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Tutor profile: Apratim M.

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Apratim M.
Software engineer in Infosys
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Questions

Subject:Trigonometry

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Question:

An airplane is approaching point A along a straight line and at a constant altitude h. At 10:00 am, the angle of elevation of the airplane is 20 degree and at 10:01 it is 60 degree. What is the altitude h of the airplane if the speed of the airplane is constant and equal to 600 miles/hour? (Round answer to 2 decimal places).

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Apratim M.

We first calculate distance d using the time and speed (1 minute = 1/60 hour) d = 600 * (1 / 60) = 10 miles We next express the tangent of the given angles of elevation as follows tan(20) = h / (d + x) and tan(60) = h / x Eliminate x in the two equations above to find a relationship between h and d h = d / [ 1 / tan(20) - 1 / tan(60) ] = 4.6 miles (rounded to 2 decimal places)

Subject:Basic Math

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Question:

Does numbers interest you then solve this I have ten boxes which I want to pack into crates. Each crate can carry a maximum of 25 kg. But I only have three crates, and the total weight of the boxes is 75kg: 15 kg, 13kg, 11 kg, 10 kg, 9 kg, 8 kg, 4 kg, 2 kg, 2kg, 1 kg How can I pack the boxes into the crates?

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Apratim M.

{Crate 1}, {Crate 2}, {Crate 3} {15,10}, {13,8,4}, {11,9,2,2,1} {15,10}, {13,11,1}, {9,8,4,2,2} {15,10}, {11,8,4,2}, {13,9,2,1} {15,10}, {11,9,4,1}, {13,8,2,2} {11,10,4}, {15,8,2}, {13,9,2,1} {11,10,4}, {15,9,1}, {13,8,2,2} {13,8,4}, {15,9,1}, {11,10,2,2} {13,10,2}, {15,8,2}, {11,9,4,1} {13,10,2}, {15,9,1}, {11,8,4,2} {13,11,1}, {15,8,2}, {10,9,4,2}

Subject:Number Theory

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Question:

As we all know one of the unique set of numbers is the prime numbers. The frequency of prime numbers decreases as we go higher. For e.g from 1-100 number of prime numbers is 25, from 1-1000 it is 168 and from 1-10000 it is 1229. So it there any highest prime number after which Primes cease to exist?

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Apratim M.

If you are looking for an answer it is no. for e.g let P be the largest Prime number after which Primes cease to exist. But if we multiply all the prime numbers before P and add 1 to it it still becomes a prime number. Let N be the new prime number. N= 2*3*5*7*11.......P+1 so N is now the largest prime. and hence there is no such thing as the largest prime number.

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