Answer:
Volume of the cone is increasing at the rate
.
Step-by-step explanation:
Given: The radius of a right circular cone is increasing at a rate of
in/s while its height is decreasing at a rate of
in/s.
To find: The rate at which volume of the cone changing when the radius is
in. and the height is
in.
Solution:
We have,
,
,
, 
Now, let
be the volume of the cone.
So,
Differentiate with respect to
.
![\frac{dv}{dt} =\frac{1}{3}\pi \left [ r^2\frac{dh}{dt}+h\left ( 2r \right )\frac{dr}{dt} \right ]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%5Cleft%20%5B%20r%5E2%5Cfrac%7Bdh%7D%7Bdt%7D%2Bh%5Cleft%20%28%202r%20%5Cright%20%29%5Cfrac%7Bdr%7D%7Bdt%7D%20%5Cright%20%5D)
Now, on substituting the values, we get
![\frac{dv}{dt} =\frac{1}{3}\pi\left [ \left ( 134 \right )^2\left ( -2.2 \right )+\left ( 136\right )\left ( 2 \right )\left ( 134 \right )\left ( 1.9 \right ) \right ]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%5Cleft%20%5B%20%5Cleft%20%28%20134%20%5Cright%20%29%5E2%5Cleft%20%28%20-2.2%20%5Cright%20%29%2B%5Cleft%20%28%20%20136%5Cright%20%29%5Cleft%20%28%202%20%5Cright%20%29%5Cleft%20%28%20134%20%5Cright%20%29%5Cleft%20%28%201.9%20%5Cright%20%29%20%5Cright%20%5D)
![\frac{dv}{dt} =\frac{1}{3}\pi\left [ 29748 \right ]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%5Cleft%20%5B%2029748%20%5Cright%20%5D)

Hence, the volume of the cone is increasing at the rate
.
Answer:
c
Step-by-step explanation:
The equation of the circle is may be expressed as,
(x - h)² + (y - k)² = r²
where h and k are the abscissa and ordinates of its center, respectively, and r is radius. Substituting the given values to the equation above,
(x - 5)² + (y - 3)² = 4²
Simplifying further the equation gives,
x² + y² - 10x - 6y + 18
Answer:
D i believe.
Step-by-step explanation:
P.S. if im wong im really sorry, its been awhile since ive had to do that. But honestly i think its D. also comment on this if im right or wrong pls, id like to know.