Respuesta :
The radius r used in below equations would equal 2in because it is half the given diameter of 4in.
The volume of the cone would be
[tex]V = \frac{1}{3} \pi r^2h \\ \\ V = \frac{1}{3} \pi (2in)^2(6in) = 8 \pi\ in^3[/tex]
The volume of a cylinder with the same dimensions would be
[tex]V = \pi r^2h \\ V = \pi (2in)^2(6in) = 24 \pi in^3[/tex]
The cylinder is 24 - 8 = 16 cubic inches greater.
The volume of the cone would be
[tex]V = \frac{1}{3} \pi r^2h \\ \\ V = \frac{1}{3} \pi (2in)^2(6in) = 8 \pi\ in^3[/tex]
The volume of a cylinder with the same dimensions would be
[tex]V = \pi r^2h \\ V = \pi (2in)^2(6in) = 24 \pi in^3[/tex]
The cylinder is 24 - 8 = 16 cubic inches greater.
Answer:The radius r used in below equations would equal 2in because it is half the given diameter of 4in.
The volume of the cone would be
The volume of a cylinder with the same dimensions would be
The cylinder is 24 - 8 = 16 cubic inches greater.
Step-by-step explanation: