Question Number 81937 by Joel578 last updated on 16/Feb/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{PDE}\:\mathrm{by}\:\mathrm{method}\:\mathrm{of}\:\mathrm{separating}\:\mathrm{variables} \\ $$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }\:+\:\mathrm{2}{t}\frac{\partial^{\mathrm{2}} {u}}{\partial{x}\partial{t}}\:−\:\mathrm{4}{u}\:=\:\mathrm{0} \\ $$ Commented by Joel578 last updated on 16/Feb/20 $$\mathrm{My}\:\mathrm{approach}…
Question Number 147469 by rexford last updated on 21/Jul/21 Commented by rexford last updated on 21/Jul/21 $${please},{help}\:{me}\:{out} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 147465 by gsk2684 last updated on 21/Jul/21 $${if}\:{n}>\mathrm{2},{n}\in{N}\:{then}\:{prove}\:{that}\: \\ $$$$\left\{\left(\mathrm{2}{n}−\mathrm{1}\right)^{{n}} +\left(\mathrm{2}{n}\right)^{{n}} \right\}<\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147467 by mathmax by abdo last updated on 21/Jul/21 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{x}} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)\:\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$…
Question Number 147466 by mathmax by abdo last updated on 21/Jul/21 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{5} \\ $$$$\mathrm{find}\:\int\:\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)}\mathrm{dx}\:\:\:\mathrm{and}\:\int\:\:\frac{\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\mathrm{dx} \\ $$ Answered by EDWIN88 last updated on…
Question Number 16392 by ajfour last updated on 21/Jun/17 Answered by Tinkutara last updated on 21/Jun/17 Commented by Tinkutara last updated on 21/Jun/17 $$\angle{AMI}\:=\:\mathrm{90}°,\:\angle{MAI}\:=\:\frac{{A}}{\mathrm{2}},\:{s}\:=\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}}, \\…
Question Number 16387 by Nayon last updated on 21/Jun/17 $${Why}\:{sinx}\:{is}\:{a}\:{power}\:{series}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147459 by liberty last updated on 21/Jul/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\sqrt[{\mathrm{3}}]{\mathrm{8}+{x}^{\mathrm{3}} }−\mathrm{2}}{{x}^{\mathrm{2}} }\:=? \\ $$ Answered by mathmax by abdo last updated on 21/Jul/21…
Question Number 81921 by M±th+et£s last updated on 16/Feb/20 Answered by MJS last updated on 16/Feb/20 $$\int\frac{\sqrt{\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}}}{\:\sqrt{{x}}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}\:\rightarrow\:{dx}=\frac{{t}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}{t}^{\mathrm{3}} }{dt}\right] \\ $$$$=\int\frac{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\:\sqrt{{t}^{\mathrm{3}} }\left({t}+\mathrm{1}\right)}{dt}=…
Question Number 147453 by EDWIN88 last updated on 21/Jul/21 $$\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{25}}\right)…=? \\ $$ Commented by gsk2684 last updated on 21/Jul/21 $$\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{25}}\right)…=? \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)… \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{3}}{\mathrm{2}}\:\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{4}}{\mathrm{3}}\:\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\mathrm{5}}{\mathrm{4}}\:… \\…