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Question Number 14878 Answers: 0 Comments: 3

Find the number of solution of equation x sin x = 2

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{equation} \\ $$$${x}\:\mathrm{sin}\:{x}\:=\:\mathrm{2} \\ $$

Question Number 14876 Answers: 0 Comments: 0

Question Number 14614 Answers: 1 Comments: 2

If 5 doesn′t divide any of n,n+1, n+2,n+3 then prove that n(n+1)(n+2)(n+3)≡24(mod100)

$$\mathrm{If}\:\:\mathrm{5}\:\:\mathrm{doesn}'\mathrm{t}\:\mathrm{divide}\:\mathrm{any}\:\mathrm{of}\:\mathrm{n},\mathrm{n}+\mathrm{1}, \\ $$$$\mathrm{n}+\mathrm{2},\mathrm{n}+\mathrm{3}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)\equiv\mathrm{24}\left(\mathrm{mod100}\right) \\ $$

Question Number 14758 Answers: 2 Comments: 4

Solve tan x + tan 2x + tan 3x = 0

$$\mathrm{Solve}\:\mathrm{tan}\:{x}\:+\:\mathrm{tan}\:\mathrm{2}{x}\:+\:\mathrm{tan}\:\mathrm{3}{x}\:=\:\mathrm{0} \\ $$

Question Number 14757 Answers: 1 Comments: 0

Solve: ((1000!)/(5×10×15×...1000))≡x(mod 10)

$$\mathrm{Solve}: \\ $$$$\frac{\mathrm{1000}!}{\mathrm{5}×\mathrm{10}×\mathrm{15}×...\mathrm{1000}}\equiv\mathrm{x}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$

Question Number 14759 Answers: 0 Comments: 2

Where is 123456 ? S/He was most senior of us and had a great knoledge of maths! S/He used to guide us when we were wrong.

$$\:\:\:\:\:\:\:\:\mathrm{Where}\:\mathrm{is} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{123456} \\ $$$$\:\:\:\:\:\:\:\:\:? \\ $$$$\mathrm{S}/\mathrm{He}\:\mathrm{was}\:\mathrm{most}\:\mathrm{senior}\:\mathrm{of}\:\mathrm{us}\: \\ $$$$\mathrm{and} \\ $$$$\mathrm{had}\:\mathrm{a}\:\mathrm{great}\:\mathrm{knoledge}\:\mathrm{of}\:\mathrm{maths}! \\ $$$$\mathrm{S}/\mathrm{He}\:\mathrm{used}\:\mathrm{to}\:\mathrm{guide}\:\mathrm{us}\:\mathrm{when} \\ $$$$\mathrm{we}\:\mathrm{were}\:\mathrm{wrong}. \\ $$$$ \\ $$

Question Number 14596 Answers: 0 Comments: 3

Question Number 14594 Answers: 1 Comments: 0

Question Number 14593 Answers: 1 Comments: 0

Question Number 14592 Answers: 0 Comments: 2

Question Number 14591 Answers: 1 Comments: 0

Question Number 14590 Answers: 1 Comments: 0

Question Number 14572 Answers: 0 Comments: 1

The number of possible pairs of successive prime numbers such that each of them is greater than 40 and their sum is atmost 100 is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{possible}\:\mathrm{pairs}\:\mathrm{of}\: \\ $$$$\mathrm{successive}\:\mathrm{prime}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{each}\:\mathrm{of}\:\mathrm{them}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{40}\:\mathrm{and}\: \\ $$$$\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{atmost}\:\mathrm{100}\:\mathrm{is} \\ $$

Question Number 14588 Answers: 1 Comments: 0

Determine the fourth roots of − 16, giving the results in the form a + jb.

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{fourth}\:\mathrm{roots}\:\mathrm{of}\:\:\:−\:\mathrm{16},\:\:\:\mathrm{giving}\:\mathrm{the}\:\mathrm{results}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\:\mathrm{a}\:+\:\mathrm{jb}. \\ $$

Question Number 14587 Answers: 1 Comments: 0

Determine the roots of the equation x^3 + 64 = 0 in the polar form a + jb, Where a and b are real.

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{64}\:=\:\mathrm{0}\:\:\mathrm{in}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{form}\:\:\mathrm{a}\:+\:\mathrm{jb}, \\ $$$$\mathrm{Where}\:\:\mathrm{a}\:\:\mathrm{and}\:\:\mathrm{b}\:\:\mathrm{are}\:\mathrm{real}. \\ $$

Question Number 14564 Answers: 0 Comments: 9

What is the last 2 digits of 2^(613)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{last}\:\mathrm{2}\:\mathrm{digits}\:\mathrm{of}\:\:\:\:\:\:\mathrm{2}^{\mathrm{613}} \\ $$

Question Number 14560 Answers: 0 Comments: 0

Question Number 14559 Answers: 1 Comments: 0

Solve for x ((6x + 2a + 3b + c )/(6x + 2a − 3b − c)) = ((2x + 6a + b + 3c)/(2x + 6a − b − 3c))

$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x} \\ $$$$\frac{\mathrm{6x}\:+\:\mathrm{2a}\:+\:\mathrm{3b}\:+\:\mathrm{c}\:}{\mathrm{6x}\:+\:\mathrm{2a}\:−\:\mathrm{3b}\:−\:\mathrm{c}}\:=\:\frac{\mathrm{2x}\:+\:\mathrm{6a}\:+\:\mathrm{b}\:+\:\mathrm{3c}}{\mathrm{2x}\:+\:\mathrm{6a}\:−\:\mathrm{b}\:−\:\mathrm{3c}} \\ $$

Question Number 14544 Answers: 0 Comments: 0

Question Number 14543 Answers: 2 Comments: 1

An open box of area 486cm^2 .If the length is twice the breadth.Find the maximum volume of the box. hence,Show the volume is maximum.

$$\mathrm{An}\:\mathrm{open}\:\mathrm{box}\:\mathrm{of}\:\mathrm{area}\:\mathrm{486cm}^{\mathrm{2}} .\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{length}\:\mathrm{is}\:\mathrm{twice}\:\mathrm{the}\:\mathrm{breadth}.\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{box}. \\ $$$$\mathrm{hence},\mathrm{Show}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{is}\:\mathrm{maximum}. \\ $$

Question Number 14541 Answers: 0 Comments: 0

Question Number 14535 Answers: 2 Comments: 6

Question Number 14523 Answers: 2 Comments: 0

Question Number 14521 Answers: 0 Comments: 0

Question Number 14513 Answers: 2 Comments: 0

If y^2 (1 + x^2 ) = 1 − x^2 Show that, ((dy/dx))^2 = ((1 − y^4 )/(1 − x^4 ))

$$\mathrm{If}\:\:\mathrm{y}^{\mathrm{2}} \left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)\:=\:\mathrm{1}\:−\:\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\:\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{1}\:−\:\mathrm{y}^{\mathrm{4}} }{\mathrm{1}\:−\:\mathrm{x}^{\mathrm{4}} } \\ $$

Question Number 14510 Answers: 0 Comments: 0

∫ (1/((x^3 − 1)^3 )) dx

$$\int\:\:\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{dx} \\ $$

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