Question 1034340
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If the width of a rectangle is *[tex \Large \frac{x\ +\ 1}{x}] and its length is *[tex \Large \frac{3x}{x^2\ -\ 1}], then its area is:


The area of a rectangle is simply the length times the width, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ +\ 1}{x}\ \cdot\ \frac{3x}{x^2\ -\ 1}]


You will find this much easier to simplify if you first factor the difference of two squares in the second denominator.


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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