Question Number 62455 by aliesam last updated on 21/Jun/19 Commented by mathmax by abdo last updated on 22/Jun/19 $${S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{A}_{{n}} \:\:\:{with}\:{A}_{{n}} =\int_{{n}−\frac{\pi}{\mathrm{2}}} ^{{n}+\frac{\pi}{\mathrm{2}}} \:{e}^{−{x}}…
Question Number 62453 by Tawa1 last updated on 21/Jun/19 $$\int\:\frac{\mathrm{x}}{\mathrm{e}^{\mathrm{x}} \:−\:\mathrm{1}}\mathrm{dx},\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\:\mathrm{x}\:>\:\mathrm{0} \\ $$ Commented by mathmax by abdo last updated on 21/Jun/19 $$\int\:\:\frac{{x}}{{e}^{{x}} −\mathrm{1}}{dx}\:=\int\:\:\frac{{x}\:{e}^{−{x}} }{\mathrm{1}−{e}^{−{x}}…
Question Number 62452 by Tawa1 last updated on 21/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\:\:\mathrm{2014}!\:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\:\mathrm{2017} \\ $$ Answered by Rasheed.Sindhi last updated on 21/Jun/19 $${Wilson}'{s}\:{Theorm}: \\ $$$$\:\:\:\:\:\:\:\left({p}−\mathrm{1}\right)!\equiv−\mathrm{1}\left({mod}\:{p}\right)\::\:{p}\in\mathbb{P} \\ $$$$\:\:\:\because\:\:\:\:\mathrm{2017}\in\mathbb{P} \\…
Question Number 62449 by Tawa1 last updated on 21/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{digit}\:\mathrm{in}\:\:\:\:\mathrm{2}^{\mathrm{50}} \\ $$ Commented by Tawa1 last updated on 21/Jun/19 $$\mathrm{And}\:\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{a}\:\mathrm{general}\:\mathrm{nth}\:\mathrm{number}\:\mathrm{of}\:\mathrm{term}\:\mathrm{in}\:\mathrm{any}\:\mathrm{number}\:\mathrm{and}\:\:\mathrm{powers} \\ $$ Answered by Rasheed.Sindhi…
Question Number 62448 by mr W last updated on 21/Jun/19 Commented by mr W last updated on 21/Jun/19 $${The}\:{distances}\:{of}\:{a}\:{point}\:{M}\:{to}\:{the} \\ $$$${vertexes}\:{of}\:{a}\:{triangle}\:{are}\:{p},{q},{r}. \\ $$$$\left({assume}\:{p}\geqslant{q}\geqslant{r}\right) \\ $$$${Find}\:{the}\:{side}\:{lengthes}\:{and}\:{thus}\:{the}…
Question Number 127982 by Dwaipayan Shikari last updated on 03/Jan/21 $${Some}\:{Values}\:.. \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}{e}^{−\pi{n}^{\mathrm{2}} } =\frac{\pi^{\frac{\mathrm{1}}{\mathrm{4}}} }{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}\pi{n}^{\mathrm{2}} } =\frac{\pi^{\frac{\mathrm{1}}{\mathrm{4}}} }{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:\frac{\sqrt[{\mathrm{4}}]{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{2}}…
Question Number 127979 by bobhans last updated on 03/Jan/21 $$\:\:\left(\mathrm{1}+{i}\right)^{\mathrm{2020}} \:=? \\ $$ Answered by liberty last updated on 03/Jan/21 $$\:\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2020}} \:=\:\left(\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\:+\:\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)\right)^{\mathrm{2020}} \\ $$$$\:=\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2020}} \:\left(\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}\right)+\mathrm{i}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}\right)\right)^{\mathrm{2020}}…
Question Number 62440 by mathsolverby Abdo last updated on 21/Jun/19 $${let}\:{h}\left({x}\right)=\:{arctan}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\left.\mathrm{1}\right){calculate}\:{h}^{\left({n}\right)} \left({x}\right)\:{and}\:{h}^{\left({n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right){developp}\:{f}\left({x}\right){at}\:{integr}\:{serie}\:{at}\:{x}_{\mathrm{0}} =\mathrm{1} \\ $$ Commented by mathmax by abdo…
Question Number 127974 by Dwaipayan Shikari last updated on 03/Jan/21 $$\overset{\bullet\bullet} {\theta}+\frac{{g}}{{l}}{sin}\theta=\mathrm{0} \\ $$$${Exact}\:{form}\:\left({May}\:{include}\:{elliptic}\:{integral}\right) \\ $$ Commented by Dwaipayan Shikari last updated on 03/Jan/21 $${My}\:{try}…
Question Number 62439 by mathsolverby Abdo last updated on 21/Jun/19 $${sove}\:{inside}\:{Z}/\mathrm{3}{Z}\:{the}\:{systeme} \\ $$$$\begin{cases}{\mathrm{5}{x}+\mathrm{7}{y}\:=\mathrm{10}}\\{\mathrm{2}{x}+\mathrm{5}{y}\:=\mathrm{8}}\end{cases} \\ $$$$ \\ $$ Commented by arcana last updated on 22/Jun/19 $$\mathbb{Z}/\mathrm{3}\mathbb{Z}=\mathbb{Z}_{\mathrm{3}}…