if indeed two functions are inverse of each other, then their composite will render a result of "x", namely, if g(x) is indeed an inverse of f(x), then
![\bf (g\circ f)(x)=x\implies g(~~f(x)~~)=x \\\\\\ \begin{cases} f(x) = 3x\\ g(x)=\cfrac{1}{3}x \end{cases}\qquad \qquad g(~~f(x)~~)=\cfrac{1}{3}[f(x)]\implies g(~~f(x)~~)=\cfrac{1}{3}(3x)](https://tex.z-dn.net/?f=%5Cbf%20%28g%5Ccirc%20f%29%28x%29%3Dx%5Cimplies%20g%28~~f%28x%29~~%29%3Dx%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20f%28x%29%20%3D%203x%5C%5C%20g%28x%29%3D%5Ccfrac%7B1%7D%7B3%7Dx%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20g%28~~f%28x%29~~%29%3D%5Ccfrac%7B1%7D%7B3%7D%5Bf%28x%29%5D%5Cimplies%20g%28~~f%28x%29~~%29%3D%5Ccfrac%7B1%7D%7B3%7D%283x%29)
Answer:
-6 and 1
Step-by-step explanation:
If you separate the different powers (^), you have -m^3 - 5m^3 and -m^2 + 2m^2.
-m^3 - 5m^3 = -6m^3
-m^2 + 2m^2 = m^2 or 1m^2
Answer:
a and c is true
Step-by-step explanation:
Answer:

U d nhkfkzzizjzjusyddfccgfcdjskxndnndkdkfkckgkgvkccxddvjvjdjsujcufjcsxjjxjscxjxjycd
Answer:
6u^3
Step-by-step explanation: