HEY CAN ANYONE HELP ME OUT IN DIS PLS

Mathematics

1 answer:

dimulka [17.4K]2 years ago

7 0

The answer is C.
You have to subtract 13 from both sides.

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-5 >3 true or false plssssss answer

Ivan

Answer:

false

Step-by-step explanation:

-5 is smaller than 3 not bigger.

8 0

2 years ago

I forgot how to do this n they didn’t put an example help me lol

lyudmila [28]

Answer:

60 $m^2$

Step-by-step explanation:

First, find the area of the square as it the missing piece wasn't missing.

8 x 9 = 72 $m^2$

Then, find the area of the missing section.

3 x 4 = 12 $m^2$

Finally, subtract 12 $m^2$ from 72 $m^2$ and you get 60 $m^2$

3 0

2 years ago

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.

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

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