
The Daily Herald charges $25 for up to 4 lines. Any line over 4 there is an additional costs $5 per a line.  The algebraic representation requires two rules and is expressed as the following:



How much does a 6 line advertisement cost?

Let's define the algebraic representation for the cost of an advertisement in terms of the number of lines.

### Two rules:
1. **If the number of lines is 4 or fewer:**  
   The cost is $25 (flat rate).
2. **If the number of lines is more than 4:**  
   The cost is $25 for the first 4 lines, plus $5 for each additional line.

Let the number of lines be represented by $ n $.  
Thus, the cost function $ C(n) $ is as follows:

\[
C(n) = \begin{cases} 
25 & \text{if } n \leq 4 \\
25 + 5(n - 4) & \text{if } n > 4 
\end{cases}
\]

### Applying this to 6 lines:
Since 6 lines are more than 4, we use the second case:

\[
C(6) = 25 + 5(6 - 4) = 25 + 5(2) = 25 + 10 = 35
\]

Thus, a 6-line advertisement costs **$35**.

Would you like more details or further explanation?

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Here are 5 related questions for practice:
1. How much would a 5-line advertisement cost?
2. What would the cost be for an 8-line advertisement?
3. What is the cost of a 3-line advertisement?
4. Can you write a function that calculates the cost for any $ n $ number of lines?
5. If the cost per extra line increased to $7, how would the equation change?

**Tip:** Break down word problems by identifying the core rules or conditions, then express each in algebraic form.

How much does a 6-line advertisement cost if the Daily Herald charges $25 for up to 4 lines, and $5 for each additional line over 4?

Learn how to calculate the cost of a 6-line advertisement for the Daily Herald, which charges $25 for up to 4 lines and $5 per line beyond that. This algebra problem involves piecewise functions and is suitable for students in Grades 6-8.

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