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See this : http://www.goiit.com/posts/list/algebra-want-answer-with-steps-38778.htm
similarly,
|Az1 - Bz2|^2 = (A^2) z1^2 + (B^2) z2^2 - 2Re(ABz1z2(bar))
and
|Bz1 + Az2|^2 = (B^2) z1^2 + (A^2) z2^2 + 2Re(ABz1z2(bar))
=> |Az1 - Bz2|^2 + |Bz1 + Az2|^2
= (A^2) z1^2 + (B^2) z2^2 + (B^2) z1^2 + (A^2) z2^2
= (A^2) (z1^2 + z2^2) + (B^2) (z1^2 + z2^2)
= ((A^2) + (B^2)) (z1^2 + z2^2) ANS
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aakash,
u r right! however, u have un-necessarily given many steps.
|z| = 1 => 2/|z| = 2 => |2/z| = 2 that's all.
=> 2/z lies on a circle of radius 2 with centre at origin
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sin(y-x)=5/13 => cos(y-x) = 12/13 => tan(y-x) = 5/12
cos(y+x)=4/5 => sin(y+x) = 3/5 => tan(y+x) = 3/4
tan 2y = tan ((x+y) + (y-x)) = (tan(x+y) + tan(y-x))/(1 - tan(x+y)*tan(y-x))
= (3/4 + 5/12)/(1 - (3/4) * (5/12))
= (14/12)/(11/16) = 56/33
(sorry for mis-calculations if any  , as ans is not matching with previous ones)
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Please note here that there can be infinite answers here if we don't put the principal arg of w and w^2.
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arg(iw) = pi/2 + arg(w) = pi/2 + 2pi/3
arg(iw^2) = pi/2 + arg(w^2) = pi/2 + (-2pi/3)
=> arg(iw)+arg(iw^2) = pi ANS
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write ur e-mail id here to get the solution.
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Aakash!
events that A is selected (say E) and event that B is selected (say F) are independent events.
=> P(E  F) = P(E) * P(F)
=> P(F) = P(E F) / P(E) <= max( P(E  F)) / P(E) = 0.3/0.5
=> P(F) <= 0.6
i.e. P(F) can't be 0.9 or it is not possible that the probability of B getting selected is 0.9.
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|z1 + z2|^2 = (z1 + z2)(z1(bar) + z2(bar)) since zz(bar) = |z|^2
= z1z1(bar) + z2z2(bar) + z1z2(bar) + z1(bar)z2
= z1z1(bar) + z2z2(bar) + (z1z2(bar) + (z1z2(bar))(bar))
= |z1|^2 + |z2|^2 + 2Re (z1z2(bar)) since z + z(bar) = 2Re(z)
= 1 + 4 + 2Re (z1z2(bar))
Now,
|z1 - z2|^2 = (z1 - z2)(z1(bar) - z2(bar)) since zz(bar) = |z|^2
= z1z1(bar) + z2z2(bar) - z1z2(bar) - z1(bar)z2
= z1z1(bar) + z2z2(bar) - (z1z2(bar) - (z1z2(bar))(bar))
= |z1|^2 + |z2|^2 - 2Re (z1z2(bar)) since z + z(bar) = 2Re(z)
= 1 + 4 - 2Re (z1z2(bar))
Now,
|z1 + z2|^2 + |z1 - z2|^2 = (1 + 4 + 2Re (z1z2(bar))) + (1 + 4 + 2Re (z1z2(bar)))
= 10 ANS
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( sin ( x + h ) - sin x ) /h = (sin x cos h + cos x sin h - sin x)/h
= (sin x cos h - sin x + cos x sin h)/h
= (sin x (cos h - 1) + cos x sin h)/h
= (sin x (1 - 2 (sin h/2)^2 - 1) + cos x sin h)/h
= - 2 sin x (sin h/2)^2)/h + cos x (sin h)/h
= - sin x (sin h/2) ((sin (h/2))/(h/2)) + cos x (sin h)/h
=> [ h] [ o] ( sin ( x + h ) - sinx ) /h
= [ h][ o] (- sin x (sin h/2) ((sin (h/2))/(h/2)) + cos x (sin h)/h)
= [ h][ o] (- sin x (sin h/2) ((sin (h/2))/(h/2)) + cos x (sin h)/h)
= [ h][ o] (- sin x (sin h/2) ((sin (h/2))/(h/2))) + [ h][ o] (cos x (sin h)/h)
= (- sinx) [ h][ o] (sin h/2) ((sin (h/2))/(h/2)) + cos x [ h][ o] (sin h)/h
= (-sin x) (0) (1) + cos x (1) since [ y][ o] (sin y)/y
= 0 + cos x
Thus [ h][ o] ( sin ( x + h ) - sinx ) /h = cos x
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why angle around a point is taken as 360 degrees?
This is just a convention. It started when a person (may be aaryabhatt or somebody like him) took one angle between two lines, when they are making all 4 angles equal made by their intersection, equal to 90 (right angle) and called 1 degree to the angle obtained by dividing right angle into 90 parts.
However, if he had divided a right angle into 25 parts, angle around a point would have been taken 100 degree.
So these are all conventions only..... some more are ........
1. giving +ve and -ve sign to charge while it doesnt mean that 1 +ve unit charge is 2 units greater than 1 -ve unit charge
2. Celsius divided freezing n boiling point in 100 parts but Fahrenheit didn't
3. Why 8 is shown as 8, not like '4'
and so forth...........
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Suppose x is the fraction of milk in the mixture after k times adding 0.5 L of water.
Now, suppose in next dk times (divide one time adding 1/2 L water into infinite times adding infinitesimal quantity of water), dx is the amount by which fraction of milk decreases,
=> dx = [(40*x - dk*(0.5*x) ) / 40] - x
(40*x is the total milk at that time, dk*(0.5*x) milk is gone again, so (40*x - dk*(0.5*x) ) / 40 becomes the fraction of milk)
=> - dx / x = dk / 80
=> - 1 0.5 dx / x = 0n dk / 80
=> ln 1 - ln 0.5 = n/80
=> 0 - ( - 0.69) = n/80
=> n = 80*0.69  55 times Answer
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I think pratik is right!
sizzle!
rates r not as important as the dedication u hav to the community.
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bharti,
you r correct, however it's understood that we r talking abt +ve integer soln, otherwise thr r infinite soln.
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sin 10 = sin (3*pi + (10 - 3*pi)) = - sin (10 - 3*pi) = sin (3*pi - 10)
=> sin-1(sin 10) = sin-1(sin (3*pi - 10)) = 3*pi - 10 = - (10 - 3*pi) ANS
lies in [-pi/2, +pi/2] i.e. range of sine inverse
cos 10 = cos (4*pi - (4*pi - 10)) = cos (4*pi - 10)
=> cos-1(cos 10) = cos-1(cos (4*pi - 10)) = 4*pi - 10 ANS
ans lies in [0, pi] i.e. range of cosine inverse
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vandya!
write ur email here. i'll send u the soln.
difficult to type here. thr were mistakes in my prev soln
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Sudakar,
z1, z2 r 2 complex numbers on z-plane, |z1| and |z2| represent modulus (magnitude) of z1 n z2.
|z1 + z2| is again a complex number on same plane.
z1, z2 and z1+z2 will make a triangle,
in triangle, difference of any two side is greater than third side.
=> |z1| - |z2| > |z1 + z2|
but |z1| - |z2| = |z1 + z2| is given => z1, z2, z1+z2 don't make a triangle. they are in one line.
Now if they are on the same side of origin, obviously |z1 + z2| = |z1| + |z2|
but we have, |z1 + z2| = |z1| - |z2|
=> z1 and z2 must be on the different side of origin.
=> if we rotate line of z1 by pi-angle we get line of z2.
=> arg z1 - arg z2 = pi
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Akshay!
in ur solution:
since 60 can be written as 4*6*5 (it is 120, not 60)
so no. of ways of selecting x1 = 4 (how do u get this?)
no. of ways of selecting x2 = 6 (how do u get this?)
no. of ways of selecting x3 = 5 (how do u get this?)
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I apologize for the typing mistake, last time forgot to write 24, so did not get ans 2
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11C4 is total no. of quadrilaterals when all points are non-collinear
but here those cases are also counted when 4 points are taken from 5 points which are collinear now, so can't make a quadrilateral (cases are 5C4)
and also those cases when 3 points were taken from 5 points which are collinear now, and 1 point from rest 6 points, so can't make a quadrilateral
(cases are 5C3 * 6C1)
so ans is 11C4 - 5C4 - 5C3 * 6C1
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tobeiitian!
No. of distinct quadrilaterals = 11C4 - 5C4 - 5C3*6C1
= 330 - 5 - 60 = 265 Ans
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Vriti Edustore
