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Blush Babe
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#1
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How do you solve these?!?!

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

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

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xAngelx
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#2
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#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

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errr............................ ...????
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Blush Babe
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(Original post by xAngelx)
errr............................ ...????
It's AS pure 1 maths
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prince_capri
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#4
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(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

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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
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xAngelx
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(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
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ladyblitzer
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#6
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(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

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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!
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Blush Babe
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#7
Report Thread starter 16 years ago
#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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