Let's break down the functions and solve each part of the question step by step.

We are given three functions:

$f(x) = 4 - 2x$ $g(x) = x^2 + 5$ $h(x) = \frac{x - 6}{7}$

Part a: $f(g(-3))$

We are told $f(g(-3)) = -24$, but let's confirm this.

1. First, find $g(-3)$: $g(x) = x^2 + 5$ Substituting $x = -3$: $g(-3) = (-3)^2 + 5 = 9 + 5 = 14$

2. Now, find $f(14)$: $f(x) = 4 - 2x$ Substituting $x = 14$: $f(14) = 4 - 2(14) = 4 - 28 = -24$

Thus, $f(g(-3)) = -24$, which is correctly given.

Part b: $g \circ h(20)$ (This means $g(h(20))$)

1. First, find $h(20)$: $h(x) = \frac{x - 6}{7}$ Substituting $x = 20$: $h(20) = \frac{20 - 6}{7} = \frac{14}{7} = 2$

2. Now, find $g(2)$: $g(x) = x^2 + 5$ Substituting $x = 2$: $g(2) = 2^2 + 5 = 4 + 5 = 9$

Thus, $g \circ h(20) = 9$.

Part c: $h(f(g(3)))$

1. First, find $g(3)$: $g(3) = 3^2 + 5 = 9 + 5 = 14$

2. Now, find $f(14)$ (which we already computed in Part a): $f(14) = -24$

3. Now, find $h(-24)$: $h(x) = \frac{x - 6}{7}$ Substituting $x = -24$: $h(-24) = \frac{-24 - 6}{7} = \frac{-30}{7}$

Thus, $h(f(g(3))) = \frac{-30}{7}$.

Final Answers:

a. $f(g(-3)) = -24$ (Given) b. $g \circ h(20) = 9$ c. $h(f(g(3))) = \frac{-30}{7}$

Would you like further details or explanations on any part of this?

Here are some related questions:

1. How would you compute $f(h(g(x)))$ for an arbitrary $x$?
2. What would $h(f(h(x)))$ be for a general input $x$?
3. Can you find the composite function $f \circ g(x)$ for a general $x$?
4. What would the graph of $h(x)$ look like, and how would its slope affect the composite functions?
5. If $g(x)$ were different, say $g(x) = x^2 - 5$, how would that change the result for part c?

Tip: When dealing with composite functions, always evaluate the innermost function first before moving to the outer ones.