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Author Topic: math doubts [NEW]  (Read 16079 times)

zara

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Re: math doubts [NEW]
« Reply #90 on: October 31, 2009, 12:53:44 pm »
Quote from: Z.J. on October 31, 2009, 11:55:29 am
Q3)Given that the equation


[sin(theta) - cos(theta)]/[sin(theta)+cos(theta)] =6sin(theta)/cos(theta),

find the exact values of tan(theta).Hence find (theta) for 0<=(theta)<=360
some one solve this ques plz...
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Offline falafail

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Re: math doubts [NEW]
« Reply #91 on: October 31, 2009, 02:17:01 pm »
Quote from: Z.J. on October 31, 2009, 12:53:44 pm
some one solve this ques plz...



i believe this is how it's done.
feel free to ask if you need me to explain anything, or to correct me if i'm wrong XD

oh, the picture's too small. here: http://i33.tinypic.com/308wg0p.jpg
« Last Edit: October 31, 2009, 02:19:57 pm by falafail »
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nid404

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Re: math doubts [NEW]
« Reply #92 on: October 31, 2009, 02:28:14 pm »
yaa...that's that

By the way if you don't want to confuse urself bt taking tan2theta and then substituting

assume tan theta=x and then solve. Get value for x then sub tan theta in it

and yeah use periodic property to get the answer within the given range
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zara

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Re: math doubts [NEW]
« Reply #93 on: October 31, 2009, 06:15:38 pm »
Q5)The straight line y=x-1 meets the curve y=x2-5x-8 at the points A and B. The curve y=p+qx-2x2 also passes through the points A and B. Find the values of p and q.
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zara

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Re: math doubts [NEW]
« Reply #94 on: October 31, 2009, 06:19:57 pm »
n yea whats the equation of a circle that has say 8cm as diameter? O_o
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Offline Ghost Of Highbury

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Re: math doubts [NEW]
« Reply #95 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
« Last Edit: October 31, 2009, 06:28:15 pm by A@di »
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zara

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Re: math doubts [NEW]
« Reply #96 on: October 31, 2009, 06:27:12 pm »
yea aadi its equation of a circle as mentioned in my worksheet.:\
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Offline Ghost Of Highbury

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Re: math doubts [NEW]
« Reply #97 on: October 31, 2009, 06:32:44 pm »
Quote from: Z.J. on October 31, 2009, 06:27:12 pm
yea aadi its equation of a circle as mentioned in my worksheet.:\

ok..check this

http://www.analyzemath.com/CircleEq/CircleEq.html

so the answer wud be

y = sqrt(16 - x^2)


By the way--did u understand the other question?
« Last Edit: October 31, 2009, 06:51:03 pm by A@di »
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nid404

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Re: math doubts [NEW]
« Reply #98 on: October 31, 2009, 06:33:01 pm »
she means equation addiii...

yeah It's

r=8cm....let centre be (h,k)
let point P be a point on the circumference of a circle with coordinates (x,y)

then

(x-h)2 + (y-k)2=r2

(x-h)2+(y-k)2=16(42)

I assume centre is (0,0) since it is not given

assuming....h and k become 0

x2+y2=16
equation is  x2+y2-16=0
« Last Edit: October 31, 2009, 06:53:58 pm by nid404 »
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nid404

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Re: math doubts [NEW]
« Reply #99 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??
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Offline Ghost Of Highbury

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Re: math doubts [NEW]
« Reply #100 on: October 31, 2009, 06:38:17 pm »
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??

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

sqrt(I am an orangutan)
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zara

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Re: math doubts [NEW]
« Reply #101 on: October 31, 2009, 06:40:53 pm »
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
wait....hows p=q?
im stuck..
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Offline Ghost Of Highbury

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Re: math doubts [NEW]
« Reply #102 on: October 31, 2009, 06:43:44 pm »
Quote from: Z.J. on October 31, 2009, 06:40:53 pm
wait....hows p=q?
im stuck..

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
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Offline astarmathsandphysics

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Re: math doubts [NEW]
« Reply #103 on: October 31, 2009, 06:48:28 pm »
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
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nid404

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Re: math doubts [NEW]
« Reply #104 on: October 31, 2009, 06:54:32 pm »
I modified my post sir.....thought radius was 8
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