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SashulF [63]
3 years ago
7

What is the circumference of the blue circle?

Mathematics
1 answer:
VladimirAG [237]3 years ago
8 0

Answer:

62.8

Step-by-step explanation:

The radius of the blue circle is 5+5=10

the circumference is 2πr so

2(3.14)10=        (using 3.14 as π)

20(3.14)=

62.8

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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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Answer:

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Step-by-step explanation:

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mina [271]
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3 0
3 years ago
A 40 oz jar of peanut butter costs 5.25. how much does a 100 oz jar cost
Karolina [17]

Answer:

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Step-by-step explanation:

5.25 x 100/40 is 13.125 which rounds up to 13.13

7 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!
alex41 [277]

the correct answer should be (b) 6 is extraneous.

If you look at the second row of the solution you see (x-6) as a factor in one of the denominators. That's a problem: if 6 were a solution this denominator would become 0. But denominators being 0 is bad (undefined), so while 6 comes out of the algebraic manipulation as a solution, it is invalid, or extraneous.

-3 is a fine solution.

6 0
3 years ago
Read 2 more answers
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