Combinatorial Enumeration. That whole class was a rollercoaster ride of mind-blowing generating functions to prove crazy things. The exam had ridiculous questions like 'count the number of cactus trees with n vertices such that etc etc etc' and you'd do three pages of terrible terrible sums and algebra. Then your final answer would be something beautiful like n/2 and you'd breath a sigh of relief and thank the math gods.
Answer:
the answer would be done like this
Step-by-step explanation:
you have to get y by itself so you would need to move the x to the right of the equal sign
once you move anything to the opposite side it switches, so positive becomes negative vice versa
-3y=-x+18
you need to make sure you have positive 1 y so you would divide y by -3 to make it 1
everything else is multiplied by -3 as well
so the answer would be y=⅓x-6
(g₀f)(x)=g(f(x))
=g(2x-2)
=5(2x-2)^2-3
=5(4x^2-8x+4)-3
=20x^2-40x+17