Help with exponents ... Question attached below.

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Help with exponents ... Question attached below.

Mathematics
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\[\huge \frac{ (2p m^{-1}q^0)^{-4} \times 2m^{-1}p^3 }{ 2pq^2}\]
tried it ?

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Other answers:

just take one variable at a time assume there's only p
\(\Large \dfrac{p^{-4}p^3}{p} = ..?\)
Actually what I did was this: \[\large (2p m^{-1}q^0)^4 = 2^{-4}p^{-4}m^{-5}q^{-4}\]
for 2 and p thats correct. for others it isn't.
\(\Large (m^{-1})^{-4} = m^{-1\times -4} = m^{???}\)
oh okay, so it's m^4
yes
\(\Large (q^{0})^{-4} = q^{0\times -4} = q^{???}\)
q^0
which is 1
correct!
so we would now have: \[\large \frac{ 2^{-4}p^{-4}m^4q^0 \times 2m^{-1}p^3 }{ 2pq^2 }\]
yes whats m^4 * m^{-1} = .. ?
Before that, is it possible to group the terms together according to the variables and exponents?
according to variables, yes!
also what would we do with the 2^{-4}?
combine like terms! \(\dfrac{2^{-4} \times 2}{2} = 2^{-4} = \dfrac{1}{2^4}\)
so the other two (without the variable), you took it out from 2m^{-1}?
\(\Large m^4 \times m^{-1}= .. \) \(\Large \dfrac{p^{-4}p^3}{p} = ..?\)
what variable would the 2 have? \[2^1\] or \[2^0\]
2 = 2^1
\[\frac{ 2^{-4} \times 2^1 }{ 2 } = 2^{-4}\]
am i correct?
yes
ok let me go on..
now if i did the other variables: \[\frac{ p^{-4}p^3 }{ p^1 } = p^{-2}\] \[m^4 \times m^{-1} = m^3\] \[\frac{ q^0 }{ q^2 } = q^{-2}\]
all correct! :)
good
okay so now I combine them?
yes
\[\frac{ m^3 }{ 2^4q^2p^2 }\]
\(\huge \checkmark \) *applause*
thank you! i have another question, can you help me on that as well?
I can try :)
\[\huge \frac{ (2hj^2k^{-2} \times h^4h^{-1}k^4)^0 }{ 2h^{-3}j^{-4}k^{-2}}\]
lol (anything except 0)^0 = 1 so there you have your numerator!
yeah that's what i was thinking.. but what what about solving it with the denominator?
|dw:1442770780050:dw|
1/h^(-3) = h^3
oh yeah! my teacher taught us that.. it just flushed out of my brain. so the answer is: \[\huge 2h^3j^4k^2\]
(h^3 j^4 k^2)/2
i don't understand where that came from
|dw:1442771044268:dw|
oh okay.. the positive 2 stayed on the bottom.
since it always was on the bottom...
okay so it's \[\frac{ h^3j^4k^2 }{ 2 }\]
yes! correct :)
thank you!
sorry but i just wanted to check an answer with you..
\[\large x^4y^3 \times (2y^2)^0\] would the answer be \[x^4y^3\]
yes correct :)
thank you tremendously!
welcome tremendously! ^_^

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