Let's solve the given problems step by step.
### Part (a)
Find the difference between the cost of a two-line ad and the cost of a three-line ad.
The piecewise function is defined as:
[tex]\[
a(x) =
\begin{cases}
45 & \text{when } x \leq 3 \\
45 + 9(x-3) & \text{when } x > 3
\end{cases}
\][/tex]
For a two-line ad ([tex]\(x = 2\)[/tex]):
[tex]\[
a(2) = 45
\][/tex]
For a three-line ad ([tex]\(x = 3\)[/tex]):
[tex]\[
a(3) = 45
\][/tex]
The difference between the cost of a two-line ad and a three-line ad is:
[tex]\[
a(3) - a(2) = 45 - 45 = 0
\][/tex]
### Part (b)
Find the cost of a 10-line ad.
For a 10-line ad ([tex]\(x = 10\)[/tex]):
[tex]\[
a(10) = 45 + 9(10 - 3) = 45 + 9 \times 7 = 45 + 63 = 108
\][/tex]
### Part (c)
Find the cost of an 11-line ad.
For an 11-line ad ([tex]\(x = 11\)[/tex]):
[tex]\[
a(11) = 45 + 9(11 - 3) = 45 + 9 \times 8 = 45 + 72 = 117
\][/tex]
### Part (d)
Find the difference between the cost of a 15-line ad and a 17-line ad, without finding out the cost of each ad first.
When [tex]\( x > 3 \)[/tex], the function is linear:
[tex]\[ a(x) = 45 + 9(x - 3). \][/tex]
We need to find the difference [tex]\( a(17) - a(15) \)[/tex].
We can simplify this directly:
[tex]\[
a(17) - a(15) = [45 + 9(17 - 3)] - [45 + 9(15 - 3)]
\][/tex]
Simplifying inside the brackets:
[tex]\[
a(17) - a(15) = [45 + 9 \times 14] - [45 + 9 \times 12]
\][/tex]
[tex]\[
= (45 + 126) - (45 + 108)
\][/tex]
[tex]\[
= 171 - 153 = 18
\][/tex]
So the difference between the cost of a 15-line ad and a 17-line ad is 18.
### Summary of Results
- (a) The difference between the cost of a two-line ad and a three-line ad is [tex]\(0\)[/tex].
- (b) The cost of a 10-line ad is [tex]\(108\)[/tex].
- (c) The cost of an 11-line ad is [tex]\(117\)[/tex].
- (d) The difference between the cost of a 15-line ad and a 17-line ad is [tex]\(18\)[/tex].
