Answer:
B
Step-by-step explanation:
Do you mean LCM (Least common multiple)
A
Volume of the Cylinder
Givens
H = 60 yards.
Diameter = 20 yards
pi = 3.14
Formula
V = pi * r^2 * h
Calculations
r = d/2
r = 32/2
r = 16
V = 3.14 * 16^2 * 60
V = 48230 cubic yards [Cylinder's Volume]
Cone
<em>Formula</em>
V = 1/3 pi r^2 H
<em>Givens</em>
pi = 3.14
r = 16 yards
h = 20 yards
<em>Sub and solve</em>
V = 1/3 3.14 * 16^2 * 20
V = 5359 cubic yards.
<em>Total Volume of the structure</em>
48230 + 5359 = 53589 Cubic Yards
<em>Water Content</em>
The answer to this part requires a proportion.
1 Cubic yard will hold 201.97 gallons.
53589 yd^3 = x
1/201.97 = 53589 /x [ You should get a pretty big answer]
x = 201.87 * 53589
x = 10 819 092 gallons can be held by the tank.
10 819 092 gallons <<<< answer
B
If the height of both the cylinder and the cone remain the same. If the radius doubles in both the cylinder and the cone then the tank will hold 4 times as much.
Total volume before doubling the radius = pi * r^2 h + 1/3 pi r^2 h
New Total Volume = pi * (2*r)^2 h + 1/3 pi * (2r)^2 h
New Total volume = pi * 4r^2 h + 1/3 pi *4 r^2 h
New Total Volume = 4 [pi r^2 h + 1/3 pi r^2 h]
but pi r^2 h + 1/3 pi r^2 h is the total volume before doubling the radius
New volume = 4 original volume. <<<<< answer to part B
