Question: (top-left) $\sqrt{49} + \sqrt{\sqrt{81}}$
(top-left) $\frac{\text{above}}{\sqrt{4}}$ (center)
1. **State the problem:** Simplify the expression $$\frac{\sqrt{49} + \sqrt{\sqrt{81}}}{\sqrt{4}}$$.
2. **Recall important formulas and rules:**
- The square root of a number $a$ is written as $\sqrt{a}$.
- The square root of a square root can be written as $\sqrt{\sqrt{a}} = a^{1/4}$.
- Simplify numerator and denominator separately before dividing.
3. **Simplify each term:**
- $\sqrt{49} = 7$
- $\sqrt{81} = 9$, so $\sqrt{\sqrt{81}} = \sqrt{9} = 3$
- $\sqrt{4} = 2$
4. **Substitute back:**
$$\frac{7 + 3}{2}$$
5. **Add numerator:**
$$\frac{10}{2}$$
6. **Simplify fraction:**
$$\frac{\cancel{10}^{5}}{\cancel{2}^{1}} = 5$$
**Final answer:** $5$