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)
It would come out to 3x = 3x so it would be infinitely many solutions
The answer is 18x^4 - 30x^3 + 66x^2 - 40x + 56
Answer:
<h2><u><em>
A) -1</em></u></h2><h2><u><em>
B) -1.5 or -3/2</em></u></h2>
Step-by-step explanation:
Solve 3-4 and (1⁄2)-2
3 - 4, 4 is greater than 3, so the result is negative (-1)
3 - 4 = -1
-----------------
1/2 is equal to 0.5, also here 2 is larger and we will have a negative result (-1.5 or -3/2)
1/2 - 2 =
0.5 - 2 = -1.5 or -3/2
This is how it looks on a graph.
