Chapter R - Section R.2 - Algebra Essentials - R.2 Assess Your Understanding - Page 29: 166

The number is 3.15

Let $x$ be the number, we are looking for. we know the following: $0 \leq x^2\leq10$ $x>\pi$ And the number can only be 2 decimal places long If we know $x^2 \leq 10$ then $x \leq \sqrt{10} \approx 3.16$ We know that $\pi=3.14$ so the only 2 decimal number between these two numbers is $3.15$, therefore the number must be $3.15$.