Question Number 130087 by bobhans last updated on 22/Jan/21 $$\frac{{dx}}{{dy}}\:=\:{a}\:+\:\frac{\left({b}−{a}\right){y}}{{c}}\:+\:\frac{\left({b}−{a}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}\pi{y}}{{c}}\right)}{\mathrm{2}\pi} \\ $$$${for}\:{a}>\mathrm{0}\:,\:{b}>\mathrm{0},\:{c}>\mathrm{0}\:{on}\:{x}\geqslant\mathrm{0}\: \\ $$ Answered by benjo_mathlover last updated on 22/Jan/21 $$\mathrm{dx}\:=\:\mathrm{a}\:\mathrm{dy}\:+\:\frac{\left(\mathrm{b}−\mathrm{a}\right)}{\mathrm{c}}\:\mathrm{y}\:\mathrm{dy}\:+\:\frac{\mathrm{b}−\mathrm{a}}{\mathrm{2}\pi}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi\mathrm{y}}{\mathrm{c}}\right)\:\mathrm{dy} \\ $$$$\mathrm{x}=\:\mathrm{ay}\:+\frac{\left(\mathrm{b}−\mathrm{a}\right)\mathrm{y}^{\mathrm{2}} }{\mathrm{2c}}\:−\frac{\left(\mathrm{b}−\mathrm{a}\right)\mathrm{c}}{\mathrm{4}\pi^{\mathrm{2}}…
Question Number 130077 by benjo_mathlover last updated on 22/Jan/21 $$\:\mathrm{If}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}\:,\:\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{x}}{\mathrm{y}+\mathrm{3}}. \\ $$ Answered by liberty last updated on 22/Jan/21 Answered by…
Question Number 64514 by Mikael last updated on 18/Jul/19 $${y}={arc}\:{tan}\left[\sqrt{\frac{\mathrm{1}−{cosx}}{\mathrm{1}+{cosx}}}\right] \\ $$$${y}^{} =? \\ $$ Commented by mathmax by abdo last updated on 18/Jul/19 $${we}\:{have}\:\frac{\mathrm{1}−{cosx}}{\mathrm{1}+{cosx}}\:=\frac{\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 129988 by mnjuly1970 last updated on 21/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}\:… \\ $$$$\:\:{evaluate}:\: \\ $$$$\:\:\:\phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dx}}{\left({x}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:=? \\ $$$$ \\ $$ Answered by Dwaipayan…
Question Number 64295 by mathmax by abdo last updated on 16/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 64287 by aliesam last updated on 16/Jul/19 Commented by mathmax by abdo last updated on 16/Jul/19 $${R}^{'} \left({t}\right)\:=\frac{\mathrm{15}.\mathrm{0},\mathrm{01}\:{e}^{−\mathrm{0},\mathrm{01}{t}} \left(\mathrm{1}+\mathrm{1},\mathrm{5}{e}^{−\mathrm{0},\mathrm{01}{t}} \right)−\mathrm{15}\left(\mathrm{1}−{e}^{−\mathrm{0},\mathrm{01}{t}} \right)\left(−\mathrm{1},\mathrm{5}\right)\mathrm{0},\mathrm{01}\:{e}^{−\mathrm{0},\mathrm{01}{t}} }{\left(\mathrm{1}+\mathrm{1},\mathrm{5}\:{e}^{−\mathrm{0},\mathrm{01}{t}} \right)^{\mathrm{2}}…
Question Number 129815 by stelor last updated on 19/Jan/21 $$\mathrm{Calculer}\:\mathrm{les}\:\mathrm{d}\acute {\mathrm{e}riv}\acute {\mathrm{e}es}\:\mathrm{n}-\mathrm{i}\grave {\mathrm{e}mes}\:\mathrm{en}\:\mathrm{0}\:\mathrm{de}\: \\ $$$$\mathrm{la}\:\mathrm{fonction}\:\mathrm{d}\acute {\mathrm{e}finie}\:\mathrm{par}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{4}} } \\ $$ Answered by mathmax by abdo…
Question Number 129806 by mnjuly1970 last updated on 19/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:\:{calculus}\:… \\ $$$$\:\:\:{please}\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{Arctan}\left({x}\right)}{\mathrm{1}+{x}}\:{dx}=\frac{\pi}{\mathrm{8}}\:{log}\left(\mathrm{2}\right)\:… \\ $$$$ \\ $$ Answered by Dwaipayan Shikari last…
Question Number 129805 by mnjuly1970 last updated on 19/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:\:\:\:\:\:{calculus}\:… \\ $$$$\:{please}\:{calculate}\:::: \\ $$$$\:\:\:\:\:\mathrm{I}:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\frac{\mathrm{1}}{\:\sqrt[{\mathrm{6}}]{{x}}}\:\right\}{dx} \\ $$$$\:\:\:\:{notice}:\:\left\{{x}\right\}\:{is}\:{the}\:{fractionl}\:{of}\:''\:{x}\:''. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……………. \\ $$ Answered by mindispower…
Question Number 129758 by I want to learn more last updated on 18/Jan/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{derivatives}\:\mathrm{of}\:\:\:\:\mathrm{log}_{\mathrm{e}} \left(\mathrm{6x}\:+\:\mathrm{8}\right)^{\mathrm{5}} \\ $$ Commented by Dwaipayan Shikari last updated on 18/Jan/21…