Answer:
The answer is 8ft
Step-by-step explanation:
hope that helps :)))
Answer:
V = one-third pi (4 squared) (7) feet cubed
Step-by-step explanation:
Volume of cone v = 1/3×π(r²h)
r = 4 and h = 7
V = 1/3 × π(4²×7)
bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.
we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.
we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.
![\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%5C%5C%20%5Ctextit%7Basymptotes%7D%5Cquad%20y%3D%20k%5Cpm%20%5Ccfrac%7Bb%7D%7Ba%7D%28x-%20h%29%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)

Answer:
Angle bisector
Step-by-step explanation:
Find the measure of the angle COA. By angle addition postulate,

From the diagram,

then

Find the measure of the angle BOA. By angle addition postulate,

From the diagram,

then

Find the measure of the angle COB. By angle addition postulate,

From the diagram,

then

This means, the measures of angles COB and BOA are the same and are equal half the measure of angle COA, so angles COB and BOA are congruent. This means, the ray OB is the angle bisector of angle COA
I think the answer is D.348