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Question Number 211812    Answers: 1   Comments: 1

prove lim_(x→∞) ( 1 + (5/x) )^(1/x) − 1 = 5

$${prove}\:\underset{{x}\rightarrow\infty} {{lim}}\:\left(\:\mathrm{1}\:+\:\frac{\mathrm{5}}{{x}}\:\right)^{\frac{\mathrm{1}}{{x}}} −\:\mathrm{1}\:=\:\mathrm{5}\: \\ $$

Question Number 198688    Answers: 1   Comments: 0

y′′ + (2/x).y′ + y = 0 y=¿

$${y}''\:+\:\frac{\mathrm{2}}{{x}}.{y}'\:+\:{y}\:=\:\mathrm{0} \\ $$$${y}=¿ \\ $$

Question Number 198684    Answers: 3   Comments: 0

ζ

$$\:\cancel{\zeta} \\ $$

Question Number 198604    Answers: 1   Comments: 1

Question Number 198692    Answers: 0   Comments: 0

Question Number 198695    Answers: 1   Comments: 0

∫_0 ^∞ ((e^(at) −e^(−at) )/(e^(πt) −e^(−πt) )) dt = ??

$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\mathrm{at}} −\mathrm{e}^{−\mathrm{at}} }{\mathrm{e}^{\pi\mathrm{t}} −\mathrm{e}^{−\pi\mathrm{t}} }\:\mathrm{dt}\:\:\:=\:\:\:?? \\ $$

Question Number 198647    Answers: 3   Comments: 1

Question Number 198646    Answers: 1   Comments: 0

Question Number 198576    Answers: 1   Comments: 10

How many numbers with a maximum of 5 digits, greater than 4000, can be formed with the digits 2, 3, 4, 5, 6; if repetition is allowed for 2 and 3 only?

How many numbers with a maximum of 5 digits, greater than 4000, can be formed with the digits 2, 3, 4, 5, 6; if repetition is allowed for 2 and 3 only?

Question Number 198575    Answers: 1   Comments: 3

Question Number 198572    Answers: 1   Comments: 0

lim_(x→(π/(2n))) ((√((sin 2nx)/(1+cos nx)))/(4n^2 x^2 −π^2 )) =?

$$\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2n}}} {\mathrm{lim}}\:\frac{\sqrt{\frac{\mathrm{sin}\:\mathrm{2nx}}{\mathrm{1}+\mathrm{cos}\:\mathrm{nx}}}}{\mathrm{4n}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} −\pi^{\mathrm{2}} }\:=? \\ $$

Question Number 198566    Answers: 3   Comments: 0

((3x^2 −1)/x)+((5x)/(3x^2 −x−1))=((119)/(18))

$$\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}}{{x}}+\frac{\mathrm{5}{x}}{\mathrm{3}{x}^{\mathrm{2}} −{x}−\mathrm{1}}=\frac{\mathrm{119}}{\mathrm{18}} \\ $$

Question Number 198565    Answers: 0   Comments: 0

Let a,b,c ∈ R^+ such that a+b+c=1. Prove that: b+c≥16abc

$${Let}\:{a},{b},{c}\:\in\:{R}^{+} \:{such}\:{that}\:{a}+{b}+{c}=\mathrm{1}. \\ $$$$\:{Prove}\:{that}:\:\:\:\:{b}+{c}\geqslant\mathrm{16}{abc} \\ $$

Question Number 198563    Answers: 1   Comments: 0

find all numbers (with any number of digits) satisfying (abcd...xyz)×2=(zyx...dcba)

$${find}\:{all}\:{numbers}\:\left({with}\:{any}\:{number}\right. \\ $$$$\left.{of}\:{digits}\right)\:{satisfying} \\ $$$$\left(\underline{{abcd}...{xyz}}\right)×\mathrm{2}=\left(\underline{{zyx}...{dcba}}\right) \\ $$

Question Number 198562    Answers: 1   Comments: 0

f(tan x)+ 2f(cot x) = 4x f ′(x)= ?

$$\:\:\:\mathrm{f}\left(\mathrm{tan}\:\mathrm{x}\right)+\:\mathrm{2f}\left(\mathrm{cot}\:\mathrm{x}\right)\:=\:\mathrm{4x}\: \\ $$$$\:\:\:\mathrm{f}\:'\left(\mathrm{x}\right)=\:? \\ $$

Question Number 198559    Answers: 0   Comments: 0

Question Number 198558    Answers: 1   Comments: 1

Question Number 198555    Answers: 1   Comments: 2

Question Number 198624    Answers: 0   Comments: 1

Post have been cleanup for users and access blocked

$$\mathrm{Post}\:\mathrm{have}\:\mathrm{been}\:\mathrm{cleanup}\:\mathrm{for}\:\mathrm{users} \\ $$$$\mathrm{and}\:\mathrm{access}\:\mathrm{blocked} \\ $$

Question Number 198644    Answers: 2   Comments: 2

Question Number 198643    Answers: 0   Comments: 0

Given an isosceles triangle ABC which ∠A= 30°, AB = AC. A point D is midpoint of BC . A point P is chosen on then segment AD and a point Q is chosen on the side AB so that BP= PQ. Find the angle PQC

$$ \\ $$$$\mathrm{Given}\:\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle}\:\mathrm{ABC} \\ $$$$\:\mathrm{which}\:\:\angle\mathrm{A}=\:\mathrm{30}°,\:\mathrm{AB}\:=\:\mathrm{AC}.\: \\ $$$$\mathrm{A}\:\mathrm{point}\:\mathrm{D}\:\mathrm{is}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{BC}\:.\: \\ $$$$\mathrm{A}\:\mathrm{point}\:\mathrm{P}\:\mathrm{is}\:\mathrm{chosen}\:\mathrm{on}\:\mathrm{then} \\ $$$$\mathrm{segment}\:\mathrm{AD}\:\mathrm{and}\:\mathrm{a}\:\mathrm{point}\:\mathrm{Q}\:\mathrm{is} \\ $$$$\mathrm{chosen}\:\mathrm{on}\:\mathrm{the}\:\mathrm{side}\:\mathrm{AB}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{BP}=\:\mathrm{PQ}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{PQC} \\ $$

Question Number 198519    Answers: 1   Comments: 0

Find a,b,c and d such that { ((abc = 1 (mod 11))),((abd = 2 (mod 11) )),((acd = 3 (mod 11) )),((bcd = 4 (mod 11) )) :}

$$\:\:\mathrm{Find}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{and}\:\mathrm{d}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{abc}\:=\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{11}\right)}\\{\mathrm{abd}\:=\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\\{\mathrm{acd}\:=\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\\{\mathrm{bcd}\:=\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\end{cases} \\ $$

Question Number 198517    Answers: 0   Comments: 0

Question Number 198516    Answers: 0   Comments: 2

To administrator Tinku Tara sir: please have a look at what the members Hridiana and HomeAlone are doing in the forum! They are rioting! Please ban them from the forum!

$${To}\:{administrator}\:{Tinku}\:{Tara}\:{sir}: \\ $$$${please}\:{have}\:{a}\:{look}\:{at}\:{what}\:{the}\: \\ $$$${members}\:\underline{{Hridiana}}\:{and}\:\underline{{HomeAlone}} \\ $$$${are}\:{doing}\:{in}\:{the}\:{forum}! \\ $$$${They}\:{are}\:{rioting}! \\ $$$${Please}\:{ban}\:{them}\:{from}\:{the}\:{forum}! \\ $$

Question Number 198497    Answers: 0   Comments: 0

∫_0 ^∞ ((e^(at) −e^(−at) )/(e^(πt) −e^(−πt) )) dt

$$\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\mathrm{at}} −\mathrm{e}^{−\mathrm{at}} }{\mathrm{e}^{\pi\mathrm{t}} −\mathrm{e}^{−\pi\mathrm{t}} }\:\mathrm{dt} \\ $$

Question Number 198496    Answers: 1   Comments: 0

A nice series If , Ω = Σ_(n=2) ^∞ (( 1)/(n^( 2) +n −1)) =(( π tan( aπ ))/( b)) ⇒ find the value of (b/a) = ?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:{A}\:{nice}\:\:{series} \\ $$$$\:\:{If}\:,\:\:\Omega\:=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\:\mathrm{1}}{{n}^{\:\mathrm{2}} \:+{n}\:−\mathrm{1}}\:=\frac{\:\pi\:{tan}\left(\:{a}\pi\:\right)}{\:{b}} \\ $$$$\:\:\:\:\:\Rightarrow\:{find}\:{the}\:{value}\:{of}\:\:\:\frac{{b}}{{a}}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$

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