Answer :
Answer:
9^(-1/2) = 1/3
27^(-2/3) = 1/9
Step-by-step explanation:
We know that:
[tex] \displaystyle \large {{a}^{ - n} = \frac{1}{ {a}^{n} } } \\ \displaystyle \large { {a}^{ \frac{m}{n} } = \sqrt[n]{ {a}^{m} } }[/tex]
If we add negative sign in m/n.
[tex] \displaystyle \large{ {a}^{ - \frac{m}{n} } = \frac{1}{ {a}^{ \frac{m}{n} } } }[/tex]
Therefore,
[tex] \displaystyle \large{ {a}^{ - \frac{m}{n} } = \frac{1}{ \sqrt[n]{ {a}^{m} } } }[/tex]
From the expression:
1. 9^(-1/2)
[tex] \displaystyle \large{ {9}^{ - \frac{1}{2} } = \frac{1}{ \sqrt{9} } }[/tex]
From above, apply the exponent rules. Simplify the expression:
[tex] \displaystyle \large{ {9}^{ - \frac{1}{2} } = \frac{1}{ 3 }}[/tex]
2. 27^(-2/3)
[tex] \displaystyle \large{ {27}^{ - \frac{2}{3} } = \frac{1}{ \sqrt[3]{ {27}^{2} } } }[/tex]
27 can be factored as 3^3.
[tex] \displaystyle \large{ {27}^{ - \frac{2}{3} } = \frac{1}{ \sqrt[3]{ { ({3}^{3}) }^{2} } } }[/tex]
From above:- Use the following exponent rules to multiply the exponents and leave the base as 3 so we can cancel the cube (3 out of the root) and the exponent.
[tex] \displaystyle \large{ {27}^{ - \frac{2}{3} } = \frac{1}{ \sqrt[3]{ {3}^{6} } }}[/tex]
Divide 6 by 3 or the cube root.
[tex] \displaystyle \large{ {27}^{ - \frac{2}{3} } = \frac{1}{ {3}^{2} }}[/tex]
Therefore:
[tex] \displaystyle \large{ {27}^{ - \frac{2}{3} } = \frac{1}{ 9}}[/tex]