a solid sphere of radius r is melted and cast into the shape of a solid cone of height r, then the radius of the base of cone is

volume of sphere = (4/3)πr^3

let the radius of the cone be x
volume of cone
= (1/3)πx^2 r

(1/3)π x^2 r = (4/3)π r^3
x^2 r = 4 r^3
x^2 = 4r^2

x = 2r

pass a plane through a cube of edge 8 inches so that the section formed will be a regular hexagon. through each side of the hexagon pass two planes , one plane containing one of the two vertices of the cube which are farthest away from the plane of the hexagon, the other plane containing the diagonally opposite vertex of the cube. find the volume of the solid bounded by these planes.

pls help me to solve the problem.the answer is 384 cubic inches..how come?

pls answer my problem..the answer is 384 inches cubic.what will be the process?

To find the radius of the base of the cone formed by melting a solid sphere, we can use the concept of volume conservation.

Let's consider the volume of the sphere and the cone:

Volume of the sphere = (4/3)πr³
Volume of the cone = (1/3)πr²h

Since the sphere is melted and cast into the shape of a cone, the volume of the sphere should be equal to the volume of the cone:

(4/3)πr³ = (1/3)πr²h

We are given that the height of the cone is equal to the radius of the sphere. Substituting h = r, we have:

(4/3)πr³ = (1/3)πr²r

Now, we can simplify and solve for the radius of the base of the cone:

(4/3)πr³ = (1/3)πr³

Dividing both sides by πr³:

(4/3) = (1/3)

We notice that the equation becomes 4 = 1, which is not possible. This suggests that the given scenario of melting a solid sphere and casting it into the shape of a solid cone of height r cannot be achieved.