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
You have $1.75 left
Step-by-step explanation:
Given,
Number of quarters already have = 20
Percent of quarters found = 40% of already have
Number of quarters found = 
Number of quarters found = 
Number of quarters found = 8
New total = 20+8 = 28 quarters
Percent of quarters spent = 75%
Number of quarters spent = 
Number of quarters spent = 
Number of quarters spent = 21
Quarters left = 28-21 = 7
1 quarter = 25 cents
7 quarters = 25*7 = 175 cents
$1 = 100 cents
175 cents = 
You have $1.75 left
Keywords: division, addition
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