Question

Algebra II Help please! Posting question shortly

Answer 1

@phi

Answer 2

I just need a refresher haha I haven't done this in awhile

Answer 3

@jim_thompson5910 @nincompoop

Answer 4

the first step is use the rule
\[ (a^b)^c = a^{bc} \]
on the bottom
\[( yx^{-3})^4 \]

Answer 5

okay... So would it be to the -12?

Answer 6

on the x, also y^4

the bottom becomes \( y^4 x^{-12} \)

Answer 7

now for each base (the x's and the y's), add their exponents up top
what do you get for the top ?

Answer 8

-2?

Answer 9

you should get x to some exponent time y to an exponent.
what is
\[ x^{-3}y^0\cdot 2 x^4 y^{-3} \]
?
this is the same as
\[ 2 x^{-3}x^4 y^0y^{-3} \]
now add the exponents for the x's. what do you get?

Answer 10

x^1 and y^-3?

Answer 11

@jim_thompson5910 can you help me please??

Answer 12

so you have

\[\large \frac{2x^{1}y^{-3}}{x^{-12}y^{4}}\]

simplify that now

Answer 13

they have to be positive exponents... Doesn't that add up to a negative?

Answer 14

when you have a fraction, it is the exponent of the top minus the exponent in the bottom
so the x's exponent is 1 - (-12) = 1+12 =13
you can do y?

Answer 15

-7?

Answer 16

so you'll get this

\[\large \frac{2x^{1}y^{-3}}{x^{-12}y^{4}}\]

\[\large \frac{2x^{13}y^{-7}}{1}\]

\[\large \frac{2x^{13}}{1*y^7}\]

\[\large \frac{2x^{13}}{y^7}\]

Answer 17

how did you get from the second picture to the third?

Answer 18

by using the idea that

\[\large x^{-k} = \frac{1}{x^k}\]

or the idea that

\[\large \left(\frac{a}{b}\right)^{-k} = \left(\frac{b}{a}\right)^{k}\]