if the sphere has a diameter of 5, then its radius is half that, or 2.5.
![\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2.5 \end{cases}\implies V=\cfrac{4\pi (2.5)^3}{3}\implies V=\cfrac{62.5\pi }{3} \\\\\\ V\approx 65.44984694978736\implies V=\stackrel{\textit{rounded up}}{65.45}](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%3D2.5%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B4%5Cpi%20%282.5%29%5E3%7D%7B3%7D%5Cimplies%20V%3D%5Ccfrac%7B62.5%5Cpi%20%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20V%5Capprox%2065.44984694978736%5Cimplies%20V%3D%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B65.45%7D)
Answer:
13.3f
Step-by-step explanation:
Answer:
x=10.5
Step-by-step explanation:

Answer:
Quadrant III ( C )
Quadrant IV ( D )
Step-by-step explanation:
Ordered pair ( a, b )
gives a point P on the coordinate plane
( a, b ) = ( x, y )
given : b = negative ( y-axis )
a ≠ 0 ( i.e. a = negative or positive ) ( x -axis )
Therefore point P is located in the : Third and fourth quadrants