Question Number 213877 by issac last updated on 20/Nov/24 $$\mathrm{evaluate}. \\ $$$$\int_{−\pi} ^{\:+\pi} \:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3cos}^{\mathrm{2}} \left({z}\right)}\:\mathrm{d}{z} \\ $$$$\mathrm{real}\:\mathrm{analysis}\:\mathrm{method}: \\ $$$$\mathrm{complex}\:\mathrm{analysis}\:\mathrm{method}: \\ $$ Terms of Service Privacy…
Question Number 213888 by mnjuly1970 last updated on 20/Nov/24 Commented by mnjuly1970 last updated on 20/Nov/24 $$\:\:\:\:\:{circle}\:{is}\:{tangant}\:{to}\:{parabola} \\ $$$$\:\:.{A}\:{is}\:{center}\:{of}\:{circle} \\ $$$$\:\:\:\:{Find}\:\:\:\:,\:\:{inf}\:\left(\:{R}\:\right)=? \\ $$ Commented by…
Question Number 213890 by Tawa11 last updated on 20/Nov/24 Answered by mr W last updated on 21/Nov/24 $${h}={u}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${t}=\frac{{u}\:\mathrm{sin}\:\theta}{{g}}\left(\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{\mathrm{2}{gh}}{{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}}\right) \\ $$$$\:\:=\frac{\mathrm{45}\:\mathrm{sin}\:\mathrm{52}°}{\mathrm{9}.\mathrm{81}}\left(\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{\mathrm{2}×\mathrm{9}.\mathrm{81}×\mathrm{12}}{\mathrm{45}^{\mathrm{2}}…
Question Number 213871 by 073 last updated on 19/Nov/24 Commented by Frix last updated on 20/Nov/24 $$\mathrm{Elliptic}\:\mathrm{Integral}: \\ $$$$\mathrm{4}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\sqrt{{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{x}\:+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{x}}\:{dx}\:=\mathrm{4}{a}\mathrm{E}\:\frac{{a}^{\mathrm{2}}…
Question Number 213861 by BaliramKumar last updated on 19/Nov/24 Answered by A5T last updated on 19/Nov/24 $${cotA}+{cotB}=\frac{\mathrm{1}}{{tanA}}+\frac{\mathrm{1}}{{tanB}}=\frac{{p}}{{tanAtanB}}={q} \\ $$$$\Rightarrow{tanAtanB}=\frac{{p}}{{q}} \\ $$$${cot}\left({A}+{B}\right)=\frac{\mathrm{1}}{{tan}\left({A}+{B}\right)}=\frac{\mathrm{1}−{tanAtanB}}{{tanA}+{tanB}}=\frac{\mathrm{1}−\frac{{p}}{{q}}}{{p}} \\ $$$$\Rightarrow{cot}\left({A}+{B}\right)=\frac{{q}−{p}}{{pq}} \\ $$…
Question Number 213862 by issac last updated on 19/Nov/24 $$\mathrm{Help}\:\mathrm{me}…..!!!\:\::\left(\:\:\right. \\ $$$$\mathrm{complex}\:\mathrm{anaylsis}\:\mathrm{problem}.. \\ $$$${f}\left({z}\right)\:\mathrm{is}\:\mathrm{entire}\:\mathrm{in}\:\mathrm{path}\:{C}\: \\ $$$$\mathrm{entire}:\:\mathrm{Differantiable}\:\mathrm{complex}\:\mathrm{function} \\ $$$$\mathrm{mean}\:{f}\left({z}\right)\:\mathrm{satisfy}\:{f}\left({z}\right)={u}\left({x},{y}\right)+\boldsymbol{{i}}\centerdot{v}\left({x},{y}\right)\:\: \\ $$$$\frac{\partial{u}}{\partial{x}}=−\frac{\partial{v}}{\partial{y}}\:\mathrm{or}\:\:\frac{\partial{u}}{\partial{y}}=−\frac{\partial{v}}{\partial{x}}\:\left(\mathrm{couchy}-\mathrm{riemann}\right) \\ $$$$\mathrm{show}\:\mathrm{that}\:\int_{\:{C}} \:\frac{{f}\left({z}\right)}{{f}'\left({z}\right)}\:\mathrm{d}{z}=\mathrm{2}\pi\boldsymbol{{i}}\underset{{h}=\mathrm{1}} {\overset{{M}} {\sum}}\:{P}_{{h}}…
Question Number 213859 by ajfour last updated on 19/Nov/24 Commented by mr W last updated on 19/Nov/24 $$\mathrm{0}<{AB}<\mathrm{1} \\ $$$${no}\:{maximum}\:{or}\:{minimum}\:{exists}. \\ $$ Commented by ajfour…
Question Number 213838 by ajfour last updated on 18/Nov/24 Commented by ajfour last updated on 18/Nov/24 $${A}\:{solid}\:{ball}\:{is}\:{released}\:{over}\:{a}\:{fixed} \\ $$$${cylindrical}\:{wedge}\:{as}\:{shown}.\:{Friction} \\ $$$${is}\:{sufficient}.\:{If}\:{just}\:{after}\:{the}\:{ball} \\ $$$${leaves}\:{the}\:{curved}\:{surface}\:{due}\:{to} \\ $$$${Normal}\:{reaction}\:{vanishing},\:{it}\:…
Question Number 213835 by BaliramKumar last updated on 18/Nov/24 Answered by mehdee7396 last updated on 18/Nov/24 $${OA}=\sqrt{\mathrm{13}}\:\:\&\:\:\:{OB}=\mathrm{3} \\ $$$${sin}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}\Rightarrow{cos}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}} \\ $$$$\Rightarrow{tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow\theta=\mathrm{2}{tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{2}}\:\:\checkmark \\ $$$$ \\…
Question Number 213844 by universe last updated on 18/Nov/24 $$\:\int_{−\mathrm{1}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \int_{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }} ^{\sqrt{\mathrm{2}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }} \sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\:{dzdydx} \\…