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:
64 yds
Step-by-step explanation:
by proportion, MN is half way to BC, so BC = 2(MN) = 64 yds
3. f(-6) = 12+1 =13
f(-2) = 4+1 = 5
f(0) =1
Range {1,5,13}
4. f(-2) = (-2)^3+1 =-7
f(-1) = (-1)^2 +1 =0
f(3) = (3)^3 +1 = 28
Range = {-7,0,28}
5.the sequence is arithmetic
d= -11+19 = 8
an = a1 + d(n-1)
an = -19 +8(n-1)
6.l =w+5
a =l*w
a(w) =(w+5) * w
a(w)= w^2 +5w
f(w) = w^2 +5w
f(8) = 8^2 +5(8)
f(8) = 64 +40
f(8) =104 in^2
I believe the answer is D
y+2=-2(x-5)
-2*-5=10
y+2=-2x+10
-2 -2
y=-2x+8
