2007-04-21 08:40:49 · 5 answers · asked by lashawnwuzhea2005 1 in Science & Mathematics ➔ Mathematics
2y= sqrt(10-3y)
square both sides:
4y^2 = 10-3y
4y^2+3y-10=0
(4y-5)(y+2)=0
y = 5/4 y =-2
2007-04-21 08:46:27 · answer #1 · answered by jaybee 4 · 0⤊ 0⤋
2y= sqrt(10-3y)
square both sides:
4y^2 = 10-3y
4y^2+3y-10=0
(4y-5)(y+2)=0
y = 5/4 y =-2
2007-04-21 15:47:32 · answer #2 · answered by Alyssa 2 · 0⤊ 0⤋
4y² = 10 - 3y
4y² + 3y - 10 = 0
(4y - 5).(y + 2) = 0
y = 5 / 4 , y = - 2
2007-04-21 15:48:32 · answer #3 · answered by Como 7 · 0⤊ 0⤋
square both sides
4y^2 = 10-3y
0= -4y^2-3y+10
0=(-4y+5)(y+2)
two ans
0=-4y+5, 0=y+2
y=5/4 or y = -2
2007-04-21 15:49:08 · answer #4 · answered by T.B. 2 · 0⤊ 0⤋
9y^2+3y-10=0
y=(-3+ or - sqrt(369))/18
and
10-3y>=0
2007-04-21 15:46:24 · answer #5 · answered by iyiogrenci 6 · 0⤊ 0⤋