1. **State the problem:** Solve the compound inequality $$d-5 < \frac{2d}{5} \leq \frac{d}{2} + \frac{1}{5}$$. 2. **Break the compound inequality into two parts:** Part 1: $$d - 5 < \frac{2d}{5}$$ Part 2: $$\frac{2d}{5} \leq \frac{d}{2} + \frac{1}{5}$$ 3. **Solve Part 1:** Start with $$d - 5 < \frac{2d}{5}$$. Multiply both sides by 5 to clear the denominator: $$5(d - 5) < 5 \times \frac{2d}{5}$$ $$5d - 25 < 2d$$ Subtract $$2d$$ from both sides: $$5d - 2d - 25 < 0$$ $$3d - 25 < 0$$ Add 25 to both sides: $$3d < 25$$ Divide both sides by 3: $$\frac{\cancel{3}d}{\cancel{3}} < \frac{25}{3}$$ $$d < \frac{25}{3}$$ 4. **Solve Part 2:** Start with $$\frac{2d}{5} \leq \frac{d}{2} + \frac{1}{5}$$. Multiply both sides by 10 (the least common multiple of 5 and 2) to clear denominators: $$10 \times \frac{2d}{5} \leq 10 \times \left(\frac{d}{2} + \frac{1}{5}\right)$$ $$4d \leq 5d + 2$$ Subtract $$5d$$ from both sides: $$4d - 5d \leq 2$$ $$-d \leq 2$$ Multiply both sides by -1 and reverse the inequality sign: $$d \geq -2$$ 5. **Combine the two parts:** $$-2 \leq d < \frac{25}{3}$$ This is the solution to the compound inequality. **Final answer:** $$\boxed{-2 \leq d < \frac{25}{3}}$$