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Question Number 72086 by wo1lxjwjdb last updated on 24/Oct/19 $$\left(\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\right)^{{v}} +\left(\mathrm{cos}^{−\mathrm{1}} \left({x}\right)^{{v}} \right) \\ $$$${for}\:{any}\:{value}\:{v}\:{what}\:{is}\:{it}\: \\ $$$${in}\:{terms}\:{of}\:{pi} \\ $$$$ \\ $$ Terms of Service…
Question Number 6549 by Rasheed Soomro last updated on 01/Jul/16 $${A}\:{table}\:{with}\:{three}\:{legs}\:{sits}\:{smoothly} \\ $$$${on}\:{every}\:{kind}\:{of}\:{surface},\:{whereas}\:{a} \\ $$$${table}\:{with}\:{four}\:{legs}\:{may}\:{not}\:{sit}\:{on} \\ $$$${rough}\:{surface}.\:{Why}? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 137618 by I want to learn more last updated on 04/Apr/21 Commented by I want to learn more last updated on 04/Apr/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\:\mathrm{nth}\:\:\mathrm{trajectory}.…
Question Number 137613 by physicstutes last updated on 04/Apr/21 $$\mathrm{Two}\:\mathrm{particles}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{and}\:\mathrm{4}{m}\:\mathrm{are}\:\mathrm{connected}\:\mathrm{by}\:\mathrm{a}\:\mathrm{light}\:\mathrm{inelastic} \\ $$$$\mathrm{string}\:\mathrm{of}\:\mathrm{length}\:\mathrm{3}{a},\:\mathrm{which}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{a}\:\mathrm{small}\:\mathrm{smooth}\:\mathrm{fixed}\:\mathrm{ring}. \\ $$$$\mathrm{The}\:\mathrm{heavier}\:\mathrm{particle}\:\mathrm{hangs}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{2}{a}\:\mathrm{beneath}\:\mathrm{the}\:\mathrm{ring}, \\ $$$$\mathrm{while}\:\mathrm{the}\:\mathrm{lighter}\:\mathrm{particle}\:\mathrm{describes}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{speed}. \\ $$$$\:\mathrm{Find} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{The}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{below}\:\mathrm{the}\:\mathrm{ring}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{angular}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lighter}\:\mathrm{particle}. \\…
Question Number 137615 by mohammad17 last updated on 04/Apr/21 $$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}+{tanx}}{\mathrm{1}−{tanx}}\right)^{{xcos}\left(\mathrm{8}{x}\right)} {dx} \\ $$ Commented by mohammad17 last updated on 04/Apr/21 $${hwo}\:{is}\:{can}\:{solve}\:{this}\:? \\ $$…
Question Number 137611 by JulioCesar last updated on 04/Apr/21 Answered by Dwaipayan Shikari last updated on 04/Apr/21 $$\gamma\:\:=\:\mathrm{0}.\mathrm{57721}\:\left({Euler}\:{Mascheroni}\:{constant}\right) \\ $$ Terms of Service Privacy Policy…
Question Number 137610 by mnjuly1970 last updated on 04/Apr/21 $$\:\:\:\:\:\:\:\:\:\:……..\:{mathematical}\:\:\:{analysis}\:\left({II}\right)…. \\ $$$$\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}{ln}\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix}}=\frac{\pi^{\mathrm{2}} }{\mathrm{18}}.. \\ $$ Terms…
Question Number 137605 by mnjuly1970 last updated on 04/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:…..{nice}\:\:\:\:{calculus}… \\ $$$$\:\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{F}_{{n}} }\right).{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{F}_{{n}+\mathrm{1}} }\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:{F}_{{n}} \:{is}\:{fibonacci}\:{sequence}…. \\…
Question Number 6535 by Temp last updated on 01/Jul/16 $$\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+… \\ $$$$\boldsymbol{\mathrm{Is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{true}}? \\ $$$$\therefore\Sigma\frac{\mathrm{1}}{{x}}=\Sigma\left(\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}{x}−\mathrm{1}+\mathrm{2}{x}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}{x}−\mathrm{2}+\mathrm{1}+\mathrm{2}{x}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{4}{x}−\mathrm{2}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\…