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AllQuestion and Answers: Page 1080

Question Number 98768    Answers: 2   Comments: 0

lim_(x→0) (((√(x+1)) sin x+ln(1+x^2 )−x)/(((1+x^2 ))^(1/(3 )) −1))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}+\mathrm{1}}\:\mathrm{sin}\:\mathrm{x}+\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\mathrm{x}}{\sqrt[{\mathrm{3}\:\:}]{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{1}} \\ $$

Question Number 98761    Answers: 0   Comments: 5

(√(x+(√x) )) −(√(x−(√x))) = m(√(x/(x+(√x)))) m is a real parameter

$$\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}\:}\:−\sqrt{\mathrm{x}−\sqrt{\mathrm{x}}}\:=\:\mathrm{m}\sqrt{\frac{\mathrm{x}}{\mathrm{x}+\sqrt{\mathrm{x}}}} \\ $$$$\mathrm{m}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{parameter} \\ $$

Question Number 98750    Answers: 0   Comments: 0

Question Number 98744    Answers: 0   Comments: 2

∫_0 ^π ∫_0 ^(2sinθ) (1+rsinθ)r dr dθ

$$\int_{\mathrm{0}} ^{\pi} \int_{\mathrm{0}} ^{\mathrm{2}{sin}\theta} \left(\mathrm{1}+{rsin}\theta\right){r}\:{dr}\:{d}\theta \\ $$

Question Number 98728    Answers: 0   Comments: 5

Currently working on enhancing this app to draw shapes. So posting a math problem realted to drawing.^ Ref. Frame1 X-Y Frame 2: Axis translated by (h,k) and rotated about point (u,v). Consider a point (x_1 ,y_1 ) on X−Y axis. 1. What will be the postion of the point on X−Y axis after it is translated and plotted in frame 2. 2. A point is moved by distance dx,dy in X−Y. How much distane will it moved in the new frame.

$$\mathrm{Currently}\:\mathrm{working}\:\mathrm{on}\:\mathrm{enhancing} \\ $$$$\mathrm{this}\:\mathrm{app}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{shapes}. \\ $$$$\mathrm{So}\:\mathrm{posting}\:\mathrm{a}\:\:\mathrm{math}\:\mathrm{problem}\:\mathrm{realted} \\ $$$$\mathrm{to}\:\mathrm{drawing}\bar {.} \\ $$$$\mathrm{Ref}.\:\mathrm{Frame1}\:\mathrm{X}-\mathrm{Y} \\ $$$$\mathrm{Frame}\:\mathrm{2}: \\ $$$$\mathrm{Axis}\:\mathrm{translated}\:\mathrm{by}\:\left({h},{k}\right)\:\mathrm{and} \\ $$$$\mathrm{rotated}\:\mathrm{about}\:\mathrm{point}\:\left({u},{v}\right). \\ $$$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{point}\:\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:\mathrm{on}\:\mathrm{X}−\mathrm{Y}\:\mathrm{axis}. \\ $$$$\mathrm{1}.\:\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{postion}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{X}−\mathrm{Y}\:\mathrm{axis}\:\mathrm{after}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{translated}\:\mathrm{and}\:\mathrm{plotted}\:\mathrm{in}\:\mathrm{frame}\:\mathrm{2}. \\ $$$$\mathrm{2}.\:\mathrm{A}\:\mathrm{point}\:\mathrm{is}\:\mathrm{moved}\:\mathrm{by}\:\mathrm{distance} \\ $$$${dx},{dy}\:\mathrm{in}\:\mathrm{X}−\mathrm{Y}.\:\mathrm{How}\:\mathrm{much}\:\mathrm{distane} \\ $$$$\mathrm{will}\:\mathrm{it}\:\mathrm{moved}\:\mathrm{in}\:\mathrm{the}\:\mathrm{new}\:\mathrm{frame}. \\ $$

Question Number 98723    Answers: 0   Comments: 4

Version 2.084 has fixes for all crashes which were reported on playstore in last week. If anyone is facing crashes please update to v2.084 and see if the problem is solved. If the problem is still present, please send us an email as the problem may be specifuc to device model.

$$\mathrm{Version}\:\mathrm{2}.\mathrm{084}\:\mathrm{has}\:\mathrm{fixes}\:\mathrm{for}\:\mathrm{all} \\ $$$$\mathrm{crashes}\:\mathrm{which}\:\mathrm{were}\:\mathrm{reported}\:\mathrm{on} \\ $$$$\mathrm{playstore}\:\mathrm{in}\:\mathrm{last}\:\mathrm{week}. \\ $$$$\mathrm{If}\:\mathrm{anyone}\:\mathrm{is}\:\mathrm{facing}\:\mathrm{crashes}\:\mathrm{please} \\ $$$$\mathrm{update}\:\mathrm{to}\:\mathrm{v2}.\mathrm{084}\:\mathrm{and}\:\mathrm{see}\:\mathrm{if}\:\mathrm{the}\: \\ $$$$\mathrm{problem}\:\mathrm{is}\:\mathrm{solved}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{still}\:\mathrm{present},\:\mathrm{please} \\ $$$$\mathrm{send}\:\mathrm{us}\:\mathrm{an}\:\mathrm{email}\:\mathrm{as}\:\mathrm{the}\:\mathrm{problem} \\ $$$$\mathrm{may}\:\mathrm{be}\:\mathrm{specifuc}\:\mathrm{to}\:\mathrm{device}\:\mathrm{model}. \\ $$

Question Number 98722    Answers: 3   Comments: 0

let f(x) =arctan((3/x)) 1) calculste f^((n)) (x) and f^((n)) (1) 2) developp f at integr seri at point x_0 =1

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculste}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{seri}\:\mathrm{at}\:\mathrm{point}\:\mathrm{x}_{\mathrm{0}} =\mathrm{1} \\ $$

Question Number 98721    Answers: 2   Comments: 0

calculate ∫_0 ^∞ (dx/(x^4 +x^2 +1)) 1) by using residue theorem 2) by using complex decomposition

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{by}\:\mathrm{using}\:\mathrm{residue}\:\mathrm{theorem} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{by}\:\mathrm{using}\:\mathrm{complex}\:\mathrm{decomposition} \\ $$

Question Number 98713    Answers: 2   Comments: 1

∫((sin(x))/x)dx

$$\int\frac{{sin}\left({x}\right)}{{x}}{dx} \\ $$$$ \\ $$

Question Number 98690    Answers: 1   Comments: 3

Question Number 98687    Answers: 1   Comments: 4

Question Number 98679    Answers: 1   Comments: 2

prove that ∫_0 ^∞ ((3+2(√x))/(x^2 +2x+5))dx=4.13049

$${prove}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{3}+\mathrm{2}\sqrt{{x}}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}{dx}=\mathrm{4}.\mathrm{13049}\: \\ $$

Question Number 98678    Answers: 2   Comments: 9

∫_0 ^∞ e^(−ax) ((sin mx)/x) dx = tan^(−1) ((m/a)), a>0

$$\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{ax}}} \frac{\mathrm{sin}\:\boldsymbol{\mathrm{mx}}}{\boldsymbol{\mathrm{x}}}\:\boldsymbol{\mathrm{dx}}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{a}}}\right),\:\boldsymbol{\mathrm{a}}>\mathrm{0} \\ $$

Question Number 98677    Answers: 1   Comments: 0

prove ∫_0 ^a ((ln(1+ax))/(1+x^2 ))dx=(1/2)ln(1+a^2 )tan^(−1) a, a>0

$$\mathrm{prove} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{a}} \frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{ax}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)\mathrm{tan}^{−\mathrm{1}} \boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{a}}>\mathrm{0} \\ $$

Question Number 98675    Answers: 0   Comments: 10

Question Number 98673    Answers: 2   Comments: 0

find a_n in terms of n (I can′t find it...) a_1 =1; a_2 =4 a_3 =a_2 ×4×((2^2 −1)/2^2 ) a_4 =a_3 ×4×((2^2 −1)/2^2 )×((3^2 −1)/3^2 ) a_5 =a_4 ×4×((2^2 −1)/2^2 )×((3^2 −1)/3^2 )×((4^2 −1)/4^2 ) ... n≥2: a_(n+1) =4a_n Π_(k=2) ^n ((k^2 −1)/k^2 )

$$\mathrm{find}\:{a}_{{n}} \:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n} \\ $$$$\left(\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\mathrm{it}...\right) \\ $$$${a}_{\mathrm{1}} =\mathrm{1};\:{a}_{\mathrm{2}} =\mathrm{4} \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{2}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{4}} ={a}_{\mathrm{3}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{5}} ={a}_{\mathrm{4}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }×\frac{\mathrm{4}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}^{\mathrm{2}} } \\ $$$$... \\ $$$${n}\geqslant\mathrm{2}:\:{a}_{{n}+\mathrm{1}} =\mathrm{4}{a}_{{n}} \underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$

Question Number 98672    Answers: 1   Comments: 0

∫_0 ^4 ∫_0 ^(x/4) e^x^2 dx dy

$$\int_{\mathrm{0}} ^{\mathrm{4}} \int_{\mathrm{0}} ^{\frac{{x}}{\mathrm{4}}} {e}^{{x}^{\mathrm{2}} } \:{dx}\:{dy} \\ $$

Question Number 98661    Answers: 0   Comments: 1

using cayley − hamilton theorem what is the inverse of matrix A= [((0 1 −1)),((1 2 2)),((0 1 −1)) ]

$$\mathrm{using}\:\mathrm{cayley}\:−\:\mathrm{hamilton} \\ $$$$\mathrm{theorem}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of} \\ $$$$\mathrm{matrix}\:\mathrm{A}=\:\begin{bmatrix}{\mathrm{0}\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\end{bmatrix}\: \\ $$

Question Number 98796    Answers: 2   Comments: 0

Question Number 98657    Answers: 2   Comments: 0

solve y^(′′) −3y^′ +2y =((sinx)/x)

$$\mathrm{solve}\:\:\mathrm{y}^{''} \:−\mathrm{3y}^{'} \:\:+\mathrm{2y}\:=\frac{\mathrm{sinx}}{\mathrm{x}} \\ $$

Question Number 98656    Answers: 2   Comments: 0

solve xy^(′′) +(2+x^2 )y^′ =xe^(−x^2 )

$$\mathrm{solve}\:\:\:\mathrm{xy}^{''} \:+\left(\mathrm{2}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{y}^{'} \:\:=\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \\ $$

Question Number 98653    Answers: 0   Comments: 0

Question Number 98640    Answers: 0   Comments: 1

Question Number 98638    Answers: 0   Comments: 2

Le plan complexe est rapporte^ a^ un repe^ re orthornorme directe (0,e_1 ^→ ,e_2 ^→ ). On note A et B les points d′affixes respectives i, et 2i. Soit f, l′application du plan prive^ de A dans lui-me^ me qui a^ tout point M d′affixe z distincte i associe le point M d′affixe z′ definie par: z′=((2z−i)/(iz+1)) 1\ Soit z≠i a\ On pose z−i=re^(iθ) . Interpreter ge^ ometriquement r et θ a^ l′aide des points A et M.

$$\mathrm{Le}\:\mathrm{plan}\:\mathrm{complexe}\:\mathrm{est}\:\mathrm{rapport}\acute {\mathrm{e}}\:\grave {\mathrm{a}}\:\mathrm{un}\:\mathrm{rep}\grave {\mathrm{e}re} \\ $$$$\mathrm{orthornorme}\:\mathrm{directe}\:\left(\mathrm{0},\overset{\rightarrow} {\mathrm{e}}_{\mathrm{1}} ,\overset{\rightarrow} {\mathrm{e}}_{\mathrm{2}} \right).\:\mathcal{O}\mathrm{n}\:\mathrm{note}\:\mathrm{A}\:\mathrm{et}\:\mathrm{B}\:\mathrm{les} \\ $$$$\mathrm{points}\:\mathrm{d}'\mathrm{affixes}\:\mathrm{respectives}\:\boldsymbol{\mathrm{i}},\:\mathrm{et}\:\mathrm{2}\boldsymbol{\mathrm{i}}.\:\mathrm{Soit}\:\mathrm{f},\:\mathrm{l}'\mathrm{application} \\ $$$$\mathrm{du}\:\mathrm{plan}\:\mathrm{priv}\acute {\mathrm{e}}\:\mathrm{de}\:\mathrm{A}\:\mathrm{dans}\:\mathrm{lui}-\mathrm{m}\hat {\mathrm{e}me}\:\mathrm{qui}\:\grave {\mathrm{a}}\:\mathrm{tout}\:\mathrm{point} \\ $$$$\mathrm{M}\:\mathrm{d}'\mathrm{affixe}\:\mathrm{z}\:\mathrm{distincte}\:\boldsymbol{\mathrm{i}}\:\mathrm{associe}\:\mathrm{le}\:\mathrm{point}\:\mathrm{M}\:\mathrm{d}'\mathrm{affixe} \\ $$$$\boldsymbol{\mathrm{z}}'\:\mathrm{definie}\:\mathrm{par}:\:\mathrm{z}'=\frac{\mathrm{2z}−\mathrm{i}}{\mathrm{iz}+\mathrm{1}} \\ $$$$\mathrm{1}\backslash\:\mathrm{Soit}\:\mathrm{z}\neq\mathrm{i} \\ $$$$\mathrm{a}\backslash\:\mathrm{On}\:\mathrm{pose}\:\mathrm{z}−\mathrm{i}=\mathrm{re}^{\mathrm{i}\theta} .\:\mathcal{I}\mathrm{nterpreter}\:\mathrm{g}\acute {\mathrm{e}ometriquement}\:\mathrm{r}\:\mathrm{et}\:\theta \\ $$$$\grave {\mathrm{a}}\:\mathrm{l}'\mathrm{aide}\:\mathrm{des}\:\mathrm{points}\:\mathrm{A}\:\mathrm{et}\:\mathrm{M}. \\ $$

Question Number 98623    Answers: 3   Comments: 0

evaluate ∫_(2/(√3)) ^2 (1/(x^2 (√(4+x^2 ))))dx using the substitution x=2tanθ

$${evaluate}\: \\ $$$$\int_{\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}} ^{\mathrm{2}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{\mathrm{4}+{x}^{\mathrm{2}} }}{dx}\:{using}\:{the}\:{substitution}\:{x}=\mathrm{2tan}\theta \\ $$$$ \\ $$

Question Number 98621    Answers: 2   Comments: 1

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