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Question-199368

Question Number 199368 by sonukgindia last updated on 02/Nov/23 Answered by qaz last updated on 02/Nov/23 $$\int_{−\infty} ^{+\infty} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx} \\ $$$$=\mathrm{2}\pi{i}\left(\left[\left({z}−{i}\right)^{−\mathrm{1}} \right]+\left[\left({z}−\mathrm{2}{i}\right)^{−\mathrm{1}}…

Question-199338

Question Number 199338 by sonukgindia last updated on 01/Nov/23 Answered by aleks041103 last updated on 01/Nov/23 $$\mathrm{5}^{\mathrm{2}^{{M}} } −\mathrm{1}=\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } \right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} =\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } +\mathrm{1}\right)\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}}…

Question-199174

Question Number 199174 by tri26112004 last updated on 29/Oct/23 Answered by ajfour last updated on 29/Oct/23 $${If}\:{i}\:{take}\:\angle{ACI}\:=\theta\:\:{then} \\ $$$$\frac{{CI}}{{HI}}=\frac{\mathrm{sin}\:\theta}{\mathrm{2}−\mathrm{sin}\:\theta} \\ $$ Commented by tri26112004 last…

Question-199175

Question Number 199175 by SANOGO last updated on 29/Oct/23 Answered by a.lgnaoui last updated on 29/Oct/23 $$\mathrm{g}_{\mathrm{m}} \left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{2x}}{\mathrm{m}}+\frac{\mathrm{1}}{\mathrm{m}^{\mathrm{2}} }\:\:\:\:\left(\mathrm{m}\neq\mathrm{0}\:\:\:\:\mathrm{x}\geqslant\mathrm{0}\right) \\ $$$$\:\:\:\:\mathrm{n}\:\mathrm{est}\:\mathrm{pas}\:\mathrm{une}\:\mathrm{suite}\:\mathrm{convergente} \\ $$$$\:\mathrm{preuve}: \\…

i-m-Calculated-gauess-law-in-Gravity-Field-S-g-dS-g-x-y-z-Gmx-x-2-y-2-z-2-e-1-Gmy-x-2-y-2-z-2-e-2-Gmz-x-2-y-2-z-2-e-3-S-is-Clo

Question Number 199157 by MathedUp last updated on 30/Oct/23 $$\mathrm{i}'\mathrm{m}\:\mathrm{Calculated}\:\:\mathrm{gauess}\:\mathrm{law}\:\mathrm{in}\:\mathrm{Gravity}\:\mathrm{Field} \\ $$$$ \\ $$$$\int\int_{\:\boldsymbol{{S}}} \:\hat {\boldsymbol{\mathrm{g}}}\centerdot\mathrm{d}\hat {\boldsymbol{\mathrm{S}}} \\ $$$$\hat {\boldsymbol{\mathrm{g}}}\left({x},{y},{z}\right)=−\frac{{Gmx}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\hat {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −\frac{{Gmy}}{\:\sqrt{{x}^{\mathrm{2}}…