​ DUE TOMORROW!!!!!

I'm not sure if I understand the question, but will take a crack at it:

Let the radius of the hemisphere  = a, then:

The radius of the balloon =a^(1/3)

So the ratio of the 2 radii =a : a^(1/3)

Volume =4/3*pi*r^3

[4/3*pi*(a^(1/3))^3] =[4/3*pi*a^3]/2, solve for a

a =+or- sqrt(2)

P.S. Somebody should look at this. Thanks.

Let  V  be the volume of the balloon.

Let  r₁  be the radius of the sphere.

Let  r₂  be the radius of the hemisphere.

\(V=\frac43\pi (r_{\small1})^3\\~\\ \frac3{4\pi}\cdot V=\frac3{4\pi}\cdot\frac43\pi (r_{\small1})^3\\~\\ \frac{3V}{4\pi}=(r_{\small1})^3\\~\\ \sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{(r_{\small1})^3}\\~\\ \sqrt[3]{\frac{3V}{4\pi}}=r_{\small1}\)                 Solve this equation for  r₁

\(V=\frac12\cdot\frac43\pi (r_{\small2})^3\\~\\ V=\frac23\pi (r_{\small2})^3\\~\\ \frac3{2\pi}\cdot V=\frac3{2\pi}\cdot\frac23\pi (r_{\small2})^3\\~\\ \frac{3V}{2\pi}=(r_{\small2})^3\\~\\ \sqrt[3]{\frac{3V}{2\pi}}=\sqrt[3]{(r_{\small2})^3}\\~\\ \sqrt[3]{\frac{3V}{2\pi}}=r_{\small2} \)                  Solve this equation for  r₂ .

The ratio of  r₁  to  r₂  can be expressed in the form   \(\sqrt[3]{a}\)   for some real number  a .

\(\dfrac{r_{\small1}}{r_{\small2}}=\sqrt[3]{a} \\~\\ \dfrac{\sqrt[3]{\frac{3V}{4\pi}}}{\sqrt[3]{\frac{3V}{2\pi}}}=\sqrt[3]{a}\\~\\ \dfrac{\frac{3V}{4\pi}}{\frac{3V}{2\pi}}=a\\~\\ \frac{3V}{4\pi}\cdot\frac{2\pi}{3V}=a\\~\\ \frac14\cdot\frac21=a\\~\\ \frac24=a \\~\\ \frac12=a\)      Plug in the equivalent expressions of  r₁  and  r₂  and solve for  a .

Thank you hectictar. Now, I understand it.