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#1 Re: Help Me ! » Trapezium » 2017-09-30 17:04:03

Assume the ratio as x:1 and equate the perimeters.The internal side being common to both can just be cancelled ,no calculation of that side is necessary.

#2 Re: Help Me ! » Help!! Fast!! » 2017-09-10 03:18:22

When we say vertex of an angle, as far as my knowledge goes,it is the point where the two arms meet. The two additional points taken on the arms are only for nomenclature. This way I do not approve bob bundy's approach. I feel the statement “If the bisectors of two angles with a common vertex are perpendicular, then the angles are supplementary.”needs additional clause. "with a common side." We may have  angle ABC and angle DAF with their bisectors perpendicular but the two angles are not supplementary. I am sorry for not showing it on a diagram.

#3 Re: Help Me ! » Geometry-Transformations » 2017-08-21 04:59:30

I overlooked "clockwise". My answer is for counterclockwise 270 degrees rotation.

#5 Re: Help Me ! » Algebra 2 » 2017-08-01 23:09:19

This  is surely a home work problem.You must be knowing the slope of line joining two points in coordinate form. Equate it to the given value and find y.

#7 Re: Help Me ! » help! Geo » 2017-07-30 19:18:39

In a right angle triangle the vertices are on a circle with hypotenuse as diameter.The segment joining midpoint of hypotenuse to the opposite vertex is a radius. If the triangle is also isosceles this segment is perpendicular to the hypotenuse. So the area of the triangle is=1/2*hypotenuse *(hypotenuse/2).

#8 Re: Help Me ! » help! Geo » 2017-07-30 06:03:25

Do you visualize if hypotenuse is taken as base the height of the triangle equals half the length of hypotenuse?

#9 Re: Maths Is Fun - Suggestions and Comments » Partial Derivatives » 2017-06-19 21:41:16

Not from Bengaluru but from Kumta of Karwar district.

#10 Re: Help Me ! » Polynomial help » 2017-06-18 03:38:39

mathicINDIA wrote:

If 1 & -1 are the zeroes of the polynomial p(x)=Lx^4+Mx^3+Nx^2+Rx+P=0, prove that L+M+P=M+R=0.    [JEE mains 2014, AIEEE 2006, IIT 1998]

I think it should be
If 1 & -1 are the zeroes of the polynomial

, prove that

#11 Re: Help Me ! » Polynomial help » 2017-06-16 03:39:54

I am sorry, bob bundy. I did not mean that.You have had so many wonderful solutions.But when we are preoccupied with some problem, we can't differentiate between tea and coffee. I only hoped you could put x=1 instead of x-2 in your post.

#12 Re: Help Me ! » Polynomial help » 2017-06-15 19:15:07

The problem is just a hoax, simply to test presence of mind.

#13 Re: Help Me ! » Seating arrangements » 2017-06-13 23:33:19

I understand Samuel's point of view.The Chinese among themselves are indistinguishable. Similarly others.It is just like saying 5Cs,5Rs and 5Ms  are to be arranged in 15 slots with the given conditions. let us see in first 5 slots.
(1) 0 russian and 5 Mexicans can be seated in only 1 way. Now  seats 6 to 10 can accommodate 5 Chinese in 1 way and last 5 seats also 1 way. It is 1*1*1=1 way.
(2) 1 russian and 4 Mexicans can be seated in 5 different ways. so 5*5*5=125 ways.
(3)2 russian and 3 Mexicans can be seated in

  different ways. so totally 10*10*10=1000 ways. Continuuing this way we get 1+125+1000+1000+125+1=
ways.

#14 Re: Help Me ! » Seating arrangements » 2017-06-12 20:37:46

What is cooking? No response from any body including Samuel!

#15 Re: Help Me ! » Polynomial help » 2017-06-12 04:51:47

The sum of coefficients , obviously is

#16 Re: Help Me ! » How to calculate and express a number as the sum of three odd primes? » 2017-06-07 18:39:37

I do not know how you express an odd number as sum of 2 odd primes.Well. For 3 numbers you choose a prime ,say 3 and subtract. The result you can express as sum of 2 primes by your method.

#17 Re: Help Me ! » Seating arrangements » 2017-05-26 18:39:49

I was away from the forum for quite some time.Was busy with Quora. The problem looks very complicated but it is quite simple. Suppose you choose 3 Russians and 2 Mexicans for the first 5 seats,no more choice is left as 2 remaining russians have to go to last 5 and remaining Mexicans have to go to middle 5. the Chinese are also split likewise. after this selection in each group people could be arranged in 5! different ways.

#20 Re: Help Me ! » Circle Problems » 2017-05-05 16:50:57

(2) is simply arithmetic not even algebra.Can you find C1A,AC2,BC2and C1B?
for (3) simply ascertain the radii of the circles first and then put the equation.

#23 Re: Help Me ! » Conditional Probabilty Question » 2017-04-22 19:10:29

bob bundy wrote:

The correct formula is

If we assume that P(A) = P(B) then this reduces to thickhead's formula and the result 3/5 follows.

But the question does not say that a class is chosen first with equal probability.

If a West High precalculus student chosen at random

The classes are unequal so P(A) is not equal to P(B).

The correct calculation is

Bob

Just for fun let us change number of boys in each class.
Abel' class  boys=20  girls=12  total 32
Bonitz's class   boys=25  girls=10  total 35

#24 Re: Help Me ! » Conditional Probabilty Question » 2017-04-18 05:35:53

P(G/Abel)=12/20=3/5=0.6
P(G/Bonitz)=10/25=2/5=0.4
P(Abel/G)=P(G/Abel)/{P(G/abel)+P(G/Bonitz)}=0.6/(0.6+0.4)=0.6
This is established rule.

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