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Ratios and Proportion

Ratios and Proportion. ALGEBRA 1 LESSON 4-1. (For help, go to Skills Handbook pages 724 and 727.). Write each fraction in simplest form. 1. 2. 3. Simplify each product. 4. 5. 6. . 49 84. 24 42. 135 180. 35 25. 40 14. 99 144. 96 88. 21 81. 108 56. . . . 4-1.

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Ratios and Proportion

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  1. Ratios and Proportion ALGEBRA 1 LESSON 4-1 (For help, go to Skills Handbook pages 724 and 727.) Write each fraction in simplest form. 1. 2. 3. Simplify each product. 4. 5. 6. 49 84 24 42 135 180 35 25 40 14 99 144 96 88 21 81 108 56    4-1

  2. Ratios and Proportion ALGEBRA 1 LESSON 4-1 1. 2. 3. 4. 5. 6. Solutions 49 7 • 7 7 84 7 • 12 12 = = 246 • 4 4 42 6 • 7 7 = = 13545 • 33 180 45 • 4 4 = = 3540 5 • 7 5 • 8 5 • 7 • 5 • 8 8 25 14 5 • 5 7 • 2 5 • 5 • 7 • 2 2   = = = = 4 99 96 9 • 11 8 • 12 9 • 11 • 8 • 12 9 3 144 88 12 • 12 8 • 11 12 • 12 • 8 • 11 12 4   = = = = 21 108 3 • 7 3 • 3 • 3 • 4 3 • 7 • 4 4 1 81 56 3 • 3 • 3 • 3 7 • 8 3 • 7 • 8 8 2   4 = = = = 4 4-1

  3. cost $1.56 ounces 48 oz = $.0325/oz Ratios and Proportion ALGEBRA 1 LESSON 4-1 Another brand of apple juice costs $1.56 for 48 oz. Find the unit rate. The unit rate is 3.25¢/oz. 4-1

  4. 40 mi 1 h 5280 ft 1 mi 1 h 60 min 1 min 60 s • • • Use appropriate conversion factors. 40 mi 1 h 5280 ft 1 mi 1 h 60 min 1 min 60 s • • • Divide the common units. = 58.6 ft/s Simplify. Ratios and Proportion ALGEBRA 1 LESSON 4-1 The fastest recorded speed for an eastern gray kangaroo is 40 mi per hour. What is the kangaroo’s speed in feet per second? The kangaroo’s speed is about 58.7 ft/s. 4-1

  5. Multiply each side by the least common multiple of 3 and 4, which is 12. y 3 3 4 • 12 = • 12 4y = 9 Simplify. 4y 4 9 4 = Divide each side by 4. y = 2.25 Simplify. Ratios and Proportion ALGEBRA 1 LESSON 4-1 y 3 3 4 Solve = . 4-1

  6. w 4.5 6 5 = – w(5) = (4.5)(–6) Write cross products. 5w = –27 Simplify. 5w 5 –27 5 = Divide each side by 5. w = –5.4 Simplify. Ratios and Proportion ALGEBRA 1 LESSON 4-1 w 4.5 6 5 Use cross products to solve the proportion = – . 4-1

  7. Define: Let t = time needed to ride 295 km. Relate: Tour de France average speed Write: equals = 295-km trip average speed kilometers hours 3630 92.5 295 t 3630 92.5 295 t = 3630t = 92.5(295) Write cross products. 92.5(295) 3630 t = Divide each side by 3630. t 7.5 Simplify. Round to the nearest tenth. Ratios and Proportion ALGEBRA 1 LESSON 4-1 In 2000, Lance Armstrong completed the 3630-km Tour de France course in 92.5 hours. Traveling at his average speed, how long would it take Lance Armstrong to ride 295 km? Traveling at his average speed, it would take Lance approximately 7.5 hours to cycle 295 km. 4-1

  8. z + 3 4 z – 4 6 = (z + 3)(6) = 4(z – 4) Write cross products. 6z + 18 = 4z – 16 Use the Distributive Property. 2z + 18 = –16 Subtract 4z from each side. 2z = –34 Subtract 18 from each side. 2z 2 –34 2 Divide each side by 2. = z = –17 Simplify. Ratios and Proportion ALGEBRA 1 LESSON 4-1 z + 3 4 z – 4 6 Solve the proportion = . 4-1

  9. Ratios and Proportion ALGEBRA 1 LESSON 4-1 Solve. 1. Find the unit rate of a 12-oz bottle of orange juice that sells for $1.29. 2. If you are driving 65 mi/h, how many feet per second are you driving? Solve each proportion. 3. 4. 5. 6. 10.75¢/oz. about 95.3 ft/s c 6 12 15 21 12 7 y 4.8 4 = = 1 2 3 + x 7 4 8 2 + x x – 4 25 35 –17 = = 4-1

  10. Proportions and Similar Figures ALGEBRA 1 LESSON 4-2 (For help, go to Skills Handbook and Lesson 4-1.) Simplify 1.2.3. Solve each proportion. 4. 5. 6. 7. 8. 9. 36 42 81 108 26 52 x 12 7 30 y 12 8 45 w 15 12 27 = = = n + 1 24 9 a 81 10 25 75 z 30 n 9 = = = 4-2

  11. x 12 7 30 = 90 81 96 45 84 30 y = x = a = 180 27 750 75 4 5 z = w = x = 2 Proportions and Similar Figures ALGEBRA 1 LESSON 4-2 1.5. 7. 9. 2. 3. 4. 6. 8. Solutions y 12 8 45 9 a 81 10 n 9 n + 1 24 36 42 6 • 66 6 • 7 7 = = = = = 45y = 12(8) 81a = 9(10) 24n = 9(n + 1) 81 108 27 • 33 27 • 4 4 = = 45y = 96 81a = 90 24n = 9n + 9 26 52 26 • 11 26 • 2 2 15n = 9 = = 9 15 2 15 1 9 n = y = 2 a = 1 3 5 n = 30x = 12(7) w 15 12 27 25 75 z 30 = = 30x = 84 27w = 15(12) 75z = 25(30) 27w = 180 75z = 750 2 3 w = 6 z = 10 4-2

  12. EF BC DE AB Relate: = Write a proportion comparing the lengths of the corresponding sides. Define: Let x = AB. 6 9 8 x Write: = Substitute 6 for EF, 9 for BC, 8 for DE, and x for AB. 6x = 9(8) Write cross products. 6x 6 72 6 = Divide each side by 6. x = 12 Simplify. Proportions and Similar Figures ALGEBRA 1 LESSON 4-2 In the figure below, ABC ~ DEF. Find AB. AB is 12 mm. 4-2

  13. 102 17 x 6 = Write a proportion. 17x = 102 • 6 Write cross products. 17x = 612 Simplify. 17x 17 612 17 = Divide each side by 17. x = 36 Simplify. Proportions and Similar Figures ALGEBRA 1 LESSON 4-2 A flagpole casts a shadow 102 feet long. A 6 ft tall man casts a shadow 17 feet long. How tall is the flagpole? The flagpole is 36 ft tall. 4-2

  14. map actual map actual Write a proportion. 1 10 2.25 x = 1 • x = 10 • 2.25 Write cross products. x = 22.5 Simplify. Proportions and Similar Figures ALGEBRA 1 LESSON 4-2 The scale of a map is 1 inch : 10 miles. The map distance from Valkaria to Gifford is 2.25 inches. Approximately how far is the actual distance? The actual distance from Valkaria to Gifford is approximately 22.5 mi. 4-2

  15. Proportions and Similar Figures ALGEBRA 1 LESSON 4-2 1. In the figure below, ABC ~ DEF. Find DF. 2. A boy who is 5.5 feet tall casts a shadow that is 8.25 feet long. The tree next to him casts a shadow that is 18 feet long. How tall is the tree? 3. The scale on a map is 1 in.: 20 mi. What is the actual distance between two towns that are 3.5 inches apart on the map? About 19.7 cm 12 ft 70 mi 4-2

  16. Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 (For help, go to skills handbook pages 727 and 728.) Find each product. 1. 0.6 • 9 2. 3.8 • 6.8 3. 4. Write each fraction as a decimal and as a percent. 5. 6. 7. 8. 9. 10. 11. 12. 23 60 20 46 17 135 5 34 • • 7 10 23 100 2 5 13 20 35 40 7 16 4 25 170 200 4-3

  17. Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 Solutions 1. 0.6 • 9 = 5.4 2. 3.8 • 6.8 = 25.84 3. 4. 5. 6. 7. 23 60 20 46 23 • 20 20 • 3 • 23 • 2 1 3 • 2 1 6 • = = = 17 135 5 34 17 • 5 5 • 27 • 17 • 2 1 27 • 2 1 54 • = = = 7 10 = 7 ÷ 10 = 0.7; 0.7(100%) = 70% 23 100 = 23 ÷ 10 = 0.23; 0.23(100%) = 23% 2 5 = 2 ÷ 5 = 0.4; 0.4(100%) = 40% 4-3

  18. 13 20 = 13 ÷ 20 =0.65; 0.65(100%) = 65% 8. 9. 10. 11. 12. 35 40 = 35 ÷ 40 = 0.875; 0.875(100%) = 87.5% 7 16 = 7 ÷ 16 = 0.4375; 0.4375(100%) = 43.75% 4 25 = 4 ÷ 25 = 0.16; 0.16(100%) = 16% 170 200 = 170 ÷ 200 = 0.85; 0.85(100%) = 85% Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 Solutions (continued) 4-3

  19. part whole n 100 27 90 percent = 90n = 2700 Find the cross products. n = 30 Divide each side by 90. Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 What percent of 90 is 27? 30% of 90 is 27. 4-3

  20. part whole 25 100 n 480 = 12,000 = 100n Find the cross products. 120 = n Divide each side by 100. Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 Find 25% of 480. 25% of 480 is 120. 4-3

  21. Relate: 70.8% of the total surface area is 361,736,000 km2. Define: Let t the total surface area. part whole 70.8 100 361,736,000 t Write: =  70.8t = 361,736,000,000  Find cross products. t = 510,926,553.7 Divide each side by 70.8. Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 Water covers about 361,736,000 km2, or about 70.8% of the earth’s surface area. Approximately what is the total surface area of the earth? The total surface area of the earth is approximately 510,926,554 km2. 4-3

  22. Relate: What percent of 140 is 84? Define: Let p = the decimal form of the percent. Write: p • 140 = 84 p = 0.6 Divide each side by 140. p = 60% Write the decimal as a percent. Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 What percent of 140 is 84? 140p = 84 60% of 140 is 84. 4-3

  23. Relate: What percent of 60 is 114? Define: Let n = the decimal form of the percent. Write: n • 60 = 114 n = 1.90 Divide each side by 60. n = 190% Write the decimal as a percent. Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 What percent of 60 is 144? 60n = 114 190% of 60 is 114. 4-3

  24. 1 5 1 5 1 5 19% = 20%. So is a good approximation of 19%. 323 325 325 and 5 are compatible numbers. 1 5 • 325 = 65 65 is approximately 19% of 323. 3 4 3 4 3 4 73% = 75%. So is a good approximation of 73%. 125 124 124 and 4 are compatible numbers. 3 4 • 124 = 93 Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 a. Estimate the number that is 19% of 323. b. What is 73% of 125? Use fractions to estimate the answer. 93 is approximately 73% of 125. 4-3

  25. Relate: What is 18% of 8056? Define: Let n = the unknown number. Write: n = 0.18 • 8056 n = 1450.08 Simplify. Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 A candidate for mayor sent out surveys to 8056 people in his city. After two weeks, about 18% of the surveys were returned. How many surveys were returned? n = 0.18 • 8056 About 1450 surveys were returned. 4-3

  26. Proportions and Percent Equations ALGEBRA 1 LESSON 4-3 1. What is 35% of 160? 2. What percent of 450 is 36? 3. 32 is 80% of what number? 4. What is 0.03% of 260,000? 5. What percent of 50 is 75? 6. Estimate 62% of 83? 56 8% 40 78 150% 51 4-3

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