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Mathematics
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1.\[15a ^{-4}b ^{-6} \over 3a ^{7}b ^{-10}\]
steps please :)
alright, let's get the negative powers positive by switching their places and we'll get :\[= (15b^{10})/(3a^7a^4b^6)\] so now we know that : \[= (3)(5)b^{10}/3a^{11}b^6\] let's cancel out the commons and simplify ^_^: \[= 5b^4/a^{11}\] get the idea now? :)

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Other answers:

we subtract the powers :)
i am confused on the top part...you but 3a^7 a^4 .... how did you get two a's
lol, okay, we want to get the powers (+) instead of (-) so in order to do that, if we have a negative powe in the top we get it down to get it positive and vice versa ^_^
power*
so I got a^-4 down to make it a^4 :)
to make your life easier ^_^
ok so you brought that down i get that
how did you get b ^10
yep, same thing goes with b too ^_^
oh since it was - you brought it up to be pos.
you want b^-10 positive, so you get it up to make it b^10 :) switch the places
exactly ^_^
ok
then we simplify and subtract, b^10 - b^6 = b^4 and the one with the greatest power takes the lead, which means the answer stays up :)
got the point? ^_^
yes thanks so much, i have two other problems i posted if you get a chance i would love your help!
np, I'll take a look ^_^

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