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12 and 4. Either one can be the length, the other is the width.

To solve algebraically:

2L + 2W = 32

L x W = 48

substitute the L value for W (W = 16-L)

L (16-L) = 48

-L2 +16L -48 = 0

L2 -16L + 48 = 0

(L-12)(L-4) = 0

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โˆ™ 2011-10-21 04:52:01
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Q: What is the length and width of a rectangle with a perimeter of 32 feet and area of 48 feet squared?
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