(5rs-1t3)(-3r2st)
-30r^2s^2t+18rst^2
Answer:
Step-by-step explanation:
To find the inverse of function y=f(x), swap variables and then solve for the original variable.
![y=(x-5)^3+1\\ \\ x=(y-5)^3+1\\ \\ x-1=(y-5)^3\\ \\ \sqrt[3]{x-1}=y-5\\ \\ y=\sqrt[3]{x-1}+5\\ \\ f^{-1}(x)=\sqrt[3]{x-1}+5](https://tex.z-dn.net/?f=y%3D%28x-5%29%5E3%2B1%5C%5C%20%5C%5C%20x%3D%28y-5%29%5E3%2B1%5C%5C%20%5C%5C%20x-1%3D%28y-5%29%5E3%5C%5C%20%5C%5C%20%5Csqrt%5B3%5D%7Bx-1%7D%3Dy-5%5C%5C%20%5C%5C%20y%3D%5Csqrt%5B3%5D%7Bx-1%7D%2B5%5C%5C%20%5C%5C%20f%5E%7B-1%7D%28x%29%3D%5Csqrt%5B3%5D%7Bx-1%7D%2B5)
Answer:
C)x^2 - 6x-6
Step-by-step explanation:
-3(x + 2) + (x – 3)x=
-3x-6+x^2-3x=
x^2 - 6x-6
<h2>Hello!</h2>
The answer is:
The relation is a function
Domain: {-5,-3,1}
Range: {6,3}
<h2>Why?</h2>
It's a function since each input (domain) has only one value at the output (range). A function exists when there is only a single value for each input, if there is more than one output for each input, the relation is not a function.
Have a nice day!
First one the one that has the blue inside