A cone and cylinder both have a radius of 3 inches and a height of 6 inches. If the cone-shaped container is filled with water and this water is then poured into the empty cylindrical container, what will be the depth, in inches, of the water in the cylindrical container?

No real calculations needed here.

Since the volume of a cone is 1/3 the volume of its corresponding cylinder, the water in the cylinder would rise 1/3 of the 6 inches or 2 inches.

To find the depth of the water in the cylindrical container, we need to understand the volume relationship between the cone and the cylinder.

The formula for the volume of a cone is given by V_cone = (1/3)πr^2h, where r is the radius of the cone's base and h is the height of the cone.

The formula for the volume of a cylinder is given by V_cylinder = πr^2h, where r is the radius of the cylinder's base and h is the height of the cylinder.

In this problem, both the cone and the cylinder have the same radius, r = 3 inches, and height, h = 6 inches.

1. First, calculate the volume of the water in the cone.
V_cone = (1/3)π(3^2)(6) = 18π cubic inches.

2. Next, pour the water from the cone into the cylindrical container.

3. The volume of the water in the cylinder will be equal to the volume of the water in the cone.

V_cylinder = V_cone = 18π cubic inches.

4. Find the depth of the water in the cylindrical container using the volume formula for a cylinder.

V_cylinder = π(3^2)depth = 18π.

Simplify the equation:

9πdepth = 18π.

5. Solve for the depth.

depth = 18/9 = 2 inches.

Therefore, the depth of the water in the cylindrical container will be 2 inches.