Answer:
Hi! The answer to your question is 2x + 12
Step-by-step explanation:
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Answer:
![f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7Bx-4%7D%20%2C%20g%28x%29%3D6x%5E%7B2%7D%5Ctextrm%7B%20or%20%7Df%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D%2Cg%28x%29%3D6x%5E%7B2%7D%20-4)
Step-by-step explanation:
Given:
The function, ![H(x)=\sqrt[3]{6x^{2}-4}](https://tex.z-dn.net/?f=H%28x%29%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D)
Solution 1:
Let ![f(x)=\sqrt[3]{x}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D)
If
, then,
![\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bg%28x%29%7D%20%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D%5C%5Cg%28x%29%3D6x%5E%7B2%7D-4)
Solution 2:
Let
. Then,
![f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}](https://tex.z-dn.net/?f=f%28g%28x%29%29%3DH%28x%29%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D%5C%5C%5Csqrt%5B3%5D%7Bg%28x%29-4%7D%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D%20%5C%5Cg%28x%29-4%3D6x%5E%7B2%7D-4%5C%5Cg%28x%29%3D6x%5E%7B2%7D)
Similarly, there can be many solutions.
Idk how to this but look up the app Socratic and scan the question it should helped
Answer:
<em>32x - 5y </em>
Step-by-step explanation:
27x - 5y + 5x = <em>32x - 5y</em>
Pi is actually one of the letters of the Greek alphabet, but it is most known in the math field as the ratio between the circumference of a circle and the circle's diameter.
There is no exact measure of pi, as it does not terminate but it starts like:
3.14159265