To determine whether each expression is odd when \( m \) and \( n \) are both odd integers, let's examine each case.
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For \( 2m + n \):
- Since \( m \) is odd, \( 2m \) is even (as multiplying an odd integer by 2 results in an even integer).
- \( n \) is odd.
- The sum of an even integer and an odd integer is odd.
- Therefore, \( 2m + n \) is odd.
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For \( (m + n)^2 \):
- The sum \( m + n \) is the sum of two odd integers, which is even (the sum of two odd numbers is always even).
- Squaring an even integer results in an even integer.
- Therefore, \( (m + n)^2 \) is even.
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For \( mn \):
- The product of two odd integers is always odd.
- Therefore, \( mn \) is odd.
In summary:
- \( 2m + n \) is odd.
- \( (m + n)^2 \) is even.
- \( mn \) is odd.
The expressions that must also be odd are:
- \( 2m + n \)
- \( mn \)
So the expressions that must be odd are:
- True: \( 2m + n \) is odd.
- False: \( (m + n)^2 \) is even.
- True: \( mn \) is odd.
Thus, the final answer is that the expressions \( 2m + n \) and \( mn \) are odd (while \( (m + n)^2 \) is not).