Question

Another problem see attached file

Answer 1

Have you tried it??

Answer 2

Simplify?

Answer 3

have you tried it? hint: the -1 indicates you reciprocate

Answer 4

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

Answer 5

alright first you need to know that

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

the property is that if you have a negative exponent it can be written by flipping it and changing making positive exponent

Answer 6

\[(3a ^{2}x/2)(4a ^{2}/x ^{3})\] do you see how I arrived at that expression?

Answer 7

my answer

Answer 8

@waterineyes I attach my answer

Answer 9

No. you cannot cancel a^2..

See here I am giving you the formula:

\[\large \frac{a^m}{a^n} = a^{m-n}\]

Use this concept..

Answer 10

so i need to try and solve again

Answer 11

yes..

Answer 12

I let you know @waterineyes @Outkast3r09 in a minuete

Answer 13

How can you solve this:

\[\large \frac{a^2}{a^{-2}} = ??\]

Use that formula here..

Answer 14

Thanks guys

Answer 15

the 4 shouldn't change

Answer 16

\[4a^{-2}\neq (4a)^{-2}\]

Answer 17

@gitahimart

Answer 18

Is @gitahimart right?