simpifiy
[ 5x^-3]^-1
_________
[8y^-3]^-1
2007-06-25 07:14:25 · 5 answers · asked by Y ask ME! 2 in Science & Mathematics ➔ Mathematics
8y^-3
_____
5x^-3
=>
8x^3
____
5y^3
2007-06-25 07:18:27 · answer #1 · answered by sharky.mark 4 · 0⤊ 0⤋
When you raise a value to a negative power, it really means "1 divided by the value to the opposite power".
For example:
2^(-1) = 1/2^1 = 1/2
x^(-3) = 1/x^3
(2x+1)^(-1/2) = 1/(2x+1)^(1/2)
So, you can turn a negative power positive by simply putting the term into the numerator if it was in the denominator, or putting it into the denominator if it was in the numerator. Hence:
(5x^-3)^(-1) / (8y^-3)^(-1)
= (8y^-3) / (5x^-3)
= (5x^3) / (8y^3)
= 5/8 * (x/y)^3
2007-06-25 14:25:34 · answer #2 · answered by lithiumdeuteride 7 · 0⤊ 1⤋
8y^-3 / 5x^-3
8x^3 / 5y^3
*Just because the x and y flip flop, doesn't mean the 8 and 5 do*
2007-06-25 14:20:38 · answer #3 · answered by Becky M 4 · 0⤊ 0⤋
[ 5x^-3]^-1=(x^3)/5 and [8y^-3]^-1=(y^3)/8
Thus,
[ 5x^-3]^-1
_________ =
[8y^-3]^-1
(x^3)/5
_________ =
(y^3)/8
8x^3
_________ =
5y^3
2007-06-25 14:22:30 · answer #4 · answered by brigitte 2 · 0⤊ 0⤋
(5x^-4) / (8y^-4)
hmmmm.... or better 8y^3/5x^3
2007-06-25 14:20:58 · answer #5 · answered by petitfemme 2 · 0⤊ 1⤋