Question

ok who can help me with this problem ill post the link

Answer 1

\[\frac{(-6)^{-4}+(-6)^{-2}}{-6^{-3}}\]

Answer 2

First of all I would write everything without a negative exponent.

Answer 3

\[a^{-n}=\frac{1}{a^{n}}\]

Answer 4

That is one way to begin the problem anyways.

Answer 5

ok

Answer 6

Write here what you get after doing that ?

Answer 7

I would then get rid of the compound fractions after that.

Answer 8

ok wouldn't it be -6^3/-6^4+-6^2 or did i do that completely wrong

Answer 9

\[\frac{1}{-6^{-3}} ((-6)^{-4}+(-6)^{-2}) \\ -6^{3}(\frac{1}{(-6)^4}+\frac{1}{(-6)^2})\]

Answer 10

now what is 6^3? (-6)^4? (-6)^2?

Answer 11

or actually we could reduce before multiplying

Answer 12

distribute the -6^3

Answer 13

a(b+c)=ab+ac

Answer 14

apply distribute property is what I'm saying

Answer 15

yep understand

Answer 16

ok what would i get for a answer I'm getting huge numbers

Answer 17

and also (-1)^even=1
so you could write (-6)^4 as 6^4 and (-6)^2 as 6^2

Answer 18

have you distribute the -6^3

Answer 19

yes

Answer 20

write here what you have so I know

Answer 21

well I'm getting -216/279936+-216/7776

Answer 22

well I wound't multiply all of that all quiet yet

Answer 23

so am i able to cancel things before multiplying this is where I'm getting lost

Answer 24

\[\frac{1}{-6^{-3}} ((-6)^{-4}+(-6)^{-2}) \\ -6^{3}(\frac{1}{(-6)^4}+\frac{1}{(-6)^2}) \\ \frac{-6^3}{6^4}+\frac{-6^3}{6^2}\]
because you can cancel some things :)

Answer 25

use the law of exponent rule that we have x^m/x^n=x^{m-n}

Answer 26

ok so then i would get -6/6+-6/6

Answer 27

not exactly

Answer 28

\[\frac{-6^3}{6^4}=-6^{3-4}=-6^{-1}=-\frac{1}{6}\]
you try the other addend

Answer 29

\[\frac{-6^3}{6^2}=?\]

Answer 30

so that would be -6^1

Answer 31

Thanks for your help i found the answer