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Question Number 154258    Answers: 1   Comments: 0

Question Number 154255    Answers: 1   Comments: 0

(5/(3^2 .7^2 ))+(9/(7^2 .11^2 ))+((13)/(11^2 .15^2 ))+…=?

$$\:\frac{\mathrm{5}}{\mathrm{3}^{\mathrm{2}} .\mathrm{7}^{\mathrm{2}} }+\frac{\mathrm{9}}{\mathrm{7}^{\mathrm{2}} .\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{13}}{\mathrm{11}^{\mathrm{2}} .\mathrm{15}^{\mathrm{2}} }+\ldots=? \\ $$

Question Number 154254    Answers: 0   Comments: 0

form a PDE from Z(x,y)=xf_1 (x−y)+2xf_2 (2x+y)

$${form}\:{a}\:{PDE}\:{from}\:{Z}\left({x},{y}\right)={xf}_{\mathrm{1}} \left({x}−{y}\right)+\mathrm{2}{xf}_{\mathrm{2}} \left(\mathrm{2}{x}+{y}\right) \\ $$

Question Number 154252    Answers: 0   Comments: 0

The {x_n } sequence is specified by the conditions { ((x_1 =1982)),((x_(n+1) =(1/(4−x_n )),n≥0)) :} find lim_(n→∞) x_n .

$$\:\:{The}\:\left\{{x}_{{n}} \right\}\:{sequence}\:{is}\:{specified} \\ $$$$\:{by}\:{the}\:{conditions}\: \\ $$$$\:\:\begin{cases}{{x}_{\mathrm{1}} =\mathrm{1982}}\\{{x}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}−{x}_{{n}} },{n}\geqslant\mathrm{0}}\end{cases} \\ $$$$\:{find}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}_{{n}} . \\ $$

Question Number 154250    Answers: 0   Comments: 0

Question Number 154240    Answers: 1   Comments: 0

{ ((tg(a+b)=7)),((tg(a−b)=5)) :} tg(a)=? easy question

$$\begin{cases}{{tg}\left({a}+{b}\right)=\mathrm{7}}\\{{tg}\left({a}−{b}\right)=\mathrm{5}}\end{cases}\:\:{tg}\left({a}\right)=?\:\:\:{easy}\:{question} \\ $$

Question Number 154235    Answers: 1   Comments: 0

prove that: Im( ψ ( i ) )= (( 1)/( 2)) + (( π)/2) coth(π ) m.n

$$ \\ $$$$\:\:\:\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\mathrm{I}{m}\left(\:\psi\:\left(\:{i}\:\right)\:\right)=\:\frac{\:\mathrm{1}}{\:\mathrm{2}}\:+\:\frac{\:\pi}{\mathrm{2}}\:{coth}\left(\pi\:\right) \\ $$$$\:\:\:\:\:\:\:\:{m}.{n} \\ $$

Question Number 154226    Answers: 0   Comments: 0

Given i(t)=25cos(ωt) ; u(t)=50[1+(ωt)^(2/(2!)) +(ωt)^(4/(4!)) +(ωt)^(6/(6!)) +...] . (1/T)∫_0 ^T u(t)×i(t)dt=??? choose the correct answer: a. 225 b. 425 c. 625 d. an other one

$${Given}\:{i}\left({t}\right)=\mathrm{25}{cos}\left(\omega{t}\right)\:; \\ $$$${u}\left({t}\right)=\mathrm{50}\left[\mathrm{1}+\left(\omega{t}\right)^{\frac{\mathrm{2}}{\mathrm{2}!}} +\left(\omega{t}\right)^{\frac{\mathrm{4}}{\mathrm{4}!}} +\left(\omega{t}\right)^{\frac{\mathrm{6}}{\mathrm{6}!}} +...\right]\:. \\ $$$$\frac{\mathrm{1}}{{T}}\underset{\mathrm{0}} {\overset{{T}} {\int}}{u}\left({t}\right)×{i}\left({t}\right){dt}=??? \\ $$$${choose}\:{the}\:{correct}\:{answer}: \\ $$$${a}.\:\mathrm{225} \\ $$$${b}.\:\mathrm{425} \\ $$$${c}.\:\mathrm{625} \\ $$$${d}.\:{an}\:{other}\:{one} \\ $$

Question Number 154223    Answers: 0   Comments: 0

∫_0 ^( 1) (x^( i+1) /(1−x)) dx = [−ln(1−x)x^( i+1) ]_0 ^( 1) + (1+i)∫_0 ^( 1) x^( i) ln (1−x )dx = (1+i ) ∫_0 ^( 1) Σ_(n=1) ^∞ (x^( n+i) /n)dx = (1+i )Σ(1/(n (n+i+1 )))= = Σ(1/(n )) −(1/(n+i+1)) = γ + ψ ( i+2 ) = γ + (1/(1+i)) +ψ (1+i) γ + ((1−i)/2) + (1/i) + ψ (i ) im = −(3/2) + (1/2) +(π/2) coth(π ) −1 +(π/2) tan h(π ) ..✓

$$ \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{\:{i}+\mathrm{1}} }{\mathrm{1}−{x}}\:{dx}\:=\:\left[−{ln}\left(\mathrm{1}−{x}\right){x}^{\:{i}+\mathrm{1}} \right]_{\mathrm{0}} ^{\:\mathrm{1}} \\ $$$$\:\:\:+\:\left(\mathrm{1}+{i}\right)\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:{i}} \:{ln}\:\left(\mathrm{1}−{x}\:\right){dx} \\ $$$$\:\:\:=\:\left(\mathrm{1}+{i}\:\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\:{n}+{i}} }{{n}}{dx} \\ $$$$\:=\:\:\left(\mathrm{1}+{i}\:\right)\Sigma\frac{\mathrm{1}}{{n}\:\left({n}+{i}+\mathrm{1}\:\right)}= \\ $$$$\:=\:\Sigma\frac{\mathrm{1}}{{n}\:}\:−\frac{\mathrm{1}}{{n}+{i}+\mathrm{1}}\:=\:\gamma\:+\:\psi\:\left(\:{i}+\mathrm{2}\:\right) \\ $$$$\:\:\:=\:\gamma\:+\:\frac{\mathrm{1}}{\mathrm{1}+{i}}\:+\psi\:\left(\mathrm{1}+{i}\right) \\ $$$$\:\:\:\:\:\gamma\:+\:\frac{\mathrm{1}−{i}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{{i}}\:+\:\psi\:\left({i}\:\right) \\ $$$$\:\:\:{im}\:=\:−\frac{\mathrm{3}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{2}}\:{coth}\left(\pi\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:+\frac{\pi}{\mathrm{2}}\:{tan}\:{h}\left(\pi\:\right)\:..\checkmark \\ $$

Question Number 154225    Answers: 1   Comments: 1

In the expansion (1+(1/2)x)^n , the coefficient of x^m is (5/4) times the coefficient of x^(m+1) . a) Show that 5n−13m=8 b) If m and n are positive integers, such that m≤n, determine the smallest values of m and n.

$$\mathrm{In}\:\mathrm{the}\:\mathrm{expansion}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}\right)^{{n}} ,\:\mathrm{the}\:\mathrm{coefficient} \\ $$$$\mathrm{of}\:{x}^{{m}} \:\mathrm{is}\:\frac{\mathrm{5}}{\mathrm{4}}\:\mathrm{times}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{m}+\mathrm{1}} . \\ $$$$\left.{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{5}{n}−\mathrm{13}{m}=\mathrm{8} \\ $$$$\left.{b}\right)\:\mathrm{If}\:{m}\:\mathrm{and}\:{n}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers},\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\:\:{m}\leqslant{n},\:\mathrm{determine}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{values}\:\mathrm{of} \\ $$$$\:\:\:\:\:{m}\:\mathrm{and}\:{n}. \\ $$

Question Number 154216    Answers: 3   Comments: 0

Solve the equation for reals ((cos^2 x−(1/5)sin^2 x+1)/(sin^2 x−(1/5)cos^2 x+1)) = 5tan^2 x

$${Solve}\:{the}\:{equation}\:{for}\:{reals}\: \\ $$$$\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{1}}\:=\:\mathrm{5tan}\:^{\mathrm{2}} {x}\: \\ $$

Question Number 154208    Answers: 0   Comments: 0

∵∴∵∴prove that ∫_0 ^∞ ∫_0 ^∞ ((log(1−e^(−x) )(yLi_2 (e^(−x−y) )+Li_3 (e^(−x−y) ))/(1−e^(x+y) ))e^(x+y) dxdy=((21)/8)ζ(6)+ζ^2 (3) ∵∴∵by MATH.AMIN∴∵∴

$$\because\therefore\because\therefore{prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{{log}\left(\mathrm{1}−{e}^{−{x}} \right)\left({yLi}_{\mathrm{2}} \left({e}^{−{x}−{y}} \right)+{Li}_{\mathrm{3}} \left({e}^{−{x}−{y}} \right)\right.}{\mathrm{1}−{e}^{{x}+{y}} }{e}^{{x}+{y}} {dxdy}=\frac{\mathrm{21}}{\mathrm{8}}\zeta\left(\mathrm{6}\right)+\zeta^{\mathrm{2}} \left(\mathrm{3}\right) \\ $$$$\because\therefore\because\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{MATH}}.\boldsymbol{\mathrm{AMIN}}\therefore\because\therefore \\ $$

Question Number 154202    Answers: 1   Comments: 0

𝛀 =∫_( 0) ^( ∞) ((ln∙(1 + x))/(x∙(x^2 + x + 1))) dx = ?

$$\boldsymbol{\Omega}\:\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{ln}\centerdot\left(\mathrm{1}\:+\:\mathrm{x}\right)}{\mathrm{x}\centerdot\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\right)}\:\mathrm{dx}\:=\:? \\ $$

Question Number 154194    Answers: 0   Comments: 1

Question Number 154195    Answers: 3   Comments: 0

∫((x+sinx)/(1+cosx))dx please,help me

$$\int\frac{{x}+{sinx}}{\mathrm{1}+{cosx}}{dx}\:\:\:\: \\ $$$${please},{help}\:{me} \\ $$

Question Number 154192    Answers: 1   Comments: 0

Ω :=∫_0 ^( 1) (( x.sin(ln(x)))/(1−x))dx method 1 Ω= Im[∫_0 ^( 1) (( x^( i+1) )/(1−x)) dx=Φ] Φ = ∫_0 ^( 1) (( x^( i+1) +x^( i+2) )/(1−x^( 2) ))dx =^(x^( 2) =t) (1/2)∫_0 ^( 1) (( t^( (i/2)) −t^((i+1)/2) )/(1−t))dt = (1/2) { ψ (1 +((i+1)/2))−ψ (1+(i/2))} = (1/2) { (2/(1+i)) + ψ ( ((1+i)/2) )−(2/i)−ψ ((i/2))} = ((−i)/(−1+i)) + (1/2) {ψ (((1+i)/2))−ψ((i/2) )} = −(1/2) +(i/(2 )) +(1/2) {ψ(((1+i)/2))−ψ((i/2))} Ω = Im (Φ )= (1/2) +(1/2) Im(ψ((1/2) +(i/2)))−(1/2) Im((i/2)) = (1/(2 )) { 1 +(π/2) tanh((π/2)) −1−(π/2) coth((π/2))} =(π/4) {((−1)/(sinh((π/2)).cosh((π/2))))}=((−π)/(2 sinh (π ))) ✓

$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\Omega\::=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}.{sin}\left({ln}\left({x}\right)\right)}{\mathrm{1}−{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:{method}\:\mathrm{1} \\ $$$$\:\:\:\:\:\Omega=\:\mathrm{I}{m}\left[\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\:{x}^{\:{i}+\mathrm{1}} }{\mathrm{1}−{x}}\:{dx}=\Phi\right] \\ $$$$\:\:\:\:\:\:\:\:\:\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}^{\:{i}+\mathrm{1}} +{x}^{\:{i}+\mathrm{2}} }{\mathrm{1}−{x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{{x}^{\:\mathrm{2}} ={t}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{t}^{\:\frac{{i}}{\mathrm{2}}} −{t}^{\frac{{i}+\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}−{t}}{dt} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\:\psi\:\left(\mathrm{1}\:+\frac{{i}+\mathrm{1}}{\mathrm{2}}\right)−\psi\:\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\:\:\frac{\mathrm{2}}{\mathrm{1}+{i}}\:+\:\psi\:\left(\:\frac{\mathrm{1}+{i}}{\mathrm{2}}\:\right)−\frac{\mathrm{2}}{{i}}−\psi\:\left(\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:=\:\frac{−{i}}{−\mathrm{1}+{i}}\:\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\psi\:\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−\psi\left(\frac{{i}}{\mathrm{2}}\:\right)\right\} \\ $$$$\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}\:}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−\psi\left(\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\Omega\:=\:\mathrm{I}{m}\:\left(\Phi\:\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{I}{m}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{I}{m}\left(\frac{{i}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}\:}\:\left\{\:\mathrm{1}\:+\frac{\pi}{\mathrm{2}}\:{tanh}\left(\frac{\pi}{\mathrm{2}}\right)\:−\mathrm{1}−\frac{\pi}{\mathrm{2}}\:{coth}\left(\frac{\pi}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\:\:\left\{\frac{−\mathrm{1}}{{sinh}\left(\frac{\pi}{\mathrm{2}}\right).{cosh}\left(\frac{\pi}{\mathrm{2}}\right)}\right\}=\frac{−\pi}{\mathrm{2}\:{sinh}\:\left(\pi\:\right)}\:\checkmark \\ $$$$ \\ $$

Question Number 154186    Answers: 1   Comments: 0

Question Number 154177    Answers: 1   Comments: 2

Question Number 154175    Answers: 2   Comments: 1

Question Number 154172    Answers: 0   Comments: 1

Question Number 154148    Answers: 1   Comments: 0

z^4 - ((50)/(2z^4 - 7)) = 14 ⇒ z = ?

$$\boldsymbol{{z}}^{\mathrm{4}} \:-\:\frac{\mathrm{50}}{\mathrm{2}\boldsymbol{{z}}^{\mathrm{4}} \:-\:\mathrm{7}}\:=\:\mathrm{14}\:\:\:\Rightarrow\:\:\:\boldsymbol{{z}}\:=\:? \\ $$

Question Number 154143    Answers: 2   Comments: 0

Question Number 154142    Answers: 2   Comments: 0

Question Number 154138    Answers: 1   Comments: 0

Question Number 154133    Answers: 2   Comments: 0

Given f(x)=((x+(√(1+x^2 ))))^(1/3) +((x−(√(1+x^2 ))))^(1/3) Find f^(−1) (x)=?

$$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{3}}]{\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}+\sqrt[{\mathrm{3}}]{\mathrm{x}−\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{Find}\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=? \\ $$

Question Number 154131    Answers: 0   Comments: 1

A long distance runner runs 14km 30°north of east, 7km 60° north of west, 6km 30° south of west, and finally 4km south. Find his final distance and direction relative to the starting point?

$$ \\ $$A long distance runner runs 14km 30°north of east, 7km 60° north of west, 6km 30° south of west, and finally 4km south. Find his final distance and direction relative to the starting point?

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