Question Number 207116 by hardmath last updated on 06/May/24 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{x}\:−\:\mathrm{2y}}\\{\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{y}\:−\:\mathrm{2x}}\end{cases}\:\:\:\:\:\Rightarrow\:\:\:\:\:\mathrm{x}\:−\:\mathrm{y}\:=\:? \\ $$ Answered by A5T last updated on 06/May/24 $${x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={x}−{y}+\mathrm{2}\left({x}−{y}\right)=\mathrm{3}\left({x}−{y}\right) \\…
Question Number 207085 by necx122 last updated on 06/May/24 $${Let}\:{f}\:{be}\:{a}\:{function}\:{with}\:{the}\:{following} \\ $$$${properties}:\:\left({i}\right)\:{f}\left(\mathrm{1}\right)\:=\mathrm{1}\:\left({ii}\right)\:{f}\left(\mathrm{2}{n}\right)={n}.{f}\left({n}\right)\:{for} \\ $$$${any}\:{positive}\:{integer}\:{n}.\:{Find}\:{the}\:{value} \\ $$$${of}\:{f}\left(\mathrm{2}^{\mathrm{10}} \right) \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left.\:{b}\right)\:\mathrm{2}^{\mathrm{10}\:} {c}\right)\:\mathrm{2}^{\mathrm{35}} \:{d}\right)\:\mathrm{2}^{\mathrm{45}} \\ $$ Answered by…
Question Number 207096 by Wuji last updated on 06/May/24 $${for}\:{the}\:{given}\:{system}\:{of}\:{ODEs},\:{calculate}\:{the} \\ $$$${eigenvalues}\:{and}\:{corresponding}\:{eigenvectors}\:{of}\:{the}\: \\ $$$${coefficient}\:{matrix} \\ $$$$\frac{{dx}}{{dt}}=\mathrm{2}{x}+{y}\:\:\:\frac{{dy}}{{dt}}={x}+\mathrm{2}{y} \\ $$ Commented by Wuji last updated on 07/May/24…
Question Number 207082 by SEKRET last updated on 06/May/24 $$\:\:\:\boldsymbol{\mathrm{Let}}\:\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)\:\:\boldsymbol{\mathrm{be}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{inverse}}\:\boldsymbol{\mathrm{function}}\:\:\:\boldsymbol{\mathrm{of}} \\ $$$$ \\ $$$$\:\:\:\:−−>\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{x}}+\mathrm{5}\:\:\:\:\:\:<−− \\ $$$$\:\: \\ $$$$ \\ $$$$\boldsymbol{\mathrm{Evaluate}}\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\boldsymbol{\mathrm{Lim}}}\:\mathrm{4}\boldsymbol{\mathrm{n}}\centerdot\left(\:\boldsymbol{\mathrm{g}}\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\right)\:−\boldsymbol{\mathrm{g}}\left(\mathrm{1}−\frac{\mathrm{2}}{\boldsymbol{\mathrm{n}}}\right)\:\right)=? \\ $$$$…
Question Number 207099 by tri26112004 last updated on 06/May/24 Answered by Berbere last updated on 06/May/24 $$=\int_{−\infty} ^{\infty} \frac{{e}^{{i}\pi{ax}} }{\left({x}^{\mathrm{2}} +\beta^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }{dx};{a}\in\mathbb{R}_{+} \\ $$$${if}\:{Imx}\geqslant\mathrm{0}\:\mid{e}^{{i}\pi{ax}}…
Question Number 207083 by HtetArkarKyaw last updated on 06/May/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 207092 by hardmath last updated on 06/May/24 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{x}\:−\:\mathrm{6y}}\\{\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{y}\:−\:\mathrm{6x}}\end{cases}\:\:\:\:\:\Rightarrow\:\:\:\:\mathrm{x}\:+\:\mathrm{y}\:=\:? \\ $$ Answered by A5T last updated on 06/May/24 $${x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={x}−{y}−\mathrm{6}{y}+\mathrm{6}{x}=\mathrm{7}{x}−\mathrm{7}{y} \\…
Question Number 207109 by MATHEMATICSAM last updated on 06/May/24 $$\mathrm{If}\:{y}\:=\:\left(\mathrm{1}\:+\:{x}\right)\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\left(\mathrm{1}\:+\:{x}^{\mathrm{4}} \right)\:….\:\left(\mathrm{1}\:+\:{x}^{\mathrm{2}{n}} \right) \\ $$$$\mathrm{then}\:\mathrm{find}\:\frac{{dy}}{{dx}}\:\mathrm{at}\:{x}\:=\:\mathrm{0}. \\ $$ Answered by Berbere last updated on 06/May/24 $${y}\left({x}\right)=\left(\mathrm{1}+{x}\right)\underset{{k}=\mathrm{1}}…
Question Number 207106 by Wuji last updated on 07/May/24 $${Calculate}\:{the}\:{generalized}\:{solution}\:{for}\:{the}\:{following} \\ $$$${system}\:{of}\:{ODEs}: \\ $$$$\frac{{dx}}{{dt}}=−\frac{\mathrm{1}}{\mathrm{2}}{x},\:\frac{{dy}}{{dt}}=\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{4}}{y},\:\:\frac{{dz}}{{dt}}=\frac{\mathrm{1}}{\mathrm{4}}{y}−\frac{\mathrm{1}}{\mathrm{6}}{z} \\ $$ Answered by mr W last updated on 07/May/24 $$\begin{vmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}−\lambda}&{\mathrm{0}}&{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{2}}}&{−\frac{\mathrm{1}}{\mathrm{4}}−\lambda}&{\mathrm{0}}\\{\mathrm{0}}&{\frac{\mathrm{1}}{\mathrm{4}}}&{−\frac{\mathrm{1}}{\mathrm{6}}−\lambda}\end{vmatrix}=\mathrm{0}…
Question Number 207054 by Ghisom last updated on 05/May/24 $$\Omega_{\alpha} =\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\alpha} \sqrt{−{x}\mathrm{ln}\:{x}}\:{dx}=? \\ $$ Answered by Frix last updated on 05/May/24 $$\alpha>−\frac{\mathrm{3}}{\mathrm{2}} \\…