1. The problem asks to find the position of $\sqrt{71}$ on the number line. 2. Recall that $\sqrt{71}$ is the positive number which when squared gives 71. 3. We know $8^2 = 64$ and $9^2 = 81$, so $\sqrt{71}$ lies between 8 and 9. 4. To estimate $\sqrt{71}$, use the formula for approximation: $$\sqrt{a} \approx b + \frac{a - b^2}{2b}$$ where $b$ is a number close to $\sqrt{a}$. 5. Here, let $b=8$ since $8^2=64$ is close to 71. 6. Calculate: $$8 + \frac{71 - 64}{2 \times 8} = 8 + \frac{7}{16} = 8 + 0.4375 = 8.4375$$ 7. This means $\sqrt{71} \approx 8.44$, which is between 8 and 9, closer to 8.5. 8. On the number line, point C is between 8 and 9, slightly left of 9, matching the approximate value. **Final answer:** Point C shows the position of $\sqrt{71}$ on the number line.