Let's break this problem down step-by-step.
### Part (a)
We need to find the cost of a 2-line ad and a 3-line ad and then determine the difference in cost.
The piecewise function given is:
[tex]\[
a(x) = \begin{cases}
45 & \text{when } x \leq 3 \\
45 + 9(x - 3) & \text{when } x > 3
\end{cases}
\][/tex]
#### Cost of a 2-line ad:
For [tex]\( x = 2 \)[/tex]:
Since [tex]\( 2 \leq 3 \)[/tex], use the first part of the piecewise function:
[tex]\[ a(2) = 45 \][/tex]
#### Cost of a 3-line ad:
For [tex]\( x = 3 \)[/tex]:
Since [tex]\( 3 \leq 3 \)[/tex], again use the first part of the piecewise function:
[tex]\[ a(3) = 45 \][/tex]
#### Difference between the cost of a 2-line ad and a 3-line ad:
[tex]\[ \text{Difference} = a(3) - a(2) = 45 - 45 = 0 \][/tex]
So, the cost difference between a 2-line ad and a 3-line ad is \[tex]$0.
### Part (b)
We need to find the cost of a 10-line ad.
Using the piecewise function for \( x = 10 \):
Since \( 10 > 3 \), use the second part of the piecewise function:
\[ a(10) = 45 + 9(10 - 3) \]
\[ a(10) = 45 + 9(7) \]
\[ a(10) = 45 + 63 \]
\[ a(10) = 108 \]
The cost of a 10-line ad is $[/tex]108.
### Part (c)
We need to find the cost of an 11-line ad.
Using the piecewise function for [tex]\( x = 11 \)[/tex]:
Since [tex]\( 11 > 3 \)[/tex], use the second part of the piecewise function:
[tex]\[ a(11) = 45 + 9(11 - 3) \][/tex]
[tex]\[ a(11) = 45 + 9(8) \][/tex]
[tex]\[ a(11) = 45 + 72 \][/tex]
[tex]\[ a(11) = 117 \][/tex]
The cost of an 11-line ad is [tex]$117.
In summary:
a. The difference between the cost of a 2-line ad and a 3-line ad is \$[/tex]0.
b. The cost of a 10-line ad is [tex]$108.
c. The cost of an 11-line ad is $[/tex]117.
