calculusxy
  • calculusxy
Help with exponents ... Question attached below.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
calculusxy
  • calculusxy
\[\huge \frac{ (2p m^{-1}q^0)^{-4} \times 2m^{-1}p^3 }{ 2pq^2}\]
calculusxy
  • calculusxy
@hartnn
hartnn
  • hartnn
tried it ?

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hartnn
  • hartnn
just take one variable at a time assume there's only p
hartnn
  • hartnn
\(\Large \dfrac{p^{-4}p^3}{p} = ..?\)
calculusxy
  • calculusxy
Actually what I did was this: \[\large (2p m^{-1}q^0)^4 = 2^{-4}p^{-4}m^{-5}q^{-4}\]
calculusxy
  • calculusxy
@hartnn
hartnn
  • hartnn
for 2 and p thats correct. for others it isn't.
hartnn
  • hartnn
\(\Large (m^{-1})^{-4} = m^{-1\times -4} = m^{???}\)
calculusxy
  • calculusxy
oh okay, so it's m^4
hartnn
  • hartnn
yes
hartnn
  • hartnn
\(\Large (q^{0})^{-4} = q^{0\times -4} = q^{???}\)
calculusxy
  • calculusxy
q^0
calculusxy
  • calculusxy
which is 1
hartnn
  • hartnn
correct!
calculusxy
  • calculusxy
so we would now have: \[\large \frac{ 2^{-4}p^{-4}m^4q^0 \times 2m^{-1}p^3 }{ 2pq^2 }\]
hartnn
  • hartnn
yes whats m^4 * m^{-1} = .. ?
calculusxy
  • calculusxy
Before that, is it possible to group the terms together according to the variables and exponents?
hartnn
  • hartnn
according to variables, yes!
calculusxy
  • calculusxy
also what would we do with the 2^{-4}?
hartnn
  • hartnn
combine like terms! \(\dfrac{2^{-4} \times 2}{2} = 2^{-4} = \dfrac{1}{2^4}\)
calculusxy
  • calculusxy
so the other two (without the variable), you took it out from 2m^{-1}?
hartnn
  • hartnn
\(\Large m^4 \times m^{-1}= .. \) \(\Large \dfrac{p^{-4}p^3}{p} = ..?\)
calculusxy
  • calculusxy
what variable would the 2 have? \[2^1\] or \[2^0\]
hartnn
  • hartnn
2 = 2^1
calculusxy
  • calculusxy
\[\frac{ 2^{-4} \times 2^1 }{ 2 } = 2^{-4}\]
calculusxy
  • calculusxy
am i correct?
hartnn
  • hartnn
yes
calculusxy
  • calculusxy
ok let me go on..
calculusxy
  • calculusxy
now if i did the other variables: \[\frac{ p^{-4}p^3 }{ p^1 } = p^{-2}\] \[m^4 \times m^{-1} = m^3\] \[\frac{ q^0 }{ q^2 } = q^{-2}\]
hartnn
  • hartnn
all correct! :)
madhu.mukherjee.946
  • madhu.mukherjee.946
good
calculusxy
  • calculusxy
okay so now I combine them?
hartnn
  • hartnn
yes
calculusxy
  • calculusxy
\[\frac{ m^3 }{ 2^4q^2p^2 }\]
hartnn
  • hartnn
\(\huge \checkmark \) *applause*
calculusxy
  • calculusxy
thank you! i have another question, can you help me on that as well?
hartnn
  • hartnn
I can try :)
calculusxy
  • calculusxy
\[\huge \frac{ (2hj^2k^{-2} \times h^4h^{-1}k^4)^0 }{ 2h^{-3}j^{-4}k^{-2}}\]
calculusxy
  • calculusxy
@hartnn
hartnn
  • hartnn
lol (anything except 0)^0 = 1 so there you have your numerator!
calculusxy
  • calculusxy
yeah that's what i was thinking.. but what what about solving it with the denominator?
hartnn
  • hartnn
|dw:1442770780050:dw|
hartnn
  • hartnn
1/h^(-3) = h^3
calculusxy
  • calculusxy
oh yeah! my teacher taught us that.. it just flushed out of my brain. so the answer is: \[\huge 2h^3j^4k^2\]
hartnn
  • hartnn
(h^3 j^4 k^2)/2
calculusxy
  • calculusxy
i don't understand where that came from
hartnn
  • hartnn
|dw:1442771044268:dw|
calculusxy
  • calculusxy
oh okay.. the positive 2 stayed on the bottom.
calculusxy
  • calculusxy
since it always was on the bottom...
calculusxy
  • calculusxy
okay so it's \[\frac{ h^3j^4k^2 }{ 2 }\]
hartnn
  • hartnn
yes! correct :)
calculusxy
  • calculusxy
thank you!
calculusxy
  • calculusxy
sorry but i just wanted to check an answer with you..
calculusxy
  • calculusxy
\[\large x^4y^3 \times (2y^2)^0\] would the answer be \[x^4y^3\]
hartnn
  • hartnn
yes correct :)
calculusxy
  • calculusxy
thank you tremendously!
hartnn
  • hartnn
welcome tremendously! ^_^

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