Question

1 multiple choice 1 point the values for variables x, y, and z are: x = 0, y = 1, z = 1 select the boolean expression that evaluates to 1. (x + \\(\overline{y}\\))z \\(\overline{z + x}+y\overline{x}\\) (\\(\overline{x + z}\\))(\\(\overline{y + z}\\)) \\(\overline{z}\\)(x + y) 2 multiple choice 1 point select the expression that is equivalent to \\(\overline{x}\\). (\\(\overline{x}+y\\))\\(\overline{y}\\) (x + \\(\overline{x}\\))(y + x) \\(\overline{x}\cdot\overline{y}+\overline{x}\\) \\(\overline{x}\cdot y + x\\) 3 multiple choice 1 point select the expression that is equivalent to x + y. x + \\(\overline{y}\\)+xy xy + y+\\(\overline{x}\cdot\overline{y}\\) (x + \\(\overline{y}\\))(x + y) y + x\\(\overline{y}\\)

Explanation:

1.

Step1: Substitute values into first option

Given $x = 0,y = 1,z = 1$. For $(x+\overline{y})z$, $\overline{y}=0$, then $x+\overline{y}=0 + 0=0$, and $(x+\overline{y})z=0\times1 = 0$.

Step2: Substitute values into second option

For $\overline{z + x}+y\overline{x}$, $z + x=1+0 = 1$, $\overline{z + x}=0$, $\overline{x}=1$, $y\overline{x}=1\times1 = 1$, so $\overline{z + x}+y\overline{x}=0 + 1=1$.

Step3: Substitute values into third option

For $(\overline{x + z})(\overline{y + z})$, $x + z=0 + 1=1$, $\overline{x + z}=0$, $y + z=1+1 = 1$, $\overline{y + z}=0$, then $(\overline{x + z})(\overline{y + z})=0\times0 = 0$.

Step4: Substitute values into fourth option

For $\overline{z}(x + y)$, $\overline{z}=0$, $x + y=0 + 1=1$, so $\overline{z}(x + y)=0\times1 = 0$.

Step1: Simplify first option

$(\overline{x}+y)\overline{y}=\overline{x}\overline{y}+y\overline{y}=\overline{x}\overline{y}+0=\overline{x}\overline{y}
eq\overline{x}$.

Step2: Simplify second option

$(x+\overline{x})(y + x)=1\times(y + x)=y + x
eq\overline{x}$.

Step3: Simplify third option

Using the distributive law, $\overline{x}\cdot\overline{y}+\overline{x}=\overline{x}(\overline{y}+1)=\overline{x}\times1=\overline{x}$.

Step4: Simplify fourth option

$\overline{x}\cdot y+x
eq\overline{x}$.

Step1: Simplify first option

$x+\overline{y}+xy=x(1 + y)+\overline{y}=x+\overline{y}
eq x + y$.

Step2: Simplify second option

$xy + y+\overline{x}\cdot\overline{y}=y(x + 1)+\overline{x}\cdot\overline{y}=y+\overline{x}\cdot\overline{y}
eq x + y$.

Step3: Simplify third option

Using the distributive law, $(x+\overline{y})(x + y)=x\cdot x+x\cdot y+\overline{y}\cdot x+\overline{y}\cdot y=x+xy+\overline{y}x+0=x(1 + y+\overline{y})=x\times1=x + y$.

Step4: Simplify fourth option

$y+x\overline{y}=y(1 + x)+\overline{y}x=y+\overline{y}x
eq x + y$.

Answer:

$\overline{z + x}+y\overline{x}$

2.