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Question Number 61178 by mathsolverby Abdo last updated on 30/May/19 $${solve}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \:+\left(\mathrm{1}−{x}^{\mathrm{2}} \right){y}\:={x}\:{e}^{−\mathrm{3}{x}} \\ $$ Commented by maxmathsup by imad last updated on 31/May/19…
Question Number 126712 by help last updated on 23/Dec/20 Commented by liberty last updated on 24/Dec/20 $$\:\left(\bullet\right)\:{x}+\frac{\mathrm{1}}{{x}}\:=\:{w}\:\Rightarrow{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:{w}^{\mathrm{2}} −\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\:=\:{w}^{\mathrm{2}} −\mathrm{2} \\…
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Question Number 126710 by slahadjb last updated on 23/Dec/20 $$\int{ln}\left({x}\right){e}^{{x}^{\mathrm{2}} } {dx}\:\:\:\:\:\:??? \\ $$ Answered by Dwaipayan Shikari last updated on 23/Dec/20 $$\int{e}^{{x}^{\mathrm{2}} } {log}\left({x}\right){dx}\:\:\:\:\:{x}^{\mathrm{2}}…
Question Number 126708 by slahadjb last updated on 23/Dec/20 $$\boldsymbol{{solve}}\:\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{x}}^{\sqrt{\mathrm{2}}} =\mathrm{1} \\ $$ Commented by Dwaipayan Shikari last updated on 23/Dec/20 $$\sim\mathrm{0}.\mathrm{559793} \\ $$ Commented…
Question Number 192241 by yaslm last updated on 12/May/23 Answered by Frix last updated on 12/May/23 $${Z}_{\mathrm{1}} =\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}{x}} }=−\frac{\mathrm{1}−\mathrm{2cos}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}−\frac{\mathrm{2sin}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}\mathrm{i} \\ $$$${Z}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{{ix}} }=\frac{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{x}\right)}{\mathrm{5}−\mathrm{4cos}\:{x}}+\frac{\mathrm{2sin}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}\mathrm{i} \\…
Question Number 126704 by Dwaipayan Shikari last updated on 23/Dec/20 $$\frac{{e}^{\pi} −\mathrm{1}}{{e}^{\pi} +\mathrm{1}}=\frac{\pi}{\mathrm{2}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}+\frac{\pi^{\mathrm{2}} }{\mathrm{10}+\frac{\pi^{\mathrm{2}} }{\mathrm{14}+….}}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 61169 by Tawa1 last updated on 29/May/19 Commented by Tawa1 last updated on 29/May/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Black} \\ $$ Answered by mr W last updated…