Answer:
4
Step-by-step explanation:
It's much easier to see the meaning if brackets are used.
h(g(x) ) = h·g(x)
What this means is that in h(x) where ever there is an x, put a g(x)
h(x) = x^2 + 4
h(g(x)) = [g(x)]^2 + 4
Now where ever you see a g(x), put 2x
h(g(x))= (2x)^2 + 4
h(g(x)) = 4x^2 + 4
Now solve for h(g(0)) which means where ever you see a x substitute a 0.
h(g(0)) = 4*(0) + 4
h(g(0)) = 4
