Given:


To find:
The rate of change in volume at 
Solution:
We know that, volume of a cone is

Differentiate with respect to t.
![\dfrac{dV}{dt}=\dfrac{1}{3}\pi\times \left[(r^2\dfrac{dh}{dt}) + h(2r\dfrac{dr}{dt})\right]](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5Cpi%5Ctimes%20%5Cleft%5B%28r%5E2%5Cdfrac%7Bdh%7D%7Bdt%7D%29%20%2B%20h%282r%5Cdfrac%7Bdr%7D%7Bdt%7D%29%5Cright%5D)
Substitute the given values.
![\dfrac{dV}{dt}=\dfrac{1}{3}\times \dfrac{22}{7}\times \left[(120)^2(-2.1) +175(2)(120)(1.4)\right]](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cdfrac%7B22%7D%7B7%7D%5Ctimes%20%5Cleft%5B%28120%29%5E2%28-2.1%29%20%2B175%282%29%28120%29%281.4%29%5Cright%5D)
![\dfrac{dV}{dt}=\dfrac{22}{21}\times \left[-30240+58800\right]](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%3D%5Cdfrac%7B22%7D%7B21%7D%5Ctimes%20%5Cleft%5B-30240%2B58800%5Cright%5D)


Therefore, the volume of decreased by 29920 cubic inches per second.
The answer is C) s+10 because Lupe's dog is 10 inches taller than Sarah's dog; therefore, all you have to do is add 10 inches to the height of Sarah's dog.
The sound intensity of the Pile Driver is 39.5
or nearly 40 times the sound intensity of the jackhammer.
Given with Loudness in dB for pile driver = 112 dB
We have to convert it in terms of sound intensity.
First,
112dB/10 = 11.2
Then we'll use this as exponent of 10
(10)^(11.2) = 1.5849 * 10 ^ 11
Then use the equation of Watts per square meter to find the intensity:
I / (10^-12 W/m^2) =1.5849 * 10 ^ 11
I = sound intensity = 0.158
Then compare:
Sound intensity of Pile Driver/ Sound intensity of Jackhammer
(0.158) / (0.004)
= 39.5
or nearly 40 times the jackhammer.

is a parabola (looks like the letter U).
The letter a represents the coefficient of

and it controls two things (1) how wide or narrow the parabola is and (2) whether it is concave up (like a U) or concave down (like an up-side-down).
The absolute value of a (the number without the sign) controls how wide or narrow it is. If the absolute value is a fraction less than 1 the graph gets wider. The smaller the absolute value of the fraction the wider the graph gets.
If the absolute value of a is greater than 1 the graph gets narrower (it gets skinnier). The bigger the absolute value the narrower the graph.
So, if all the graphs look like a U (concave up) then the one with the smallest a is the one that is the widest.
The a also controls whether the graph is concave up or concave down. If a is negative
If a is negative the graph is concave down so any graph that is concave down has a smaller value of a than any graph that is concave up. However, if the graph is concave down the one with the smallest a would be the most narrow one.
So to find the one with the smallest a...
If they are all concave up (like a U) pick the widest one
and
If they are not all concave up pick the narrowest one that is concave down (looks like an upside down U)