Answer:
picture? answers? need more information on this
![\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\ -----\\ V=972\pi \end{cases}\implies 972\pi =\cfrac{4\pi r^3}{3}\implies \cfrac{3\cdot 972\pi }{4\pi }=r^3 \\\\\\ 729=r^3\implies \sqrt[3]{729}=r\implies \boxed{9=r}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20sphere%7D%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D~~%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0A-----%5C%5C%0AV%3D972%5Cpi%20%0A%5Cend%7Bcases%7D%5Cimplies%20972%5Cpi%20%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D%5Cimplies%20%5Ccfrac%7B3%5Ccdot%20972%5Cpi%20%7D%7B4%5Cpi%20%7D%3Dr%5E3%0A%5C%5C%5C%5C%5C%5C%0A729%3Dr%5E3%5Cimplies%20%5Csqrt%5B3%5D%7B729%7D%3Dr%5Cimplies%20%5Cboxed%7B9%3Dr%7D)
now, recall that a diameter is twice as long as the radius, so the diameter is 2r.
I believe you should do as much work as you can on each of these problems and only then ask for help. You'll almost always learn more by becoming involved in the thinking and problem solving you encounter here.
|x| + 5 = 13 reduces to |x| = 13, so x could be either -13 or +13.
The answer to this question is: 17