How do you get the answer to the math problem The perimeter of a rectangle is 58m and the length is 2m more than twice the width find the dimension for the length and the width?
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Answer:
Perimeter of a rectangle is 2 times (length + width) P = 2 (L+W) In this case the length is 2m more than twice the width. That is written L = 2 + 2W P = 2 (L + W) 58 = 2 (2+ 2W + W) 58 = 2 (2 + 3W) 58/2 = (2(2+3W))/2 29 = 2 + 3W 29 - 2 = 2 + 3W - 2 27 = 3W 9 = W L = 2 + 2W L = 2 +2(9) L = 2 + 18 L = 20 For a rectangle with a perimeter of 58 m and a length that is 2 m more than twice the width, the length is 20 m and the width is 9 m.
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