plz help

- anonymous

plz help

- chestercat

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- anonymous

hi

- anonymous

whats the question

- anonymous

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

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## More answers

- anonymous

write each expression so that it contains only positive exponents

- anonymous

I dunno this 1 im so sorry

- anonymous

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

- anonymous

how did you do that

- anonymous

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

- anonymous

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

- anonymous

how did you get b

- anonymous

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

- anonymous

oh ok

- anonymous

satellite are you there:(

- anonymous

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

- anonymous

idk i just dont know how to do it

- anonymous

what happened to the -1

- anonymous

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

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

no

- anonymous

i thought x-1=1/x1

- anonymous

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

- anonymous

ok

- anonymous

but isnt negative 1

- anonymous

One thing first. Whenever you want to say that x has an exponential value of -1, you should type x^-1 or x to the power of -1 rather than x-1 because x-1 is just subtraction.

- anonymous

oh ok sorry

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

oh ok

- anonymous

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

- anonymous

yeah i know

- anonymous

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

- anonymous

uh \[1\over x\]

- anonymous

i think

- anonymous

am i right

- anonymous

No. General rule of thumb. If you see a negative number as an exponent, flip the term(with the exponent) over the fraction bar and make the exponent positive

- anonymous

?

- anonymous

so I flip x^-1 over to the numerator and get \[x^{-1}\]
then I take away the negative sign which gets me \[x^{1}\] or \[x\]

- anonymous

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

- anonymous

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

- anonymous

so I fliped x^21 over to the numerator and get
xâˆ’2
then I take away the negative sign which gets me
x2

- anonymous

-2

- anonymous

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

- anonymous

\[1\over x^2\]

- anonymous

Are you sure?

- anonymous

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

- anonymous

Might be because the number was so small.

- anonymous

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

- anonymous

\[1 \over x^3\]

- anonymous

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

- anonymous

idk 4/x^2

- anonymous

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

- anonymous

so it would be 4x^2

- anonymous

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

- anonymous

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

- anonymous

im still thinking

- anonymous

3x/4

- anonymous

i got it wrong

- anonymous

Then change it :P

- anonymous

change what

- anonymous

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

- anonymous

1/x^2

- anonymous

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

- anonymous

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

- anonymous

but 3 is a numerator too it can turn to 1

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

But you agree with it, right?

- anonymous

yes

- anonymous

i should've looked at it as a whole

- anonymous

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

- anonymous

yes

- anonymous

so if you substitute the (3x)^-1 with \[\frac{ 1 }{ 3x}\] in the expression, you get \[\frac{ 1 }{ 4(3x) }\]

- anonymous

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

- anonymous

But if you think about it, all I actually did was flip the (3x)^-1 over to the denominator and take away the negative sign|dw:1363833108473:dw|

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

you know what your right its really easy well thank you very much i truly get thank you very much you should be a teacher and again thank you very much 100x:)

- anonymous

lolwut. Just understand it. and REMEMBER IT.

- anonymous

yes sir

- anonymous

lol

- anonymous

hahahadont make me take out my shoe

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