Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Given the conditions:
1. [tex]\( R + S \)[/tex] is an odd number.
2. [tex]\(-S\)[/tex] is also an odd number.
We need to determine which among the given expressions [tex]\( R+T \)[/tex], [tex]\( S+T \)[/tex], [tex]\( R \)[/tex], [tex]\( S \)[/tex], or [tex]\( T \)[/tex] must be an even number.
First, let's analyze the conditions and constraints provided:
#### Step 1: Understanding [tex]\( R + S \)[/tex]
- An integer [tex]\( R + S \)[/tex] being odd means one integer must be even and the other must be odd.
- This can be written as:
- [tex]\( R \)[/tex] is even and [tex]\( S \)[/tex] is odd, or
- [tex]\( R \)[/tex] is odd and [tex]\( S \)[/tex] is even.
#### Step 2: Understanding [tex]\(-S\)[/tex]
- [tex]\(-S\)[/tex] is odd. Importantly, the negative of an integer does not change its property of being even or odd.
- Therefore, if [tex]\(-S\)[/tex] is odd, [tex]\( S \)[/tex] must be odd.
We have just concluded:
- [tex]\( S \)[/tex] must be odd.
Since [tex]\( S \)[/tex] is odd and [tex]\( R + S \)[/tex] is odd, it follows that:
- [tex]\( R \)[/tex] must be even (because the sum of an even and an odd number is odd).
Now, let's evaluate the given expressions to identify which must be even.
#### Evaluating [tex]\(R + T\)[/tex]:
- Let's consider the parity of [tex]\( T \)[/tex] (even or odd).
- If [tex]\( T \)[/tex] is even, then [tex]\( R + T \)[/tex] is even (because even + even = even).
- If [tex]\( T \)[/tex] is odd, then [tex]\( R + T \)[/tex] is odd (because even + odd = odd).
Since [tex]\( T \)[/tex] can be either even or odd, [tex]\( R + T \)[/tex] can be either even or odd.
#### Evaluating [tex]\(S + T\)[/tex]:
- Since [tex]\( S \)[/tex] is established to be odd:
- If [tex]\( T \)[/tex] is even, [tex]\( S + T \)[/tex] is odd (odd + even = odd).
- If [tex]\( T \)[/tex] is odd, [tex]\( S + T \)[/tex] is even (odd + odd = even).
Since [tex]\( T \)[/tex] can be either even or odd, [tex]\( S + T \)[/tex] can be either odd or even.
#### Evaluating [tex]\( R \)[/tex]:
- [tex]\( R \)[/tex] is confirmed to be even based on the conditions provided.
#### Evaluating [tex]\( S \)[/tex]:
- [tex]\( S \)[/tex] is confirmed to be odd.
#### Evaluating [tex]\( T \)[/tex]:
- There is no restriction given explicitly for [tex]\( T \)[/tex], so [tex]\( T \)[/tex] can be either even or odd.
Considering all possible conditions, the only integer out of the provided options that must be even is:
[tex]\[ \boxed{R} \][/tex]
1. [tex]\( R + S \)[/tex] is an odd number.
2. [tex]\(-S\)[/tex] is also an odd number.
We need to determine which among the given expressions [tex]\( R+T \)[/tex], [tex]\( S+T \)[/tex], [tex]\( R \)[/tex], [tex]\( S \)[/tex], or [tex]\( T \)[/tex] must be an even number.
First, let's analyze the conditions and constraints provided:
#### Step 1: Understanding [tex]\( R + S \)[/tex]
- An integer [tex]\( R + S \)[/tex] being odd means one integer must be even and the other must be odd.
- This can be written as:
- [tex]\( R \)[/tex] is even and [tex]\( S \)[/tex] is odd, or
- [tex]\( R \)[/tex] is odd and [tex]\( S \)[/tex] is even.
#### Step 2: Understanding [tex]\(-S\)[/tex]
- [tex]\(-S\)[/tex] is odd. Importantly, the negative of an integer does not change its property of being even or odd.
- Therefore, if [tex]\(-S\)[/tex] is odd, [tex]\( S \)[/tex] must be odd.
We have just concluded:
- [tex]\( S \)[/tex] must be odd.
Since [tex]\( S \)[/tex] is odd and [tex]\( R + S \)[/tex] is odd, it follows that:
- [tex]\( R \)[/tex] must be even (because the sum of an even and an odd number is odd).
Now, let's evaluate the given expressions to identify which must be even.
#### Evaluating [tex]\(R + T\)[/tex]:
- Let's consider the parity of [tex]\( T \)[/tex] (even or odd).
- If [tex]\( T \)[/tex] is even, then [tex]\( R + T \)[/tex] is even (because even + even = even).
- If [tex]\( T \)[/tex] is odd, then [tex]\( R + T \)[/tex] is odd (because even + odd = odd).
Since [tex]\( T \)[/tex] can be either even or odd, [tex]\( R + T \)[/tex] can be either even or odd.
#### Evaluating [tex]\(S + T\)[/tex]:
- Since [tex]\( S \)[/tex] is established to be odd:
- If [tex]\( T \)[/tex] is even, [tex]\( S + T \)[/tex] is odd (odd + even = odd).
- If [tex]\( T \)[/tex] is odd, [tex]\( S + T \)[/tex] is even (odd + odd = even).
Since [tex]\( T \)[/tex] can be either even or odd, [tex]\( S + T \)[/tex] can be either odd or even.
#### Evaluating [tex]\( R \)[/tex]:
- [tex]\( R \)[/tex] is confirmed to be even based on the conditions provided.
#### Evaluating [tex]\( S \)[/tex]:
- [tex]\( S \)[/tex] is confirmed to be odd.
#### Evaluating [tex]\( T \)[/tex]:
- There is no restriction given explicitly for [tex]\( T \)[/tex], so [tex]\( T \)[/tex] can be either even or odd.
Considering all possible conditions, the only integer out of the provided options that must be even is:
[tex]\[ \boxed{R} \][/tex]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
