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7-1 Multiplication Properties of Exponents Determine whether each expression is a monomial. Write yes or no. Explain your reasoning. 1. 15 SOLUTION: 15 is a monomial. It is a constant and all constants are monomials. 2. 2− 3a SOLUTION: 2− 3a is not a monomial. There is subtraction and more than one term. 4 2 8. m (m ) SOLUTION: 2 4 9. 2q (9q ) SOLUTION: 4 4 3 10. (5u v)(7u v ) SOLUTION: 3. SOLUTION: is not a monomial. There is a variable in the denominator. 4. −15g 2 2 2 11. [(3 ) ] SOLUTION: 2 SOLUTION: 2 −15g is a monomial. It is the product of a number and variables. 5. SOLUTION: is a monomial. It is the product of a number and a 4 6 12. (xy ) SOLUTION: variable. 6. 7b + 9 SOLUTION: No; there is addition and more than one term. Simplify each expression. 3 7. k(k ) SOLUTION: 4 9 13. (4a b c) 2 SOLUTION: 2 3 2 3 4 2 eSolutions 8. m (mManual ) - Powered by Cognero SOLUTION: 14. (−2f g h ) Page 1 SOLUTION: SOLUTION: b. 7-1 Multiplication Properties of Exponents So the surface area is 69,984 square units. 2 3 2 3 14. (−2f g h ) Simplify each expression. 2 SOLUTION: 2 3 3 17. (5x y) (2xy z) (4xyz) SOLUTION: 56 4 15. (−3p t ) SOLUTION: 2 3 2 2 3 2 18. (−3d f g) [(−3d f ) ] 16. GEOMETRY The formula for the surface area of 2 a cube is SA = 6s , where SA is the surface area and s is the length of any side. SOLUTION: a. Express the surface area of the cube as a monomial. b. What is the surface area of the cube if a = 3 and b = 4? SOLUTION: 3 4 2 19. (−2g h)(−3gj ) (−ghj) 2 SOLUTION: a. 4 3 2 2 3 20. (−7ab c) [(2a c) ] SOLUTION: b. So the surface area is 69,984 square units. Simplify each expression. 2 2 3 3 17. (5x y) (2xy z) (4xyz) SOLUTION: eSolutions Manual - Powered by Cognero Determine whether each expression is a monomial. Write yes or no. Explain your reasoning. 21. 122 Page 2 7-1 Multiplication Properties of Exponents Determine whether each expression is a monomial. Write yes or no. Explain your reasoning. 21. 122 2 6 28. (−2u )(6u ) SOLUTION: SOLUTION: 122 is a monomial. A constant is a monomial. 2 8 6 4 29. (9w x )(w x ) 22. 3a 4 SOLUTION: SOLUTION: 4 3a is a monomial. It is the product of a number and variables. 6 9 4 2 30. (y z )(6y z ) 23. 2c + 2 SOLUTION: 2c + 2 is not a monomial. It involves addition and has more than one term. SOLUTION: 8 6 5 24. 6 2 31. (b c d )(7b c d) SOLUTION: SOLUTION: is not a monomial. The expression has a variable in the denominator. 2 2 4 2 2 32. (14fg h )(−3f g h ) SOLUTION: 25. SOLUTION: is a monomial. It can be written as the product of a number and variable. 5 7 4 33. (j k ) SOLUTION: 26. 6m + 3n SOLUTION: 6m + 3n is not a monomial. In the expression, there is addition and more than one term. Simplify each expression. 2 4 27. (q )(2q ) SOLUTION: 3 34. (n p ) 4 SOLUTION: 2 2 2 2 6 28. (−2uManual )(6u -)Powered by Cognero eSolutions SOLUTION: 35. [(2 ) ] SOLUTION: Page 3 7-1 Multiplication Properties of Exponents 2 2 2 GEOMETRY Express the area of each triangle as a monomial. 35. [(2 ) ] SOLUTION: 39. SOLUTION: 2 2 4 36. [(3 ) ] SOLUTION: 40. SOLUTION: 2 3 2 37. [(4r t) ] SOLUTION: Simplify each expression. 3 4 3 3 41. (2a ) (a ) SOLUTION: 2 3 2 38. [(−2xy ) ] SOLUTION: 3 2 5 2 42. (c ) (−3c ) SOLUTION: GEOMETRY Express the area of each triangle as a monomial. 4 3 4 3 2 43. (2gh ) [(−2g h) ] 39. eSolutions Manual - Powered by Cognero SOLUTION: SOLUTION: Page 4 7-1 Multiplication Properties of Exponents 4 3 4 3 2 2 3 4 3 4 2 43. (2gh ) [(−2g h) ] 47. (5a b c )(6a b c ) SOLUTION: SOLUTION: 5 3 4 6 3 48. (10xy z )(3x y z ) SOLUTION: 2 3 3 2 4 2 2 49. (0.5x ) 44. (5k m) [(4km ) ] SOLUTION: SOLUTION: 5 3 50. (0.4h ) SOLUTION: 5 2 4 3 4 2 3 45. (p r ) (−7p r ) (6pr ) SOLUTION: 51. SOLUTION: 2 2 3 3 46. (5x y) (2xy z) (4xyz) SOLUTION: 52. 2 3 4 3 4 2 SOLUTION: 47. (5a b c )(6a b c ) SOLUTION: eSolutions Manual - Powered by Cognero Page 5 SOLUTION: 7-1 Multiplication Properties of Exponents 57. FINANCIAL LITERACY Cleavon has money in an account that earns 3% simple interest. The formula for computing simple interest is I = Prt, where I is the interest earned, P represents the principal that he put into the account, r is the interest rate (in decimal form), and t represents time in years. a. Cleavon makes a deposit of $2c and leaves it for 2 years. Write a monomial that represents the interest earned. b. If c represents a birthday gift of $250, how much will Cleavon have in this account after 2 years? 52. SOLUTION: SOLUTION: a. 3 2 2 53. (8y )(−3x y ) b. SOLUTION: Cleavon will make $30 interest, so he will have 250 + 30 = $280 in his account. 54. CCSS TOOLS Express the volume of each solid as a monomial. (49m)(17p ) SOLUTION: 58. SOLUTION: 3 4 3 2 2 3 2 3 2 3 4 4 2 3 2 4 5 2 55. (−3r w ) (2rw) (−3r ) (4rw ) (2r w ) SOLUTION: 2 2 2 4 2 3 2 4 3 56. (3ab c) (−2a b ) (a c ) (a b c ) (2a b c ) SOLUTION: 57. FINANCIAL LITERACY Cleavon has money in an account that earns 3% simple interest. The formula for computing simple interest is I = Prt, eSolutions Manual - Powered by Cognero where I is the interest earned, P represents the principal that he put into the account, r is the interest rate (in decimal form), and t represents time in years. 59. SOLUTION: Page 6 7-1 Multiplication Properties of Exponents 61. PACKAGING For a commercial art class, Aiko must design a new container for individually wrapped pieces of candy. The shape that she chose is a cylinder. The formula for the volume of a cylinder is 59. 2 V = πr h. a. The radius that Aiko would like to use is 2p 3, and SOLUTION: 3 the height is 4p . Write a monomial that represents the volume of her container. b. Make a table of five possible measures for the radius and height of a cylinder having the same volume. c. What is the volume of Aiko’s container if the height is doubled? SOLUTION: 60. SOLUTION: a. b. The product of the square of the radius’ coefficient and the height’s coefficient must be 16. The exponents of the radius and the height must have a sum of 9. Sample answer: 61. PACKAGING For a commercial art class, Aiko must design a new container for individually wrapped pieces of candy. The shape that she chose is a cylinder. The formula for the volume of a cylinder is c. If the height is doubled, h = 2(4p 3) = 8p 3 2 V = πr h. a. The radius that Aiko would like to use is 2p 3, and 3 the height is 4p . Write a monomial that represents the volume of her container. b. Make a table of five possible measures for the radius and height of a cylinder having the same volume. c. What is the volume of Aiko’s container if the height is doubled? SOLUTION: eSolutions Manual - Powered by Cognero 9 So, the volume of Aiko’s container is 32πp cubic units. 2 62. ENERGY Albert Einstein’s formula E = mc shows that if mass is accelerated enough, it can be converted into usable energy. Energy E is measured in joules, mass m in kilograms, and the speed of light is about 300 million meters per second. Page 7 a. Complete the calculations to convert 3 kilograms of gasoline completely into energy. 9 So, the volume of Aiko’s container is 32πp cubic 7-1 Multiplication Properties of Exponents units. 2 62. ENERGY Albert Einstein’s formula E = mc shows that if mass is accelerated enough, it can be converted into usable energy. Energy E is measured in joules, mass m in kilograms, and the speed of light is about 300 million meters per second. 63. MULTIPLE REPRESENTATIONS In this problem, you will explore exponents. a. TABULAR Copy and use a calculator to complete the table. a. Complete the calculations to convert 3 kilograms of gasoline completely into energy. b. What is the energy if the amount of gasoline is doubled? SOLUTION: a. b. ANALYTICAL What do you think the values −1 0 of 5 and 5 are? Verify your conjecture using a calculator. c. ANALYTICAL Complete: For any nonzero −n number a and any integer n, a = ______. d. VERBAL Describe the value of a nonzero number raised to the zero power. SOLUTION: a. So, 3 kilograms of gasoline converts to 270,000,000,000,000,000 joules of energy. b. If the amount of gasoline is doubled to 6 kilograms, then the energy is also doubled. 63. MULTIPLE REPRESENTATIONS In this problem, you will explore exponents. a. TABULAR Copy and use a calculator to complete the table. 0 −1 0 of 5 and 5 are? Verify your conjecture using a calculator. c. ANALYTICAL Complete: For any nonzero −n number a and any integer n, a = ______. d. VERBAL Describe the value of a nonzero number raised to the zero power. SOLUTION: a. 0 0 -1 d. Any nonzero number raised to the zero power is 1. 64. CCSS PERSEVERANCE For any nonzero real numbers a and b and any integers m and t, simplify the expression b. ANALYTICAL What do you think the values 0 b. If 3 = 1, then 5 may also equal 1. If 3 equals -1 one-third, then 5 may equal one-fifth. c. From the table, it seems that a number raised to a negative power is the same the same as its reciprocal raised to the same power, only positive. So, and describe each step. SOLUTION: Move the negative sign to the numerator. Since raising the fraction to the 2t power means multiplying both the numerator and the denominator by themselves 2t times, we can rewrite the expression as the power of a power for both the numerator and the denominator. -1 b. If 3 = 1, then 5 may also equal 1. If 3 equals -1 one-third, then 5 may equal one-fifth. c. From the table, it seems that a number raised to a negative power is the same the same as its reciprocal eSolutions Manual - Powered by Cognero raised to the same power, only positive. So, To find the power of the power, multiply the exponents. Page 8 negative power is the same the same as its reciprocal raised to the same power, only positive. So, 7-1 Multiplication Properties of Exponents d. Any nonzero number raised to the zero power is 1. 64. CCSS PERSEVERANCE For any nonzero real numbers a and b and any integers m and t, simplify the expression 65. REASONING Copy the table below. and describe each step. SOLUTION: Move the negative sign to the numerator. Since raising the fraction to the 2t power means multiplying both the numerator and the denominator by themselves 2t times, we can rewrite the expression as the power of a power for both the numerator and the denominator. a. For each equation, write the related expression and record the power of x. b. Graph each equation using a graphing calculator. c. Classify each graph as linear or nonlinear. d. Explain how to determine whether an equation, or its related expression, is linear or nonlinear without graphing. SOLUTION: a. b. To find the power of the power, multiply the exponents. Regroup the numerator to isolate the negative base. Simplify the numerator and denominator. c. Use the power of a power rule one last time to simplify. 65. REASONING Copy the table below. a. For each equation, write the related expression and record the power of x. b. Graph each equation using a graphing calculator. eSolutions Manual - Powered by Cognero c. Classify each graph as linear or nonlinear. d. Explain how to determine whether an equation, or its related expression, is linear or nonlinear without d. If the power of x is 1, the equation or its related expression is linear. 66. OPEN ENDED Write three different expressions 6 that can be simplified to x . SOLUTION: 4 2 5 Sample answer: x ⋅ x ; x ⋅ x; For these, the exponents must have a sum of 6. 3 2 (x ) ; For this type, the product of 3 and 2 is 6. 67. WRITING IN MATH Write two formulas that have monomial expressions in them. Explain howPage 9 each is used in a real-world situation. SOLUTION: 4 2 5 Sample answer: x ⋅ x ; x ⋅ x; For these, the exponents must have a sum of 6. 7-1 Multiplication Properties of Exponents 3 2 (x ) ; For this type, the product of 3 and 2 is 6. 67. WRITING IN MATH Write two formulas that have monomial expressions in them. Explain how each is used in a real-world situation. SOLUTION: 2 Sample answer: The area of a circle or A = πr , where r is the radius, can be used to find the area of any circle. The area of a rectangle or A = w ⋅ ℓ, where w is the width and ℓ is the length, can be used to find the area of any rectangle. 68. Which of the following is not a monomial? A −6xy B Because the original triangle got smaller in the transformation, it is a dilation. None of the other types of transformations involve changing the size of a shape (they only affect location). Therefore, choice F is the correct answer. 70. CARS In 2002, the average price of a new domestic car was $19,126. In 2008, the average price was $28,715. Based on a linear model, what is the predicted average price for 2014? A $45,495 B $38,304 C $35,906 D $26,317 SOLUTION: Let x be the year and let y be the average price of a new domestic car. Since all the points are part of a linear model, the slope between any two points will be the same.Find the slope of the line containing the data points (2002, 19,126) and (2008, 28,715) . C 4 D 5gh SOLUTION: The only example that has a negative exponent is C. This means that the term has a variable in the denominator. Therefore it is not a monomial and choice C is the correct answer. Use this slope and the points (2002, 19,126) and (2014, y 2) to determine the average cost of a new domestic car in 2014. 69. GEOMETRY The accompanying diagram shows the transformation of ΔXYZ to ΔX′Y′Z′. The predicted average price of a new domestic car for 2014 is about $38,304. Therefore, the correct choice is B. This transformation is an example of a F dilation G line reflection H rotation J translation 71. SHORT RESPONSE If a line has a positive slope and a negative y–intercept, what happens to the x– intercept if the slope and the y–intercept are doubled? SOLUTION: Because the original triangle got smaller in the transformation, it is a dilation. None of the other types of transformations involve changing the size of a shape (they only affect location). Therefore, choice F is the correct answer. 70. CARS In 2002, the average price of a new domestic car was $19,126. In 2008, the average price was $28,715. Based on a linear model, what is the predicted average price for 2014? A $45,495 eSolutions Manual - Powered by Cognero B $38,304 C $35,906 D $26,317 SOLUTION: The y–intercept of the graphed equation is –4 and the slope is 2, so the equation would be . From the graph, we see that the x–intercept is 2. Now, double the slope and intercept, and the new equation is . Substitute y = 0, to find the x–10 Page intercept. The predicted average price of a new domestic car 7-1 Multiplication Properties of Exponents for 2014 is about $38,304. Therefore, the correct choice is B. 71. SHORT RESPONSE If a line has a positive slope and a negative y–intercept, what happens to the x– intercept if the slope and the y–intercept are doubled? Notice that the x–intercept does not change. Solve each system of inequalities by graphing. 72. y < 4x 2x + 3y ≥ −21 SOLUTION: Write equation 2 in slope, intercept form. SOLUTION: The y–intercept of the graphed equation is –4 and the slope is 2, so the equation would be . From the graph, we see that the x–intercept is 2. Now, double the slope and intercept, and the new equation is . Substitute y = 0, to find the x– intercept. Graph each inequality. The graph of y < 4x is dashed and is not included in the graph of the solution. The graph of 2x + 3y ≥ −21 is solid and is included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y < 4x and 2x + 3y ≥ −21. This region is darkly shaded in the graph below. Notice that the x–intercept does not change. Solve each system of inequalities by graphing. 72. y < 4x 2x + 3y ≥ −21 SOLUTION: Write equation 2 in slope, intercept form. Graph each inequality. The graph of y < 4x is dashed and is not included in the graph of the solution. The graph of 2x + 3y ≥ −21 is solid and is included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y < 4x and 2x + 3y ≥ −21. This region is darkly shaded in the graph below. eSolutions Manual - Powered by Cognero 73. y ≥ 2 2y + 2x ≤ 4 SOLUTION: Write equation 2 in slope-intercept form. Graph each inequality. The graph of y ≥ 2 is solid and is not included in the graph of the solution. The graph of 2y + 2x ≤ 4 is also solid and is included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y ≥ 2 and 2y + 2x ≤ 4. This region is darkly shaded in the graph below. Page 11 7-1 Multiplication Properties of Exponents 73. y ≥ 2 2y + 2x ≤ 4 74. y > −2x − 1 2y ≤ 3x + 2 SOLUTION: Write equation 2 in slope-intercept form. SOLUTION: Rewrite equation 2 in slope-intercept form. Graph each inequality. The graph of y ≥ 2 is solid and is not included in the graph of the solution. The graph of 2y + 2x ≤ 4 is also solid and is included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y ≥ 2 and 2y + 2x ≤ 4. This region is darkly shaded in the graph below. Graph each inequality. The graph of y > −2x − 1 is dashed and is not included in the graph of the solution. The graph of 2y ≤ 3x + 2 is solid and is included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y > −2x − 1 and 2y ≤ 3x + 2. This region is darkly shaded in the graph below. 74. y > −2x − 1 2y ≤ 3x + 2 SOLUTION: Rewrite equation 2 in slope-intercept form. Graph each inequality. The graph of y > −2x − 1 is dashed and is not included in the graph of the solution. The graph of 2y ≤ 3x + 2 is solid and is included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y > −2x − 1 and 2y ≤ 3x + 2. This region is darkly shaded in the graph below. eSolutions Manual - Powered by Cognero 75. 3x + 2y < 10 2x + 12y < –6 75. 3x + 2y < 10 2x + 12y < –6 SOLUTION: Write each equation in slope-intercept form. Equation 1: Equation 2: Graph each inequality. The graph of 3x + 2y < 10 is dashed and is not included in the graph of the solution. The graph of 2x + 12y < –6 is also dashed and is not included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of 3x + 2y < 10 and 2x + 12y < –6. This region is darkly shaded in the graph below. Page 12 7-1 Multiplication Properties of Exponents 75. 3x + 2y < 10 2x + 12y < –6 SOLUTION: Write each equation in slope-intercept form. Equation 1: 76. SPORTS In the 2006 Winter Olympic Games, the total number of gold and silver medals won by the U.S. was 18. The total points scored for gold and silver medals was 45. Write and solve a system of equations to find how many gold and silver medals were won by the U.S. Equation 2: Graph each inequality. The graph of 3x + 2y < 10 is dashed and is not included in the graph of the solution. The graph of 2x + 12y < –6 is also dashed and is not included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of 3x + 2y < 10 and 2x + 12y < –6. This region is darkly shaded in the graph below. SOLUTION: Each gold medal is worth 3 points and each silver medal is worth 2 points. So, if the U.S. won g gold medals and s silver medals, then they won 3g + 2s points. The total number of gold and silver medals, g + s, is 18. So, we have 2 equations: Solve the 2nd equation for g. Substitute for g in the other equation. 76. SPORTS In the 2006 Winter Olympic Games, the total number of gold and silver medals won by the U.S. was 18. The total points scored for gold and silver medals was 45. Write and solve a system of equations to find how many gold and silver medals were won by the U.S. g = 18 – 9 = 9 g = 9; s = 9 77. DRIVING Tires should be kept within 2 pounds per square inch (psi) of the manufacturer’s recommended tire pressure. If the recommendation for a tire is 30 psi, what is the range of acceptable pressures? SOLUTION: Each gold medal is worth 3 points and each silver medal is worth 2 points. So, if the U.S. won g gold medals and s silver medals, then they won 3g + 2s points. eSolutions Manual - Powered by Cognero The total number of gold and silver medals, g + s, is SOLUTION: 30 + 2 = 32 30 - 2 = 28 The pressure will range between 28 and 32 psi,Page 13 inclusive. 78. BABYSITTING Alexis charges $10 plus $4 per g = 18 – 9 = 9 7-1 Multiplication Properties of Exponents g = 9; s = 9 77. DRIVING Tires should be kept within 2 pounds per square inch (psi) of the manufacturer’s recommended tire pressure. If the recommendation for a tire is 30 psi, what is the range of acceptable pressures? SOLUTION: 30 + 2 = 32 30 - 2 = 28 The pressure will range between 28 and 32 psi, inclusive. 78. BABYSITTING Alexis charges $10 plus $4 per hour to babysit. Alexis needs at least $40 more to buy a television for which she is saving. Write an inequality for this situation. Will she be able to get her television if she babysits for 5 hours? SOLUTION: Let h represent the number of hours that Alexis has to babysit. If she babysits 5 hours, substitute h = 5 into the inequality. SOLUTION: The quotient of a negative number and a positive number is negative. −78 ÷ 1.3 = –60 81. 42.3 ÷ (−6) SOLUTION: The quotient of a positive number and a negative number is negative. 42.3 ÷ (−6) = –7.05 82. −23.94 ÷ 10.5 SOLUTION: The quotient of a negative number and a positive number is negative. −23.94 ÷ 10.5 = –2.28 83. −32.5 ÷ (−2.5) SOLUTION: The quotient of two negative numbers is positive. −32.5 ÷ (−2.5) = 13 84. −98.44 ÷ 4.6 SOLUTION: The quotient of a negative number and a positive number is negative. −98.44 ÷ 4.6 = –21.4 This inequality is not true, so she will not be able to afford her television if she babysits for 5 hours. Find each quotient. 79. −64 ÷ (−8) SOLUTION: The quotient of two negative integers is positive. −64 ÷ (−8) = 8 80. −78 ÷ 1.3 SOLUTION: The quotient of a negative number and a positive number is negative. −78 ÷ 1.3 = –60 81. 42.3 ÷ (−6) SOLUTION: The quotient of a positive number and a negative number is negative. 42.3 ÷ (−6) = –7.05 eSolutions Manual - Powered by Cognero 82. −23.94 ÷ 10.5 SOLUTION: Page 14 7-2 Division Properties of Exponents Simplify each expression. Assume that no denominator equals zero. 6. 1. SOLUTION: SOLUTION: 7. 2. SOLUTION: SOLUTION: 3. 8. SOLUTION: SOLUTION: 4. 9. SOLUTION: SOLUTION: 5. SOLUTION: 10. SOLUTION: 6. eSolutions Manual - Powered by Cognero SOLUTION: Page 1 SOLUTION: 7-2 Division Properties of Exponents 10. 15. SOLUTION: SOLUTION: 11. 16. SOLUTION: SOLUTION: 17. SOLUTION: 12. SOLUTION: A value to the zero power is 1. 13. SOLUTION: A value to the zero power is 1. 14. 18. FINANCIAL LITERACY The gross domestic product (GDP) for the United States in 2008 was $14.204 trillion, and the GDP per person was $47,580. Use order of magnitude to approximate the population of the United States in 2008. SOLUTION: Since there are 12 zeros in a trillion, the order of SOLUTION: 12 magnitude for 14.204 trillion is 10 . Since there are 4 zeros in a ten-thousand, the order 4 of magnitude for 45,780 is 10 . Divide the GDP for the U.S. by the GDP per person to find the approximate population of the U.S. 15. SOLUTION: The approximate population of the U.S. in 2008 was 8 eSolutions Manual - Powered by Cognero 10 or 100,000,000. Simplify each expression. Assume that no denominator equals zero. Page 2 7-2 Division Properties of Exponents 18. FINANCIAL LITERACY The gross domestic product (GDP) for the United States in 2008 was $14.204 trillion, and the GDP per person was $47,580. Use order of magnitude to approximate the population of the United States in 2008. 21. SOLUTION: SOLUTION: Since there are 12 zeros in a trillion, the order of 12 magnitude for 14.204 trillion is 10 . Since there are 4 zeros in a ten-thousand, the order 4 of magnitude for 45,780 is 10 . Divide the GDP for the U.S. by the GDP per person to find the approximate population of the U.S. 22. SOLUTION: The approximate population of the U.S. in 2008 was 8 10 or 100,000,000. Simplify each expression. Assume that no denominator equals zero. 23. 19. SOLUTION: SOLUTION: 20. SOLUTION: 24. SOLUTION: A value to the zero power is 1. 21. SOLUTION: 25. SOLUTION: eSolutions Manual - Powered by Cognero 22. Page 3 24. SOLUTION: 7-2 Division Properties of Exponents A value to the zero power is 1. 25. 29. SOLUTION: SOLUTION: 26. 30. SOLUTION: SOLUTION: 27. SOLUTION: 31. SOLUTION: A value to the zero power is 1. 28. SOLUTION: 32. SOLUTION: 33. SOLUTION: 29. eSolutions Manual - Powered by Cognero SOLUTION: Page 4 7-2 Division Properties of Exponents 33. 37. SOLUTION: SOLUTION: 38. 34. SOLUTION: SOLUTION: 39. SOLUTION: 35. SOLUTION: 40. SOLUTION: 36. SOLUTION: 41. 37. SOLUTION: eSolutions Manual - Powered by Cognero SOLUTION: Page 5 zeros in a hundred-million, the order of magnitude for 8 208 million is 10 .The numbers differ by an order of 2 magnitude of 10 , so there were 100 times as many Internet users as Internet hosts. 7-2 Division Properties of Exponents 44. PROBABILITY The probability of rolling a die 41. and getting an even number is SOLUTION: . If you roll the die twice, the probability of getting an even number twice is or . a. What does represent? b. Write an expression to represent the probability of rolling a die d times and getting an even number every time. Write the expression as a power of 2. 42. SOLUTION: a. probability of all evens on 4 rolls b. SOLUTION: ; 43. INTERNET In a recent year, there were approximately 3.95 million Internet hosts. Suppose there were 208 million Internet users. Determine the order of magnitude for the Internet hosts and Internet users. Using the orders of magnitude, how many Internet users were there compared to Internet hosts? SOLUTION: Since there are 6 zeros in a million, the order of 6 magnitude for 3.95 million is: 10 . Since there are 8 zeros in a hundred-million, the order of magnitude for Simplify each expression. Assume that no denominator equals zero. 45. SOLUTION: 8 208 million is 10 .The numbers differ by an order of 2 magnitude of 10 , so there were 100 times as many Internet users as Internet hosts. 44. PROBABILITY The probability of rolling a die and getting an even number is . If you roll the die twice, the probability of getting an even number twice is or 46. SOLUTION: . a. What does represent? b. Write an expression to represent the probability of rolling a die d times and getting an even number every time. Write the expression as a power of 2. SOLUTION: eSolutions Manual - Powered by Cognero a. probability of all evens on 4 rolls 47. Page 6 SOLUTION: 7-2 Division Properties of Exponents 50. 47. SOLUTION: SOLUTION: 51. SOLUTION: 48. SOLUTION: 52. 49. SOLUTION: SOLUTION: 50. 53. SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero Page 7 7-2 Division Properties of Exponents 53. 56. SOLUTION: 54. SOLUTION: 57. CCSS SENSE-MAKING The processing speed of 8 SOLUTION: an older computer is about 10 instructions per 10 second. BA new computer can process about 10 instructions per second. The newer computer is how many times as fast as the older one? SOLUTION: 10 8 2 10 and 10 differ by an order of magnitude of 10 , so the newer computer is 100 times faster than the older computer. 55. SOLUTION: 58. ASTRONOMY The brightness of a star is measured in magnitudes. The lower the magnitude, the brighter the star. A magnitude 9 star is 2.51 times as bright as a magnitude 10 star. A magnitude 8 star 2 is 2.51 ⋅ 2.51 or 2.51 times as bright as a magnitude 10 star. a. How many times as bright is a magnitude 3 star as a magnitude 10 star? b. Write an expression to compare a magnitude m star to a magnitude 10 star. c. A full moon is considered to be approximately magnitude –13. Does your expression make sense for this magnitude? Explain. SOLUTION: a. The levels of the magnitude of the brightness of a star are given as a multiple of 2.51 (starting with the brightness of a magnitude 10 star). So a magnitude 9 56. star is eSolutions Manual - Powered by Cognero SOLUTION: times as Page 8 bright as a magnitude 10 star. A magnitude 8 star is times as bright as a 1,557,742,231 times as bright as a magnitude 10 star. Since we know that the lower the magnitude the brighter the object, it follows that a magnitude -13 object is significantly brighter than a magnitude 10 object. The expression in part b does make sense. SOLUTION: 10 8 2 10 and 10 differ by an order of magnitude of 10 , 7-2 Division Properties so the newer computerofisExponents 100 times faster than the older computer. 58. ASTRONOMY The brightness of a star is measured in magnitudes. The lower the magnitude, the brighter the star. A magnitude 9 star is 2.51 times as bright as a magnitude 10 star. A magnitude 8 star 2 is 2.51 ⋅ 2.51 or 2.51 times as bright as a magnitude 10 star. a. How many times as bright is a magnitude 3 star as a magnitude 10 star? b. Write an expression to compare a magnitude m star to a magnitude 10 star. c. A full moon is considered to be approximately magnitude –13. Does your expression make sense for this magnitude? Explain. SOLUTION: a. The levels of the magnitude of the brightness of a star are given as a multiple of 2.51 (starting with the brightness of a magnitude 10 star). So a magnitude 9 59. PROBABILITY The probability of rolling a die and getting a 3 is . If you roll the die twice, the probability of getting a 3 both times is or . a. Write an expression to represent the probability of rolling a die d times and getting a 3 each time. b. Write the expression as a power of 6. SOLUTION: a. b. times as star is bright as a magnitude 10 star. A magnitude 8 star is 60. MULTIPLE REPRESENTATIONS To find the 2 area of a circle, use A = πr . The formula for the 2 area of a square is A = s . times as bright as a magnitude 10 star. It follows that a magnitude 3 star times as is bright as a magnitude 10 star. b. A magnitude m star would be times as bright as a magnitude 10 star. a. ALGEBRAIC Find the ratio of the area of the circle to the area of the square. b. ALGEBRAIC If the radius of the circle and the length of each side of the square is doubled, find the ratio of the area of the circle to the square. c. TABULAR Copy and complete the table. c. According to the expression, a full Moon would be or 1,557,742,231 times as bright as a magnitude 10 star. Since we know that the lower the magnitude the brighter the object, it follows that a magnitude -13 object is significantly brighter than a magnitude 10 object. The expression in part b does make sense. 59. PROBABILITY The probability of rolling a die and getting a 3 is . If you roll the die twice, the probability of getting a 3 both times is or d. ANALYTICAL What conclusion can be drawn from this? . a. Write an expression to represent the probability of rolling a die d times and getting a 3 each time. eSolutions Manual - Powered by Cognero b. Write the expression as a power of 6. SOLUTION: SOLUTION: a. Use the diagram to write an equation for s in terms of r to substitute into the equation for the area of a square. The value of s is the length of one side Page 9 of the square, which is equal to the diameter of the circle. Therefore, s = 2r. d. ANALYTICAL What conclusion can be drawn from this? SOLUTION: 7-2 Division Properties of Exponents a. Use the diagram to write an equation for s in terms of r to substitute into the equation for the area of a square. The value of s is the length of one side of the square, which is equal to the diameter of the circle. Therefore, s = 2r. d. The ratio of the area of the circle to the area of the square will always be y z . yz 61. REASONING Is x · x = x sometimes, always, or never true? Explain. SOLUTION: Sometimes; sample answer: The equation is true whenever x=1. The equation is false when x=2, y=3, and z=4. 62. OPEN ENDED Name two monomials with a 2 3 quotient of 24a b . SOLUTION: You need to find two monomials, I and II, such that the terms in monomial I divided by the corresponding 2 3 b. Replace r with 2r in the equations. s = 2 · 2r = 4r terms in monomial II equal 24a b . The easiest way to do this is to first create monomial II using all of the available terms. 2 3 Monomial II = a b . 2 3 Now, multiply this monomial by 24a b and the result will be monomial I. 2 3 2 3 4 6 a b × 24a b = 24a b Check your answer. c. 63. CHALLENGE Use the Quotient of Powers Property to explain why . SOLUTION: d. The ratio of the area of the circle to the area of the square will always be y z . yz 61. REASONING Is x · x = x sometimes, always, or never true? Explain. SOLUTION: Sometimes; sample answer: The equation is true whenever x=1. The equation is false when x=2, y=3, eSolutions Manual - Powered by Cognero and z=4. 64. CCSS REGULARITY Write a convincing 0 argument to show why 3 = 1. SOLUTION: We can use the Quotient of Powers Property to 0 Page 10 show that 3 = 1. 62. OPEN ENDED Name two monomials with a 2 3 7-2 Division Properties of Exponents 64. CCSS REGULARITY Write a convincing 66. What is the perimeter of the figure in meters? 0 argument to show why 3 = 1. SOLUTION: We can use the Quotient of Powers Property to 0 show that 3 = 1. A 40x B 80x C 160x D 400x SOLUTION: The perimeter is the distance around the figure. 65. WRITING IN MATH Explain how to use the Quotient of Powers property and the Power of a Quotient property. SOLUTION: The Quotient of Powers Property is used when dividing two powers with the same base. The exponents are subtracted. Consider the following example of the Quotient of Powers Property. The Power of a Quotient Property is used to find the power of a quotient. You find the power of the numerator and the power of the denominator. Consider the following example of the Power of a Quotient Property. 66. What is the perimeter of the figure in meters? eSolutions Manual - Powered by Cognero A 40x No values are provided for horizontal sides a, b and c. However,the sum of a, b, and c is 20x. There are also no values provided for the vertical sides d and e. However, the sum of d and e is 12x + 8x or 20x. The correct answer is B. 67. In researching her science project, Leigh learned that light travels at a constant rate and that it takes 500 seconds for light to travel the 93 million miles from the Sun to Earth. Mars is 142 million miles from the Sun. About how many seconds will it take for light to travel from the Sun to Mars? F 235 seconds G 327 seconds H 642 seconds J 763 seconds SOLUTION: Set up a proportion of seconds to millions of miles. Page 11 b. The number of possible combinations is 4 • 4, or 16. There are three possible spins that would produce a product of 4: 1, 4; 4, 1; 2, 2. 20x. 7-2 Division Properties of Exponents The correct answer is B. . The probability of the product of 4 occurring is 67. In researching her science project, Leigh learned that light travels at a constant rate and that it takes 500 seconds for light to travel the 93 million miles from the Sun to Earth. Mars is 142 million miles from the Sun. About how many seconds will it take for light to travel from the Sun to Mars? F 235 seconds G 327 seconds H 642 seconds J 763 seconds −2 69. Simplify (4 0 3 • 5 • 64) . A B 64 C 320 D 1024 SOLUTION: SOLUTION: Set up a proportion of seconds to millions of miles. The correct answer is B. The correct answer is J. 68. EXTENDED RESPONSE Jessie and Jonas are playing a game using the spinners shown. Each spinner is equally likely to stop on any of the four numbers. In the game, a player spins both spinners and calculates the product of the two numbers on which the spinners have stopped. 70. GEOMETRY A rectangular prism has a width of 3 2 7x units. a length of 4x units, and a height of 3x units. What is the volume of the prism? SOLUTION: Solve each system of inequalities by graphing. 71. y ≥ 1 x < −1 a. What product has the greatest probability of occurring? b. What is the probability of that product occurring? SOLUTION: a. The possible products are 1, 2, 3, 4, 6, 8, 9, 12, and 16. The only product that can be made by two different sets of numbers is 4 (2 • 2 and 1 • 4). So, 4 is the product with the greatest probability of occurring. b. The number of possible combinations is 4 • 4, or 16. There are three possible spins that would produce a product of 4: 1, 4; 4, 1; 2, 2. The probability of the product of 4 occurring is 69. Simplify (4 −2 0 . 3 • 5 • 64) . A B 64 C 320 D 1024 SOLUTION: Graph each inequality. The graph of y ≥ 1 is solid and is included in the graph of the solution. The graph of x < −1 is dashed and is not included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y ≥ 1 and x < −1. This region is darkly shaded in the graph below. eSolutions Manual - Powered by Cognero 72. y ≥ −3 y −x<1 SOLUTION: Rewire inequality 2 in slope-intercept form. Page 12 7-2 Division Properties of Exponents 72. y ≥ −3 y −x<1 SOLUTION: Rewire inequality 2 in slope-intercept form. Graph each inequality. The graph of y ≥ −3 is solid and is included in the graph of the solution. The graph of y − x < 1 is dashed and is not included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y ≥ −3 and y − x < 1. This region is darkly shaded in the graph below. 74. y − 2x < 2 y − 2x > 4 SOLUTION: Rewrite each inequality in slope-intercept form first. Equation 1: Equation 2: Graph each inequality. The graph of y − 2x < 2 is dashed and is not included in the graph of the solution. The graph of y − 2x > 4 is also dashed and is not included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y − 2x < 2 and y − 2x > 4. There is no shared area, so there is no solution. 73. y < 3x + 2 y ≥ −2x + 4 SOLUTION: Graph each inequality. The graph of y < 3x + 2 is dashed and is not included in the graph of the solution. The graph of y ≥ −2x + 4 is solid and is included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y < 3x + 2 and y ≥ −2x + 4. This region is darkly shaded in the graph below. 74. y − 2x < 2 y − 2x > 4 SOLUTION: Rewrite each inequality in slope-intercept form first. Equation 1: Solve each inequality. Check your solution. 75. 5(2h − 6) > 4h SOLUTION: To check, substitute a value greater than 5 for h in the inequality. eSolutions Manual - Powered by Cognero Page 13 Equation 2: The solution checks. 7-2 Division Properties of Exponents Solve each inequality. Check your solution. 75. 5(2h − 6) > 4h The solution checks. 76. 22 ≥ 4(b − 8) + 10 SOLUTION: SOLUTION: To check, substitute a value greater than 5 for h in the inequality. To check, substitute a value less than or equal to 11 for b in the inequality. The solution checks. The solution checks. 76. 22 ≥ 4(b − 8) + 10 77. 5(u – 8) ≤ 3(u + 10) SOLUTION: SOLUTION: To check, substitute a value less than or equal to 11 for b in the inequality. To check, substitute a value less than or equal to 35 for u in the inequality. The solution checks. The solution checks. eSolutions Manual - Powered by Cognero 77. 5(u – 8) ≤ 3(u + 10) SOLUTION: 78. 8 + t ≤ 3(t + 4) + 2 SOLUTION: Page 14 7-2 Division Properties of Exponents The solution checks. The solution checks. 78. 8 + t ≤ 3(t + 4) + 2 80. −6(b + 5) > 3(b − 5) SOLUTION: SOLUTION: To check, substitute a value greater than or equal to –3 for t in the inequality. To check, substitute a value less than for b in the inequality. The solution checks. The solution checks. 79. 9n + 3(1 − 6n) ≤ 21 81. GRADES In a high school science class, a test is worth three times as much as a quiz. What is the student’s average grade? SOLUTION: To check, substitute a value greater than or equal to –2 for n in the inequality. SOLUTION: Since the tests are worth three times as much as a quiz, each test is like 3 quizzes. So, in order to find the average grade, we have to count each test 3 times. That makes a total of nine grades. Find the average of the 9 grades. The solution checks. 80. −6(b + 5) > 3(b − 5) SOLUTION: eSolutions Manual - Powered by Cognero The student’s average grade is 87. Evaluate each expression. 2 82. 9 SOLUTION: 2 9 = 9 • 9 = 81 Page 15 7-2 Division Properties of Exponents The student’s average grade is 87. Evaluate each expression. 2 82. 9 SOLUTION: 2 9 = 9 • 9 = 81 2 83. 11 SOLUTION: 2 11 = 11 • 11 = 121 6 84. 10 SOLUTION: 6 10 = 10 • 10 • 10 • 10 • 10 • 10 = 1,000,000 4 85. 10 SOLUTION: 4 10 = 10 • 10 • 10 • 10 = 10,000 5 86. 3 SOLUTION: 5 3 = 3 • 3 • 3 • 3 • 3 = 243 3 87. 5 SOLUTION: 3 5 = 5 • 5 • 5 • = 125 3 88. 12 SOLUTION: 3 12 = 12 • 12 • 12 • = 1728 6 89. 4 SOLUTION: 6 4 = 4 • 4 • 4 • 4 • 4 • 4 = 4096 eSolutions Manual - Powered by Cognero Page 16 SOLUTION: 7-3 Rational Exponents Write each expression in radical form, or write each radical in exponential form. 7. SOLUTION: 1. SOLUTION: 2. SOLUTION: 8. SOLUTION: 3. SOLUTION: 4. 9. SOLUTION: SOLUTION: Simplify. 5. SOLUTION: 10. SOLUTION: 6. SOLUTION: 11. 7. SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero 8. Page 1 12. Therefore, the solution is 1. 7-3 Rational Exponents 15. 11. SOLUTION: SOLUTION: Therefore, the solution is 5.5. 12. SOLUTION: 16. CCSS TOOLS A weir is used to measure water flow in a channel. (Refer to the photo on page 410) For a rectangular broad crested weir, the flow Q in cubic feet per second is related to the weir length L in feet and height H of the water by . Find the water height for a weir that is 3 feet long and has flow of 38.4 cubic feet per second. Solve each equation. SOLUTION: 13. SOLUTION: Therefore, the solution is 4. 14. SOLUTION: Therefore, the water height for the weir is 4 feet. Write each expression in radical form, or write each radical in exponential form. 17. Therefore, the solution is 1. SOLUTION: 15. SOLUTION: 18. SOLUTION: eSolutions Manual - Powered by Cognero Page 2 17. SOLUTION: SOLUTION: 7-3 Rational Exponents 18. 26. SOLUTION: SOLUTION: 19. SOLUTION: 27. SOLUTION: 20. SOLUTION: 28. SOLUTION: 21. SOLUTION: 29. SOLUTION: 22. SOLUTION: 30. 23. SOLUTION: SOLUTION: 24. SOLUTION: 31. SOLUTION: Simplify. 25. SOLUTION: 32. SOLUTION: 26. eSolutions Manual - Powered by Cognero SOLUTION: Page 3 7-3 Rational Exponents 32. 37. SOLUTION: SOLUTION: 33. SOLUTION: 38. SOLUTION: 34. SOLUTION: 39. SOLUTION: 35. SOLUTION: 40. SOLUTION: 36. 41. SOLUTION: SOLUTION: 37. SOLUTION: eSolutions Manual - Powered by Cognero 42. Page 4 SOLUTION: Therefore, the solution is 5. 7-3 Rational Exponents 46. 42. SOLUTION: SOLUTION: Therefore, the solution is 2. 47. SOLUTION: 43. SOLUTION: Therefore, the solution is . 44. SOLUTION: 48. SOLUTION: Solve each equation. 45. Therefore, the solution is . SOLUTION: 49. SOLUTION: Therefore, the solution is 5. 46. SOLUTION: Therefore, the solution is . Therefore, the solution is 2. eSolutions Manual - Powered by Cognero 50. Page 5 SOLUTION: 47. 7-3 Rational Exponents Therefore, the solution is . 50. Therefore, the solution is 8. 54. SOLUTION: SOLUTION: Therefore, the solution is . Therefore, the solution is . 51. 55. SOLUTION: SOLUTION: Therefore, the solution is 8. Therefore, the solution is . 52. SOLUTION: 56. SOLUTION: Therefore, the solution is 2. 53. SOLUTION: Therefore, the solution is . 57. CONSERVATION Water collected in a rain barrel can be used to water plants and reduce city water use. Water flowing from an open rain barrel has Therefore, the solution is 8. 54. SOLUTION: eSolutions Manual - Powered by Cognero velocity , where v is in feet per second and h is the height of the water in feet. Find the height of the water if it is flowing at 16 feet per second. SOLUTION: Page 6 59. SOLUTION: 7-3 Rational Exponents Therefore, the solution is . 57. CONSERVATION Water collected in a rain barrel can be used to water plants and reduce city water use. Water flowing from an open rain barrel has 60. SOLUTION: velocity , where v is in feet per second and h is the height of the water in feet. Find the height of the water if it is flowing at 16 feet per second. 61. SOLUTION: SOLUTION: 62. SOLUTION: The height of the water is 4 feet. 58. ELECTRICITY The radius r in millimeters of a platinum wire L centimeters long with resistance 0.1 ohm is . 63. SOLUTION: How long is a wire with radius 0.236 millimeter? SOLUTION: 64. SOLUTION: So, the wire has a length of 16 centimeters. 65. SOLUTION: Write each expression in radical form, or write each radical in exponential form. 59. 66. SOLUTION: SOLUTION: 60. Simplify. SOLUTION: eSolutions Manual - Powered by Cognero 67. Page 7 SOLUTION: SOLUTION: 7-3 Rational Exponents Simplify. 67. 72. SOLUTION: SOLUTION: 73. 68. SOLUTION: SOLUTION: 69. SOLUTION: 74. SOLUTION: 70. SOLUTION: 75. SOLUTION: 71. SOLUTION: 76. SOLUTION: eSolutions Manual - Powered by Cognero 72. Page 8 Therefore, the solution is 12. 7-3 Rational Exponents 80. SOLUTION: 76. SOLUTION: Therefore, the solution is 3. 81. SOLUTION: 77. SOLUTION: Therefore, the solution is –5. 78. SOLUTION: 82. SOLUTION: Solve each equation. 79. SOLUTION: Therefore, the solution is . 83. Therefore, the solution is 12. SOLUTION: 80. SOLUTION: eSolutions Manual - Powered by Cognero Therefore, the solution is Page 9 . 7-3 Rational Exponents Therefore, the solution is Therefore, the solution is 11. . 85. CCSS MODELING The frequency f in hertz of 83. the nth key on a piano is SOLUTION: . a. What is the frequency of Concert A? b. Which note has a frequency of 220 Hz? Therefore, the solution is SOLUTION: a. Replace n with 49 in the equation to determine the frequency of Concert A. . 84. SOLUTION: So, the frequency of Concert A is 440 hertz. b. Replace f with 220 in the equation to determine the value of n. Therefore, the solution is 11. 85. CCSS MODELING The frequency f in hertz of the nth key on a piano is . a. What is the frequency of Concert A? b. Which note has a frequency of 220 Hz? SOLUTION: a. Replace n with 49 in the equation to determine the frequency of Concert A. So, the frequency of Concert A is 440 hertz. b. Replace f with 220 in the equation to determine the value of n. eSolutions Manual - Powered by Cognero So, the note with a frequency of 220 corresponds to the 37th note on the piano or the A below middle C. 86. RANDOM WALKS Suppose you go on a walk where you choose the direction of each step at random. The path of a molecule in a liquid or a gas, the path of a foraging animal, and a fluctuating stock price are all modeled as random walks. The number of possible random walks w of n steps where you choose one of d directions at each step is . a. How many steps have been taken in a 2-direction random walk if there are 4096 possible walks? Page 10 b. How many steps have been taken in a 4-direction random walk if there are 65,536 possible walks? c. If a walk of 7 steps has 2187 possible walks, how So, the note with a frequency of 220 corresponds to 7-3 Rational Exponents the 37th note on the piano or the A below middle C. 86. RANDOM WALKS Suppose you go on a walk where you choose the direction of each step at random. The path of a molecule in a liquid or a gas, the path of a foraging animal, and a fluctuating stock price are all modeled as random walks. The number of possible random walks w of n steps where you choose one of d directions at each step is . a. How many steps have been taken in a 2-direction random walk if there are 4096 possible walks? b. How many steps have been taken in a 4-direction random walk if there are 65,536 possible walks? c. If a walk of 7 steps has 2187 possible walks, how many directions could be taken at each step? So, in 2187 possible walks of 7 steps there could be 3 directions taken at each step. 87. SOCCER The radius r of a ball that holds V cubic units of air is modeled by . What are the possible volumes of each size soccer ball? SOLUTION: Size 3: d = 7.3 in.. r = 3.65 SOLUTION: a. Replace w with 4096 and d with 2 in the equation to determine the value of n. So, in a 2-direction random walk having 4096 possible walks there have been 12 steps taken. b. Replace w with 65,536 and d with 4 in the equation to determine the value of n. So, in a 4-direction random walk having 65,536 possible walks there have been 8 steps taken. c. Replace n with 7 and w with 2187 in the equation to determine the value of d. So, in 2187 possible walks of 7 steps there could be 3 directions taken at each step. 87. SOCCER The radius r of a ball that holds V cubic units of air is modeled by eSolutions Manual - Powered by Cognero Size 3: d= 7.6 in., r = 3.8 Size 4: d = 8.0 in.. r = 4.0 Size 4: d= 8.3 in., r = 4.15 . Page 11 7-3 Rational Exponents Volume of size 3 soccer balls vary from 204.0 to 230.2 cubic inches. Volume of size 4 soccer balls vary from 268.5 to 299.9 cubic inches. Volume of size 5 soccer balls vary from 333.6 to 382.4 cubic inches. Size 4: d= 8.3 in., r = 4.15 88. MULTIPLE REPRESENTATIONS In this problem, you will explore the graph of an exponential function. a. TABULAR Copy and complete the table below. Size 5: d = 8.6 in.. r = 4.3 b. GRAPHICAL Graph f (x) by plotting the points and connecting them with a smooth curve. c. VERBAL Describe the shape of the graph of f (x). What are its key features? Is it linear? SOLUTION: a. b. Size 5: d= 9.0 in., r = 4.5 x Volume of size 3 soccer balls vary from 204.0 to 230.2 cubic inches. Volume of size 4 soccer balls vary from 268.5 to 299.9 cubic inches. Volume of size 5 soccer balls vary from 333.6 to 382.4 cubic inches. 88. MULTIPLE REPRESENTATIONS In this problem, you will explore the graph of an exponential function. a. TABULAR Copy and complete the table below. b. GRAPHICAL Graph f (x) by plotting the points and connecting them with a smooth curve. c. VERBAL Describe the shape of the graph of f eSolutions Manual - Powered by Cognero (x). What are its key features? Is it linear? SOLUTION: c. The graph of f (x) = 4 is a curve. It has no xintercept, a y-intercept of 1, the domain is all real numbers, the range is all positive real numbers, it is increasing over the entire domain, as x approaches infinity f (x) approaches infinity, as x approaches negative infinity f (x) approaches 0. The graph is not linear. 89. OPEN ENDED Write two different expressions with rational exponents equal to . SOLUTION: Sample answer: First: Second: Page 12 numbers, the range is all positive real numbers, it is increasing over the entire domain, as x approaches infinity f (x) approaches infinity, as x approaches negative Exponents infinity f (x) approaches 0. The graph is not 7-3 Rational linear. 89. OPEN ENDED Write two different expressions with rational exponents equal to . SOLUTION: Sample answer: First: Thus can be represented by and . 90. CCSS ARGUMENTS Determine whether each statement is always, sometimes, or never true. Assume that x is a nonnegative real number. Explain your reasoning. a. b. c. Second: d. e. f. SOLUTION: Thus can be represented by and . 90. CCSS ARGUMENTS Determine whether each statement is always, sometimes, or never true. Assume that x is a nonnegative real number. Explain your reasoning. a. Sometimes = 1. is true. It is only true when x b. Sometimes x = 1. is true. It is only true when c. Sometimes x = 1. d. is true. It is only true when is always true by definition of . a. b. e. c. x or x. d. f. Sometimes when x = 1. is always true since = = 1 is true. It is only true e. 91. CHALLENGE For what values of x is f. SOLUTION: SOLUTION: a. Sometimes = 1. is true. It is only true when x b. Sometimes x = 1. is true. It is only true when c. Sometimes x = 1. ? is true. It is only true when eSolutions by Cognero d. Manual - Powered is always true by definition of . Thus, the expressions will be equal for x = –1, 0, 1. Page 13 92. ERROR ANALYSIS Anna and Jamal are solving . Is either of them correct? Explain your reasoning. 7-3 Rational Exponents Thus, the expressions will be equal for x = –1, 0, 1. 92. ERROR ANALYSIS Anna and Jamal are solving . Is either of them correct? Explain your reasoning. Therefore, the correct choice is D. 95. At a movie theater, the costs for various numbers of popcorn and hot dogs are shown. Which pair of equations can be used to find p , the cost of a box of popcorn, and h, the cost of a hot dog? F SOLUTION: G Anna is correct. Jamal did not write the expressions with equal bases before applying the Power Property of Equality. 93. WRITING IN MATH Explain why 2 is the principal fourth root of 16. SOLUTION: Sample answer: 2 is the principal fourth root of 16 4 because 2 is positive and 2 = 16. 94. What is the value of A5 B 11 C 25 D 35 ? H J SOLUTION: Let p be the cost of a box of popcorn and h the cost of a hot dog. Then p + h = 8.50 and 2p + 4h = 21.6. The correct choice is G. 96. SHORT RESPONSE Find the dimensions of the rectangle if its perimeter is 52 inches. SOLUTION: Therefore, the correct choice is D. SOLUTION: 95. At a movie theater, the costs for various numbers of popcorn and hot dogs are shown. Which pair of equations can be used to find p , the cost of a box of popcorn, and h, the cost of a hot eSolutions Manual - Powered by Cognero dog? F Therefore, the dimensions of the rectangle are 17.5 Page 14 inches by 8.5 inches. 4 x 97. If 3 = 9 , then x = SOLUTION: Let p be the cost of a box of popcorn and h the cost of a hot dog. Then p + h = 8.50 and 2p + 4h = 21.6. The correct choice is G. 7-3 Rational Exponents So, the correct choice is B. 96. SHORT RESPONSE Find the dimensions of the rectangle if its perimeter is 52 inches. Simplify each expression. Assume that no denominator equals zero. 98. SOLUTION: SOLUTION: 99. SOLUTION: Therefore, the dimensions of the rectangle are 17.5 inches by 8.5 inches. 4 100. x SOLUTION: 97. If 3 = 9 , then x = A1 B2 C4 D5 SOLUTION: 101. SOLUTION: So, the correct choice is B. 102. SOLUTION: Simplify each expression. Assume that no denominator equals zero. 98. SOLUTION: eSolutions Manual - Powered by Cognero Page 15 103. So, in slope-intercept form the equation is y = 3x – 1. 7-3 Rational Exponents 106. SOLUTION: 103. SOLUTION: 104. GARDENING Felipe is planting a flower garden that is shaped like a trapezoid as shown. 3 Use the formula to find the area of the garden. So, in slope-intercept form the equation is y = 6x + 11. 107. SOLUTION: So, in slope-intercept form the equation is y = –2x – 12. SOLUTION: Replace h with 3a, b 1 with 6a, and b 2 with 4a in 108. the formula to determine the value of A. 2 Therefore, the area of the garden is 15a square units. SOLUTION: So, in slope-intercept form the equation is . 109. SOLUTION: Write each equation in slope-intercept form. 105. SOLUTION: So, in slope-intercept form the equation is . So, in slope-intercept form the equation is y = 3x – 1. 110. 106. SOLUTION: So, in slope-intercept form the equation is y = 6x + 11. eSolutions Manual - Powered by Cognero SOLUTION: So, in slope-intercept form the equation is . Page 16 So, in slope-intercept form the equation is 7-3 Rational Exponents . 110. SOLUTION: So, in slope-intercept form the equation is . Find each power. 3 111. 10 SOLUTION: 5 112. 10 SOLUTION: –1 113. 10 SOLUTION: 114. SOLUTION: eSolutions Manual - Powered by Cognero Page 17 13 billion = 13,000,000,000 → 1.3 The decimal point moved 10 places to the left, so n = 10. 7-4 Scientific Notation 13,000,000,000 = 1.3 × 10 Express each number in scientific notation. 1. 185,000,000 SOLUTION: 185,000,000 → 1.85000000 The decimal point moved 8 places to the left, so n = 8. 185,000,000 = 1.85 × 10 8 2. 1,902,500,000 6. Teens have an influence on their families’ spending habit. They control about $1.5 billion of discretionary income. SOLUTION: 1.5 billion = 1,500,000,000 → 1.5 The decimal point moved 9 places to the left, so n = 9. 9 1,500,000,000 = 1.5 × 10 Express each number in standard form. SOLUTION: 1,902,500,000 → 1.9025000000 The decimal point moved 9 places to the left, so n = 9. 9 1,902,500,000 = 1.9025 × 10 7. 1.98 × 10 7 SOLUTION: 7 1.98 × 10 has an exponent of 7, so n = 7. Move the decimal point 7 places to the right. 7 1.98 × 10 = 19,800,000 3. 0.000564 SOLUTION: 0.000564 → 5.64 The decimal point moved 4 places to the right, so n = –4. 0.000564 = 5.64 × 10 –4 4. 0.00000804 SOLUTION: 0.00000804 → 8.04 The decimal point moved 6 places to the right, so n = –6. –6 0.00000804 = 8.04 × 10 8. 4.052 × 10 MONEY Express each number in scientific notation. 5. Teens spend $13 billion annually on clothing. SOLUTION: 13 billion = 13,000,000,000 → 1.3 The decimal point moved 10 places to the left, so n = 10. 13,000,000,000 = 1.3 × 10 6 SOLUTION: 6 4.052 × 10 has an exponent of 6, so n = 6. Move the decimal point 6 places to the right. 6 4.052 × 10 = 4,052,000 9. 3.405 × 10 −8 SOLUTION: –8 3.405 × 10 has an exponent of −8, so n = –8. Move the decimal point 8 places to the left. 3.405 × 10 –8 = 0.00000003405 −5 10. 6.8 × 10 SOLUTION: –5 6.8 × 10 has an exponent of −5, so n = –5. Move the decimal point 5 places to the left. –5 6.8 × 10 = 0.000068 10 6. Teens have an influence on their families’ spending habit. They control about $1.5 billion of discretionary income. SOLUTION: 1.5 billion = 1,500,000,000 → 1.5 The decimal point moved 9 places to the left, so n = 9. 9 1,500,000,000 = 1.5 × 10 Evaluate each product. Express the results in both scientific notation and standard form. 3 12 11. (1.2 × 10 )(1.45 × 10 ) SOLUTION: 1,740,000,000,000,000 14 Express each number in standard form. 7 −9 12. (7.08 × 10 )(5 × 10 ) eSolutions Manual - Powered by Cognero 7. 1.98 × 10 10 Page 1 SOLUTION: SOLUTION: 7-4 Scientific Notation 1,740,000,000,000,000 14 620,000,000,000,000 −9 12. (7.08 × 10 )(5 × 10 ) 17. SOLUTION: SOLUTION: 3,540,000 2 −5 13. (5.18 × 10 )(9.1 × 10 ) 0.00000000000085 SOLUTION: 18. SOLUTION: 0.047138 −2 2 14. (2.18 × 10 ) SOLUTION: 0.0000005125 0.0047524 Evaluate each quotient. Express the results in both scientific notation and standard form. 15. SOLUTION: 4,500 16. 19. CCSS PRECISION Salvador bought an air purifier to help him deal with his allergies. The filter in the purifier will stop particles as small as one hundredth of a micron. A micron is one millionth of a millimeter. a. Write one hundredth and one micron in standard form. b. Write one hundredth and one micron in scientific notation. c. What is the smallest size particle in meters that the filter will stop? Write the result in both standard form and scientific notation. SOLUTION: a. 0.01, 0.000001 b. 0.01 → 1.0 The decimal point moved 2 places to the right, so n = −2. −2 SOLUTION: 0.01 = 1 × 10 One micron is one millionth of a millimeter. 0.000001 → 1.0 The decimal point moved 6 places to the left, so n = −6. −6 0.000001 = 1 × 10 c. The filter in the purifier will stop particles as small −2 620,000,000,000,000 17. eSolutions Manual - Powered by Cognero SOLUTION: as one hundredth of a micron or 1 × 10 times the size of a micron. A micron is one millionth of a −6 millimeter or 1 × 10 times the size of a millimeter. Multiply to find the smallest size particle in Page 2 millimeters that the filter will stop. So, the smallest size particle in meters that the filter will stop is or 0.00000000001 m. 7-4 Scientific Notation 0.0000005125 19. CCSS PRECISION Salvador bought an air purifier to help him deal with his allergies. The filter in the purifier will stop particles as small as one hundredth of a micron. A micron is one millionth of a millimeter. a. Write one hundredth and one micron in standard form. b. Write one hundredth and one micron in scientific notation. c. What is the smallest size particle in meters that the filter will stop? Write the result in both standard form and scientific notation. SOLUTION: a. 0.01, 0.000001 b. 0.01 → 1.0 The decimal point moved 2 places to the right, so n = −2. −2 0.01 = 1 × 10 One micron is one millionth of a millimeter. 0.000001 → 1.0 The decimal point moved 6 places to the left, so n = −6. −6 0.000001 = 1 × 10 c. The filter in the purifier will stop particles as small −2 as one hundredth of a micron or 1 × 10 times the size of a micron. A micron is one millionth of a −6 millimeter or 1 × 10 times the size of a millimeter. Multiply to find the smallest size particle in millimeters that the filter will stop. Express each number in scientific notation. 20. 1,220,000 SOLUTION: 1,220,000 → 1.220000 The decimal point moved 6 places to the left, so n = 6. 6 1,220,000 = 1.22 × 10 21. 58,600,000 SOLUTION: 58,600,000 → 5.8600000 The decimal point moved 7 places to the left, so n = 7. 7 58,600,000 = 5.86 × 10 22. 1,405,000,000,000 SOLUTION: 1,405,000,000,000 → 1.405000000000 The decimal point moved 12 places to the left, so n = 12. 12 1,405,000,000,000 = 1.405 × 10 23. 0.0000013 SOLUTION: 0.0000013 → 1.3 The decimal point moved 6 places to the right, so n = –6. –6 0.0000013 = 1.3 × 10 24. 0.000056 Since the question asked for the smallest size particle in meters that the filter will stop, and 1,000 millimeters = 1 meter, use dimensional analysis to write in meters. SOLUTION: 0.000056 → 5.6 The decimal point moved 5 places to the right, so n = –5. 0.000056 = 5.6 × 10 –5 25. 0.000000000709 So, the smallest size particle in meters that the filter will stop is or 0.00000000001 m. Express each number in scientific notation. 20. 1,220,000 SOLUTION: 1,220,000 → 1.220000 The decimal point moved 6 places to the left, so n = 6. SOLUTION: 0.000000000709 → 7.09 The decimal point moved 10 places to the right, so n = –10. –10 0.000000000709 = 7.09 × 10 EMAIL Express each number in scientific notation. 26. Approximately 100 million emails sent to the President are put into the National Archives. eSolutions Manual - Powered by Cognero Page 3 SOLUTION: 100 million = 100,000,000 → 1.0 SOLUTION: 0.000000000709 → 7.09 The decimal point moved 10 places to the right, so n = –10. Notation 7-4 Scientific –10 0.000000000709 = 7.09 × 10 EMAIL Express each number in scientific notation. 26. Approximately 100 million emails sent to the President are put into the National Archives. SOLUTION: 100 million = 100,000,000 → 1.0 The decimal point moved 7 places to the left, so n = 8. 100,000,000 = 1.0 × 10 SOLUTION: –4 5 × 10 has an exponent of −4, so n = –4. Move the decimal point 4 places to the left. –4 5 × 10 = 0.0005 11 32. 8.73 × 10 SOLUTION: 11 8.73 × 10 has an exponent of 11, so n = 11. Move the decimal point 11 places to the right. 11 8.73 × 10 = 873,000,000,000 −6 33. 6.22 × 10 8 SOLUTION: –6 6.22 × 10 has an exponent of −6, so n = –6. Move the decimal point 6 places to the left. –6 6.22 × 10 = 0.00000622 27. By 2010, the email security market will generate $5.5 billion. SOLUTION: 5.5 billion = 5,500,000,000 → 5.5 The decimal point moved 9 places to the left, so n = 9. 9 5,500,000,000 = 5.5 × 10 INTERNET Express each number in standard form. 7 34. About 2.1 × 10 people, aged 12 to 17, use the Internet. SOLUTION: Express each number in standard form. 7 12 2.1 × 10 has an exponent of 7, so n = 7. Move the decimal point 7 places to the right. 28. 1 × 10 SOLUTION: 7 2.1 × 10 = 21,000,000 12 1 × 10 has an exponent of 12, so n = 12. Move the decimal point 12 places to the right. 12 1 × 10 = 1,000,000,000,000 7 35. Approximately 1.1 × 10 teens go online daily. SOLUTION: 7 7 1.1 × 10 has an exponent of 7, so n = 7. Move the decimal point 7 places to the right. 7 1.1 × 10 → 11,000,000 29. 9.4 × 10 SOLUTION: 7 9.4 × 10 has an exponent of 7, so n = 7. Move the decimal point 7 places to the right. 7 9.4 × 10 = 94,000,000 Evaluate each product or quotient. Express the results in both scientific notation and standard form. 3 −3 2 36. (3.807 × 10 )(5 × 10 ) 30. 8.1 × 10 SOLUTION: SOLUTION: –3 8.1 × 10 has an exponent of −3, so n = –3. Move the decimal point 3 places to the left. –3 8.1 × 10 = 0.0081 1,903,500 −4 31. 5 × 10 SOLUTION: –4 5 × 10 has an exponent of −4, so n = –4. Move the decimal point 4 places to the left. –4 5 × 10 = 0.0005 37. SOLUTION: 11 32. 8.73 × 10 eSolutions Manual - Powered by Cognero SOLUTION: 8.73 × 10 11 has an exponent of 11, so n = 11. Page 4 80,000,000 7-4 Scientific Notation 1,903,500 22,000,000 37. 42. SOLUTION: SOLUTION: 80,000,000 0.00015 6 2 43. (1.4 × 10 ) 38. SOLUTION: SOLUTION: 1,960,000,000,000 240,000,000 2 6 44. (2.58 × 10 )(3.6 × 10 ) −2 7 SOLUTION: 39. (6.5 × 10 )(7.2 × 10 ) SOLUTION: 928,800,000 4,680,000 45. −18 40. (9.5 × 10 9 )(9 × 10 ) SOLUTION: SOLUTION: 0.0000000855 41. 689,000 46. SOLUTION: SOLUTION: 22,000,000 4,700,000,000 42. 3 SOLUTION: –7 47. (5 × 10 )(1.8 × 10 ) SOLUTION: eSolutions Manual - Powered by Cognero 0.00015 Page 5 0.0009 7-4 Scientific Notation 4,700,000,000 5,184,000,000,000 –7 3 47. (5 × 10 )(1.8 × 10 ) 52. SOLUTION: SOLUTION: 0.0009 −3 2 48. (2.3 × 10 ) SOLUTION: 43,000,000,000 −5 2 53. (6.3 × 10 ) SOLUTION: 0.00000529 49. SOLUTION: 0.000005 50. SOLUTION: 0.000000003969 54. ASTRONOMY The distance between Earth and the Sun varies throughout the year. Earth is closest to the Sun in January when the distance is 91.4 million miles. In July, the distance is greatest at 94.4 million miles. a. Write 91.4 million in both standard form and in scientific notation. b. Write 94.4 million in both standard form and in scientific notation. c. What is the percent increase in distance from January to July? Round to the nearest tenth of a percent. SOLUTION: SOLUTION: a. 91.4 million = 91,400,000 → 9.14 The decimal point moved 7 places to the left, so n = 7. 7 91,400,000 = 9.14 × 10 b. 94.4 million = 94,400,000 → 9.44 The decimal point moved 7 places to the left, so n = 7. 5,184,000,000,000 94,400,000 = 9.44 × 10 c. Percent increase is the amount of increase divided by the original quantity. 0.000025 7 2 51. (7.2 × 10 ) 7 52. SOLUTION: So, the percent increase is about 3.3%. eSolutions Manual - Powered by Cognero 43,000,000,000 −5 2 53. (6.3 × 10 ) Evaluate each product or quotient. ExpressPage the 6 results in both scientific notation and standard form. SOLUTION: 7-4 Scientific Notation So, the percent increase is about 3.3%. Evaluate each product or quotient. Express the results in both scientific notation and standard form. −2 6 55. (4.65 × 10 )(5 × 10 ) 0.0000017889 60. SOLUTION: SOLUTION: 232,500 0.0000000072 6 −4 61. (9.04 × 10 )(5.2 × 10 ) 56. SOLUTION: SOLUTION: 4700.8 62. 9,100,000 SOLUTION: 57. SOLUTION: 1.2 LIGHT The speed of light is approximately 3 × 108 meters per second. 63. Write an expression to represent the speed of light in kilometers per second. 0.000000061 −2 2 58. (3.16 × 10 ) SOLUTION: SOLUTION: 0.00099856 −4 −3 59. (2.01 × 10 )(8.9 × 10 ) SOLUTION: 64. Write an expression to represent the speed of light in kilometers per hour. SOLUTION: 0.0000017889 eSolutions Manual - Powered by Cognero 60. 65. Make a table to show how many kilometers lightPage 7 travels in a day, a week, a 30-day month, and a 365day year. Express your results in scientific notation. SOLUTION: 7-4 Scientific Notation 65. Make a table to show how many kilometers light travels in a day, a week, a 30-day month, and a 365day year. Express your results in scientific notation. SOLUTION: For 1 day: 66. CCSS MODELING A recent cell phone study showed that company A’s phone processes up to 5 7.95 × 10 bits of data every second. Company B’s 6 phone processes up to 1.41 × 10 bits of data every second. Evaluate and interpret . . SOLUTION: For 1 week: For 1 month: For 1 year: The phone from company B is about 1.774 times as fast as the phone from company A. 67. EARTH The population of Earth is about 6.623 × 9 8 10 . The land surface of Earth is 1.483 × 10 square kilometers. What is the population density for the land surface area of Earth? 66. CCSS MODELING A recent cell phone study showed that company A’s phone processes up to SOLUTION: The population density is the population divided by the surface area. 5 7.95 × 10 bits of data every second. Company B’s 6 phone processes up to 1.41 × 10 bits of data every second. Evaluate and interpret . . So, the population density is about 44.7 persons per square kilometer. SOLUTION: 68. RIVERS A drainage basin separated from adjacent basins by a ridge, hill, or mountain, is known as a watershed. The watershed of the Amazon River is 2,300,000 square miles. The watershed of the Mississippi River is 1,200,000 square miles. a. Write each of these numbers in scientific notation. b. How many times as large is the Amazon River watershed as the Mississippi River watershed? eSolutions Manual - Powered by Cognero The phone from company B is about 1.774 times as fast as the phone from company A. SOLUTION: a. 2,300,000 → 2.300000 The decimal point moved 6 places to the left, so Page n= 8 6. 6 2,300,000 = 2.3 × 10 So, the population 7-4 Scientific Notationdensity is about 44.7 persons per square kilometer. 68. RIVERS A drainage basin separated from adjacent basins by a ridge, hill, or mountain, is known as a watershed. The watershed of the Amazon River is 2,300,000 square miles. The watershed of the Mississippi River is 1,200,000 square miles. a. Write each of these numbers in scientific notation. b. How many times as large is the Amazon River watershed as the Mississippi River watershed? SOLUTION: a. 2,300,000 → 2.300000 The decimal point moved 6 places to the left, so n = 6. 6 2,300,000 = 2.3 × 10 1,200,000 → 1.200000 The decimal point moved 6 places to the left, so n = 6. 6 1,200,000 = 1.2 × 10 b. Divide the area of Amazon River watershed by the area of the Mississippi River watershed. So, the Amazon River watershed is about 1.9 times as large as the Mississippi River watershed. 69. AGRICULTURE In a recent year, farmers planted approximately 92.9 million acres of corn. They also planted 64.1 million acres of soybeans and 11.1 million acres of cotton. a. Write each of these numbers in scientific notation and in standard form. b. How many times as much corn was planted as soybeans? Write your results in standard form and in scientific notation. Round your answer to four decimal places. c. How many times as much corn was planted as cotton? Write your results in standard form and in scientific notation. Round your answer to four decimal places. SOLUTION: a. Corn: 92.9 million = 92,900,000 → 9.2900000 The decimal point moved 7 places, so n = 7. 7 92,900,000 = 9.29 × 10 Soybeans: 64.1 million = 64,100,000 → 6.4100000 The decimal point moved 7 places, so n = 7. 7 So, the Amazon River watershed is about 1.9 times as large as the Mississippi River watershed. 69. AGRICULTURE In a recent year, farmers planted approximately 92.9 million acres of corn. They also planted 64.1 million acres of soybeans and 11.1 million acres of cotton. a. Write each of these numbers in scientific notation and in standard form. b. How many times as much corn was planted as soybeans? Write your results in standard form and in scientific notation. Round your answer to four decimal places. c. How many times as much corn was planted as cotton? Write your results in standard form and in scientific notation. Round your answer to four decimal places. SOLUTION: a. Corn: 92.9 million = 92,900,000 → 9.2900000 The decimal point moved 7 places, so n = 7. 7 92,900,000 = 9.29 × 10 Soybeans: 64.1 million = 64,100,000 → 6.4100000 eSolutions Manual - Powered by Cognero The decimal point moved 7 places, so n = 7. 64,100,000 = 6.41 × 10 7 64,100,000 = 6.41 × 10 Cotton: 11.1 million = 11,100,000 → 1.1100000 The decimal point moved 7 places, so n = 7. 7 11,100,000 = 1.11 × 10 b. Divide the amount of corn planted by the amount of soybeans planted. So, there was about 1.4493 times more corn planted than soybeans. c. Divide the amount of corn planted by the amount of cotton planted. So, there was about 8.3694 times more corn planted than cotton. 10 70. REASONING Which is greater, 100 Explain your reasoning. 100 or 10 ? Page 9 SOLUTION: 10 100 2 10 = (10 ) 20 or 10 100 and 10 20 > 10 , so 10 100 > So, there was about 8.3694 times more corn planted 7-4 Scientific Notation than cotton. 70. REASONING Which is greater, 100 Explain your reasoning. 10 100 or 10 72. CHALLENGE Order these numbers from least to greatest without converting them to standard form. 3 4 −3 −4 5.46 × 10 , 6.54 × 10 , 4.56 × 10 , −5.64 × 10 , ? SOLUTION: 10 100 2 10 = (10 ) or 10 20 100 and 10 20 > 10 , so 10 100 In step 4, Syreeta moved the decimal point in the wrong direction. > 10 100 71. ERROR ANALYSIS Syreeta and Pete are solving a division problem with scientific notation. Is either of them correct? Explain your reasoning. −4.65 × 10 SOLUTION: The numbers are all in standard form and all of the exponents are different, so we can compare the numbers by analyzing the exponents. The least number is the negative number with the greatest exponent of 10. The other negative number would come next. To order the positive numbers, order their exponents from least to greatest. 5 4 −4 −3 −4.65 × 10 , −5.64 × 10 , 4.56 × 10 , 5.46 × 10 , 6.54 × 10 SOLUTION: Pete is correct. 5 3 73. CCSS ARGUMENTS Determine whether the statement is always, sometimes, or never true. Give examples or a counterexample to verify your reasoning. When multiplying two numbers written in scientific notation, the resulting number can have no more than two digits to the left of the decimal point. SOLUTION: m Sample answer: Always; if the numbers are a × 10 n and b × 10 in scientific notation, then 1 ≤ a < 10 and 1 ≤ b < 10. So 1 ≤ ab < 100. In step 4, Syreeta moved the decimal point in the wrong direction. 72. CHALLENGE Order these numbers from least to greatest without converting them to standard form. 3 4 −3 −4 5.46 × 10 , 6.54 × 10 , 4.56 × 10 , −5.64 × 10 , −4.65 × 10 5 SOLUTION: The numbers are all in standard form and all of the exponents are different, so we can compare the numbers by analyzing the exponents. The least number is the negative number with the greatest exponent of 10. The other negative number would come next. To order the positive numbers, order their exponents from least to greatest. 5 4 −4 −3 −4.65 × 10 , −5.64 × 10 , 4.56 × 10 , 5.46 × 10 , 74. OPEN ENDED Write two numbers in scientific −3 notation with a product of 1.3 × 10 . Then name two numbers in scientific notation with a quotient of −3 1.3 × 10 . SOLUTION: Sample answer: Let the two numbers in scientific n m notation be given by a × 10 and b × 10 . -3 If the product is to be 1.3 × 10 , then choose numbers for a and b such that ab = 13 and 1 ≤ a < 10 and 1 ≤ b < 10. Select integers for n and m such -3 -4 that n + m = –4, since 1.3 × 10 = 13 × 10 . Since 2.5 × 5.2 = 13 and 5 + (–9) = –4, let the numbers be −9 5 2.5 × 10 and 5.2 × 10 . Find the product to check. 3 6.54 × 10 eSolutions Manual - Powered by Cognero 73. CCSS ARGUMENTS Determine whether the statement is always, sometimes, or never true. Give Page 10 So, two numbers in scientific notation that have a -3 5 SOLUTION: m Sample answer: Always; if the numbers are a × 10 n and b × 10Notation 7-4 Scientific in scientific notation, then 1 ≤ a < 10 and 1 ≤ b < 10. So 1 ≤ ab < 100. 74. OPEN ENDED Write two numbers in scientific −3 notation with a product of 1.3 × 10 . Then name two numbers in scientific notation with a quotient of −3 So, two numbers in scientific notation that have a -3 3 6 quotient of 1.3 × 10 could be 2.6 × 10 and 2 × 10 . 75. WRITING IN MATH Write the steps that you would use to divide two numbers written in scientific notation. Then describe how you would write the 1.3 × 10 . results in standard form. Demonstrate by finding SOLUTION: Sample answer: Let the two numbers in scientific n m notation be given by a × 10 and b × 10 . for a = 2 × 10 and b = 4 × 10 . 3 -3 If the product is to be 1.3 × 10 , then choose numbers for a and b such that ab = 13 and 1 ≤ a < 10 and 1 ≤ b < 10. Select integers for n and m such -3 -4 that n + m = –4, since 1.3 × 10 = 13 × 10 . Since 2.5 × 5.2 = 13 and 5 + (–9) = –4, let the numbers be −9 5 2.5 × 10 and 5.2 × 10 . Find the product to check. 5 SOLUTION: Sample answer: Divide the numbers to the left of the × symbols. Then divide the powers of 10. If necessary, rewrite the results in scientific notation. To convert that to standard form, check to see if the exponent is positive or negative. If positive, move the decimal point to the right, and if negative, to the left. The number of places to move the decimal point is the absolute value of the exponent. Fill in with zeros as needed. For example: So, two numbers in scientific notation that have a -3 5 product of 1.3 × 10 could be 2.5 × 10 and 5.2 × −9 10 . -3 If the quotient is to be 1.3 × 10 , then choose numbers for a and b such that = 1.3 and 1 ≤ a < 10 and 1 ≤ b < 10. Select integers for n and m such that n – m = –3. Since 2.6 ÷ 2 = 1.3 and 3 – 6 = –3, 3 6 let the numbers be 2.6 × 10 and 2 × 10 . Find the quotient to check. 8 76. Which number represents 0.05604 × 10 written in standard form? A 0.0000000005604 B 560,400 C 5,604,000 D 50,604,000 SOLUTION: The exponent is 8, which means that the decimal point should be moved to the right. This means that choice A can be eliminated. So, two numbers in scientific notation that have a -3 3 6 quotient of 1.3 × 10 could be 2.6 × 10 and 2 × 10 . 75. WRITING IN MATH Write the steps that you would use to divide two numbers written in scientific notation. Then describe how you would write the results in standard form. Demonstrate by finding 3 5 for a = 2 × 10 and b = 4 × 10 . SOLUTION: Sample answer: Divide the numbers to the left of the eSolutions Manual - Powered by Cognero × symbols. Then divide the powers of 10. If necessary, rewrite the results in scientific notation. To convert that to standard form, check to see if the When you move the decimal point 8 spaces to the right you obtain 5,604,000, so C is the correct choice. 77. Toni left school and rode her bike home. The graph below shows the relationship between her distance from the school and time. Page 11 When youNotation move the decimal point 8 spaces to the 7-4 Scientific right you obtain 5,604,000, so C is the correct choice. 77. Toni left school and rode her bike home. The graph below shows the relationship between her distance from the school and time. not work either. Toni’s distance remains the same from time t = 30 to t = 40. This means that she did not move during this time, so choice H is the correct choice. 78. SHORT RESPONSE In his first four years of coaching football, Coach Delgato’s team won 5 games the first year, 10 games the second year, 8 games the third year, and 7 games the fourth year. How many games does the team need to win during the fifth year to have an average of 8 wins per year? SOLUTION: Let x represent the number of wins in the fifth year. Which explanation could account for the section of the graph from t = 30 to t = 40? F Toni rode her bike down a hill. G Toni ran all the way home. H Toni stopped at a friend’s house on her way home. J Toni returned to school to get her mathematics book. SOLUTION: Going down a hill and running all the way home would both show a change in distance, so options F and G, respectively, do not work. If she returned to school, the distance would actually go down and we would expect the graph to show that (and it does not). So, option J does not work either. Toni’s distance remains the same from time t = 30 to t = 40. This means that she did not move during this time, so choice H is the correct choice. 78. SHORT RESPONSE In his first four years of coaching football, Coach Delgato’s team won 5 games the first year, 10 games the second year, 8 games the third year, and 7 games the fourth year. How many games does the team need to win during the fifth year to have an average of 8 wins per year? SOLUTION: Let x represent the number of wins in the fifth year. So, Coach Delgato needs to win 10 games in his fifth year to have an average of 8 wins per year. 79. The table shows the relationship between Calories and grams of fat contained in an order of fried chicken from various restaurants. Assuming that the data can best be described by a linear model, how many grams of fat would you expect to be in a 275-Calorie order of fried chicken? A 22 B 25 C 28 D 30 SOLUTION: Find the slope between two points. Now find the y–intercept using one of the points. So, Coach Delgato needs to win 10 games in his fifth year to have an average of 8 wins per year. eSolutions Manual - Powered by Cognero 79. The table shows the relationship between Calories and grams of fat contained in an order of fried Page 12 So, CoachNotation Delgato needs to win 10 games in his fifth 7-4 Scientific year to have an average of 8 wins per year. 79. The table shows the relationship between Calories and grams of fat contained in an order of fried chicken from various restaurants. The answer obtained is closest to 25, so the correct choice is B. 80. HEALTH A ponderal index p is a measure of a person’s body based on height h in centimeters and mass m in kilograms. One such formula is . If a person who is 182 centimeters tall has a ponderal index of about 2.2, how much does the person weigh in kilograms? Assuming that the data can best be described by a linear model, how many grams of fat would you expect to be in a 275-Calorie order of fried chicken? A 22 B 25 C 28 D 30 SOLUTION: Replace p with 2.2 and h with 182 in the formula to determine the value of m. SOLUTION: Find the slope between two points. So, the person will weigh about 64.2 kilograms. Simplify. Assume that no denominator is equal to zero. Now find the y–intercept using one of the points. 81. SOLUTION: So, to find the fat grams for a 275-Calorie order of fried chicken, substitute 275 for x. 82. SOLUTION: The answer obtained is closest to 25, so the correct choice is B. 80. HEALTH A ponderal index p is a measure of a person’s body based on height h in centimeters and mass m in kilograms. 83. SOLUTION: eSolutions - Poweredisby Cognero OneManual such formula . If a person who is 182 centimeters tall has a ponderal index of about 2.2, how much does the person weigh Page 13 7-4 Scientific Notation 83. 86. SOLUTION: SOLUTION: 84. SOLUTION: 2 87. CHEMISTRY Lemon juice is 10 times as acidic 3 as tomato juice. Tomato juice is 10 times as acidic as egg whites. How many times as acidic is lemon juice as egg whites? SOLUTION: 85. SOLUTION: 5 Lemon juice is 10 times more acidic than egg whites. Evaluate a(bx) for each of the given values. 88. a = 1, b = 2, x = 4 SOLUTION: 86. 89. a = 4, b = 1, x = 7 SOLUTION: SOLUTION: 90. a = 5, b = 3, x = 0 SOLUTION: eSolutions Manual - Powered by Cognero Page 14 7-4 Scientific Notation 90. a = 5, b = 3, x = 0 SOLUTION: 91. a = 0, b = 6, x = 8 SOLUTION: 92. a = –2, b = 3, x = 1 SOLUTION: 93. a = –3, b = 5, x = 2 SOLUTION: eSolutions Manual - Powered by Cognero Page 15 The graph crosses the y-axis at 1. The domain is all real numbers, and the range is all real numbers greater than 0. 7-5 Exponential Functions Graph each function. Find the y-intercept and state the domain and range. x 2. y = −5 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y –(5 ) x 1. y = 2 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y 2 –2 –2 0.25 –1 0.5 0 1 1 2 2 4 2 –1 2 0 2 1 2 2 2 –2 –0.04 –1 –0.2 0 –1 1 –5 2 –25 –2 –(5 ) –1 –(5 ) 0 –(5 ) 1 –(5 ) 2 –(5 ) The graph crosses the y-axis at –1. The domain is all real numbers, and the range is all real numbers less than 0. The graph crosses the y-axis at 1. The domain is all real numbers, and the range is all real numbers greater than 0. x 2. y = −5 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y –(5 ) –2 –2 –0.04 –1 –0.2 0 –1 1 –5 2 –25 –(5 ) –1 –(5 ) 0 –(5 ) 1 –(5 ) 2 –(5 ) 3. SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x y –2 –25 –1 –-5 0 –1 1 –0.2 2 –0.04 eSolutions Manual - Powered by Cognero Page 1 The graph crosses the y-axis at –1. The domain is all real numbers, and the range is all real numbers less than 0. 7-5 Exponential Functions The graph crosses the y-axis at –1. The domain is all real numbers, and the range is all real numbers less than 0. 3. 4. SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x y –2 –25 –1 –-5 0 –1 1 –0.2 2 –0.04 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x y –2 48 –1 12 0 3 1 0.75 2 0.188 The graph crosses the y-axis at –1. The domain is all real numbers, and the range is all real numbers less than 0. 4. SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x y –2 48 –1 12 0 1 3 0.75 eSolutions Manual - Powered by Cognero 2 0.188 The graph crosses the y-axis at 3. The domain is all real numbers, and the range is all real numbers greater than 0. x 5. f (x) = 6 + 3 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x y 6 +3 –2 –2 6 –1 6 0 6 +3 1 6 +3 2 6 +3 –1 +3 3.03 +3 3.17 0 4 1 9 2 39 Page 2 The graph crosses the y-axis at 3. The domain is all real numbers, and the range is all real numbers greater than 0.Functions 7-5 Exponential x x 6. f (x) = 2 − 2 5. f (x) = 6 + 3 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x y 6 +3 –2 1.75 –1 1.5 0 1 1 0 2 –2 –2 2–2 +3 3.17 –1 2–2 0 2–2 –1 6 0 6 +3 0 4 1 9 2 39 6 +3 2 –2 3.03 6 –1 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x y 2–2 +3 –2 1 The graph crosses the y-axis at 4. The domain is all real numbers, and the range is all real numbers greater than 3. 6 +3 The graph crosses the y-axis at 4. The domain is all real numbers, and the range is all real numbers greater than 3. x 6. f (x) = 2 − 2 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x y 2–2 –2 1.75 –1 1.5 0 1 1 0 2 –2 –2 2–2 –1 2–2 0 2–2 1 2–2 2 2–2 1 2–2 2 2–2 The graph crosses the y-axis at 1. The domain is all real numbers, and the range is all real numbers less than 2. t 7. BIOLOGY The function f (t) = 100(1.05) models the growth of a fruit fly population, where f (t) is the number of flies and t is time in days. a. What values for the domain and range are reasonable in the context of this situation? Explain. b. After two weeks, approximately how many flies are in this population? SOLUTION: a. D = {t| t ≥ 0}, the number of days is greater than or equal to 0; The y-intercept is (0, 100). Sincet ≥ 0, then R = {y| y ≥ 100}, the number of fruit flies is greater than or equal to 100. [-5, 15] scl: 2 by [-5, 145] scl: 15 b. eSolutions Powered the by Cognero The Manual graph -crosses y-axis at 1. The domain is all real numbers, and the range is all real numbers less than 2. After two weeks, there will be about 198 fruit flies in Page 3 this population. Determine whether the set of data shown below The data shown does not display exponential behavior. The domain values are at regular intervals. The graph crosses the y-axis at 1. The domain is all 7-5 Exponential real numbers,Functions and the range is all real numbers less than 2. However the range values have a common difference of 2. t 7. BIOLOGY The function f (t) = 100(1.05) models the growth of a fruit fly population, where f (t) is the number of flies and t is time in days. a. What values for the domain and range are reasonable in the context of this situation? Explain. b. After two weeks, approximately how many flies are in this population? 9. SOLUTION: SOLUTION: a. D = {t| t ≥ 0}, the number of days is greater than or equal to 0; The y-intercept is (0, 100). Sincet ≥ 0, then R = {y| y ≥ 100}, the number of fruit flies is greater than or equal to 100. The data in the table displays exponential behavior. The domain values are at regular intervals, and the range values have a common factor of 4. Graph each function. Find the y-intercept and state the domain and range. x 10. y = 2 ⋅ 8 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y 2 • 8 [-5, 15] scl: 2 by [-5, 145] scl: 15 b. –2 0.03 –1 0.25 0 2 1 16 2 128 After two weeks, there will be about 198 fruit flies in this population. –2 2 • 8 –1 2 • 8 Determine whether the set of data shown below displays exponential behavior. Write yes or no. Explain why or why not. 0 2 • 8 1 2 • 8 2 2 • 8 8. SOLUTION: The graph crosses the y-axis at 2. The domain is all real numbers, and the range is all real numbers greater than 0. The data shown does not display exponential behavior. The domain values are at regular intervals. However the range values have a common difference of 2. 9. eSolutions Manual - Powered by Cognero SOLUTION: 11. SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. Page 4 x y The graph crosses the y-axis at 2. The domain is all real numbers, and the range is all real numbers greater than 0. 7-5 Exponential Functions 11. The graph crosses the y-axis at 2. The domain is all real numbers, and the range is all real numbers greater than 0. 12. SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x y SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x y –2 72 –2 144 –1 12 –1 12 0 2 0 1 1 0.33 1 0.08 2 0.06 2 0.01 The graph crosses the y-axis at 2. The domain is all real numbers, and the range is all real numbers greater than 0. The graph crosses the y-axis at 1. The domain is all real numbers, and the range is all real numbers greater than 0. x 12. 13. y = −3 ⋅ 9 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x y –2 eSolutions Manual - Powered by Cognero –1 144 12 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y –3 • 9 –2 –0.04 –1 –0.33 0 –3 1 –27 –2 –3 • 9 –1 –3 • 9 0 –3 • 9 1 –3 • 9 2 2 –3 • 9 –243 Page 5 The graph crosses the y-axis at 1. The domain is all real numbers, and the range is all real numbers greater than 0.Functions 7-5 Exponential The graph crosses the y-axis at –3. The domain is all real numbers, and the range is all real numbers less than 0. x x 14. y = −4 ⋅ 10 13. y = −3 ⋅ 9 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y –3 • 9 –2 –0.04 –1 –0.33 0 –3 1 –27 2 –243 –2 –3 • 9 –1 –3 • 9 0 –3 • 9 1 –3 • 9 2 –3 • 9 The graph crosses the y-axis at –3. The domain is all real numbers, and the range is all real numbers less than 0. 0 –0.04 –1 –0.4 0 –4 1 –40 2 –400 –4 • 10 –1 –4 • 10 0 –4 • 10 1 –4 • 10 2 –4 • 10 The graph crosses the y-axis at –4. The domain is all real numbers, and the range is all real numbers less than 0. 15. y = 3 ⋅ 11 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y –4 • 10 –1 –2 –2 x x 14. y = −4 ⋅ 10 –2 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y –4 • 10 SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y 3 • 11 0.02 –1 0.27 0 3 1 33 2 363 3 • 11 –2 –2 –2 –0.04 –1 3 • 11 –1 –0.4 0 3 • 11 0 –4 1 3 • 11 1 –40 2 3 • 11 –4 • 10 –4 • 10 –4 • 10 1 –4 • 10 2 –4 • 10 2 eSolutions Manual - Powered by Cognero –400 Page 6 The graph crosses the y-axis at 3. The domain is all real numbers, and the range is all real numbers greater than 0. The graph crosses the y-axis at –4. The domain is all 7-5 Exponential real numbers,Functions and the range is all real numbers less than 0. 15. y = 3 ⋅ 11 x x 16. y = 4 + 3 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y 3 • 11 –2 –1 –2 0.02 –1 0.27 0 3 1 33 2 363 3 • 11 3 • 11 0 3 • 11 1 3 • 11 2 3 • 11 The graph crosses the y-axis at 3. The domain is all real numbers, and the range is all real numbers greater than 0. x y –2 3.06 –1 3.25 0 4 1 7 2 19 x 4 +3 –2 4 +3 –1 4 +3 0 4 +3 1 4 +3 2 4 +3 The y-intercept is (0, 4). The domain is all real numbers, and the range is all real numbers greater than 3. x 16. y = 4 + 3 17. SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x y –2 3.06 –1 3.25 0 4 1 7 2 19 x 4 +3 –2 4 +3 –1 4 +3 0 4 +3 1 4 +3 2 4 +3 eSolutions Manual - Powered by Cognero SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x y –2 –3.9 –1 –3.75 0 –3.5 1 –3 2 –2 Page 7 The y-intercept is (0, 4). The domain is all real 7-5 Exponential numbers, andFunctions the range is all real numbers greater than 3. The y-intercept is (0, –3.5). The domain is all real numbers, and the range is all real numbers greater than –4. x 18. y = 5(3 ) + 1 17. SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x y –2 –3.9 –1 –3.75 0 –3.5 1 –3 2 –2 x x 5(3 ) + 1 –2 5(3 ) + 1 –1 5(3 ) + 1 0 5(3 ) + 1 1 5(3 ) + 1 2 5(3 ) + 1 y –2 1.5 –1 2.7 0 6 1 16 2 46 The y-intercept is (0, 6). The domain is all real numbers, and the range is all real numbers greater than 1. x 19. y = −2(3 ) + 5 The y-intercept is (0, –3.5). The domain is all real numbers, and the range is all real numbers greater than –4. SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x 18. y = 5(3 ) + 1 −2(3 ) + 5 –2 −2(3 ) + 5 –2 4.8 –1 −2(3 ) + 5 –1 4.3 0 −2(3 ) + 5 0 3 y 1 −2(3 ) + 5 1 –1 1.5 2 −2(3 ) + 5 2 –13 x x 5(3 ) + 1 –2 5(3 ) + 1 –2 –1 2.7 0 6 1 16 2 46 –1 5(3 ) + 1 0 5(3 ) + 1 1 5(3 ) + 1 2 5(3 ) + 1 eSolutions Manual - Powered by Cognero x x SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. y Page 8 The y-intercept is (0, 6). The domain is all real 7-5 Exponential numbers, andFunctions the range is all real numbers greater than 1. x The y-intercept is (0, 3). The domain is all real numbers, and the range is all real numbers less than 5. 20. CCSS MODELING A population of bacteria in a culture increases according to the model p = 300(2.7) 19. y = −2(3 ) + 5 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. 0.02t , where t is the number of hours and t = 0 corresponds to 9:00 A.M. a. Use this model to estimate the number of bacteria at 11 A.M. b. Graph the function and name the p -intercept. Describe what the p -intercept represents, and describe a reasonable domain and range for this situation. x x −2(3 ) + 5 y –2 −2(3 ) + 5 –2 4.8 –1 −2(3 ) + 5 –1 4.3 0 −2(3 ) + 5 0 3 1 −2(3 ) + 5 1 –1 2 −2(3 ) + 5 2 –13 SOLUTION: a. There will be about 312 bacteria at 11 A.M. b. t at 0 is the p -intercept. The p -intercept is 300. This represents 300 bacteria in the culture at 9:00 A.M. Enter the function as Y1 on a graphing calculator and adjust the window to show 24 hours worth of bacteria growth. The y-intercept is (0, 3). The domain is all real numbers, and the range is all real numbers less than 5. 20. CCSS MODELING A population of bacteria in a culture increases according to the model p = 300(2.7) 0.02t , where t is the number of hours and t = 0 corresponds to 9:00 A.M. a. Use this model to estimate the number of bacteria at 11 A.M. b. Graph the function and name the p -intercept. Describe what the p -intercept represents, and describe a reasonable domain and range for this situation. The domain is all real numbers greater than or equal to 0, because the time elapsed cannot be negative. The range is all real numbers greater than or equal to 300, because the number of bacteria in the culture starts at 300 and increases over time. SOLUTION: a. Determine whether the set of data shown below displays exponential behavior. Write yes or no. Explain why or why not. There will be about 312 bacteria at 11 A.M. b. t at 0 is the p -intercept. 21. eSolutions Manual - Powered by Cognero The p -intercept is 300. This represents 300 bacteria SOLUTION: Page 9 The domain is all real numbers greater than or equal to 0, because the time elapsed cannot be negative. The range is all real numbers greater than or equal to 7-5 Exponential 300, because Functions the number of bacteria in the culture starts at 300 and increases over time. Determine whether the set of data shown below displays exponential behavior. Write yes or no. Explain why or why not. The data shown does not display exponential behavior. The domain values are at regular intervals. However, the range values have a common difference of 5. 23. SOLUTION: 21. SOLUTION: The data shown does display exponential behavior. The domain values are at regular intervals, and the range values have a common factor of 2. The data shown, does not display exponential behavior. The domain values are at regular intervals. However, the range values do not have a positive common factor. 24. SOLUTION: 22. SOLUTION: The data shown displays exponential behavior. The domain values are at regular intervals, and the range values have a common factor of 0.4. The data shown does not display exponential behavior. The domain values are at regular intervals. However, the range values have a common difference of 5. 25. PHOTOGRAPHY Jameka is enlarging a photograph to make a poster for school. She will enlarge the picture repeatedly at 150%. The function x P = 1.5 models the new size of the picture being enlarged, where x is the number of enlargements. How many times as big is the picture after 4 enlargements? SOLUTION: 23. SOLUTION: The picture is about 506% bigger than the original. 26. FINANCIAL LITERACY Daniel deposited $500 into a savings account and after 8 years, his investment is worth $807.07. The equation A = d 12t eSolutions - Powered by Cognero The Manual data shown does display exponential behavior. The domain values are at regular intervals, and the range values have a common factor of 2. (1.005) models the value of Daniel’s investment A Page 10 after t years with an initial deposit d. a. What would the value of Daniel’s investment be if 12t (1.005) , which has a value greater than 1. Since Daniel’s investment is the product of d and a factor not depending on d, it varies directly as the original deposit. That is, the greater the deposit, the greater the amount of the investment over time. 7-5 Exponential Functions The picture is about 506% bigger than the original. Identify each function as linear, exponential, or neither. 26. FINANCIAL LITERACY Daniel deposited $500 into a savings account and after 8 years, his investment is worth $807.07. The equation A = d 12t (1.005) models the value of Daniel’s investment A after t years with an initial deposit d. a. What would the value of Daniel’s investment be if he had deposited $1000? b. What would the value of Daniel’s investment be if he had deposited $250? c. Interpret d(1.005)12t to explain how the amount of the original deposit affects the value of Daniel’s investment. 27. SOLUTION: The graph is not a line, so the function is nonlinear. The graph contains the point (0, 1), is always positive, and increases rapidly as x increases. The function represented by the graph is exponential. SOLUTION: a. Replace d with 1000 and t with 8 in the formula A 12t = d(1.005) to determine the value of the investment. 28. So, if he deposits $1000, after 8 years the investment will be worth about $1614.14. b. Replace d with 250 and t with 8 in the formula A 12t = d(1.005) to determine the value of the investment. So, if he deposits $250, after 8 years the investment will be worth about $403.54. c. Sample answer: In the formula the amount deposited d is always multiplied by the same amount 12t (1.005) , which has a value greater than 1. Since Daniel’s investment is the product of d and a factor not depending on d, it varies directly as the original deposit. That is, the greater the deposit, the greater the amount of the investment over time. SOLUTION: The graph is not a line, so the function is nonlinear. The graph does not contain the point (0, 1), is not always positive, or increases rapidly as x increases. The function represented by the graph is not exponential. 29. SOLUTION: The graph displays a line with a constant slope. The function represented by the graph is linear. x 30. y = 4 SOLUTION: x Identify each function as linear, exponential, or neither. The function is in the form y = ab with a = 1 and b = 4. The function is exponential. 31. y = 2x(x − 1) eSolutions Manual - Powered by Cognero SOLUTION: Page 11 x 30. y = 4 SOLUTION: x 7-5 Exponential Functions The function is in the form y = ab with a = 1 and b = 4. The function is exponential. About 198 students will graduate in 2012. Describe the graph of each equation as a x transformation of the graph of y = 2 . 31. y = 2x(x − 1) SOLUTION: x 34. y = 2 + 6 SOLUTION: x The function is in the form of a linear or exponential function. The graph of f (x) = 2 + c represents a vertical translation of the parent graph. The value of c is 6, x and 6 > 0. If c > 0, the graph of f (x) = 2 is translated 32. 5x + y = 8 x units up. Therefore, the graph of y = 2 + 6 is a SOLUTION: x translation of the graph of y = 2 6 units up. The function is now in y = mx + b where m = -5 and b = 8. This is the form of a linear function. 33. GRADUATION The number of graduates at a high school has increased by a factor of 1.055 every year since 2001. In 2001, 110 students graduated. The t function N = 110(1.055) models the number of students N expected to graduate t years after 2001. How many students will graduate in 2012? SOLUTION: x 35. y = 3(2) SOLUTION: x The graph of f (x) = a(2) stretches or compresses x the graph of f (x) = 2 vertically. The value of a is 3, x and 3 > 1. If a > 1, the graph of f (x) = 2 is stretched About 198 students will graduate in 2012. Describe the graph of each equation as a x transformation of the graph of y = 2 . x vertically. Therefore, the graph of y = 3(2) is the x graph of y = 2 vertically stretched by a factor of 3. x 34. y = 2 + 6 SOLUTION: x The graph of f (x) = 2 + c represents a vertical translation of the parent graph. The value of c is 6, x and 6 > 0. If c > 0, the graph of f (x) = 2 is translated x units up. Therefore, the graph of y = 2 + 6 is a x translation of the graph of y = 2 6 units up. 36. eSolutions Manual x - Powered by Cognero 35. y = 3(2) SOLUTION: SOLUTION: When f (x) or the variable x is multiplied by –1, the graph is reflected over the x- or y-axis. The graph of x the function –f (x) reflects the graph of f (x) = 2 across the x-axis. x The graph of f (x) = a(2) stretches or compresses Page 12 x the graph of f (x) = 2 vertically. The value of a is , 7-5 Exponential Functions 36. 38. SOLUTION: When f (x) or the variable x is multiplied by –1, the graph is reflected over the x- or y-axis. The graph of x the function –f (x) reflects the graph of f (x) = 2 across the x-axis. x The graph of f (x) = a(2) stretches or compresses x the graph of f (x) = 2 vertically. The value of a is x In the graph of f (x) = 2 , a > 1, so y increases as x increases. In the graph of . If 0 < a < 1, the graph of f (x) = 2 is , 0 < a < 1, so y decreases as x increases. Therefore, the graph of x is the graph of y = 2 reflected over the y- , x and SOLUTION: axis. compressed vertically. Therefore, the graph of x is the graph of y = 2 reflected across the x-axis and vertically compressed. x 39. y = −5(2) SOLUTION: When f (x) or the variable x is multiplied by –1, the graph is reflected over the x- or y-axis. The graph of x the function –f (x) reflects the graph of f (x) = 2 across the x-axis. x The graph of f (x) = a(2) stretches or compresses x 37. y = −3 + 2 SOLUTION: x The graph of f (x) = 2 + c represents a vertical translation of the parent graph. The value of c is –3, x and –3 < 0. If c < 0, the graph of f (x) = 2 is translated units down. Therefore, the graph of y = x x –3 + 2 is a translation of the graph of y = 2 shifted 3 units down. x the graph of f (x) = 2 vertically. The value of a is 5, x and 5 > 1. If a > 1, the graph of f (x) = 2 is stretched x vertically. Therefore, the graph of y = −5(2) is the x graph of y = 2 reflected across the x-axis and vertically stretched by a factor of 5. 40. DEER A deer population at a national park doubles every year. In 2000, there were 25 deer in the park. t 38. eSolutions Manual - Powered by Cognero SOLUTION: x The function N = 25(2) models the number of deer N Page 13 in the national park t years after 2000. Determine how many deer there will be in the park in 2015. SOLUTION: reasoning. SOLUTION: The graph will never have an x-intercept because the graph will approach the x-axis, but never crosses it. 7-5 Exponential Functions 40. DEER A deer population at a national park doubles every year. In 2000, there were 25 deer in the park. t The function N = 25(2) models the number of deer N in the national park t years after 2000. Determine how many deer there will be in the park in 2015. 43. OPEN ENDED Find an exponential function that represents a real-world situation, and graph the function. Analyze the graph, and explain why the situation is modeled by an exponential function rather than a linear function. SOLUTION: The number of teams competing in a basketball SOLUTION: x There will be 819,200 deer in the park in 2015. 41. CCSS PERSEVERANCE Write an exponential function for which the graph passes through the points at (0, 3) and (1, 6). tournament can be represented by y = 2 , where the number of teams competing is y and the number of rounds is x. Make a table representing the number of rounds x and the number of teams y needed for the tournament. SOLUTION: x An exponential function is of the form y = ab . Substitute (0, 3) into the equation and solve for a. x y 0 1 1 2 2 4 3 8 4 16 x 2 0 2 1 2 2 2 3 2 4 2 Now use a = 3 and the point (1, 6) to solve for b. Plot the points from the table and connect them with a smooth curve. x So, f(x) = 3(2) represents an exponential function that passes through the points (0, 3) and (1, 6). 42. REASONING Determine whether the graph of y = x ab , where a ≠ 0, b > 0, and b ≠ 1, sometimes, always, or never has an x-intercept. Explain your reasoning. SOLUTION: The graph will never have an x-intercept because the graph will approach the x-axis, but never crosses it. 43. OPEN ENDED Find an exponential function that represents a real-world situation, and graph the function. Analyze the graph, and explain why the situation is modeled by an exponential function rather than a linear function. SOLUTION: The number of teams competing in a basketball x tournament can be represented by y = 2 , where the number of teams competing is y and the number of rounds is x. Make a table representing the number of rounds x and the number of teams y needed for the eSolutions Manual - Powered by Cognero tournament. x x 2 y The y-intercept of the graph is 1. The graph increases rapidly for x > 0. The domain for this scenario is {x | x ≥ 0} indicating the the number of rounds are greater than or equal to 0 . With an exponential model, each team that joins the tournament will play all of the other teams. If the scenario were modeled with a linear function, each team that joined would play a fixed number of teams. 44. REASONING Use tables and graphs to compare x and contrast an exponential function f (x) = ab + c, where a ≠ 0, b > 0, and b ≠ 1, and a linear function g x (x) = a + c. Include intercepts, intervals where the functions are increasing, decreasing, positive, or Page 14 negative, relative maxima and minima, symmetry, and end behavior. SOLUTION: rounds are greater than or equal to 0 . With an exponential model, each team that joins the tournament will play all of the other teams. If the scenario wereFunctions modeled with a linear function, each 7-5 Exponential team that joined would play a fixed number of teams. 44. REASONING Use tables and graphs to compare x and contrast an exponential function f (x) = ab + c, where a ≠ 0, b > 0, and b ≠ 1, and a linear function g x (x) = a + c. Include intercepts, intervals where the functions are increasing, decreasing, positive, or negative, relative maxima and minima, symmetry, and end behavior. numbers, while those for g(x) are positive for x > and are negative for x < . Neither f (x) nor g (x) have maximum or minimum points, and neither has symmetry. 45. WRITING IN MATH Explain how to determine whether a set of data displays exponential behavior. SOLUTION: First, look for a pattern by making sure that the domain values are at regular intervals and the range values differ by a common factor. Consider the data set in the table. SOLUTION: Sample answer: Let a = 3, b = 2 and c = 1. Make a table using x = –5 to 5 showing the values of the x functions f (x) = 3(2) + 1 and g(x) = 3x + 1. The domain values are at regular intervals of 1. The range values have a common factor of 4. Thus the data in the data displays exponential behavior. 46. SHORT RESPONSE What are the zeros of the function graphed below? Use the values from the table to plot the points for each function f (x) and g(x ). Connect the points to draw the graph for each function. SOLUTION: The points where the graph crosses the x-axis are 2 and 4. 47. Hinto invested $300 into a savings account. The 12t The y-intercept of f (x) is 4 and the y-intercept of g(x is 1. Both f (x) and g(x) increase as x increases. The function values for f (x) are positive for all real numbers, while those for g(x) are positive for x > and are negative for x < . Neither f (x) nor g (x) have maximum or minimum points, and neither has symmetry. 45. WRITING IN MATH Explain how to determine whether a set of data displays exponential behavior. eSolutions Manual - Powered by Cognero SOLUTION: equation A = 300(1.005) models the amount in Hinto’s account A after t years. How much will be in Hinto’s account after 7 years? A $25,326 B $456.11 C $385.01 D $301.52 SOLUTION: There will be $456.11 in Hinto’s account after 7 years. So, the correct choice is B. on a15 48. GEOMETRY Ayana placed a circular picturePage square picture as shown below. If the picture extends 4 inches beyond the circle on each side, There will be Functions $456.11 in Hinto’s account after 7 7-5 Exponential years. So, the correct choice is B. 48. GEOMETRY Ayana placed a circular picture on a square picture as shown below. If the picture extends 4 inches beyond the circle on each side, what is the perimeter of the square picture? The perimeter of the square paper is 112 in. So, the correct choice is J. 49. The points with coordinates (0, –3) and (2, 7) are on line l. Line p contains (3, –1) and is perpendicular to line l. What is the x-coordinate of the point where l and p intersect? A B C D F 64 in. G 80 in. H 94 in. J 112 in. SOLUTION: First, find the slope and equation of line . SOLUTION: Since the line contains (0, –3), write the equation of the line in slope-intercept form using a m = 5 and b = –3. The perimeter of the square paper is 112 in. So, the correct choice is J. 49. The points with coordinates (0, –3) and (2, 7) are on line l. Line p contains (3, –1) and is perpendicular to line l. What is the x-coordinate of the point where l and p intersect? Next, find the slope and equation of line p . Since line p is perpendicular to line , their slopes are negative reciprocals. Use the point-slope form of a line and m = and (x, y) = (3, –1) to write the equation for line p . A B C D SOLUTION: First, find the slope and equation of line Lastly, set the equation for lines and p equal and solve for x to find the x-coordinate of the point where the lines intersect. . Since the line contains (0, –3), write the equation of the line in slope-intercept form using a m = 5 and b = –3. Next, find the slope and equation of line p . Since line p is perpendicular to line , their slopes are negative reciprocals. The x-coordinate of the point where and p intersect is . Thus, the correct choice is A. Evaluate each product. Express the results in both scientific notation and standard form. 2 6 50. (1.9 × 10 )(4.7 × 10 ) eSolutions Manual - Powered by Cognero Page 16 SOLUTION: Use the point-slope form of a line and m = and 54. SOLUTION: The x-coordinate of the point where and p 7-5 Exponential Functions intersect is . Thus, the correct choice is A. Evaluate each product. Express the results in both scientific notation and standard form. 2 6 50. (1.9 × 10 )(4.7 × 10 ) 55. SOLUTION: SOLUTION: −3 4 51. (4.5 × 10 )(5.6 × 10 ) 56. SOLUTION: SOLUTION: −4 −8 52. (3.8 × 10 )(6.4 × 10 ) SOLUTION: 57. SOLUTION: Simplify. 53. SOLUTION: 58. SOLUTION: 54. SOLUTION: 55. SOLUTION: 59. DEMOLITION DERBY When a car hits an object, the damage is measured by the collision impact. For a certain car the collision impact I is 2 given by I = 2v , where v represents the speed in kilometers per minute. What is the collision impact if the speed of the car is 4 kilometers per minute? SOLUTION: eSolutions Manual - Powered by Cognero 56. Page 17 7-5 Exponential Functions The solution is (–5, 20). 59. DEMOLITION DERBY When a car hits an object, the damage is measured by the collision impact. For a certain car the collision impact I is 2 given by I = 2v , where v represents the speed in kilometers per minute. What is the collision impact if the speed of the car is 4 kilometers per minute? 62. 3x − 5y = 16 −3x + 2y = −10 SOLUTION: SOLUTION: The collision impact is 32. Use elimination to solve each system of equations. 60. x + y = −3 x −y = 1 SOLUTION: The solution is (2, –2). Find the next three terms of each arithmetic sequence. 63. 1, 3, 5, 7, … SOLUTION: 3 – 1 = 2; 5 – 3 = 2; 7 – 5 = 2. The common difference is 2. The next three terms will be: 7+2=9 9 + 2 = 11 11 + 2 = 13 The solution is (–1, –2). 64. −6, −4, −2, 0, … 61. 3a + b = 5 2a + b = 10 SOLUTION: SOLUTION: –2 – (–6) = 2; –2 – (–4) = 2; 0 – (–2) = 2. The common difference is 2. The next three terms will be: 0+2=2 2+2=4 4+2=6 65. 6.5, 9, 11.5, 14, … The solution is (–5, 20). 62. 3x − 5y = 16 −3x + 2y = −10 SOLUTION: 9 – 6.5 = 2.5; 11.5 – 9 = 2.5; 14 – 11.5 = 2.5. The common difference is 2.5. The next three terms will be: 14 + 2.5 = 16.5 16.5 + 2.5 = 19 19 + 2.5 = 21.5 SOLUTION: 66. 10, 3, −4, −11, … eSolutions Manual - Powered by Cognero SOLUTION: 3 – 10 = – 7; –4 – 3 = – 7; –11 – (–4) = – 7. The common difference is –7. Page 18 The next three terms will be: –11 + (–7) = –18 –18 + (–7) = –25 common difference is 2.5. The next three terms will be: 14 + 2.5 = 16.5 7-5 Exponential Functions 16.5 + 2.5 = 19 19 + 2.5 = 21.5 66. 10, 3, −4, −11, … 68. SOLUTION: 3 – 10 = – 7; –4 – 3 = – 7; –11 – (–4) = – 7. The common difference is –7. The next three terms will be: –11 + (–7) = –18 –18 + (–7) = –25 –25 + (–7) = –32 SOLUTION: 67. The common difference is SOLUTION: The common difference is . The next three terms will be: . The next three terms will be: 68. SOLUTION: The common difference is . The next three terms will be: eSolutions Manual - Powered by Cognero Page 19 7-6 Growth and Decay So, Paul’s investment will be worth about $620.46 in 8 years. 1. SALARY Ms. Acosta received a job as a teacher with a starting salary of $34,000. According to her contract, she will receive a 1.5% increase in her salary every year. How much will Ms. Acosta earn in 7 years? SOLUTION: Using the equation for exponential growth, let a = 34,000 and let r = 1.5% = 0.015. 3. ENROLLMENT In 2000, 2200 students attended Polaris High School. The enrollment has been declining 2% annually. a. Write an equation for the enrollment of Polaris High School t years after 2000. b. If this trend continues, how many students will be enrolled in 2015? SOLUTION: a. Using the equation for exponential decay, let a = 2200 and r = 2% = 0.02. Let t = 7 in the salary equation above. b. Using the equation from part a, let t = 15. So, Ms. Acosta will earn about $37,734.73 in 7 years. 2. MONEY Paul invested $400 into an account with a 5.5% interest rate compounded monthly. How much will Paul’s investment be worth in 8 years? SOLUTION: Using the equation for compound interest, let P = 400, r = 5.5% = 0.055, n = 12, and t = 8. So, the enrollment of Polaris High School will be about 1624 in 2015. 4. MEMBERSHIPS The Work-Out Gym sold 550 memberships in 2001. Since then the number of memberships sold has increased 3% annually. a. Write an equation for the number of memberships sold at Work-Out Gym t years after 2001. b. If this trend continues, predict how many memberships the gym will sell in 2020. SOLUTION: a. Using the equation for exponential growth, let a = 550 and let r = 3% = 0.03. So, Paul’s investment will be worth about $620.46 in 8 years. 3. ENROLLMENT In 2000, 2200 students attended Polaris High School. The enrollment has been declining 2% annually. a. Write an equation for the enrollment of Polaris High School t years after 2000. b. If this trend continues, how many students will be enrolled in 2015? SOLUTION: a. Using the equation for exponential decay, let a = 2200 and r = 2% = 0.02. eSolutions Manualthe - Powered by Cognero b. Using equation from part a, let t = 15. b. Using the equation from part a, let t = 19. So, the gym will sell about 964 memberships in 2020. 5. COMPUTERS The number of people who own computers has increased 23.2% annually since 1990. If half a million people owned a computer in 1990, predict how many people will own a computer in 2015. SOLUTION: Using the equation for exponential growth, let a = Page 1 500,000, r = 23.2% = 0.232, and t = 25. 7-6 Growth and Decay So, the gym will sell about 964 memberships in 2020. 5. COMPUTERS The number of people who own computers has increased 23.2% annually since 1990. If half a million people owned a computer in 1990, predict how many people will own a computer in 2015. SOLUTION: Using the equation for exponential growth, let a = 500,000, r = 23.2% = 0.232, and t = 25. So, about 92,095,349 people will own a computer in 2015. 6. COINS Camilo purchased a rare coin from a dealer for $300. The value of the coin increases 5% each year. Determine the value of the coin in 5 years. SOLUTION: Using the equation for exponential growth, let a = 300, r = 5% = 0.05, and t = 5. So, the value of Theo’s investment in 4 years is about $7898.97. 8. FINANCE Paige invested $1200 at an interest rate of 5.75% compounded quarterly. Determine the value of her investment in 7 years. SOLUTION: Using the equation for compound interest, let P = 1200, r = 5.75% = 0.0575, n = 4, and t = 7. So, the value of Paige’s investment in 7 years is about $1789.54. 9. CCSS PRECISION Brooke is saving money for a trip to the Bahamas that costs $295.99. She puts $150 into a savings account that pays 7.25% interest compounded quarterly. Will she have enough money in the account after 4 years? Explain. SOLUTION: Using the equation for compound interest, let P = 150, r = 7.25% = 0.0725, n = 4, and t = 4. So, the value of the coin will be about $382.88 in 5 years. 7. INVESTMENTS Theo invested $6600 at an interest rate of 4.5% compounded monthly. Determine the value of his investment in 4 years. SOLUTION: Using the equation for compound interest, let P = 6600, r = 4.5% = 0.045, n = 12, and t = 4. No, Brooke will have about $199.94 in the account in 4 years. 10. INVESTMENTS Jin’s investment of $4500 has been losing its value at a rate of 2.5% each year. What will his investment be worth in 5 years? SOLUTION: Using the equation for exponential decay, let a = 4500, r = 2.5% = 0.025, and t = 5. So, the value of Theo’s investment in 4 years is about $7898.97. 8. FINANCE Paige invested $1200 at an interest rate of 5.75% compounded quarterly. Determine the value of her investment in 7 years. eSolutions Manual - Powered by Cognero SOLUTION: Using the equation for compound interest, let P = 1200, r = 5.75% = 0.0575, n = 4, and t = 7. So, Jin’s investment will be about $3964.93 in 5 years. 11. POPULATION In the years from 2010 to 2015, the Page 2 population of the District of Columbia is expected to decrease about 0.9% annually. In 2010, the population was about 530,000. What is the population 7-6 Growth Decay will be about $3964.93 in 5 So, Jin’s and investment years. 11. POPULATION In the years from 2010 to 2015, the population of the District of Columbia is expected to decrease about 0.9% annually. In 2010, the population was about 530,000. What is the population of the District of Columbia expected to be in 2015? No, Leonardo should not sell the car for $4500. The car is worth about $5774.61. 13. HOUSING The median house price in the United States increased an average of 1.4% each year between 2005 and 2007. Assume that this pattern continues. SOLUTION: This is an exponential decay problem since the population is decreasing. Use the equation for exponential decay, let a = 530,000, r = 0.9% = 0.009, and t = 5. The population in 2015 will be about 506,575. 12. CARS Leonardo purchases a car for $18,995. The car depreciates at a rate of 18% annually. After 6 years, Manuel offers to buy the car for $4500. Should Leonardo sell the car? Explain. a. Write an equation for the median house price for t years after 2004. b. Predict the median house price in 2018. SOLUTION: Using the equation for exponential decay, let a = 18,995, r = 18% = 0.18, and t = 6. SOLUTION: a. This is an exponential growth problem since prices are increasing. Use the equation for exponential growth, let a = 247,900 and r = 1.4% = 0.014. No, Leonardo should not sell the car for $4500. The car is worth about $5774.61. b. To find the price in 2018, let t = 11. 13. HOUSING The median house price in the United States increased an average of 1.4% each year between 2005 and 2007. Assume that this pattern continues. Thus in 2018, median house price will be about $288,864. 14. ELEMENTS A radioactive element’s half-life is the time it takes for one half of the element’s quantity to decay. The half-life of Plutonium-241 is 14.4 years. The number of grams A of Plutonium-241 left after t years can be modeled by , where p is the original amount of the element. a. How much of a 0.2-gram sample remains after 72 years? b. How much of a 5.4-gram sample remains after 1095 days? a. Write an equation for the median house price for t yearsManual after-2004. eSolutions Powered by Cognero b. Predict the median house price in 2018. SOLUTION: SOLUTION: a. Let p = 0.2 and t = 72. Page 3 7-6 Growth and Decay Thus in 2018, median house price will be about $288,864. 14. ELEMENTS A radioactive element’s half-life is the time it takes for one half of the element’s quantity to decay. The half-life of Plutonium-241 is 14.4 years. The number of grams A of Plutonium-241 left after t pumped into the pool. c. Find C(t) = p (t) + w(t). What does this new function represent? d. Use the graph of C(t) to determine how long the hose must run to fill the pool to its maximum amount. SOLUTION: a. Since the amount of water is decreasing, use the formula for exponential decay and replace a with 19,000 and r with 0.005. t years can be modeled by , where p is the original amount of the element. a. How much of a 0.2-gram sample remains after 72 years? b. How much of a 5.4-gram sample remains after 1095 days? SOLUTION: a. Let p = 0.2 and t = 72. So, 0.00625 gram of Plutonium-241 remains after 72 years. b. Let p = 5.4 and t = = 3. y = a(1 – r) t w(t) = 19,000(1 – 0.005) t w(t) = 19,000(0.995) b. Since the hose is running at a constant rate, use a linear function in slope-intercept form to express the amount of water pumped into the pool. At t = 0 no water is being put into the pool, so the y-intercept is 0. The hose is running at a rate of 300 gallons per hour, so replace m with 300. y = mx + b p(t) = 300x + 0 p(t) = 300t c. C(t) = p (t) + w(t) t C(t) = 300t + 19,000(0.995) ; The function C(t) represents the number of gallons of water in the pool at any time after the hose is turned on. d. Use a graphing calculator to graph Y1= 300x + 19000(0.995)x and Y2= 20500. Select the intersect option on the 2nd [CALC] function. So, about 4.7 grams of Plutonium-241 remains after 1095 days. 15. COMBINING FUNCTIONS A swimming pool holds a maximum of 20,500 gallons of water. The water is evaporating at a rate of 0.5% per hour. The pool currently contains 19,000 gallons of water. a. Write an exponential function w(t) to express the amount of water remaining in the pool after time t where t is the number of hours after the pool reached 19,000 gallons. b. At this same time, a hose is turned on to refill the pool at a rate of 300 gallons per hour. Write a function p (t) where t is time in hours the hose is running to express the amount of water that is pumped into the pool. c. Find C(t) = p (t) + w(t). What does this new function represent? d. Use the graph of C(t) to determine how long the hose must run to fill the pool to its maximum amount. SOLUTION: a. Since the amount of water is decreasing, use the formula for exponential decay and replace a with 19,000 and r with 0.005. eSolutions Manual - Powered by Cognero Therefore, the hose will fill the pool to its maximum after about 7.3 hours. 16. REASONING Determine the growth rate (as a percent) of a population that quadruples every year. Explain. SOLUTION: Since the population is increasing, use a exponential growth model. If the population starts at 1, a = 1. If the population quadruples, it will be at 4, or y = 4. Let t = 1. Page 4 Therefore, theDecay hose will fill the pool to its maximum 7-6 Growth and after about 7.3 hours. 16. REASONING Determine the growth rate (as a percent) of a population that quadruples every year. Explain. From the table, 2500 is between 9 and 10. Adjust the step on the table to narrow to intervals. SOLUTION: Since the population is increasing, use a exponential growth model. If the population starts at 1, a = 1. If the population quadruples, it will be at 4, or y = 4. Let t = 1. The investment reaches $2500 after 9.2 years. Use the intersect function from the 2nd CALC menu to find the exact value. Then r = 3 or 300%. The rate for the population to quadruple is 300%. 17. CCSS PRECISION Santos invested $1200 into an account with an interest rate of 8% compounded monthly. Use a calculator to approximate how long it will take for Santos’s investment to reach $2500. SOLUTION: Using the equation for compound interest, let P = 1200 and r = 8% = 0.08. [-15, 10] scl: 1 by [-2, 2998] scl: 120 So, it will take Santos’ investment about 9.2 years to reach $2500. Enter the equation in to Y1. and 2500 for Y2. Adjust the viewing window. 18. REASONING The amount of water in a container doubles every minute. After 8 minutes, the container is full. After how many minutes was the container half full? Explain. SOLUTION: 7; Sample answer: Since the amount of water doubles every minute, the container would be half full a minute before it was full. 19. WRITING IN MATH What should you consider when using exponential models to make decisions? [-15, 10] scl: 1 by [-2, 2998] scl: 120 Use a table to find a value of t such that A = 2500. SOLUTION: Sample answer: Exponential models can grow without bound, which is usually not the case of the situation that is being modeled. For instance, a population cannot grow without bound due to space and food constraints. Therefore, when using a model, the situation that is being modeled should be carefully considered when used to make decisions. 20. WRITING IN MATH Compare and contrast the exponential growth formula and the exponential decay formula. From the table, 2500 is between 9 and 10. Adjust eSolutions Manual - Powered by Cognero the step on the table to narrow to intervals. SOLUTION: t Page 5 The exponential growth formula is y = a(1 + r) , where a is the initial amount, t is time, y is the final situation that is being modeled. For instance, a population cannot grow without bound due to space and food constraints. Therefore, when using a model, the situation is being modeled should be carefully 7-6 Growth and that Decay considered when used to make decisions. 20. WRITING IN MATH Compare and contrast the exponential growth formula and the exponential decay formula. SOLUTION: t The exponential growth formula is y = a(1 + r) , where a is the initial amount, t is time, y is the final amount, and r is the rate of change expressed as a decimal. The exponential decay formula is basically the same except the rate is subtracted from 1 and r represents the rate of decay. Consider an exponential growth model with r = 5% and a = 2. The height cannot be negative, so the height is 5 inches. The correct choice is C. 22. Which is greater than F ? G H J SOLUTION: 2 2 = 2 × 2 or 4; ; ; and ; Only or 8 is greater than 4. Therefore, the correct choice is H. Consider an exponential decay model with r = 5% and a = 2. 21. GEOMETRY The parallelogram has an area of 35 square inches. Find the height h of the parallelogram. 23. Thi purchased a car for $22,900. The car depreciated at an annual rate of 16%. Which of the following equations models the value of Thi’s car after 5 years? A A = 22,900(1.16)5 5 B A = 22,900(0.16) C A = 16(22,900)5 5 D A = 22,900(0.84) SOLUTION: Using the equation for exponential decay, let a = 22,900, r = 16% = 0.16, and t = 5. A 3.5 inches B 4 inches C 5 inches D 7 inches So, the correct choice is D. SOLUTION: Using the formula for the area of the parallelogram, let A = 35 and b = 2h − 3. 24. GRIDDED RESPONSE A deck measures 12 feet by 18 feet. If a painter charges $2.65 per square foot, including tax, how much will it cost in dollars to have the deck painted? SOLUTION: To find the cost of painting the deck C, first find the area of the deck. The height cannot be negative, so the height is 5 inches. The correct choice is C. eSolutions Manual - Powered by Cognero 22. Which is greater than ? Then multiply the area by the cost per square foot. C = 216(2.65) = 572.4 So, the cost to have the deck painted is $572.40. Grid in 572.4. Page 6 Graph each function. Find the y-intercept, and state the domain and range. The graph crosses the y-axis at 1, so the y-intercept is 1. The domain is all real numbers, and the range is all positive real numbers. 7-6 Growth and Decay So, the correct choice is D. 24. GRIDDED RESPONSE A deck measures 12 feet by 18 feet. If a painter charges $2.65 per square foot, including tax, how much will it cost in dollars to have the deck painted? SOLUTION: To find the cost of painting the deck C, first find the area of the deck. 26. SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. Then multiply the area by the cost per square foot. C = 216(2.65) = 572.4 So, the cost to have the deck painted is $572.40. Grid in 572.4. Graph each function. Find the y-intercept, and state the domain and range. x 25. y = 3 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x y −2 4 0 1 2 Graph the ordered pairs from the table, and connect the points with a smooth curve. x x 3 −2 3 −1 3 0 3 1 3 y −2 −1 0 1 1 3 2 9 3 Graph the ordered pairs from the table, and connect the points with a smooth curve. The graph crosses the y-axis at 1, so the y-intercept is 1. The domain is all real numbers, and the range is all positive real numbers. 2 x 27. y = 6 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. The graph crosses the y-axis at 1, so the y-intercept is 1. The domain is all real numbers, and the range is all positive real numbers. x 6x −2 6 −1 6 0 6 1 6 26. −2 −1 0 1 1 6 2 36 6 Graph the ordered pairs from the table, and connect Page 7 the points with a smooth curve. 2 eSolutions Manual - Powered by Cognero y SOLUTION: The graph crosses the y-axis at 1, so the y-intercept is 1. Theand domain is all real numbers, and the range is 7-6 Growth Decay all positive real numbers. x 5 2 30. (7 × 10 ) 27. y = 6 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. SOLUTION: x 6x −2 6 −1 6 y −2 -2 2 −1 31. (1.1 × 10 ) 0 1 1 6 0 6 1 6 2 36 6 Graph the ordered pairs from the table, and connect the points with a smooth curve. SOLUTION: 2 –2 –7 32. (9.1 × 10 )(4.2 × 10 ) SOLUTION: 2 –3 33. (3.14 × 10 )(6.1 × 10 ) SOLUTION: The graph crosses the y-axis at 1, so the y-intercept is 1. The domain is all real numbers, and the range is all positive real numbers. Evaluate each product. Express the results in both scientific notation and standard form. 3 10 28. (4.2 × 10 )(3.1 × 10 ) SOLUTION: 34. EVENT PLANNING A hall does not charge a rental fee as long as at least $4000 is spent on food. For the prom, the hall charges $28.95 per person for a buffet. How many people must attend the prom to avoid a rental fee for the hall? SOLUTION: Let x be the number of people who attend the prom. 23 -14 29. (6.02 × 10 )(5 × 10 ) SOLUTION: 5 2 - Powered by Cognero eSolutions Manual 30. (7 × 10 ) SOLUTION: So, at least 139 people must attend the prom to avoid the rental fee. Determine whether the graphs of each pair of equations are parallel , perpendicular , or neither. 35. y = −2x + 11 y + 2x = 23 SOLUTION: Write both lines in slope-intercept form. y = −2x + 11 y = −2x + 23 Page 8 Let x be the number of people who attend the prom. So, at least people must attend the prom to avoid 7-6 Growth and139 Decay the rental fee. Determine whether the graphs of each pair of equations are parallel , perpendicular , or neither. 35. y = −2x + 11 y + 2x = 23 SOLUTION: Write both lines in slope-intercept form. y = −2x + 11 y = −2x + 23 The slope of both lines is −2, so the lines are parallel. 36. 3y = 2x + 14 −3x − 2y = 2 SOLUTION: The slopes of the lines are −5 and 5. Because these slopes are not the same and are not negative reciprocals, the lines are neither parallel nor perpendicular. 38. AGES The table shows equivalent ages for horses and humans. Write an equation that relates human age to horse age and find the equivalent horse age for a human who is 16 years old. SOLUTION: According to the table, human age y is always 3 times the horse age x. This can be represented as a direct variation. y = 3x Let y = 16. SOLUTION: Write both lines in slope-intercept form. y= y= x+ x−1 The slopes of the lines are negative reciprocals of each other, so the lines are perpendicular. 37. y = −5x y = 5x − 18 SOLUTION: The slopes of the lines are −5 and 5. Because these slopes are not the same and are not negative reciprocals, the lines are neither parallel nor perpendicular. 38. AGES The table shows equivalent ages for horses and humans. Write an equation that relates human age to horse age and find the equivalent horse age for a human who is 16 years old. SOLUTION: According to the table, human age y is always 3 times the horse age x. This can be represented as a direct variation. y = 3x Let y = 16. The equivalent horse age for a human who is 16 years old is years, or 5 years 4 months. Find the total price of each item. 39. umbrella: $14.00 tax: 5.5% SOLUTION: Find the tax. Add the tax to the original price. $14.00 + $0.77 = $14.77 So, the total price of the umbrella is $14.77. 40. sandals: $29.99 tax: 5.75% SOLUTION: Find the tax. Add the tax to the original price. $29.99 + $1.72 = $31.71 So, the total price of the sandals is $31.71. 41. backpack: $35.00 tax: 7% SOLUTION: Find the tax. The equivalent horse age for a human who is 16 yearsManual old is- Powered years, or 5 years 4 months. eSolutions by Cognero Find the total price of each item. Add the tax to the original price. $35.00 + $2.45 = $37.45 So, the total price of the backpack is $37.45. Page 9 Add the tax to the original price. 7-6 Growth Decay $29.99 +and $1.72 = $31.71 So, the total price of the sandals is $31.71. 41. backpack: $35.00 tax: 7% 44. (2, 2), (−2, −3), (−3, −6) SOLUTION: SOLUTION: Find the tax. Add the tax to the original price. $35.00 + $2.45 = $37.45 So, the total price of the backpack is $37.45. Graph each set of ordered pairs. 42. (3, 0), (0, 1), (−4, −6) SOLUTION: 43. (0, −2), (−1, −6), (3, 4) SOLUTION: 44. (2, 2), (−2, −3), (−3, −6) SOLUTION: eSolutions Manual - Powered by Cognero Page 10 7-7 Geometric Sequences as Exponential Functions Determine whether each sequence is arithmetic, geometric, or neither. Explain. 1. 200, 40, 8, … Since the differences are constant, the sequence is arithmetic. The common difference is 3. 4. 1, −1, 1, −1, … SOLUTION: SOLUTION: Since the ratios are constant, the sequence is geometric. The common ratio is –1. Since the ratios are constant, the sequence is geometric. The common ratio is . Find the next three terms in each geometric sequence. 5. 10, 20, 40, 80, … SOLUTION: 2. 2, 4, 16, … SOLUTION: The ratios are not constant, so the sequence is not geometric. There is no common difference, so the sequence is not arithmetic. Thus, the sequence is neither geometric nor arithmetic. The common ratio is 2. Multiply each term by the common ratio to find the next three terms. 80 × 2 = 160 160 × 2 = 320 320 × 2 = 640 The next three terms of the sequence are 160, 320, and 640. 6. 100, 50, 25, … SOLUTION: Calculate common ratio. 3. −6, −3, 0, 3, … SOLUTION: The ratios are not constant, so the sequence is not geometric. Since the differences are constant, the sequence is arithmetic. The common difference is 3. 4. 1, −1, 1, −1, … SOLUTION: Since the ratios are constant, the sequence is geometric. The common ratio is –1. The common ratio is 0.5. Multiply each term by the common ratio to find the next three terms. 25 × 0.5 = 12.5 12.5 × 0.5 = 6.25 6.25 × 0.5 = 3.125 The next three terms of the sequence are 12.5, 6.25, and 3.125. 7. 4, −1, ,… SOLUTION: Calculate the common ratio. eSolutions Manual - Powered by Cognero Find the next three terms in each geometric Page 1 The common ratio is . Multiply each term by the common ratio to find the next three terms. –63 × –3 = 189 189 × –3 = –567 –567 × –3 = 1701 The next three terms of the sequence are 189, −567, and 1701. 25 × 0.5 = 12.5 12.5 × 0.5 = 6.25 6.25 × 0.5 = 3.125 7-7 Geometric Sequences The next three terms of as theExponential sequence are Functions 12.5, 6.25, and 3.125. 7. 4, −1, Write an equation for the nth term of the geometric sequence, and find the indicated term. 9. Find the fifth term of −6, −24, −96, … ,… SOLUTION: Calculate the common ratio. The common ratio is SOLUTION: Calculate the common ratio. n –1 Use the formula a n = a 1r to write an equation for the nth term of the geometric series. The common ratio is 4, so r = 4. The first term is –6, so . Multiply each term by the common ratio to find the next three terms. × n−1 = × × a 1 = –6. Then, a n = −6 • (4) . = = The next three terms of the sequence are , , The 5th term of the sequence is –1536. and . 10. Find the seventh term of −1, 5, −25, … 8. −7, 21, −63, … SOLUTION: Calculate the common ratio. SOLUTION: Calculate the common ratio. n –1 The common ratio is –3. Multiply each term by the common ratio to find the next three terms. –63 × –3 = 189 189 × –3 = –567 –567 × –3 = 1701 The next three terms of the sequence are 189, −567, and 1701. Write an equation for the nth term of the geometric sequence, and find the indicated term. 9. Find the fifth term of −6, −24, −96, … Use the formula a n = a 1r to write an equation for the nth term of the geometric series. The common ratio is –5, so r = –5. The first term is –1, n−1 so a 1 = –1. Then, a n = −1 • (–5) . The 7th term of the sequence is –15,625. 11. Find the tenth term of 72, 48, 32, … SOLUTION: Calculate the common ratio. SOLUTION: Calculate the common ratio. n –1 Use the formula a n = a 1r to write an equation for the nth term of the geometric series. The eSolutions Manual - Powered by Cognero common ratio is 4, so r = 4. The first term is –6, so n−1 a 1 = –6. Then, a n = −6 • (4) . n –1 Use the formula a n = a 1r to write an equation Page 2 for the nth term of the geometric series. The 7-7 Geometric Sequences as Exponential Functions The 7th term of the sequence is –15,625. 11. Find the tenth term of 72, 48, 32, … The 10th term of the sequence is . 12. Find the ninth term of 112, 84, 63, … SOLUTION: Calculate the common ratio. SOLUTION: Calculate the common ratio. n –1 n –1 Use the formula a n = a 1r to write an equation for the nth term of the geometric series. The Use the formula a n = a 1r to write an equation for the nth term of the geometric series. The common ratio is common ratio is , so r = . The first term is 72, so a 1 = 72. Then, a n = 72 • . . The 9th term of the sequence is 12. Find the ninth term of 112, 84, 63, … n –1 Use the formula a n = a 1r to write an equation for the nth term of the geometric series. The . The first term is 112, so a 1 = 112. Then, a n = 112 • The 9th term of the sequence is eSolutions Manual - Powered by Cognero . . 13. EXPERIMENT In a physics class experiment, Diana drops a ball from a height of 16 feet. Each bounce has 70% the height of the previous bounce. Draw a graph to represent the height of the ball after each bounce. SOLUTION: Calculate the common ratio. , so r = . The first term is 112, so a 1 = 112. Then, a n = 112 • The 10th term of the sequence is common ratio is , so r = . SOLUTION: Make a table of values. Bounce Ball Height 1 0.7(16) = 11.2 2 0.7(11.2) = 7.84 3 0.7(7.84) = 5.488 4 0.7(5.488) = 3.8416 5 0.7(3.8416) = 2.68912 6 0.7(2.68912) = 1.882384 7 0.7(1.882384) =1.3176688 Graph the bounce on the x-axis and the ball height on the y-axis. . 13. EXPERIMENT In a physics class experiment, Diana drops a ball from a height of 16 feet. Each bounce has 70% the height of the previous bounce. Page 3 7-7 Geometric as Exponential The 9th termSequences of the sequence is .Functions 13. EXPERIMENT In a physics class experiment, Diana drops a ball from a height of 16 feet. Each bounce has 70% the height of the previous bounce. Draw a graph to represent the height of the ball after each bounce. SOLUTION: Make a table of values. Bounce Ball Height 1 0.7(16) = 11.2 2 0.7(11.2) = 7.84 3 0.7(7.84) = 5.488 4 0.7(5.488) = 3.8416 5 0.7(3.8416) = 2.68912 6 0.7(2.68912) = 1.882384 7 0.7(1.882384) =1.3176688 Graph the bounce on the x-axis and the ball height on the y-axis. Determine whether each sequence is arithmetic, geometric, or neither. Explain. 14. 4, 1, 2, … SOLUTION: Find the ratios of consecutive terms. The ratios are not constant, so the sequence is not geometric. Find the ratios of the differences of consecutive terms There is no common difference, so the sequence is not arithmetic. Thus, the sequence is neither geometric nor arithmetic. 15. 10, 20, 30, 40 … SOLUTION: Find the ratios of consecutive terms. Determine whether each sequence is arithmetic, geometric, or neither. Explain. 14. 4, 1, 2, … SOLUTION: Find the ratios of consecutive terms. The ratios are not constant, so the sequence is not geometric. Find the differences of consecutive terms. Since the differences are constant, the sequence is arithmetic. The common difference is 10. 16. 4, 20, 100, … The ratios are not constant, so the sequence is not geometric. SOLUTION: Find the ratios of consecutive terms. Find the ratios of the differences of consecutive terms There is no common difference, so the sequence is not arithmetic. eSolutions Manual - Powered by Cognero Thus, the sequence is neither geometric nor arithmetic. Since the ratios are constant, the sequence is geometric. The common ratio is 5. 17. 212, 106, 53, … SOLUTION: Find the ratios of consecutive terms. Page 4 7-7 Geometric Sequences as Exponential Since the ratios are constant, the sequenceFunctions is geometric. The common ratio is 5. 17. 212, 106, 53, … SOLUTION: Find the ratios of consecutive terms. There is no common difference, so the sequence is not arithmetic. Thus, the sequence is neither geometric nor arithmetic. Find the next three terms in each geometric sequence. 20. 2, −10, 50, … SOLUTION: Calculate the common ratio. Since the ratios are constant, the sequence is geometric. The common ratio is . 18. −10, −8, −6, −4 … SOLUTION: Find the ratios of consecutive terms. The ratios are not constant, so the sequence is not geometric. Find the differences of consecutive terms. Since the differences are constant, the sequence is arithmetic. The common difference is 2. The common ratio is –5. Multiply each term by the common ratio to find the next three terms. 50 × –5 = –250 –250 × –5 = 1250 1250 × –5 = –6250 The next three terms of the sequence are −250, 1250, and −6250. 21. 36, 12, 4, … SOLUTION: Calculate the common ratio. The common ratio is . Multiply each term by the common ratio to find the next three terms. 4 × = 19. 5, −10, 20, 40, … SOLUTION: Find the ratios of consecutive terms. × = × = The next three terms of the sequence are , , and . The ratios are not constant, so the sequence is not geometric. Find the differences of consecutive terms. There is no common difference, so the sequence is not arithmetic. Thus, the sequence is neither geometric nor arithmetic. Find the next three terms in each geometric sequence. 20. 2, −10, 50, … SOLUTION: Calculate the common ratio. eSolutions Manual - Powered by Cognero 22. 4, 12, 36, … SOLUTION: Calculate the common ratio. The common ratio is 3. Multiply each term by the common ratio to find the next three terms. 36 × 3 = 108 108 × 3 = 324 324 × 3 = 972 The next three terms of the sequence are 108, 324, and 972. 23. 400, 100, 25, … SOLUTION: Calculate the common ratio. Page 5 36 × 3 = 108 108 × 3 = 324 324 × 3 = 972 7-7 Geometric Sequences The next three terms of as theExponential sequence are Functions 108, 324, and 972. 23. 400, 100, 25, … –294 × 7 = –2058 –2058 × 7 = –14,406 –14,406 × 7 = –100,842 The next three terms of the sequence are −2058, −14,406, and −100,842. 25. 1024, −128, 16, … SOLUTION: Calculate the common ratio. SOLUTION: Calculate the common ratio. The common ratio is . Multiply each term by the common ratio to find the next three terms. The common ratio is 25 × = common ratio to find the next three terms. × = 16 × = –2 × = –2 × = × = The next three terms of the sequence are and , . Multiply each term by the , . The next three terms of the sequence are −2, 24. −6, −42, −294, … SOLUTION: Calculate the common ratio. , and . 26. The first term of a geometric series is 1 and the common ratio is 9. What is the 8th term of the sequence? The common ratio is 7. Multiply each term by the common ratio to find the next three terms. –294 × 7 = –2058 –2058 × 7 = –14,406 –14,406 × 7 = –100,842 The next three terms of the sequence are −2058, −14,406, and −100,842. SOLUTION: The 8th term of the sequence is 4,782,969. 25. 1024, −128, 16, … SOLUTION: Calculate the common ratio. 27. The first term of a geometric series is 2 and the common ratio is 4. What is the 14th term of the sequence? SOLUTION: The common ratio is . Multiply each term by the common ratio to find the next three terms. 16 × = –2 The 14th term of the sequence is 134,217,728. –2 × = × 28. What is the 15th term of the geometric sequence −9, 27, −81, …? = The next three terms of the sequence are −2, eSolutions Manual - Powered by Cognero . , and SOLUTION: Calculate the common ratio. Page 6 7-7 Geometric Sequences as Exponential Functions The 14th term of the sequence is 134,217,728. 28. What is the 15th term of the geometric sequence −9, 27, −81, …? SOLUTION: Calculate the common ratio. The common ratio is –3. The 15th term of the sequence is –43,046,721. The 10th term of the sequence is –1,572,864. 30. PENDULUM A pendulum swings with an arc length of 24 feet on its first swing. On each swing after the first swing, the arc length is 60% of the length of the previous swing. Draw a graph that represents the arc length after each swing. SOLUTION: Make a table of values. Swing Arc Length 1 24 2 0.6(24) = 14.4 3 0.6(14.4) = 8.64 4 0.6(8.64) = 5.184 5 0.6(5.184) = 3.1104 6 0.6(3.1104) = 1.86624 Graph the swing on the x-axis and the arc length on the y-axis. 29. What is the 10th term of the geometric sequence 6, −24, 96, …? SOLUTION: Calculate the common ratio. The common ratio is –4. The 10th term of the sequence is –1,572,864. 30. PENDULUM A pendulum swings with an arc length of 24 feet on its first swing. On each swing after the first swing, the arc length is 60% of the length of the previous swing. Draw a graph that represents the arc length after each swing. SOLUTION: Make a table of values. Swing Arc Length 1 24 2 0.6(24) = 14.4 3 0.6(14.4) = 8.64 4 0.6(8.64) = 5.184 5 0.6(5.184) = 3.1104 6 0.6(3.1104) = 1.86624 Graph the swing on the x-axis and the arc length on the y-axis. eSolutions Manual - Powered by Cognero 31. Find the eighth term of a geometric sequence for which a 3 = 81 and r = 3. SOLUTION: Because a 3 = 81, the third term in the sequence is 81. To find the eighth term of the sequence, you need to find the 1st term of the sequence. Use the nth term of a Geometric Sequence formula. Then a 1 is 9. Use a 1 to find the eighth term of the sequence. Page 7 7-7 Geometric Sequences as Exponential Functions The eighth term of the geometric sequence is 19,683. 31. Find the eighth term of a geometric sequence for which a 3 = 81 and r = 3. SOLUTION: Because a 3 = 81, the third term in the sequence is 81. To find the eighth term of the sequence, you need to find the 1st term of the sequence. Use the nth term of a Geometric Sequence formula. 32. CCSS REASONING At an online mapping site, Mr. Mosley notices that when he clicks a spot on the map, the map zooms in on that spot. The magnification increases by 20% each time. a. Write a formula for the nth term of the geometric sequence that represents the magnification of each zoom level. (Hint: The common ratio is not just 0.2.) b. What is the fourth term of this sequence? What does it represent? SOLUTION: a. Because the magnification increases by 20% with each click, the total magnification after each click is 120%. The common ratio is 1.2. To find the nth term of the geometric sequence that represents the magnification of each zoom level, use the formula n Then a 1 is 9. Use a 1 to find the eighth term of the sequence. The eighth term of the geometric sequence is 19,683. 32. CCSS REASONING At an online mapping site, Mr. Mosley notices that when he clicks a spot on the map, the map zooms in on that spot. The magnification increases by 20% each time. a. Write a formula for the nth term of the geometric sequence that represents the magnification of each zoom level. (Hint: The common ratio is not just 0.2.) b. What is the fourth term of this sequence? What does it represent? SOLUTION: a. Because the magnification increases by 20% with each click, the total magnification after each click is 120%. The common ratio is 1.2. To find the nth term of the geometric sequence that represents the magnification of each zoom level, use the formula n 1.2 . b. The fourth term of the sequence is 2.0736. It the fourth click. So, the map will be magnified at approximately 207% of the original size after the fourth click. represents magnification eSolutions Manual -the Powered by Cognero after 1.2 . b. The fourth term of the sequence is 2.0736. It represents the magnification after the fourth click. So, the map will be magnified at approximately 207% of the original size after the fourth click. 33. ALLOWANCE Danielle’s parents have offered her two different options to earn her allowance for a 9week period over the summer. She can either get paid $30 each week or $1 the first week, $2 for the second week, $4 for the third week, and so on. a. Does the second option form a geometric sequence? Explain. b. Which option should Danielle choose? Explain. SOLUTION: a. Calculate the common ratio. There is a common ratio of 2. So, the second option does form a geometric sequence. b. Calculate how much Danielle would earn with each option. Option 1 9(30) = 270 Option 2 1 + 2 + 4 + 8+ 16 + 32 + 64 + 128 + 256 or 511 In nine weeks, Danielle would earn $270 with the first option and $511 with the second option. So, she should choose the second option. Page 8 34. SIERPINSKI’S TRIANGLE Consider the inscribed equilateral triangles shown. The perimeter of each triangle is one half of the perimeter of the The fourth term of the sequence is 2.0736. It represents the magnification after the fourth click. 7-7 Geometric as Exponential Functions So, the map Sequences will be magnified at approximately 207% of the original size after the fourth click. 33. ALLOWANCE Danielle’s parents have offered her two different options to earn her allowance for a 9week period over the summer. She can either get paid $30 each week or $1 the first week, $2 for the second week, $4 for the third week, and so on. a. Does the second option form a geometric sequence? Explain. b. Which option should Danielle choose? Explain. Option 2 1 + 2 + 4 + 8+ 16 + 32 + 64 + 128 + 256 or 511 In nine weeks, Danielle would earn $270 with the first option and $511 with the second option. So, she should choose the second option. 34. SIERPINSKI’S TRIANGLE Consider the inscribed equilateral triangles shown. The perimeter of each triangle is one half of the perimeter of the next larger triangle. What is the perimeter of the smallest triangle? SOLUTION: a. Calculate the common ratio. SOLUTION: This is a geometric sequence. The first term is 3(40) There is a common ratio of 2. So, the second option does form a geometric sequence. b. Calculate how much Danielle would earn with each option. Option 1 9(30) = 270 Option 2 1 + 2 + 4 + 8+ 16 + 32 + 64 + 128 + 256 or 511 In nine weeks, Danielle would earn $270 with the first option and $511 with the second option. So, she should choose the second option. 34. SIERPINSKI’S TRIANGLE Consider the inscribed equilateral triangles shown. The perimeter of each triangle is one half of the perimeter of the next larger triangle. What is the perimeter of the smallest triangle? or 120 and the common ratio is . To find the perimeter of the smallest triangle, find the 5th term of the sequence. The perimeter of the smallest triangle is 7.5 centimeters. 35. If the second term of a geometric sequence is 3 and the third term is 1, find the first and fourth terms of the sequence. SOLUTION: Divide the 3rd term by the 2nd term to find the common ratio. The common ratio is . Substitute 2 for n and for r to find the first term. SOLUTION: This is a geometric sequence. The first term is 3(40) or 120 and the common ratio is . To find the perimeter of the smallest triangle, find the 5th term of the sequence. The first term is 9. Find the 4th term. eSolutions Manual - Powered by Cognero The perimeter of the smallest triangle is 7.5 centimeters. Page 9 7-7 Geometric Sequences as Exponential Functions The perimeter of the smallest triangle is 7.5 centimeters. 35. If the second term of a geometric sequence is 3 and the third term is 1, find the first and fourth terms of the sequence. SOLUTION: Divide the 3rd term by the 2nd term to find the common ratio. The common ratio is . Substitute 2 for n and The fourth term is . 36. If the third term of a geometric sequence is −12 and the fourth term is 24, find the first and fifth terms of the sequence. SOLUTION: Divide the 4th term by the 3rd term to find the common ratio. for r to find the first term. The common ratio is or –2. Substitute 3 for n and –2 for r to find the first term. The first term is 9. Find the 4th term. The first term is –3. Find the fifth term. The fifth term is 48. The fourth term is . 36. If the third term of a geometric sequence is −12 and the fourth term is 24, find the first and fifth terms of the sequence. SOLUTION: Divide the 4th term by the 3rd term to find the common ratio. The common ratio is or –2. Substitute 3 for n and –2 for r to find the first term. The first term is –3. Find the fifth term. eSolutions Manual - Powered by Cognero 37. EARTHQUAKES The Richter scale is used to measure the force of an earthquake. The table shows the increase in magnitude for the values on the Richter scale. Richter Increase in Rate of Number Magnitude Change (x) (y) (slope) 1 1 − 2 10 9 3 100 4 1000 5 10,000 a. Copy and complete the table. Remember that the rate of change is the change in y divided by the change in x. b. Plot the ordered pairs (Richter number, increase in magnitude). c. Describe the graph that you made of the Richter scale data. Is the rate of change between any two points the same? d. Write an exponential equation that represents the Richter scale. SOLUTION: a. Richter Increase in Number Magnitude (x) (y) Rate of Change Page 10 (slope) points the same? d. Write an exponential equation that represents the Richter scale. 7-7 Geometric Sequences as Exponential Functions SOLUTION: a. Richter Increase in Rate of Change Number Magnitude (slope) (x) (y) 1 1 − 2 10 9 3 100 4 1000 5 10,000 geometric sequence. The common difference is 0, making it an arithmetic sequence as well. This can be done for any value n. n, n, n, ... is arithmetic and geometric. 39. CCSS CRITIQUE Haro and Matthew are finding the ninth term of the geometric sequence −5, 10, −20, … . Is either of them correct? Explain your reasoning. b. Graph the Richter number on the x-axis and the increase in magnitude on the y-axis. SOLUTION: The common ratio of the sequence is –2. c. The graph appears to be exponential. The rate of change between any two points does not match any others. d. Calculate the common ratio. There is a common ratio of 10, so this is situation can be modeled by a geometric sequence. The exponential equation that represents the Richter scale x−1 is y = 1 • (10) . 38. CHALLENGE Write a sequence that is both geometric and arithmetic. Explain your answer. SOLUTION: The sequence 1, 1, 1, 1, … is both geometric and arithmetic. The common ratio is 1 making it a geometric sequence. The common difference is 0, making it an arithmetic sequence as well. This can be done for any value n. n, n, n, ... is arithmetic and geometric. 39. CCSS CRITIQUE Haro and Matthew are finding eSolutions Manualterm - Powered Cognero the ninth of thebygeometric sequence −5, 10, −20, … . Is either of them correct? Explain your reasoning. The ninth term of the sequence is –1280. Neither Haro nor Matthew is correct. Haro calculated the exponent incorrectly. Matthew did not enclose the common ratio in parentheses which caused him to make a sign error. 40. REASONING Write a sequence of numbers that form a pattern but are neither arithmetic nor geometric. Explain the pattern. SOLUTION: The sequence 1, 4, 9, 16, 25, 36, … has a pattern, because each number is a perfect square. However, there is no common ratio which means it is not a geometric sequence. There is no common difference, which means it is not an arithmetic sequence. 41. WRITING IN MATH How are graphs of geometric sequences and exponential functions similar? different? SOLUTION: Sample answer: When graphed, the terms of a geometric sequence lie on a curve that can be Page 11 represented by an exponential function. They are different in that the domain of a geometric sequence 41. WRITING IN MATH How are graphs of geometric sequences and exponential functions similar? different? 7-7 Geometric Sequences as Exponential Functions SOLUTION: Sample answer: When graphed, the terms of a geometric sequence lie on a curve that can be represented by an exponential function. They are different in that the domain of a geometric sequence is the set of natural numbers, while the domain of an exponential function is all real numbers. Thus, geometric sequences are discrete, while exponential functions are continuous. For example, the geometric sequence 1, 2, 4, 8, ... has a = 1, r = 2, and the nth term given by an = 1 (2)n–1, where n is any positive integer. The graph of the function an = 1(2)n–1 would be as follows. Even though, the two graphs contain many of the same points, the graph of the geometric sequence is discrete while the graph of the exponential function is continuous. 42. WRITING IN MATH Summarize how to find a specific term of a geometric sequence. SOLUTION: Sample answer: First, find the common ratio. Then, n−1 use the formula a n = a 1 • r . Substitute the first term of the sequence for a 1 and the common ratio for r. Let n be equal to the number of the term you are finding. Then, solve the equation. 43. Find the eleventh term of the sequence 3, −6, 12, −24, … A 1024 B 3072 C 33 D −6144 SOLUTION: Calculate the common ratio. The common ratio is –2. The exponential function given by y = 1(2)x–1will generate similar values but the domain of x is all real numbers. The graph of this function is below. The eleventh term of the sequence is 3072. Choice B is the correct answer. 44. What is the total amount of the investment shown in the table below if interest is compounded monthly? F $613.56 G $616.00 H $616.56 J $718.75 Even though, the two graphs contain many of the same points, the graph of the geometric sequence is discrete while the graph of the exponential function is continuous. 42. WRITING IN MATH Summarize how to find a specific term of a geometric sequence. eSolutions Manual - Powered by Cognero SOLUTION: Sample answer: First, find the common ratio. Then, SOLUTION: Use the equation for compound interest, with P = 500, r = 0.0525, n = 12, and t = 4. Page 12 The eleventhSequences term of theassequence is 3072. Choice B 7-7 Geometric Exponential Functions is the correct answer. 44. What is the total amount of the investment shown in the table below if interest is compounded monthly? F $613.56 G $616.00 H $616.56 J $718.75 SOLUTION: Use the equation for compound interest, with P = 500, r = 0.0525, n = 12, and t = 4. The total amount of the investment is about $616.56. Choice H is the correct answer. 45. SHORT RESPONSE Gloria has $6.50 in quarters and dimes. If she has 35 coins in total, how many of each coin does she have? SOLUTION: Let q = the number of quarters and let d = the number of dimes. Then, q + d = 35 and 0.25q + 0.10d = 6.50. Solve the first equation for d. Substitute 35 – q for d in the second equation and solve for q. Use the value of q and either equation to find the value of d. The total amount of the investment is about $616.56. Choice H is the correct answer. 45. SHORT RESPONSE Gloria has $6.50 in quarters and dimes. If she has 35 coins in total, how many of each coin does she have? SOLUTION: Let q = the number of quarters and let d = the number of dimes. Then, q + d = 35 and 0.25q + 0.10d = 6.50. Solve the first equation for d. Substitute 35 – q for d in the second equation and solve for q. Gloria has 15 dimes and 20 quarters. 46. What are the domain and range of the function y = 4 x (3 ) – 2? A D = {all real numbers}, R = {y | y > –2} B D = {all real numbers}, R = {y | y > 0} C D = {all integers}, R = {y | y > –2} D D = {all integers}, R = {y | y > 0} SOLUTION: Use a graphing calculator to graph the function Y1= 4(3x) – 2. Use the value of q and either equation to find the value of d. eSolutions Manual - Powered by Cognero Gloria has 15 dimes and 20 quarters. 46. What are the domain and range of the function y = 4 The graph is continuous from left to right and increases from –2 to infinity.Thus, the domain is all real numbers and the range is all real numbers Page 13 greater than –2. Therefore, the correct choice is A. Find the next three terms in each geometric 54 × 3 = 162 162 × 3 = 486 486 × 3 = 1458 The next three terms of the sequence are 162, 486, and 1458 value of d. 7-7 Geometric Sequences as Exponential Functions Gloria has 15 dimes and 20 quarters. 46. What are the domain and range of the function y = 4 48. −5, −10, −20, −40, … x (3 ) – 2? SOLUTION: Calculate the common ratio. A D = {all real numbers}, R = {y | y > –2} B D = {all real numbers}, R = {y | y > 0} C D = {all integers}, R = {y | y > –2} D D = {all integers}, R = {y | y > 0} The common ratio is 2. Multiply each term by the common ratio to find the next three terms. –40 × 2 = –80 –80 × 2 = –160 –160 × 2 = –320 The next three terms of the sequence are −80, −160, and −320. SOLUTION: Use a graphing calculator to graph the function Y1= 4(3x) – 2. 49. SOLUTION: Calculate the common ratio. The graph is continuous from left to right and increases from –2 to infinity.Thus, the domain is all real numbers and the range is all real numbers greater than –2. Therefore, the correct choice is A. Find the next three terms in each geometric sequence. 47. 2, 6, 18, 54, … SOLUTION: Calculate the common ratio. The common ratio is common ratio to find the next three terms. × = × = × The common ratio is 3. Multiply each term by the common ratio to find the next three terms. 54 × 3 = 162 162 × 3 = 486 486 × 3 = 1458 The next three terms of the sequence are 162, 486, and 1458 48. −5, −10, −20, −40, … . Multiply each term by the = The next three terms of the sequence are and , , . 50. −3, 1.5, −0.75, 0.375, … SOLUTION: Calculate the common ratio. SOLUTION: Calculate the common ratio. The common ratio is 2. Multiply each term by the common ratio to find the next three terms. –40 × 2 = –80 eSolutions Manual - Powered by Cognero –80 × 2 = –160 –160 × 2 = –320 The next three terms of the sequence are −80, −160, The common ratio is –0.5. Multiply each term by the common ratio to find the next three terms. 0.375 × –0.5 = –0.1875 –0.1875 × –0.5 = 0.09375 0.09375 × –0.5 = –0.046875 Page 14 The next three terms of the sequence are −0.1875, 0.09375, and −0.046875. The next three terms of the sequence are , 13.5 × 1.5 = 20.25 20.25 × 1.5 = 30.375 30.375 × 1.5 = 45.5625 The next three terms of the sequence are 20.25, 30.375, and 45.5625. , 7-7 Geometric and . Sequences as Exponential Functions Graph each function. Find the y-intercept and state the domain and range. 50. −3, 1.5, −0.75, 0.375, … SOLUTION: Calculate the common ratio. 53. The common ratio is –0.5. Multiply each term by the common ratio to find the next three terms. 0.375 × –0.5 = –0.1875 –0.1875 × –0.5 = 0.09375 0.09375 × –0.5 = –0.046875 The next three terms of the sequence are −0.1875, 0.09375, and −0.046875. 51. 1, 0.6, 0.36, 0.216, … SOLUTION: Calculate the common ratio. SOLUTION: x y –2 11 –1 –1 0 –4 1 2 The common ratio is 0.6. Multiply each term by the common ratio to find the next three terms. 0.216 × 0.6 = 0.1296 0.1296 × 0.6 = 0.7776 0.7776 × 0.6 = 0.046656 The next three terms of the sequence are 0.1296, 0.07776, and 0.046656. 52. 4, 6, 9, 13.5, … SOLUTION: Calculate the common ratio. The function crosses the y-axis at –4. The domain is all real numbers, and the range is all real numbers greater than –5. The common ratio is 1.5. Multiply each term by the common ratio to find the next three terms. 13.5 × 1.5 = 20.25 20.25 × 1.5 = 30.375 30.375 × 1.5 = 45.5625 The next three terms of the sequence are 20.25, 30.375, and 45.5625. Graph each function. Find the y-intercept and state the domain and range. 53. SOLUTION: x SOLUTION: x x (4) –2 –1 –2 (4) –1 (4) y = = 0 (4) = 1 0 2 1 1 (4) = 4 8 2 (4) = 16 2 32 y eSolutions Manual - Powered by Cognero –2 x 54. y = 2(4) Page 15 11 The function crosses the y-axis at –4. The domain is all real numbers, and the range is all real numbers 7-7 Geometric as Exponential Functions greater thanSequences –5. The function crosses the y-axis at 2. The domain is all real numbers, and the range is all real numbers greater than 0. x 54. y = 2(4) 55. SOLUTION: x x (4) –2 –1 –2 (4) –1 (4) y = –2 = –1 0 (4) = 1 2 (4) = 4 1 8 2 32 0 1 2 (4) = 16 SOLUTION: x x (3) –2 (3) –1 (3) y = = 0 0 (3) = 1 1 (3) = 3 2 (3) = 9 1 2 The function crosses the y-axis at 2. The domain is all real numbers, and the range is all real numbers greater than 0. 55. The function crosses the y-axis at SOLUTION: x x (3) –2 –1 –2 (3) –1 (3) y = = 0 0 (3) = 1 1 (3) = 3 2 (3) = 9 1 2 . The domain is all real numbers, and the range is all real numbers greater than 0. 56. LANDSCAPING A blue spruce grows an average of 6 inches per year. A hemlock grows an average of 4 inches per year. If a blue spruce is 4 feet tall and a hemlock is 6 feet tall, write a system of equations to represent their growth. Find and interpret the solution in the context of the situation. SOLUTION: Let x = the number of years the tree grows and let y = the height of the tree after x years. So, y = 48 + 6x, and y = 72 + 4x. Substitute 48 + 6x for y in the second equation. Use the value for x and either equation to find the value for y. eSolutions Manual - Powered by Cognero Page 16 The function crosses the y-axis at . The domain is all real numbers, and theas range is all real numbers 7-7 Geometric Sequences Exponential Functions greater than 0. 56. LANDSCAPING A blue spruce grows an average of 6 inches per year. A hemlock grows an average of 4 inches per year. If a blue spruce is 4 feet tall and a hemlock is 6 feet tall, write a system of equations to represent their growth. Find and interpret the solution in the context of the situation. SOLUTION: Let x = the number of years the tree grows and let y = the height of the tree after x years. So, y = 48 + 6x, and y = 72 + 4x. Substitute 48 + 6x for y in the second equation. Mr. Hayashi should start with at least $3747. Write an equation in slope-intercept form of the line with the given slope and y-intercept. 58. slope: 4, y-intercept: 2 SOLUTION: The slope-intercept form of a line is y = mx + b, where m = the slope and b = the y-intercept. So, y = 4x + 2. 59. slope: −3, y-intercept: SOLUTION: The slope-intercept form of a line is y = mx + b, where m = the slope and b = the y-intercept. So, . Use the value for x and either equation to find the value for y. 60. slope: The solution is (12, 120). This means that in 12 years the trees will be the same height, 120 inches or 10 feet. 57. MONEY City Bank requires a minimum balance of $1500 to maintain free checking services. If Mr. Hayashi is going to write checks for the amounts listed in the table, how much money should he start with in order to have free checking? , y-intercept: −5 SOLUTION: The slope-intercept form of a line is y = mx + b, where m = the slope and b = the y-intercept. So, . 61. slope: , y-intercept: −9 SOLUTION: The slope-intercept form of a line is y = mx + b, where m = the slope and b = the y-intercept. So, . SOLUTION: Let x = the amount Mr. Hayashi should start with in order to have free checking. Then, x – (1300 + 947) ≥ 1500. 62. slope: , y-intercept: SOLUTION: The slope-intercept form of a line is y = mx + b, where m = the slope and b = the y-intercept. So, . Mr. Hayashi should start with at least $3747. Write an equation in slope-intercept form of the line with the given slope and y-intercept. 58. slope: 4, y-intercept: 2 SOLUTION: The slope-intercept form of a line is y = mx + b, eSolutions Manual - Powered by Cognero where m = the slope and b = the y-intercept. So, y = 4x + 2. 63. slope: −6, y-intercept: −7 SOLUTION: The slope-intercept form of a line is y = mx + b, where m = the slope and b = the y-intercept. So, y = −6x − 7. Page 17 Simplify each expression. If not possible, write simplified. SOLUTION: The slope-intercept form of a line is y = mx + b, where m = the slope and b = the y-intercept. So, can be simplified. 7-7 Geometric Sequences as Exponential Functions . 63. slope: −6, y-intercept: −7 SOLUTION: The slope-intercept form of a line is y = mx + b, where m = the slope and b = the y-intercept. So, y = −6x − 7. Simplify each expression. If not possible, write simplified. 64. 3u + 10u SOLUTION: Since 3u and 10u are like terms, this expression can be simplified. 68. 13(5 + 4a) SOLUTION: Since this expression has indicated multiplication, the Distributive Property can be used to simplify the expression. 69. (4t – 6)16 SOLUTION: Since this expression contains indicated multiplication, the Distributive Property can be used to simplify the expression. 65. 5a – 2 + 6a SOLUTION: Since 5a and 6a are like terms, the expression can be simplified. 2 66. 6m – 8m SOLUTION: 2 6m and 8m are not like terms. Therefore, this expression is already simplified. 2 2 67. 4w + w + 15w SOLUTION: 2 2 Since 4w and 15w are like terms, this expression can be simplified. 68. 13(5 + 4a) SOLUTION: Since this expression has indicated multiplication, the Distributive Property can be used to simplify the expression. eSolutions Manual - Powered by Cognero 69. (4t – 6)16 SOLUTION: Page 18 7-8 Recursive Formulas Find the first five terms of each sequence. 1. SOLUTION: Use a 1 = 16 and the recursive formula to find the The first five terms are 16, 13, 10, 7, and 4. 2. SOLUTION: Use a 1 = –5 and the recursive formula to find the next four terms. next four terms. The first five terms are 16, 13, 10, 7, and 4. 2. SOLUTION: Use a 1 = –5 and the recursive formula to find the next four terms. The first five terms are –5, –10, –30, –110, and – 430. Write a recursive formula for each sequence. 3. 1, 6, 11, 16, ... SOLUTION: Subtract each term from the term that follows it. 6 – 1 = 5; 11 – 6 = 5, 16 – 11 = 5 There is a common difference of 5. The sequence is arithmetic. Use the formula for an arithmetic sequence. The first term a 1 is 1, and n ≥ 2. A recursive formula for the sequence 1, 6, 11, 16, … is a 1 = 1, a n = a n – 1 + 5, n ≥ 2. 4. 4, 12, 36, 108, ... eSolutions Manual - Powered by Cognero The first five terms are –5, –10, –30, –110, and – 430. SOLUTION: Subtract each term from the term that follows it. 12 – 4 = 8; 36 – 12 = 24, 108 – 36 = 72 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. Page 1 = 3 ; =3; =3 There is a common ratio of 3. The sequence is geometric. The first term a 1 is 1, and n ≥ 2. The first term a 1 is 4, and n ≥ 2. A recursive formula A recursive formula for the sequence 1, 6, 11, 16, … 7-8 Recursive is a 1 = 1, a nFormulas = a n – 1 + 5, n ≥ 2. 4. 4, 12, 36, 108, ... SOLUTION: Subtract each term from the term that follows it. 12 – 4 = 8; 36 – 12 = 24, 108 – 36 = 72 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. = 3 ; =3; =3 There is a common ratio of 3. The sequence is geometric. Use the formula for a geometric sequence. for the sequence 4, 12, 36, 108, … is a 1 = 4, a n = 3a n – 1, n ≥ 2. 5. BALL A ball is dropped from an initial height of 10 feet. The maximum heights the ball reaches on the first three bounces are shown. a. Write a recursive formula for the sequence. b. Write an explicit formula for the sequence. SOLUTION: a. The sequence of heights is 10, 6, 3.6, and 2.16. Subtract each term from the term that follows it. 6 – 10 = 4; 3.6 – 6 = -2.4, 2.16 – 3.6 = -1.44 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. ; ; There is a common ratio of 0.6. The sequence is geometric. Use the formula for a geometric sequence. The first term a 1 is 4, and n ≥ 2. A recursive formula for the sequence 4, 12, 36, 108, … is a 1 = 4, a n = 3a n – 1, n ≥ 2. 5. BALL A ball is dropped from an initial height of 10 feet. The maximum heights the ball reaches on the first three bounces are shown. a. Write a recursive formula for the sequence. b. Write an explicit formula for the sequence. The first term a 1 is 10, and n ≥ 2. A recursive formula for the sequence 10, 6, 3.6, and 2.16, … is a 1 = 10, a n = 0.6a n – 1, n ≥ 2. SOLUTION: a. The sequence of heights is 10, 6, 3.6, and 2.16. Subtract each term from the term that follows it. 6 – 10 = 4; 3.6 – 6 = -2.4, 2.16 – 3.6 = -1.44 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. ; ; There is a common ratio of 0.6. The sequence is geometric. Use the formula for a geometric sequence. The first term a 1 is 10, and n ≥ 2. A recursive formula for the sequence 10, 6, 3.6, and 2.16, … is eSolutions Cognero a 1 =Manual 10, a n- =Powered 0.6a n by – 1, n ≥ 2. b. Use the formula for the nth terms of a geometric b. Use the formula for the nth terms of a geometric sequence. The explicit formula is a n = 10(0.6) n– 1 . For each recursive formula, write an explicit formula. For each explicit formula, write a recursive formula. 6. SOLUTION: The common difference is 16. Use the formula for the nth terms of an arithmetic sequence. Page 2 The first term a 1 is 13, and n ≥ 2. A recursive The explicit formula is a = 10(0.6) 7-8 Recursive Formulas n n– 1 formula for the explicit formula an = 5n + 8 is a 1 = . 13, a n = a n – 1 + 5, n ≥ 2. For each recursive formula, write an explicit formula. For each explicit formula, write a recursive formula. 8. a n = 15(2)n – 1 SOLUTION: Write out the first 4 terms. 15, 30, 60, 120 Subtract each term from the term that follows it. 30 – 15 = 30; 60 – 30 = 30, 120 – 60 = 60 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. There is a common ratio of 2. The sequence is geometric. Use the formula for a geometric sequence. 6. SOLUTION: The common difference is 16. Use the formula for the nth terms of an arithmetic sequence. The explicit formula is a n = 16n – 12. The first term a 1 is 15, and n ≥ 2. A recursive 7. a n = 5n + 8 SOLUTION: Write out the first 4 terms. 13, 18, 23, 28 Subtract each term from the term that follows it. 18 – 13 = 5; 23 – 18 = 5, 28 – 23 = 5 There is a common difference of 5. The sequence is arithmetic. Use the formula for an arithmetic sequence. n 1 formula for the explicit formula an = 15(2) - is a 1 = 15, a n = 2a n – 1, n ≥ 2. 9. SOLUTION: The common ratio is 4. Use the formula for the nth terms of a geometric sequence. The first term a 1 is 13, and n ≥ 2. A recursive n– 1 The explicit formula is a n = 22(4) . formula for the explicit formula an = 5n + 8 is a 1 = 13, a n = a n – 1 + 5, n ≥ 2. 8. a n = 15(2)n – 1 SOLUTION: Write out the first 4 terms. 15, 30, 60, 120 Subtract each term from the term that follows it. 30 – 15 = 30; 60 – 30 = 30, 120 – 60 = 60 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. There is a common ratio of 2. The sequence is geometric. Use the formula for a geometric sequence. Find the first five terms of each sequence. 10. SOLUTION: Use a 1 = 23 and the recursive formula to find the next four terms. eSolutions Manual - Powered by Cognero The first term a 1 is 15, and n ≥ 2. A recursive n 1 formula for the explicit formula an = 15(2) - is a 1 Page 3 7-8 Recursive Formulas n– 1 The explicit formula is a n = 22(4) . Find the first five terms of each sequence. The first five terms are 23, 30, 37, 44, and 51. 11. 10. SOLUTION: Use a 1 = 48 and the recursive formula to find the SOLUTION: Use a 1 = 23 and the recursive formula to find the next four terms. next four terms. The first five terms are 48, –16, 16, 0, and 8. The first five terms are 23, 30, 37, 44, and 51. 12. SOLUTION: Use a 1 = 8 and the recursive formula to find the next four terms. 11. SOLUTION: Use a 1 = 48 and the recursive formula to find the next four terms. The first five terms are 8, 20, 50, 125, and 312.5.Page 4 eSolutions Manual - Powered by Cognero The first five terms are 48, –16, 16, 0, and 8. 13. The first five terms are 8, 20, 50, 125, and 312.5. 7-8 Recursive Formulas The first five terms are 48, –16, 16, 0, and 8. 13. 12. SOLUTION: Use a 1 = 8 and the recursive formula to find the next four terms. SOLUTION: Use a 1 = 12 and the recursive formula to find the The first five terms are 8, 20, 50, 125, and 312.5. The first five terms are 12, 15, 24, 51, and 132. 13. next four terms. 14. SOLUTION: Use a 1 = 12 and the recursive formula to find the next four terms. SOLUTION: Use a 1 = 13 and the recursive formula to find the next four terms. eSolutions Manual - Powered by Cognero The first five terms are 12, 15, 24, 51, and 132. 14. The first five terms are 13, –29, 55, –113, and 223. Page 5 15. 7-8 Recursive Formulas The first five terms are 12, 15, 24, 51, and 132. 14. The first five terms are 13, –29, 55, –113, and 223. 15. SOLUTION: Use a 1 = 13 and the recursive formula to find the next four terms. SOLUTION: Use and the recursive formula to find the next four terms. The first five terms are 13, –29, 55, –113, and 223. 15. SOLUTION: Use and the recursive formula to find the next four terms. The first five terms are . Write a recursive formula for each sequence. 16. SOLUTION: Subtract each term from the term that follows it. –1 – 12 = –13; –14 – (–1) = –13, –27 – (–14) = –13 There is a common difference of –13. The sequence is arithmetic. Use the formula for an arithmetic sequence. The first term a 1 is 12, and n ≥ 2. A recursive formula for the sequence 12, –1, –14, –27, … is a 1 = 12, a n = a n– 1 – 13, n ≥ 2. eSolutions Manual - Powered by Cognero Page 6 17. 27, 41, 55, 69, ... SOLUTION: The first term a 1 is 2, and n ≥ 2. A recursive formula for the sequence 2, 11, 20, 29, … is a 1 = 2, a n = a n – 7-8 Recursive Formulas The first five terms are 1 + 9, n ≥ . Write a recursive formula for each sequence. 19. 100, 80, 64, 51.2, ... 16. SOLUTION: Subtract each term from the term that follows it. 80 – 100 = –20; 64 – 80 = –16, 51.2 – 64 = 12.8 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. There is a common ratio of 0.8. The sequence is geometric. Use the formula for a geometric sequence. SOLUTION: Subtract each term from the term that follows it. –1 – 12 = –13; –14 – (–1) = –13, –27 – (–14) = –13 There is a common difference of –13. The sequence is arithmetic. Use the formula for an arithmetic sequence. The first term a 1 is 12, and n ≥ 2. A recursive formula for the sequence 12, –1, –14, –27, … is a 1 = 12, a n = a n– 1 – 13, n ≥ 2. The first term a 1 is 100, and n ≥ 2. A recursive formula for the sequence 100, 80, 64, 51.2, … is a 1 17. 27, 41, 55, 69, ... SOLUTION: Subtract each term from the term that follows it. 41 – 27 = 14; 55 – 41 = 14, 69 – 55 = 14 There is a common difference of 14. The sequence is arithmetic. Use the formula for an arithmetic sequence. The first term a 1 is 27, and n ≥ 2. A recursive formula for the sequence 27, 41, 55, 69, … is a 1 = 27, a n = a n – 1 + 14, n ≥ 2. 2. = 100, a n = 0.8a n – 1, n ≥ 2. 20. SOLUTION: Subtract each term from the term that follows it. –60 – 40 = –100; 90 – (–60) = 30, –135 – 90 = –225 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. There is a common ratio of –1.5. The sequence is geometric. Use the formula for a geometric sequence. 18. 2, 11, 20, 29, ... SOLUTION: Subtract each term from the term that follows it. 11 – 2 = 9; 20 – 11 = 9, 29 – 20 = 9 There is a common difference of 9. The sequence is arithmetic. Use the formula for an arithmetic sequence. The first term a 1 is 2, and n ≥ 2. A recursive formula for the sequence 2, 11, 20, 29, … is a 1 = 2, a n = a n – 1 + 9, n ≥ 2. 19. 100, 80, 64, 51.2, ... eSolutions Manual - Powered by Cognero SOLUTION: Subtract each term from the term that follows it. 80 – 100 = –20; 64 – 80 = –16, 51.2 – 64 = 12.8 The first term a 1 is 40, and n ≥ 2. A recursive formula for the sequence 40, –60, 90, –135, … is a 1 = 40, a n = –1.5a n– 1, n ≥ 2. 21. 81, 27, 9, 3, ... SOLUTION: Subtract each term from the term that follows it. 27 – 81 = –54; 9 – 27 = –18, 3 – 9 = –6 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. There is a common ratio of . The sequence is Page 7 geometric. Use the formula for a geometric sequence. a. Write a recursive formula for the sequence. b. Write an explicit formula for the sequence. The first term a 1 is 40, and n ≥ 2. A recursive formula for the sequence 40, –60, 90, –135, … is a 1 7-8 Recursive Formulas = 40, a n = –1.5a n– 1, n ≥ 2. SOLUTION: 21. 81, 27, 9, 3, ... SOLUTION: Subtract each term from the term that follows it. 27 – 81 = –54; 9 – 27 = –18, 3 – 9 = –6 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. For each recursive formula, write an explicit formula. For each explicit formula, write a recursive formula. 23. SOLUTION: Write out the first 4 terms: 3, 12, 48, 192. Subtract each term from the term that follows it. 12 – 3 = 9; 48 – 12 = 36, 192 – 48 = 144 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. There is a common ratio of . The sequence is geometric. Use the formula for a geometric sequence. There is a common ratio of 4. The sequence is geometric. Use the formula for a geometric sequence. The first term a 1 is 81, and n ≥ 2. A recursive formula for the sequence 81, 27, 9, 3, … is a 1 = 81, a n = a n– 1, n ≥ 2. The first term a 1 is 3, and n ≥ 2. A recursive formula n –1 for the explicit formula an = 15(2) 4a n – 1, n ≥ 2. 22. CCSS MODELING A landscaper is building a brick patio. Part of the patio includes a pattern constructed from triangles. The first four rows of the pattern are shown. is a 1 = 3, a n = 24. SOLUTION: The common difference is –12. Use the formula for the nth terms of an arithmetic sequence. a. Write a recursive formula for the sequence. b. Write an explicit formula for the sequence. SOLUTION: The explicit formula is a n = –12n + 10. For each recursive formula, write an explicit formula. For each explicit formula, write a recursive formula. 25. SOLUTION: 23. The common ratio is . Use the formula for the nth terms of a geometric sequence. SOLUTION: Write out the first 4 terms: 3, 12, 48, 192. Subtract each term from the term that follows it. 12 – 3 = 9; 48 – 12 = 36, 192 – 48 = 144 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. There is a common ratio of 4. eSolutions Manual - Powered by Cognero The explicit formula is The sequence is geometric. Use the formula for a geometric sequence. 26. . Page 8 The first term a 1 is 45, and n ≥ 2. A recursive formula for the explicit formula an = –7n + 52 is a 1 = 7-8 Recursive Formulas The explicit formula is a n = –12n + 10. 45, a n = a n – 1 – 7, n ≥ 2. 25. SOLUTION: The common ratio is . Use the formula for the nth terms of a geometric sequence. The explicit formula is . 26. SOLUTION: Write out the first 4 terms. 45, 38, 31, 24 Subtract each term from the term that follows it. 38 – 45 = –7; 31 – 38 = –7, 24 – 31 = –7 There is a common difference of –7. The sequence is arithmetic. Use the formula for an arithmetic sequence. 27. TEXTING Barbara received a chain text that she forwarded to five of her friends. Each of her friends forwarded the text to five more friends, and so on. a. Find the first five terms of the sequence representing the number of people who receive the text in the nth round. b. Write a recursive formula for the sequence. c. If Barbara represents a 1, find a 8. SOLUTION: a. Then the first 5 terms of the sequence would be 1, 5, 25, 125, 625. b. The first term a 1 is 1, and n ≥ 2. A recursive formula is a 1 = 1, a n = 5a n – 1, n ≥ 2. c. The first term a 1 is 45, and n ≥ 2. A recursive formula for the explicit formula an = –7n + 52 is a 1 = 45, a n = a n – 1 – 7, n ≥ 2. 27. TEXTING Barbara received a chain text that she forwarded to five of her friends. Each of her friends forwarded the text to five more friends, and so on. a. Find the first five terms of the sequence representing the number of people who receive the text in the nth round. b. Write a recursive formula for the sequence. c. If Barbara represents a 1, find a 8. SOLUTION: a. Then the first 5 terms of the sequence would be 1, 5, 25, 125, 625. b. The first term a 1 is 1, and n ≥ 2. A recursive eSolutions Manual by Cognero formula is a- Powered 1 = 1, a n = 5a n – 1, n ≥ 2. c. On the 8th round, 78,125 would receive the chain text message. 28. GEOMETRY Consider the pattern below. The number of blue boxes increases according to a specific pattern. a. Write a recursive formula for the sequence of the number of blue boxes in each figure. b. If the first box represents a 1, find the number of blue boxes in a 8. SOLUTION: a. The sequence of blue boxes is 0, 4, 8, and 12. Subtract each term from the term that follows it. 4 – 0 = 4; 8 – 4 = 4, 12 – 8 = 4 There is a common difference of 4. The sequence is arithmetic. Use the formula for an arithmetic sequence. Page 9 On the 8th round, 78,125 would receive the chain text 7-8 Recursive Formulas message. 28. GEOMETRY Consider the pattern below. The number of blue boxes increases according to a specific pattern. a. Write a recursive formula for the sequence of the number of blue boxes in each figure. b. If the first box represents a 1, find the number of blue boxes in a 8. SOLUTION: a. The sequence of blue boxes is 0, 4, 8, and 12. Subtract each term from the term that follows it. 4 – 0 = 4; 8 – 4 = 4, 12 – 8 = 4 There is a common difference of 4. The sequence is arithmetic. Use the formula for an arithmetic sequence. The first term a 1 is 0, and n ≥ 2. A recursive formula When n = 8, there will be 28 blue boxes. 29. TREE The growth of a certain type of tree slows as the tree continues to age. The heights of the tree over the past four years are shown. a. Write a recursive formula for the height of the tree. b. If the pattern continues, how tall will the tree be in two more years? Round your answer to the nearest tenth of a foot. SOLUTION: a. The sequence of heights is 10, 11, 12.1, and 13.31. Subtract each term from the term that follows it. 11 – 10 = 11; 12.1 – 11 = 1.1, 13.31 – 12.1 = 1.21 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. There is a common ratio of 1.1. The sequence is geometric. Use the formula for a geometric sequence. is a 1 = 0, a n = a n – 1 + 4, n ≥ 2. b. Use the formula for the nth terms of an arithmetic sequence. When n = 8, there will be 28 blue boxes. The first term a 1 is 10, and n ≥ 2. A recursive formula for the sequence 10, 11, 12.1, 13.31, … is a 1 = 10, a n = 1.1a n – 1, n ≥ 2. b. Use the formula for the nth terms of a geometric sequence. 29. TREE The growth of a certain type of tree slows as the tree continues to age. The heights of the tree over the past four years are shown. In two more years, the tree will be 16.1 feet tall. a. Write a recursive formula for the height of the tree. b. If the pattern continues, how tall will the tree be in two more years? Round your answer to the nearest tenth of a foot. SOLUTION: a. The sequence of heights is 10, 11, 12.1, and 13.31. eSolutions Manual - Powered by Cognero Subtract each term from the term that follows it. 11 – 10 = 11; 12.1 – 11 = 1.1, 13.31 – 12.1 = 1.21 There is no common difference. Check for a 30. MULTIPLE REPRESENTATIONS The Fibonacci sequence is neither arithmetic nor geometric and can be defined by a recursive formula. The first terms are 1, 1, 2, 3, 5, 8, … a. Logical Determine the relationship between the terms of the sequence. What are the next five terms in the sequence? Page 10 b. Algebraic Write a formula for the nth term if a 1 = 1, a 2 = 1, and n ≥ 3. a 14 or 610. In two more years, the tree will be 16.1 feet tall. 7-8 Recursive Formulas 30. MULTIPLE REPRESENTATIONS The Fibonacci sequence is neither arithmetic nor geometric and can be defined by a recursive formula. The first terms are 1, 1, 2, 3, 5, 8, … a. Logical Determine the relationship between the terms of the sequence. What are the next five terms in the sequence? b. Algebraic Write a formula for the nth term if a 1 = 1, a 2 = 1, and n ≥ 3. c. Algebraic Find the 15th term. d. Analytical Explain why the Fibonacci sequence is not an arithmetic sequence. SOLUTION: a. Sample answer: The first two terms are 1. Starting with the third term, the two previous terms are added together to get the next term. So, the next 5 terms after 8 is 5 + 8 or 13, 8 + 13 or 21, 13 + 21 or 34, 21 + 34 or 55, and 34 + 55 or 89. b. The first term a 1 is 1 and the second term a2 is 1, and n ≥ 3. A recursive formula for Fibonacci sequence is a 1 = 1, a 2 = 1, a n = a n ÿ 2 + a n ÿ 1, n ≥ 3. d. Sample answer: Fibonacci sequence is not an arithmetic sequence since there is no common difference. 31. ERROR ANALYSIS Patrick and Lynda are working on a math problem that involves the sequence 2, –2, 2, –2, 2, … . Patrick thinks that the sequence can be written as a recursive formula. Lynda believes that the sequence can be written as an explicit formula. Is either of them correct? Explain. SOLUTION: Both; sample answer: The sequence can be written as the recursive formula a 1 = 2, a n = (–1)a n – 1, n ≥ 2. The sequence can also be written as the explicit n–1 formula a n = 2(–1) . 32. CHALLENGE Find a 1 for the sequence in which a 4 = 1104 and a n = 4a n – 1 + 16. SOLUTION: Find a 3 first. c. First, find the 13th and 14th terms by writing out 14 terms of the sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377. Then, a 13 = 233 and a 14 = 377. Thus, a 15 = a 13 + Find a 2 next. a 14 or 610. d. Sample answer: Fibonacci sequence is not an arithmetic sequence since there is no common difference. 31. ERROR ANALYSIS Patrick and Lynda are working on a math problem that involves the sequence 2, –2, 2, –2, 2, … . Patrick thinks that the sequence can be written as a recursive formula. Lynda believes that the sequence can be written as an explicit formula. Is either of them correct? Explain. SOLUTION: Both; sample answer: The sequence can be written as the recursive formula a 1 = 2, a n = (–1)a n – 1, n ≥ 2. The sequence can also be written as the explicit n–1 formula a n = 2(–1) . eSolutions Manual - Powered by Cognero 32. CHALLENGE Find a 1 for the sequence in which a 4 = 1104 and a n = 4a n – 1 + 16. Find a 1. Therefore, a1 is 12. 33. CCSS ARGUMENTS Determine whether the following statement is true or false . Justify your reasoning. There is only one recursive formula for every sequence. Page 11 SOLUTION: False; sample answer: A recursive formula for the sequence 1, 2, 3, … can be written as a = 1, a = a Therefore, a1 is 12. 7-8 Recursive Formulas 33. CCSS ARGUMENTS Determine whether the following statement is true or false . Justify your reasoning. There is only one recursive formula for every sequence. SOLUTION: False; sample answer: A recursive formula for the sequence 1, 2, 3, … can be written as a 1 = 1, a n = a n - 1 + 1, n ≥ 2 or as a 1 = 1, a 2 = 2, a n = a n - 2 + 2, n ≥ 3. 34. CHALLENGE Find a recursive formula for 4, 9, 19, 39, 79, … SOLUTION: Subtract each term from the term that follows it. 9 – 4 = 5; 19 – 9 = 10, 39 – 19 = 20 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. There is no common ratio. Therefore the sequence must be a combination of both. From the difference above, you can see each is twice as big as the previous. So r is 2. From the ratios, if each numerator was one less, the ratios would be 2. Thus, the common difference is 1. So, if the first term a 1 is 4, and n ≥ 2 a recursive formula for the sequence 4, 9, 19, 39, 79, … is a 1 = 4, a n = 2a n – 1 + 1, n ≥ 2. 35. WRITING IN MATH Explain the difference between an explicit formula and a recursive formula. Sample answer: In an explicit formula, the nth term a n is given as a function of n. In a recursive formula, the nth term a n is found by performing operations to one or more of the terms that precede it. 36. Find a recursive formula for the sequence 12, 24, 36, 48, … . A B C D SOLUTION: Subtract each term from the term that follows it. 24 – 12 = 12; 36 – 24 = 12, 48 – 36 = 12 There is a common difference of 12. The sequence is arithmetic. Use the formula for an arithmetic sequence. The first term a 1 is 12, and n ≥ 2. A recursive formula for the sequence 12, 24, 36, 48, … is a 1 = 12, a n = a n – 1 + 12, n ≥ 2. Therefore, the correct choice is C. 4 6 37. GEOMETRY The area of a rectangle is 36 m n 3 3 square feet. The length of the rectangle is 6 m n feet. What is the width of the rectangle? 7 9 F 216m n ft G 6mn3 ft 7 3 H 42m n ft J 30mn3 ft SOLUTION: Use the area formula for a rectangle to determine the measure of the width. SOLUTION: Sample answer: In an explicit formula, the nth term a n is given as a function of n. In a recursive formula, the nth term a n is found by performing operations to one or more of the terms that precede it. 36. Find a recursive formula for the sequence 12, 24, 36, 48, … . A B C D SOLUTION: Subtract each term from the term that follows it. 24 – 12 = 12; 36 – 24 = 12, 48 – 36 = 12 There is a common difference of 12. The sequence eSolutions Manual - Powered by Cognero Therefore, the correct choice is G. 38. Find an inequality for the graph shown. Page 12 7-8 Recursive Formulas Therefore, the correct choice is G. 38. Find an inequality for the graph shown. Therefore, the correct choice is F. Find the next three terms in each geometric sequence. 40. 675, 225, 75, ... SOLUTION: Find the common ratio. A B C D SOLUTION: The y-intercept of the line is –4. The slope is 2. The graph is shaded below, so the inequality should use a < or ≤ symbol. Since the line is dashed, the inequality does not include the equals. Therefore, the correct choice is C. 39. Write an equation of the line that passes through (–2, –20) and (4, 58). F y = 13x + 6 G y = 19x – 18 H y = 19x + 18 J y = 13x – 6 SOLUTION: First find the slope. The common ratio is . Multiply by the common ratio to find next three terms. They are . 41. 16, –24, 36, ... SOLUTION: Find the common ratio. The common ratio is . Multiply by the common ratio to find next three terms. They are -54, 81, 121.5. 42. 6, 18, 54, ... SOLUTION: Find the common ratio. The common ratio is 3. Use the common ratio to find next three terms. They are 162, 324, 972. 43. 512, –256, 128, ... SOLUTION: Find the common ratio. Use the point-slope formula to find the equation. The common ratio is . Use the common ratio to find next three terms. They are -64, 32, -16. 44. 125, 25, 5, ... Therefore, the correct choice is F. Find the next three terms in each geometric sequence. 40. 675, 225, 75, ... SOLUTION: eSolutions Manual - Powered by Cognero Find the common ratio. SOLUTION: Find the common ratio. The common ratio is . Use the common ratio to find next three terms. They are 45. 12, 60, 300, ... . Page 13 The common ratio is . Use the common ratio to 7-8 Recursive Formulas find next three terms. They are -64, 32, -16. 44. 125, 25, 5, ... 47. TOURS The Snider family and the Rollins family are traveling together on a trip to visit a candy factory. The number of people in each family and the total cost are shown in the table below. Find the adult and children’s admission prices. SOLUTION: Find the common ratio. The common ratio is After 5 years, Nicholas will have $2664.35. . Use the common ratio to find next three terms. They are . 45. 12, 60, 300, ... SOLUTION: Let a be the admissions cost for adults and c the admission cost of children. Then 2a + 3c = 58 and 2a + c = 38. SOLUTION: Find the common ratio. The common ratio is 5. Multiply by the common ratio to find next three terms. They are 1500, 7500, 37,500. 46. INVESTMENT Nicholas invested $2000 with a 5.75% interest rate compounded monthly. How much money will Nicholas have after 5 years? So, tickets for adults cost $14 and tickets for children SOLUTION: cost $10. Write each equation in standard form. 48. After 5 years, Nicholas will have $2664.35. 47. TOURS The Snider family and the Rollins family are traveling together on a trip to visit a candy factory. The number of people in each family and the total cost are shown in the table below. Find the adult and children’s admission prices. SOLUTION: 49. SOLUTION: SOLUTION: Let a be the admissions cost for adults and c the admission cost of children. Then 2a + 3c = 58 and 2a + c = 38. eSolutions Manual - Powered by Cognero 50. SOLUTION: Page 14 SOLUTION: 7-8 Recursive Formulas 50. 55. SOLUTION: SOLUTION: 56. SOLUTION: 51. SOLUTION: 57. SOLUTION: There are no like terms, so the expression is simplified. 58. 52. SOLUTION: SOLUTION: 59. SOLUTION: There are no like terms, so the expression is simplified. 53. SOLUTION: Simplify each expression. If not possible, write simplified. 54. SOLUTION: 55. SOLUTION: eSolutions Manual - Powered by Cognero Page 15 SOLUTION: Mid-Chapter Quiz Simplify each expression. 3 5 1. (x )(4x ) 5. MULTIPLE CHOICE Express the volume of the solid as a monomial. SOLUTION: 9 A 6x B 8x9 24 C 8x D 7x24 2 5 3 2. (m p ) SOLUTION: SOLUTION: 3 2 3 So, the correct choice is B. 3. [(2xy ) ] SOLUTION: Simplify each expression. Assume that no denominator equals 0. 6. SOLUTION: 3 4 2 3 4. (6ab c )( −3a b c) SOLUTION: 5. MULTIPLE CHOICE Express the volume of the solid as a monomial. 7. SOLUTION: 9 A 6x B 8x9 24 C 8x D 7x24 SOLUTION: 8. eSolutions Manual - Powered by Cognero Page 1 SOLUTION: 12. SOLUTION: Mid-Chapter Quiz 8. 13. SOLUTION: SOLUTION: 14. SOLUTION: 9. 15. SOLUTION: SOLUTION: 16. SOLUTION: 10. ASTRONOMY Physicists estimate that the number of stars in the universe has an order of 21 magnitude of 10 . The number of stars in the Milky Way galaxy is around 100 billion. Using orders of magnitude, how many times as many stars are there in the universe as the Milky Way? Simplify. 17. SOLUTION: SOLUTION: 100 billion is equal to 100,000,000,000. This can be 11 written as 10 . 10 So, there are 10 times more stars in the universe than the Milky Way galaxy. 18. SOLUTION: Write each expression in radical form, or write each radical in exponential form. 11. SOLUTION: 19. SOLUTION: 12. SOLUTION: 20. 13. eSolutions Manual - Powered by Cognero SOLUTION: Page 2 SOLUTION: Mid-Chapter Quiz 20. 24. SOLUTION: SOLUTION: 21. SOLUTION: Solve each equation. 25. SOLUTION: Therefore, the solution is 6. 22. SOLUTION: 26. SOLUTION: Therefore, the solution is 1. 23. 27. SOLUTION: SOLUTION: Therefore, the solution is 6.5. 24. eSolutions Manual - Powered by Cognero SOLUTION: Express each number in scientific notation. 28. 0.00000054 Page 3 SOLUTION: 418,000,000→ 4.18 The decimal moved 8 places to the left, so n = 8. 8 418,000,000 = 4.18 × 10 Mid-Chapter Quiz Therefore, the solution is 1. Express each number in standard form. 27. −3 32. 4.1 × 10 SOLUTION: SOLUTION: −3 4.1 × 10 Since n = −3, move the decimal point 3 places to the left. −3 4.1 × 10 33. 2.74 × 10 Therefore, the solution is 6.5. 5 2.74 × 10 Since n = 5, move the decimal point 5 places to the right. 5 2.74 × 10 = 274,000 9 34. 3 × 10 SOLUTION: −7 0.00000054 = 5.4 × 10 9 3 × 10 Since n = 9, move the decimal point 9 places to the right. 29. 0.0042 SOLUTION: 0.0042→ 4.2 The decimal point moved 3 places to the right, so n = −3. −3 0.0042 = 4.2 × 10 9 3 × 10 = 3,000,000,000 −5 35. 9.1 × 10 SOLUTION: −5 9.1 × 10 Since n = -5, move the decimal point 5 places to the left. −5 9.1 × 10 = 0.000091 30. 234,000 SOLUTION: 234,000 → 2.34 The decimal point moved 5 places to the left, so n = 5. 5 234,000 = 2.34 × 10 Evaluate each product or quotient. Express the results in scientific notation. 2 5 36. (2.13 × 10 )(3 × 10 ) SOLUTION: 31. 418,000,000 SOLUTION: 418,000,000→ 4.18 The decimal moved 8 places to the left, so n = 8. 8 418,000,000 = 4.18 × 10 5 SOLUTION: Express each number in scientific notation. 28. 0.00000054 SOLUTION: 0.00000054→ 5.4 The decimal point moved 7 places to the right, so n = −7. = 0.0041 6 −2 37. (7.5 × 10 )(2.5 × 10 ) SOLUTION: Express each number in standard form. −3 32. 4.1 × 10 SOLUTION: −3 4.1 × 10 Since n = −3, movebythe decimal point 3 places to the eSolutions Manual - Powered Cognero left. −3 4.1 × 10 = 0.0041 38. Page 4 SOLUTION: Mid-Chapter Quiz 38. SOLUTION: 39. SOLUTION: 40. MAMMALS A blue whale has been caught that 5 was 4.2 × 10 pounds. The smallest mammal is a bumblebee bat, which is about 0.0044 pound. a. Write the whale’s weight in standard form. b. Write the bat’s weight in scientific notation. c. How many orders of magnitude as big is a blue whale as a bumblebee bat? SOLUTION: a. In 4.2 × 105 the exponent is 5, so move the decimal point 5 places to the right. The whale’s weight is 420,000 pounds. b. To change 0.0044 to 0004.4, the decimal point is moved 3 places to the right, so n = –3. –3 0.0044 = 4.4 × 10 –3 So, the bat's weight is 4.4 × 10 pounds. c. The order of magnitude of the whale’s weight is 5 10 and the order of magnitude of the bat’s weight is -3 10 . Compare by dividing. 8 So, the blue whale is 10 or 100,000,000 times as big as a bumblebee bat. eSolutions Manual - Powered by Cognero Page 5 SOLUTION: Practice Test - Chapter 7 Simplify each expression. 2 8 5. 1. (x )(7x ) SOLUTION: 7 2 2 SOLUTION: A value to the zero power equals 1. 5 6. 2. (5a bc )(−6a bc ) SOLUTION: SOLUTION: 3. MULTIPLE CHOICE Express the volume of the solid as a monomial. Simplify. 7. SOLUTION: 3 A x B 6x C 6x3 D x 6 SOLUTION: 8. SOLUTION: The correct answer is A. 9. Simplify each expression. Assume that no denominator equals 0. SOLUTION: 4. SOLUTION: 10. SOLUTION: 5. SOLUTION: A value to the zero power equals 1. eSolutions Manual - Powered by Cognero Page 1 Practice Test - Chapter 7 10. 14. SOLUTION: SOLUTION: Solve each equation. 11. SOLUTION: 15. SOLUTION: So, the solution is 3. 12. 16. SOLUTION: 13. SOLUTION: 17. SOLUTION: SOLUTION: Express each number in scientific notation. 18. 0.00021 14. SOLUTION: SOLUTION: 0.00021→2.1 The decimal moved 4 places to the right, so n = –4. −4 2.1 × 10 19. 58,000 eSolutions Manual - Powered by Cognero Page 2 SOLUTION: 58,000→5.8 The decimal moved 4 places to the left, so n = 4. SOLUTION: 0.00021→2.1 The decimal moved 4 places to the right, so n = –4. −4 Practice 2.1 ×Test 10 - Chapter 7 19. 58,000 SOLUTION: 58,000→5.8 The decimal moved 4 places to the left, so n = 4. 4 5.8 × 10 Express each number in standard form. −5 20. 2.9 × 10 SOLUTION: n = –5, so move the decimal 5 places to the left. 2.9→0.000029 6 21. 9.1 × 10 SOLUTION: 35,980,000→3.598 The decimal moved 7 places to the left, so n = 7. 7 3.598 × 10 Graph each function. Find the y-intercept, and state the domain and range. x 25. y = 2(5) SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y 2(5) –2 0.08 –1 0.4 0 2 1 10 2 50 –2 2(5) –1 2(5) 0 2(5) 1 2(5) 2 2(5) SOLUTION: n = 6, so move the decimal 6 places to the right. 9.1→9,100,000 Evaluate each product or quotient. Express the results in scientific notation. 3 4 22. (2.5 × 10 )(3 × 10 ) SOLUTION: The y-intercept is 2. The domain is all real numbers, and the range is all real numbers greater than 0. x 26. y = –3(11) 23. SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y 3(11) SOLUTION: 24. ASTRONOMY The average distance from Mercury to the Sun is 35,980,000 miles. Express this distance in scientific notation. SOLUTION: 35,980,000→3.598 The decimal moved 7 places to the left, so n = 7. 3.598 × 10 –2 –0.02 –1 –0.27 0 –3 1 –33 2 –363 –2 3(11) –1 3(11) 0 3(11) 1 3(11) 2 3(11) 7 Graph each function. Find the y-intercept, and state the domain and range. x 25. y = 2(5) SOLUTION: Complete table by of Cognero values for – 2 < x < 2. eSolutions Manual - aPowered Connect the points on the graph with a smooth curve. x Page 3 The y-intercept is –3. The domain is all real numbers, and the range is all real numbers less than 0. The y-intercept is 2. The domain is all real numbers, Practice Test - Chapter 7 numbers greater than 0. and the range is all real 26. y = –3(11) x SOLUTION: Complete a table of values for – 2 < x < 2. Connect the points on the graph with a smooth curve. x x y 3(11) –2 –0.02 –1 –0.27 0 –3 1 –33 2 –363 –2 3(11) –1 3(11) 0 3(11) 1 2 3(11) 3(11) The y-intercept is 2. The domain and range are both all real numbers. Find the next three terms in each geometric sequence. 28. 2, −6, 18, … SOLUTION: 18(–3) = –54 –54(–3) = 162 162(–3) = –486 29. 1000, 500, 250, … SOLUTION: The y-intercept is –3. The domain is all real numbers, and the range is all real numbers less than 0. 27. y = 3x + 2 SOLUTION: To graph, plot the y-intercept at (0, 2). Then use the slope (m = 3) to plot additional points. Move up 3 units and right 1 unit or down three units and left 1 unit. Connect the points. The y-intercept is 2. The domain and range are both all real numbers. Find the next three terms in each geometric sequence. 28. 2, −6, 18, … SOLUTION: eSolutions Manual - Powered by Cognero 250(0.5) = 125 125(0.5) = 62.5 62.5(0.5) = 31.25 30. 32, 8, 2, … SOLUTION: 31. MONEY Lynne invested $500 into an account with a 6.5% interest rate compounded monthly. How much will Lynne’s investment be worth in 10 years? F $600.00 G $938.57 H $956.09 J $957.02 Page 4 SOLUTION: Practice Test - Chapter 7 Her investment will be worth $2498.92 in 6 years. 31. MONEY Lynne invested $500 into an account with a 6.5% interest rate compounded monthly. How much will Lynne’s investment be worth in 10 years? F $600.00 G $938.57 H $956.09 J $957.02 Find the first five terms of each sequence. 33. SOLUTION: Use a 1 = 18 and the recursive formula to find the next four terms. SOLUTION: The correct choice is H. 32. INVESTMENTS Shelly’s investment of $3000 has been losing value at a rate of 3% each year. What will her investment be worth in 6 years? SOLUTION: Thus, the first five terms are 18, 14, 10, 6, and 2. Her investment will be worth $2498.92 in 6 years. Find the first five terms of each sequence. 33. SOLUTION: Use a 1 = 18 and the recursive formula to find the next four terms. 34. SOLUTION: Use a 1 = –2 and the recursive formula to find the next four terms. eSolutions Manual - Powered by Cognero Page 5 Practice - Chapter 7 are 18, 14, 10, 6, and 2. Thus,Test the first five terms 34. SOLUTION: Use a 1 = –2 and the recursive formula to find the next four terms. So, the first five terms are –2, –3, –7, –23, and –87. eSolutions Manual - Powered by Cognero Page 6 Preparing for Standardized Tests - Chapter 7 The correct answer is J. Read each problem. Identify what you need to know. Then use the information in the problem to solve. 1. Since its creation 5 years ago, approximately 2.504 × 7 10 items have been sold or traded on a popular online website. What is the average daily number of items sold or traded over the 5-year period? A about 9640 items per day B about 13,720 items per day C about 1,025,000 items per day D about 5,008,000 items per day SOLUTION: Find the number of days in a 5-year period: 365 · 5 = 1825. Express the number in scientific notation. 3 1825 = 1.825 × 10 Then divide. 3. The population of the United States is about 3.034 × 8 10 people. The land area of the country is about 3.54 6 × 10 square miles. What is the average population density (number of people per square mile) of the United States? A about 136.3 people per square mile B about 112.5 people per square mile C about 94.3 people per square mile D about 85.7 people per square mile SOLUTION: Divide to find the population density. So, the average population density of the United States is about 85.7 people per square mile. The correct answer is D. So, the average daily number of items sold or traded over the 5-year period is about 13,720 items per day. The correct answer is B. 2. Evaluate if a = 121 and b = 23. F about 5.26 G about 9.90 H about 12 J about 52.75 4. Eleece is making a cover for the marching band’s bass drum.The drum has a diameter of 20 inches. Estimate the area of the face of the bass drum. F 31.41 square inches G 62.83 square inches H 78.54 square inches J 314.16 square inches SOLUTION: 2 and r = 10, so A = 3.1416 × 10 , or 314.16. The correct answer is J. SOLUTION: The correct answer is J. 3. The population of the United States is about 3.034 × 8 10 people. The land area of the country is about 3.54 6 × 10 square miles. What is the average population density (number of people per square mile) of the United States? A about 136.3 people per square mile B about 112.5 people per square mile C about 94.3 people per square mile D about 85.7 people per square mile eSolutions Manual - Powered by Cognero SOLUTION: Divide to find the population density. Page 1 Standardized Test Practice, Chapters 17 The correct answer is D. 1. Express the area of the triangle below as a monomial. 2. Simplify the following expression. F G A B C D H SOLUTION: SOLUTION: J The correct answer is D. 2. Simplify the following expression. The correct answer is G. 3. Which equation of a line is perpendicular to F ? G A H B J C D SOLUTION: SOLUTION: In order for the lines to be perpendicular, the slopes must have opposite reciprocals. The slope of the given line is so the opposite reciprocal is only line given with this slope is eSolutions Manual - Powered by Cognero correct answer is A. . The so the Page 1 4. Write a recursive formula for the sequence of the only line given with this slope is so the correct answer is A. Standardized Test Practice, Chapters 17 The correct answer is G. 3. Which equation of a line is perpendicular to 4. Write a recursive formula for the sequence of the number of squares in each figure. ? A B F G H J C D SOLUTION: In order for the lines to be perpendicular, the slopes must have opposite reciprocals. The slope of the given line is so the opposite reciprocal is only line given with this slope is . The so the SOLUTION: The sequence of the number of squares in the four figures is 1, 5, 9, 13. Subtract each term from the term that follows it. 5 – 1 = 4; 9 – 5 = 4, 13 – 9 = 4 There is a common difference of 4. The sequence is arithmetic. Use the formula for an arithmetic sequence. correct answer is A. 4. Write a recursive formula for the sequence of the number of squares in each figure. The first term a 1 is 1, and n ≥ 2. A recursive formula for the sequence 1, 5, 9, 13 is a 1 = 1, a n = a n– 1 + 4, n ≥ 2. Thus, the correct answer is H. 6 8 5. Evaluate (4.2 × 10 )(5.7 × 10 ). F G H J A 2.394 × 1015 14 SOLUTION: The sequence of the number of squares in the four figures is 1, 5, 9, 13. Subtract each term from the term that follows it. 5 – 1 = 4; 9 – 5 = 4, 13 – 9 = 4 There is a common difference of 4. The sequence is arithmetic. Use the formula for an arithmetic sequence. B 23.94 × 10 C 9.9 × 1014 48 D 2.394 × 10 SOLUTION: Thus, the correct answer is A. 6. Which inequality is shown in the graph? The first term a 1 is 1, and n ≥ 2. A recursive formula for the sequence 1, 5, 9, 13 is a 1 = 1, a n = a n– 1 + 4, n ≥ 2. Thus, the correct answer is H. eSolutions Manual - Powered by Cognero 6 Page 2 8 5. Evaluate (4.2 × 10 )(5.7 × 10 ). F equation would be .Because it is shaded below and the line is solid the inequality would be Standardized Test Practice, Chapters 17 Thus, the correct answer is A. 6. Which inequality is shown in the graph? . The correct answer is H. 7. Jaden created a Web site for the Science Olympiad team. The total number of hits the site has received is shown. a. Find an equation for the regression line. b. Predict the number of total hits that the Web site will have received on day 46. F G SOLUTION: a. On a graphing calculator enter the number of the days in L1 and the number of hits in L2. Then H J perform the regression by pressing and choosing the CALC option. Scroll down to LinReg SOLUTION: The boundary of the inequality is a line with a yintercept of (0, 1) and a slope of equation would be (ax + b) and press twice. . Therefore the .Because it is shaded below and the line is solid the inequality would be . The correct answer is H. 7. Jaden created a Web site for the Science Olympiad team. The total number of hits the site has received is shown. a. Find an equation for the regression line. b. Predict the number of total hits that the Web site will have received on day 46. SOLUTION: a. On a graphing calculator enter the number of the days in L1 and the number of hits in L2. Then The equation of the regression line is y = 1.67x – 2.64. b. Evaluate the linear regression when x = 46. Therefore, the number of hits the Web site will receive on day 46 will be about 74. 8. Find the value of x so that the figures have the same area. perform the regression by pressing and CALC LinReg choosing the option. Scroll down to (ax + b) and press twice. eSolutions Manual - Powered by Cognero Page 3 SOLUTION: Therefore, Test the number of hits the Web17 site will Standardized Practice, Chapters receive on day 46 will be about 74. 8. Find the value of x so that the figures have the same area. Since , the correct answer is no solution. 10. GRIDDED RESPONSE At a family fun center, the Wilson and Sanchez families each bought video game tokens and batting cage tokens as shown in the table. What is the cost in dollars of a batting cage token at the family fun center? SOLUTION: Find the area of each figure. SOLUTION: Let x = the cost of a video game token Let y = the cost of a batting cage token Wilson: Sanchez: Use multiplication to solve the system of equations. Set the areas equal and solve for x. The video game tokens cost $0.50. Substitute this back into one of the original equations to find the cost of a batting cage token. So, when x has a value of 4, the figures will have the same area. 9. What is the solution to the following system of equations? Show your work. SOLUTION: Using substitution. Therefore, the cost of a batting cage token is $1.75. 11. The table below shows the distances from the Sun to mercury, Earth, Mars, and Saturn. Use the data to answer each question. Since , the correct answer is no solution. 10. GRIDDED RESPONSE At a family fun center, the Wilson and Sanchez families each bought video eSolutions - Powered by Cognero gameManual tokens and batting cage tokens as shown in the table. Page 4 a. Of the planets listed, which one is the closest to the Sun? b. About how many times as far from the Sun is Standardized Test Practice, Chapters 17 Therefore, the cost of a batting cage token is $1.75. 11. The table below shows the distances from the Sun to mercury, Earth, Mars, and Saturn. Use the data to answer each question. a. Of the planets listed, which one is the closest to the Sun? b. About how many times as far from the Sun is Mars as Earth? SOLUTION: Because each of the distances is written in scientific notation, the exponent can be used to determine the smallest distance. Because the exponent in the scientific notation for Mercury is the smallest, its distance is the smallest and it is the closest planet to the Sun. (Earth’s distance)(how many times?) = Mar’s distance Mars is 1.52 times farther from the Sun than the Earth. eSolutions Manual - Powered by Cognero Page 5 SOLUTION: A formula that gives the first term of a sequence and tells you how to find the next term when you know the preceding term is called a recursive formula. Study Guide and Review - Chapter 7 Choose the word or term that best completes each sentence. 4 1. 7xy is an example of a(n)___________ . SOLUTION: A product of a number and variables is a monomial. SOLUTION: An equation in which the variable occurs as exponent is an exponential equation. t 9. The equation for _____________ is y = C(1 – r) . 5 2. The ___________ of 95,234 is 10 . SOLUTION: 5 3x – 1 8. 2 = 16 is an example of a(n) _______________ . 5 95,234 is almost 100,000 or 10 , so 10 is its order of magnitude. SOLUTION: Since 0 < 1 – r < 1, this is an example of exponential decay. n 10. If a = b for a positive integer n, then a is a(n) _______________ of b. 3. 2 is a(n) __________ of 8. SOLUTION: SOLUTION: 3 Since 8 = 2 , 2 is the cube root of 8. n If a = b for a positive integer n, then a is an nth root of b. 4. The rules for operations with exponents can be extended to apply to expressions with a(n) ___________ such as Simplify each expression. 3 . 11. x ⋅ x ⋅ x SOLUTION: 5 SOLUTION: The exponent is a rational exponent. n 5. A number written in is of the form a × 10 , where 1 ≤ a < 10 and n is an integer. SOLUTION: n A number written as a × 10 , where 1 ≤ a < 10 and n is an integer, is written in scientific notation. 2 5 12. (2xy)( −3x y ) x 6. f (x) = 3 is an example of a(n) ______________ . SOLUTION: SOLUTION: x A function in the form y = ab , where a ≠ 0, b > 0, and b ≠ 1 is an exponential function. 4 7. 5 2 13. (−4ab )(−5a b ) is a(n) ______________ for the sequence . SOLUTION: SOLUTION: A formula that gives the first term of a sequence and tells you how to find the next term when you know the preceding term is called a recursive formula. 3 2 2 14. (6x y ) SOLUTION: 3x – 1 8. 2 = 16 is an example of a(n) _______________ . SOLUTION: eSolutions Manual - Powered by Cognero An equation in which the variable occurs as exponent is an exponential equation. Page 1 3 3 2 15. [(2r t) ] t 13. (−4ab )(−5a b ) SOLUTION: Study Guide and Review - Chapter 7 3 2 2 2 19. GEOMETRY Use the formula V = πr h to find the volume of the cylinder. 14. (6x y ) SOLUTION: SOLUTION: 3 3 2 15. [(2r t) ] SOLUTION: Simplify each expression. Assume that no denominator equals zero. 3 16. (−2u )(5u) 20. SOLUTION: SOLUTION: 2 3 3 3 17. (2x ) (x ) SOLUTION: 21. 18. SOLUTION: 3 3 (2x ) SOLUTION: 2 19. GEOMETRY Use the formula V = πr h to find the volume of the cylinder. eSolutions Manual - Powered by Cognero 22. SOLUTION: Page 2 Study Guide and Review - Chapter 7 22. 26. SOLUTION: SOLUTION: −3 0 6 23. a b c SOLUTION: 27. SOLUTION: 24. SOLUTION: 2 4 28. GEOMETRY The area of a rectangle is 25x y square feet. The width of the rectangle is 5xy feet. What is the length of the rectangle? 25. SOLUTION: SOLUTION: 3 The length of the rectangle is 5xy ft. Simplify. 26. 29. SOLUTION: eSolutions Manual - Powered by Cognero SOLUTION: Page 3 Study Guide and Review - Chapter 7 3 The length of the rectangle is 5xy ft. Simplify. 34. 29. SOLUTION: SOLUTION: 30. SOLUTION: 35. SOLUTION: 31. SOLUTION: 36. SOLUTION: 32. SOLUTION: Solve each equation. x 37. 6 = 7776 SOLUTION: 33. SOLUTION: Therefore, the solution is 5. 4x – 1 38. 4 = 32 SOLUTION: 34. SOLUTION: eSolutions Manual - Powered by Cognero Page 4 Study Guide and 7 Therefore, the Review solution -isChapter 5. 4x – 1 38. 4 SOLUTION: 0.0000543 → 5.43 The decimal point moved 5 places to the right, so n = –5. −5 0.0000543 = 5.43 × 10 41. ASTRONOMY Earth has a diameter of about 8000 miles. Jupiter has a diameter of about 88,000 miles. Write in scientific notation the ratio of Earth’s diameter to Jupiter’s diameter. = 32 SOLUTION: SOLUTION: Earth: 8000 = 8.0 × 10 3 Jupiter: 88,000 = 8.8 × 10 4 Therefore, the solution is . Express each number in scientific notation. 39. 2,300,000 SOLUTION: 2,300,000 → 2.300000 The decimal point moved 6 places to the left, so n = 6. 6 2,300,000 = 2.3 × 10 In scientific notation, the ratio of Earth’s diameter to −2 Jupiter’s diameter is about 9.1 × 10 . Graph each function. Find the y-intercept, and state the domain and range. x 42. y = 2 40. 0.0000543 SOLUTION: 0.0000543 → 5.43 The decimal point moved 5 places to the right, so n = –5. −5 0.0000543 = 5.43 × 10 41. ASTRONOMY Earth has a diameter of about 8000 miles. Jupiter has a diameter of about 88,000 miles. Write in scientific notation the ratio of Earth’s diameter to Jupiter’s diameter. SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x y 2 2 –1 2 0 2 1 2 SOLUTION: Earth: 8000 = 8.0 × 10 2 3 Jupiter: 88,000 = 8.8 × 10 –2 –2 –1 0 1 1 2 2 4 2 4 In scientific notation, the ratio of Earth’s diameter to −2 Jupiter’s diameter is about 9.1 × 10 . The graph crosses the y-axis at 1. The domain is all real numbers, and the range is all real numbers greater than 0. eSolutions Manual - Powered by Cognero Graph each function. Find the y-intercept, and state the domain and range. Page 5 x 43. y = 3 + 1 In Guide scientific notation, ratio of Earth’s diameter to Study and Reviewthe - Chapter 7 −2 Jupiter’s diameter is about 9.1 × 10 . Graph each function. Find the y-intercept, and state the domain and range. x 42. y = 2 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x y 2 +1 –1 0 3 +1 1 3 +1 2 3 +1 0 1 1 2 2 4 2 2 2 2 The graph crosses the y-axis at 1. The domain is all real numbers, and the range is all real numbers greater than 0. x 43. y = 3 + 1 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x y 3 +1 0 –1 3 1 –2 +1 –1 +1 3 +1 –1 2 –1 –2 –2 –1 3 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x y 3 +1 3 2 –2 x 43. y = 3 + 1 –2 –2 0 The graph crosses the y-axis at 1. The domain is all real numbers, and the range is all real numbers greater than 0. 0 2 1 4 2 10 3 +1 1 3 +1 2 3 +1 eSolutions Manual - Powered by Cognero 0 2 1 4 2 10 The graph crosses the y-axis at 2. The domain is all real numbers, and the range is all real numbers greater than 1. x 44. y = 4 + 2 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x y 4 +2 –2 –2 4 –1 4 0 4 +2 1 4 +2 2 4 +2 –1 +2 +2 0 3 1 6 2 18 Page 6 The graph crosses the y-axis at 2. The domain is all real numbers, the range is all real Study Guide and and Review - Chapter 7 numbers greater than 1. x x 45. y = 2 − 3 44. y = 4 + 2 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x y 4 +2 –2 –1 The graph crosses the y-axis at 3. The domain is all real numbers, and the range is all real numbers greater than 2. –2 4 –1 4 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. +2 +2 0 3 1 6 2 18 0 4 +2 1 4 +2 2 4 +2 The graph crosses the y-axis at 3. The domain is all real numbers, and the range is all real numbers greater than 2. x x x 2 –3 –2 2 –1 2 0 2 –3 1 2 –3 2 2 –3 –2 –3 –1 –3 y 0 –2 1 –1 2 1 The graph crosses the y-axis at –2. The domain is all real numbers, and the range is all real numbers greater than –3. 45. y = 2 − 3 SOLUTION: Complete a table of values for –2 < x < 2. Connect the points on the graph with a smooth curve. x x 2 –3 –2 2 –1 2 0 2 –3 1 2 –3 2 2 –3 –2 –3 –1 –3 y 0 –2 1 –1 2 1 46. BIOLOGY The population of bacteria in a petri dish increases according to the model p = 550(2.7) 0.008t , where t is the number of hours and t = 0 corresponds to 1:00 P.M. Use this model to estimate the number of bacteria in the dish at 5:00 P.M. SOLUTION: If t = 0 corresponds to 1:00 P.M, then t = 4 represents 5:00 P.M. There will be about 568 bacteria in the dish at 5:00 P.M. 47. Find the final value of $2500 invested at an interest rate of 2% compounded monthly for 10 years. eSolutions Manual - Powered by Cognero Page 7 SOLUTION: Use the equation for compound interest, with P = 2500, r = 0.02, n = 12, and t = 10. There willand be about 568- bacteria Study Guide Review Chapterin7the dish at 5:00 P.M. After 5 years, Zita’s computer value is about $1030.48. 47. Find the final value of $2500 invested at an interest rate of 2% compounded monthly for 10 years. Find the next three terms in each geometric sequence. 49. −1, 1, −1, 1, ... SOLUTION: Use the equation for compound interest, with P = 2500, r = 0.02, n = 12, and t = 10. The final value of the investment is about $3053.00. 48. COMPUTERS Zita’s computer is depreciating at a rate of 3% per year. She bought the computer for $1200. a. Write an equation to represent this situation. b. What will the computer’s value be after 5 years? SOLUTION: a. Use the equation for exponential decay, with a = 1200 and r = 0.03. The equation that represents the depreciation of t Zita’s computer is y = 1200(1 − 0.03) . b. Substitute 5 for t and solve. SOLUTION: Calculate the common ratio. The common ratio is –1. Multiply each term by the common ratio to find the next three terms. 1 × –1 = –1 –1 × –1 = 1 1 × –1 = –1 The next three terms of the sequence are –1, 1, and –1. 50. 3, 9, 27 ... SOLUTION: Calculate the common ratio. The common ratio is 3. Multiply each term by the common ratio to find the next three terms. 27 × 3 = 81 81 × 3 = 243 243 × 3 = 729 The next three terms of the sequence are 81, 243, and 729. 51. 256, 128, 64, ... SOLUTION: Calculate the common ratio. After 5 years, Zita’s computer value is about $1030.48. Find the next three terms in each geometric sequence. 49. −1, 1, −1, 1, ... SOLUTION: Calculate the common ratio. The common ratio is –1. Multiply each term by the common ratio to find the next three terms. 1 × –1 = –1 –1 × –1 = 1 eSolutions Manual - Powered by Cognero The common ratio is . Multiply each term by the common ratio to find the next three terms. 64 × = 32 32 × = 16 16 × = 8 The next three terms of the sequence are 32, 16, and Page 8 8. Write the equation for the nth term of each 27 × 3 = 81 81 × 3 = 243 243 × 3 = 729 The next three terms of- the sequence Study Guide and Review Chapter 7 are 81, 243, and 729. The common ratio is 3. So, r = 3. 54. 256, 128, 64, ... 51. 256, 128, 64, ... SOLUTION: Calculate the common ratio. SOLUTION: The first term of the sequence is 256. So, a 1 = 256. Calculate the common ratio. The common ratio is . Multiply each term by the common ratio to find the next three terms. The common ratio is . So, r = . 64 × = 32 32 × = 16 16 × = 8 The next three terms of the sequence are 32, 16, and 8. Write the equation for the nth term of each geometric sequence. 52. −1, 1, −1, 1, ... SOLUTION: The first term of the sequence is –1. So, a 1 = –1. 55. SPORTS A basketball is dropped from a height of 20 feet. It bounces to its height each bounce. Draw a graph to represent the situation. SOLUTION: Compared to the previous bounce, the ball returns to its height. So, the common ratio is . Use common ratio to find next y term Calculate the common ratio. The common ratio is –1. So, r = –1. 53. 3, 9, 27, ... SOLUTION: The first term of the sequence is 3. So, a 1 = 3. Calculate the common ratio. The common ratio is 3. So, r = 3. 54. 256, 128, 64, ... eSolutions Manual - Powered by Cognero SOLUTION: The first term of the sequence is 256. So, a 1 = 256. Page 9 Study Guide and Review - Chapter 7 Find the first five terms of each sequence. 56. SOLUTION: Use a 1 = 11 and the recursive formula to find the next four terms. Therefore, the geometric sequence that models this situation is 20, 10, 5, , , , , , and so forth. The first five terms are 11, 7, 3, –1, and –5. 57. Find the first five terms of each sequence. 56. SOLUTION: Use a 1 = 3 and the recursive formula to find the next four terms. SOLUTION: Use a 1 = 11 and the recursive formula to find the next four terms. eSolutions Manual - Powered by Cognero Page 10 The first term a 1 is 2, and n ≥ 2. A recursive formula Study Guide andterms Review Chapter The first five are -11, 7, 3, –1,7 and –5. for the sequence 2, 7, 12, 17, … is a 1 = 2, a n = a n – 1 + 5, n ≥ 2. 59. 32, 16, 8, 4, ... 57. SOLUTION: Use a 1 = 3 and the recursive formula to find the next four terms. SOLUTION: Subtract each term from the term that follows it. 16 – 32 = –16; 8 – 16 = –8, 4 – 8 = –4 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. There is a common ratio of 0.5. The sequence is geometric. Use the formula for a geometric sequence. The first term a 1 is 32, and n ≥ 2. A recursive formula for the sequence 32, 16, 8, 4, … is a 1 = 32, a n = 0.5a n – 1, n ≥ 2. The first five terms are 3, 12, 30, 66, and 138. Write a recursive formula for each sequence. 58. 2, 7, 12, 17, ... 60. 2, 5, 11, 23, ... SOLUTION: Subtract each term from the term that follows it. 5 – 2 = 3; 11 – 5 = 6, 23 – 11 = 12 There is no common difference. Check for a common ratio by dividing each term by the term that precedes it. SOLUTION: Subtract each term from the term that follows it. 7 – 2 = 5; 12 – 7 = 5, 17 – 12 = 5 There is a common difference of 5. The sequence is arithmetic. Use the formula for an arithmetic sequence. There is no common ratio. Therefore the sequence must be a combination of both. From the difference above, you can see each is twice as big as the previous. So r is 2. From the ratios, if each denominator was one less, the ratios would be 0.5. Thus, the common difference is 1. Then, if the first term a 1 is 2, and n ≥ 2, a recursive formula for the sequence 2, 5, 11, 23, … is a 1 = 2, a n = 2a n– 1 + 1, n ≥ 2. The first term a 1 is 2, and n ≥ 2. A recursive formula for the sequence 2, 7, 12, 17, … is a 1 = 2, a n = a n – 1 + 5, n ≥ 2. 59. 32, 16, 8, 4, ... SOLUTION: Subtract each term from the term that follows it. 16 – 32 = –16; 8 – 16 = –8, 4 – 8 = –4 There is no common difference. Check for a common ratio by dividing each term by the term that eSolutions Manual - Powered by Cognero precedes it. Page 11