of this following question (using boolean algebra simplify following (A) \overline{(A}+B)(A+C) (B) A (overline B) + A (overline B) * C +A (overline B) CD+A( overline B) CDE (C) BC+\overline{BCD}+B (D) (B + (overlineB))(BC + BC(overlineD)) (E) BC+\overline{(B}+\overline{C})D+BC ) the provided answer is correct if not olz provide correct answer ((A) ( � + � ) ‾ ( � + � ) (A+B) ​ (A+C) Applying De Morgan's theorem and the distributive law: = ( � ‾ ⋅ � ‾ ) ( � + � ) = � ‾ � + � ‾ � ⋅ � ‾ =( A ⋅ B )(A+C)= A A+ A C⋅ B = 0 + � ‾ � ⋅ � ‾ = � ‾ � ⋅ � ‾ =0+ A C⋅ B = A C⋅ B (B) � � ‾ + � � ‾ � + � � ‾ � � + � � ‾ � � � A B +A B C+A B CD+A B CDE Notice that � � ‾ A B is common in all terms, so we factor it out: = � � ‾ ( 1 + � + � � + � � � ) =A B (1+C+CD+CDE) Since 1 + 1+ anything = 1 =1: = � � ‾ =A B (C) � � + � � � ‾ + � BC+ BCD +B Using the law of idempotence and considering � � � ‾ BCD does not simplify directly with � � BC or � B without further context, the expression is already in its simplified form if there are no additional constraints or identities to apply: = � � + � � � ‾ + � =BC+ BCD +B One might argue to simplify further based on specific use cases, but with the given information, this is the simplified form. (D) ( � + � ‾ ) ( � � + � � � ‾ ) (B+ B )(BC+BC D ) Using the law of complementarity ( � + � ‾ = 1 B+ B =1) and idempotence ( � � + � � = � � BC+BC=BC): = 1 ( � � + � � � ‾ ) = � � + � � � ‾ = � � ( 1 + � ‾ ) = � � =1(BC+BC D )=BC+BC D =BC(1+ D )=BC (E) � � + ( � + � ‾ ) ‾ � + � � BC+ (B+ C ) ​ D+BC Applying De Morgan's theorem to ( � + � ‾ ) ‾ (B+ C ) ​ : = � � + ( � ‾ ⋅ � ) � + � � =BC+( B ⋅C)D+BC Using idempotence law ( � � + � � = � � BC+BC=BC): = � � + � ‾ � � =BC+ B CD Since � ‾ � � B CD does not directly combine with � � BC, this is the simplified form given the information: = � � + � ‾ � � =BC+ B CD)