Question 186683
Since B = {1,2,3} and U = {1,2,3,4,5,6,7}, this means that B'= everything in U that is NOT in B = {4,5,6,7}



Also, because C = {2,3,4,5,6} and U = {1,2,3,4,5,6,7}, this means C' = everything in U that is NOT in C = {1,7}



Now let's evaluate *[Tex \LARGE C' \cap A]. So simply find the elements that BOTH C' and A have in common:



*[Tex \LARGE C' \cap A=\left\{1,7\right\}] Note: the elements 1 and 7 are BOTH in C' and A



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Now  evaluate *[Tex \LARGE C' \cap B']. So simply find the elements that BOTH C' and B' have in common:



*[Tex \LARGE C' \cap B'=\left\{7\right\}] Note: 7 is the only element that is BOTH in C' and B



So


*[Tex \LARGE \left(C' \cap A \right) \cup \left(C' \cap B' \right)]



then becomes


*[Tex \LARGE \left\{1,7\right\} \cup \left\{7\right\}]



combine them to get 



*[Tex \LARGE \left\{1,7\right\}]



Answer:


So this means that 


*[Tex \LARGE \left(C' \cap A \right) \cup \left(C' \cap B' \right)=\left\{1,7\right\}]