Question Number 124796 by Engr_Jidda last updated on 06/Dec/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{interms} \\ $$$$\mathrm{bessels}\:\mathrm{function}. \\ $$$$\mathrm{y}^{''} +\left(\mathrm{1}+\frac{\mathrm{1}−\mathrm{4k}^{\mathrm{2}} }{\mathrm{4x}^{\mathrm{2}} }\right)\mathrm{y}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 124779 by TITA last updated on 05/Dec/20 Commented by TITA last updated on 05/Dec/20 $${please}\:{help} \\ $$ Answered by Lordose last updated on…
Question Number 59220 by Aditya789 last updated on 06/May/19 $$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{y}=\:\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \mathrm{x}} \\ $$ Answered by MJS last updated on 06/May/19…
Question Number 124723 by mnjuly1970 last updated on 05/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\:{nice}\:\:\:{calculus}…. \\ $$$$\:\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:{challanging}\:\:\:{integral}:: \\ $$$$\:\:\:\:\Omega=\:\:\int_{\mathrm{1}} ^{\:\:\infty} \:\left(\frac{\left\{{x}\right\}−\frac{\mathrm{1}}{\mathrm{2}}}{{x}}\right){dx}\overset{???} {=}{ln}\left(\sqrt{\mathrm{2}\pi}\:\right)−\mathrm{1} \\ $$$$\:\:\:\left\{{x}\right\}\:{is}\:{fractional}\:{part}\:{of}\:\:'{x}' \\ $$ Answered by…
Question Number 59171 by maxmathsup by imad last updated on 05/May/19 $${let}\:{f}\left({x},{y}\right)\:\:\frac{{arctan}\left({x}+\mathrm{2}{y}\right)}{{x}\:+{y}^{\mathrm{2}} } \\ $$$${calculate}\:\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:\:,\:\frac{\partial{f}}{\partial{y}}\left({x},{y}\right),\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }\left({x},{y}\right),\:\frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }\left({x},{y}\right)\:,\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}\left({x},{y}\right) \\ $$$$\frac{\partial^{\mathrm{2}} {f}}{\partial{y}\partial{x}}\left({x},{y}\right) \\ $$…
Question Number 124698 by Algoritm last updated on 05/Dec/20 Answered by Olaf last updated on 05/Dec/20 $$\mathrm{Let}\:{f}\left({x}\right)\:=\:\mathrm{ln}\left(\mathrm{3}+{x}^{\mathrm{2}} \right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{ln}\mid{x}−\sqrt{\mathrm{3}}{i}\mid+\mathrm{ln}\mid{x}+\sqrt{\mathrm{3}}{i}\mid \\ $$$${f}'\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{3}}{i}}+\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{3}}{i}} \\ $$$${f}''\left({x}\right)\:=\:−\frac{\mathrm{1}}{\left({x}−\sqrt{\mathrm{3}}{i}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({x}+\sqrt{\mathrm{3}}{i}\right)^{\mathrm{2}}…
Question Number 124666 by benjo_mathlover last updated on 05/Dec/20 $$\:{Find}\:{the}\:{volume}\:{that}\:{remains} \\ $$$${after}\:{the}\:{hole}\:{of}\:{radius}\:\mathrm{1}\:{bored}\: \\ $$$${through}\:{the}\:{center}\:{of}\:{a}\:{solid} \\ $$$${sphere}\:{of}\:{radius}\:\mathrm{3}. \\ $$$$\left({a}\right)\:\mathrm{18}\pi\:\:\:\:\left({b}\right)\:\frac{\mathrm{28}}{\mathrm{3}}\pi\:\:\:\:\left({c}\right)\:\mathrm{36}\pi\:\:\:\:\left({d}\right)\:\frac{\mathrm{56}\pi}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$ Commented…
Question Number 124662 by rs4089 last updated on 05/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 124651 by benjo_mathlover last updated on 05/Dec/20 Answered by liberty last updated on 05/Dec/20 $${for}\:−\mathrm{3}\leqslant{x}\leqslant\mathrm{3}\:\rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{9}\:;\:{g}\left({x}\right)=\sqrt{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$$${for}\:\mathrm{3}\leqslant{x}\leqslant\mathrm{6}\rightarrow\left({x}−\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\:\frac{\mathrm{9}}{\mathrm{4}} \\…
Question Number 124558 by rs4089 last updated on 04/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com