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Consider the sets below:

[tex]\[
\begin{array}{l}
U=\{x \mid x \text{ is a real number} \} \\
A=\{x \mid x \text{ is an odd integer} \} \\
R=\{x \mid x=3,7,11,27\}
\end{array}
\][/tex]

Is [tex]\( R \subset A \)[/tex]?

A. Yes, because all the elements of set [tex]\(A\)[/tex] are in set [tex]\(R\)[/tex].
B. Yes, because all the elements of set [tex]\(R\)[/tex] are in set [tex]\(A\)[/tex].
C. No, because each element in set [tex]\(A\)[/tex] is not represented in set [tex]\(R\)[/tex].
D. No, because each element in set [tex]\(R\)[/tex] is not represented in set [tex]\(A\)[/tex].



Answer :

GinnyAnswer
To determine whether [tex]\( R \subset A \)[/tex], we must verify whether every element in [tex]\( R \)[/tex] is also an element of [tex]\( A \)[/tex].

Given:
- [tex]\( U \)[/tex] is the set of all real numbers.
- [tex]\( A \)[/tex] is the set of all odd integers.
- [tex]\( R = \{3, 7, 11, 27\} \)[/tex].

Let's analyze each element in [tex]\( R \)[/tex] to see if it is in [tex]\( A \)[/tex].

1. [tex]\( 3 \)[/tex]:
- [tex]\( 3 \)[/tex] is an integer and it's odd.
- Therefore, [tex]\( 3 \in A \)[/tex].

2. [tex]\( 7 \)[/tex]:
- [tex]\( 7 \)[/tex] is an integer and it's odd.
- Therefore, [tex]\( 7 \in A \)[/tex].

3. [tex]\( 11 \)[/tex]:
- [tex]\( 11 \)[/tex] is an integer and it's odd.
- Therefore, [tex]\( 11 \in A \)[/tex].

4. [tex]\( 27 \)[/tex]:
- [tex]\( 27 \)[/tex] is an integer and it's odd.
- Therefore, [tex]\( 27 \in A \)[/tex].

Since all elements of [tex]\( R \)[/tex] (which are [tex]\( 3, 7, 11, \)[/tex] and [tex]\( 27 \)[/tex]) are elements of [tex]\( A \)[/tex] (the set of odd integers), we can conclude that [tex]\( R \subset A \)[/tex].

Therefore, the correct answer is:
Yes, because all the elements of set [tex]\( R \)[/tex] are in set [tex]\( A \)[/tex].

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