Answered

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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].

Sagot :

GinnyAnswer
To determine if [tex]\( R \subset A \)[/tex], we need to check if every element in set [tex]\( R \)[/tex] is also an element in set [tex]\( A \)[/tex].

Here are the sets given:
[tex]\[ U = \{ x \mid x \text{ is a real number} \} \][/tex]
[tex]\[ A = \{ x \mid x \text{ is an odd integer} \} \][/tex]
[tex]\[ R = \{ x \mid x = 3, 7, 11, 27 \} \][/tex]

Let's examine the elements of set [tex]\( R \)[/tex]:
- The elements of [tex]\( R \)[/tex] are [tex]\( \{ 3, 7, 11, 27 \} \)[/tex].

Next, let's define set [tex]\( A \)[/tex]:
- Set [tex]\( A \)[/tex] contains all odd integers. So [tex]\( A \)[/tex] includes numbers like [tex]\(-1, 1, 3, 5, 7, 9, 11, \dots, \)[/tex].

Now, we check if each element of [tex]\( R \)[/tex] is in [tex]\( A \)[/tex]:
- [tex]\( 3 \)[/tex] is an odd integer and is in [tex]\( A \)[/tex].
- [tex]\( 7 \)[/tex] is an odd integer and is in [tex]\( A \)[/tex].
- [tex]\( 11 \)[/tex] is an odd integer and is in [tex]\( A \)[/tex].
- [tex]\( 27 \)[/tex] is an odd integer and is in [tex]\( A \)[/tex].

Since all the elements [tex]\( 3, 7, 11, 27 \)[/tex] in set [tex]\( R \)[/tex] are also in set [tex]\( A \)[/tex], we can conclude that [tex]\( R \subset A \)[/tex].

Therefore, the correct statement is:
[tex]\[ \text{Yes, because all the elements of set } R \text{ are in set } A. \][/tex]
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