The volume of a cube is found by raising the side length to the third power. Since 4 cubed is 64, the correct volume is 64 cubic units. This straightforward calculation reinforces the cube volume formula.

The volume of a rectangular prism is determined by multiplying its length, width, and height. Multiplying 3, 4, and 5 gives 60 cubic units. This problem helps to solidify the basic formula for the volume of a prism.

The formula for a cylinder's volume is πr²h. With a radius of 3 and a height of 7, the calculation becomes π × 9 × 7 = 63π cubic units. This reinforces the correct use of the cylinder volume formula.

A sphere's volume is calculated using the formula (4/3)πr³. Substituting r = 2 gives (4/3)π(8) = (32/3)π cubic units. This question tests application of the sphere volume formula.

The volume of a cube is calculated by multiplying the side length by itself three times, which is expressed as side³. This precise formula distinguishes volume from area formulas. It reinforces the fundamental concept of three-dimensional measurement.

Multiplying the length, width, and height (8 × 3 × 5) gives the volume of the prism. This straightforward application yields 120 cubic centimeters. It reinforces using multiplication to compute three-dimensional space.

For a cube, the side length is the cube root of the volume. Since 7³ equals 343, the side length is 7 cm. This problem tests understanding of cube roots in volume calculations.

Using the cylinder volume formula, V = πr²h, substituting the given values yields 100π = πr²×10. Dividing by 10π gives r² = 10, so the radius is √10 units. This problem requires algebraic manipulation of the formula.

The volume of a cone is calculated as one-third the product of the base area and the height, expressed as (1/3)πr²h. This distinguishes it from the volume formulas for cylinders and spheres. It tests the student's ability to identify different volume formulas.

Substitute the given volume into the sphere formula: (4/3)πr³ = 288π. Dividing by π and multiplying by 3/4 gives r³ = 216, so the cube root of 216 is 6. This question reinforces reverse application of volume formulas.

First, calculate the cylinder's volume using πr²h, which gives 160π cubic units. Next, calculate the cone's volume using (1/3)πr²h, which gives (80/3)π cubic units. Adding these volumes results in a total of (560/3)π cubic units.

The volume of a pyramid is calculated as (1/3) multiplied by the area of the base and the height. A square with side 6 has an area of 36, and multiplying (1/3)*36*9 yields 108 cubic units. This verifies the pyramid volume formula.

Starting with the cone volume formula, V = (1/3)πr²h, substituting the given values leads to (1/3)πr²(10) = 150. Solving for πr² gives (150×3)/10 = 45. This tests algebraic manipulation of the volume formula.

Since the volume of a cylinder is directly proportional to its height, doubling the height will double the volume. The radius remaining constant means no change in the base area. This problem emphasizes understanding of proportional relationships in volume.

The cube's volume is calculated as 6³ = 216 cubic units, and the cylinder's volume is π×3²×6 = 54π cubic units. Dividing 216 by 54π simplifies to 4/π, establishing the ratio of their volumes. This challenges students to compare volumes of different shapes.

For a cone, the volume is one-third that of a cylinder with the same dimensions; hence, the difference is two-thirds of the cylinder's volume. Setting (2/3)×(cylinder volume) equal to 96π leads to a cylinder volume of 144π cubic units. This problem reinforces the relationship between the volumes of cones and cylinders.

The cylinder's volume is πr²(2r) = 2πr³ and the hemisphere's volume is half of (4/3)πr³, which is (2/3)πr³. Their sum is (8/3)πr³, and setting this equal to 576π leads to r³ = 216, so r = 6. This combines volume formulas for different solids.

The cylinder's volume is π×5²×12 = 300π cubic units. For the cone with the same base, its volume is (1/3)π×5²×h. Setting these equal and solving for h gives h = 36 units. This problem requires solving for an unknown dimension using volume equality.

The base of the pyramid is square, so its area is 4² = 16 square units. Using the formula (1/3)×16×h = 80 and solving for h gives h = 15 units. This challenges the student to rearrange the pyramid volume formula.

The prism's volume is calculated as 10×4×6 = 240 cubic meters. The half-cylinder's volume is half of π×3²×10, which approximates to 141.3 cubic meters when π ≈ 3.14. Adding these volumes results in approximately 381.3 cubic meters.