SOLUTION: perimiter of rectangle is 64cm the , the length is 4cm less than twice the width find the l&w I tried 4/64=32 2/32=16

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Question 573941: perimiter of rectangle is 64cm the , the length is 4cm less than twice the width find the l&w I tried 4/64=32 2/32=16
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


I just did this one but here it is again:

If a rectangle with perimeter has a length that is units longer (or shorter if ) than times the width, ,

Then knowing that the Perimeter is given by:

,

make the substitution:



Then solve for







For your problem: , , and . Do the arithmetic.

John

My calculator said it, I believe it, that settles it
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