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math doubts [NEW]

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Ghost Of Highbury:

--- Quote from: nid404 on October 31, 2009, 06:35:31 pm ---equation of the circle has the general form

x2+y2+2gx+2fy+c=0

where centre is (-g,-f) and radius is root of (g2+f2-c)

@adi...how do u use the square root sign??

--- End quote ---

its latex

click that "pi" symbol next to "sub" "sup" and others..

for square root type ... sqrt(type here) in the latex form.. i.e after clicking "pi"

like this

zara:

--- Quote from: A@di on October 31, 2009, 06:24:18 pm ---first find where they intersect

x-1 = x2 - 5x -8

simplifies to

-x2 + 6x + 7 =0

x = -1                 x = 7
y = -2                 y= 6

------------------------------

y = -2x2 + qx + p

substitute the values

x = -1 ; u get p=q

substitute x=7 and 'q' for 'p' (because they are equal) ; u get q = 13

so the curve is

-2x2 + 13q + 13

@Z.J - equation of a circle?..or  area

--- End quote ---
wait....hows p=q?
im stuck..

Ghost Of Highbury:

--- Quote from: Z.J. on October 31, 2009, 06:40:53 pm ---wait....hows p=q?
im stuck..

--- End quote ---

wen u substitute x=-1 ad y=-2 in the equation

y = -2x2 + qx + p

u get

-2 = -2 -q + p

-2 + 2 = p - q

p-q =0

p=q

astarmathsandphysics:
The equation of a curcle with centre and radius r is (a,b) is (x-1)^2+(y-b)^2=r^2
r=4 if the diameter is 8

nid404:
I modified my post sir.....thought radius was 8

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