\(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\)
= \(\frac{3^{-1}}{3^2}\) = \(\frac{1}{27}\)
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 \(\frac{1}{27}\).
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