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(D) Let the output of the $NOR$ gate be $Y$. The $NOR$ gate performs the operation $Y = \overline{A+B}$. The $NAND$ gate then takes $Y$ and $C$ as inp

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$1$ and $1$

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(D) Let the output of the $NOR$ gate be $Y$. The $NOR$ gate performs the operation $Y = \overline{A+B}$. The $NAND$ gate then takes $Y$ and $C$ as inputs to produce output $D = \overline{Y \cdot C}$.Case $1$: $A=0, B=0, C=0$$Y = \overline{0+0} = \overline{0} = 1$$D = \overline{Y \cdot C} = \overline{1 \cdot 0} = \overline{0} = 1$Case $2$: $A=1, B=0, C=1$$Y = \overline{1+0} = \overline{1} = 0$$D = \overline{Y \cdot C} = \overline{0 \cdot 1} = \overline{0} = 1$Thus, the outputs are $1$ and $1$ respectively.

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