Answer:
.... thats alot there buddy ...
Step-by-step explanation:
um is the end a 3 and 2 or 32 or 3*2?
I think C but probably try to find something to back it up
All you gotta do is divide the pound by the cost
Let
. Then
![|z|=\sqrt{(-\sqrt3)^2+1^2}=2](https://tex.z-dn.net/?f=%7Cz%7C%3D%5Csqrt%7B%28-%5Csqrt3%29%5E2%2B1%5E2%7D%3D2)
lies in the second quadrant, so
![\arg z=\pi+\tan^{-1}\left(-\dfrac1{\sqrt3}\right)=\dfrac{5\pi}6](https://tex.z-dn.net/?f=%5Carg%20z%3D%5Cpi%2B%5Ctan%5E%7B-1%7D%5Cleft%28-%5Cdfrac1%7B%5Csqrt3%7D%5Cright%29%3D%5Cdfrac%7B5%5Cpi%7D6)
So we have
![z=2e^{i5\pi/6}](https://tex.z-dn.net/?f=z%3D2e%5E%7Bi5%5Cpi%2F6%7D)
and the fourth roots of
are
![2^{1/4}e^{i(5\pi/6+k\pi)/4}](https://tex.z-dn.net/?f=2%5E%7B1%2F4%7De%5E%7Bi%285%5Cpi%2F6%2Bk%5Cpi%29%2F4%7D)
where
. In particular, they are
![2^{1/4}e^{i(5\pi/6)/4}=2^{1/4}e^{i5\pi/24}](https://tex.z-dn.net/?f=2%5E%7B1%2F4%7De%5E%7Bi%285%5Cpi%2F6%29%2F4%7D%3D2%5E%7B1%2F4%7De%5E%7Bi5%5Cpi%2F24%7D)
![2^{1/4}e^{i(5\pi/6+2\pi)/4}=2^{1/4}e^{i17\pi/24}](https://tex.z-dn.net/?f=2%5E%7B1%2F4%7De%5E%7Bi%285%5Cpi%2F6%2B2%5Cpi%29%2F4%7D%3D2%5E%7B1%2F4%7De%5E%7Bi17%5Cpi%2F24%7D)
![2^{1/4}e^{i(5\pi/6+4\pi)/4}=2^{1/4}e^{i29\pi/24}](https://tex.z-dn.net/?f=2%5E%7B1%2F4%7De%5E%7Bi%285%5Cpi%2F6%2B4%5Cpi%29%2F4%7D%3D2%5E%7B1%2F4%7De%5E%7Bi29%5Cpi%2F24%7D)
![2^{1/4}e^{i(5\pi/6+6\pi)/4}=2^{1/4}e^{i41\pi/24}](https://tex.z-dn.net/?f=2%5E%7B1%2F4%7De%5E%7Bi%285%5Cpi%2F6%2B6%5Cpi%29%2F4%7D%3D2%5E%7B1%2F4%7De%5E%7Bi41%5Cpi%2F24%7D)
Answer:
It is C. 233 because if you multiply 5x5x5 thats 125
If you multiply 12x3x3, its 108. So add 108+125 and that will be your final answer. Hope this helps and if so then please mark as brainliest. Remember that volume= length x width x height.