Answer:
1008
Step-by-step explanation:
like the 2nd picture we split it in 3 rectangular prism
=6*7*8=336
=6*7*8=336
=6*7*8=336
V=336+336+336=1008
Answer:
Step-by-step explanation:
We have volume of cone as

and for a cone always r/h = constant
Given that r' = rate of change of radius = -7 inches/sec
(Negative sign because decresing)
V' =- 948 in^3/sec
Radius = 99 inches and volume = 525 inches
Height at this instant = 
Let us differentiate the volume equation with respect to t using product rule
![V=\frac{1}{3} \pi r^2 h\\V' = \frac{1}{3} \pi[2rhr'+r^2 h']\\-948 = \frac{1}{3} \pi[2(99)(-7)(\frac{0.1607}{\pi})+99^2 h']\\](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20r%5E2%20h%5C%5CV%27%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%5B2rhr%27%2Br%5E2%20h%27%5D%5C%5C-948%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%5B2%2899%29%28-7%29%28%5Cfrac%7B0.1607%7D%7B%5Cpi%7D%29%2B99%5E2%20h%27%5D%5C%5C)
![-948 = \frac{1}{3} \pi[2(99)(-7)(\frac{0.1607}{\pi})+99^2 h']\\-948 = 33(3.14)(-2.25/3.14 + 99 h')\\-9.149=-0.72+99h'\\-8.429 = 99h'\\h' = 0.08514](https://tex.z-dn.net/?f=-948%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%5B2%2899%29%28-7%29%28%5Cfrac%7B0.1607%7D%7B%5Cpi%7D%29%2B99%5E2%20h%27%5D%5C%5C-948%20%3D%2033%283.14%29%28-2.25%2F3.14%20%20%2B%2099%20h%27%29%5C%5C-9.149%3D-0.72%2B99h%27%5C%5C-8.429%20%3D%2099h%27%5C%5Ch%27%20%3D%200.08514)
Rate of change of height = 0.08514 in/sec
The answer is this i’m sure of it
Answer:
c
did the test got it right.
Step-by-step explanation:
The first graph function has a constant additive rate of change of -1/4.
From the figure:
First of all when we talk about a rate of change that means a linear relation between variables, which is proper of a linear function represented by the first function showed in the image.
In the first graph we can observe that the x variable increases and y - variable decrease.
rate of change = slope = -1/4
from first graph take any two points.
(2,1) and (-2,2)
slope = 2 - 1 / -2-2
= 1/-4
= -1/4
Learn more about the function here:
brainly.com/question/5975436
#SPJ4
Full question:
Is in the image uploaded.