Respuesta :
Let S and L be the volumes of the smaller and larger cans respectively...
The volume of a cylinder is: V=hπr^2 so
S=hπr^2 and we are told that the larger can's dimensions are 3h and 3r respectively so:
L=(3h)π(3r)^2
L=3hπ(9r^2)
L=27hπr^2
Then S/L is:
S/L=(hπr^2)/(27hπr^2)
S/L=1/27 so we can say that:
L=27S, we are told that the volume of the smaller can is 35.28 in^3 so
L=27(35.28) in^3
L=952.56 in^3
The volume of a cylinder is: V=hπr^2 so
S=hπr^2 and we are told that the larger can's dimensions are 3h and 3r respectively so:
L=(3h)π(3r)^2
L=3hπ(9r^2)
L=27hπr^2
Then S/L is:
S/L=(hπr^2)/(27hπr^2)
S/L=1/27 so we can say that:
L=27S, we are told that the volume of the smaller can is 35.28 in^3 so
L=27(35.28) in^3
L=952.56 in^3