Download Determine whether each expression is a monomial. Write yes or no

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