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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.
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.
I overlooked "clockwise". My answer is for counterclockwise 270 degrees rotation.
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.
Nice.
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).
Do you visualize if hypotenuse is taken as base the height of the triangle equals half the length of hypotenuse?
Not from Bengaluru but from Kumta of Karwar district.
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
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.
The problem is just a hoax, simply to test presence of mind.
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
What is cooking? No response from any body including Samuel!
The sum of coefficients , obviously is
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.
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.
4 times hotter is meaningless.
(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.
My answer for EF is
Due to symmetry.
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
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.
(1) My answer is 4/3.