anonymous
  • anonymous
plz help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
hi
anonymous
  • anonymous
whats the question
anonymous
  • anonymous
\[(3x)^-1 \over 4\]

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anonymous
  • anonymous
write each expression so that it contains only positive exponents
anonymous
  • anonymous
I dunno this 1 im so sorry
anonymous
  • anonymous
\[\frac{(3x)^{-1}}{ 4}=\frac{1}{4(3x)}\]
anonymous
  • anonymous
how did you do that
anonymous
  • anonymous
\(b^{-1}=\frac{1}{b}\)
anonymous
  • anonymous
in general, \[b^{-n}=\frac{1}{b^n}\]
anonymous
  • anonymous
how did you get b
anonymous
  • anonymous
i picked \(b\) as a variable you can use anyone you prefer
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
satellite are you there:(
anonymous
  • anonymous
I don't know what else there is to say. What don't you understand?
anonymous
  • anonymous
idk i just dont know how to do it
anonymous
  • anonymous
what happened to the -1
anonymous
  • anonymous
This is what your problem looks like, right? \[\frac{ (3x)^{-1} }{ 4 }\]
anonymous
  • anonymous
This is what your problem looks like, right? \[\frac{ (3x)^{-1} }{ 4 }\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
Did you know that \[x^{-1}=\frac{ 1 }{ x }\]?
anonymous
  • anonymous
no
anonymous
  • anonymous
i thought x-1=1/x1
anonymous
  • anonymous
Same thing. \[x^1=x\] just like \[3^{1}=3\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
but isnt negative 1
anonymous
  • 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
  • anonymous
oh ok sorry
anonymous
  • anonymous
Is this what you're saying?\[x^{-1}=\frac{ 1 }{ x^{1} }\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
Then you are right. But, as I said, \[x^{1}=x\] So you can just change the \[x^{1}\] to \[x\]
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
This means that \[x^{−1}=\frac{ 1 }{ x }\]
anonymous
  • anonymous
yeah i know
anonymous
  • anonymous
Rewrite this so that there are only positive exponents:\[\frac{ 1 }{ x^{-1} }\]
anonymous
  • anonymous
uh \[1\over x\]
anonymous
  • anonymous
i think
anonymous
  • anonymous
am i right
anonymous
  • 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
?
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
  • anonymous
so every time there an equation like this \[1 \over x ^-1\] you have to flip
anonymous
  • anonymous
Yes. But ONLY if the exponent is negative. Next problem:\[\frac{ 1 }{ x^{-2} }\]
anonymous
  • anonymous
so I fliped x^21 over to the numerator and get x−2 then I take away the negative sign which gets me x2
anonymous
  • anonymous
-2
anonymous
  • anonymous
x^2 is right. Next one:\[x ^{-3}\]
anonymous
  • anonymous
\[1\over x^2\]
anonymous
  • anonymous
Are you sure?
anonymous
  • anonymous
I think it's a silly mistake, but it was x^3, not x^2
anonymous
  • anonymous
Might be because the number was so small.
anonymous
  • anonymous
oh yeah it was a little as i looked closer it was a 3
anonymous
  • anonymous
\[1 \over x^3\]
anonymous
  • anonymous
Next one(and the important one to help you with your problem):\[\frac{ 4 }{ x^{-2} }\]
anonymous
  • anonymous
idk 4/x^2
anonymous
  • anonymous
Why don't you bring x^-2 to the numerator and take away the negative sign?
anonymous
  • anonymous
so it would be 4x^2
anonymous
  • anonymous
Exactly. Now can you do this?\[\frac{ (3x)^{-1} }{ 4 }\]
anonymous
  • anonymous
If you really don't understand, no is a good answer too.
anonymous
  • anonymous
im still thinking
anonymous
  • anonymous
3x/4
anonymous
  • anonymous
i got it wrong
anonymous
  • anonymous
Then change it :P
anonymous
  • anonymous
change what
anonymous
  • anonymous
remember what we did with \[x^{-2}\]
anonymous
  • anonymous
1/x^2
anonymous
  • anonymous
We flipped it into the denominator and took away the negative sign to get \[\frac{ 1 }{ x^{2} }\]
anonymous
  • anonymous
Can you apply this to your problem or should I just explain?
anonymous
  • anonymous
but 3 is a numerator too it can turn to 1
anonymous
  • anonymous
yeah you should explain i get everything you taught me i just can put it all together
anonymous
  • anonymous
Do you agree that \[(3x)^{-1}=\frac{ 1 }{ (3x)^{1} }\]or \[\frac{ 1 }{ (3x) }\]
anonymous
  • anonymous
\[1\over (3x)^1\]
anonymous
  • anonymous
Errr I mean that it equals both but the lower one without the exponent of 1 is simplified
anonymous
  • anonymous
But you agree with it, right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i should've looked at it as a whole
anonymous
  • anonymous
Now do you agree that \[(3x)^{−1}\times \frac{ 1 }{ 4 }=\frac{ (3x)^{−1} }{ 4 }\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
so if you substitute the (3x)^-1 with \[\frac{ 1 }{ 3x}\] in the expression, you get \[\frac{ 1 }{ 4(3x) }\]
anonymous
  • anonymous
and that simplifies to\[\frac{ 1 }{ 12x }\]
anonymous
  • 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
  • anonymous
is it the same thing as multiplying 1/3x and 1/4
anonymous
  • anonymous
Yes, but the point is that I just flipped it over and that it's incredibly simple.
anonymous
  • anonymous
Everyone else does it like:|dw:1363833370136:dw|
anonymous
  • 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
  • anonymous
lolwut. Just understand it. and REMEMBER IT.
anonymous
  • anonymous
yes sir
anonymous
  • anonymous
lol
anonymous
  • anonymous
hahahadont make me take out my shoe

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