Question 494756
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First note that the perimeter is given in yards and the difference between the length and the width is also given in yards.  Hence there are no difficulties in this problem regarding changing units.


The perimeter of a rectangle is given by:


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


If you let *[tex \Large w] represent the width, and you are given that the length is then *[tex \Large w\ +\ 30], you can write:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ 2(w\ +\ 30)\ +\ 2w\ =\ 300]


Solve for *[tex \Large w] and then calculate *[tex \Large w\ +\ 30]


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