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Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 38 feet

Respuesta :

texaschic101
P = 2(L + W)
P = 38
L = 2W + 1

38 = 2(2W + 1 + W)
38 = 2(3W + 1)
38 = 6W + 2
38 - 2 = 6W
36 = 6W
36/6 = W
6 = W <=== width

L = 2W + 1
L = 2(6) + 1
L = 12 + 1
L = 13 <=== length

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