Question Number 183420 by mokys last updated on 25/Dec/22 $${prove}\:{that}\:\underset{{x}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{4}^{{x}} \:.\:{x}}{{x}!}\:=\:\mathrm{4}\:{e}^{\mathrm{4}} \\ $$ Answered by Ar Brandon last updated on 25/Dec/22 $$\underset{{x}=\mathrm{0}} {\overset{\infty}…
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Question Number 183419 by aurpeyz last updated on 25/Dec/22 Commented by Ml last updated on 25/Dec/22 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{problum}\:\mathrm{in}\:\mathrm{this}\:\mathrm{contiliver}\:\mathrm{beam} \\ $$ Commented by aurpeyz last updated on…
Question Number 183411 by aurpeyz last updated on 25/Dec/22 Commented by aurpeyz last updated on 26/Dec/22 $${Determine}\:{the}\:{equivalent}\:{resistance} \\ $$$${relative}\:{to}\:{the}\:{nodes}\:\left({points}\right)\:\ll{A}\gg\:{and}\: \\ $$$$\ll{B}\gg\:{with}\:{excluded}\:{sources}\:{if}\:{R}_{\mathrm{1}} =\mathrm{6}\Omega \\ $$$${R}_{\mathrm{2}} ={R}_{\mathrm{3}}…
Question Number 183409 by aurpeyz last updated on 25/Dec/22 Commented by Yhusuph last updated on 25/Dec/22 $${F}_{{x}} ={F}\bullet{Cos}\mathrm{30}° \\ $$ Commented by mr W last…
Question Number 117866 by aurpeyz last updated on 14/Oct/20 $${how}\:{many}\:{four}\:{digit}\:{numbers}\:{can}\:{be} \\ $$$${formed}\:{with}\:{the}\:{digits}\:\mathrm{0}\:\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{6}\:\mathrm{7}\:\mathrm{8} \\ $$$$\mathrm{9}? \\ $$ Commented by mr W last updated on 14/Oct/20 $${with}\:{repetition}:…
Question Number 183405 by aurpeyz last updated on 25/Dec/22 Commented by mr W last updated on 25/Dec/22 $${what}'{s}\:{your}\:{question}? \\ $$ Commented by aurpeyz last updated…
Question Number 117862 by aurpeyz last updated on 14/Oct/20 Commented by aurpeyz last updated on 15/Mar/21 $${pls}\:{help} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 117865 by aurpeyz last updated on 14/Oct/20 $${discuss}\:{the}\:{lim}_{{n}\rightarrow\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \\ $$ Answered by 1549442205PVT last updated on 14/Oct/20 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{Lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{n}} =\mathrm{e}\approx\mathrm{2}.\mathrm{718} \\ $$$$\mathrm{e}^{\mathrm{x}}…
Question Number 117863 by aurpeyz last updated on 14/Oct/20 $${if}\:{y}=\frac{{x}+\mathrm{2}}{\:\sqrt{{x}+\mathrm{1}}}\:{find}\:\frac{{dy}}{{dx}}\:{from}\:{first} \\ $$$${principle}. \\ $$ Answered by bemath last updated on 14/Oct/20 $$\mathrm{y}=\frac{\mathrm{x}+\mathrm{1}+\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{1}}}\:=\:\sqrt{\mathrm{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{1}}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}+\mathrm{h}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{h}+\mathrm{1}}}−\sqrt{\mathrm{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{1}}}}{\mathrm{h}}…