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stich3 [128]
3 years ago
12

Please please answer! The area of a rectangle is 45 square feet. The length is 4 feet longer than the width. Which equation coul

d you use to find the length of the rectangle in feet?

Mathematics
2 answers:
AURORKA [14]3 years ago
8 0
The answer is 11.25
I hope this helps!!
Mademuasel [1]3 years ago
7 0

Let the length be x


<u>Define Length and Width:</u>

Length =x

Width = x - 4 [Because the length is 4 feet longer, therefore the width is 4 feet shorter]



<u>Given that Length x Width = Area:</u>

x(x - 4) = 45


Answer: x(x - 4) = 45

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so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

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so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

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Step-by-step explanation:

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