first off, let's recall that a cube is just 6 squares stacked up to each other, like in the picture below. Since we know its volume, we can find how long each side is.
part A)
![\bf \textit{volume of a cube}\\\\V=x^3~~\begin{cases}x=side\\[-0.5em]\hrulefill\\V=64\end{cases}\implies 64=x^3\implies \sqrt[3]{64}=x\implies 4=x](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cube%7D%5C%5C%5C%5CV%3Dx%5E3~~%5Cbegin%7Bcases%7Dx%3Dside%5C%5C%5B-0.5em%5D%5Chrulefill%5C%5CV%3D64%5Cend%7Bcases%7D%5Cimplies%2064%3Dx%5E3%5Cimplies%20%5Csqrt%5B3%5D%7B64%7D%3Dx%5Cimplies%204%3Dx%20)
part B)
they have a painting with an area of 12 ft², will the painting fit flat against a side? Well, it can only fit flat if the sides of the painting are the same length or smaller than the sides of the crate, we know the crate is a 4x4x4, so are the painting's sides 4 or less?
![\bf \textit{area of a square}\\\\A=s^2~~\begin{cases}s=side\\[-0.5em]\hrulefill\\A=12\end{cases}\implies 12=s^2\implies \sqrt{12}=s\implies \stackrel{yes}{3.464\approx s}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Barea%20of%20a%20square%7D%5C%5C%5C%5CA%3Ds%5E2~~%5Cbegin%7Bcases%7Ds%3Dside%5C%5C%5B-0.5em%5D%5Chrulefill%5C%5CA%3D12%5Cend%7Bcases%7D%5Cimplies%2012%3Ds%5E2%5Cimplies%20%5Csqrt%7B12%7D%3Ds%5Cimplies%20%5Cstackrel%7Byes%7D%7B3.464%5Capprox%20s%7D%20)
Answer=2x+21 do you want the whole solution?
Answer:
Sorry, I don't speak confusion.
Step-by-step explanation:
Answer:
8.2 minutes
Step-by-step explanation:

492 seconds = 8.2 minutes
Answer:
12!!
Step-by-step explanation:
since 8 x 12 = 96 :)