Answer:
![\bold{\huge{\fbox{\color{Green}{☑Verified\:Answer}}}}](https://tex.z-dn.net/?f=%5Cbold%7B%5Chuge%7B%5Cfbox%7B%5Ccolor%7BGreen%7D%7B%E2%98%91Verified%5C%3AAnswer%7D%7D%7D%7D)
Step-by-step explanation:
10%
![( 1 = 0.9 + x) \times 100x = 100\% - 90\% \: x = 10\%](https://tex.z-dn.net/?f=%28%201%20%3D%200.9%20%2B%20x%29%20%5Ctimes%20100x%20%3D%20100%5C%25%20-%2090%5C%25%20%5C%3A%20x%20%3D%2010%5C%25)
Fям:-
Answer:
![\frac{dV}{dt}=1120 \pi cm^3/min](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D1120%20%5Cpi%20cm%5E3%2Fmin)
Step-by-step explanation:
We are given that
Radius of sphere expanding at the rate=![\frac{dr}{dt}=70 cm/min](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D%3D70%20cm%2Fmin)
Volume of sphere=![V=\frac{4}{3}\pi r^3](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3)
Surface area of sphere=![S=4\pi r^2](https://tex.z-dn.net/?f=S%3D4%5Cpi%20r%5E2)
We have to determine the rate at which the volume is changing with respect to time at r= 2 cm.
![V=\frac{4}{3}\pi r^3](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3)
Differentiate w.r.t time
![\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D4%5Cpi%20r%5E2%5Cfrac%7Bdr%7D%7Bdt%7D)
Substitute the values then we get
![\frac{dV}{dt}=4\pi (2)^2(70)=1120 \pi cm^3/min](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D4%5Cpi%20%282%29%5E2%2870%29%3D1120%20%5Cpi%20cm%5E3%2Fmin)
Hence, the rate at which the volume of sphere is changing is given by
![\frac{dV}{dt}=1120 \pi cm^3/min](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D1120%20%5Cpi%20cm%5E3%2Fmin)
V = 3.14(4^2) × 4
V = 3.14(16) × 4
V = 50.24 × 4
V = 200.96
According to this, answer is E.
Answer:
ayo ion get the equation. type up it again?
Step-by-step explanation: