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Section 2.3 Applications

Blueprint for Problem Solving. Read the problem, and then mentally list the items that are known and the items that are unknown.Assign a variable to one of the unknown items. (In most cases this will amount to letting x = the item that is asked for in the problem.) Then translate the other inform

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Section 2.3 Applications

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    1. Section 2.3 Applications

    2. Blueprint for Problem Solving Read the problem, and then mentally list the items that are known and the items that are unknown. Assign a variable to one of the unknown items. (In most cases this will amount to letting x = the item that is asked for in the problem.) Then translate the other information in the problem to expressions involving the variable. Reread the problem, and then write an equation, using the items and variables listed in Steps 1 and 2, that describes the situation.

    3. Blueprint for Problem Solving, continued Solve the equation found in Step 3 Write your answer in a complete sentence. Reread the problem and check your solution with the original words in the problem.

    4. Geometry, Problem 2 on page 104 The length of a rectangle is 5 times the width. The perimeter is 48 inches. Find the dimensions. List the unknowns: length and width Let W = the width of the rectangle 5W = the length of the rectangle

    5. Geometry, Problem 2 on page 104 Recall the formula for Perimeter: P = 2L + 2W 2(5W) + 2(W) = 48 2(5W) + 2(W) = 48 10W + 2W = 48 12W = 48 W = 4 Length = 5(4) = 20

    6. Geometry, Problem 2 on page 104 The length of a rectangle is 5 times the width. The perimeter is 48 inches. Find the dimensions. The width of the rectangle is 4 inches and the length is 20 inches.

    7. Geometry, Problem 2 on page 104 The length of a rectangle is 5 times the width. The perimeter is 48 inches. Find the dimensions. If the width of the rectangle is 4 inches, then the length is 5(4 inches) = 20 inches. The perimeter is 2(20) + 2(4) = 40 + 8 = 48 inches.

    8. Geometry, Problem 2 on page 104

    9. Problem 10, page 105 A garden is in the shape of a square with a perimeter of 42 feet. The garden is surrounded by two fences. One fence is around the perimeter of the garden, wile the second fence is 3 feet from the first fence, as the figure indicates. If the material used to build the two fences costs $1.28 per foot, what was the total cost of the fences?

    10. Problem 10, page 105 s = length of side of inner fence. Perimeter = 4s 4s = 42 s = 10.5 s + 6 = 16.5 The perimeter of the inner fence + the perimeter of the outer fence = total amount of fencing needed

    11. Problem 10, page 105 4(s) + 4(s + 6) = 42 + 66 = 108 Cost of the fence per foot X # feet of fencing = Total Cost ($1.28)(108) = $138.24 The total cost of the fencing is $138.24.

    12. Problem 16, page 105 Suppose a college bookstore buys a textbook from a publishing company, then marks up the price they paid for the book 33%, and sells it to a student at the marked-up price. If the student pays $45.00 for the textbook, what did the bookstore pay for it? Round your answer to the nearest cent.

    13. Problem 16, page 105 The amount the bookstore paid for the book + the amount of the marked-up price = the amount the student pays for the book Let x = the amount the bookstore paid for the book 0.33x = the amount of the marked-up price x + 0.33x = 45.00

    14. Problem 16, page 105 x + 0.33x = 45 1.33x = 45 x = 33.83458646617 The bookstore paid $33.83 for the textbook.

    15. Problem 16, page 105 33.83 + 0.33(33.83) = 44.9939 The price of the textbook paid by the student was $45.00

    16. Geometry, Problem 16 on page 105

    17. Problem 20, page 106 Two complementary angles are such that one is five times as large as the other. Find the two angles. Let a = measure of the smaller angle 5a = measure of the larger angle

    18. Problem 20, page 106 Recall that complementary angles add to 90 degrees. a + 5a = 90 6a = 90 a = 15 5a = 75 The two angles are 15? and 75?.

    19. Problem 20, page 106 15 + 75 = 90 The angles are complementary. 75 = 5(15) The larger angle is 5 times the smaller angle.

    20. Problem 20, page 106

    21. Problem 24, page 106 The smallest angle in a triangle is half of the largest angle. The third angle is 15? less than the largest angle. Find the measure of all three angles. Let x = measure of the largest angle. 0.5x = measure of the smallest angle x - 15 = measure of the third angle

    22. Problem 24, page 106 Recall that the sum of the three angles of a triangle equals 180?. x + 0.5x + x - 15 = 180 2.5x = 195 x = 78 0.5x = 39 x - 15 = 63

    23. Problem 24, page 106 The angles of the triangle are 78?, 39°, and 63°. 78 + 39 + 63 = 180 The smallest angle (39?) is half of the largest angle (78?) and the third angle (63?) is 15? less than the largest angle (78?).

    24. Problem 24, page 106

    25. Problem 32, page 106 The length of a rectangle is 4 yards more than twice the width. If the area is 70 square yards, find the width and length. W = width of rectangle 2W + 4 = length of rectangle Recall that the area of a rectangle = length X width.

    26. Problem 32, page 106 (2W + 4)W = 70 2W2 + 4W = 70 2W2 + 4W - 70 = 0 W2 + 2W - 35 = 0 (W + 7)(W - 5) = 0 W + 7 = 0 or W - 5 = 0 W = -7 or W = 5 2W + 4 = -10 2W + 4 = 14

    27. Problem 32, Page 106 The width cannot be -7. The width is 5 yards. The length is 14 yards. 5 yards x 14 yards = 70 yards The length is 4 yards more than twice the width. ( 14 = 4 + 2(5))

    28. Problem 32, page 106

    29. Problem 36, Page 106 A Girl Scout troop sells 62 tickets to their mother and daughter dinner for a total of $216. If the tickets cost $4.00 for mothers and $3.00 for daughters, how many of each ticket did they sell? Let x = # daughters 62 - x = # mothers

    30. Problem 36, page 106 # daughters x Cost for each daughter + # mothers x Cost for each mother = total cost x(3) + (62 - x)(4) = 216 3x + 248 - 4x = 216 -x = -32 x = 32 62 - x = 30

    31. Problem 36, page 106 There were 32 daughters and 30 mothers attending the dinner. 32(3) + 30(4) = 96 + 120 = 216

    32. Problem 36, page 106

    33. Problem 40, page 107 A person makes a long distance person-to-person call to Santa Barbara, California. The telephone company charges 41 cents for the first minute and 32 cents for each additional minute. Because the call is person-to-person, there is also a service charge of $3.00. If the cost of the call is $6.29, how many minutes did the person talk?

    34. Problem 40, page 107 Let m = the number of minutes the call lasted m - 1 = the number of additional minutes The cost for the first minute + the cost for the additional minutes + the service charge = total cost of call 0.41 + 0.32(m - 1) + 3.00 = 6.29

    35. Problem 40, page 107 0.41 + 0.32(m - 1) + 3.00 = 6.29 0.41 + 0.32m - 0.32 + 3.00 = 6.29 0.32m + 3.09 = 6.29 0.32m = 3.2 m = 10 The phone call lasted 10 minutes.

    36. Problem 40, page 107 0.41 + 0.32(9) + 3.00 = 0.41 + 2.88 + 3.00 = 6.29

    37. Problem 40, page 107

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