# 3-5 HW: Pg. 165-167 #6-42eoe, 50-51, 56-57. 56. C 57. D.

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3-5 HW: Pg. 165-167 #6-42eoe, 50-51, 56-57

56. C 57. D

Wir2.5 LESSON

3-6 Solve Proportions Using Cross Products denominator of the other ratio. 2dimensional drawing of an object where dimensions product of numerator of 1 ratio and the of drawing are in proportion to the real object

Use the cross products property EXAMPLE 1 8 15 = x 6 Solve the proportion =. 8 x 6 15 Cross products property Simplify. 120 = 6x Divide each side by 6. 20 = x The solution is 20. Check by substituting 20 for x in the original proportion. ANSWER Substitute 20 for x. CHECK Cross products property Simplify. Solution checks. 8 20 6 15 = ? 8 15 = 20 6 ? 120 = 120

EXAMPLE 2 Standardized Test Practice What is the value of x in the proportion = ? 4 x 8 3x – A – 6 B – 3 C 3 D 6 SOLUTION 4 x = 8 x– 3 Write original proportion. Cross products property 4(x – 3) = x 8 4x – 12 = 8x Simplify. Subtract 4x from each side. –12 = 4x Divide each side by 4. –3 = x The value of x is –3. The correct answer is B. ANSWER ABCD

Write and solve a proportion EXAMPLE 3 x 280 = 8 amount of food 100 weight of seal SOLUTION STEP 1 Write a proportion involving two ratios that compare the amount of food with the weight of the seal. Each day, the seals at an aquarium are each fed 8 pounds of food for every 100 pounds of their body weight. A seal at the aquarium weighs 280 pounds. How much food should the seal be fed per day ? Seals STEP 2 Solve the proportion. 8 280 = 100 x Cross products property 2240 = 100x Simplify. 22.4 = x Divide each side by 100. ANSWER A 280 pound seal should be fed 22.4 pounds of food per day.

EXAMPLE 1 Solve the proportion. Check your solution. GUIDED PRACTICE for Examples 1, 2, and 3 = 4 a 24 30 1. 5 ANSWER 3 x = 2 x– 6 2. 18 ANSWER 4 m 5 = m – 6 3. 30 ANSWER 20.8 ANSWER WHAT IF? In Example 3, suppose the seal weighs 260 pounds. How much food should the seal be fed per day ? 4.

EXAMPLE 4 Use the scale on a map Maps Use a metric ruler and the map of Ohio to estimate the distance between Cleveland and Cincinnati. SOLUTION From the map’s scale, 1 centimeter represents 85 kilometers. On the map, the distance between Cleveland and Cincinnati is about 4.2 centimeters. Write and solve a proportion to find the distance d between the cities. = 3 d 1 centimeters 85 kilometers Cross products property d = 255 Simplify. ANSWERThe actual distance between Cleveland and Cincinnati is about 255 kilometers. 1 d = 85 3

EXAMPLE 4 Use the scale on a map GUIDED PRACTICE for Example 4 5. Use a metric ruler and the map in Example 4 to estimate the distance (in kilometers) between Columbus and Cleveland. about 144.5 kmANSWER 6.6. The ship model kits sold at a hobby store have a scale of 1 ft : 600 ft. A completed model of the Queen Elizabeth II is 1.6 feet long. Estimate the actual length of the Queen Elizabeth II. Model ships ANSWERabout 960 ft

Summary How do you solve proportions using cross products? Ans: Cross products of a proportion are equal. Multiply the numerator of each ratio by the denominator of the other ratio and write an equal sign in between the 2 products. Then solve for the variable. If the distance between Columbus and Cincinnati is about 170 km, and the map’s scale is drawn so 1 cm = 85 km. Can you estimate the distance between the two cities? Ans:

Check Yourself Pg. 171-173 #4-24eoe, 33-34, 44-45 and Quiz Pg. 173 #2-16e

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