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Vikki [24]
4 years ago
13

A sphere has a volume of 457.3 in^3. What is the radius of the sphere?

Mathematics
1 answer:
Ne4ueva [31]4 years ago
4 0

Answer:

\large\boxed{R\approx4.78\ in}

Step-by-step explanation:

The formula of a volume of a sphere:

V=\dfrac{4}{3}\pi R^3

R - radius

We have the volume V = 457.3 in³. Substitute:

\dfrac{4}{3}\pi R^3=457.3\qquad\text{multiply both sides by 3}\\\\4\pi R^3=1371.9\qquad\text{divide both sides by 4}\\\\\pi R^3=342.975\qquad\text{divide both sides by}\ \pi\\\\R^3=\dfrac{342.975}{\pi}\qquad/\pi\approx3.14/\\\\R^3\approx\dfrac{342.975}{3.14}\\\\R^3\approx109.2277\to R\approx\sqrt{109.2277}\\\\R\approx4.78\ in

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4 years ago
8/x-5 - 9/x-4 = 5/x^2-9x+20
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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2752942

_______________


Solve the equation:

\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}


Reduce the fractions at the left side so that they have the same denominator:

\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}


Numerators must be equal:

\mathsf{8x-32-(9x-45)=5}\\\\
\mathsf{8x-32-9x+45=5}\\\\
\mathsf{8x-9x=5+32-45}\\\\
\mathsf{-x=-8}\\\\
\mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}


I hope this helps. =)


Tags:  <em>rational equation fraction solution algebra</em>

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3 years ago
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