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) (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:
\ +\ 2w)
Then solve for
w\ =\ \frac{P\ -\ 2b}{2})
For your problem: 
, 
, and 
. Do the arithmetic.
John
My calculator said it, I believe it, that settles it
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