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hi

whats the question

\[(3x)^-1 \over 4\]

write each expression so that it contains only positive exponents

I dunno this 1 im so sorry

\[\frac{(3x)^{-1}}{ 4}=\frac{1}{4(3x)}\]

how did you do that

\(b^{-1}=\frac{1}{b}\)

in general,
\[b^{-n}=\frac{1}{b^n}\]

how did you get b

i picked \(b\) as a variable
you can use anyone you prefer

oh ok

satellite are you there:(

I don't know what else there is to say. What don't you understand?

idk i just dont know how to do it

what happened to the -1

This is what your problem looks like, right?
\[\frac{ (3x)^{-1} }{ 4 }\]

This is what your problem looks like, right?
\[\frac{ (3x)^{-1} }{ 4 }\]

yes

Did you know that \[x^{-1}=\frac{ 1 }{ x }\]?

no

i thought x-1=1/x1

Same thing. \[x^1=x\] just like \[3^{1}=3\]

ok

but isnt negative 1

oh ok sorry

Is this what you're saying?\[x^{-1}=\frac{ 1 }{ x^{1} }\]

yes

Then you are right. But, as I said, \[x^{1}=x\]
So you can just change the \[x^{1}\] to \[x\]

oh ok

This means that \[x^{âˆ’1}=\frac{ 1 }{ x }\]

yeah i know

Rewrite this so that there are only positive exponents:\[\frac{ 1 }{ x^{-1} }\]

uh \[1\over x\]

i think

am i right

so every time there an equation like this \[1 \over x ^-1\] you have to flip

Yes. But ONLY if the exponent is negative. Next problem:\[\frac{ 1 }{ x^{-2} }\]

-2

x^2 is right. Next one:\[x ^{-3}\]

\[1\over x^2\]

Are you sure?

I think it's a silly mistake, but it was x^3, not x^2

Might be because the number was so small.

oh yeah it was a little as i looked closer it was a 3

\[1 \over x^3\]

Next one(and the important one to help you with your problem):\[\frac{ 4 }{ x^{-2} }\]

idk 4/x^2

Why don't you bring x^-2 to the numerator and take away the negative sign?

so it would be 4x^2

Exactly. Now can you do this?\[\frac{ (3x)^{-1} }{ 4 }\]

If you really don't understand, no is a good answer too.

im still thinking

3x/4

i got it wrong

Then change it :P

change what

remember what we did with \[x^{-2}\]

1/x^2

We flipped it into the denominator and took away the negative sign to get \[\frac{ 1 }{ x^{2} }\]

Can you apply this to your problem or should I just explain?

but 3 is a numerator too it can turn to 1

yeah you should explain i get everything you taught me i just can put it all together

Do you agree that \[(3x)^{-1}=\frac{ 1 }{ (3x)^{1} }\]or \[\frac{ 1 }{ (3x) }\]

\[1\over (3x)^1\]

Errr I mean that it equals both but the lower one without the exponent of 1 is simplified

But you agree with it, right?

yes

i should've looked at it as a whole

Now do you agree that \[(3x)^{âˆ’1}\times \frac{ 1 }{ 4 }=\frac{ (3x)^{âˆ’1} }{ 4 }\]

yes

and that simplifies to\[\frac{ 1 }{ 12x }\]

is it the same thing as multiplying 1/3x and 1/4

Yes, but the point is that I just flipped it over and that it's incredibly simple.

Everyone else does it like:|dw:1363833370136:dw|

lolwut. Just understand it. and REMEMBER IT.

yes sir

lol

hahahadont make me take out my shoe