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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.?

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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.?


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Answer #1 | 15/05 2017 20:08
length is a foot longer than twice its width l = 2w + 1 perimeter is 20 feet p = 2l + 2w = 20 substitute for l = 2w + 1 2(2w + 1) + 2w = 20 Distribute 4w + 2 + 2w = 20 Combine like terms 6w + 2 = 20 Subtract 2 from both sides 6w = 18 Divide both sides by 6 w = 3 <––––– Substitute w = 3 into l = 2w + 1 l = 2(3) + 1 l = 7 <––––– the rectangle is 3 feet by 7 feet <––––– http://goo.gl/y49sTR
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