Question Number 137177 by mnjuly1970 last updated on 30/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:……{advanced}\:\:\:\:\:….\:\:\:\:\:{calculus}…. \\ $$$$\:\:\:\:\:\:\:\Phi=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi'\left({k}\right)}{{k}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}}{{n}^{{b}} } \\ $$$$\:\:\:\:\:\:{a}\:,\:{b}\:=??\:\left({adapted}\:{from}\:{brilliant}\right) \\ $$$$\:\:\:\:\:\:\:……………. \\ $$$$\:\:\:\:\:\:\:\psi\left({k}\right)\overset{??} {=}−\gamma+\int_{\mathrm{0}} ^{\:\mathrm{1}}…
Question Number 6107 by gourav~ last updated on 14/Jun/16 $$\int\frac{{x}}{{x}^{\mathrm{3}} +{x}+\mathrm{2}}{dx}\:=? \\ $$ Commented by Yozzii last updated on 14/Jun/16 $${x}^{\mathrm{3}} +{x}+\mathrm{2}={x}^{\mathrm{3}} +\mathrm{1}+{x}+\mathrm{1} \\ $$$$=\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}}…
Question Number 71643 by ajfour last updated on 18/Oct/19 Commented by ajfour last updated on 23/Oct/19 $${Assuming}\:{perfectly}\:{inelastic} \\ $$$${collision}\:{takes}\:{place}\:{and}\:{sphere} \\ $$$${climbs}\:{up}\:{the}\:{embankment}. \\ $$$${Find}\:{u}_{{min}} \:{for}\:{this}\:{to}\:{be}\:{possible}. \\…
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Question Number 6105 by sanusihammed last updated on 13/Jun/16 $${Solve}\:{the}\:{system}\:{of}\:{equation}\: \\ $$$$ \\ $$$${x}^{\mathrm{2}} \:−\:{yz}\:=\:−\:\mathrm{52}\:\:\:…………..\:\left({i}\right) \\ $$$${y}^{\mathrm{2}} \:−\:{xz}\:=\:−\:\mathrm{6}\:…………….\:\left({ii}\right) \\ $$$${z}^{\mathrm{2}} \:−\:{xy}\:=\:\mathrm{86}\:\:\:…………..\:\left({iii}\right) \\ $$ Commented by…
Question Number 137173 by JulioCesar last updated on 30/Mar/21 Commented by mathmax by abdo last updated on 31/Mar/21 $$\mathrm{not}\:\mathrm{defined}\:\:\:\mathrm{arcsin}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{on}\:\left[−\mathrm{1},\mathrm{1}\right]\:\:\mathrm{but}\:\mathrm{x}^{\mathrm{2}} −\mathrm{1}\leqslant\mathrm{0}\:! \\ $$ Answered by Ñï=…
Question Number 6099 by LMTV last updated on 13/Jun/16 $$\int_{\mathrm{0}} ^{\infty} {e}^{−{st}} \mathrm{sinh}\:{atdt}=? \\ $$$$\mathrm{or}… \\ $$$$−\frac{\mathrm{1}}{{s}}\left[{e}^{−{st}} \mathrm{sinh}\:{at}\right]_{\mathrm{0}} ^{\infty} =? \\ $$$$−\frac{{a}}{{s}^{\mathrm{2}} }\left[{e}^{−{st}} \mathrm{coth}\:{at}\right]_{\mathrm{0}} ^{\infty}…
Question Number 6098 by Rasheed Soomro last updated on 13/Jun/16 $$\mathrm{A}\:\mathrm{party}\:\mathrm{of}\:\:\mathrm{7}\:\:\:\mathrm{members}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{chosen} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{group}\:\mathrm{of}\:\:\mathrm{6}\:\:\mathrm{ladies}\:\mathrm{and}\:\:\mathrm{5}\:\:\mathrm{gents}. \\ $$$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{the}\:\mathrm{party}\:\mathrm{be} \\ $$$$\mathrm{formed}\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\mathrm{to}\:\mathrm{contain} \\ $$$$\left(\mathrm{i}\right)\:\:\:\mathrm{exactly}\:\mathrm{4}\:\:\mathrm{ladies} \\ $$$$\left(\mathrm{i}\right)\:\:\:\mathrm{at}\:\mathrm{least}\:\mathrm{4}\:\:\mathrm{ladies} \\ $$$$\left(\mathrm{i}\right)\:\:\:\mathrm{at}\:\mathrm{most}\:\mathrm{4}\:\:\mathrm{ladies}\:? \\ $$…
Question Number 137171 by mnjuly1970 last updated on 30/Mar/21 Answered by Dwaipayan Shikari last updated on 30/Mar/21 $${log}\left(\mathrm{2}\right)=\mathrm{1}+\mathrm{1}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)+… \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}+..}}}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{1}+…}}}}} \\ $$$$\frac{\mathrm{1}}{{log}\left(\mathrm{2}\right)}=\mathrm{1}+\frac{\mathrm{1}^{\mathrm{2}}…
Question Number 71633 by mind is power last updated on 18/Oct/19 $$\mathrm{let}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\in\mathrm{IR}_{+} \\ $$$$\mathrm{show}\:\mathrm{that}\:\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}+\mathrm{c}\right)\geqslant\mathrm{2}\sqrt{\mathrm{abc}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)} \\ $$$$ \\ $$ Answered by MJS last updated on 18/Oct/19…