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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

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

Question Number 14507 Answers: 0 Comments: 0

Question Number 14502 Answers: 1 Comments: 3

Question Number 14491 Answers: 1 Comments: 0

(√(25))

$$\sqrt{\mathrm{25}} \\ $$$$ \\ $$

Question Number 14486 Answers: 0 Comments: 0

S=1−2+3−4+... ∴S=Σ_(n=1) ^∞ (−1)^(n+1) n S=lim_(s→0) (Σ_(n=1) ^∞ (−1)^(n+1) n^(1−s) ) Prove S=(1/4)

$${S}=\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+... \\ $$$$\therefore{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n} \\ $$$$\: \\ $$$${S}=\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n}^{\mathrm{1}−{s}} \right) \\ $$$$\: \\ $$$$\mathrm{Prove}\:{S}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Question Number 14483 Answers: 0 Comments: 8

x^y +y^x =3.....(1) x+y=3.....(2) solve the equation

$$\mathrm{x}^{\mathrm{y}} +\mathrm{y}^{\mathrm{x}} =\mathrm{3}.....\left(\mathrm{1}\right) \\ $$$$\mathrm{x}+\mathrm{y}=\mathrm{3}.....\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$

Question Number 14481 Answers: 0 Comments: 0

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