Question Number 214402 by golsendro last updated on 07/Dec/24 $$\:\:\:\mathrm{If}\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}\:=\:\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}\:\mathrm{then}\:\mathrm{find}\: \\ $$$$\:\:\:\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:. \\ $$ Answered by efronzo1 last updated on 07/Dec/24 $$\:\:\:\:\mathrm{Let}\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}\:=\:\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}\:=\:\mathrm{m}\: \\ $$$$\:\:\:\:\:\mathrm{then}\:\mathrm{a}+\mathrm{b}\:=\:\mathrm{mc}\:;\:\mathrm{b}+\mathrm{c}\:=\:\mathrm{ma}\:;\:\mathrm{a}+\mathrm{c}\:=\:\mathrm{mb} \\…
Question Number 214366 by issac last updated on 06/Dec/24 $${f}\left(\alpha\right)=\int_{−\infty} ^{\:\infty} \:{e}^{−\alpha{x}^{\mathrm{2}} } \mathrm{d}{x} \\ $$$$\int\:\:{e}^{−\alpha{t}^{\mathrm{2}} } \mathrm{d}{t}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{\alpha}}\centerdot\mathrm{erf}\left(\sqrt{\alpha}{t}\right)+\mathrm{Const} \\ $$$$\therefore\int_{−\infty} ^{\:\infty} \:{e}^{−\alpha{t}^{\mathrm{2}} } \mathrm{d}{t}=\sqrt{\frac{\pi}{\alpha}}\:,\:\alpha\in\left(\mathrm{0},\infty\right) \\…
Question Number 214360 by shunmisaki007 last updated on 06/Dec/24 $$\mathrm{Find}\:\underset{−\infty} {\overset{\infty} {\int}}{e}^{−{ax}^{\mathrm{2}} } {dx}\:\mathrm{when}\:{a}\:\mathrm{is}\:\mathrm{constant}\:\mathrm{without}\:\mathrm{changing}\:\mathrm{the}\:\mathrm{coordinate}. \\ $$ Answered by mathmax last updated on 06/Dec/24 $$=\mathrm{2}\int_{\mathrm{0}} ^{\infty}…
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