Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
