Question Number 126744 by bemath last updated on 24/Dec/20 $$\:\:{Given}\:{f}\left({x}\right)=\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\left(\mathrm{6}{x}−\pi\right)^{\mathrm{2}} } \\ $$$$\:{Find}\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:{f}\left({x}\right)\:. \\ $$ Answered by liberty last updated on 24/Dec/20 $$\:\underset{{x}\rightarrow\pi/\mathrm{6}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{\mathrm{12}\left(\mathrm{6}{x}−\pi\right)}\:=\:\underset{{x}\rightarrow\pi/\mathrm{6}}…
Question Number 192282 by universe last updated on 14/May/23 $$\:\:\:{find}\:{the}\:{value}\:{of} \\ $$$$\:\:\:\:\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\:+\:\mathrm{4sin}\:\frac{\pi}{\mathrm{9}}\:=\:? \\ $$ Answered by som(math1967) last updated on 14/May/23 $${tan}\mathrm{20}+\mathrm{4}{sin}\mathrm{20} \\ $$$$\frac{{sin}\mathrm{20}+\mathrm{4}{sin}\mathrm{20}{cos}\mathrm{20}}{{cos}\mathrm{20}} \\…
Question Number 61208 by Tawa1 last updated on 30/May/19 Commented by maxmathsup by imad last updated on 30/May/19 $${its}\:{only}\:{a}\:{try}\:{let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\mathrm{1}+{tx}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:\:{with}\:{t}\:\geqslant\mathrm{0} \\ $$$${we}\:{have}\:{f}^{'}…
Question Number 192277 by York12 last updated on 13/May/23 $$ \\ $$$${Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,…,\:{a}_{{n}} \:{be}\:{real}\:{numbers}\:{such}\:{that}: \\ $$$$\sqrt{{a}_{\mathrm{1}} }\:+\:\sqrt{{a}_{\mathrm{2}} −\mathrm{1}\:\:}\:+\sqrt{{a}_{\mathrm{3}} −\mathrm{2}\:}\:+…+\sqrt{{a}_{{n}} −\left({n}−\mathrm{1}\right)\:}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{n}}…
Question Number 192276 by York12 last updated on 13/May/23 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left[\frac{{k}\left({n}−{k}\right)!+\left({k}+\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)!\left({n}−{k}\right)!}\right]\right) \\ $$ Answered by witcher3 last updated on 14/May/23 $$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}!\left(\mathrm{n}−\mathrm{k}\right)!}=\frac{\mathrm{1}}{\mathrm{n}!}\underset{\mathrm{k}=\mathrm{0}}…
Question Number 61205 by necx1 last updated on 30/May/19 $${A}\:{cubical}\:{block}\:{of}\:{ice}\:{of}\:{mass}\:{m}\:{and} \\ $$$${edge}\:{L}\:{is}\:{placed}\:{in}\:{a}\:{large}\:{tray}\:{of}\:{mass} \\ $$$${M}.{If}\:{the}\:{ice}\:{block}\:{melts},{how}\:{far}\:{does} \\ $$$${the}\:{centre}\:{of}\:{mass}\:{of}\:{the}\:{system}\:“{ice}\:+\:{tray}'' \\ $$$${come}\:{down}\:? \\ $$$$ \\ $$$$\left.{a}\left.\right)\left.\frac{{ml}}{{m}+{M}}\left.\:\:{b}\right)\frac{\mathrm{2}{ml}}{{m}+{M}}\:\:{c}\right)\frac{{ml}}{\mathrm{2}\left({m}+{M}\right)}\:\:{d}\right){none} \\ $$ Commented…
Question Number 192278 by manxsol last updated on 13/May/23 $${S}={arctan}\left(\frac{\mathrm{2}}{\mathrm{1}^{\mathrm{2}} }\right)+{artan}\left(\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }\right)+…….. \\ $$ Commented by Frix last updated on 14/May/23 $$\mathrm{As}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{before}\:\left(\mathrm{but}\:\mathrm{it}\:\mathrm{was}\:\mathrm{deleted}\right): \\ $$$$\mathrm{I}\:\mathrm{approximated}\:\mathrm{it}\:\mathrm{using}\:\mathrm{software}\:\mathrm{and}\:\mathrm{got} \\…
Question Number 61202 by aanur last updated on 30/May/19 Commented by aanur last updated on 30/May/19 $${a}=\mathrm{73}\frac{\mathrm{5}}{\mathrm{7}}\:\:\:\:\:{b}=\mathrm{31}\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${sir}\:{help}\:{me}\:{plz} \\ $$ Answered by Kunal12588 last…
Question Number 192275 by York12 last updated on 13/May/23 $$\mathrm{2009}^{\mathrm{3}^{\mathrm{2016}{n}+\mathrm{2013}} } +\mathrm{2010}^{\mathrm{2}^{\mathrm{2016}{n}+\mathrm{2013}} } \equiv{x}\:{mod}\left(\mathrm{11}\right)\:{where}\:{n}\:{is}\:{any}\:{integer}\:\geq\mathrm{0} \\ $$$$ \\ $$ Answered by BaliramKumar last updated on 14/May/23…
Question Number 192274 by York12 last updated on 13/May/23 $${prove}\:{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}^{\mathrm{2}} }{\:\sqrt{{n}^{\mathrm{6}} +{k}}}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$ Answered by aleks041103 last…