Question Number 214350 by ajfour last updated on 06/Dec/24 Answered by mr W last updated on 06/Dec/24 $$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\sqrt{\mathrm{5}}−\mathrm{2} \\ $$$$\mathrm{2}+{r}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\sqrt{\left({r}+\mathrm{1}\right)^{\mathrm{2}} −\left({r}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}+\left(\sqrt{\mathrm{5}}−\mathrm{2}\right){r}=\mathrm{2}\sqrt{{r}} \\…
Question Number 214351 by MATHEMATICSAM last updated on 06/Dec/24 Commented by MATHEMATICSAM last updated on 06/Dec/24 $$\mathrm{Find}\:{a} \\ $$ Commented by mr W last updated…
Question Number 214372 by ajfour last updated on 06/Dec/24 Commented by ajfour last updated on 06/Dec/24 $${Find}\:{maximum}\:{r}.\:{Outer}\:{figure}\:{is}\:{a} \\ $$$${rhombus}.{Circle}\:{is}\:{inscribed}\:{in}\:{an}\: \\ $$$${isosceles}\:{triangle}\:{as}\:{shown}\:{above}. \\ $$ Answered by…
Question Number 214340 by issac last updated on 06/Dec/24 $$\mathrm{evaluate} \\ $$$$\frac{\oint_{{C}} \:\frac{{z}}{\mathrm{2}\centerdot\mathrm{sin}\left({z}\right)−\mathrm{2}{z}\centerdot\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}}{\oint_{\:{C}} \:\frac{\mathrm{2}}{\mathrm{2}\centerdot\mathrm{sin}\left({z}\right)−\mathrm{2}{z}\centerdot\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}} \\ $$$$\mathrm{where}\:{C}\:\mathrm{is}\:\mathrm{the}\:\mathrm{circle}\mid{z}−\frac{\mathrm{3}\pi}{\mathrm{4}}\mid=\frac{\pi}{\mathrm{4}}. \\ $$ Answered by MrGaster last updated on 24/Dec/24…
Question Number 214341 by liuxinnan last updated on 06/Dec/24 $$\int\frac{{dx}}{\mathrm{3}+{cosx}}=? \\ $$ Answered by chhaythean last updated on 06/Dec/24 $$\mathrm{let},\:\mathrm{t}\:=\:\mathrm{tan}\frac{{x}}{\mathrm{2}}\:\Rightarrow\:\mathrm{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\mathrm{d}{x} \\ $$$$\mathrm{or}\:\mathrm{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\mathrm{d}{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{d}{x}…
Question Number 214342 by issac last updated on 06/Dec/24 $$\mathrm{why} \\ $$$$\mathrm{differantiable}\:{f}\:\rightarrow\:{f}\:\mathrm{is}\:\mathrm{continious}\: \\ $$$$\mathrm{but}\:{f}\:\mathrm{is}\:\mathrm{continous}\:\nrightarrow\:\mathrm{differantiable}\:?? \\ $$ Commented by mr W last updated on 06/Dec/24 $${a}\:{smooth}\:{line}\:{is}\:{always}\:{continous},…
Question Number 214369 by ajfour last updated on 06/Dec/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 214317 by muallimRiyoziyot last updated on 05/Dec/24 $${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{11}{x}^{\mathrm{2}} +{x}−\mathrm{12}={f}\left({x}\right)×{g}\left({x}\right) \\ $$$${f}\left({x}\right)=?\:\:\:\:{g}\left({x}\right)=? \\ $$ Answered by A5T last updated on 05/Dec/24 $${x}^{\mathrm{4}}…
Question Number 214335 by depressiveshrek last updated on 05/Dec/24 $$\mathrm{For}\:\mathrm{what}\:\mathrm{values}\:\mathrm{of}\:{k}\:\mathrm{does}\:\mathrm{the}\:\mathrm{equation} \\ $$$${e}^{{kx}} =\mathrm{3}\sqrt{{x}}\:\mathrm{have}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{R}? \\ $$ Answered by mr W last updated on 06/Dec/24 $${y}={e}^{{kx}} \:{and}\:{y}=\mathrm{3}\sqrt{{x}}\:{should}\:{have}\:{only}…
Question Number 214319 by golsendro last updated on 05/Dec/24 $$\:\:\:\:\mathrm{f}\left(\frac{\mathrm{10x}+\mathrm{3}}{\mathrm{10x}−\mathrm{3}}\:\right)=\:\frac{\mathrm{10}}{\mathrm{3}}\:\mathrm{x} \\ $$$$\:\:\:\mathrm{f}\left(\mathrm{4}\right).\mathrm{f}\left(\mathrm{6}\right).\mathrm{f}\left(\mathrm{8}\right).\mathrm{f}\left(\mathrm{10}\right)…\mathrm{f}\left(\mathrm{2024}\right)=? \\ $$ Answered by A5T last updated on 05/Dec/24 $$\frac{\mathrm{10}{y}+\mathrm{3}}{\mathrm{10}{y}−\mathrm{3}}={x}\Rightarrow\mathrm{10}{y}+\mathrm{3}=\mathrm{10}{xy}−\mathrm{3}{x} \\ $$$${y}=\frac{−\mathrm{3}−\mathrm{3}{x}}{\mathrm{10}−\mathrm{10}{x}} \\…