Question Number 201401 by MrGHK last updated on 05/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201393 by mokys last updated on 05/Dec/23 $${by}\:{diffention}\:{find}\:{f}\:'\left({z}\right)\:{of}\:{f}\left({z}\right)\:=\:\sqrt[{\mathrm{3}}]{{z}} \\ $$ Commented by mokys last updated on 05/Dec/23 $${how}\:{can}\:{solve}\:{this} \\ $$ Answered by aleks041103…
Question Number 201373 by sonukgindia last updated on 05/Dec/23 Answered by aleks041103 last updated on 05/Dec/23 Commented by aleks041103 last updated on 05/Dec/23 $${AQ}.{BQ}={DQ}.{FQ} \\…
Question Number 201374 by MrGHK last updated on 05/Dec/23 Answered by witcher3 last updated on 05/Dec/23 $$\mathrm{xy}=\mathrm{z} \\ $$$$\Rightarrow\mathrm{ydx}=\mathrm{z} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{y}} \frac{\mathrm{tan}^{−\mathrm{1}}…
Question Number 201357 by sonukgindia last updated on 05/Dec/23 Commented by mahdipoor last updated on 05/Dec/23 $$\left(\mathrm{1}−\mathrm{4}{i}\right){z}=\mathrm{4}{ni}\Rightarrow{z}=\frac{\mathrm{4}{ni}}{\mathrm{1}−\mathrm{4}{i}}=\frac{\mathrm{4}{n}\left(\mathrm{4}+{i}\right)}{\mathrm{17}} \\ $$$${im}\left({z}\right)=\frac{\mathrm{4}{n}}{\mathrm{17}}=\mathrm{164}\Rightarrow{n}=\mathrm{17}×\mathrm{41}=\mathrm{697} \\ $$$${z}=\mathrm{656}+\mathrm{164}{i} \\ $$ Terms of…
Question Number 201370 by sonukgindia last updated on 05/Dec/23 Commented by mr W last updated on 05/Dec/23 $${is}\:{it}\:{the}\:{same}\:{question}\:{as}\:{Q}\mathrm{201346}? \\ $$ Terms of Service Privacy Policy…
Question Number 201353 by sonukgindia last updated on 05/Dec/23 Commented by mr W last updated on 05/Dec/23 $${do}\:{you}\:{have}\:{the}\:{answer}? \\ $$ Answered by mr W last…
Question Number 201355 by MrGHK last updated on 05/Dec/23 Answered by witcher3 last updated on 05/Dec/23 $$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{xy}\right)\mathrm{y}}{\left(\mathrm{xy}+\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)}\mathrm{dxdy}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\: \\ $$ Terms…
Question Number 201348 by sonukgindia last updated on 05/Dec/23 Answered by aleks041103 last updated on 05/Dec/23 $${I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({sin}\left({x}\right)\right){dx} \\ $$$${u}=\pi/\mathrm{2}−{x}\Rightarrow{du}=−{dx} \\ $$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} {ln}\left({cos}\left({u}\right)\right){du}…
Question Number 201349 by sonukgindia last updated on 05/Dec/23 Answered by Sutrisno last updated on 05/Dec/23 $${misal}\:: \\ $$$${A}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({cosx}\right)\:{dx} \\ $$$${B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({sinx}\right)\:{dx}…