1. Stating the problem: Evaluate the expression $$6^0 + 2^{-3} \cdot \left(\frac{1}{4}\right)^{-1}$$. 2. Recall the rules: - Any number raised to the power 0 is 1, so $$6^0 = 1$$. - Negative exponents mean reciprocal, so $$a^{-n} = \frac{1}{a^n}$$. 3. Calculate each part: - $$6^0 = 1$$. - $$2^{-3} = \frac{1}{2^3} = \frac{1}{8}$$. - $$\left(\frac{1}{4}\right)^{-1} = 4$$ because raising to -1 flips the fraction. 4. Substitute back: $$1 + \frac{1}{8} \cdot 4$$ 5. Multiply: $$\frac{1}{8} \cdot 4 = \frac{4}{8} = \frac{1}{2}$$ 6. Add: $$1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2}$$ Final answer: $$\frac{3}{2}$$ or 1.5.