Question 1045707
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Let *[tex \Large x] represent the measure of the width.  Then *[tex \Large x\ +\ 1] must be the measure of the length.  Since the area of a rectangle is the length times the width:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x\ +\ 1)\ =\ 30]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ x\ -\ 30\ =\ 0]


Solve for *[tex \Large x] then calculate *[tex \Large x\ +\ 1]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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