\(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) = \(\frac{3^{-1}}{3^2}\) = \(\frac{1}{27}\) In full: To simplify the expression : \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) We start by rewriting the bases in terms of powers of 3. We know that: \(9 = 3^2 \quad \text{and} \quad 27 = 3^3\) Thus, we can express \(9^{-\frac{1}{2}}\) as follows: \(9^{-\frac{1}{2}} = (3^2)^{-\frac{1}{2}} = 3^{-1}\) Next, we rewrite \(27^{\frac{2}{3}}\): \(27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^{2}\) Substitute back into the expression: Now we substitute these values back into the original expression: \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}} = \frac{3^{-1}}{3^{2}}\) This simplifies to: \(\frac{1/3}{9} = \frac{1}{3} \cdot \frac{1}{9} = \frac{1}{27}\) Thus, the simplified form of \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) is \(\frac{1}{27}\).