Question

|A∆B|=5,|A|=12 & |B|=13 so |A intersection B|= ? ​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{|A\,{\triangle}\,B|=5,\;|A|=12,\;|B|=13}$

$\underline{\textbf{To find:}}$

$\mathsf{|A\,{\cap}\,B|}$

$\underline{\textbf{Solution:}}$

$\textsf{We know that,}$

$\mathsf{|A\,\triangle\,B|=|A|+|B|-2|A\{\cap}\,B|}$

$\mathsf{5=12+13-2|A\,{\cap}\,B|}$

$\mathsf{5=25-2|A\,{\cap}\,B|}$

$\mathsf{2|A\,{\cap}\,B|=25-5}$

$\mathsf{2|A\,{\cap}\,B|=20}$

$\mathsf{|A\,{\cap}\,B|=\dfrac{20}{2}}$

$\implies\boxed{\mathsf{|A\,{\cap}\,B|=10}}$

$\underline{\textbf{Find more:}}$

If U={x:x in N x<=30} A={x:x is prime <5} B={x:x is a perfect square <=10} and C={x:x is a perfect cube <=30} then verify the following results : (i) (A uu B)'=A'nn B' (ii)(A nn B)'=A'uu B' (iii) (A nn B)nn C=A nn(B nn C) (iv) A'-B'=B-A

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