jmc

Since the square of $x-4$ is at most 9, the value of $x-4$ must be between $-3$ and $3$ (or equal to either). So, we have $-3\le x-4\le 3$. Thus, $1\le x\le 7$. Therefore, our answer is $\boxed{6}$.

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If $(x-4{)}^2\le 9$, then $x$ can be no more than 3 away from 4. Therefore, the values of $x$ from 1 to 7 satisfy the inequality, and our answer is $\boxed{6}$.