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If the lengths of sides of a,b,c of a triangle satisfy
2(bc2+ca2+ab2)=b2 c+c2a+a2b+3abc or bc(2c-b) + ca(2a - c) + ab(2b - a) = 3abc or bc(2c-b) -abc + ca(2a - c) -abc + ab(2b - a) - abc = 0 or bc(2c-b-a) + ca(2a - c - b) + ab(2b - a - c) = 0 so, (2c-b-a) = 0, (2a - c - b) = 0 and (2b - a - c) = 0 which further implies 2c = a + b 2a = c + b 2b = a + c solving these three equations we obtain a = b = c hence the triangle is equilateral NOTE: since the equation is symmetric in a, b, and c. thus without deriving even we can conclude that the triangle is symmetric.
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Solution to your second problem is explained below: If A  B = A C and A  B = A C. To Prove that B=C. We know that n(A B) = nA + nB - n(A B)..............(1) Similarly, n(A C) = nA + nC - n(A C)..............(2) LHS of eqns. (1) and (2) are equal thus equating their RHS's we have nA + nB - n(A B) = nA + nC - n(A C) or nA + nB = nA + nC [as (A B) = A C) ] or B = C
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| difference between A  B and A  B A subset of a set contains any or all of the elements of the set. A proper (denoted by ) subset contains any of the elements of the set but not all of them |
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Problem) If y = log(x/a+bx)^x,prove that x^3 d^2y/dx^2=(x dy/dx-y)^2 Solution) y = log(x/a+bx)x = x log(x/a+bx) or dy/dx = log(x/a+bx) + 1
and d2y/d2x = (1/a + b) /(x/a+bx) so x3 d2y/d2x = x2 (x/a+bx)/(x/a+bx) = x2 ...........(1) Now, (x dy/dx-y) = x [log(x/a+bx) + 1] - x log(x/a+bx) or (x dy/dx-y) = x or (x dy/dx-y)2 = x2 ................(2) Hence from eq. (1) and (2)
x3 d2y/d2x = (x dy/dx-y)2 , hence proved.
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You have given the function f(x) = (x^2-1) /x^2-3x+2 I cos (IxI) Is it f(x) = (x2-1) /(x2-3x+2 I cosIxI I), I mean is cosine term also within I I ?
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Well done polymath keep it up. The way u cracked the problem is fantastic and appreciated
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I agree upon the logic suggested by Puneet1622, please mention whether m/n is the simplest ratio or any other ratio.
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Problem) Suppose a piece of wood is floating inside a closed waterbottle and through a pump, air is pumped into the bottle. Will the wood float or sink? Explanation) When air is pumped into the bottle, the air column just above the surface of water exerts more and more pressure on the surface of bottle as well as water. Hence the exerted pressure tries to compress the water by exerting force =( Pressure x area) in the downward direction, but since the water is incompressible and the force that is also acting on the wooden piece will increase the effective weight of the wooden piece consequently more buoyant force is required to balance the weight of floating piece. This implies that wooden piece starts to submerge more and more inside the water depending upon the force due to air column.
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Dear nilesh, i am unable to imagine shape of the track and figure you want to explain. I further request Administrator to kindly make available the editing of figures to avoid inconvinience, if any, from students and experts as well.
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Value of g at height h from the earth's surface is given by gh = GM/(R + h)2 Where, G= gravitational constant R = radius of the earth h = height h from the surface of earth. or gh = GM/R2(1 + h/R)2 = g/(1 + h/R)2 where, g = acceleration due to gravity on the earth's surface. so height say 'h' at which gh = 1% of g = g/100 is given by g/100 = g/(1 + h/R)2 or 100 = (1 + h/R)2 now substitute R to find the value of 'h' Similarly, to work out for second problem take gh = g - (g/1000) and proceed.
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the kinetic energy at the highest point will be due the horizontal component of velocity of the body as vertical component of velocity at the highest point will be zero. Mass of the body = m Kinetic energy = (1/2)mv2 = E so, v =  (2mE) Angle of projection is = 30 deg. So horizontal component of velocity = v cos30 = ( 2mE)cos30 or v x = (2mE) 3/2 As there is no force along the horizontal direction so horizontal velocity remains the same throughout the motion. So horizontal velocity on the highest point = (2mE) 3/2 So kinetic energy at the highest point is given by Kh = 1/2 m (vx)2, substitute the value to get the results.
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the degree of equation is '6' hence the number of roots of this equation is the degree itself that is 6.
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Hope the solution is clear to u by following the approach suggested by Puneet and myself
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True, but infact it is the horse's feet that exerts force on the ground and the reaction force due to this makes the cart wheel to move.
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Always remember that the induced current produced in the conductor always flows in such a direction that the magnetic field it produces will oppose the change that produces it. EXPLANATION Lenz's Law states that in a given circuit with an induced EMF caused by a change in a magnetic flux, the induced EMF causes a current to flow in the direction that opposes the change in flux. That is, if a decreasing magnetic flux induces an EMF, the resulting current will oppose a further decrease in magnetic flux. Likewise, for an EMF induced by an increasing magnetic flux, the resulting current flows in a direction that opposes a further increase in magnetic flux. It is important to note that the induced current will always flow in a direction which opposes any change of magnetic flux, but it does not oppose the magnetic flux itself. Lenz's law can be derived from Faraday's law of induction, simply by noting the minus sign on the right side of the equation
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Let t = xy log t = y log x or (dt/dy)/t = log x + (y/x)dx/dy This is the case when x is a function of y, otherwise if x is indpendent variable then dx/dy is taken as zero and cosidering dx/dy = 0 we have, or dt/dy = t(log x) = (xy) log x
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4 x + 1.5 + 9 x + 0.5 = 10 * ( 6 x ) or 4x 4 1.5 + 9x 90.5 = 10 * ( 6 x ) or 22x 8 + 3 2x 3= 10 * ( 3x 2x ) Let and 2x = u and 3x = v Therefore, above equation becomes as 8u2 + 3v2 = 10uv or 8u2 - 10uv + 3v2 = 0 or 8u2 - 6uv - 4uv + 3v2 = 0 or 2u( 4u - 3v) - v( 4u - 3v) = 0 or ( 4u - 3v) ( 2u - v) = 0 or 4u = 3v and 2u = v or 4*2x = 3*3x and 2*2x = 3x Now find the value of 'x'.
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To find mean velocity use the following expression Mean velocity = Total distance travelled/ total time taken
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sin2A=cos3A or sin2A=sin(  /2 - 3A) or 2A = (  /2 - 3A) or 5A = /2 or A = /10 so, sinA = sin /10 |
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eix = cos x + i sin x (1) where i denotes  -1. This is an equation which allows you to interpret the exponentiation of an imaginary number ix as having a real part, cos x, and an imaginary part, i sin x. This was an especially useful observation in the solution of differential equations. Because of this and other uses of i, it became quite acceptable for use in mathematics. Euler, recommended the general use of these imaginary numbers now, e-ix = cos x - i sin x (2) From (1) and (2) eix e-ix = (cos x + i sin x )(cos x - i sin x ) or, 1 = cos2x - i2 sin2x or, 1 - cos2x = - i2 sin2x or, sin2x = - i2 sin2x or, i2 = -1 or, i = -1
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