Question Number 144057 by lapache last updated on 21/Jun/21 $${En}\:{utilisant}\:{la}\:{transforme}\:{de}\:{laplace} \\ $$$${Calculer} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{{tsin}\left({xt}\right)}{{a}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt}\:\:\:\forall{a},{x}\in\mathbb{R}^{\ast} \\ $$ Answered by qaz last updated…
Question Number 78522 by jagoll last updated on 18/Jan/20 $${what}\:{is}\:{the}\: \\ $$$${line}\:{passing}\:{through}\:\left(\mathrm{2},\mathrm{2},\mathrm{1}\right) \\ $$$${and}\:{parallel}\:{to}\:\mathrm{2}\hat {{i}}\:−\:\hat {{j}}\:−\:\hat {{k}}\:? \\ $$ Commented by mr W last updated…
Question Number 144059 by 0731619 last updated on 21/Jun/21 Answered by Olaf_Thorendsen last updated on 21/Jun/21 $$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{sin}^{\mathrm{2}} {x}}{{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}+{a}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{tan}^{\mathrm{2}} {x}}{{b}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}}…
Question Number 144053 by mnjuly1970 last updated on 21/Jun/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……..\:{Calculus}…….. \\ $$$$\:\:\:\:\:\:\:\Omega:={lim}\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{sin}\left({kx}\right)}{\:\sqrt{\mathrm{2}^{{k}} }}\right)^{\mathrm{2}} {dx}=? \\ $$$$ \\ $$ Answered…
Question Number 144052 by physicstutes last updated on 21/Jun/21 $$\mathrm{Evaluate}\: \\ $$$$\:\int\:\frac{\sqrt{{x}}}{\mathrm{sinh}\:{x}}\:{dx} \\ $$ Answered by mindispower last updated on 21/Jun/21 $${x}>\mathrm{0} \\ $$$$=\int\frac{\mathrm{2}{e}^{−{x}} \sqrt{{x}}{dx}}{\mathrm{1}−{e}^{−\mathrm{2}{x}}…
Question Number 144049 by physicstutes last updated on 20/Jun/21 $$\:\mathrm{Given}\:\mathrm{the}\:\mathrm{equation}\:\:\mathrm{1000}\:=\:\mathrm{2000}\left(\frac{\mathrm{1}−\left(\mathrm{1}+{t}\right)^{−{n}} }{{t}}\right) \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{t}. \\ $$ Answered by Olaf_Thorendsen last updated on 21/Jun/21 $$\mathrm{1000}\:=\:\mathrm{2000}\left(\frac{\mathrm{1}−\left(\mathrm{1}+{t}\right)^{−{n}} }{{t}}\right) \\…
Question Number 144050 by ArielVyny last updated on 21/Jun/21 $${in}\:{a}\:{triangle}\:{ABC}\:{we}\:{have} \\ $$$$\begin{cases}{\mathrm{2}{sin}\hat {{A}}+\mathrm{4}{cos}\hat {{B}}=\mathrm{6}}\\{\mathrm{4}{sin}\hat {{B}}+\mathrm{3}{cos}\hat {{A}}=\mathrm{1}}\end{cases} \\ $$$${determine}\:\hat {{C}} \\ $$ Commented by MJS_new last…
Question Number 78511 by ajfour last updated on 18/Jan/20 Commented by ajfour last updated on 18/Jan/20 $${The}\:{cylinder}\:{has}\:{half}\:{the}\:{volume} \\ $$$${of}\:{the}\:{cuboid}.\:{If}\:{its}\:{circular}\:{faces} \\ $$$${touch}\:{the}\:{adjoining}\:{walls},\:{find} \\ $$$${coordinates}\:{of}\:{points}\:{A},\:{B},\:{C} \\ $$$${in}\:{terms}\:{of}\:{a},\:{b},\:{c}.…
Question Number 144042 by maryxxxx last updated on 20/Jun/21 Answered by mindispower last updated on 20/Jun/21 $$\int\frac{{dx}}{{x}^{\mathrm{3}} \sqrt{\mathrm{1}−\left(\frac{\mathrm{10}}{{x}}\right)^{\mathrm{2}} }} \\ $$$${y}=\frac{\mathrm{100}}{{x}^{\mathrm{2}} },{y}<\mathrm{1}\Rightarrow{dy}=\frac{−\mathrm{200}}{{x}^{\mathrm{3}} }{dx} \\ $$$$\Rightarrow\frac{−\mathrm{1}}{\mathrm{200}}\int\frac{{dy}}{\:\sqrt{\mathrm{1}−{y}}}{dy}=\frac{\mathrm{1}}{\mathrm{100}}\sqrt{\mathrm{1}−{y}}+{c}=\frac{\sqrt{\mathrm{1}−\frac{\mathrm{100}}{{x}^{\mathrm{2}}…
Question Number 12968 by @ANTARES_VY last updated on 08/May/17 $$\left(\boldsymbol{\mathrm{x}}+\mathrm{3}\right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{y}}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{45}\:\:\:\boldsymbol{\mathrm{A}}\left(\mathrm{0};\mathrm{11}\right) \\ $$$$\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{circle}}\:\:\boldsymbol{\mathrm{to}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{point}}\:\:\boldsymbol{\mathrm{of}} \\ $$$$\boldsymbol{\mathrm{trying}}\:\:\boldsymbol{\mathrm{to}}\:\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{angular}}\:\:\boldsymbol{\mathrm{coefficient}}. \\ $$ Answered by 433 last updated on 08/May/17…