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Question Number 160035 by HongKing last updated on 23/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94479 by i jagooll last updated on 19/May/20 $$\mathrm{find}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{C}\: \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$ Answered by john santu last updated on…
Question Number 160008 by HongKing last updated on 23/Nov/21 $$\mathrm{x}_{\mathrm{1}} =\mathrm{3}\:;\:\mathrm{n}\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +…+\mathrm{x}_{\boldsymbol{\mathrm{n}}} \right)=\mathrm{x}_{\boldsymbol{\mathrm{n}}} \:;\:\mathrm{n}\in\mathbb{N}\:;\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\mathrm{Find}: \\ $$$$\Omega\:=\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\left(-\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:\mathrm{x}_{\boldsymbol{\mathrm{n}}} \\ $$ Commented…
Question Number 160009 by HongKing last updated on 23/Nov/21 $$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{n}!}\:\centerdot\underset{\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:…\:+\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }} {\overset{\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}}} {\int}}\:\mathrm{e}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\mathrm{dx}\right) \\ $$ Answered by…
Question Number 160006 by HongKing last updated on 23/Nov/21 $$\mathrm{Evaluate}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\underset{\boldsymbol{\mathrm{n}}} {\overset{\boldsymbol{\mathrm{n}}+\mathrm{1}} {\int}}\:\mathrm{e}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} \:\mathrm{dx}\:=\:? \\ $$$$ \\ $$ Commented by kowalsky78 last updated…
Question Number 160007 by HongKing last updated on 23/Nov/21 $$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{n}\left(\left(\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{n}}\right)^{\boldsymbol{\mathrm{n}}} -\:\mathrm{e}\:-\:\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} -\:\mathrm{e}^{-\:\frac{\mathrm{e}}{\mathrm{2}}} \right)\right) \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 94463 by Abdulrahman last updated on 18/May/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{6x}^{\mathrm{k}−\mathrm{2}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{10}}{\mathrm{3x}^{\mathrm{5}} +\mathrm{x}+\mathrm{20}} \\ $$$$\mathrm{then}\:\mathrm{k}^{\mathrm{2}} +\mathrm{1}=? \\ $$ Commented by john santu last updated…
Question Number 94456 by Abdulrahman last updated on 18/May/20 $$\mathrm{log}_{\mathrm{x}} \mathrm{16}+\mathrm{log}_{\mathrm{16}} \mathrm{x}=\mathrm{log}_{\mathrm{512}} \mathrm{x}+\mathrm{log}_{\mathrm{x}} \mathrm{512} \\ $$$$\mathrm{x}=? \\ $$ Answered by john santu last updated on…
Question Number 28921 by 803jaideep@gmail.com last updated on 01/Feb/18 Answered by Joel578 last updated on 01/Feb/18 $$\mathrm{10cos}\:\left(\mathrm{200}{t}\right)\:=\:\mathrm{8} \\ $$$$\mathrm{cos}\:\left(\mathrm{200}{t}\right)\:=\:\mathrm{0}.\mathrm{8} \\ $$$$\mathrm{200}{t}\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{8}\right)\:\approx\:\mathrm{37}° \\ $$ Terms…