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 [ -39503.2+69251.2 \right ]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%5Cleft%20%5B%20%20-39503.2%2B69251.2%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:
D) f(x) = 6x
Step-by-step explanation:
Let's remember that V = lwh. If you are given two other measurements, you can multiply those two and divide V by that quantity to solve for the missing variable. Same process no matter the value:)
Answer:
25
Step-by-step explanation:
To calculate rate, divide the number of cans by the time to get "Cans per minute".
300/15min = 20/min
Let m represent minutes and c for cans.
We write an equation for the problem:
c = 20m
We want to know the time for 500 cans, so substitute 500 for c.
500 = 20m
Isolate m and solve:
m = 500/20
m = 25
It will take 25 minutes to put 500 labels.
