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

- anonymous

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- schrodinger

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- anonymous

uhhhh i think so i belive so

- 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

so, your expression is:
\(m^3n^{-6}p^0\)
is that correct?

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## More answers

- anonymous

i may be wrong my b

- 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

you are still doing 2 parts incorrectly, not just the n^(-6), but p^0 is incorrect as well-:(

- SolomonZelman

lets start from \(p^0\), ok?

- anonymous

Okay.

- 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

So if any number (except 0) raised to the power of 0, gives a result of 1, then
\(p^0={\rm what?}\)

- 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

Alright.

- anonymous

So p^1?

- SolomonZelman

p^1 ? you don't have that in your expression, do you?

- anonymous

No, so do i leave the p^0 alone?

- SolomonZelman

you have p^0, and you know that:
(anything)\(^0\) = ?

- anonymous

x?

- SolomonZelman

(anything)\(^0\)=1
(as long as this anything is NOT zero)

- SolomonZelman

therefore, you can say that
p\(^0\)=?

- anonymous

But p has a zer exponent?

- SolomonZelman

yes, so p\(^0\)= what?

- anonymous

Zero* And is it 1

- SolomonZelman

yes, p\(^0\)=1

- 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

is everything making sense till this point?

- anonymous

Alright. Yeah so far.

- anonymous

Because the p is = to 1 right?

- SolomonZelman

yes, because p\(^0\)=1

- 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

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

Makes sense. Thanks for your help.

- SolomonZelman

we aren't done yet.

- anonymous

Alright

- 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

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

Yeah

- SolomonZelman

same, but n instead of a, and 6 instead of b.
good...

- anonymous

n^-6/1/n^6

- SolomonZelman

n\(^{-6}\) = 1 / n\(^6\)
like this, you mean?

- anonymous

Yeah forgot to put in my equal.

- SolomonZelman

lol.

- anonymous

:P

- 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

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

if you have any questions regarding any of the rules or steps that were aplied or mentioned, please ask...

- anonymous

No questions, because you timing it by one which would make it the same number.

- SolomonZelman

alright.
G☼☼d luck

- anonymous

Thanks :)

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