Answer:
See below for all the cube roots
Step-by-step explanation:
<u>DeMoivre's Theorem</u>
Let 
be a complex number in polar form, where 
is an integer and 
. If 
, then 
.
<u>Nth Root of a Complex Number</u>
If is any positive integer, the nth roots of 
are given by ![\sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Brcis%5Ctheta%7D%3D%28rcis%5Ctheta%29%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
where the nth roots are found with the formulas:
![\sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr]](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Br%7D%5Cbiggr%5Bcis%28%5Cfrac%7B%5Ctheta%2B360%5E%5Ccirc%20k%7D%7Bn%7D%29%5Cbiggr%5D)
for degrees (the one applicable to this problem)![\sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr]](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Br%7D%5Cbiggr%5Bcis%28%5Cfrac%7B%5Ctheta%2B2%5Cpi%20k%7D%7Bn%7D%29%5Cbiggr%5D)
for radians
for 
<u>Calculation</u>
<u />![z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)](https://tex.z-dn.net/?f=z%3D27%28cos330%5E%5Ccirc%2Bisin330%5E%5Ccirc%29%5C%5C%5C%5C%5Csqrt%5B3%5D%7Bz%7D%20%3D%5Csqrt%5B3%5D%7B27%28cos330%5E%5Ccirc%2Bisin330%5E%5Ccirc%29%7D%5C%5C%5C%5Cz%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%3D%2827%28cos330%5E%5Ccirc%2Bisin330%5E%5Ccirc%29%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C%5C%5Cz%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%3D27%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%28cos%28%5Cfrac%7B1%7D%7B3%7D%5Ccdot330%5E%5Ccirc%29%2Bisin%28%5Cfrac%7B1%7D%7B3%7D%5Ccdot330%5E%5Ccirc%29%29%5C%5C%5C%5Cz%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%3D3%28cos110%5E%5Ccirc%2Bisin110%5E%5Ccirc%29)
<u>First cube root where k=2</u>
<u />![\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27%7D%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%2B360%5E%5Ccirc%282%29%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%2B720%5E%5Ccirc%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28%5Cfrac%7B1050%5E%5Ccirc%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28350%5E%5Ccirc%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcos%28350%5E%5Ccirc%29%2Bisin%28350%5E%5Ccirc%29%5Cbiggr%5D)
<u>Second cube root where k=1</u>
![\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27%7D%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%2B360%5E%5Ccirc%281%29%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%2B360%5E%5Ccirc%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28%5Cfrac%7B690%5E%5Ccirc%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28230%5E%5Ccirc%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcos%28230%5E%5Ccirc%29%2Bisin%28230%5E%5Ccirc%29%5Cbiggr%5D)
<u>Third cube root where k=0</u>
<u />![\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27%7D%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%2B360%5E%5Ccirc%280%29%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28110%5E%5Ccirc%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcos%28110%5E%5Ccirc%29%2Bisin%28110%5E%5Ccirc%29%5Cbiggr%5D)
The second choice is the answer
<span>To calculate the average; add up all the numbers, then divide by how many numbers there are. In this case, there are 5 numbers; personal, small, medium, large, and extra large.
</span>8 + 10 + 12 + 14 + 16 / 5 --> 60/5 = 12
Simplified value of subtraction of 7x+27x+2 from -5x^2-2x−5x 2 −2x. is 
.
Given Expressions 7x+27x+2 and -5x^2-2x−5x 2 −2x.
Expressions refers to the combination of numbers, variables and symbols. It represents the relationship between variables. They are solved for finding out the value of variables. It is mainly not in equal to form. Subtraction is possible when subtracting the coefficients of same variable.
We have to subtract 7x+27x+2 from -5x^2-2x−5x 2 −2x. and to do this we need to first solve the expressions itself.
7x+27x+2=34x+2------------------1
=
--------------------2
Now subtract 1 from 2
=-34x+2
=.
Hence the simplified value of subtraction is .
Learn more about expressions at brainly.com/question/723406
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