Let R = {4, 8, 12, 16} and S = {2, 6, 10, 14, 20}. What is R ∩ S? Choices: (A){2, 4, 6, 8, 10, 12, 14, 16, 20} (B) ∅ (C){2, 4, 6, 8, 10, 12} (D){4, 20}

Given sets R and S: Given sets R = {4, 8, 12, 16} and S = {2, 6, 10, 14, 20}, we need to find the intersection of R and S, which is the set of elements that are common to both R and S.Find intersection of R and S: To find the intersection, we compare each element of set R with each element of set S to see if there are any elements that appear in both sets.Compare elements of R and S: Comparing the elements, we see that: - 4 is not in set S. - 8 is not in set S. - 12 is not in set S. - 16 is not in set S.Intersection is empty set: Since none of the elements of set R are in set S, the intersection of R and S is the empty set, denoted by ∅.

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Let $R = \{4, 8, 12, 16\}$ and $S = \{2, 6, 10, 14, 20\}$ . What is $R \cap S$ ? $\newline$ Choices: $\newline$ (A) $\{2, 4, 6, 8, 10, 12, 14, 16, 20\}$ $\newline$ (B) $\emptyset$ $\newline$ (C) $\{2, 4, 6, 8, 10, 12\}$ $\newline$ (D) $\{4, 20\}$

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Q. Let

R = \{4, 8, 12, 16\}

and

S = \{2, 6, 10, 14, 20\}

. What is

R \cap S

\newline

Choices:

\newline

(A)

\{2, 4, 6, 8, 10, 12, 14, 16, 20\}

\newline

(B)

\emptyset

\newline

(C)

\{2, 4, 6, 8, 10, 12\}

\newline

(D)

\{4, 20\}

Given sets $R$ and $S$ : Given sets $R = \{4, 8, 12, 16\}$ and $S = \{2, 6, 10, 14, 20\}$ , we need to find the intersection of $R$ and $S$ , which is the set of elements that are common to both $R$ and $S$ .
Find intersection of $R$ and $S$ : To find the intersection, we compare each element of set $R$ with each element of set $S$ to see if there are any elements that appear in both sets.
Compare elements of R and S: Comparing the elements, we see that: $\newline$ - $4$ is not in set $S$ . $\newline$ - $8$ is not in set $S$ . $\newline$ - $12$ is not in set $S$ . $\newline$ - $16$ is not in set $S$ .
Intersection is empty set: Since none of the elements of set $R$ are in set $S$ , the intersection of $R$ and $S$ is the empty set, denoted by $\emptyset$ .