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serious [3.7K]
3 years ago
6

Solve the following problems. Be sure to indicate if a solution is to be rejected based on the contextual situation.

Mathematics
1 answer:
aleksklad [387]3 years ago
4 0

Answer:

1. 5/3 cm

2. 10 in

Step-by-step explanation:

1. The length of a rectangle is 4 cm more than 3 times its width.

Let the length be L and the width B

Then,

L = 4 + 3B

If the area of the rectangle is 15 cm2

Area = LB = 15

B(4 +3B) = 15

3B² + 4B - 15 = 0

3B² - 5B + 9B - 15 = 0

B(3B - 5) + 3(3B - 5) = 0

(3B - 5)(B + 3) = 0

3B - 5 =  or B + 3 = 0

B = 5/3 or -3

since B cannot be negative, B = 5/3 cm

2. The ratio of length to width in a rectangle is 2 ∶ 3.

Let the length be L and the width B

Then,

L/B = 2/3

2B = 3L, B = (3/2)L

when the area is 150 in2.

Area = LB = 150

(3/2)L² = 150

L² = 150×2÷3

L² = 100

L = √100

L = ±10

since the length cannot be negative, L = 10 in

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