Question

Easy Question??

Answer 1

if it's easy, makes one wonder how come you haven't done it

Answer 2

then explain it to me please? you're not helping...

Answer 3

-1/2 i think. it is definitely one of the negative ones

Answer 4

Can you explain it to me?

Answer 5

so i think the last one is the right answer 64

Answer 6

ohhh I see, so it's NOT easy then... as you see it, well then

one sec

Answer 7

first off, things to keep in mind

\(\bf a^{-\frac{{\color{blue} n}}{{\color{red} m}}} =
\cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}} \implies \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}\qquad\qquad

% radical denominator
\cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}}\)

so.. let's see yours... one sec

Answer 8

hmmm actually, I meant to say
\(\bf a^{-{\color{red} n}} \implies \cfrac{1}{a^{\color{red} n}}\qquad \qquad

\cfrac{1}{a^{\color{red} n}}\implies a^{-{\color{red} n}}
\\ \quad \\
% negative exponential denominator
a^{{\color{red} n}} \implies \cfrac{1}{a^{-\color{red} n}}\)

anyhow, one sec, let's see yours

Answer 9

\(\bf \left( \cfrac{jk^{-2}}{j^{-1}k^{-3}} \right)^3
\\ \quad \\
\left( j^1\cdot k^{-2}\cdot \cfrac{1}{j^{-1}}\cdot \cfrac{1}{k^{-3}}\right)^3 \implies \left( j^1 \cdot \cfrac{1}{\cancel{k^2}}\cdot j^1\cdot \cancel{k^3}\right)^3
\\ \quad \\
\left( j^1j^1k^1 \right)^3\implies (j^2k)^3\)

pretty sure you can take it from there