6x^3/ 8x
and
y^2 - 81/ (y - 9)^2
2007-06-28 03:37:59 · 6 answers · asked by lushay 2 in Science & Mathematics ➔ Mathematics
6x^3/ 8x = 3x²/4
y^2 - 81/ (y - 9)^2
= (y+9)(y-9) / (y-9)(y-9)
= (y+9) / (y-9)
2007-06-28 03:41:17 · answer #1 · answered by MamaMia © 7 · 2⤊ 0⤋
take out the common factors
6x^3/8x = (2x)(3x^2)/(2x)(4) = 3x^2/4
y^2-81 = (y+9)(y-9)
(y^2-81)/(y-9)^2 = (y+9)/(y-9) = 1 + 18/(y-9)
2007-06-28 10:42:19 · answer #2 · answered by Alhazi 2 · 0⤊ 0⤋
6x^3 / 8x
= 3x^2 / 4
y^2-81 / (y-9)^2
(y - 9) (y + 9) / (y - 9) (y - 9) = (y + 9) / (y - 9)
2007-06-28 10:42:49 · answer #3 · answered by Doctor Q 6 · 0⤊ 1⤋
6x^3/ 8x
= 3x^2/4
y^2 - 81/ (y - 9)^2
=(y +9)(y-9) / (y-9)(y-9)
=(y+9) / (y-9)
2007-06-28 10:39:50 · answer #4 · answered by ignoramus 7 · 2⤊ 3⤋
6x^3/8x
3x^3/4x
3x^2/4
y^2-81/(y-9)^2
(y-9)(y+9)/(y-9)^2
(y+9)/(y-9)
2007-06-28 10:41:47 · answer #5 · answered by Anonymous · 1⤊ 0⤋
Please do your own work or you will never learn it.
2007-06-28 10:40:07 · answer #6 · answered by blue_girl_05us 3 · 0⤊ 2⤋