Question Number 7585 by Tawakalitu. last updated on 04/Sep/16 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx} \\ $$ Commented by sou1618 last updated on 05/Sep/16 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}}…
Question Number 138658 by ajfour last updated on 16/Apr/21 Commented by ajfour last updated on 16/Apr/21 $${Find}\:{maximum}\:{area}\:{of}\:\bigtriangleup{AOB}. \\ $$$${a}=\mathrm{1}. \\ $$ Commented by mr W…
Question Number 7582 by A WLAN last updated on 04/Sep/16 Commented by A WLAN last updated on 04/Sep/16 $${How}\:{to}\:{solve}\:{this}\:{problem}? \\ $$ Answered by Yozzia last…
Question Number 73117 by aliesam last updated on 06/Nov/19 Commented by mathmax by abdo last updated on 06/Nov/19 $${we}\:{have}\:{cos}\left(\mathrm{3}{x}\right)\sim\mathrm{1}−\frac{\left(\mathrm{3}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{cos}\left(\mathrm{3}{x}\right)−\mathrm{1}\:\sim−\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{1}−{cos}\left(\mathrm{3}{x}\right)\sim\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}−{cos}\left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} }\sim\frac{\mathrm{9}}{\mathrm{2}}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}}…
Question Number 7579 by ajy last updated on 04/Sep/16 $$\mathrm{2}×\mathrm{2} \\ $$ Answered by FilupSmith last updated on 04/Sep/16 $$\mathrm{4} \\ $$ Terms of Service…
Question Number 7577 by Tawakalitu. last updated on 04/Sep/16 $${p}\:=\:\frac{{a}\left[\mathrm{1}\:−\:\left(\mathrm{1}\:+\:{r}\right)^{−{n}} \right]}{{r}} \\ $$$$ \\ $$$${Make}\:\:\:{r}\:\:\:{the}\:{subject}\:{of}\:{the}\:{fomular}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 73113 by TawaTawa last updated on 06/Nov/19 Commented by mathmax by abdo last updated on 06/Nov/19 $$\left.\mathrm{1}\right)\:{we}\:{have}\:{arg}\left({z}\right)={arg}\left(\mathrm{7}−\mathrm{3}\sqrt{\mathrm{3}}{i}\right)+{arg}\left(−\mathrm{1}−{i}\right)\left[\mathrm{2}\pi\right] \\ $$$$\mid\mathrm{7}−\mathrm{3}\sqrt{\mathrm{3}}\mid\:=\sqrt{\mathrm{49}+\mathrm{27}}=\sqrt{\mathrm{76}}\:\Rightarrow\mathrm{7}−\mathrm{3}\sqrt{\mathrm{3}}=\sqrt{\mathrm{76}}{e}^{{iarctan}\left(\frac{−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{7}}\right)} \:\Rightarrow \\ $$$${arg}\left(\mathrm{7}−\mathrm{3}\sqrt{\mathrm{3}}\right)\:=−{arctan}\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{7}}\right) \\…
Question Number 73111 by oyemi kemewari last updated on 06/Nov/19 Answered by mr W last updated on 07/Nov/19 $$\mathrm{5}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$$\Rightarrow{t}=\mathrm{10}\:{s} \\ $$$${horizontal}\:{distance}\:=\mathrm{0}.\mathrm{8}×\mathrm{10}=\mathrm{8}\:{m}\: \\ $$…
Question Number 138641 by soudo last updated on 15/Apr/21 Answered by MJS_new last updated on 16/Apr/21 $$\left.\mathrm{f}\left.\mathrm{rom}\:\mathrm{1}\right)\:\Rightarrow\:\mathrm{2}\right) \\ $$$${x}^{\mathrm{2}} \pm{xy}+{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{x}^{\mathrm{2}} \pm\mathrm{2}{xy}+{y}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=…
Question Number 138643 by ArielVyny last updated on 15/Apr/21 Answered by mnjuly1970 last updated on 15/Apr/21 Commented by Dwaipayan Shikari last updated on 16/Apr/21 $${Great}\:{sir}!\:{but}\:{in}\:{wrong}\:{place}…