Question Number 197794 by universe last updated on 28/Sep/23 $$\:\:\:\:\mathrm{if}\:\mathrm{x}\:\:\:=\:\:\:\mathrm{log}\:\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{y}}{\mathrm{2}}\right),\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\mathrm{y}\:\:\:\:=\:\:\:−{i}\mathrm{log}\:\mathrm{tan}\left(\frac{{ix}}{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{4}}\right)\:\:\:\:\:\mathrm{here}\:{i}\:\:=\:\sqrt{−\mathrm{1}} \\ $$ Answered by Frix last updated on 29/Sep/23 $${x}=\mathrm{ln}\:\mathrm{tan}\:\frac{\mathrm{2}{y}+\pi}{\mathrm{4}}\:\Leftrightarrow\:{y}=−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}} \:\mathrm{e}^{{x}} \\ $$$$−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}}…
Question Number 197708 by karowan last updated on 26/Sep/23 $$ \\ $$$${x}^{{a}} ={x}^{{a}+\mathrm{4}} \:\:\mathrm{where}\:{a}\in\mathbb{Z} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x}\:\mathrm{by}\:\mathrm{showing}\:\mathrm{steps} \\ $$ Answered by Frix last updated on 26/Sep/23…
Question Number 197706 by universe last updated on 26/Sep/23 Answered by witcher3 last updated on 27/Sep/23 $$\left(\frac{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\Leftrightarrow\frac{\mathrm{a}+\mathrm{b}}{\mathrm{4}}\geqslant\frac{\sqrt{\mathrm{ab}}}{\mathrm{2}}.\mathrm{Am}−\mathrm{GM} \\ $$$$\left.\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }}\leqslant\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\right.}\right) \\…
Question Number 197623 by hardmath last updated on 24/Sep/23 $$\mathrm{x},\mathrm{y}\in\mathbb{N} \\ $$$$\mathrm{162}\:\centerdot\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}^{\mathrm{3}} \\ $$$$\mathrm{min}\left(\mathrm{x}+\mathrm{y}\right)=? \\ $$ Commented by SANOGO last updated on 25/Sep/23 merci bien monsieur…
Question Number 197618 by srijanGuha last updated on 24/Sep/23 Answered by Rasheed.Sindhi last updated on 24/Sep/23 $$\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{2} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}}…
Question Number 197609 by dimentri last updated on 24/Sep/23 $$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}\:+\:\frac{{x}}{\mathrm{2025}}\:=\:−\mathrm{4} \\ $$ Answered by Rasheed.Sindhi last updated on 24/Sep/23 $$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}\:+\:\frac{{x}}{\mathrm{2025}}\:=\:−\mathrm{4} \\ $$$$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}+\mathrm{1}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}+\mathrm{1}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}+\mathrm{1}\:+\:\frac{{x}}{\mathrm{2025}}+\mathrm{1}\:=\:−\mathrm{4}+\mathrm{4} \\ $$$$\:\:\:\frac{{x}+\mathrm{2025}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2024}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2025}}\:=\:\mathrm{0} \\…
Question Number 197567 by hardmath last updated on 21/Sep/23 Answered by Frix last updated on 22/Sep/23 $$\mathrm{Obviously}\:{x}={y}=\mathrm{1} \\ $$ Commented by Frix last updated on…
Question Number 197530 by Erico last updated on 20/Sep/23 $$\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{f}''\left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=? \\ $$ Commented by AST last updated on 20/Sep/23 $${f}\left({x}\right)={x}^{\mathrm{2}} \:{works}. \\…
Question Number 197476 by universe last updated on 19/Sep/23 $$ \\ $$$$\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\:\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}}{{x}^{\mathrm{4}} +\mathrm{1}}+………+\frac{\mathrm{2}^{{n}} }{{x}^{\mathrm{2}^{{n}} } +\mathrm{1}}\:\:=\:?? \\ $$ Commented by witcher3 last…
Question Number 197455 by mathlove last updated on 18/Sep/23 $${f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{6}}{\mathrm{2}}+\mathrm{6}−{x} \\ $$$${g}\left({x}\right)=\sqrt{{x}^{\mathrm{99}} +{x}^{\mathrm{98}} +{x}^{\mathrm{97}} +…..+{x}} \\ $$$${f}\left({g}\left(\mathrm{1}\right)\right)+{f}\left({g}\left(\mathrm{2}\right)\right)+………+{f}\left({g}\left(\mathrm{100}\right)\right)=? \\ $$ Answered by hmr last updated on…