Archive for category Maths

Ultimate concept 14(binomial theorem)

Note that

For |x|<1

Now for |x|<1, we have

By using the identity and equating coefficients we can solve several problems of sequence and binomial.

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Challenge 49(permutation combination)

See ultimate concept 13

The number of ways 3 children can distribute 10 tickets

out of 15 consecutively numbered tickets among themselves

such that they get consecutive blocks of 5, 3, and 2 ticket is:

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Challenge 55(inverse trigonometry)

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Challenge 54(differential equation)

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Challenge 52(definite integration)

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Challenge 51(basic algebra)

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Challenge 50(sequence)

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Ultimate concept 13(solution of triangle)

From a point M inside an equilateral triangle ABC perpendiculars MP, MQ, MR are

dropped to sides BC, CA, AB respectively.

Then

Proof:

By pythagores theorem

 

 

 

 

 

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Ultimate concept 12(solution of triangle)

A triangle cuts off from the circum circle three circular segments.

The largest altitudes of the segments with in radius and circum radius

of the triangle has relation

Where K, L, M are the length of the largest altitudes and R, r have usual meaning.

Proof:

Let O be the circum centre of the triangle. Let D be the midpoint of BC

OD=RcosA

K=(R-RcosA)

.

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Challenge 7

P(-3,-4) is a point on circle and let circle = intersect it at Q,R. PQ and PR intersect at A and B. find slope of line AB.

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