Question Number 197238 by sonukgindia last updated on 10/Sep/23 Commented by Frix last updated on 12/Sep/23 $$\mathrm{No}\:\mathrm{useful}\:\mathrm{exact}\:\mathrm{solution}. \\ $$$$\mathrm{Set}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quarter}\:\mathrm{circle}\:=\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\mathrm{Radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{semi}\:\mathrm{circle}\:\approx.\mathrm{717910784} \\ $$$${r}\approx.\mathrm{205269551}…
Question Number 197234 by sonukgindia last updated on 10/Sep/23 Answered by HeferH last updated on 10/Sep/23 Commented by HeferH last updated on 11/Sep/23 $$\:\left({m}+{n}\right)^{\mathrm{2}} +{n}^{\mathrm{2}}…
Question Number 197228 by sonukgindia last updated on 10/Sep/23 Answered by cherokeesay last updated on 10/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 197225 by sonukgindia last updated on 10/Sep/23 Answered by Frix last updated on 10/Sep/23 $${x}=\sqrt[{\mathrm{2}}]{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{10}}{\mathrm{9}}{x}}} \\ $$$${x}^{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{10}}{\mathrm{9}}{x}} \\ $$$$\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{10}}{\mathrm{9}}{x}} \\ $$$$\mathrm{27}\left({x}^{\mathrm{2}}…
Question Number 197229 by sonukgindia last updated on 10/Sep/23 Commented by Sachinkhar last updated on 10/Sep/23 $$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{fourier}}\:\boldsymbol{\mathrm{inverse}}\:\boldsymbol{\mathrm{transform}}\:\boldsymbol{\mathrm{of}}\: \\ $$$$\boldsymbol{\mathrm{F}}\left(\boldsymbol{\xi}\right)=\boldsymbol{\mathrm{cosat}\xi}^{\mathrm{2}} \\ $$ Answered by HeferH last…
Question Number 197226 by sonukgindia last updated on 10/Sep/23 Answered by Frix last updated on 10/Sep/23 $$\int\frac{{dx}}{\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:\overset{{t}=\frac{\mathrm{2}{x}+\mathrm{1}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{\:\sqrt{\mathrm{3}}}} {=} \\ $$$$=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}\left({t}^{\mathrm{2}} +\frac{\mathrm{4}{t}}{\:\sqrt{\mathrm{3}}}+\mathrm{1}\right)}{dt}=\int\left(\frac{\mathrm{1}}{{t}}+\frac{\mathrm{2}}{{t}+\sqrt{\mathrm{3}}}−\frac{\mathrm{6}}{\mathrm{3}{t}+\sqrt{\mathrm{3}}}\right){dt}= \\…
Question Number 197227 by sonukgindia last updated on 10/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 197196 by sonukgindia last updated on 10/Sep/23 Answered by mahdipoor last updated on 10/Sep/23 $${blue}\equiv \\ $$$$\pi{r}^{\mathrm{2}} \left(\frac{\alpha_{\mathrm{1}} }{\mathrm{360}}+\frac{\alpha_{\mathrm{2}} }{\mathrm{360}}+\frac{\alpha_{\mathrm{3}} }{\mathrm{360}}…+\frac{\alpha_{{n}} }{\mathrm{360}}\right) \\…
Question Number 197194 by sonukgindia last updated on 10/Sep/23 Answered by witcher3 last updated on 10/Sep/23 $$\mathrm{k}^{\mathrm{1}−\mathrm{n}} +\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{1}−\mathrm{n}} =\frac{\left(\mathrm{1}+\mathrm{k}\right)^{\mathrm{n}−\mathrm{1}} +\mathrm{k}^{\mathrm{n}−\mathrm{1}} }{\left(\mathrm{k}\left(\mathrm{1}+\mathrm{k}\right)\right)^{\mathrm{n}−\mathrm{1}} } \\ $$$$\Leftrightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{50}}…
Question Number 197188 by sonukgindia last updated on 10/Sep/23 Answered by witcher3 last updated on 10/Sep/23 $$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(−\mathrm{x}\right)\left(\pi−\mathrm{arcos}\left(\mathrm{x}\right)\right)\mathrm{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{arccos}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{arccos}\left(\mathrm{x}\right)\left(−\mathrm{ln}\left(−\mathrm{x}\right)+\mathrm{ln}\left(\mathrm{x}\right)\right)+\pi\int_{\mathrm{0}}…