Answer:
No, 6 is not a solution to the equation x2-4=5x
Step-by-step explanation:
6(2)-4=5(6)
12-4=30
8=30
The question is basically asking if you replace x with 6, will the equation equal each other on both sides, for example 6=6 or 3.5=3 1/2. In this case the equation gives us 8=30 which means 6 would not be a solution since both sides aren't equal to one another, 30 is bigger than 8.
Answer:2 b
Step-by-step explanation:3 h
Answer:

![g(x)=\sqrt[3]{x}-5](https://tex.z-dn.net/?f=g%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D-5)
Step-by-step explanation:
Consider the given function is
![f(x)=\sqrt[3]{-2x+4}-5](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7B-2x%2B4%7D-5)
It is given that
and neither g(x) nor h(x) is solely x.
![f(x)=\sqrt[3]{(-2x+4)}-5](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7B%28-2x%2B4%29%7D-5)
Let
, then we get
![f(x)=g(h(x))=\sqrt[3]{h(x)}-5](https://tex.z-dn.net/?f=f%28x%29%3Dg%28h%28x%29%29%3D%5Csqrt%5B3%5D%7Bh%28x%29%7D-5)
Substitute h(x)=x in the above function.
![g(x)=\sqrt[3]{x}-5](https://tex.z-dn.net/?f=g%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D-5)
Therefore, the required functions are
and
.
Check the solutions.
![[\because h(x)=-2x+4]](https://tex.z-dn.net/?f=%5B%5Cbecause%20h%28x%29%3D-2x%2B4%5D)
![[\because g(x)=\sqrt[3]{x}-5]](https://tex.z-dn.net/?f=%5B%5Cbecause%20g%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D-5%5D)

Therefore, our solution is correct.
Answer:
The second option, 3x^2(x+3)(4x+1)
Step-by-step explanation:
Factor out the common term, which is 3x^2
This then becomes 3x^2(4x^2+13x+3)
Factor the inside term and it becomes
(3x^2)(x+3)(4x+1)
You cant solve y but you can form an equation in terms of y
which is y=7-2x
if you were given another equation that has the term y and x than you can solve it.also that is a simultaneous equation.
Hope this helps,dont forget to put this as the brainliest answer if it does to help me out too. xD