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We are given that the average cost $ c(x) $ for printing a book is represented by the function:

\[
c(x) = \frac{5.5x + k}{x}
\]

where $ x $ is the number of books printed, and $ k $ is a constant. We are asked to find the value of $ k $ if the average cost per book is $9 when 200 books are printed (i.e., $ x = 200 $).

### Step-by-Step Solution:

1. Plug in the given values: 
   - $ c(x) = 9 $ (the average cost),
   - $ x = 200 $ (the number of books).

\[
9 = \frac{5.5(200) + k}{200}
\]

2. Simplify the equation:

\[
9 = \frac{1100 + k}{200}
\]

3. Multiply both sides by 200 to eliminate the denominator:

\[
9 \times 200 = 1100 + k
\]

\[
1800 = 1100 + k
\]

4. Solve for $ k $:

\[
k = 1800 - 1100
\]

\[
k = 700
\]

Thus, the value of $ k $ is **700**.

Would you like any further details on this solution?

Here are 5 related questions to expand on this topic:

1. How would the average cost change if the number of books printed was 500 instead of 200?
2. What is the minimum possible average cost as $ x $ increases?
3. How would you interpret the constant $ k $ in the context of this cost function?
4. If the cost per book must remain under $8, what is the minimum number of books that need to be printed?
5. How would the equation change if the variable cost per book (currently 5.5) increased by 10%?

**Tip:** When solving problems like these, always remember to check the units and ensure your calculations align with the context of the problem (e.g., costs per book).

Find the value of k if, on the day when 200 books were printed, the average cost was $9 per book. The average cost function is c(x) = (5.5x + k) / x, where x is the number of books printed and k is a constant.

Learn how to find the constant k in an average cost function when given the number of books printed and the cost per book. This algebra problem involves using a linear equation to solve for k, suitable for high school students in Grades 9-12.

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