anonymous
  • anonymous
Did i do this right? Simplify each expression. Use Positive exponents m3 n^-6 p^0. 1n^-6/1 = 1/n^-6 now we put the positive n^6 back in it's original place. m^3 n^6 p
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
uhhhh i think so i belive so
SolomonZelman
  • SolomonZelman
The rule for negative exponents is: \(\Large\color{black}{ \displaystyle \color{blue}{\rm a}^{-\color{red}{\rm b}}=\frac{1 }{\color{blue}{\rm a}^{\color{red}{\rm b}}} }\)
SolomonZelman
  • SolomonZelman
so, your expression is: \(m^3n^{-6}p^0\) is that correct?

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anonymous
  • anonymous
i may be wrong my b
anonymous
  • anonymous
Okay 1n^-6/1 = 1/n^-6 = 1/n^6 now we put the positive n^6 back in it's original place. m^3 n^6 p How about now
SolomonZelman
  • SolomonZelman
you are still doing 2 parts incorrectly, not just the n^(-6), but p^0 is incorrect as well-:(
SolomonZelman
  • SolomonZelman
lets start from \(p^0\), ok?
anonymous
  • anonymous
Okay.
SolomonZelman
  • SolomonZelman
The rule for an exponent of 0 is: \(\Large\color{black}{ \displaystyle \color{blue}{\rm x}^{\color{red}{\rm 0}}=1 }\) This is true for any value of x, but provided that x is NOT zero. ------------------------------------------------------ ADDITIONALLY: The prove for this rule is: \(\Large\color{black}{ \displaystyle \color{blue}{\rm x}^{\color{red}{\rm 0}}=\color{blue}{\rm x}^{w-w} }\) (i am using w, but you can replace w with 1, 3, or any other number and the statement will still hold) \(\Large\color{black}{ \displaystyle \color{blue}{\rm x}^{\color{red}{\rm 0}}=\color{blue}{\rm x}^{w-w} =\dfrac{x^w}{x^w}=1}\) (anything divided by itself gives a result of 1, but if x was 0 then we have 0/0 which is not defined. This is why I said every x but NOT x=0)
SolomonZelman
  • SolomonZelman
So if any number (except 0) raised to the power of 0, gives a result of 1, then \(p^0={\rm what?}\)
SolomonZelman
  • SolomonZelman
((Another NOTE: they didn't state that \(p\ne0\), it is their fault, but you have to make this assumption, to get the answer, or else it would be indeterminate))
anonymous
  • anonymous
Alright.
anonymous
  • anonymous
So p^1?
SolomonZelman
  • SolomonZelman
p^1 ? you don't have that in your expression, do you?
anonymous
  • anonymous
No, so do i leave the p^0 alone?
SolomonZelman
  • SolomonZelman
you have p^0, and you know that: (anything)\(^0\) = ?
anonymous
  • anonymous
x?
SolomonZelman
  • SolomonZelman
(anything)\(^0\)=1 (as long as this anything is NOT zero)
SolomonZelman
  • SolomonZelman
therefore, you can say that p\(^0\)=?
anonymous
  • anonymous
But p has a zer exponent?
SolomonZelman
  • SolomonZelman
yes, so p\(^0\)= what?
anonymous
  • anonymous
Zero* And is it 1
SolomonZelman
  • SolomonZelman
yes, p\(^0\)=1
SolomonZelman
  • SolomonZelman
So lets re-write our expression \({\rm m}^3\cdot {\rm n}^{-6} \cdot {\rm p}^0\) this kis what it is used to be, but we know p\(^0\)=1, so we can say: \({\rm m}^3\cdot {\rm n}^{-6} \cdot 1\) and multiplying times 1, is not changing the value, so we can leave the "•1" part out. (right?) We have: \({\rm m}^3\cdot {\rm n}^{-6} \)
SolomonZelman
  • SolomonZelman
is everything making sense till this point?
anonymous
  • anonymous
Alright. Yeah so far.
anonymous
  • anonymous
Because the p is = to 1 right?
SolomonZelman
  • SolomonZelman
yes, because p\(^0\)=1
anonymous
  • anonymous
So yeah i get it not sure why we need to leave the 1 out after solving the problem though. Is it because it's exponent was 0?
SolomonZelman
  • SolomonZelman
why we have to leave the "•1" out? Not that we must, but it does matter. For any number, if you multiply it times 1 you still have that same number. Multiplying times 1 is same, and just as good as, not multiplying by anything at all.
anonymous
  • anonymous
Makes sense. Thanks for your help.
SolomonZelman
  • SolomonZelman
we aren't done yet.
anonymous
  • anonymous
Alright
SolomonZelman
  • SolomonZelman
So our new expression is: \({\rm m}^3\cdot {\rm n}^{-6} \) And, this can be simplified more. ------------------------------ Now, if we can recall, I mentioned another important property, and it refers to negative exponents. This property is: \(\Large\color{black}{ \displaystyle \color{blue}{\rm a}^{-\color{red}{\rm b}}=\frac{1 }{\color{blue}{\rm a}^{\color{red}{\rm b}}} }\)
SolomonZelman
  • SolomonZelman
according to this property, \(\Large\color{black}{ \displaystyle \color{blue}{\rm n}^{-\color{red}{\rm 6}}=\frac{1 }{\color{blue}{\rm n}^{\color{red}{\rm 6}}} }\) right?
anonymous
  • anonymous
Yeah
SolomonZelman
  • SolomonZelman
same, but n instead of a, and 6 instead of b. good...
anonymous
  • anonymous
n^-6/1/n^6
SolomonZelman
  • SolomonZelman
n\(^{-6}\) = 1 / n\(^6\) like this, you mean?
anonymous
  • anonymous
Yeah forgot to put in my equal.
SolomonZelman
  • SolomonZelman
lol.
anonymous
  • anonymous
:P
SolomonZelman
  • SolomonZelman
so, we had: \({\rm m}^3\cdot {\rm n}^{-6} \) and now we know that: \( {\rm n}^{-6}=\dfrac{1}{{\rm n}^6} \) So, we can re-write our expression the following way: \({\rm m}^3\cdot\dfrac{1}{{\rm n}^{6}} \)
SolomonZelman
  • SolomonZelman
then you can make it as 1 fraction, just the following way: \({\rm m}^3\cdot\dfrac{1}{{\rm n}^{6}} \) \(\dfrac{{\rm m}^3}{1}\cdot\dfrac{1}{{\rm n}^{6}} \) \(\dfrac{{\rm m}^3\cdot 1}{1\cdot {\rm n}^{6}} \) \(\dfrac{{\rm m}^3}{{\rm n}^{6}} \)
SolomonZelman
  • SolomonZelman
if you have any questions regarding any of the rules or steps that were aplied or mentioned, please ask...
anonymous
  • anonymous
No questions, because you timing it by one which would make it the same number.
SolomonZelman
  • SolomonZelman
alright. G☼☼d luck
anonymous
  • anonymous
Thanks :)

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