Ratios and Proportion    Write each fraction in simplest form. 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    7-1

Ratios and Proportion       1. 2. 3. 4. 5. 6. Solutions = = = = ALGEBRA 1 LESSON 4-1 1. 2. 3. 4. 5. 6. Solutions 49 7 • 7 7 84 7 • 12 12 = = 24 6 • 4 4 42 6 • 7 7 = = 135 45 • 3 3 180 45 • 4 4 = = 35 40 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 7-1

Ratios and Proportion ALGEBRA 1 LESSON 4-1 Another brand of apple juice costs $1.56 for 48 oz. Find the unit rate. cost $1.56 ounces 48 oz = $.0325/oz The unit rate is 3.25¢/oz. 7-1

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? 40 mi 1 h 5280 ft 1 mi 60 min 1 min 60 s • Use appropriate conversion factors. 40 mi 1 h 5280 ft 1 mi 60 min 1 min 60 s • Divide the common units. = 58.6 ft/s Simplify. The kangaroo’s speed is about 58.7 ft/s. 7-1

Ratios and Proportion Solve = . • 12 = ALGEBRA 1 LESSON 4-1 Solve = . y 3 3 4 y 3 • 12 = 4 Multiply each side by the least common multiple of 3 and 4, which is 12. • 12 4y = 9 Simplify. 4y 4 = 9 Divide each side by 4. y = 2.25 Simplify. 7-2

Ratios and Proportion Use cross products to solve the proportion = – . ALGEBRA 1 LESSON 4-1 w 4.5 6 5 Use cross products to solve the proportion = – . w 4.5 = – 6 5 w(5) = (4.5)(–6) Write cross products. 5w = –27 Simplify. 5w 5 = –27 Divide each side by 5. w = –5.4 Simplify. 7-2

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? Define: Let t = time needed to ride 295 km. Relate: Tour de France average speed Write: equals = 295-km trip 3630 92.5 295 t kilometers hours 3630 92.5 = 295 t 3630t = 92.5(295) Write cross products. t = Divide each side by 3630. 92.5(295) 3630 t 7.5 Simplify. Round to the nearest tenth. Traveling at his average speed, it would take Lance approximately 7.5 hours to cycle 295 km. 7-2

Ratios and Proportion Solve the proportion = . = (z + 3)(6) = 4(z – 4) ALGEBRA 1 LESSON 4-1 z + 3 4 z – 4 6 Solve the proportion = . 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. Divide each side by 2. = 2z 2 –34 z = –17 Simplify. 7-2

Ratios and Proportion Solve. 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 3 + x 7 4 8 1 2 2 + x x – 4 25 35 = = –17 7-2

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 = = = 9 a 81 10 25 75 z 30 n 9 n + 1 24 = = = 7-2

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

Proportions and Similar Figures ALGEBRA 1 LESSON 4-2 In the figure below, ABC ~ DEF. Find AB. Write: = Relate: = EF BC DE AB Define: Let x = AB. 6 9 8 x Write a proportion comparing the lengths of the corresponding sides. Substitute 6 for EF, 9 for BC, 8 for DE, and x for AB. 6x = 9(8) Write cross products. = Divide each side by 6. 6x 6 72 x = 12 Simplify. AB is 12 mm. 7-2

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? = 102 17 x 6 Write a proportion. 17x = 102 • 6 Write cross products. 17x = 612 Simplify. Divide each side by 17. = 17x 17 612 x = 36 Simplify. The flagpole is 36 ft tall. 7-2

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? = map actual 1 10 2.25 x Write a proportion. 1 • x = 10 • 2.25 Write cross products. x = 22.5 Simplify. The actual distance from Valkaria to Gifford is approximately 22.5 mi. 7-2

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 7-2