Question 483060
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Let *[tex \Large w] represent the width.  Then the length must be *[tex \Large 6w].


The perimeter of a rectangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2l\ +\ 2w]


Substitute what you know:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(6w)\ +\ 2w\ =\ 52]


And solve for *[tex \Large w], then calculate *[tex \Large 6w] for the length.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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