Question

A sphere has a radius of 16 in. Which statements about the sphere are true? Check all that apply.

Answer 1

Answer:

The True statements are:

The sphere has a diameter of 32 in.

The radius’s length is one-half the length of the diameter.

The volume of the sphere is StartFraction 16,384 Over 3 EndFraction pi inches cubed.

Step-by-step explanation:

• The sphere has a diameter of 8 in. False

      This is false because a radius is half of a diameter. Half of      the given diameter of 8 in is 4 in whereas the radius in the question is 16 in.

• The volume of the sphere is StartFraction 2,048 Over 3 EndFraction pi inches cubed. False

        This is false because the volume of the sphere is 4πr³/3 = (16,384/3) π cubic inches.

• The sphere has a diameter of 32 in. True

This is True because a radius is half of a diameter. Half of the given diameter of 32 in is 16 in and  the radius in the question is also 16 in.

• The radius’s length is one-half the length of the diameter. True

This is True because a radius is half of a diameter

• The volume of the sphere is StartFraction 16,384 Over 3 EndFraction pi inches cubed. True

This is True because the volume of the sphere is 4πr³/3 = (16,384/3) π cubic inches.

• The diameter’s length is one-half the length of the radius. False

It is the radius’s length that is one-half the length of the diameter

Answer 2

Answer:

• The sphere has a diameter of 32 in. CORRECT: the diameter of a sphere doubles its radio, so if the radio equals 16,the diameter measures the double, in this case, 32.
• The radius' lenght is one half the length of the diameter. CORRECT: , by definition, the diameter is the maximum distance between two opposite points on the surface of the sphere, and its lenght equals twice the radius because the radius is the distance between the center of the sphere and any point of the frontier of the sphere (then, one colud imagine that the distance between two opposite points in a sphere can be united by two radius, or a diameter).
• The volume (V) of a sphere can be calculated as $\frac{4\pi r^3}{3}$, where r is the radio. In this case, the volume of the sphere will be $\frac{4\pi\times16^3}{3}=\frac{4\pi\times4096}{3}$, which is also equal to $\frac{16384\times\pi}{3}$.