Answer:
1π
Step-by-step explanation:
suppose the radius of semicircle P is r,
then the radius of semicircle Q = (r+d)/2 ... d≤r
radius of semicircle R = (r-d)/2
area P = 1/2 (r)²π
area Q = 1/2 ((r+d)/2)² π = 1/8 (r² + 2rd + d²)π
area R = 1/2 ((r-d)/2)² π = 1/8 (r² - 2rd + d²)π
shaded area = P-Q-R = 1/2 r²π - 1/4 (r² + d²)π
= ((r² - d²)/4) * π
because there is no constant r value in the question and d value changes with the r change, when the vertical segment length equal the semicircle P radius (r), r=2 and d = 0
therefore the shaded area = ((2² -0²)/4)*π = 1π
Answer: i dont get what u mean explain
Step-by-step explanation:
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.
80.64
because 16 times 12 is 192
if every tile is 2 feet divide 192 by 2
you should get 96 meaning there are 96 tiles
then take 96 and multiply by it by .84 because every tile is 84 cents
you then get 80.64