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Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 20 feet.

a. Length = 6 feet, Width = 4 feet
b. Length = 8 feet, Width = 6 feet
c. Length = 7 feet, Width = 3 feet
d. Length = 5 feet, Width = 2 feet

User Ezhil
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Final answer:

The rectangle's dimensions, where the length is one foot longer than twice the width and the perimeter is 20 feet, are found to be Length = 7 feet and Width = 3 feet.

Step-by-step explanation:

To find the dimensions of a rectangle where the length is a foot longer than twice its width and the perimeter is 20 feet, we need to first define the variables for length (L) and width (W). According to the problem, L = 2W + 1. The perimeter of a rectangle is 2 times the length plus 2 times the width (P = 2L + 2W).

Since the perimeter is given as 20 feet, we can set up the equation 2(2W + 1) + 2W = 20. Simplifying, we get 4W + 2 + 2W = 20, which further simplifies to 6W + 2 = 20. Subtracting 2 from both sides gives us 6W = 18, and dividing both sides by 6 gives us W = 3 feet.

Plugging the width back into the length equation, we find the length to be L = 2(3) + 1 = 7 feet. Therefore, the dimensions of the rectangle are Length = 7 feet, Width = 3 feet, which corresponds to option c.

User Thepoosh
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