1. Problem b: Simplify $5^{\frac{1}{4}} \cdot 5^{\frac{11}{4}}$. 2. Use the rule of exponents: $a^m \cdot a^n = a^{m+n}$. 3. Add the exponents: $\frac{1}{4} + \frac{11}{4} = \frac{12}{4} = 3$. 4. So, $5^{\frac{1}{4}} \cdot 5^{\frac{11}{4}} = 5^3 = 125$. --- 1. Problem c: Simplify $\sqrt{\frac{64}{9}}$. 2. Use the property: $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$. 3. Calculate $\sqrt{64} = 8$ and $\sqrt{9} = 3$. 4. So, $\sqrt{\frac{64}{9}} = \frac{8}{3}$. --- 1. Problem d: Simplify $\sqrt[5]{64x^5y^{11}}$. 2. Rewrite inside the fifth root: $64 = 2^6$, so $64x^5y^{11} = 2^6 x^5 y^{11}$. 3. Use the property: $\sqrt[5]{a^m} = a^{\frac{m}{5}}$. 4. Apply to each term: $$\sqrt[5]{2^6} = 2^{\frac{6}{5}}, \quad \sqrt[5]{x^5} = x^{\frac{5}{5}} = x, \quad \sqrt[5]{y^{11}} = y^{\frac{11}{5}}$$ 5. Combine: $2^{\frac{6}{5}} x y^{\frac{11}{5}}$. --- 1. Problem 5a: Rationalize and simplify $\frac{5}{\sqrt{7}}$. 2. Multiply numerator and denominator by $\sqrt{7}$: $$\frac{5}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{5\sqrt{7}}{\cancel{\sqrt{7}}\cancel{\sqrt{7}}} = \frac{5\sqrt{7}}{7}$$ --- 1. Problem 6: Find the distance between points $(8, -3)$ and $(3, 1)$. 2. Use the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. Substitute values: $$d = \sqrt{(3 - 8)^2 + (1 - (-3))^2} = \sqrt{(-5)^2 + (4)^2} = \sqrt{25 + 16} = \sqrt{41}$$ Final answers: b. $125$ c. $\frac{8}{3}$ d. $2^{\frac{6}{5}} x y^{\frac{11}{5}}$ 5a. $\frac{5\sqrt{7}}{7}$ 6. $\sqrt{41}$