13) The functions [tex]\( q \)[/tex] and [tex]\( r \)[/tex] are defined as follows.

[tex]\[
\begin{array}{l}
q(x) = x^2 + 6 \\
r(x) = \sqrt{x + 9}
\end{array}
\][/tex]

Find the following.

[tex]\[
\begin{array}{l}
(r \circ q)(7) \\
(q \circ r)(7)
\end{array}
\][/tex]



Answer :

Certainly! Let's break down the problem and find [tex]\((r \circ q)(7)\)[/tex] and [tex]\((q \circ r)(7)\)[/tex] step by step.

### Definitions
The functions [tex]\(q(x)\)[/tex] and [tex]\(r(x)\)[/tex] are defined as:
[tex]\[ q(x) = x^2 + 6 \][/tex]
[tex]\[ r(x) = \sqrt{x + 9} \][/tex]

### Step 1: Calculate [tex]\( q(7) \)[/tex]
First, evaluate the function [tex]\( q \)[/tex] at [tex]\( x = 7 \)[/tex]:
[tex]\[ q(7) = 7^2 + 6 = 49 + 6 = 55 \][/tex]

### Step 2: Calculate [tex]\( r(q(7)) = r(55) \)[/tex]
Next, substitute this result into the function [tex]\( r \)[/tex]:
[tex]\[ r(55) = \sqrt{55 + 9} = \sqrt{64} = 8.0 \][/tex]

Thus,
[tex]\[ (r \circ q)(7) = r(q(7)) = 8.0 \][/tex]

### Step 3: Calculate [tex]\( r(7) \)[/tex]
Now, evaluate the function [tex]\( r \)[/tex] at [tex]\( x = 7 \)[/tex]:
[tex]\[ r(7) = \sqrt{7 + 9} = \sqrt{16} = 4.0 \][/tex]

### Step 4: Calculate [tex]\( q(r(7)) = q(4) \)[/tex]
Finally, substitute this result into the function [tex]\( q \)[/tex]:
[tex]\[ q(4) = 4^2 + 6 = 16 + 6 = 22.0 \][/tex]

Thus,
[tex]\[ (q \circ r)(7) = q(r(7)) = 22.0 \][/tex]

### Summary of Results
[tex]\[ \begin{array}{l} (r \circ q)(7) = 8.0 \\ (q \circ r)(7) = 22.0 \end{array} \][/tex]