In the expression $f(x)=|x-k|+1$, the minimum value occurs when $x=k$, because $|x-k|$ is smallest when it equals $0$. From the table, the smallest value of $f(x)$ is $1$, and that happens at $x=0$. Therefore, $k=0$.

If $|x-4|=3$, then the distance between $x$ and $4$ is $3$. So $x$ can be $4+3=7$ or $4-3=1$. Therefore, the statement that must be true is that $x$ is either $1$ or $7$.

Distance on a number line is represented with absolute value. If point $P$ is at location $x$, then its distance from $7$ is $|x-7|$. Since that distance is $4$ centimeters, the equation is $|x-7|=4$.

An absolute value equation of the form $|2x-5|=9$ means the expression inside can be either $9$ or $-9$. So solve

From $2x-5=9$, $2x=14$, so $x=7$. From $2x-5=-9$, $2x=-4$, so $x=-2$. The greater of the two possible values is $7$.

The absolute value of a real expression is equal to the principal square root of its square. So

This is true for every real value of $x$, so $\sqrt{(x-4{)}^2}$ is the equivalent expression.