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The Maths Thread! |
k4m!k4ze
Joined: Oct 02, 2004
Posts: > 500
From: CBE
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x = -1 and x = -1/2
Same technique as above
y=2x+1
y^2=2x^2+x
subst y value in eqn 1 in eqn 2
(2x+1)^2=2x^2+x
=> 2x^2+3x+1=0
=> 2x^2 + 2x + x + 1 = 0
=> (2x + 1) ( x +1 ) = 0
so u get x, subst both values in eqn 1 to get y values
If at first you don't succeed, destroy all evidence of your attempt !! |
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50Cent
Joined: Nov 08, 2003
Posts: > 500
From: ...whatever psycho!!!
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Quote:
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On 2004-10-29 12:31:08, kllr_dude wrote:
x = -1 and x = -1/2
Same technique as above
y=2x+1
y^2=2x^2+x
subst y value in eqn 1 in eqn 2
(2x+1)^2=2x^2+x
=> 2x^2+3x+1=0
=> 2x^2 + 2x + x + 1 = 0
=> (2x + 1) ( x +1 ) = 0
so u get x, subst both values in eqn 1 to get y values
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ah right so
y=(2 x -1) +1
y=-1
and
y=(2 x -1/2)+1
y=0
i think.
Trusted Trader: 50Cent (+6, -0)
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fatreg
Joined: Jul 26, 2003
Posts: > 500
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This so isnt the maths i did at school!
and i went to a grammar school? what went wrong!?
we were all about algebra, but saying that i never did get all the z,y's so maybe i aint got a hope in hell of getting calculus?
fatreg
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kimcheeboi
Joined: Dec 19, 2003
Posts: > 500
From: Abducted by hot blondes to Les
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help me with series, peeps!
power, taylor, and maclaurin series! 
[addsig] |
maddav
Joined: Dec 01, 2002
Posts: 356
From: Nottingham, UK
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I need help, i just can't work out what to do, I know about half of this problem, and the answer, but I need to know how to finish it, I'm sure it's really simple but I just can't get my head round it, here it is:
I need the gradient for the curve f(x) at point A (2,13)
f(x) = 3x² + 2x¯¹
f¹(x) = 6x - 2x¯²
I'm sure for the gradient, i need to do something like substituting the x co-ordinate value, but then I don't know what to do next:
f¹(2) =
Help me please 
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kimcheeboi
Joined: Dec 19, 2003
Posts: > 500
From: Abducted by hot blondes to Les
PM |
since f'(x) is he slope of the tangent line try finding the slope of the line using subistitution and use the equation of a line (y=mx+b) to find the tangent line at the point A
gradient = tangent line right?
[addsig] |
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Posted: 2004-10-29 12:31
