All in parentheses, 3m^1/6 n^1/3 over 4n^-2/3 all to ^2. I'm having trouble simplifing the problem. If someone could explain on how to do it, it would help me out a lot. I'm studying for a test and it's been awhile since we've done this in class, so I don't remember how to do it. Thanks in advance...
2006-11-12 07:45:47 · 3 answers · asked by Belkel 1 in Science & Mathematics ➔ Mathematics
3m^1/6 n^1/3 over 4n^-2/3 all to ^2
= ( 3m^1/6 n^1/3 / 4n^-2/3 )^2
= (3 m^1/6 n^{1/3+2/3} /4 )^2
= (9/16) m^{2/6} n^2
= (9/16) m^{1/3} n^2 '
2006-11-12 08:26:55 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The rules are as follows m^a * m^b = m^(a+b)
if you have 1/(m^a) = m^(-a)
and finally (m^a)^b = m^(a*b)
This should give you enough information to simplify your equation.
2006-11-12 15:50:35 · answer #2 · answered by rscanner 6 · 0⤊ 0⤋
im sorry i cant help...i think my teacher was trying to show sumthing like that to me the otha day.
2006-11-12 16:05:59 · answer #3 · answered by erikkaa2002 2 · 0⤊ 1⤋