Question

A=\ 3,5,7,9,11\ and B=\ 7,9,11,13\ , C=\ 11,13,15\

Find

\bf{A \cap \: B) \cap (B \: \cup \: c)​}A∩B)∩(B∪c)​
​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:A = \{3,5,7,9,11\}$

$\rm :\longmapsto\:B = \{7,9,11,13\}$

$\rm :\longmapsto\:C = \{11,13,15\}$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:(A\cap B)\cap (B\cup C)$

$\green{\large\underline{\sf{Solution-}}}$

Given that,

$\rm :\longmapsto\:A = \{3,5,7,9,11\}$

$\rm :\longmapsto\:B = \{7,9,11,13\}$

$\rm :\longmapsto\:C = \{11,13,15\}$

Now, Consider

$\rm :\longmapsto\:A\cap B$

$\rm \:  =  \: \{3,5,7,9,11\}\cap \{7,9,11,13\}$

$\rm \:  =  \: \{7,9,11\}$

Now, Consider

$\rm :\longmapsto\:B\cup C$

$\rm \:  =  \: \{7,9,11,13\}\cup \{11,13,15\}$

$\rm \:  =  \: \{7,9,11,13,15\}$

Now, Consider

$\purple{\rm :\longmapsto\:(A\cap B)\cap (B\cup C) \: }$

$\rm \:  =  \: \{7,9,11\}\cap \{7,9,11,13,15\}$

$\rm \:  =  \: \{7,9,11\}$

Hence,

$\rm\implies \:\boxed{\tt{ (A\cap B)\cap (B\cup C) = \{7,9,11\} \: }}$

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LEARN MORE :-

1. Commutative Law

$\boxed{\tt{A \cap B = B\cap A \: }}$

$\boxed{\tt{A \cup B = B\cup A \: }}$

2. Associative Law

$\boxed{\tt{ (A\cup B)\cup C = A\cup (B\cup C) \: }}$

$\boxed{\tt{ (A\cap B)\cap C = A\cap (B\cap C) \: }}$

3. Distributive Law

$\boxed{\tt{ A\cup (B\cap C) = (A\cup B)\cap (A\cup C) \: }}$

$\boxed{\tt{ A\cap (B\cup C) = (A\cap B)\cup (A\cap C) \: }}$

4. De Morgan's Law

$\boxed{\tt{ (A\cup B)' = A'\cap B' \: }}$

$\boxed{\tt{ (A\cap B)' = A'\cup B' \: }}$

Answer 2

Step-by-step explanation:

This is the answer

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