Answer:
that's the correct answers
Step-by-step explanation:
thanks for your
So you plug in g(x) for x in h(x)
Since g(x) = 4x + 9, substitution that in
4x + 9 + 2
4x + 11
Solution: 4x + 11
Answer:
who ever line crosses the y-axis at -4.
Step-by-step explanation:
![\dfrac{\sqrt[3]{x+h}-\sqrt[3]x}h\times\dfrac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}=\dfrac{(\sqrt[3]{x+h})^3-(\sqrt[3]x)^3}\cdots](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7Bx%2Bh%7D-%5Csqrt%5B3%5Dx%7Dh%5Ctimes%5Cdfrac%7B%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7Bx%28x%2Bh%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7Bx%28x%2Bh%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%3D%5Cdfrac%7B%28%5Csqrt%5B3%5D%7Bx%2Bh%7D%29%5E3-%28%5Csqrt%5B3%5Dx%29%5E3%7D%5Ccdots)

The

s then cancel, leaving you with the
![\cdots=\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Ccdots%3D%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7Bx%28x%2Bh%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D)
term.
If it's not clear what I did above, consider the substitution
![a=\sqrt[3]{x+h}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7Bx%2Bh%7D)
and
![b=\sqrt[3]x](https://tex.z-dn.net/?f=b%3D%5Csqrt%5B3%5Dx)
. Then
