Question 333119
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If the ratio of the measures of the length and width of a rectangle is 3:2, then there exists a real number *[tex \Large k] such that the length is equal to *[tex \Large 3k] and the width is equal to *[tex \Large 2k]


Using the formula for the perimeter:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l\ +\ 2w\ =\ 55]


and substituting what we know about the length and width:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(3k)\ +\ 2(2k)\ =\ 55]


Solve for *[tex \Large k], then multiply by 3 to get the length and by 2 to get the width.


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