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# Urgent!!! watch

1. How do you solve these?!?!

1) y + 4 = [-1/[4x(x-2)(x+2)]] [x-1]

2) [-160/v^2] + [v/50] = 0

BB
2. (Original post by Blush Babe)
How do you solve these?!?!

1) y + 4 = [-1/[4x(x-2)(x+2)]] [x-1]

2) [-160/v^2] + [v/50] = 0

BB

errr............................ ...????
3. (Original post by xAngelx)
errr............................ ...????
It's AS pure 1 maths
4. (Original post by Blush Babe)
How do you solve these?!?!

1) y + 4 = [-1/[4x(x-2)(x+2)]] [x-1]

2) [-160/v^2] + [v/50] = 0

BB
well,
1) this will be only valid for x=1, for which, y=-4;

2) taking lcm.....(8000+v^3)/(50*v^3) = 0
So, v^3=-8000.
so, v=-20
5. (Original post by prince_capri)
well,
1) this will be only valid for x=1, for which, y=-4;

2) taking lcm.....(8000+v^3)/(50*v^3) = 0
So, v^3=-8000.
so, v=-20

er............i'm still trying to read it
6. (Original post by Blush Babe)
How do you solve these?!?!

1) y + 4 = [-1/[4x(x-2)(x+2)]] [x-1]

2) [-160/v^2] + [v/50] = 0

BB
In the 1st number, I think you have to use the method in solving for line equations. In the second, since there is an exponent in the variable, use the quadratic equation. Hope that helps!
7. (Original post by prince_capri)
well,
1) this will be only valid for x=1, for which, y=-4;

2) taking lcm.....(8000+v^3)/(50*v^3) = 0
So, v^3=-8000.
so, v=-20
Cheers , except it's a wee bit too late cuz I just did my maths exam (and failed)...the first question was so easy it got my hopes up, until I turned to next page.
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