if the diameter is 10, its radius is half that, or 5.
![\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=5 \end{cases}\implies V=\cfrac{4\pi (5)^3}{3}\implies V=\cfrac{500\pi }{3} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill V\approx 523.599~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20sphere%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D5%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B4%5Cpi%20%285%29%5E3%7D%7B3%7D%5Cimplies%20V%3D%5Ccfrac%7B500%5Cpi%20%7D%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20V%5Capprox%20523.599~%5Chfill)
Answer: <em><u>The </u></em><em><u>product </u></em><em><u>of </u></em><em><u>8</u></em><em><u>5</u></em><em><u> </u></em><em><u>and </u></em><em><u>2</u></em><em><u>5</u></em><em><u> </u></em><em><u>is </u></em><em><u>2</u></em><em><u>1</u></em><em><u>2</u></em><em><u>5</u></em><em><u> </u></em><em><u>.</u></em><em><u> </u></em>
<em><u>8</u></em><em><u>5</u></em><em><u>x</u></em><em><u>2</u></em><em><u>5</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>2</u></em><em><u>1</u></em><em><u>2</u></em><em><u>5</u></em>
Step-by-step explanation:
<h3><em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> works</u></em><em><u> out</u></em><em><u> </u></em><em><u>!</u></em></h3>
Y= 2x + 1, hope this helps
