1. **Problem 23:** Find the Boolean algebra expression for the given circuit. 2. The circuit has inputs $x$, $y$, and $z$. The input $x$ passes through a NOT gate producing $x'$. Then, $x'$ and $y$ go into an OR gate producing $(x' + y)$. Another OR gate takes $x$ and $y$ producing $(x + y)$. An AND gate takes $y$ and $z$ producing $(yz)$. 3. The outputs $(x' + y)$, $(x + y)$, and $(yz)$ are combined by AND gates. So the overall expression is: $$ (x' + y)(x + y)(yz) $$ 4. Comparing with options, this matches option A. 5. **Problem 24:** Find the Boolean algebra expression for the given circuit. 6. Inputs $x$, $y$, and $z$ with $x$ passing through NOT gate to get $x'$. Then $x'$ and $y$ go into an OR gate $(x' + y)$, and $x$ and $y$ go into an AND gate $(xy)$. 7. The outputs $(x' + y)$ and $(xy)$ are combined by an OR gate producing $[(x' + y) + xy]$. 8. Input $z$ and $y$ go into an OR gate producing $(y + z)$. 9. Finally, these two outputs are combined by an AND gate: $$ [(x' + y) + xy](y + z) $$ 10. This matches option A. 11. **Problem 25:** Find the Boolean algebra expression for the given circuit. 12. Inputs $x$, $y$, and $z$ with $x$ passing through NOT gate to get $x'$. Then $x'$ and $y$ go into an OR gate $(x' + y)$. 13. Inputs $x$ and $y$ go into an AND gate $(xy)$. 14. Inputs $y$ and $z$ go into an AND gate $(yz)$. 15. The outputs $(x' + y)$ and $(xy)$ are combined by an AND gate, and the output of this AND gate is combined with $(yz)$ by an OR gate. 16. So the expression is: $$ (x' + y)(xy) + (yz) $$ 17. This matches option C. **Final answers:** - Problem 23: A - Problem 24: A - Problem 25: C