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
Answer:
X is equal to 20
Step-by-step explanation:
Angles of a triangle equal 180.
110+x+x+30=180
2x+140=180
2x=40
x=20
If you input 20 as X back into the equation, then you can check that it is the right answer.
There was 32 total pupils and 2 obtained a level 5, so you subtract 2 from 32 and get 30. Then you divide 30 by 5 because to give you the the number of students which got a level 4 which is 6. 6+2=8 32-8=24
Level 4; 6pupils
Level 3; 24
Level 5:2
