Let A = {10, 11, 12, 13, 14} and B = {-13, 13}. What is A ∩ B? Choices: (A){13} (B){-13, 10, 11, 12, 13, 14} (C){-13, 14} (D){10, 12, 13, 14}

Define Sets A and B: Define the sets A and B. Set A = {10, 11, 12, 13, 14} Set B = {-13, 13} We need to find the intersection of sets A and B, which is the set of elements that are common to both A and B.Find Common Elements: Find the common elements in sets A and B. Looking at both sets, we can see that the number 13 is the only element that appears in both sets A and B.Write Intersection: Write the intersection of sets A and B. The intersection of sets A and B, denoted by A ∩ B, is {13}.Match with Choices: Match the result with the given choices. The intersection we found is {13}, which corresponds to choice (A).

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Let $A = \{10, 11, 12, 13, 14\}$ and $B = \{-13, 13\}$ . What is $A \cap B$ ? $\newline$ Choices: $\newline$ (A) $\{13\}$ $\newline$ (B) $\{-13, 10, 11, 12, 13, 14\}$ $\newline$ (C) $\{-13, 14\}$ $\newline$ (D) $\{10, 12, 13, 14\}$

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Math Problems

Grade 8

Write variable expressions for arithmetic sequences

Full solution

Q. Let

A = \{10, 11, 12, 13, 14\}

and

B = \{-13, 13\}

. What is

A \cap B

\newline

Choices:

\newline

(A)

\{13\}

\newline

(B)

\{-13, 10, 11, 12, 13, 14\}

\newline

(C)

\{-13, 14\}

\newline

(D)

\{10, 12, 13, 14\}

Define Sets $A$ and $B$ : Define the sets $A$ and $B$ . $\newline$ Set $A = \{10, 11, 12, 13, 14\}$ $\newline$ Set $B = \{-13, 13\}$ $\newline$ We need to find the intersection of sets $A$ and $B$ , which is the set of elements that are common to both $A$ and $B$ .
Find Common Elements: Find the common elements in sets $A$ and $B$ . Looking at both sets, we can see that the number $13$ is the only element that appears in both sets $A$ and $B$ .
Write Intersection: Write the intersection of sets $A$ and $B$ . The intersection of sets $A$ and $B$ , denoted by $A \cap B$ , is $\{13\}$ .
Match with Choices: Match the result with the given choices. $\newline$ The intersection we found is ${13}$ , which corresponds to choice $(A)$ .