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Let $E = \{2, 6\}$ and $F = \{1, 3, 4, 5, 7\}$ . What is $E \cap F$ ? $\newline$ Choices: $\newline$ (A) $\emptyset$ $\newline$ (B) $\{1, 2, 3, 4, 5, 6, 7\}$ $\newline$ (C) $\{1, 2, 5, 6\}$ $\newline$ (D) $\{1, 2, 5, 6, 7\}$

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Full solution

Q. Let

E = \{2, 6\}

and

F = \{1, 3, 4, 5, 7\}

. What is

E \cap F

?

\newline

Choices:

\newline

(A)

\emptyset

\newline

(B)

\{1, 2, 3, 4, 5, 6, 7\}

\newline

(C)

\{1, 2, 5, 6\}

\newline

(D)

\{1, 2, 5, 6, 7\}

Define Sets $E$ and $F$ : $E = \{2, 6\}$ and $F = \{1, 3, 4, 5, 7\}$ are given. To find the intersection $E \cap F$ , we need to find the elements that are common to both sets $E$ and $F$ .
Find Common Elements: We compare each element of set $E$ with each element of set $F$ to determine if there are any common elements.
Check Element $2$ : The element $2$ from set $E$ is not present in set $F$ , which contains \{$\(1\), \(3\), \(4\), \(5\), \(7\)\}.
Check Element \(6\): The element \(6\) from set \(E\) is also not present in set \(F\), which contains \{\(1, 3, 4, 5, 7\)\}.
Identify Intersection: Since there are no common elements between set \(E\) and set \(F\), the intersection \(E \cap F\) is the empty set, denoted by \(\emptyset\).

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