1. Stating the problem: Calculate $$\left( \frac{3}{4^{-1}} + \frac{5}{3^{-2}} \right) \cdot 19^{-1}$$ and determine which option (A, B, C, or D) matches the result. 2. Recall the rule for negative exponents: $$a^{-n} = \frac{1}{a^n}$$. 3. Simplify each term inside the parentheses: $$\frac{3}{4^{-1}} = 3 \times 4^{1} = 3 \times 4 = 12$$ $$\frac{5}{3^{-2}} = 5 \times 3^{2} = 5 \times 9 = 45$$ 4. Add the two results: $$12 + 45 = 57$$ 5. Simplify the multiplication by $$19^{-1}$$: $$57 \cdot 19^{-1} = 57 \times \frac{1}{19} = \frac{57}{19}$$ 6. Simplify the fraction $$\frac{57}{19}$$: Since $$19 \times 3 = 57$$, we have $$\frac{57}{19} = 3$$ 7. Final answer: $$3$$, which corresponds to option B.