruler to draw a line equal to the length of each object in the space provided.

Question 1. a. A dinner fork is _____ inches. Answer: A dinner fork is __ 7 ___ inches.

$Engage-NY-Eureka-Math-2nd-Grade-Module-7-Lesson-15-Answer-Key-Eureka-Math-Grade-2-Module-7-Lesson-15-Homework-Answer-Key-Question-1-b$

Question 2. a. A tablespoon is ______ inches. Answer: A tablespoon is __8____ inches.

$Engage-NY-Eureka-Math-2nd-Grade-Module-7-Lesson-15-Answer-Key-Eureka-Math-Grade-2-Module-7-Lesson-15-Homework-Answer-Key-Question-2-b$

Measure two other household objects.

Question 3. a. ______ is ______ inches. Answer: A Tea Glass is 3 inches.

$Engage-NY-Eureka-Math-2nd-Grade-Module-7-Lesson-15-Answer-Key-Eureka-Math-Grade-2-Module-7-Lesson-15-Homework-Answer-Key-Question-3-b$

Question 4. a. _________ is ______ inches. Answer: The Curry Spoon is 10 inches.

$Engage-NY-Eureka-Math-2nd-Grade-Module-7-Lesson-15-Answer-Key-Eureka-Math-Grade-2-Module-7-Lesson-15-Homework-Answer-Key-Question-4-b$

Question 5. a. What was the longest object you measured? Answer: The longest object i measured is Curry Spoon.

b. What was the shortest object you measured? Answer: The shortest object i measured is the Tea Glass.

c. The difference between the longest object and the shortest object is _______ inches. Answer: The difference between the shortest and longest object is 10 inches – 3 inches = 7 inches.

Question 6. Measure and label the length of each side of each shape in inches using your ruler.

$Eureka Math Grade 2 Module 7 Lesson 15 Homework Answer Key 5$

a. The longer side of the rectangle is _______ inches. Answer: The longer side of the rectangle is __4_____ inches.

b. The shorter side of the rectangle is _______ inches. Answer: The shorter side of the rectangle is __ 2 _____ inches.

$Engage-NY-Eureka-Math-2nd-Grade-Module-7-Lesson-15-Answer-Key-Eureka-Math-Grade-2-Module-7-Lesson-15-Homework-Answer-Key-Question-6-a,b,c$

d. The shortest side of the trapezoid is _______ inches. Answer: The shortest side of the trapezoid is __ 1 _____ inches.

e. The longest side of the trapezoid is _______ inches. Answer: The longest side of the trapezoid is __ 3_ ____ inches.

f. The longest side of the trapezoid is _______ inches longer than the shortest side. Answer: The longest side of the trapezoid is __ 3 – 1 = 2 _____ inches longer than the shortest side.

$Eureka Math Grade 2 Module 7 Lesson 15 Homework Answer Key 7$

g. Each side of the hexagon is _______ inches. Answer: Each side of the hexagon is __1_____ inches.

h. The total length around the hexagon is _______ inches. Answer: The total length around the hexagon is __ 6 _____ inches. 1 + 1 + 1 +1 + 1 +1 = 6 inches 6 sides .