1. **State the problem:** Solve the compound inequality $$9f - 2 > 16$$ or $$\frac{f + 20}{2} \leq 10$$. 2. **Solve the first inequality:** $$9f - 2 > 16$$ Add 2 to both sides: $$9f - \cancel{2} + \cancel{2} > 16 + 2$$ $$9f > 18$$ Divide both sides by 9: $$\frac{9f}{\cancel{9}} > \frac{18}{\cancel{9}}$$ $$f > 2$$ 3. **Solve the second inequality:** $$\frac{f + 20}{2} \leq 10$$ Multiply both sides by 2: $$\cancel{2} \times \frac{f + 20}{\cancel{2}} \leq 10 \times 2$$ $$f + 20 \leq 20$$ Subtract 20 from both sides: $$f + \cancel{20} - \cancel{20} \leq 20 - 20$$ $$f \leq 0$$ 4. **Combine the solutions:** The compound inequality uses "or," so the solution is all values of $f$ such that $$f > 2 \quad \text{or} \quad f \leq 0$$. 5. **Interpretation:** This means $f$ can be any number greater than 2 or any number less than or equal to 0. **Final answer:** $$f > 2 \quad \text{or} \quad f \leq 0$$