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The length of a rectangular poster is 10 more inches than three times its width. The area of the poster is 88 square inches. Solve for the dimensions (length and width) of the poster.

The dimensions are ___ inches by___ inches.

User Tjvr
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2 Answers

6 votes

Answer:

It's just 22

Explanation:

22

User Andreas Rehm
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2 votes

Let's assume that the width of the poster is "x" inches.

According to the problem, the length of the poster is 10 more inches than three times its width. So, the length is (3x + 10) inches.

The area of a rectangle is given by the formula A = length x width. We are given that the area of the poster is 88 square inches. So, we can write:

(3x + 10) x x = 88

Simplifying the above equation, we get:

3x^2 + 10x - 88 = 0

We can solve this quadratic equation by factoring or using the quadratic formula. Let's use the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / 2a

where a = 3, b = 10, and c = -88

Plugging in the values, we get:

x = [-10 ± sqrt(10^2 - 4(3)(-88))] / 2(3)

x = [-10 ± sqrt(1000)] / 6

x = [-10 ± 31.62] / 6

x = 3.27 or x = -8.94

Since the width cannot be negative, we ignore the negative solution. Therefore, the width of the poster is approximately 3.27 inches.

Using this value, we can find the length:

length = 3x + 10 = 3(3.27) + 10 = 19.81 inches (rounded to two decimal places)

Therefore, the dimensions of the poster are approximately 3.27 inches by 19.81 inches.

User Vesperknight
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