Question

why is (1/3)^-2 equals to 9? but (1/3)^2 gives you this decimal answer?

Answer 1

When you take (1/3)^-2 you get 9 because (1/3)^-2 is the same thing (or can be written as) 3^2 which is 9.
(1/3)^2 is a decimal answer because that's the same thing as multiplying 1/3 by 1/3. Which would give you 1/9 or .111111...

If you need a further explanation please say so!
Otherwise, I hope this helps!

Answer 2

thankyou that explained a lot

Answer 3

\(\bf \large {a^{-\frac{n}{m}} = \cfrac{1}{a^{\frac{n}{m}}}\\ \quad \\
a^{-\frac{n}{m}} = \cfrac{1}{a^{\frac{n}{m}}} \implies \cfrac{1}{\sqrt[m]{a^n}}}\qquad thus\\ \quad \\
\left(\frac{1}{3}\right)^{-2}\implies \cfrac{1}{\left(\frac{1}{3}\right)^{2}}\implies \cfrac{1}{\frac{1}{9}}\implies \cfrac{1}{1}\cdot \cfrac{9}{1}\implies 9\)

Answer 4

\(\bf \left(\cfrac{1}{3}\right)^{2}\implies \cfrac{1^2}{3^2}\implies \cfrac{1}{9}\)

Answer 5

well... didn't need the radicals there.... so what I meant was \(\bf \large a^{-n} = \cfrac{1}{a^n}\)