Answer:
The rate of change of radius is 0.0530 ft/sec.
Step-by-step explanation:
Given : Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second.
To find : How fast is the radius increasing after 2 minutes?
Solution :
The volume of the sphere is 
Differentiating w.r.t. r,

We have given, Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second.
i.e. 
Integrate w.r.t. t,

Assuming V=0 when t=0 then C=0
So, 

![r=\sqrt[3]{\frac{t}{4\pi}}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7Bt%7D%7B4%5Cpi%7D%7D)
Now, we apply chain rule
i.e. 

The radius increasing after 2 minutes i.e. at 




Therefore, the rate of change of radius is 0.0530 ft/sec.
a. 1^5 = 1 × 1 × 1 × 1 × 1 = 1
b. 2 × 2 × 2 × 2 × 2 = 2^5 = 32
c. 20^3 = 20 × 20 × 20 = 8.000
d. 7 × 7 × 7 × 7 = 7^4 = 2.401
Answer:
48f^2-110f+63
Step-by-step explanation:
We multiple every part of first sum with every prt of the 2nd sum:
(6f-7)(8f-9)=48f^2-54f-56f+63=
48f^2-110f+63