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The expression provided is:

\[
p(x) = \left\lfloor \frac{60}{x} \right\rfloor
\]

where \(\left\lfloor \cdot \right\rfloor\) denotes the **floor function**, which rounds down to the nearest integer. We are asked to evaluate \(p(7)\).

Substituting \(x = 7\) into the function:

\[
p(7) = \left\lfloor \frac{60}{7} \right\rfloor
\]

First, calculate \(\frac{60}{7}\):

\[
\frac{60}{7} \approx 8.5714
\]

Now, applying the floor function to 8.5714, we round down to the nearest integer:

\[
p(7) = \left\lfloor 8.5714 \right\rfloor = 8
\]

Thus, \(p(7) = 8\).

Would you like more details on this, or do you have any questions?

Here are 5 related questions to expand on this topic:
1. How does the floor function differ from the ceiling function?
2. Can you evaluate \(p(5)\) for the same function?
3. What is the difference between the floor and the round function?
4. How does the function behave as \(x\) approaches infinity?
5. What happens to \(p(x)\) when \(x = 1\)?

**Tip:** The floor function always rounds down, so even if the result is very close to the next integer, it will still go to the lower one.

Explore the evaluation of the function p(x) = floor(60/x) at x=7 with a detailed step-by-step solution. This example illustrates the use of the floor function, rounding down to the nearest integer. Suitable for students in Grades 8-10, enhancing their understanding of algebraic functions.

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