Question Number 150601 by mathdanisur last updated on 13/Aug/21 Answered by Olaf_Thorendsen last updated on 13/Aug/21 $${u}_{\mathrm{1}} \:=\:\frac{\mathrm{sin}\:\mathrm{ln}{i}^{{i}} }{{i}}\:=\:\frac{\mathrm{sin}\left({i}\mathrm{ln}{i}\right)}{{i}}\:=\:\frac{\mathrm{sin}\left({i}\mathrm{ln}{e}^{{i}\frac{\pi}{\mathrm{2}}} \right)}{{i}} \\ $$$${u}_{\mathrm{1}} \:=\:\frac{\mathrm{sin}\left({i}\left({i}\frac{\pi}{\mathrm{2}}\right)\right)}{{i}}=\:\frac{−\mathrm{1}}{{i}}\:=\:{i} \\ $$$${u}_{\mathrm{2}}…
Question Number 19531 by mondodotto@gmail.com last updated on 12/Aug/17 Answered by Tinkutara last updated on 12/Aug/17 Commented by Tinkutara last updated on 12/Aug/17 $$\mathrm{Total}\:\mathrm{who}\:\mathrm{have}\:\mathrm{taken}\:\mathrm{atleast}\:\mathrm{one}\:\mathrm{of} \\…
Question Number 150600 by mathdanisur last updated on 13/Aug/21 $$\mathrm{xyz}\:=\:\mathrm{10} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:-\:\mathrm{7} \\ $$$$\mathrm{xy}\:+\:\mathrm{xz}\:+\:\mathrm{yz}\:=\:\mathrm{2} \\ $$$$\mathrm{Find}\:\:\frac{\mathrm{xy}}{\mathrm{z}}\:+\:\frac{\mathrm{xz}}{\mathrm{y}}\:+\:\frac{\mathrm{yz}}{\mathrm{x}}\:=\:? \\ $$ Answered by amin96 last updated on 13/Aug/21…
Question Number 150603 by mathdanisur last updated on 13/Aug/21 $$\mathrm{Compare}: \\ $$$$\boldsymbol{\mathrm{x}}=\mathrm{sin}\left(\mathrm{165}°\right) \\ $$$$\boldsymbol{\mathrm{y}}=\mathrm{cos}\left(\mathrm{165}°\right) \\ $$$$\boldsymbol{\mathrm{z}}=\mathrm{tan}\left(\mathrm{165}°\right) \\ $$ Answered by ajfour last updated on 13/Aug/21…
Question Number 150597 by mathdanisur last updated on 13/Aug/21 Answered by Olaf_Thorendsen last updated on 13/Aug/21 $$\frac{\overline {{nnn}…{nn}}}{{n}+{n}+{n}…{n}}\:=\:\frac{{n}\underset{{p}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\mathrm{10}^{{p}} }{{kn}}\:=\:\frac{\frac{\mathrm{1}−\mathrm{10}^{{k}} }{\mathrm{1}−\mathrm{10}}}{{k}} \\ $$$$=\:\frac{\mathrm{10}^{{k}} −\mathrm{1}}{\mathrm{9}{k}}…
Question Number 150596 by mr W last updated on 14/Aug/21 Commented by mr W last updated on 14/Aug/21 $${a}\:{more}\:{challenging}\:{case}:\:\mathrm{3}{D}\:{case} \\ $$$${three}\:{vertex}\:{of}\:{a}\:{tetrahedron}\:{lie}\:{on} \\ $$$${the}\:{coordinate}\:{axes}.\:{the}\:{fourth}\:{one} \\ $$$${lies}\:{on}\:{the}\:{sphere}\:{with}\:{radius}\:{R}\:{and}…
Question Number 150599 by mathdanisur last updated on 13/Aug/21 $$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \:=\:? \\ $$ Commented by n0y0n last updated on 13/Aug/21 $$\mathrm{limit}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$$$ \\…
Question Number 85061 by M±th+et£s last updated on 18/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}\:{tan}\mathrm{2}{x}−\mathrm{2}{x}\:{tan}\left({x}\right)}{\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} } \\ $$ Commented by mathmax by abdo last updated on 18/Mar/20 $${let}\:{f}\left({x}\right)=\frac{{xtan}\left(\mathrm{2}{x}\right)−\mathrm{2}{x}\:{tan}\left({x}\right)}{\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} }\:\:{we}\:{have}\:…
Question Number 85059 by Power last updated on 18/Mar/20 Answered by MJS last updated on 18/Mar/20 $${x}^{\mathrm{5}} +\mathrm{1}= \\ $$$$=\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}+\mathrm{1}\right) \\ $$$$\mathrm{you}'\mathrm{ll}\:\mathrm{have}\:\mathrm{to}\:\mathrm{decompose}\:\mathrm{and}\:\mathrm{then}\:\mathrm{use} \\…
Question Number 150594 by liberty last updated on 13/Aug/21 $$\mathrm{The}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{x}} +\mathrm{x}\:\mathrm{being} \\ $$$$\mathrm{differentiable}\:\mathrm{and}\:\mathrm{one}\:\mathrm{to}\:\mathrm{one}\:, \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{differentiable}\:\mathrm{inverse}\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right). \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\frac{{d}}{{dx}}\:\left({f}^{−\mathrm{1}} \right)\:\mathrm{at}\:\mathrm{point}\: \\ $$$$\mathrm{f}\left(\mathrm{ln}\:\mathrm{2}\right)\:\mathrm{is}\:\_\_ \\ $$ Answered by…