We actually don't need to do any computation. By definition, the inverse function
changes the role of input and output. So, if a function f maps x onto y, the inverse function maps y onto x.
You have to think like this: if the function makes a step further, the inverse function makes that same step back.
This means that the composition
is always the identity function
. In fact,
![(f\circ f^{-1})(x) = f(f^{-1}(x)) = x](https://tex.z-dn.net/?f=%20%28f%5Ccirc%20f%5E%7B-1%7D%29%28x%29%20%3D%20f%28f%5E%7B-1%7D%28x%29%29%20%3D%20x%20)
So, for every function, you have
![(f\circ f^{-1})(4) = f(f^{-1}(4)) = 4](https://tex.z-dn.net/?f=%20%28f%5Ccirc%20f%5E%7B-1%7D%29%284%29%20%3D%20f%28f%5E%7B-1%7D%284%29%29%20%3D%204%20)
I think it d. I’m not sure I could be wrong
Answer:
![26 \frac{1}{4}](https://tex.z-dn.net/?f=26%20%5Cfrac%7B1%7D%7B4%7D)
Step-by-step explanation:
First we turn them into improper fractions
![-\frac{7}{2}(-\frac{45}{4})](https://tex.z-dn.net/?f=-%5Cfrac%7B7%7D%7B2%7D%28-%5Cfrac%7B45%7D%7B4%7D%29)
then we multiply it
![-\frac{7}{2}*-\frac{45}{4}=\frac{315}{12}=26\frac{1}{4}](https://tex.z-dn.net/?f=-%5Cfrac%7B7%7D%7B2%7D%2A-%5Cfrac%7B45%7D%7B4%7D%3D%5Cfrac%7B315%7D%7B12%7D%3D26%5Cfrac%7B1%7D%7B4%7D)
(3,3) is the point in which the 2 lines intersect
Answer:
Answer is 7cm,8cm,10cm
Step-by-step explanation:
A^2+B^2=C^2