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
2.77 × 10 ^ ( -3 ) is the answer
The answer to your question is 6
<h2>
Answer:</h2>
The ratio of the area of region R to the area of region S is:

<h2>
Step-by-step explanation:</h2>
The sides of R are in the ratio : 2:3
Let the length of R be: 2x
and the width of R be: 3x
i.e. The perimeter of R is given by:

( Since, the perimeter of a rectangle with length L and breadth or width B is given by:
)
Hence, we get:

i.e.

Also, let " s " denote the side of the square region.
We know that the perimeter of a square with side " s " is given by:

Now, it is given that:
The perimeters of square region S and rectangular region R are equal.
i.e.

Now, we know that the area of a square is given by:

and

Hence, we get:

and

i.e.

Hence,
Ratio of the area of region R to the area of region S is:
