Question Number 203826 by necx122 last updated on 29/Jan/24 $${Evaluate} \\ $$$$\left(\mathrm{26}\:+\:\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}} +\:\left(\mathrm{26}\:−\:\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$ Answered by AST last updated on 29/Jan/24 $${Let}\:{a}=\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}};{b}=\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}}\Rightarrow{ab}=\mathrm{1} \\ $$$${a}^{\mathrm{3}}…
Question Number 203807 by aba last updated on 28/Jan/24 $$\mathrm{proof}\::\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{f}^{\mathrm{2}} \left(\mathrm{t}\right)\mathrm{dt}=\mathrm{0}\:\Rightarrow\:\mathrm{f}=\mathrm{0} \\ $$ Answered by JDamian last updated on 28/Jan/24 $$ \\ $$$$\mathrm{f}^{\mathrm{2}}…
Question Number 203809 by Fabricista15 last updated on 28/Jan/24 $${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{33} \\ $$ Answered by Fabricista15 last updated on 28/Jan/24 $$\mathrm{Ver}\:\mathrm{E}{quacao}\:{Diofantina} \\ $$…
Question Number 203774 by Calculusboy last updated on 27/Jan/24 $$\boldsymbol{{determine}}\:\boldsymbol{{whether}}\:\boldsymbol{{the}}\:\boldsymbol{{series}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{convergent}}\:\boldsymbol{{or}}\:\boldsymbol{{divergent}} \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\boldsymbol{{n}}}{\:\sqrt{\mathrm{4}\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{1}}} \\ $$ Answered by witcher3 last updated on…
Question Number 203742 by Calculusboy last updated on 27/Jan/24 Answered by mr W last updated on 27/Jan/24 $${x}^{\mathrm{2024}} +{x}^{\mathrm{2024}} −\mathrm{2024}×\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2023}} +…=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2024}} −\mathrm{506}{x}^{\mathrm{2023}} +…=\mathrm{0}…
Question Number 203771 by Calculusboy last updated on 27/Jan/24 Answered by DwaipayanShikari last updated on 27/Jan/24 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\begin{pmatrix}{{n}+\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}!}{\left({n}+\mathrm{3}\right)!\mathrm{3}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}!}\underset{{n}=\mathrm{0}}…
Question Number 203638 by thean0000 last updated on 24/Jan/24 Commented by thean0000 last updated on 24/Jan/24 $${help}\:{me} \\ $$$$ \\ $$ Commented by thean0000 last…
Question Number 203570 by Frix last updated on 22/Jan/24 $$\mathrm{Suggested}\:\mathrm{solution}\:\mathrm{method}\:\mathrm{to} \\ $$$$\mathrm{question}\:\mathrm{203502} \\ $$$$ \\ $$$${z}\mathrm{e}^{{z}} =\mathrm{1} \\ $$$$\mathrm{Obviously}\:\mathrm{the}\:\mathrm{only}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{is} \\ $$$${z}={W}\left(\mathrm{1}\right)\approx.\mathrm{567143290} \\ $$$$ \\ $$$${z}={a}+{b}\mathrm{i}\wedge{b}\neq\mathrm{0}…
Question Number 203474 by ajfour last updated on 20/Jan/24 $$\frac{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}}=\frac{{z}+\mathrm{1}}{{z}−\mathrm{1}} \\ $$$${Find}\:{z}\in\mathbb{R}. \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 203502 by Frix last updated on 20/Jan/24 $${z}\mathrm{e}^{{z}} =\mathrm{e} \\ $$$$\mathrm{Obviously}\:{z}=\mathrm{1} \\ $$$$\mathrm{Now}\:\mathrm{find}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$ Commented by Frix last updated on 22/Jan/24 $$\mathrm{See}\:\mathrm{question}\:\mathrm{203570}…