Question Number 66084 by F_Nongue last updated on 09/Aug/19 $${How}\:{to}\:{solve}\:{this}\:{limit}? \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\left(\mathrm{7}{x}+\frac{\mathrm{2}}{{x}}\right)^{{x}} \\ $$ Commented by mathmax by abdo last updated on 09/Aug/19 $${let}\:{A}\left({x}\right)\:=\left(\mathrm{7}{x}+\frac{\mathrm{2}}{{x}}\right)^{{x}}…
Question Number 131622 by rydasss last updated on 06/Feb/21 $${for}\:-\pi\:<\:{x}\:<\:-\:\frac{\pi}{\mathrm{2}}\:{find}\:{the}\:{solution}\:{of} \\ $$$${sin}\:{x}\:>\:\:\frac{{sin}\:{x}\:+\:\mathrm{2}}{\mathrm{2}\:{sin}\:{x}\:+\:\mathrm{1}} \\ $$ Answered by mr W last updated on 07/Feb/21 $$−\pi<{x}<−\frac{\pi}{\mathrm{2}}\:\Rightarrow\:−\mathrm{1}<\mathrm{sin}\:{x}<\mathrm{0} \\ $$$${case}\:\mathrm{1}:\:\mathrm{2sin}\:{x}+\mathrm{1}>\mathrm{0},\:{i}.{e}.\:\mathrm{sin}\:{x}>−\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 66085 by mathmax by abdo last updated on 09/Aug/19 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}}}{dx} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 548 by 123456 last updated on 25/Jan/15 $${how}\:{many}\:{digits}\:{are}\:{in}\:{periodic}\:{part} \\ $$$${of}\:\frac{\mathrm{1}}{\mathrm{60}^{\mathrm{30}} } \\ $$ Answered by prakash jain last updated on 25/Jan/15 $$\frac{\mathrm{1}}{\left(\mathrm{60}\right)^{\mathrm{30}} }=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{30}}…
Question Number 66082 by aliesam last updated on 08/Aug/19 Commented by mathmax by abdo last updated on 09/Aug/19 $${let}\:{I}\:=\int_{−\infty} ^{+\infty} \:{xsin}\left({x}^{\mathrm{3}} \right){dx}\:\Rightarrow{I}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{x}\:{sin}\left({x}^{\mathrm{3}} \right){dx}…
Question Number 131619 by help last updated on 06/Feb/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 545 by 123456 last updated on 25/Jan/15 $${if}\:\tau=\underset{\mathrm{0}} {\overset{+\infty} {\int}}{e}^{−\frac{{t}}{\mathrm{2}}} {dt}\underset{−\infty} {\overset{+\infty} {\int}}\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} },\:{then}\:\frac{\tau}{\mathrm{2}}=? \\ $$ Answered by prakash jain last updated on…
Question Number 131618 by physicstutes last updated on 06/Feb/21 $$\:\left(\mathrm{1}\right)\:\mathrm{A}\:\mathrm{cell}\:\mathrm{has}\:\mathrm{an}\:\mathrm{Emf}\:\mathrm{of}\:\mathrm{2}.\mathrm{5}\:\mathrm{V}\:.\mathrm{It}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{balanced}\:\mathrm{by}\:\mathrm{a}\: \\ $$$$\mathrm{potentiometer}\:\mathrm{of}\:\mathrm{lenght}\:\mathrm{100}.\mathrm{0}\:\mathrm{cm}\:\mathrm{when}\:\mathrm{a}\:\mathrm{driver}\:\mathrm{cell}\:\mathrm{of}\:\mathrm{of}\:\mathrm{2}.\mathrm{0V} \\ $$$$\mathrm{is}\:\mathrm{connected}\:\mathrm{to}\:\mathrm{it}.\:\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{best}\:\mathrm{explains}\:\mathrm{why}\:\mathrm{there} \\ $$$$\mathrm{is}\:\mathrm{no}\:\mathrm{balance}\:\mathrm{lenght}? \\ $$$$\:\mathrm{A}.\:\mathrm{The}\:\mathrm{current}\:\mathrm{in}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{is}\:\mathrm{too}\:\mathrm{low}.\:\:\:\:\:\:\:\mathrm{B}.\:\mathrm{the}\:\mathrm{balance}\:\mathrm{lenght}\:\mathrm{is}\:\mathrm{too}\:\mathrm{small} \\ $$$$\:\mathrm{C}.\:\mathrm{the}\:\mathrm{emf}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cell}\:\mathrm{is}\:\mathrm{too}\:\mathrm{low}.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{D}.\:\mathrm{The}\:\mathrm{voltae}\:\mathrm{accross}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{is}\:\mathrm{too}\:\mathrm{low}. \\ $$$$\mathrm{please}\:\mathrm{site}\:\mathrm{a}\:\mathrm{little}\:\mathrm{explanation}. \\ $$ Terms…
Question Number 544 by 123456 last updated on 26/Jan/15 $$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\mathrm{sin}\:{x}\:\mathrm{sinh}\:{x}\:{dx}=? \\ $$ Answered by prakash jain last updated on 26/Jan/15 $$\int\mathrm{sin}\:{x}\:\mathrm{sinh}\:{xdx} \\ $$$$=\mathrm{sin}\:{x}\:\mathrm{cosh}\:{x}\:−\int\mathrm{cos}\:{x}\:\mathrm{cosh}\:{x}\:{dx}…
Question Number 131613 by mohammad17 last updated on 06/Feb/21 $$\frac{{z}^{\mathrm{3}} +\mathrm{8}}{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{2}} +\mathrm{16}} \\ $$ Answered by Dwaipayan Shikari last updated on 06/Feb/21 $$\frac{{z}^{\mathrm{3}} +\mathrm{8}}{{z}^{\mathrm{4}}…