Question Number 220540 by hardmath last updated on 14/May/25 $$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\mathrm{n}^{\mathrm{3}} \centerdot\left(\mathrm{n}\:+\:\mathrm{1}\right)^{\mathrm{3}} \centerdot\left(\mathrm{2n}\:+\:\mathrm{1}\right)^{\mathrm{2}} }\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 220526 by Larry last updated on 14/May/25 Answered by Frix last updated on 14/May/25 $$=\int{x}^{−\frac{\mathrm{2}}{\mathrm{7}}} {dx}=\frac{\mathrm{7}}{\mathrm{8}}{x}^{\frac{\mathrm{5}}{\mathrm{7}}} +{C} \\ $$ Answered by SdC355 last…
Question Number 220511 by mehdee7396 last updated on 14/May/25 $${AB}=\mathrm{2}{CE}\:\:\&\:\:{DE}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}\:\:\: \\ $$$${CE}\:\bot{AB}\:\:\:\&\:\:\:{AD}\bot{BC}\:\:\&\:\:{AB}={AC}\:\:\&\:{EF}\:\bot{BC}\: \\ $$$${BF}=? \\ $$$$ \\ $$ Answered by mehdee7396 last updated on 14/May/25…
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Question Number 220502 by SdC355 last updated on 14/May/25 $$\mathrm{each}\:{J}_{\nu} \left({z}\right),{Y}_{\nu} \left({z}\right)\:\mathrm{are}\:\mathrm{linear}\:\mathrm{independent}….?? \\ $$$${W}_{\mathrm{Ronskian}} \left\{{J}_{\nu} ^{\:} \left({z}\right),{Y}_{\nu} \left({z}\right)\right\}=\begin{vmatrix}{{J}_{\nu} \left({z}\right)}&{\:{Y}_{\nu} \left({z}\right)}\\{{J}_{\nu} '\left({z}\right)}&{{Y}_{\nu} '\left({z}\right)}\end{vmatrix} \\ $$$$={J}_{\nu} ^{\left(\mathrm{1}\right)}…
Question Number 220519 by hardmath last updated on 14/May/25 $$\mathrm{Solve}\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$\frac{\mathrm{15}}{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{3x}\:+\:\mathrm{4}}\:\:+\:\:\frac{\mathrm{7}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7x}}\:\:+\:\:\frac{\mathrm{10}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:-\:\mathrm{21}}\:\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$ Commented by Nicholas666 last updated on 14/May/25 $$…
Question Number 220473 by mehdee7396 last updated on 13/May/25 $${lim}_{{n}\rightarrow\infty} \left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)\right)^{{n}} \:\:\:\:\:\overset{{t}=\frac{\mathrm{1}}{{n}}} {\rightarrow} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)−\mathrm{1}\right]×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{tant}}{\mathrm{1}−{tant}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{tant}}{\mathrm{1}−{tant}}\right)×\frac{\mathrm{1}}{{t}}=\mathrm{2}…
Question Number 220474 by Rojarani last updated on 13/May/25 Commented by Ghisom last updated on 13/May/25 $${a}_{{n}} =\mathrm{2}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\sqrt{{n}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}}}= \\ $$$$=\mathrm{4}{n}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}\sqrt{\mathrm{4}{n}^{\mathrm{4}} +\mathrm{1}} \\…
Question Number 220468 by Rojarani last updated on 13/May/25 $$\:{a}+{b}+{c}=\mathrm{1},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}\:\:\left({a},{b},{c}\:\in{R}\right) \\ $$$$\:{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$ Answered by…
Question Number 220469 by Nicholas666 last updated on 13/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{6}{x}\left(\mathrm{1}\:−\:{x}\right)}{\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:\mathrm{ln}\:\left({x}\:+\:\mathrm{1}\right)}\:{dx} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…