Question Number 3035 by 123456 last updated on 03/Dec/15 $$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left[{z}+\left({z}+\mathrm{1}\right)+…+\left({z}+{n}\right)\right]}{{z}\left({z}+\mathrm{1}\right)\left({z}+\mathrm{2}\right)…\left({z}+{n}\right)} \\ $$$$\omega\left(\mathrm{1}\right)=? \\ $$ Commented by prakash jain last updated on 04/Dec/15 $${w}\left(\mathrm{1}\right)=\underset{{n}\rightarrow\infty}…
Question Number 134102 by mnjuly1970 last updated on 27/Feb/21 $$\:\:\:\:\:\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left(\sqrt{{x}}\:\right)}{{e}^{\mathrm{2}\pi\sqrt{{x}}\:} −\mathrm{1}}{dx}=\mathrm{1}−\frac{{e}}{\left({e}−\mathrm{1}\right)_{} ^{\mathrm{2}} } \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 134097 by bobhans last updated on 27/Feb/21 $$\mathrm{In}\:\mathrm{a}\:\mathrm{square}\:\mathrm{ABCD}\:,\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{APQ}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{it}.\:\mathrm{AP}=\mathrm{4}\:\mathrm{cm}, \\ $$$$\mathrm{PQ}=\mathrm{3}\:\mathrm{cm}\:\mathrm{and}\:\mathrm{AQ}=\mathrm{5}\:\mathrm{cm}.\:\mathrm{Point} \\ $$$$\mathrm{P}\:\mathrm{is}\:\mathrm{on}\:\mathrm{the}\:\mathrm{side}\:\mathrm{BC}\:\mathrm{and}\:\mathrm{point}\:\mathrm{Q} \\ $$$$\mathrm{is}\:\mathrm{on}\:\mathrm{side}\:\mathrm{CD}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{square}\:\mathrm{ABCD}. \\ $$ Answered by mr…
Question Number 3027 by Rasheed Soomro last updated on 03/Dec/15 $${Determine} \\ $$$$\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} +… \\ $$ Commented by Filup last updated on 03/Dec/15 $$=\underset{{i}=\mathrm{1}}…
Question Number 134093 by mohammad17 last updated on 27/Feb/21 $${where}\:{is}\:{the}\:{function}\:{derivable}\: \\ $$$$ \\ $$$${f}\left({z}\right)={zRez}+\overset{−} {{z}imz}+\overset{−} {{z}} \\ $$$$ \\ $$$${help}\:{me}\:{sir}\:{please} \\ $$ Terms of Service…
Question Number 3023 by Rasheed Soomro last updated on 03/Dec/15 $$\mathrm{Pl}\:\mathrm{notice}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{of}\:\underset{−} {\mathrm{Q2771}},\mathrm{I}\:\mathrm{have}\:\mathrm{updated}\:\mathrm{it}. \\ $$ Answered by prakash jain last updated on 04/Dec/15 $$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be} \\ $$$$\frac{\mathrm{number}\:\mathrm{of}\:\mathrm{rectangles}\:\mathrm{in}\:\mathrm{7}×\mathrm{7}\:\mathrm{grid}\:\left(\mathrm{8}×\mathrm{8}\:\mathrm{points}\right)}{\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{select}\:\mathrm{4}\:\mathrm{points}=^{\mathrm{64}}…
Question Number 3022 by Rasheed Soomro last updated on 03/Dec/15 $$\mathrm{Determine} \\ $$$$\mathrm{1}+\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} +… \\ $$ Commented by prakash jain last updated on 04/Dec/15…
Question Number 68554 by Mikael last updated on 13/Sep/19 Commented by Prithwish sen last updated on 13/Sep/19 $$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{1}}{\mathrm{n}}\left[\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{0}}{\mathrm{n}}}+\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}\:+……+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}}\right] \\ $$$$=\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\underset{\boldsymbol{\mathrm{r}}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}−\mathrm{1}} {\sum}}\:\:\boldsymbol{\mathrm{lim}}\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\:}\frac{\mathrm{1}}{\mathrm{1}+\frac{\boldsymbol{\mathrm{r}}}{\boldsymbol{\mathrm{n}}}}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}}…
Question Number 3017 by Filup last updated on 03/Dec/15 $${A}=\int_{\mu} ^{\:\mu+\epsilon} \:\frac{{dx}}{{x}+{n}} \\ $$$$ \\ $$$${A}=\mathrm{ln}\left({x}+{n}\right)\:\mid_{\mu} ^{\mu+\epsilon} \\ $$$$=\mathrm{ln}\left(\mu+\epsilon+{n}\right)−\mathrm{ln}\left(\mu+{n}\right) \\ $$$$=\mathrm{ln}\left(\frac{\mu+\epsilon+{n}}{\mu+{n}}\right) \\ $$$$=\mathrm{ln}\left(\frac{\mu+{n}}{\mu+{n}}+\frac{\epsilon}{\mu+{n}}\right) \\ $$$$…
Question Number 68549 by gunawan last updated on 13/Sep/19 $${y}=\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${y}^{−\mathrm{1}} =… \\ $$ Commented by mathmax by abdo last updated on…