If A = {x : x text{ is a multiple of } 4} and | vedclass

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(B) Given sets are $A = \{4, 8, 12, 16, 20, 24, \dots\}$ and $B = \{6, 12, 18, 24, 30, \dots\}$.The intersection $A \cap B$ contains elements that are

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$12$

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(B) Given sets are $A = \{4, 8, 12, 16, 20, 24, \dots\}$ and $B = \{6, 12, 18, 24, 30, \dots\}$.The intersection $A \cap B$ contains elements that are common to both sets $A$ and $B$.Since $A$ contains multiples of $4$ and $B$ contains multiples of $6$,the common elements are multiples of the least common multiple $(LCM)$ of $4$ and $6$.$LCM(4, 6) = 12$.Thus,$A \cap B = \{12, 24, 36, \dots\}$,which represents all multiples of $12$.

If $A = \{x : x \text{ is a multiple of } 4\}$ and $B = \{x : x \text{ is a multiple of } 6\}$,then $A \cap B$ consists of all multiples of:

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