To find the diameter of the sphere, we need to find the radius.

The volume of a sphere is \(\displaystyle V=\frac{4}{3}\pi r^3\).

Plug in our given values.

\(\displaystyle 2 88\pi=\frac{4}{3}\pi r^3\)

Notice that the \(\displaystyle \pi\)'s cancel out.

\(\displaystyle 288=\frac{4}{3}r^3\)

\(\displaystyle \frac{3}{4}*288=r^3\)

\(\displaystyle 216=r^3\)

\(\displaystyle \sqrt[3]{216}=\sqrt[3]{r^3}\)

\(\displaystyle 6= r\)

The diameter is twice the radius, so \(\displaystyle d=2r\).

\(\displaystyle d=2(6)\)

\(\displaystyle d=12\)

The standard equation to find the surface area of a sphere is 

\(\displaystyle SA=4\pi r^2\) where \(\displaystyle r\) denotes the radius. Rearrange this equation in terms of \(\displaystyle r\):

\(\displaystyle r=\sqrt{\frac{SA}{4\pi }}\)

Substitute the given surface area into this equation and solve for the radius and then double the radius to get the diameter:

\(\displaystyle r=\sqrt{\frac{64\Pi }{4\Pi }}=\sqrt{16}=4\)

\(\displaystyle diameter=2\cdot r=2\cdot 4=8\)

The standard equation to find the volume of a sphere is 

\(\displaystyle V=\frac{4}{3}\pi r^3\) where \(\displaystyle r\) denotes the radius. Rearrange this equation in terms of \(\displaystyle r\):

\(\displaystyle r=\sqrt[3]{\frac{3V}{4\pi }}\)

Substitute the given volume into this equation and solve for the radius. Double the radius to find the diameter:

\(\displaystyle r=\sqrt[3]{\frac{3V}{4\pi }}=\sqrt[3]{\frac{3\cdot 4\pi }{4\pi }}=\sqrt[3]{\frac{12\pi }{4\pi }}=\sqrt[3]{3 }\)

\(\displaystyle diameter=2\cdot r=2\sqrt[3]{3}\)

The volume of a sphere is determined by the following equation:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle 36\pi=\frac{4}{3}\pi r^3\)

\(\displaystyle (\frac{3}{4})36\pi=(\frac{3}{4})(\frac{4}{3})\pi r^3\)

\(\displaystyle \frac{27\pi}{\pi}=\frac{\pi r^3}{\pi}\)

\(\displaystyle 27=r^3\)

\(\displaystyle \sqrt[3]{27}=\sqrt[3]{r^3}\)

\(\displaystyle 3=r\)

\(\displaystyle d=2r\)

\(\displaystyle d=2(3)=6\)

We know that the surface area of the spere is \(\displaystyle SA=100cm^{2}\).

\(\displaystyle SA=4\pi r^{2}\)

\(\displaystyle 100cm^{2}=4\pi r^{2}\)

Rearrange and solve for \(\displaystyle r\).

\(\displaystyle r^{2}=\frac{100cm^{2}}{4\pi}\)

\(\displaystyle \dpi{100} r=\sqrt{\frac{100cm^{2}}{4\pi}}\)

\(\displaystyle \dpi{100}\rightarrow 2.82cm\)

The standard equation to find the area of a sphere is \(\displaystyle SA=4\pi r^2\) where \(\displaystyle r\) denotes the radius. Rearrange this equation in terms of \(\displaystyle r\):

\(\displaystyle r=\sqrt{\frac{SA}{4\pi }}\)

To find the answer, substitute the given surface area into this equation and solve for the radius:

\(\displaystyle r=\sqrt{\frac{16\pi }{4\pi }}=\sqrt{4}=2\)

The standard equation to find the volume of a sphere is 

\(\displaystyle V=\frac{4}{3}\pi r^3\) where \(\displaystyle r\) denotes the radius. Rearrange this equation in terms of \(\displaystyle r\):

\(\displaystyle r=\sqrt[3]{\frac{3V}{4\pi }}\)

Substitute the given volume into this equation and solve for the radius:

\(\displaystyle r=\sqrt[3]{\frac{3V}{4\pi }}=\sqrt[3]{\frac{3\cdot 12\pi }{4\pi }}=\sqrt[3]{\frac{36\pi }{4\pi }}=\sqrt[3]{9 }\)

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle 288\pi=\frac{4}{3}\pi r^3\)

\(\displaystyle \frac{288\pi}{\pi}=\frac{4\pi r^3}{3\pi}\)

\(\displaystyle 288=\frac{4r^3}{3}\)

\(\displaystyle (288)(\frac{3}{4})=\frac{3}{4}(\frac{4r^3}{3})\)

\(\displaystyle 216=r^3\)

\(\displaystyle \sqrt[3]{216}=\sqrt[3]{r^3}\)

\(\displaystyle 6=r\)