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SOLUTION: means the intersection of C and F. Identify
the elements that belong to both C and F. C = {0, 1,
2, 3, 4} and F = {0, 8}. The only element found in
both sets is 0. So
= {0}.
Pretest
Use set notation to write the elements of each
set. Then determine whether the statement
about the set is true or false .
1. L is the set of whole number multiples of 2 that are
less than 22. 18 ∈ L
6. SOLUTION: means the union of D and E. Identify the
elements that belong to D, E, or to both sets. D = {3,
5, 7, 8} and E = {0, 1, 2}. The elements found in D,
E, or both sets are 0, 1, 2, 3, 5, 7, and 8. So
=
{0, 1, 2, 3, 5, 7, 8}.
SOLUTION: L = {0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20}; 18 is an
element of this set. Therefore, the statement 18 ∈ L
is true.
Simplify.
7. (6 + 5i) + (–3 + 2i)
SOLUTION: 2. S is the set of integers that are less than 5 but
greater than -6. -8 ∈ S
SOLUTION: S = {−6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}; -8 is not
an element of this set since -8 is not greater than -6.
Therefore, the statement -8 ∈ S is false.
8. (–3 + 4i) – (4 – 5i)
SOLUTION: Let C = {0, 1, 2, 3, 4}, D = {3, 5, 7, 8}, E = {0,
1, 2}, and F = {0, 8}. Find each of the following.
3. SOLUTION: 9. (1 + 8i)(6 + 2i)
means the intersection of D and E. Identify
the elements that belong to D and E. D = {3, 5, 7, 8}
and E = {0, 1, 2}. Since there are no elements
common to both D and E,
is the empty set. That is,
= .
SOLUTION: 4. SOLUTION: means the intersection of C and E. Identify
the elements that belong to both C and E. C = {0, 1,
2, 3, 4} and E = {0, 1, 2}. The elements found in
both sets are 0, 1, and 2. So
= {0, 1, 2}.
10. (–3 + 3i)(–2 + 2i)
SOLUTION: 5. SOLUTION: means the intersection of C and F. Identify
the elements that belong to both C and F. C = {0, 1,
2, 3, 4} and F = {0, 8}. The only element found in
both sets is 0. So
= {0}.
11. 6. SOLUTION: means the union of D and E. Identify the
elements that belong to D, E, or to both sets. D = {3,
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5, 7, 8} and E = {0, 1, 2}. The elements found in D,
E, or both sets are 0, 1, 2, 3, 5, 7, and 8. So
=
{0, 1, 2, 3, 5, 7, 8}.
SOLUTION: Page 1
Determine whether each function has a
maximum or minimum value. Then find the
value of the maximum or minimum, and state
the domain and range of the function.
Pretest
11. SOLUTION: 13. SOLUTION: 2
For the function f (x) = x – 5x + 4, a = 1. Because a
is positive, the graph opens up and the function has a
minimum value.
The minimum value of the function is the ycoordinate of the vertex. To find the vertex, first find
equation of the axis of symmetry.
12. SOLUTION: Because the equation of the axis of symmetry is x =
, the x-coordinate of the vertex is
. Find the y-
coordinate of the vertex.
Determine whether each function has a
maximum or minimum value. Then find the
value of the maximum or minimum, and state
the domain and range of the function.
or Therefore, f (x) has a minimum value at
.
The domain of the function is all real numbers, so the
domain is R. The range of the function is all real
numbers greater than or equal to the minimum value
, so the range is y ≥
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SOLUTION: 2
For the function f (x) = x – 5x + 4, a = 1. Because a
, for y
R.
Page 2
domain is R. The range of the function is all real
numbers greater than or equal to the minimum value
Pretest , so the range is y ≥
, for y
R.
The domain of the function is all real numbers, so the
domain is R. The range of the function is all real
numbers less than or equal to the maximum value
, so the range is y ≤
, for y
R.
Solve each equation.
15. x2 – x – 20 = 0
SOLUTION: Factor the quadratic expression on the left side of
the equation and then apply the Zero Product
Property.
14. SOLUTION: 2
For the function f (x) = −2x – 3x + 2, a = -2.
Because a is negative, the graph opens down and
the function has a maximum value.
The maximum value of the function is the ycoordinate of the vertex. To find the vertex, first find
equation of the axis of symmetry.
16. x2 – 3x + 5 = 0
SOLUTION: 2
In the equation x – 3x + 5 = 0, a = 1, b = -3, and c
= 5. Apply the Quadratic Formula.
Because the equation of the axis of symmetry is x =
, the x-coordinate of the vertex is
. Find the
y-coordinate of the vertex.
17. x2 + 2x – 1 = 0
SOLUTION: Solve by completing the square.
Therefore, f (x) has a maximum value at
or .
The domain of the function is all real numbers, so the
domain is R. The range of the function is all real
numbers less than or equal to the maximum value
, so the range is y ≤
, for y
R.
Solve each equation.
15. x2 – x – 20 = 0
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Factor the quadratic expression on the left side of
the equation and then apply the Zero Product
18. x2 + 11x + 24 = 0
SOLUTION: Factor the quadratic expression on the left side of
the equation and then apply the Zero Product
Property.
Page 3
Because the result involves two even powers and
outside the radical, it is not necessary to use absolute value.
Pretest
18. x2 + 11x + 24 = 0
21. SOLUTION: SOLUTION: Factor the quadratic expression on the left side of
the equation and then apply the Zero Product
Property.
Because the index is odd, it is not necessary to use
absolute value.
22. SOLUTION: 19. CARS The current value V and the original value v
n
of a car are related by V = v(1 – r) , where r is the
rate of depreciation per year and n is the number of
years. If the original value of a car is $10,000, what
would be the current value of the car after 30
months at an annual depreciation rate of 10%?
Because you are taking an even root of an even
power and the result is an odd power, you must use
the absolute value of u.
SOLUTION: From the information given, we know that v = 10,000
and r = 0.1. The time is given in months but the
depreciation formula requires that the time be in
years. Since 30 months is equivalent to
23. SOLUTION: or 2.5 years, n = 2.5. Evaluate the given equation to find
the value of the car V.
Simplify.
24. The value of the car would be $7684.33.
SOLUTION: Simplify each expression.
20. SOLUTION: Because the result involves two even powers and
outside the radical, it is not necessary to use absolute value.
25. SOLUTION: 21. SOLUTION: eSolutions Manual - Powered by Cognero
Because the index is odd, it is not necessary to use
absolute value.
Page 4
Pretest
28. JOBS Destiny mows lawns for $8 per lawn and
25. weeds gardens for $10 per garden. If she had 8 jobs
and made $72, how many of the jobs were mowing?
How many were weeding?
SOLUTION: SOLUTION: Use the information given to write a system of
equations. Let x be the number of lawn mowing jobs
and y be the number of garden weeding jobs. If
Destiny had a total of 8 jobs, then x + y = 8. If she
makes $8 per lawn and $10 per garden and made a
total of $72, then 8x + 10y = 72. Therefore, a system
of equations representing this situation is
.
26. To solve this system, solve the first equation for y.
SOLUTION: Then substitute this expression for y into the other
equation and solve for x.
Substitute this value for x into the equation you
solved for y to find the value of y.
27. SOLUTION: The solution is (4, 4), which means that Destiny
worked 4 jobs mowing and 4 jobs weeding.
Solve each system of equations. State whether
the system is consistent and independent,
consistent and dependent, or inconsistent.
29. SOLUTION: To solve the system
by substitution, first solve one equation for x or y. In this case, x is
easiest to solve for in the second equation.
28. JOBS Destiny mows lawns for $8 per lawn and
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weeds gardens for $10 per garden. If she had 8 jobs
and made $72, how many of the jobs were mowing?
How many were weeding?
Then substitute this expression for x into the other
equation and solve for y.
Page 5
Because 0 = 0 is always true, there are an infinite
number of solutions. Therefore, the system is
consistent and dependent.
The solution is (4, 4), which means that Destiny
Pretest
worked 4 jobs mowing and 4 jobs weeding.
Solve each system of equations. State whether
the system is consistent and independent,
consistent and dependent, or inconsistent.
31. 29. SOLUTION: Eliminate one variable in two pairs of the system
SOLUTION: To solve the system
by substitution, first .
solve one equation for x or y. In this case, x is
easiest to solve for in the second equation.
Multiply the first equation by -1 and then add it to the
second equation.
Then substitute this expression for x into the other
equation and solve for y.
Multiply the third equation by 2 and then add it to the
first equation.
Solve this system of two equations by multiplying the
second by -3 and then adding the two equations
together.
Substitute this value for y into the equation you
solved for y to find the value of x.
The solution is (4, -8).
30. Substitute these two values into one of the original
equations to find z.
SOLUTION: Eliminate the variable y in the system
by adding multiplying the first equation by 2 and then adding the two equations
together.
The solution is (-1, 2, 7). The system is consistent
and independent because it has exactly one solution.
Because 0 = 0 is always true, there are an infinite
number of solutions. Therefore, the system is
consistent and dependent.
32. SOLUTION: 31. Eliminate one variable in two pairs of the system
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SOLUTION: Eliminate one variable in two pairs of the system
Page 6
.
Because 0 = 3 is not a true statement, this system
has no solutions. Therefore, the system is
inconsistent.
The solution is (-1, 2, 7). The system is consistent
Pretest
and independent because it has exactly one solution.
Solve each system of inequalities. If the system
has no solution, state no solution.
32. 33. SOLUTION: SOLUTION: Eliminate one variable in two pairs of the system
Use a solid line to graph each related equation,
and
, since each inequality
contains either ≥ or ≤.
.
Multiply the first equation by -1 and then add it to the
second equation.
Multiply the first equation by 2, the third equation by
3, and then add these equations.
Points on or above the line
make the
inequality
true, so shade the region above
the line
. Points on or to the right of the line
true, so
make the inequality shade the region to the right of the line
.
Because 0 = 3 is not a true statement, this system
has no solutions. Therefore, the system is
inconsistent.
Solve each system of inequalities. If the system
has no solution, state no solution.
33. SOLUTION: Use a solid line to graph each related equation,
and
, since each inequality
contains either ≥ or ≤.
The solution of
is Regions 1 and 3.
The solution of
is Regions 2 and 3.
Region 3 contains points that are solutions to both
inequalities, so this region is the solution to the
system.
Points on or above the line
make the
inequality
true, so shade the region above
the line
. Points on or to the right of the line
true, so
make the inequality shade the region to the right of the line
.
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34. SOLUTION: Use a dashed line to graph each related equation,
and
, since each inequality
contains either < or >. To graph
, first
rewrite the equation in slope intercept form,
Page 7
.
The solution of
is Regions 1 and 3.
The solution of
is Regions 2 and 3.
Region 3 contains points that are solutions to both
inequalities, so this region is the solution to the
system.
The solution of
is Regions 2 and 3.
Region 3 contains points that are solutions to both
inequalities, so this region is the solution to the
Pretest
system.
34. 35. SOLUTION: SOLUTION: Use a dashed line to graph each related equation,
and
, since each inequality
contains either < or >. To graph
, first
rewrite the equation in slope intercept form,
.
Use a dashed line to graph each related equation,
and
, since each inequality
contains either < or >. To graph
and
, first rewrite each equation in slope
intercept form,
Points to the left of the line
make the
inequality
true, so shade the region or to the
left of
. Points to the right of the line true, so
make the inequality shade the region to the right of
.
and
.
Points to the left of the line
make the
inequality
true, so shade the region or to
the left of
. Points below the line
true, so
make the inequality shade the region below
.
The solution of
is Regions 1 and 3.
The solution of
is Regions 2 and 3.
Region 3 contains points that are solutions to both
inequalities, so this region is the solution to the
system.
The solution of
is Regions 1 and 3.
The solution of
is Regions 2 and 3.
Region 3 contains points that are solutions to both
inequalities, so this region is the solution to the
system.
35. 36. SOLUTION: Use a dashed line to graph each related equation,
and
, since each inequality
contains either < or >. To graph
and
, first rewrite
each equation in slope
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by Cognero
intercept form,
and
.
SOLUTION: Graph each related equation,
and , using a solid line since each inequality
Page 8
contains ≤ or ≥. To graph
, first rewrite
The solution of
is Regions 1 and 3.
The solution of
is Regions 2 and 3.
Region 3 contains points that are solutions to both
inequalities, so this region is the solution to the
Pretest
system.
36. The solution of
is Region 1.
Since these solutions do not contain points that are
common to both inequalities, this system of
inequalities has no solution.
Find each of the following for D =
,E
SOLUTION: =
and Graph each related equation,
, using a solid line since each inequality
contains ≤ or ≥. To graph
, first rewrite
the equation in slope intercept form,
make the inequality
so shade the region above
The solution of
The solution of
.
37. D – F
SOLUTION: .
Points on or below the line
make the
inequality
true, so shade the region below
. Points on or above the
line
, and F =
38. D + 2F
SOLUTION: true,
.
is Region 2.
is Region 1.
Since these solutions do not contain points that are
common to both inequalities, this system of
inequalities has no solution.
39. 2D – E
SOLUTION: Find each of the following for D =
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=
, and F =
,E
Page 9
.
Pretest
39. 2D – E
SOLUTION: 41. 3D – 2E
SOLUTION: 42. D – 3E + 3F
40. D + E + F
SOLUTION: SOLUTION: 41. 3D – 2E
SOLUTION: eSolutions Manual - Powered by Cognero
Find each permutation or combination.
43. 9C5
SOLUTION: Page 10
Pretest
Find each permutation or combination.
43. 9C5
46. 5C5
SOLUTION: SOLUTION: 47. 4P2
SOLUTION: 44. 9P5
SOLUTION: 48. 4C2
SOLUTION: 45. 5P5
SOLUTION: 46. 5C5
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49. CARDS Three cards are randomly drawn from a
standard deck of 52 cards. Find each probability.
a. P(all even)
b. P(two clubs and one heart)
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SOLUTION: a. The even cards in each suit are 2, 4, 6, 8, and 10
49. CARDS Three cards are randomly drawn from a
Pretest
standard deck of 52 cards. Find each probability.
a. P(all even)
b. P(two clubs and one heart)
b. There are 13 clubs and 13 hearts in a standard
deck of cards. Since the order of the cards is not
important, use combinations and the Fundamental
Counting Principle to find the number of ways to
choose 2 out of 13 clubs and 1 out of 13 hearts.
SOLUTION: a. The even cards in each suit are 2, 4, 6, 8, and 10
for a total of 5 even cards per suit. Since there are 4
suits, there are a total of 4(5) or 20 even cards in
standard deck of cards. Since the order of the cards
is not important, use combinations to find the number
of ways to choose 3 out of 20 even cards.
.
Then use combinations to find the number of ways to
choose 3 cards out of 52 from the deck.
.
Then use combinations to find the number of ways to
choose 3 cards out of 52 from the deck.
To find the probability that 2 cards are clubs and 1
card is a heart, divided the number of ways to
choose 2 clubs cards and 1 heart by the number of
ways to choose 3 cards.
To find the probability that both cards are an even
number, divided the number of ways to choose 3
even cards by the number of ways to choose 3
cards.
The probability that 2 cards are clubs and 1 card is a
heart is
or about 4.6%.
Find the mean, median, and mode for each set
of data. Then find the range, variance, and
standard deviation for each population.
50. {7, 7, 8, 10, 10, 10}
SOLUTION: The probability that all three cards drawn are even
numbers is
or about 5.2%.
Mean To find the mean of the set of data, divide
the sum of the data by the number of pieces of data.
b. There are 13 clubs and 13 hearts in a standard
deck of cards. Since the order of the cards is not
important, use combinations and the Fundamental
Counting Principle to find the number of ways to
choose 2 out of 13 clubs and 1 out of 13 hearts.
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.
So the mean of the data is about 8.7.
Median To find the median, order the data and find
the middle number in the set of data or the average
of the two middle numbers.
The middle two numbers of this set are 8 and 10.
Page 12
The mean of these two numbers is
Therefore, the median of the set of data is 9.
.
The probability that 2 cards are clubs and 1 card is a
Pretest
heart is
The standard deviation of the population data is
about 1.4.
or about 4.6%.
Find the mean, median, and mode for each set
of data. Then find the range, variance, and
standard deviation for each population.
50. {7, 7, 8, 10, 10, 10}
SOLUTION: 51. {0.5, 0.4, 0.2, 0.5, 0.2}
SOLUTION: Mean To find the mean of the set of data, divide
the sum of the data by the number of pieces of data.
Mean To find the mean of the set of data, divide
the sum of the data by the number of pieces of data.
So the mean of the data is about 8.7.
Median To find the median, order the data and find
the middle number in the set of data or the average
of the two middle numbers.
The middle two numbers of this set are 8 and 10.
The mean of these two numbers is
.
Therefore, the median of the set of data is 9.
Mode To find the mode, determine which piece or
pieces of data appear most often. Since 10 appears
the most often, 10 is the mode.
Range The range is the difference between the greatest and least data values, so the range of the
data is 10 - 7 or 3.
Variance The population variance is calculated by
taking the mean of the sum of the squares of the
deviations from the population mean.
So the mean of the data is 0.38.
Median To find the median, order the data and find
the middle number in the set of data or the average
of the two middle numbers. When arranged in
ascending order, the data are as follows.
0.2, 0.2, 0.4, 0.5, 0.5
The middle number of this set is 0.4. Therefore, the
median of the set of data is 0.4.
Mode To find the mode, determine which piece or
pieces of data appear most often. Since 0.2 and 0.5
each appear twice, the data set has two modes, 0.2
and 0.5.
Range The range is the difference between the greatest and least data values, so the range of the
data is 0.5 - 0.2 or 0.3.
Variance The population variance is calculated by
taking the mean of the sum of the squares of the
deviations from the population mean.
The variance of the population data is about 0.02.
Standard Deviation The standard deviation of the population data is the square root of the variance.
The variance of the population data is about 1.9.
Standard Deviation The standard deviation of the population data is the square root of the variance.
The standard deviation of the population data is
about 0.14.
The standard deviation of the population data is
about 1.4.
51. {0.5, 0.4, 0.2, 0.5, 0.2}
SOLUTION: Mean To find the mean of the set of data, divide
the sum of the data by the number of pieces of data.
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