6y^-2 + y = 0
2007-09-05 20:50:53 · 7 answers · asked by i<3WL 2 in Science & Mathematics ➔ Mathematics
Hi
We have
6·x² + x = 0
We can factor out an x to get
x·(6·x + 1) = 0
So, one solution is x = 0
6·x + 1 = 0 when x = -1/6, which is the other solution.
James :-)
2007-09-05 21:01:54 · answer #1 · answered by ? 3 · 0⤊ 0⤋
6y^-2 + y=0
6/y^2 + y=0
(6+y^3)/y^2=0
6+y^3=0
y^3= -6
y= -1.8171
2007-09-06 04:04:55 · answer #2 · answered by xietra 1 · 0⤊ 0⤋
6 / y ² + y = 0
6 + y ³ = 0
y ³ = - 6
y ³ = ( i ² ) 6
y = ( 6 i ² )^( 1/3 )
2007-09-10 03:06:14 · answer #3 · answered by Como 7 · 0⤊ 0⤋
6y^-2 = 6/y^2 + y= 0
Multiply both sides by y^2
6 + y^3 = 0
y^3 = -6
-y = cuberoot 6
y = - cuberoot 6
2007-09-06 04:29:54 · answer #4 · answered by nothingtodo007 2 · 0⤊ 0⤋
You can factor out y^-2, so:
6y^-2 + y = 0
y^-2(6 + y^3) = 0
Thus, either:
y^-2 = 0 (Sol'n: y =0)
Or:
6 + y^3 = 0 (Sol'n: y = cube root of -6 = -1.81)
are solutions for y...
2007-09-06 04:07:47 · answer #5 · answered by taq2007 2 · 0⤊ 0⤋
Factor:
y(6y + 1) = 0
1st solution:
y = 0
2nd solution:
6y = - 1
y = - 1/6
Answer: either y = 0 or y = - 1/6
2007-09-06 04:31:32 · answer #6 · answered by Jun Agruda 7 · 3⤊ 0⤋
6 / y^2 + y = 0
6 + y^3 = 0
y = -1.81712
2007-09-06 04:15:59 · answer #7 · answered by CPUcate 6 · 0⤊ 0⤋