Question Number 198443 by lapache last updated on 20/Oct/23 $${Proove}\: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\left({xi}+{yj}\right)={m}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{xi}+{n}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}{yj} \\ $$ Answered by mr…
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Question Number 198413 by mokys last updated on 19/Oct/23 Commented by mokys last updated on 19/Oct/23 $${find}\:{log}_{\mathrm{12}} \mathrm{60} \\ $$ Answered by cortano12 last updated…
Question Number 198352 by sonukgindia last updated on 18/Oct/23 Commented by Rasheed.Sindhi last updated on 18/Oct/23 $${Q}#\mathrm{198313} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 198355 by lapache last updated on 18/Oct/23 $${solve}\:{the}\:{EDP} \\ $$$$\mathrm{1}−\:\:\:{x}\frac{\partial{f}}{\partial{x}}−{y}\frac{\partial{f}}{\partial{y}}=\mathrm{0} \\ $$$$\mathrm{2}−\:\:\:{x}\frac{\partial{f}}{\partial{x}}−\frac{\partial{f}}{\partial{y}}=\frac{{z}^{\mathrm{2}} }{{x}} \\ $$$$ \\ $$$$\mathrm{3}−\:\:{x}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }−{y}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }=\mathrm{0}…
Question Number 198377 by sonukgindia last updated on 18/Oct/23 Answered by witcher3 last updated on 19/Oct/23 $$=\int_{\mathrm{0}} ^{\infty} \mathrm{xe}^{−\mathrm{4x}^{\mathrm{2}} } .\mathrm{xe}^{\frac{−\mathrm{9}}{\mathrm{x}^{\mathrm{2}} }} \mathrm{dx}\:\:\:\mathrm{IBP} \\ $$$$=\left[−\frac{\mathrm{e}^{−\mathrm{4x}^{\mathrm{2}}…
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Question Number 198302 by sonukgindia last updated on 17/Oct/23 Answered by mr W last updated on 17/Oct/23 $${z}_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{\mathrm{3}}=\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} =\mathrm{2}{e}^{{i}\pi} =−\mathrm{2} \\…
Question Number 198326 by MrGHK last updated on 17/Oct/23 $$\boldsymbol{\mathrm{express}}\:\boldsymbol{\Gamma}\left(\boldsymbol{{k}}−\boldsymbol{{n}}\right)\:\boldsymbol{{in}}\:\boldsymbol{{pocchammer}}\:\boldsymbol{{symbol}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com