Question Number 214366 by issac last updated on 06/Dec/24 $${f}\left(\alpha\right)=\int_{−\infty} ^{\:\infty} \:{e}^{−\alpha{x}^{\mathrm{2}} } \mathrm{d}{x} \\ $$$$\int\:\:{e}^{−\alpha{t}^{\mathrm{2}} } \mathrm{d}{t}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{\alpha}}\centerdot\mathrm{erf}\left(\sqrt{\alpha}{t}\right)+\mathrm{Const} \\ $$$$\therefore\int_{−\infty} ^{\:\infty} \:{e}^{−\alpha{t}^{\mathrm{2}} } \mathrm{d}{t}=\sqrt{\frac{\pi}{\alpha}}\:,\:\alpha\in\left(\mathrm{0},\infty\right) \\…
Question Number 214351 by MATHEMATICSAM last updated on 06/Dec/24 Commented by MATHEMATICSAM last updated on 06/Dec/24 $$\mathrm{Find}\:{a} \\ $$ Commented by mr W last updated…
Question Number 214341 by liuxinnan last updated on 06/Dec/24 $$\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} \\ $$$$\mathrm{or}\:\mathrm{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\mathrm{d}{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{d}{x}…
Question Number 214340 by issac last updated on 06/Dec/24 $$\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 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 $${a}\:{smooth}\:{line}\:{is}\:{always}\:{continous},…
Question Number 214317 by muallimRiyoziyot last updated on 05/Dec/24 $${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 $${x}^{\mathrm{4}}…
Question Number 214321 by MathematicsExpert last updated on 05/Dec/24 $$\mathrm{15}{x}\left(\frac{\mathrm{6}{x}}{\mathrm{10}}+\frac{\mathrm{12}{x}}{\mathrm{20}}+\frac{\mathrm{15}{x}}{\mathrm{30}}+…+\frac{\mathrm{60}{x}}{\mathrm{100}}\right)=\mathrm{160} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x} \\ $$ Answered by MATHEMATICSAM last updated on 06/Dec/24 $$\mathrm{15}{x}\left(\frac{\mathrm{6}{x}}{\mathrm{10}}\:+\:\frac{\mathrm{12}{x}}{\mathrm{20}}\:+\:\frac{\mathrm{15}{x}}{\mathrm{30}}\:+\:…\:+\:\frac{\mathrm{60}{x}}{\mathrm{100}}\right)\:=\:\mathrm{160} \\ $$$$\Rightarrow\:\mathrm{15}{x}\left(\frac{\mathrm{3}{x}}{\mathrm{5}}\:+\:\frac{\mathrm{3}{x}}{\mathrm{5}}\:+\:\frac{\mathrm{3}{x}}{\mathrm{5}}\:+\:…\:+\:\frac{\mathrm{3}{x}}{\mathrm{5}}\right)\:=\:\mathrm{160} \\…
Question Number 214293 by muallimRiyoziyot last updated on 04/Dec/24 $${P}\left({x}\right)\:\:\:\:\:\:\vdots\left({x}^{\mathrm{2}} +\mathrm{3}\right)\:\:\:\:\:{mod}\left(\mathrm{5}{x}−\mathrm{1}\right) \\ $$$${P}\left({x}\right)\:\:\:\:\:\:\vdots\left({x}−\mathrm{2}\right)\:\:\:\:\:\:\:\:\:{mod}\left(\mathrm{16}\right) \\ $$$${P}\left({x}\right)\:\:\:\:\:\:\vdots\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left({x}−\mathrm{2}\right)\:\:\:{mod}\left(?\right) \\ $$ Answered by MrGaster last updated on 24/Dec/24…
Question Number 214279 by issac last updated on 03/Dec/24 $$\mathrm{evaluate} \\ $$$$\rho=\frac{\oint_{\:\mathcal{C}} \:\frac{{z}}{\mathrm{2sin}\left({z}\right)−\mathrm{2}{z}\mathrm{cos}\left({z}\right)−\pi}\:\mathrm{d}{z}}{\oint_{\:\mathcal{C}} \:\frac{\mathrm{2}}{\mathrm{2sin}\left({z}\right)−\mathrm{2}{z}\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}} \\ $$$$\mathcal{C}=\mid{z}−\frac{\mathrm{3}\pi}{\mathrm{4}}\mid=\frac{\pi}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 214244 by daniella last updated on 02/Dec/24 $$\mathrm{4}=\sqrt{{x}−\mathrm{7}} \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated on 02/Dec/24 $${x}−\mathrm{7}=\mathrm{16} \\ $$$${x}=\mathrm{23} \\…