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)
Total=14+20+16=50
P(blue)=20/50=2/5
so, answer=(2/5)*40=16
The improper fraction to that is 167/3
Answer:
b
Step-by-step explanation:
Given K is the midpoint of JL, then
JK = KL ← substitute values
6x = 3x + 3 ( subtract 3x from both sides )
3x = 3 ( divide both sides by 3 )
x = 1
Hence
JK = 6x = 6 × 1 = 6
KL = 3x + 3 = (3 × 1) + 3 = 3 + 3 = 6
JL = 6 + 6 = 12
Answer:dwqdwdqw
Step-by-step explanation:
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