Factoring is the process of finding the factors of polynomials
For our quadratic:

I hope that
helps!
I am with you if you faced any difficulties!
It’s the second answer. 2•10•2 + 2•10•3 + 2•2•3
Answer: There will enough to paint the outside of a typical spherical water tower.
Step-by-step explanation:
1. Solve for the radius r from the formula for calculate the volume of a sphere. as following:
![V=\frac{4}{3}r^{3}\pi\\\frac{3V}{4\pi}=r^{3}\\r=\sqrt[3]{\frac{3V}{4\pi}}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7Dr%5E%7B3%7D%5Cpi%5C%5C%5Cfrac%7B3V%7D%7B4%5Cpi%7D%3Dr%5E%7B3%7D%5C%5Cr%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%7D%7D)
2. Substitute values:
![r=\sqrt[3]{\frac{3(66,840.28ft^{3})}{4\pi}}=25.17ft](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3%2866%2C840.28ft%5E%7B3%7D%29%7D%7B4%5Cpi%7D%7D%3D25.17ft)
3. Substitute the value of the radius into the equation fo calculate the surface area of a sphere, then you obtain that the surface area of a typical spherical water tower is:

3. If a city has 25 gallons of paint available and one gallon of paint covers 400 square feet of surface area, you must multiply 25 by 400 square feet to know if there will be enough to paint the outside of a typical spherical water tower.

As you can see, there will enough to paint the outside of a typical spherical water tower.
Answer:
A
Step-by-step explanation:
Bc they are very diff
(i think )