PRE CALC QUESTION, PLEASE HELP

Mathematics

1 answer:

strojnjashka [21]2 years ago

4 0

For applies= associative and distributive
and for doesnt apply = commutative.

im not 100 sure for the distributive answer but ik the other ones are right

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Calculate the total area of the shaded region.

LUCKY_DIMON [66]

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

$\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}$

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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1 year ago

You are carpeting a room in a house for your job. You have a boss that likes to play jokes, so instead of just giving you the di

forsale [732]

Answer:

$x^2-4x+3$

$63\ \text{sq. units}$

Step-by-step explanation:

l = Length = $\dfrac{x^2-x-6}{x-5}$

w = Width = $\dfrac{x^2-6x+5}{x+2}$

Area is given by

$A=lw\\\Rightarrow A=\dfrac{x^2-x-6}{x-5}\times\dfrac{x^2-6x+5}{x+2}\\\Rightarrow A=\dfrac{x^2-3x+2x-6}{x-5}\times\dfrac{x^2-5x-1x+5}{x+2}\\\Rightarrow A=\dfrac{(x-3)(x+2)}{x-5}\times\dfrac{(x-5)(x-1)}{x+2}\\\Rightarrow A=(x-3)(x-1)\\\Rightarrow A=x^2-4x+3$