To determine which of the given expressions must be even if [tex]\(9p + 5\)[/tex] is even, let's analyze the given condition step by step.
First, understand that for [tex]\(9p + 5\)[/tex] to be even, [tex]\(9p\)[/tex] itself must satisfy certain conditions. Here's how we can reason through the problem:
1. Given condition: [tex]\(9p + 5\)[/tex] is even.
2. If we subtract 5 from both sides of the given condition, we get:
[tex]\[
9p + 5 - 5 = \text{even number} - 5 \implies 9p = \text{even number} - 5
\][/tex]
3. An even number minus an odd number results in an odd number. Thus:
[tex]\[
9p = \text{odd number}
\][/tex]
4. Since [tex]\(9p\)[/tex] is odd, [tex]\(p\)[/tex] itself must be odd. This is because 9 is an odd number, and for the product of two numbers to be odd, both numbers must be odd.
With [tex]\(p\)[/tex] confirmed to be odd, let's investigate each of the given options:
Option A: [tex]\(p + 5\)[/tex]
If [tex]\(p\)[/tex] is odd, adding 5 (which is also odd) to it will result in an even number since:
[tex]\[
\text{odd} + \text{odd} = \text{even}
\][/tex]
Hence, [tex]\(p + 5\)[/tex] is even when [tex]\(p\)[/tex] is odd.
Option B: [tex]\(2p + 5\)[/tex]
If [tex]\(p\)[/tex] is odd, multiplying it by 2 will result in an even number since:
[tex]\[
2 \times \text{odd} = \text{even}
\][/tex]
However, adding 5 (an odd number) to an even number will result in an odd number since:
[tex]\[
\text{even} + \text{odd} = \text{odd}
\][/tex]
Therefore, [tex]\(2p + 5\)[/tex] is odd when [tex]\(p\)[/tex] is odd.
Option C: [tex]\(5p\)[/tex]
If [tex]\(p\)[/tex] is odd, multiplying it by 5 (an odd number) will result in an odd number since:
[tex]\[
\text{odd} \times \text{odd} = \text{odd}
\][/tex]
Hence, [tex]\(5p\)[/tex] is odd when [tex]\(p\)[/tex] is odd.
Option D: [tex]\(9p\)[/tex]
We already established that [tex]\(9p\)[/tex] is odd if [tex]\(p\)[/tex] is odd, since [tex]\(9\)[/tex] (an odd number) multiplied by another odd number results in an odd number:
[tex]\[
\text{odd} \times \text{odd} = \text{odd}
\][/tex]
Hence, [tex]\(9p\)[/tex] is odd.
In summary, among the given options, the expression that must be even is:
[tex]\[
\boxed{p+5}
\][/tex]
