3-5 hw: pg. 165-167 #6-42eoe, 50-51, 56-57. 56. c 57. d
TRANSCRIPT
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3-5 HW: Pg. 165-167 #6-42eoe, 50-51, 56-57
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56. C 57. D
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Wir2.5• LESSON
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3-6 Solve Proportions Using Cross Products
Cross Product: __________________________________
________________________
Scale Drawing: __________________________________________
_______________________________________
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
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Use the cross products property
EXAMPLE 1
8 15 = x 6
Solve the proportion = .8 x
615
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.
820
615
=?
8 15 = 20 6?
120 = 120
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EXAMPLE 2 Standardized Test Practice
What is the value of x in the proportion = ?4x
83x –
A – 6 B – 3 C 3 D 6
SOLUTION
4x =
8x – 3
Write original proportion.
Cross products property4(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
A B C D
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Write and solve a proportion
EXAMPLE 3
x280
= 8 amount of food100 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.
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EXAMPLE 1
Solve the proportion. Check your solution.
GUIDED PRACTICE for Examples 1, 2, and 3
=4 a
2430
1. 5ANSWER
3x =
2x – 6
2.18ANSWER
4m5 =
m – 6 3. 30ANSWER
20.8ANSWER
WHAT IF? In Example 3, suppose the seal weighs 260 pounds. How much food should the seal be fed per day?
4.
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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 centimeters85 kilometers
Cross products property
d = 255 Simplify.ANSWER The actual distance between Cleveland and
Cincinnati is about 255 kilometers.
1 d = 85 3
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EXAMPLE 4 Use the scale on a mapGUIDED 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.
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
ANSWER about 960 ft
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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: 1
85 17085 170
= 2, so distance between Columbus
and Cincinnati is about 2 kil
8
omete s.
5
r
85
x
x
x
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Check Yourself
Pg. 171-173 #4-24eoe, 33-34, 44-45 and
Quiz Pg. 173 #2-16e