1. **State the problem:** Simplify the algebraic expression and solve the given fractions and polynomial expressions. 2. **Given expressions:** $$\frac{k^2 - 9k}{2} + \frac{10}{1}$$ $$\frac{\frac{k}{8} (k+5)}{16}$$ $$\frac{k^2 - 9k + 20}{16} \div (k-5)$$ 3. **Step 1: Simplify the first expression** $$\frac{k^2 - 9k}{2} + 10 = \frac{k^2 - 9k}{2} + \frac{20}{2} = \frac{k^2 - 9k + 20}{2}$$ 4. **Step 2: Simplify the second expression** $$\frac{\frac{k}{8} (k+5)}{16} = \frac{k(k+5)}{8 \times 16} = \frac{k(k+5)}{128}$$ 5. **Step 3: Simplify the third expression** $$\frac{k^2 - 9k + 20}{16} \div (k-5) = \frac{k^2 - 9k + 20}{16} \times \frac{1}{k-5} = \frac{k^2 - 9k + 20}{16(k-5)}$$ 6. **Step 4: Factor the numerator in the third expression** $$k^2 - 9k + 20 = (k-5)(k-4)$$ 7. **Step 5: Cancel common factors** $$\frac{\cancel{(k-5)}(k-4)}{16\cancel{(k-5)}} = \frac{k-4}{16}$$ **Final answers:** - First expression simplified: $$\frac{k^2 - 9k + 20}{2}$$ - Second expression simplified: $$\frac{k(k+5)}{128}$$ - Third expression simplified: $$\frac{k-4}{16}$$