Question Number 71824 by mr W last updated on 20/Oct/19 $${solve}\: \\ $$$${x}^{{x}^{{x}^{\mathrm{2019}} } } =\mathrm{2019} \\ $$ Commented by mr W last updated on…
Question Number 6289 by sanusihammed last updated on 22/Jun/16 Commented by FilupSmith last updated on 22/Jun/16 $${a}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{6}}+…+\frac{\mathrm{2015}}{\mathrm{2016}} \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{4}}{\mathrm{6}}×…×\frac{\mathrm{2015}}{\mathrm{2016}} \\ $$$$\frac{{a}}{{b}}+\frac{{b}}{{a}}=?? \\ $$$$ \\ $$$${a}=\underset{{n}=\mathrm{1}}…
Question Number 137359 by rexford last updated on 01/Apr/21 Answered by Ar Brandon last updated on 01/Apr/21 $$\mathcal{I}=\int_{\sqrt[{\mathrm{3}}]{\mathrm{log3}}} ^{\sqrt[{\mathrm{3}}]{\mathrm{log4}}} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{sinx}^{\mathrm{3}} }{\mathrm{sinx}^{\mathrm{3}} +\mathrm{sin}\left(\mathrm{log12}−\mathrm{x}^{\mathrm{3}} \right)}\mathrm{dx} \\…
Question Number 137353 by mey3nipaba last updated on 01/Apr/21 $$\mathrm{If}\:\mathrm{75\%}\:\mathrm{of}\:\mathrm{68}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{85\%}\:\mathrm{of}\:\mathrm{n},\:\mathrm{find}\:\mathrm{n}. \\ $$ Answered by Rasheed.Sindhi last updated on 01/Apr/21 $$\mathrm{75\%}\:\mathrm{of}\:\mathrm{68}=\mathrm{85\%}\:\mathrm{of}\:\mathrm{n} \\ $$$$\mathrm{68}×\frac{\mathrm{75}}{\mathrm{100}}={n}×\frac{\mathrm{85}}{\mathrm{100}} \\ $$$$\mathrm{85}{n}=\mathrm{68}×\mathrm{75} \\…
Question Number 71819 by Maclaurin Stickker last updated on 20/Oct/19 $${There}\:{are}\:\mathrm{3}\:{tangent}\:{circumferences} \\ $$$${inscribed}\:{in}\:{an}\:{isosceles}\:{right}\:{triangle} \\ $$$${Two}\:{of}\:{these}\:{circumferences}\:{have} \\ $$$${radius}\:\boldsymbol{{R}}\:{and}\:{are}\:{tangent}\:{to}\:{the}\: \\ $$$${hypotenuse}\:{and}\:{to}\:{the}\:{two}\:{cathetus}. \\ $$$${The}\:{smaller}\:{circumference}\:{has}\: \\ $$$${radius}\:\boldsymbol{{r}}\:{and}\:{is}\:{tangent}\:{to}\:{the}\:{two} \\ $$$${cathetus}.\:{How}\:{can}\:{I}\:{find}\:{the}\:{radius}…
Question Number 71816 by psyche last updated on 20/Oct/19 $${suppose}\:{that}\:{f}\:{is}\:{continuous}\:{and}\:{differentiable}\:{in}\:\left({a},{b}\right)\:{if}\:{f}'\left({x}\right)\:=\mathrm{0}\:,\forall\:{x}\in\left({a},{b}\right)\:{then}\:{show}\:{that}\:{f}\:{is}\:{constant}\:{on}\:\left[{a},{b}\right]. \\ $$ Answered by mind is power last updated on 20/Oct/19 $$\mathrm{assum}\:\mathrm{f}\:\mathrm{is}\:\mathrm{not}\:\mathrm{consrante} \\ $$$$\exists\mathrm{x},\mathrm{y}\:\mathrm{suche}\:\mathrm{that}\:\mathrm{1}\geqslant\mathrm{y}\neq\mathrm{x}\geqslant\mathrm{0}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\neq\mathrm{f}\left(\mathrm{y}\right) \\…
Question Number 6280 by prakash jain last updated on 21/Jun/16 $$\mathrm{52}\::\:\mathrm{9}\:::\:\mathrm{48}\::\:\mathrm{31}\:::\:\mathrm{27}\::\:\mathrm{13}\:::\:\mathrm{65}\::\:? \\ $$ Commented by prakash jain last updated on 21/Jun/16 $$\mathrm{The}\:\mathrm{question}\:\mathrm{from}\:\mathrm{a}\:\mathrm{mental}\:\mathrm{ability}\:\mathrm{test}. \\ $$ Answered…
Question Number 137349 by mohammad17 last updated on 01/Apr/21 Commented by mohammad17 last updated on 01/Apr/21 $${help}\:{me}\:{sir}\:{please}\:? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 71814 by Aman Arya last updated on 20/Oct/19 $$\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cot}\:{x}}{dx} \\ $$ Commented by mathmax by abdo last updated on 20/Oct/19 $${let}\:{I}=\int\:\frac{{dx}}{\mathrm{1}+{cotanx}}\:\Rightarrow{I}=\int\:\frac{{dx}}{\mathrm{1}+\frac{{cosx}}{{sinx}}}\:=\int\frac{{sinx}}{{sinx}\:+{cosx}}{dx} \\ $$$${changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give}\:{I}=\int\:\:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 71813 by mathmax by abdo last updated on 20/Oct/19 $$\left.\mathrm{1}\right){calculate}\:{F}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {arctan}\left(\mathrm{1}+{a}\:{cosx}\right){dx} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{valeur}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{arctan}\left(\mathrm{1}+\sqrt{\mathrm{2}}{cosx}\right){dx} \\ $$ Terms of Service Privacy Policy…