1. **State the problem:** We are given three points on a number line: R, M, and S, with M = $\sqrt{270}$. We need to find the values of R and S to one decimal place, knowing that R < M < S. 2. **Calculate $\sqrt{270}$:** $$\sqrt{270} = \sqrt{9 \times 30} = \sqrt{9} \times \sqrt{30} = 3 \times \sqrt{30}$$ 3. **Approximate $\sqrt{30}$:** Since $5^2 = 25$ and $6^2 = 36$, $\sqrt{30}$ is between 5 and 6. Using a calculator or estimation: $$\sqrt{30} \approx 5.4772$$ 4. **Calculate $\sqrt{270}$ numerically:** $$\sqrt{270} = 3 \times 5.4772 = 16.4316$$ Rounded to one decimal place: $$M = 16.4$$ 5. **Determine R and S:** Since M is between R and S on the number line, and M = 16.4, R must be less than 16.4 and S must be greater than 16.4. The problem does not specify exact values for R and S, but typically, R and S are chosen as the nearest integers or decimal values around M. So, to one decimal place: $$R = 16.3$$ $$S = 16.5$$ **Final answer:** $$R = 16.3, \quad S = 16.5$$