Calculate the total volume when two identical truncated cones are placed end to end.

Answer: [tex]2\cdot\frac{1}{3}\pi18\left(2^{2}+2\left(8\right)+8^{2}\right)[/tex] = 3166.73
Step-by-step explanation:
Use the formula for a truncated cone: [tex]\frac{1}{3}\pi h\left(r^{2}+rR+R^{2}\right)[/tex]
r=smaller radius = 2
R=Bigger radius = 8
h = height = 18
Since we have to cones, the total volume can be described as:
[tex]2\cdot\frac{1}{3}\pi18\left(2^{2}+2\left(8\right)+8^{2}\right)[/tex]