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Texas Go Math Grade 5 Lesson 2.5 Answer Key Estimate with 2-Digit Divisors

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 2.5 Answer Key Estimate with 2-Digit Divisors.

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Estimate Money Miriam has saved $650 to spend during her 18-day trip to Chicago. She doesn’t want to run out of money before the trip is over, so she plans to spend about the same amount each day. Estimate how much she can spend each day.

Given Miriam’s situation, which estimate do you think is the better one for her to use? Explain your reasoning. Answer: Given Miriam’s situation, The estimate of $600 is a better one for her to use because $600 is close to $650 than $800 Hence, from the above, We can conclude that Given Miriam’s situation, the estimate of $600 is a better one for her

Math Talk Mathematical Processes

Would it be more reasonable to have an estimate or an exact answer for this example? Explain your reasoning. Answer: From the above example, It is given that Miriam has saved $650 to spend during her 18-day trip to Chicago. She doesn’t want to run out of money before the trip is over, so she plans to spend about the same amount each day So, According to the given information, We can’t get the exact answer Hence, It would be more reasonable to have an estimate than an exact for the given example

Try This! Use compatible numbers.

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Use compatible numbers to find two estimates.

Problem-Solving

Use compatible numbers to estimate the quotient.

Practice! Copy and Solve Use compatible numbers to estimate the quotient.

Question 10. 462 ÷ 83 Answer: The given division expression is: 462 ÷ 83 Now, Estimate: 480 ÷ 80 = 6 400 ÷ 80 = 5 Now, Since 480 is close to 462, the estimate of 480 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 6

Question 11. 9,144 ÷ 27 Answer: The given division expression is: 9,144 ÷ 27 Now, Estimate: 9,000 ÷ 30 = 300 10,000 ÷ 25 = 400 Now, Since 9,000 is close to 9,144, the estimate of 9,000 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 300

Go Math Grade 5 Answer Key Estimate with 2-Digit Divisors Lesson 2.5 Question 12. 710 ÷ 68 Answer: The given division expression is: 710 ÷ 68 Now, Estimate: 700 ÷ 70 = 10 630 ÷ 70 = 9 Now, Since 700 is close to 710, the estimate of 700 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 10

Question 13. 1,607 ÷ 36 Answer: The given division expression is: 1,607 ÷ 36 Now, Estimate: 1,600 ÷ 40 = 40 2,000 ÷ 40 = 50 Now, Since 1,600 is close to 1,607, the estimate of 1,600 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 40

Question 14. Write Math Explain how you know whether the quotient of 298 ÷ 31 is closer to 9 or to 10. Answer: The given division expression is: 298 ÷ 31 Now, Estimate: 300 ÷ 30 = 10 270 ÷ 30 = 9 Now, Since 300 is close to 298, the estimate of 300 will be more reasonable Hence, from the above, We can conclude that The quotient of 298 ÷ 31 will be closer to 10

Problem Solving

Question 17. H.O.T. Eli needs to save $235. He plans to mow lawns and charge $21 for each. Write two estimates for the number of lawns he needs to mow. Decide which estimate you think is the better one for Eli to use. Explain your reasoning. Answer: It is given that Eli needs to save $235. He plans to mow lawns and charge $21 for each. Write two estimates for the number of lawns he needs to mow Now, According to the given situation, The number of lawns Eli needs to mow = $235 ÷ $21 Now, Estimate: $240 ÷ $20 = 12 $200 ÷ $20 = 10 Now, Since $240 is close to $235, the estimate of $240 will be more reasonable Hence, from the above, We can conclude that The number of lawns Eli needs to mow is about 12 lawns

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Texas Go Math Grade 5 Lesson 2.5 Homework and Practice Answer Key

Question 4. 7,233 ÷ 84 Answer: The given division expression is: 7,233 ÷ 84 Now, Estimate: 7,200 ÷ 80                    (or)            8,000 ÷ 80 = 90                             (or)            = 100 Now, Since 7,200 is close to 7,233, the estimate of 7,200 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 90

Question 5. 568 ÷ 34 Answer: The given division expression is: 568 ÷ 34 Now, Estimate: 540 ÷ 30                    (or)            570 ÷ 30 = 18                           (or)            = 19 Now, Since 568 is close to 5700, the estimate of 570 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 19

Question 6. 938 ÷ 57 Answer: The given division expression is: 938 ÷ 57 Now, Estimate: 900 ÷ 60                    (or)            900 ÷ 50 = 15                             (or)            = 18 Now, Since 938 is close to 900, the estimate of 900 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 15

Question 7. 4,479 ÷ 89 Answer: The given division expression is: 4,479 ÷ 89 Now, Estimate: 4,500 ÷ 90                    (or)            3,600 ÷ 90 = 50                             (or)            = 40 Now, Since 4,479 is close to 4,500, the estimate of 4,500 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 50

Question 8. 1,238 ÷ 57 Answer: The given division expression is: 1,238 ÷ 57 Now, Estimate: 1,200 ÷ 60                    (or)            900 ÷ 60 = 20                             (or)            = 15 Now, Since 1,238 is close to 1,200, the estimate of 1,200 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 20

Question 9. 5,587 ÷ 77 Answer: The given division expression is: 5,587 ÷ 77 Now, Estimate: 5,600 ÷ 80                    (or)            4,800 ÷ 80 = 70                             (or)            = 60 Now, Since 5,587 is close to 5,600, the estimate of 5,600 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 70

Question 10. 4,192 ÷ 55 Answer: The given division expression is: 4,192 ÷ 55 Now, Estimate: 4,200 ÷ 60                    (or)            3,600 ÷ 60 = 70                             (or)            = 960 Now, Since 4,192 is close to 4,200, the estimate of 4,200 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 70

Question 11. 1,847 ÷ 24 Answer: The given division expression is: 1,847 ÷ 24 Now, Estimate: 1,500 ÷ 25                    (or)            2,000 ÷ 525 = 60                             (or)            = 80 Now, Since 1,847 is close to 2,000, the estimate of 2,000 will be more reasonable Hence, from the above, We can conclude that The value of the given expression is about 60

Question 12. Alec needs to save $376. He plans to rake fall leaves for S12 per lawn. Using compatible numbers, write an estimate that shows the number of lawns he will need to rake. Answer: It is given that Alec needs to save $376. He plans to rake fall leaves for S12 per lawn Now, According to the given information, The number of lawns Alec will need to make = $376 ÷ $12 Now, Estimate: The number of lawns Alec will need to make = $360 ÷ $12 = 30 The number of lawns Alec will need to make = $480 ÷ $12 = 40 Hence, from the above, We can conclude that The number of lawns Alec will need to make is about 30 to 40 lawns

Question 13. Mr. Rodriguez’s construction company built an office building that is 1,848 feet tall. Each floor of the building is 14 feet high. About how many floors are in the building? Write an estimate. Answer: It is given that Mr. Rodriguez’s construction company built an office building that is 1,848 feet tall. Each floor of the building is 14 feet high Now, According to the given information, The number of floors that are present in the building = 1,848 ÷ 14 Now, Estimate: The number of floors that are present in the building = 1,600 ÷ 16 = 100 The number of floors that are presen in the building = 2,400 ÷ 16 = 150 Hence, from the above, We can conclude that The number of buildings that are in the building is about 100 to 150 buildings

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Algebra 2 Worksheets with answer keys

Enjoy these free printable math worksheets . Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Plus each one comes with an answer key.

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Answer Key 2.5

• [latex]x=\pm 8[/latex]
• [latex]n=\pm 7[/latex]
• [latex]b=\pm 1[/latex]
• [latex]x=\pm 2[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrr} 5&+&8a&=&53 \\ -5&&&&-5 \\ \hline &&\dfrac{8a}{8}&=&\dfrac{48}{8} \\ \\ &&a&=&6 \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrl} 5&+&8a&=&-53 \\ -5&&&&-5 \\ \hline &&\dfrac{8a}{8}&=&\dfrac{-58}{8} \\ \\ &&a&=&-\dfrac{58}{8}\text{ or }-7\dfrac{1}{4} \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrl} 9n&+&8&=&46 \\ &-&8&&-8 \\ \hline &&\dfrac{9n}{9}&=&\dfrac{38}{9} \\ \\ &&n&=&\dfrac{38}{9}\text{ or }4\dfrac{2}{9} \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrr} 9n&+&8&=&-46 \\ &-&8&&-8 \\ \hline &&\dfrac{9n}{9}&=&\dfrac{-54}{9} \\ \\ &&n&=&-6 \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrr} 3k&+&8&=&2 \\ &-&8&&-8 \\ \hline &&\dfrac{3k}{3}&=&\dfrac{-6}{3} \\ \\ &&k&=&-2 \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrr} 3k&+&8&=&-2 \\ &-&8&&-8 \\ \hline &&\dfrac{3k}{3}&=&\dfrac{-10}{3} \\ \\ &&k&=&-\dfrac{10}{3} \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrl} 3&-&x&=&\phantom{-}6 \\ -3&&&&-3 \\ \hline &&(-x&=&\phantom{-}3)(-1) \\ &&x&=&-3 \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrl} 3&-&x&=&-6 \\ -3&&&&-3 \\ \hline &&(-x&=&-9)(-1) \\ &&x&=&\phantom{-}9 \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} \dfrac{-7}{-7}\left| -3-3r \right|&=&\dfrac{-21}{-7} \\ |-3-3r|&=&3 \end{array}[/latex] [latex]\phantom{1}[/latex] [latex]\begin{array}{ll} \begin{array}{rrrrr} -3&-&3r&=&3 \\ +3&&&&+3 \\ \hline &&\dfrac{-3r}{-3}&=&\dfrac{6}{-3} \\ \\ &&r&=&-2 \end{array} & \hspace{0.5in} \begin{array}{rrrrr} -3&-&3r&=&-3 \\ +3&&&&+3 \\ \hline &&\dfrac{-3r}{-3}&=&\dfrac{0}{-3} \\ \\ &&r&=&0 \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrr} |2+2b|&+&1&=&3 \\ &-&1&&-1 \\ \hline |2+2b|&&&=&2 \\ \end{array}[/latex] [latex]\begin{array}[t]{ll} \\ \begin{array}[t]{rrrrr} 2&+&2b&=&2 \\ -2&&&&-2 \\ \hline &&2b&=&0 \\ &&b&=&0 \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrr} 2&+&2b&=&-2 \\ -2&&&&-2 \\ \hline &&\dfrac{2b}{2}&=&\dfrac{-4}{2} \\ \\ &&b&=&-2 \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} \dfrac{7}{7}|-7x-3|&=&\dfrac{21}{7} \\ |-7x-3|&=&3 \end{array}[/latex] [latex]\begin{array}[t]{ll}\\ \begin{array}[t]{rrrrr} -7x&-&3&=&3 \\ &+&3&&+3 \\ \hline &&\dfrac{-7x}{-7}&=&\dfrac{6}{-7} \\ \\ &&x&=&-\dfrac{6}{7} \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrr} -7x&-&3&=&-3 \\ &+&3&&+3 \\ \hline &&-7x&=&0 \\ &&x&=&0 \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrr} -4&-&3n&=&2 \\ +4&&&&+4 \\ \hline &&\dfrac{-3n}{-3}&=&\dfrac{6}{-3} \\ \\ &&n&=&-2 \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrr} -4&-&3n&=&-2 \\ +4&&&&+4 \\ \hline &&\dfrac{-3n}{-3}&=&\dfrac{2}{-3} \\ \\ &&n&=&-\dfrac{2}{3} \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrr} 8|5p &+&8|&-&5&=&11 \\ &&&+&5&&+5 \\ \hline &&\dfrac{8}{8}|5p &+&8|&=&\dfrac{16}{8} \\ &&|5p &+&8|&=&2 \end{array}[/latex] [latex]\begin{array}[t]{ll}\\ \begin{array}[t]{rrrrr} 5p&+&8&=&2 \\ &-&8&&-8 \\ \hline &&\dfrac{5p}{5}&=&\dfrac{-6}{5} \\ \\ &&p&=&-\dfrac{6}{5} \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrr} 5p&+&8&=&-2 \\ &-&8&&-8 \\ \hline &&\dfrac{5p}{5}&=&\dfrac{-10}{5} \\ \\ &&p&=&-2 \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrl} 3&-&|6n&+&7|&=&-40 \\ -3&&&&&&-3 \\ \hline &&(-|6n&+&7|&=&-43)(-1) \\ &&|6n&+&7|&=&43 \end{array}[/latex] [latex]\begin{array}[t]{ll}\\ \begin{array}[t]{rrrrr} 6n&+&7&=&43 \\ &-&7&&-7 \\ \hline &&\dfrac{6n}{6}&=&\dfrac{36}{6} \\ \\ &&n&=&6 \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrr} 6n&+&7&=&-43 \\ &-&7&&-7 \\ \hline &&\dfrac{6n}{6}&=&\dfrac{-50}{6} \\ \\ &&n&=&-\dfrac{25}{3} \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrr} 5|3&+&7m|&+&1&=&51 \\ &&&-&1&&-1 \\ \hline &&\dfrac{5}{5}|3&+&7m|&=&\dfrac{50}{5} \\ &&|3&+&7m|&=&10 \end{array}[/latex] [latex]\begin{array}[t]{ll}\\ \begin{array}[t]{rrrrr} 3&+&7m&=&10 \\ -3&&&&-3 \\ \hline &&\dfrac{7m}{7}&=&\dfrac{7}{7} \\ \\ &&m&=&1 \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrr} 3&+&7m&=&-10 \\ -3&&&&-3 \\ \hline &&\dfrac{7m}{7}&=&\dfrac{-13}{7} \\ \\ &&m&=&-\dfrac{13}{7} \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrr} 4|r&+&7|&+&3&=&59 \\ &&&-&3&&-3 \\ \hline &&\dfrac{4}{4}|r&+&7|&=&\dfrac{56}{4} \\ &&|r&+&7|&=&14 \end{array}[/latex] [latex]\begin{array}[t]{ll}\\ \begin{array}[t]{rrrrr} r&+&7&=&14 \\ &-&7&&-7 \\ \hline &&r&=&7 \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrr} r&+&7&=&-14 \\ &-&7&&-7 \\ \hline &&r&=&-21 \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrr} -7&+&8|-7x&-&3|&=&73 \\ +7&&&&&&+7 \\ \hline &&\dfrac{8}{8}|-7x&-&3|&=&\dfrac{80}{8} \\ &&|-7x&-&3|&=&10 \end{array}[/latex] [latex]\phantom{1}[/latex] [latex]\begin{array}{ll} \begin{array}{rrrrr} -7x&-&3&=&10 \\ &+&3&&+3 \\ \hline &&\dfrac{-7x}{-7}&=&\dfrac{13}{-7} \\ \\ &&x&=&-\dfrac{13}{7} \end{array} & \hspace{0.5in} \begin{array}{rrrrr} -7x&-&3&=&-10 \\ &+&3&&+3 \\ \hline &&\dfrac{-7x}{-7}&=&\dfrac{-7}{-7} \\ \\ &&x&=&1 \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrr} 8|3&-&3n|&-&5&=&91 \\ &&&+&5&&+5 \\ \hline &&\dfrac{8}{8}|3&-&3n|&=&\dfrac{96}{8} \\ &&|3&-&3n|&=&12 \end{array}[/latex] [latex]\begin{array}[t]{ll}\\ \begin{array}[t]{rrrrr} 3&-&3n&=&12 \\ -3&&&&-3 \\ \hline &&\dfrac{-3n}{-3}&=&\dfrac{9}{-3} \\ \\ &&n&=&-3 \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrr} 3&-&3n&=&-12 \\ -3&&&&-3 \\ \hline &&\dfrac{-3n}{-3}&=&\dfrac{-15}{-3} \\ \\ &&n&=&5 \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrrrr} 5x&+&3&=&2x&-&1 \\ -2x&-&3&&-2x&-&3 \\ \hline &&\dfrac{3x}{3}&=&\dfrac{-4}{3}&& \\ \\ &&x&=&-\dfrac{4}{3}&& \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrrrr} 5x&+&3&=&-2x&+&1 \\ +2x&-&3&&+2x&-&3 \\ \hline &&\dfrac{7x}{7}&=&\dfrac{-2}{7}&& \\ \\ &&x&=&-\dfrac{2}{7}&& \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrrrr} 2&+&3x&=&4&-&2x \\ -2&+&2x&&-2&+&2x \\ \hline &&\dfrac{5x}{5}&=&\dfrac{2}{5}&& \\ \\ &&x&=&\dfrac{2}{5}&& \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrrrr} 2&+&3x&=&-4&+&2x \\ -2&-&2x&&-2&-&2x \\ \hline &&x&=&-6&& \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrrrr} 3x&-&4&=&2x&+&3 \\ -2x&+&4&&-2x&+&4 \\ \hline &&x&=&7&& \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrlrr} 3x&-&4&=&-2x&-&3 \\ +2x&+&4&&+2x&+&4 \\ \hline &&\dfrac{5x}{5}&=&\dfrac{1}{5}&& \\ \\ &&x&=&\dfrac{1}{5}&& \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrrrr} 2x&-&5&=&3x&+&4 \\ -3x&+&5&&-3x&+&5 \\ \hline &&-x&=&9&& \\ &&x&=&-9&& \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrlrr} 2x&-&5&=&-3x&-&4 \\ +3x&+&5&&+3x&+&5 \\ \hline &&\dfrac{5x}{5}&=&\dfrac{1}{5}&& \\ \\ &&x&=&\dfrac{1}{5}&& \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrrrr} 4x&-&2&=&6x&+&3 \\ -6x&+&2&&-6x&+&2 \\ \hline &&\dfrac{-2x}{-2}&=&\dfrac{5}{-2}&& \\ \\ &&x&=&-\dfrac{5}{2}&& \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrrrr} 4x&-&2&=&-6x&-&3 \\ +6x&+&2&&+6x&+&2 \\ \hline &&\dfrac{10x}{10}&=&\dfrac{-1}{10}&& \\ \\ &&x&=&-\dfrac{1}{10}&& \end{array} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ll} \begin{array}[t]{rrrrrrr} 3x&+&2&=&2x&-&3 \\ -2x&-&2&&-2x&-&2 \\ \hline &&x&=&-5&& \end{array} & \hspace{0.5in} \begin{array}[t]{rrrrlrr} 3x&+&2&=&-2x&+&3 \\ +2x&-&2&&+2x&-&2 \\ \hline &&\dfrac{5x}{5}&=&\dfrac{1}{5}&& \\ \\ &&x&=&\dfrac{1}{5}&& \end{array} \end{array}[/latex]

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Home > CCA2 > Chapter 5 > Lesson 5.2.5

Lesson 5.1.1, lesson 5.1.2, lesson 5.1.3, lesson 5.2.1, lesson 5.2.2, lesson 5.2.3, lesson 5.2.4, lesson 5.2.5.

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Be Prepared

ⓐ 8; ⓑ −1; ⓒ -729

ⓐ x 6 x 6 ; ⓑ y 6 y 6 ; ⓒ z 12 z 12

1 y 8 1 y 8

53 60 53 60

64 x 6 y 15 64 x 6 y 15

1 125 1 125

2 x 2 + 11 x − 8 2 x 2 + 11 x − 8

8 + 2 a − a 2 8 + 2 a − a 2

81 − 90 y + 25 y 2 81 − 90 y + 25 y 2

49 − 9 x 2 49 − 9 x 2

y 2 − 6 y + 9 y 2 − 6 y + 9

x = 5 2 x = 5 2

n = 2 or n = 4 n = 2 or n = 4

( − ∞ , 1 2 ] ( − ∞ , 1 2 ]

f ( 2 ) = 2 , f ( −1 ) = −7 , f ( 0 ) = −4 f ( 2 ) = 2 , f ( −1 ) = −7 , f ( 0 ) = −4

domain: [ 0 , ∞ ) ; range: [ 0 , ∞ ) domain: [ 0 , ∞ ) ; range: [ 0 , ∞ )

ⓐ −4 , 0. 5 – , 7 3 , 3 , 81 ; −4 , 0. 5 – , 7 3 , 3 , 81 ; ⓑ 7 ; 7 ; ⓒ −4 , − 7 , 0. 5 – , 7 3 , 3 , 81 −4 , − 7 , 0. 5 – , 7 3 , 3 , 81

2 x 2 − x − 15 2 x 2 − x − 15

5 + 15 2 5 + 15 2

ⓐ −8 −8 ⓑ 15

ⓐ 10 ⓑ −11 −11

ⓐ not a real number ⓑ −9 −9

ⓐ −7 −7 ⓑ not a real number

ⓐ 3 ⓑ 4 ⓒ 3

ⓐ 10 ⓑ 2 ⓒ 4

ⓐ −3 −3 ⓑ not real ⓒ −2 −2

ⓐ −6 −6 ⓑ not real ⓒ −4 −4

ⓐ 6 < 38 < 7 6 < 38 < 7 ⓑ 4 < 93 3 < 5 4 < 93 3 < 5

ⓐ 9 < 84 < 10 9 < 84 < 10 ⓑ 5 < 152 3 < 6 5 < 152 3 < 6

ⓐ ≈ 3.32 ≈ 3.32 ⓑ ≈ 4.14 ≈ 4.14 ⓒ ≈ 3.36 ≈ 3.36

ⓐ ≈ 3.61 ≈ 3.61 ⓑ ≈ 4.38 ≈ 4.38 ⓒ ≈ 3.15 ≈ 3.15

ⓐ | b | | b | ⓑ w ⓒ | m | | m | ⓓ q

ⓐ | y | | y | ⓑ p ⓒ | z | | z | ⓓ q

ⓐ | y 9 | | y 9 | ⓑ z 6 z 6

ⓐ m 2 m 2 ⓑ | b 5 | | b 5 |

ⓐ | u 3 | | u 3 | ⓑ v 5 v 5

ⓐ c 4 c 4 ⓑ d 4 d 4

ⓐ 8 | x | 8 | x | ⓑ −10 | p | −10 | p |

ⓐ 13 | y | 13 | y | ⓑ −11 | y | −11 | y |

ⓐ 3 x 9 3 x 9 ⓑ 3 | q 7 | 3 | q 7 |

ⓐ 5 q 3 5 q 3 ⓑ 3 q 5 3 q 5

ⓐ 10 | a b | 10 | a b | ⓑ 12 p 6 q 10 12 p 6 q 10 ⓒ 2 x 10 y 4 2 x 10 y 4

ⓐ 15 | m n | 15 | m n | ⓑ 13 | x 5 y 7 | 13 | x 5 y 7 | ⓒ 3 w 12 z 5 3 w 12 z 5

ⓐ 12 2 12 2 ⓑ 3 3 3 3 3 3 ⓒ 2 4 4 2 4 4

ⓐ 12 3 12 3 ⓑ 5 5 3 5 5 3 ⓒ 3 9 4 3 9 4

ⓐ b 2 b b 2 b ⓑ | y | y 2 4 | y | y 2 4 ⓒ z z 2 3 z z 2 3

ⓐ p 4 p p 4 p ⓑ y y 3 5 y y 3 5 ⓒ q 2 q 6 q 2 q 6

ⓐ 4 y 2 2 y 4 y 2 2 y ⓑ 3 p 3 2 p 3 3 p 3 2 p 3 ⓒ 2 q 2 4 q 2 4 2 q 2 4 q 2 4

ⓐ 5 a 4 3 a 5 a 4 3 a ⓑ 4 m 3 2 m 2 3 4 m 3 2 m 2 3 ⓒ 3 | n | 2 n 3 4 3 | n | 2 n 3 4

ⓐ 7 | a 3 | b 2 2 a b 7 | a 3 | b 2 2 a b ⓑ 2 x y 7 x 2 y 3 2 x y 7 x 2 y 3 ⓒ 2 | x | y 2 2 x 4 2 | x | y 2 2 x 4

ⓐ 6 m 4 | n 5 | 5 m n 6 m 4 | n 5 | 5 m n ⓑ 2 x 2 y 9 y 2 3 2 x 2 y 9 y 2 3 ⓒ 2 | x y | 5 x 3 4 2 | x y | 5 x 3 4

ⓐ −4 −4 ⓑ no real number no real number

ⓐ −5 5 3 −5 5 3 ⓑ no real number

ⓐ 5 + 5 3 5 + 5 3 ⓑ 2 − 3 2 − 3

ⓐ 2 + 7 2 2 + 7 2 ⓑ 2 − 5 2 − 5

ⓐ 5 4 5 4 ⓑ 3 5 3 5 ⓒ 2 3 2 3

ⓐ 7 9 7 9 ⓑ 2 5 2 5 ⓒ 1 3 1 3

ⓐ | a | | a | ⓑ | x | | x | ⓒ y 3 y 3

ⓐ x 2 x 2 ⓑ m 2 m 2 ⓒ n 2 n 2

2 | p | 6 p 7 2 | p | 6 p 7

2 x 2 3 x 5 2 x 2 3 x 5

ⓐ 4 | m | 5 m | n 3 | 4 | m | 5 m | n 3 | ⓑ 3 c 3 4 c 3 d 2 3 c 3 4 c 3 d 2 ⓒ 2 x 2 5 x 2 4 | y | 2 x 2 5 x 2 4 | y |

ⓐ 3 u 3 6 u v 4 3 u 3 6 u v 4 ⓑ 2 r 5 3 s 2 2 r 5 3 s 2 ⓒ 3 | m 3 | 2 m 2 4 | n 3 | 3 | m 3 | 2 m 2 4 | n 3 |

ⓐ 5 | y | x 6 5 | y | x 6 ⓑ 2 x y y 2 3 3 2 x y y 2 3 3 ⓒ | a b | a 4 2 | a b | a 4 2

ⓐ 2 | m | 3 5 | n 3 | 2 | m | 3 5 | n 3 | ⓑ 3 x y x 2 3 5 3 x y x 2 3 5 ⓒ 2 | a b | a 2 4 3 2 | a b | a 2 4 3

ⓐ 7 z 2 7 z 2 ⓑ −5 2 3 −5 2 3 ⓒ 3 | m | 2 m 2 4 3 | m | 2 m 2 4

ⓐ 8 m 4 8 m 4 ⓑ −4 −4 ⓒ 3 | n | 2 4 3 | n | 2 4

ⓐ t t ⓑ m 3 m 3 ⓒ r 4 r 4

ⓐ b 6 b 6 ⓑ z 5 z 5 ⓒ p 4 p 4

ⓐ ( 10 m ) 1 2 ( 10 m ) 1 2 ⓑ ( 3 n ) 1 5 ( 3 n ) 1 5 ⓒ 3 ( 6 y ) 1 4 3 ( 6 y ) 1 4

ⓐ ( 3 k ) 1 7 ( 3 k ) 1 7 ⓑ ( 5 j ) 1 4 ( 5 j ) 1 4 ⓒ 8 ( 2 a ) 1 3 8 ( 2 a ) 1 3

ⓐ 6 ⓑ 2 ⓒ 2

ⓐ 10 ⓑ 3 ⓒ 3

ⓐ No real solution ⓑ −8 −8 ⓒ 1 8 1 8

ⓐ No real solution ⓑ −4 −4 ⓒ 1 4 1 4

ⓐ x 5 2 x 5 2 ⓑ ( 3 y ) 3 4 ( 3 y ) 3 4 ⓒ ( 2 m 3 n ) 5 2 ( 2 m 3 n ) 5 2

ⓐ a 2 5 a 2 5 ⓑ ( 5 a b ) 5 3 ( 5 a b ) 5 3 ⓒ ( 7 x y z ) 3 2 ( 7 x y z ) 3 2

ⓐ 9 ⓑ 1 729 1 729 ⓒ 1 8 1 8

ⓐ 8 ⓑ 1 9 1 9 ⓒ 1 125 1 125

ⓐ −64 −64 ⓑ − 1 64 − 1 64 ⓒ not a real number

ⓐ −729 −729 ⓑ − 1 729 − 1 729 ⓒ not a real number

ⓐ x 3 2 x 3 2 ⓑ x 8 x 8 ⓒ 1 x 1 x

ⓐ y 11 8 y 11 8 ⓑ m 2 m 2 ⓒ 1 d 1 d

ⓐ 8 x 1 5 8 x 1 5 ⓑ x 1 2 y 1 3 x 1 2 y 1 3

ⓐ 729 n 3 5 729 n 3 5 ⓑ a 2 b 2 3 a 2 b 2 3

ⓐ m 2 m 2 ⓑ 5 n m 1 4 5 n m 1 4

ⓐ u 3 u 3 ⓑ 3 x 1 5 y 1 3 3 x 1 5 y 1 3

ⓐ − 2 − 2 ⓑ 11 x 3 11 x 3 ⓒ 3 x 4 − 5 y 4 3 x 4 − 5 y 4

ⓐ −4 3 −4 3 ⓑ 8 y 3 8 y 3 ⓒ 5 m 4 − 2 m 3 5 m 4 − 2 m 3

ⓐ −2 7 x −2 7 x ⓑ − 5 x y 4 − 5 x y 4

ⓐ − 3 y − 3 y ⓑ 3 7 m n 3 3 7 m n 3

ⓐ 9 2 9 2 ⓑ 2 2 3 2 2 3 ⓒ 3 3 3 3

ⓐ 7 3 7 3 ⓑ −10 5 3 −10 5 3 ⓒ −3 2 3 −3 2 3

ⓐ − m 3 2 m − m 3 2 m ⓑ x 2 5 x 3 x 2 5 x 3

ⓐ − p 3 p − p 3 p ⓑ 4 y 4 y 2 3 − 2 y 4 n 2 3 4 y 4 y 2 3 − 2 y 4 n 2 3

ⓐ 12 15 12 15 ⓑ −18 4 3 −18 4 3

ⓐ 27 2 27 2 ⓑ −36 2 3 −36 2 3

ⓐ 288 x 3 5 288 x 3 5 ⓑ 8 y 6 y 2 4 8 y 6 y 2 4

ⓐ 144 y 2 5 y 144 y 2 5 y ⓑ −36 a 3 a 4 −36 a 3 a 4

ⓐ 18 + 6 18 + 6 ⓑ −2 4 3 − 2 3 3 −2 4 3 − 2 3 3

ⓐ −40 + 4 2 −40 + 4 2 ⓑ −3 − 18 3 −3 − 18 3

ⓐ −66 + 15 7 −66 + 15 7 ⓑ x 2 3 − 5 x 3 + 6 x 2 3 − 5 x 3 + 6

ⓐ 41 − 14 11 41 − 14 11 ⓑ x 2 3 + 4 x 3 + 3 x 2 3 + 4 x 3 + 3

1 + 9 21 1 + 9 21

−12 − 20 3 −12 − 20 3

ⓐ 102 + 20 2 102 + 20 2 ⓑ 55 + 6 6 55 + 6 6

ⓐ 41 − 12 5 41 − 12 5 ⓑ 121 − 36 10 121 − 36 10

ⓐ 5 s 8 5 s 8 ⓑ 2 a 2 a

ⓐ 5 q 2 6 5 q 2 6 ⓑ 2 b 2 b

ⓐ 9 x 2 y 2 9 x 2 y 2 ⓑ −4 x y −4 x y

ⓐ 10 n 3 m 10 n 3 m ⓑ −3 p q 2 −3 p q 2

4 x y 2 x 4 x y 2 x

4 a b 3 b 4 a b 3 b

ⓐ 5 3 3 5 3 3 ⓑ 6 8 6 8 ⓒ 2 x x 2 x x

ⓐ 6 5 5 6 5 5 ⓑ 14 6 14 6 ⓒ 5 x x 5 x x

ⓐ 49 3 7 49 3 7 ⓑ 90 3 6 90 3 6 ⓒ 5 3 y 2 3 3 y 5 3 y 2 3 3 y

ⓐ 4 3 2 4 3 2 ⓑ 150 3 10 150 3 10 ⓒ 2 5 n 2 3 5 n 2 5 n 2 3 5 n

ⓐ 27 4 3 27 4 3 ⓑ 12 4 4 12 4 4 ⓒ 3 5 x 3 4 5 x 3 5 x 3 4 5 x

ⓐ 125 4 5 125 4 5 ⓑ 14 4 4 14 4 4 ⓒ 2 4 x 3 4 x 2 4 x 3 4 x

− 3 ( 1 + 5 ) 4 − 3 ( 1 + 5 ) 4

4 + 6 5 4 + 6 5

5 ( x − 2 ) x − 2 5 ( x − 2 ) x − 2

10 ( y + 3 ) y − 3 10 ( y + 3 ) y − 3

( p + 2 ) p − 2 2 ( p + 2 ) p − 2 2

( q − 10 ) q − 10 2 ( q − 10 ) q − 10 2

m = 23 3 m = 23 3

z = 3 10 z = 3 10

no solution no solution

x = 2 , x = 3 x = 2 , x = 3

y = 5 , y = 6 y = 5 , y = 6

x = −6 x = −6

x = −9 x = −9

x = 8 x = 8

x = 6 x = 6

m = 7 m = 7

n = 3 n = 3

a = 63 a = 63

b = 311 b = 311

x = 3 x = 3

x = − 6 5 x = − 6 5

x = 4 x = 4

x = 9 x = 9

x = 5 x = 5

x = 0 x = 4 x = 0 x = 4

3.5 3.5 seconds

42.7 42.7 feet

54.1 54.1 feet

ⓐ f ( 6 ) = 4 f ( 6 ) = 4 ⓑ no value at x = 0 x = 0

ⓐ g ( 4 ) = 5 g ( 4 ) = 5 ⓑ no value at f ( −3 ) f ( −3 )

ⓐ g ( 4 ) = 2 g ( 4 ) = 2 ⓑ g ( 1 ) = −1 g ( 1 ) = −1

ⓐ h ( 2 ) = 2 h ( 2 ) = 2 ⓑ h ( −5 ) = −3 h ( −5 ) = −3

ⓐ f ( 4 ) = 2 f ( 4 ) = 2 ⓑ f ( −1 ) = 1 f ( −1 ) = 1

ⓐ g ( 16 ) = 3 g ( 16 ) = 3 ⓑ g ( 3 ) = 2 g ( 3 ) = 2

[ 5 6 , ∞ ) [ 5 6 , ∞ )

( − ∞ , 4 5 ] ( − ∞ , 4 5 ]

( −3 , ∞ ) ( −3 , ∞ )

( 5 , ∞ ) ( 5 , ∞ )

( − ∞ , ∞ ) ( − ∞ , ∞ )

ⓐ domain: [ −2 , ∞ ) [ −2 , ∞ ) ⓑ

ⓒ range: [ 0 , ∞ ) [ 0 , ∞ )

ⓐ domain: [ 2 , ∞ ) [ 2 , ∞ ) ⓑ

ⓐ domain: ( − ∞ , ∞ ) ( − ∞ , ∞ ) ⓑ

ⓒ range: ( − ∞ , ∞ ) ( − ∞ , ∞ )

ⓐ 9 i 9 i ⓑ 5 i 5 i ⓒ 3 2 i 3 2 i

ⓐ 6 i 6 i ⓑ 3 i 3 i ⓒ 3 3 i 3 3 i

6 2 i 6 2 i

7 3 i 7 3 i

ⓐ 6 + 5 i 6 + 5 i ⓑ 6 − 3 i 6 − 3 i

ⓐ −2 − 6 i −2 − 6 i ⓑ 2 + 9 i 2 + 9 i

12 + 20 i 12 + 20 i

12 − 6 i 12 − 6 i

−11 − 7 i −11 − 7 i

−5 − 10 i −5 − 10 i

−21 + 21 i −21 + 21 i

9 − 40 i 9 − 40 i

−12 − 22 3 i −12 − 22 3 i

6 + 12 2 i 6 + 12 2 i

4 17 + 16 17 i 4 17 + 16 17 i

2 5 + 4 5 i 2 5 + 4 5 i

3 2 − 3 2 i 3 2 − 3 2 i

4 5 − 2 5 i 4 5 − 2 5 i

Section 8.1 Exercises

ⓐ 8 ⓑ −9 −9

ⓐ 14 ⓑ −1 −1

ⓐ 2 3 2 3 ⓑ −0.1 −0.1

ⓐ not real number ⓑ −17 −17

ⓐ −15 −15 ⓑ not real number

ⓐ 8 ⓑ 3 ⓒ 1

ⓐ −2 −2 ⓑ not real not real ⓒ −2 −2

ⓐ −5 −5 ⓑ not real not real ⓒ −4 −4

ⓐ 8 < 70 < 9 8 < 70 < 9 ⓑ 4 < 71 3 < 5 4 < 71 3 < 5

ⓐ 14 < 200 < 15 14 < 200 < 15 ⓑ 5 < 137 3 < 6 5 < 137 3 < 6

ⓐ ≈ 4.36 ≈ 4.36 ⓑ ≈ 4.46 ≈ 4.46 ⓒ ≈ 3.14 ≈ 3.14

ⓐ ≈ 7.28 ≈ 7.28 ⓑ ≈ 5.28 ≈ 5.28 ⓒ ≈ 4.61 ≈ 4.61

ⓐ u ⓑ | v | | v |

ⓐ | y | | y | ⓑ m m

ⓐ | x 3 | | x 3 | ⓑ y 8 y 8

ⓐ x 12 x 12 ⓑ | y 11 | | y 11 |

ⓐ x 3 x 3 ⓑ | y 3 | | y 3 |

ⓐ m 2 m 2 ⓑ n 4 n 4

ⓐ 7 | x | 7 | x | ⓑ −9 | x 9 | −9 | x 9 |

ⓐ 11 m 10 11 m 10 ⓑ −8 | a | −8 | a |

ⓐ 2 x 2 2 x 2 ⓑ 2 y 2 2 y 2

ⓐ 6 a 2 6 a 2 ⓑ 2 b 4 2 b 4

ⓐ 12 | x y | 12 | x y | ⓑ 13 w 4 | y 5 | 13 w 4 | y 5 | ⓒ 2 a 17 b 2 2 a 17 b 2

ⓐ 11 | a b | 11 | a b | ⓑ 3 c 4 d 6 3 c 4 d 6 ⓒ 4 x 5 y 22 4 x 5 y 22

Answers will vary.

Section 8.2 Exercises

ⓐ 2 2 4 2 2 4 ⓑ 2 2 5 2 2 5

ⓐ 2 4 4 2 4 4 ⓑ 4 4 3 4 4 3

ⓐ | y 5 | y | y 5 | y ⓑ r r 2 3 r r 2 3 ⓒ s 2 s 2 4 s 2 s 2 4

ⓐ n 10 n n 10 n ⓑ q 2 q 2 3 q 2 q 2 3 ⓒ | n | n 2 8 | n | n 2 8

ⓐ 5 r 6 5 r 5 r 6 5 r ⓑ 3 x 4 x 2 3 3 x 4 x 2 3 ⓒ 2 | y | 3 y 2 4 2 | y | 3 y 2 4

ⓐ 11 | m 11 | 2 m 11 | m 11 | 2 m ⓑ 3 m 2 5 m 2 4 3 m 2 5 m 2 4 ⓒ 2 n 5 n 3 5 2 n 5 n 3 5

ⓐ 7 | m 3 n 5 | 3 m n 7 | m 3 n 5 | 3 m n ⓑ 2 x 2 y 2 6 y 3 2 x 2 y 2 6 y 3 ⓒ 2 | x y | 2 x 4 2 | x y | 2 x 4

ⓐ 8 | q r 3 | 3 q r 8 | q r 3 | 3 q r ⓑ 3 m 3 n 3 2 n 3 3 m 3 n 3 2 n 3 ⓒ 3 a 2 b 2 a 4 3 a 2 b 2 a 4

ⓐ −6 4 3 −6 4 3 ⓑ not real

ⓐ −2 −2 ⓑ not real

ⓐ 5 + 2 3 5 + 2 3 ⓑ 5 − 6 5 − 6

ⓐ 1 + 3 5 1 + 3 5 ⓑ 1 + 10 1 + 10

ⓐ 3 4 3 4 ⓑ 2 3 2 3 ⓒ 1 3 1 3

ⓐ 5 3 5 3 ⓑ 3 5 3 5 ⓒ 1 4 1 4

ⓐ x 2 x 2 ⓑ p 3 p 3 ⓒ | q | | q |

ⓐ 1 y 2 1 y 2 ⓑ u 2 u 2 ⓒ | v 3 | | v 3 |

4 | x 3 | 6 x 11 4 | x 3 | 6 x 11

5 m 2 3 m 4 5 m 2 3 m 4

7 r 2 2 r 10 7 r 2 2 r 10

2 | q 3 | 7 15 2 | q 3 | 7 15

ⓐ 5 r 4 3 r s 4 5 r 4 3 r s 4 ⓑ 3 a 2 2 a 2 3 b 3 a 2 2 a 2 3 b ⓒ 2 | c | 4 c 4 | d | 2 | c | 4 c 4 | d |

ⓐ 2 | p 3 | 7 p | q | 2 | p 3 | 7 p | q | ⓑ 3 s 2 3 s 2 3 t 3 s 2 3 s 2 3 t ⓒ 2 | p 3 | 4 p 3 4 | q 3 | 2 | p 3 | 4 p 3 4 | q 3 |

ⓐ 4 | x y | 3 4 | x y | 3 ⓑ y 2 x 3 2 y 2 x 3 2 ⓒ | a b | a 4 2 | a b | a 4 2

ⓐ 1 2 | p q | 1 2 | p q | ⓑ 2 c d d 2 3 5 2 c d d 2 3 5 ⓒ | m n | 2 | m n | 2

ⓐ 3 p 4 p | q | 3 p 4 p | q | ⓑ 2 2 4 2 2 4 ⓒ 2 x 2 x 5 2 x 2 x 5

ⓐ 5 | m 3 | 5 | m 3 | ⓑ 5 5 3 5 5 3 ⓒ 3 | y | 3 y 2 4 3 | y | 3 y 2 4

Section 8.3 Exercises

ⓐ x x ⓑ y 3 y 3 ⓒ z 4 z 4

ⓐ u 5 u 5 ⓑ v 9 v 9 ⓒ w 20 w 20

ⓐ x 1 7 x 1 7 ⓑ y 1 9 y 1 9 ⓒ f 1 5 f 1 5

ⓐ ( 7 c ) 1 3 ( 7 c ) 1 3 ⓑ ( 12 d ) 1 7 ( 12 d ) 1 7 ⓒ 2 ( 6 b ) 1 4 2 ( 6 b ) 1 4

ⓐ ( 21 p ) 1 2 ( 21 p ) 1 2 ⓑ ( 8 q ) 1 4 ( 8 q ) 1 4 ⓒ 4 ( 36 r ) 1 6 4 ( 36 r ) 1 6

ⓐ 9 ⓑ 5 ⓒ 8

ⓐ 2 ⓑ 4 ⓒ 5

ⓐ −6 −6 ⓑ −6 −6 ⓒ 1 6 1 6

ⓐ not real ⓑ −3 −3 ⓒ 1 3 1 3

ⓐ not real ⓑ −6 −6 ⓒ 1 6 1 6

ⓐ not real ⓑ −10 −10 ⓒ 1 10 1 10

ⓐ m 5 2 m 5 2 ⓑ ( 3 y ) 7 3 ( 3 y ) 7 3 ⓒ ( 4 x 5 y ) 3 5 ( 4 x 5 y ) 3 5

ⓐ u 2 5 u 2 5 ⓑ ( 6 x ) 5 3 ( 6 x ) 5 3 ⓒ ( 18 a 5 b ) 7 4 ( 18 a 5 b ) 7 4

ⓐ 32,768 ⓑ 1 729 1 729 ⓒ 9

ⓐ 4 ⓑ 1 9 1 9 ⓒ not real

ⓐ −27 −27 ⓑ − 1 27 − 1 27 ⓒ not real

ⓐ c 7 8 c 7 8 ⓑ p 9 p 9 ⓒ 1 r 1 r

ⓐ y 5 4 y 5 4 ⓑ x 8 x 8 ⓒ 1 m 1 m

ⓐ 81 q 2 81 q 2 ⓑ a 1 2 b a 1 2 b

ⓐ 8 u 1 4 8 u 1 4 ⓑ 8 p 1 2 q 3 4 8 p 1 2 q 3 4

ⓐ r 7 2 r 7 2 ⓑ 6 s t 6 s t

ⓐ c 2 c 2 ⓑ 2 x 3 y 2 x 3 y

Section 8.4 Exercises

ⓐ 3 2 3 2 ⓑ 7 m 3 7 m 3 ⓒ 6 m 4 6 m 4

ⓐ 9 5 9 5 ⓑ 12 a 3 12 a 3 ⓒ 6 2 z 4 6 2 z 4

ⓐ 4 2 a 4 2 a ⓑ 0

ⓐ 3 c 3 c ⓑ 4 p q 3 4 p q 3

ⓐ −2 3 −2 3 ⓑ −2 5 3 −2 5 3 ⓒ 3 2 4 3 2 4

ⓐ 7 3 7 3 ⓑ 7 2 3 7 2 3 ⓒ 3 5 4 3 5 4

ⓐ a 2 2 a a 2 2 a ⓑ 0

ⓐ 2 c 3 5 c 2 c 3 5 c ⓑ 14 r 2 2 r 2 4 14 r 2 2 r 2 4

4 y 2 4 y 2

ⓐ −18 6 −18 6 ⓑ −64 9 3 −64 9 3

ⓐ −30 2 −30 2 ⓑ 6 2 4 6 2 4

ⓐ 72 z 2 3 72 z 2 3 ⓑ 45 x 2 2 3 45 x 2 2 3

ⓐ −42 z 5 2 z −42 z 5 2 z ⓑ −8 y 6 y 4 −8 y 6 y 4

ⓐ 14 + 5 7 14 + 5 7 ⓑ 4 6 3 + 3 4 3 4 6 3 + 3 4 3

ⓐ 44 − 3 11 44 − 3 11 ⓑ 3 2 4 + 54 4 3 2 4 + 54 4

60 + 2 3 60 + 2 3

ⓐ 30 + 18 2 30 + 18 2 ⓑ x 2 3 − 2 x 3 − 3 x 2 3 − 2 x 3 − 3

ⓐ −55 + 13 10 −55 + 13 10 ⓑ 2 x 2 3 + 8 x 3 + 6 2 x 2 3 + 8 x 3 + 6

23 + 3 30 23 + 3 30

−439 − 2 77 −439 − 2 77

ⓐ 14 + 6 5 14 + 6 5 ⓑ 79 − 20 3 79 − 20 3

ⓐ 87 − 18 6 87 − 18 6 ⓑ 163 + 60 7 163 + 60 7

9 x 2 3 − 4 9 x 2 3 − 4

− 5 4 − 5 4

10 c 2 3 − 9 c 3 3 10 c 2 3 − 9 c 3 3

17 q 2 17 q 2

−42 9 3 −42 9 3

29 − 7 17 29 − 7 17

54 − 36 2 54 − 36 2

6 + 3 2 3 6 + 3 2 3

Section 8.5 Exercises

ⓐ 4 3 4 3 ⓑ 4 3 4 3

ⓐ 10 m 2 7 10 m 2 7 ⓑ 3 y 3 y

ⓐ 5 6 r 2 5 6 r 2 ⓑ 2 x 3 2 x 3

ⓐ 6 p q 2 6 p q 2 ⓑ − 2 a 2 b − 2 a 2 b

ⓐ 8 m 4 3 n 4 8 m 4 3 n 4 ⓑ − 2 x 2 3 y 2 − 2 x 2 3 y 2

2 x 2 7 y 2 x 2 7 y

2 a b 2 a 3 2 a b 2 a 3

ⓐ 5 6 3 5 6 3 ⓑ 2 3 9 2 3 9 ⓒ 2 5 x x 2 5 x x

ⓐ 6 7 7 6 7 7 ⓑ 2 10 15 2 10 15 ⓒ 4 3 p p 4 3 p p

ⓐ 25 3 5 25 3 5 ⓑ 45 3 6 45 3 6 ⓒ 2 6 a 2 3 3 a 2 6 a 2 3 3 a

ⓐ 121 3 11 121 3 11 ⓑ 28 3 6 28 3 6 ⓒ 9 x 3 x 9 x 3 x

ⓐ 343 4 7 343 4 7 ⓑ 40 4 4 40 4 4 ⓒ 2 4 x 2 4 x 2 4 x 2 4 x

ⓐ 9 4 3 9 4 3 ⓑ 50 4 4 50 4 4 ⓒ 2 3 a 3 4 a 2 3 a 3 4 a

−2 ( 1 + 5 ) −2 ( 1 + 5 )

3 ( 3 + 7 ) 3 ( 3 + 7 )

3 ( m + 5 ) m − 5 3 ( m + 5 ) m − 5

2 ( x + 6 ) x − 6 2 ( x + 6 ) x − 6

( r + 5 ) r − 5 2 ( r + 5 ) r − 5 2

( x + 2 2 ) x − 8 2 ( x + 2 2 ) x − 8 2

Section 8.6 Exercises

x = 14 x = 14

no solution

x = −4 x = −4

m = 14 m = 14

v = 17 v = 17

m = 7 2 m = 7 2

u = 3 , u = 4 u = 3 , u = 4

r = 1 , r = 2 r = 1 , r = 2

x = 10 x = 10

x = −8 x = −8

x = 7 x = 7

z = 21 z = 21

x = 42 x = 42

r = 3 r = 3

u = 3 u = 3

r = −2 r = −2

x = 1 x = 1

x = −8 , x = 2 x = −8 , x = 2

a = 0 a = 0

u = 9 4 u = 9 4

a = 4 a = 4

x = 1 x = 5 x = 1 x = 5

8.7 8.7 feet

4.7 4.7 seconds

Section 8.7 Exercises

ⓐ f ( 5 ) = 4 f ( 5 ) = 4 ⓑ no value at x = 0 x = 0

ⓐ g ( 4 ) = 5 g ( 4 ) = 5 ⓑ g ( 8 ) = 7 g ( 8 ) = 7

ⓐ F ( 1 ) = 1 F ( 1 ) = 1 ⓑ F ( −11 ) = 5 F ( −11 ) = 5

ⓐ G ( 5 ) = 2 6 G ( 5 ) = 2 6 ⓑ G ( 2 ) = 3 G ( 2 ) = 3

ⓐ g ( 6 ) = 2 g ( 6 ) = 2 ⓑ g ( −2 ) = −2 g ( −2 ) = −2

ⓐ h ( −2 ) = 0 h ( −2 ) = 0 ⓑ h ( 6 ) = 2 4 3 h ( 6 ) = 2 4 3

ⓐ f ( 0 ) = 0 f ( 0 ) = 0 ⓑ f ( 2 ) = 2 f ( 2 ) = 2

ⓐ g ( 1 ) = 0 g ( 1 ) = 0 ⓑ g ( −3 ) = 2 g ( −3 ) = 2

[ 1 3 , ∞ ) [ 1 3 , ∞ )

( − ∞ , 2 3 ] ( − ∞ , 2 3 ]

( 2 , ∞ ) ( 2 , ∞ )

( − ∞ , −3 ] ∪ ( 2 , ∞ ) ( − ∞ , −3 ] ∪ ( 2 , ∞ )

[ − 3 8 , ∞ ) [ − 3 8 , ∞ )

ⓐ domain: [ −1 , ∞ ) [ −1 , ∞ ) ⓑ

ⓒ [ 0 , ∞ ) [ 0 , ∞ )

ⓐ domain: [ −4 , ∞ ) [ −4 , ∞ ) ⓑ

ⓐ domain: [ 0 , ∞ ) [ 0 , ∞ ) ⓑ

ⓒ [ 2 , ∞ ) [ 2 , ∞ )

ⓐ domain: ( − ∞ , 3 ] ( − ∞ , 3 ] ⓑ

ⓒ ( − ∞ , 0 ] ( − ∞ , 0 ]

ⓒ ( − ∞ , ∞ ) ( − ∞ , ∞ )

Section 8.8 Exercises

ⓐ 4 i 4 i ⓑ 11 i 11 i ⓒ 2 2 i 2 2 i

ⓐ 10 i 10 i ⓑ 13 i 13 i ⓒ 3 5 i 3 5 i

9 3 i 9 3 i

8 2 i 8 2 i

8 + 7 i 8 + 7 i

14 + 2 i 14 + 2 i

−2 + 2 i −2 + 2 i

8 + 5 i 8 + 5 i

7 − 13 i 7 − 13 i

25 − 2 2 i 25 − 2 2 i

−12 + 18 i −12 + 18 i

−38 + + 9 i −38 + + 9 i

27 + 15 i 27 + 15 i

−7 + 24 i −7 + 24 i

−5 + 12 i −5 + 12 i

−44 − 4 i 3 −44 − 4 i 3

−20 − 2 2 i −20 − 2 2 i

2 25 + 11 25 i 2 25 + 11 25 i

6 13 + 9 13 i 6 13 + 9 13 i

− 12 13 − 8 13 i − 12 13 − 8 13 i

4 3 − 1 3 i 4 3 − 1 3 i

− 3 4 + 1 2 i − 3 4 + 1 2 i

Review Exercises

ⓐ 15 ⓑ −4 −4

ⓐ 2 ⓑ 3 ⓒ 3

ⓐ 8 < 68 < 9 8 < 68 < 9 ⓑ 4 < 84 3 < 5 4 < 84 3 < 5

ⓐ 5 5 3 5 5 3 ⓑ 2 2 6 2 2 6

ⓐ 4 | s 7 | 5 s 4 | s 7 | 5 s ⓑ 2 a 3 a 2 5 2 a 3 a 2 5 ⓒ 2 | b | 2 b 6 2 | b | 2 b 6

ⓐ 6 7 6 7 ⓑ 2 3 2 3 ⓒ 1 2 1 2

ⓐ 1 2 | p q | 1 2 | p q | ⓑ 2 c d 5 d 2 3 2 c d 5 d 2 3 ⓒ | m n | 2 | m n | 2

ⓐ r r ⓑ s 3 s 3 ⓒ t 4 t 4

ⓐ 5 ⓑ 3 ⓒ 2

ⓐ −2 −2 ⓑ 1 3 1 3 ⓒ −5 −5

ⓐ 125 ⓑ 1 27 1 27 ⓒ 16

ⓐ 6 3 6 3 ⓑ b 9 b 9 ⓒ 1 w 1 w

ⓐ 4 2 4 2 ⓑ 9 p 3 9 p 3 ⓒ 2 x 3 2 x 3

37 y 3 37 y 3

ⓐ 126 x 2 x 126 x 2 x ⓑ 48 a 5 a 2 3 48 a 5 a 2 3

ⓐ 71 − 22 7 71 − 22 7 ⓑ x 2 3 − 8 x 3 + 15 x 2 3 − 8 x 3 + 15

ⓐ 27 + 8 11 27 + 8 11 ⓑ 29 − 12 5 29 − 12 5

− 7 ( 2 + 6 ) 2 − 7 ( 2 + 6 ) 2

64.8 64.8 feet

− ∞ , 10 7 − ∞ , 10 7

23 + 14 i 23 + 14 i

Practice Test

5 x 3 5 x 3

2 x 2 y 9 x 2 y 3 2 x 2 y 9 x 2 y 3

ⓐ 1 4 1 4 ⓑ −343 −343

x 7 4 x 7 4

− x 2 3 x − x 2 3 x

36 x 4 2 36 x 4 2

2 − 7 3 2 − 7 3

7 x 2 x 3 | y 3 | y 7 x 2 x 3 | y 3 | y

3 ( 2 − 3 ) 3 ( 2 − 3 )

−12 + 8 i −12 + 8 i

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• Authors: Lynn Marecek, Andrea Honeycutt Mathis
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• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/chapter-8

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Into Math Grade 7 Module 2 Lesson 5 Answer Key Simple Interest

We included H MH Into Math Grade 7 Answer Key PDF Module 2 Lesson 5 Simple Interest to make students experts in learning maths.

HMH Into Math Grade 7 Module 2 Lesson 5 Answer Key Simple Interest

Step It Out

1. Big Money Bank loans $12,000 to Carlotta. This initial amount borrowed is called the principal. Carlotta has to repay the loan to the bank at a rate of 5.5% simple interest per year over 8 years. What is the total amount of interest she will have to pay on her loan?

Connect to Vocabulary Simple interest is a fixed percent of the principal. It is calculated using the formula I = Prt, where P represents the principal, r the rate of interest, and t the time.

Turn and Talk Before she took out her loan, a different bank offered Carlotta a $12,000 loan at 5% interest to be paid back over 9 years. Would that loan cost Carlotta more interest than the loan she took, or less? How much is the difference? Answer: Given that, Principal = $12000 Interest rate = 5% Number of years = 9 Simple interest = PTR = $12000 x 5% x 9 = $5400. Over 8 years Carlotta will pay $5280 in simple interest on her loan. Over 9 years Carlotta will pay $5400 in simple interest on her loan. The simple interest of the loan for 8 years is less than the 9 years of a loan. The difference of the two loans is $5400 – $5280 = $120

B. Write an equation for the amount of interest l Emilio’s account earns in t years. I = P. r ∙ ___ I = ___ t Answer: P = principle = $5,000 R = interest rate = 2.5% = 0.025 Time = t I = P x r x t I = $125t

C. How much interest will be added to Emilio’s account after 12 years? Show your calculation. I = ___t I = ____ ∙ ____ = ____ Answer: I = $125t T = 12 years. Substitute t in the above equation. I = $125(12) = $1500 Interest added to Emilio’s account after 12 years is $1500

3. Melanie deposits $8,200.00 in a bank account paying 4.4% simple interest. How much money will be in her account after 5 years and 6 months? A. Find the amount Melanie’s account earns in interest in 5.5 years. I = P ∙ r ∙ t I = 8,200 ∙ ____ ∙ ____ I = ____ The amount of interest in 5.5 years is $_____ Answer: Given that, Principal = P = $8200 Interest rate = 4.4% Time = 5.5 years = 5 years and 6 months. I = P x r x t I = $8200 x 4.4% x 5.5 = $1984.4 The amount of interest in 5.5 years is $1984.4.

B. Find the total amount in Melanie’s account after 5.5 years. Total amount = P + I = ___ + ___ = ____ Melanie’s account will contain $____ after 5 years and 6 months. Answer: Total amount = P + I Principal = P = $8200 Simple interest I = $1984.4 Total amount = $8200 + $1984.4 = $10,184.4.

B. Find the time t in years that it took Gregory to pay off his loan. I = P ∙ r ∙ t 3,150 = ___ ∙ ___ ∙ t 3,150 = ___ ∙ t ____ = t Gregory spent ___ years paying back his loan. Answer: I = p x r x t Principal P = 7500 Interest rate = 6% 3,150 = $7500 x 6% x t 3150 = $450 x t 3150/450 = t 7 = t Gregory spent 7 years paying back his loan.

Turn and Talk Describe the key steps that were taken to solve the problem.

Check Understanding

Question 1. Arisia puts $500.00 into a savings account with an annual simple interest rate of 4.5%. A. How much interest does she earn per year? ____ Answer: Simple interest rate = PTR Principal = P = $500. Simple interest rate = 4.5% Time = 1 year. Simple interest rate = PTR = $500 x 4.5% x 1 = $2250 Interest she earns per year = $2,250.

B. If the interest rate stayed the same, how much interest would Arisia’s account earn after 15 years? ____ Answer: Simple interest rate = PTR Principal = P = $500. Simple interest rate = 4.5% Time = 15 year. Simple interest rate = PTR = $500 x 4.5% x 15 = $33,750 Interest she earns per 15 years = $33,750.

C. If the interest rate stayed the same, how much money would be in Arisia’s account after 20 years? _____ Answer: Simple interest rate = PTR Principal = P = $500. Simple interest rate = 4.5% Time = 20 year. Simple interest rate = PTR = $500 x 4.5% x 20 = $45000 Interest she earns per 20 years = $45000.

On Your Own

Question 2. Model with Mathematics Write an equation to find the amount of interest $450,000 earns in 1 year at a simple interest rate of 3.5%. ___________ Answer: Given that, Principal P = $450,000. Simple interest rate = $3.5% Time = 1 year. The equation to find the simple interest is PTR. = $450,000 x 3.5% x 1 Simple interest I = $15,750

Question 4. Inez opens a savings account with $2,400. The account pays her 2.4% annual simple interest. A. Find the amount of simple interest that the account will earn per year. Answer: Given that, Principal = $2,400 Simple interest rate = 2.4% Time = 1 years. Simple interest I = $2400 x 2.4% x 1 Simple interest I = $57.6 The simple interest that the account will earn per year is $57.6

B. Calculate the total interest Inez would earn in 10 years. Answer: Given that, Principal = $2,400 Simple interest rate = 2.4% Time = 1 years. Simple interest I = $2400 x 2.4% x 10 Simple interest I = $576. The simple interest that the account will earn per year is $576. Total interest = $2400 + $576 = $2976. The total interest Inez would earn in 10 years is $2976.

Question 5. Green Bank has the ad shown. Barry opens a savings account after seeing the ad. He deposits $1,300. A. Write an equation that relates the amount of time t in years that Barry holds his account to the amount of simple interest I that he earns. Answer: Given that, Principal P = $1300 Interest rate = $2.2% Time = t. The equation for the simple interest is PTR = $1300 x 2.2% x t = $28.6t.

B. How much interest does Barry earn in 7 years? ________________ Answer: The equation for the simple interest I = PTR. Principal P = $1300 Interest rate = $2.2% Time = 7 years The equation for the simple interest is PTR = $1300 x 2.2% x 7. = $200.2 The interest Barry earns in 7 years is $200.2

C. At the Blue Bank, Barry would earn $159.25 in simple interest in 7 years after depositing $1,300. What rate of simple interest is offered at the Blue Bank? Answer: The equation for the simple interest I = PTR. Principal P = $1300 Interest rate = r% Time = 7 years Simple interest I = $159.25 The equation for the simple interest = PTR $159.25 = $1300 x r% x 7. $159.25 = $9100 x r% $159.25/$9100 = r% 0.0175 = r% 1.75% = r The rate of simple interest offered at the Blue Bank is 1.75%

B. How much money will Avram have to repay in all? Answer: Total money = P + I Principal = $14,500 I = $7,047. Total money = $14,500 + $7,047 = $21547 Avram has to repay all $21547.

A. Regina calculates that to buy the car in Ad A, the loan would ultimately cost her $41,503. Over how many years is the loan in Ad A to be paid back? Answer: Given that, Principal P = $24,200. Interest rate = 6.5% The loan cost = $41,503 Interest paid = $41,503 – $24,200 = $17303 The simple interest per 1 year = PTR $24200 x 6.5% x 1 = $1573. Number of years = $17303/$1573 = 11 years The number of years the loan in Ad A to be paid back is 11 years.

B. Regina calculates that to buy the car in Ad B, the loan would ultimately cost her $38,402. What is the price of the car offered in Ad B? Answer:

C. In either case, Regina would be paying back the dealer in equal monthly installments over the lifetime of the loan. What would her monthly payments be if she bought the car shown in Ad A? _________________ What would her monthly payments be if she bought the car shown in Ad B? Answer:

D. Open Ended In your opinion, which ad is offering the better deal? Explain why you think so. _________________ Answer:

Question 8. It costs $36,736 to repay a loan of $20,500 at 6.6% annual simple interest. A. How much interest would you pay each year? ______________________ Answer: Simple interest per 1 year = PTR Principal = P = $20,500 Simple interest rate = 6.6% Time = 1 year. Simple interest I = $20,500 x 6.6% x 1 = $1353. The simple interest per 1 year =$1353.

B. How many years does it take to repay the loan? Answer: Principal = P = $20,500 Simple interest rate = 6.6% The simple interest per 1 year =$1353. Time =? Interest paid = $36736 – $20,500 = $16,236 Time = $16,236/$1353 = 12 The number of years it takes to repay the loan is 12 years.

Question 9. A savings account pays an annual simple interest rate of 1.5%.

A. How much interest would you earn in 1 year on $2,000? Answer: Given that, Principal = $2,000. Interest rate = 1.5% Time = 1 year. Formula fo the simple interest I = PTR I = $2000 x 1.5% x 1 = $30 The interest per 1 year = $30.

B. What would be the balance in your account after 5 years? Answer: Given that, Principal = $2,000. Interest rate = 1.5% Time = 1 year.Formula fo the simple interest I = PTR I = $2000 x 1.5% x 5 = $150 The interest per 5 years = $150. The balance in your account after 5 years = $2000 + $150 = $2150.

C. How long would it take to earn $500 or more in interest? Answer: Given that, Principal = $2,000. Interest rate = 1.5% Time = 1 year. Formula for the simple interest I = PTR I = $2000 x 1.5% x 1 = $30 The interest per 1 year = $30. To earn 500 or more in interest = $500/$30 = 16.6 It means to earn 500 takes 16 years and 6 months.

Question 10. A loan of $9,000 has an annual simple interest rate of 3.5%. A. If you repay the loan in 6 years, how much total interest will you pay? Answer: Given that, Loan = Principal P = $9000. Simple interest rate = r = 3.5% Time = 6 years Simple interest I = PTR = $9000 x 3.5% x 6 = $1890 Total interest = $1890 The total interest you will pay is $1890.

B. What is the total cost of the loan? Answer: Loan = Principal P = $9000. Simple interest = $1890 Total cost = P + I Total cost = $9000 + $1890 = $10890. The total cost will be $10890.

C. What would be the simple interest rate if a loan of $9,000 that is repaid in 6 years has a total cost of $11,052? Answer: Loan = Principal P = $9000. Total cost = $11,052 Amount of interest = PTR $11052 = $9000 x 6 x t $11052 = 54000 x t $11052/$54000 = 0.204 = 20.4% The simple interest rate if a loan of $9,000 that is repaid in 6 years has a total cost of $11,052 is 20.4%

Question 11. A loan of $25,000 has an annual simple interest rate of 5.5%. The total cost of the loan is $38,750. A. How much total interest will you pay? Answer: Given that, Loan rate = $25,000 Simple interest rate = 5.5% The total cost of the loan = is $38,750. The total interest you will pay = $38,750 – $25,000. = $13750. The total interest you will pay is $13750.

B. How long does it take to repay the loan? Answer: Given that, Loan rate = $25,000 Simple interest rate = 5.5% Simple interest per 1 year = $25,000 x 5.5% x 1 = $1375. The total interest you will pay is $13750. $1375 x 10 = $13750. He has to repay the loan for 10 years.

Lesson 2.5 Practice/Homework

Question 2. Big Money Bank has an offer for new customers: if you deposit $5,000 in a savings account, you will earn 6.5% simple interest over the first 10 years. A. How much interest will the account earn over this period? Answer: I = Prt P = 5000 r = 6.5% t = 10 I = 5000 × 6.5% × 10 I = 5000 × 6.5/100 × 10 I = 10 × (5000 × 0.065) I = 10 × 325 = 3250 Interest for 10 years = $3250

B. How much will be in the account after the 10-year period? Answer:

C. At the Town Savings Bank, a new customer earns $2,950 in simple interest on a $5,000 deposit over the first 10 years. What rate of interest does that bank pay? Answer: At the Town Savings Bank, a new customer earns $2,950 in simple interest on a $5,000 deposit over the first 10 years. The interest for 1 year 2950/10 = $295 The rate of interest 295/5000 × 100% = 5.9%

Question 3. Dream Loan Bank offers loans. Carrie borrows $10,500 to help start a business. The loan must be repaid at 4.5% simple interest over 10 years. How much money will Carrie have to pay back? Answer: Given, At the Town Savings Bank, a new customer earns $2,950 in simple interest on a $5,000 deposit over the first 10 years. P = 10500 r = 4.5% t = 10 years F = P + Prt F = 10500 + 10500 × 4.5/100 × 10 10500(1 + 0.45) 10500 × 1.45 = 15225 Carrie have to pay back $15225.

Question 4. Financial Literacy Kevin is going to open a savings account with $4,000. Two different banks offer him two different options: Bank A offers an account that will pay 6% simple interest for 6 years. Bank B offers a special account for new customers that will pay 7% simple interest for 3 years. After the 3 years, Kevin would have to transfer all his earnings to a regular account that will pay 5% simple interest on the new transferred principal. Which offer will leave Kevin with more money after 6 years? Explain. Answer: Given, Kevin is going to open a savings account with $4,000. Bank A offers an account that will pay 6% simple interest for 6 years. 4000 × 6% × 6 40 00 × 6/1 00 × 6 = 1440 Second option: 4000 × 7% × 3 + 4000 × (7% × 3 + 1) × 5% × 3 = 840 + 4840 × 5% × 3 = 840 + 726 = 1566 1440 is less than 1556 Hence the second option will leave Kevin with more money after 6 years.

Question 5. Delilah deposits $8,255 in an account that pays 4.2% simple interest. How much money will be in her account after 8 years? A. $2,773.69 B. $11,028.68 C. $12,254.66 D. $27,736.80 Answer: Given, Delilah deposits $8,255 in an account that pays 4.2% simple interest. P = 8255 T = 8 years R = 4.2% F = P(1 + Rt) 8255 (1 + (4.2%)8) 8255(1 + 0.336) 8255 × 1.336 = 11028.68 Thus option B is the correct answer.

Question 6. Edgar takes out a loan of $5,540, to be repaid over 7 years at 8.5% Simple interest. How much interest will Edgar have to pay on the loan? A. $470.90 B. $3,296.30 C. $3,775.60 D. $8,836.30 Answer: Given, Edgar takes out a loan of $5,540, to be repaid over 7 years at 8.5% Simple interest. Simple Interest is the interest of each period that does not generate new interest. 5540 × 8.5% × 7 = 5540 × 85/10 × 7 = 3296.30 Thus option B is the correct answer.

Question 7. Gregoria borrowed $2,450, to be paid back at 3.5% simple interest. She spent $3,221.75 in all to repay the loan. How many years did Gregoria take to repay the loan? Answer: Given, Gregoria borrowed $2,450, to be paid back at 3.5% simple interest. She spent $3,221.75 in all to repay the loan. 3221.75 – 2450 = 771.75 2450 × 35% = 2450 × 35/100 = 85.75 771.75 ÷ 85.75 = 9 Thus it takes 9 years to repay the loan.

Spiral Review

Question 9. What number is 3 units to the left of 2? _________________ Answer: 2 Units to the left of 3 is 3 – 2 = 1

Question 10. Bianca had a weekly allowance of $8.50 two years ago. Last year, her weekly allowance was $9.75. This year, Bianca’s weekly allowance is $12.00. Does it make sense to represent the relationship between the amount of her allowance and the year with a constant rate? Why or why not? _________________ Answer: Given, Bianca had a weekly allowance of $8.50 two years ago. Last year, her weekly allowance was $9.75. This year, Bianca’s weekly allowance is $12.00. 9.75 – 8.50 = 1.25 1.25/8.50 = 0.147 or 14.7 % increase Between the second and third year, there was a change of 12 – 9.75 = 2.25 dollar change 2.25/9.75 = 0.23 or 23% increase Hence, each year the percent and dollar value increase is increasing more and more which would not be a constant rate.

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