and
. So we have a remainder of
![(x^3+1)-(x^3-x^2)=x^2+1](https://tex.z-dn.net/?f=%28x%5E3%2B1%29-%28x%5E3-x%5E2%29%3Dx%5E2%2B1)
and
. Subtracting this from the previous remainder gives a new remainder
![(x^2+1)-(x^2-x)=x+1](https://tex.z-dn.net/?f=%28x%5E2%2B1%29-%28x%5E2-x%29%3Dx%2B1)
and
. Subtracting this from the previous remainder gives a new one of
![(x+1)-(x-1)=2](https://tex.z-dn.net/?f=%28x%2B1%29-%28x-1%29%3D2)
and we're done since 2 does not divide
. So we have
![\dfrac{x^3+1}{x-1}=x^2+x+1+\dfrac2{x-1}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E3%2B1%7D%7Bx-1%7D%3Dx%5E2%2Bx%2B1%2B%5Cdfrac2%7Bx-1%7D)
i’m pretty sure the answer would be 1) linear :D
Hey there! :)
Answer:
![g^{-1} (g(10))=10](https://tex.z-dn.net/?f=g%5E%7B-1%7D%20%28g%2810%29%29%3D10)
Step-by-step explanation:
Begin by calculating g(10):
g(x) = 3x - 6
Substitute in 10 for x:
g(10) = 3(10) - 6
g(10) = 30-6
g(10) = 24.
Plug '24' into 'x' into ![g^{-1} (x)](https://tex.z-dn.net/?f=g%5E%7B-1%7D%20%28x%29)
![g^{-1} (24)=\frac{(24) + 6}{3}](https://tex.z-dn.net/?f=g%5E%7B-1%7D%20%2824%29%3D%5Cfrac%7B%2824%29%20%2B%206%7D%7B3%7D)
Simplify:
![g^{-1} (24)=\frac{30}{3}](https://tex.z-dn.net/?f=g%5E%7B-1%7D%20%2824%29%3D%5Cfrac%7B30%7D%7B3%7D)
![g^{-1} (24)=10](https://tex.z-dn.net/?f=g%5E%7B-1%7D%20%2824%29%3D10)
The traveled distance is 40.819 feet (downward)
Answer:
what are we trying to do with the problem tho?
Step-by-step explanation: