jmc

We have $10^2=100$, so $10$ is the smallest positive integer which satisfies the inequalities. From here, we can compute the next few perfect squares: $$\begin{array}{rl}11^2& =121,\\ 12^2& =144,\\ 13^2& =169,\\ 14^2& =196,\\ 15^2& =225.\end{array}$$ The last $x$ for which $x^2\le 200$ is $x=14$. In all, our solutions in positive integers are $$x=10,11,12,13,14,$$ so there are $\boxed{5}$ such $x$.