(For full answer you might have to go to the comments)
Answer: 28x+30
Explanation: we divide (3x^3-2x^2+4x-3) by (x^2+3x+3) Using long division
3x-11
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(x^2+3x+3) 3x^3-2x^2+4x-3
-(3x^3+9x^2+9x)
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-11x^2-5x-3
-(-11x^2-33x-33)
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28x+30
So our remainder will be 28x+30
Step-by-step explanation:
we know,

Therefore, nth term of a given gp is
16a
<h3>
Answer: E) x^5</h3>

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Explanation:
We simply take half of the exponent 10 to get 5. This applies to square roots only.
So the rule is 
A more general rule is
![\sqrt[n]{a^b} = a^{b/n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Eb%7D%20%3D%20a%5E%7Bb%2Fn%7D)
If n = 2, then we're dealing with square roots like with this problem. In this case, a = x and b = 10.