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Question Number 14882 Answers: 0 Comments: 6

Two vectors a^→ and b^→ are parallel and have same magnitude. Then they (1) have same direction, but they are not equal (2) are equal (3) are not equal (4) may or may not be equal

$$\mathrm{Two}\:\mathrm{vectors}\:\overset{\rightarrow} {{a}}\:\mathrm{and}\:\overset{\rightarrow} {{b}}\:\mathrm{are}\:\mathrm{parallel}\:\mathrm{and} \\ $$$$\mathrm{have}\:\mathrm{same}\:\mathrm{magnitude}.\:\mathrm{Then}\:\mathrm{they} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{have}\:\mathrm{same}\:\mathrm{direction},\:\mathrm{but}\:\mathrm{they}\:\mathrm{are} \\ $$$$\mathrm{not}\:\mathrm{equal} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{are}\:\mathrm{equal} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{are}\:\mathrm{not}\:\mathrm{equal} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{may}\:\mathrm{or}\:\mathrm{may}\:\mathrm{not}\:\mathrm{be}\:\mathrm{equal} \\ $$

Question Number 14879 Answers: 1 Comments: 0

Find the value of x, satisfying sin^2 x + sin x − 2 ≥ 0

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x},\:\mathrm{satisfying} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}\:+\:\mathrm{sin}\:{x}\:−\:\mathrm{2}\:\geqslant\:\mathrm{0} \\ $$

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

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