Question

Help!!

Answer 1

@freckles

Answer 2

I would first apply the law:
\[(b^{q}x^{r}y^{s})^n=b^{q \cdot n}x^{r \cdot n} y^{s \cdot n }\]

Answer 3

we see we can apply this law in both numerator and denominator

Answer 4

I did. my answer was 1/x^3

Answer 5

actually that is kinda close

Answer 6

let's look at just the x part
\[\frac{(x^2)^3}{(x)^{-9}}\]

Answer 7

what do you get on top?
and on bottom?

Answer 8

x^6/x^-9

Answer 9

ok and what can you do with negative exponents

Answer 10

like the law is that you can do this:
\[\frac{1}{x^{-n}}=x^{n} \text{ right ? }\]

Answer 11

And thats how i got x^3.

Answer 12

or you can even apply quotient rule

your answer still is off

Answer 13

Oh, so is it x^15?

Answer 14

yes! :)

Answer 15

\[x^{6} x^{9} \text{ since we brought that factor \to the \top }\]

Answer 16

and 6+9 as you say is 15
so you have x^(15)

Answer 17

Y is canceled out right?

Answer 18

o\[\frac{x^6}{x^{-9}} =x^{6-(-9)}=x^{6+9}\]r quotient rule

Answer 19

we can also look at that part if you wish
\[\frac{(y^{-3})^3}{(y)^{-9}}\]

Answer 20

we know that first law I mentioned we can apply to the top
which means the top is going to be ?

Answer 21

y^-9

Answer 22

\[\frac{(y^{-3})^3}{(y)^{-9}}=\frac{y^{-9}}{y^{-9}}=1 \]
definitely cancels in your fraction then

Answer 23

so does the constant part
\[\frac{(3)^3}{27}=\frac{27}{27}=1\]

Answer 24

Thanks! You just got a new fan! So btw, all i have to enter is x^15 right. or should i enter 1/x^15

Answer 25

definitely x^(15)
recall that we had
\[\frac{(x^2)^3}{x^{-9}}=\frac{x^6}{x^{-9}}=x^{6-(-9)}=x^{6+9}=x^{15}\]

Answer 26

Alright great. :) Thanks a bunch!

Answer 27

np