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The given problem is asking for the composition of two functions, \( q(x) = 4x \) and \( r(x) = 5x + 2 \), with the notation \( (q \circ r)(x) \). 

To find \( (q \circ r)(x) \), we need to substitute \( r(x) \) into \( q(x) \). This means evaluating \( q(r(x)) \).

Here’s how we solve it step by step:

1. **Substitute \( r(x) \) into \( q(x) \):**
   \[
   q(r(x)) = q(5x + 2)
   \]
2. **Apply the function \( q(x) = 4x \) to the expression \( 5x + 2 \):**
   Since \( q(x) = 4x \), replace the input \( x \) with \( 5x + 2 \):
   \[
   q(5x + 2) = 4(5x + 2)
   \]
3. **Simplify the expression:**
   \[
   4(5x + 2) = 4 \cdot 5x + 4 \cdot 2 = 20x + 8
   \]

Thus, the solution is:
\[
(q \circ r)(x) = 20x + 8
\]

Would you like further details or have any other questions?

Here are 5 related questions for further practice:
1. What is the composition of \( r \circ q(x) \) for the same functions?
2. How do you find the inverse of a function like \( q(x) = 4x \)?
3. What happens if you compose \( q(x) = 4x \) with itself, i.e., \( (q \circ q)(x) \)?
4. How would you solve \( (q \circ r)(x) = 0 \)?
5. What is the domain of the composition \( (q \circ r)(x) \)?

**Tip**: When composing functions, always substitute the inner function into the outer function carefully to avoid mistakes!

Find (q o r)(x) where q(x) = 4x and r(x) = 5x + 2.

Learn how to find the composition of two linear functions: q(x) = 4x and r(x) = 5x + 2. Follow a step-by-step solution to simplify (q o r)(x) to 20x + 8, using function composition and the distributive property. Suitable for students in Grades 10-11.

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