Question 1021108
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The area of the rectangle is 160 square yards, so if *[tex \Large l] is the length, and *[tex \Large w] is the width, we can write *[tex \Large lw\ =\ 160].


But since we know that the length is 6 yards greater that the width, we can say *[tex \Large l\ =\ w\ +\ 6]


Substituting, we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w(w\ +\ 6)\ =\ 160]


which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ 6w\ -\ 160\ =\ 0]


Solve the factorable quadratic for *[tex \Large w].  The lesser root will be the width and the greater root will be the length.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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