Question 348019
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The area of a rectangle is given by:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ A\ =\ lw]


and the Perimeter is given by:


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


Where *[tex \Large l] represents the measure of the length and *[tex \Large w] represents the measure of the width.


But we are given that


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ P\ =\ \frac{3A}{2}]


So:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ \frac{3lw}{2}\ =\ 2l\ +\ 2w]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ 3lw\ =\ 4l\ +\ 4w]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ 3lw\ -\ 4l\ =\ 4w]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ l(3w\ -\ 4)\ =\ 4w]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ l\ =\ \frac{4w}{3w\ -\ 4}]



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