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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, eSolutions Manual - Powered by Cognero 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 ≥ 13. eSolutions Manual - Powered by Cognero 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 SOLUTION: eSolutions Manual - Powered by Cognero 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 eSolutions Manual - Powered by Cognero 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 eSolutions Manual - Powered by Cognero 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 . eSolutions Manual - Powered by Cognero 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 eSolutions Manual - Powered 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 = eSolutions Manual - Powered by Cognero = , 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 SOLUTION: eSolutions Manual - Powered by Cognero 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) Page 11 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. eSolutions Manual - Powered by Cognero . 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. eSolutions Manual - Powered by Cognero Page 13