For this case we have a function of the form
, where
To find the real zeros we must equal zero and clear the variable "x".

We add 10 to both sides of the equation

We apply cube root to both sides of the equation:
![\sqrt[3]{(x-12)^3} = \sqrt[3] {10}\\x-12 = \sqrt[3] {10}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%28x-12%29%5E3%7D%20%3D%20%5Csqrt%5B3%5D%20%7B10%7D%5C%5Cx-12%20%3D%20%5Csqrt%5B3%5D%20%7B10%7D)
We add 12 to both sides of the equation:
![x-12 + 12 = \sqrt[3] {10} +12\\x = \sqrt[3] {10} +12](https://tex.z-dn.net/?f=x-12%20%2B%2012%20%3D%20%5Csqrt%5B3%5D%20%7B10%7D%20%2B12%5C%5Cx%20%3D%20%5Csqrt%5B3%5D%20%7B10%7D%20%2B12)
Answer:
Option D
x = 3g + 2
- 2 - 2
x - 2 = 3g
÷ 3 ÷ 3
x - 2 ÷ 3 = g
3.64 × 10^3<span>
Hope this helps
</span><span>multiply by 10, 3 times because you move the decimal 3 times
The ^ is an exponet</span>