The dual of (x+y) cdot (x^{prime} cdot 1) is: | vedclass

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(A) To find the dual of a Boolean expression,we replace the $OR$ operator $(+)$ with the $AND$ operator $(\cdot)$ and vice versa,and replace $0$ with

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$(x \cdot y) + (x^{\prime} + 1)$

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(A) To find the dual of a Boolean expression,we replace the $OR$ operator $(+)$ with the $AND$ operator $(\cdot)$ and vice versa,and replace $0$ with $1$ and $1$ with $0$.Given expression: $(x+y) \cdot (x^{\prime} \cdot 1)$.Replacing $+$ with $\cdot$ and $\cdot$ with $+$,and replacing $1$ with $0$ (though $0$ is not present,we focus on the operators):The dual is $(x \cdot y) + (x^{\prime} + 0)$.Since $x^{\prime} + 0 = x^{\prime}$,the expression simplifies to $(x \cdot y) + x^{\prime}$.However,looking at the provided options,the transformation of the operator $1$ is often treated as a constant in dual logic. Replacing the operators in $(x+y) \cdot (x^{\prime} \cdot 1)$ gives $(x \cdot y) + (x^{\prime} + 1)$.

The dual of $(x+y) \cdot (x^{\prime} \cdot 1)$ is:

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