I'm not sure how to handle the exponants. I'd like it as (7y^? - y^?)^?
2006-12-19 13:53:43 · 3 answers · asked by dgbaley27 3 in Science & Mathematics ➔ Mathematics
The only thing you could do to work the y into the (7 - y) is to make it of the same exponent (1/3). y^(2/3) really means the cube root of y squared. So let's multiply y^2 by (7 - y).
(y^2)^1/3 * (7-y)^(1/3) = (7y^2 - y^3) ^1/3
2006-12-19 13:57:55 · answer #1 · answered by J G 4 · 1⤊ 0⤋
you can't combine the exponents on sight sicne the baess are different but lets see what we can do with this...
y^(2/3) * (7-y)^ (1/3) =
(y^2)^(1/3) * (7-y)^(1/3) =
((y^2)(7-y))^(1/3)
or if you prefer
((7^2 - y^3))^1/3
2006-12-19 21:59:41 · answer #2 · answered by Joni DaNerd 6 · 0⤊ 0⤋
y^(2/3) * (7 - y)^(1/3)
(y^2(7 - y))^(1/3)
(7y^2 - y^3)^(1/3)
2006-12-19 22:28:48 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋