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This is Algebra section dear.
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1) EDIT
2) 16800, Is it right?
Decoder plz tell me the answer for the first answer.I think I have a doubt in this divisor thingy concept.
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^^
pi/6 instead of pi/3 dude.
Rest is allright.Nice
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Is the answer for 2) , 10/root27
@ elastiboy: sin@ =0 So angle between them is 90?
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1)
Area of triangle can be given as 1/2 Pm1 = 1/2 Qm2 = 1/2 R m3 where m1,m2,m3 are medians.
So m1 = 2delta/P
m2 = 2delta/Q
m3 = 2delta /R
But P Q R are in AP , hence m1 m2 m3 are in HP
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Dammit yes! I should have done that.Nice
EDIT: raulrag are u sure your answer is correct?Mine is lil diff.
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OK
Putting the value of y in the curve we get:
3x^2 - (3x-3/y1)^2 = 3
This gives 3x^2(y1^2 - 3) -9 + 9x = 3y1^2
=> 3x^2(y1^2 -3 ) + 18x -3(y1^2 +3) = 0
Putting D =0
324 + 36(y1^4 -9) =0
y1^4 = 36
So y1 = +- root6
So tangent is 3x + - 3/root6 =y
Man whats the answer?
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2root3 -root3cosx + 4sin2x = sinx
=> root3 + 2sin2x = sinx/2 + root3cosx/2
=> root3 + 2sin2x = sin(30+x)
=> 2(sin60 + sin2x) = sin (30 +x)
=> 4sin (30+x)cos(30-x) = sin(30+x)
=> sin(30+x)(1 - 4cos(30-x)) =0
=> sin(30+x) =0 OR cos(30-x) = 1/4
After this is easy but hectic to type.
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Thumb Rule dude.
Equation of tangent is given by 3xx1 - yy1 = 3
=> 3xx1/y1 -3/y1 = y
Slope is 3x1/y1
Now 3x1/y1 * -1/3 = -1
So x1/y1 =1
So equation reduces to 3x -3/y1 =y
NOw equate this equation with the curve and put D =0 to get the value of y1.
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ONLY those waves whose motion is SIMPLE HARMONIC are functions of sin and cos.
Travelling waves should be reducible to the form f(t-x/v) or f(x-vt) and also their displacement is finite for any value of x or t or v .
In this case, the first one is a wave due to finite displacement and for the same reason second is not.These are neither SH nor travelling waves but they are still waves. :)
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OK lets see..
The unit vector perpendicular to yz plane is i^ and the unit vector perpendicular to zx plane is j^ . So equations of line perp. to yz & zx plane and passing through P are
x-f = y-g/0 = z-h/0 and x-f/0 = y-g = z-h/0
So any points on these lines are (  + f , g , h ) and (f,  + g , h)
Now since these points intersect the above given planes, we get  = -f and = -g ( Because in yz plane x=0 and in zx plane y=0)
So finally the points are ( 0, g, h) and ( f , 0 ,h)
And you have the origin as (0,0,0).So equation of plane passing through 3 points can be found out easily now.
Answer is coming out to be OPTION A)
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OK lets see:
Initially pressure in the tube is po and volume is 0.05A, where A is area of crossection of the tube.
After the tube is closed and pulled ,let mercury rise by x meters.
Temperature remains constant in such cases, so we get PV = constant
Hence po*0.05A = p*(0.48-x)A
So we get p = (po*0.05)/(0.48-x)
Now by fig we get po = p + x ( In meters of mercury ofc)
Equating we get po-x = po*0.05/(0.48 - x)
This quadratic gives x =.38 and so 0.48-x = 0.1m
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Considering q and r to be rational, another root is a-ib( complex roots occur in conjugate pairs)
Sum of roots = 0 ( Coefficient of x^2 is 0)
So third root is -2a.
Since we need an equation whose one root is 2a i.e negative of -2a, we replace x by -x.
And we get the equation as x^3 + qx - r =0
PS: Please post algebra related questions in Algebra sub-forum.Thnx
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EDIT
WILL POST IN THE MORNING SRRY..
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