Question Number 73080 by oyemi kemewari last updated on 06/Nov/19 $$\mathrm{y}''=\mathrm{e}^{\mathrm{y}} \\ $$$$\mathrm{pls}\:\mathrm{solve} \\ $$ Answered by mind is power last updated on 06/Nov/19 $$\mathrm{y}''.\mathrm{y}'=\mathrm{y}'\mathrm{e}^{\mathrm{y}}…
Question Number 7545 by Tawakalitu. last updated on 03/Sep/16 $${y}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} \:=\:\mathrm{2}^{{x}} \:+\:\mathrm{2}^{{y}} \\ $$$${Find}\:{the}\:{possible}\:{greatest}\:{value}\:{of}\:\mid{x}\:−\:{y}\mid \\ $$ Commented by Yozzia last updated on 03/Sep/16 $$\mathrm{2}^{{x}−{y}}…
Question Number 7544 by Tawakalitu. last updated on 03/Sep/16 Commented by sou1618 last updated on 03/Sep/16 $${x}={OA} \\ $$$${OB}=\sqrt{{x}^{\mathrm{2}} +\mathrm{144}} \\ $$$$\Delta{ABO}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{12}{x} \\ $$$$\Delta{AOB}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{3}\left(\mathrm{12}+{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{144}}\right)…
Question Number 138613 by byaw last updated on 15/Apr/21 Answered by physicstutes last updated on 15/Apr/21 $$\mathrm{This}\:\mathrm{question}\:\mathrm{is}\:\mathrm{real}\:\mathrm{lol}! \\ $$$$\mathrm{2}.\:\left(\mathrm{a}\right)\:\mathrm{Total}\:\mathrm{length}\:=\:\frac{\mathrm{9}}{\mathrm{4}}\:\mathrm{m}\: \\ $$$$\Rightarrow\:\mathrm{lenght}\:\mathrm{used}\:\mathrm{for}\:\mathrm{a}\:\mathrm{job}\:=\:\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{9}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{m} \\ $$$$\mathrm{length}\:\mathrm{used}\:=\:\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\mathrm{m} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{lenght}\:\mathrm{used}\:\mathrm{for}\:\mathrm{job}\:=\:\frac{\mathrm{1}}{\mathrm{8}}×\frac{\mathrm{3}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{32}}…
Question Number 138612 by SOMEDAVONG last updated on 15/Apr/21 $$\mathrm{If}\:\mathrm{z}+\:\frac{\mathrm{1}}{\mathrm{z}}\:=\mathrm{2cos}\beta\:.\mathrm{show}\:\mathrm{that}\:\mathrm{z}^{\mathrm{m}} +\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{m}} }\:=\mathrm{2cosm}\beta. \\ $$ Answered by Ar Brandon last updated on 15/Apr/21 $$\mathrm{z}=\mathrm{e}^{\mathrm{i}\beta} \\ $$$$\mathrm{z}^{\mathrm{m}}…
Question Number 7541 by 314159 last updated on 03/Sep/16 Commented by prakash jain last updated on 03/Sep/16 $${a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} +{r}^{\mathrm{4}} \right)=\mathrm{211} \\ $$$${a}\:{and}\:{r}\:{integers} \\ $$$$\mathrm{211}=\mathrm{1}×\mathrm{211}\Rightarrow{a}=\mathrm{1}…
Question Number 138608 by Dwaipayan Shikari last updated on 15/Apr/21 $${Prove}\:{or}\:{disprove} \\ $$$$\int_{\mathrm{0}} ^{{a}} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\mathrm{1}}{{e}^{{a}^{\mathrm{2}} } }\left(\frac{{a}^{\mathrm{3}} }{\mathrm{1}.\mathrm{3}}+\frac{\mathrm{2}{a}^{\mathrm{5}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}}+\frac{\mathrm{2}{a}^{\mathrm{7}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}+\frac{\mathrm{2}{a}^{\mathrm{9}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}+{ad}\:{inf}..\right) \\…
Question Number 7538 by 2402@gmail.com last updated on 02/Sep/16 $${Q}.\:\int\frac{\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}{{x}^{\mathrm{2}} −\mathrm{1}}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 138611 by liberty last updated on 15/Apr/21 $${Given}\:{a}\:{function}\:{f}\:{where}\: \\ $$$${f}\left({x}\right)\geqslant\:\mathrm{0}{for}\:\forall{x}\in\mathbb{R}.\:{If}\:{the}\:{area} \\ $$$${U}\:=\:\left\{\:\left({x},{y}\right)\mid\mathrm{0}\leqslant\mathrm{2}{y}\leqslant{f}\left({x}\right),\:−\mathrm{6}\leqslant{x}\leqslant−\mathrm{2}\right\} \\ $$$${is}\:{u}\:{and}\:{the}\:{area}\:{V}=\left\{\left({x},{y}\right)\mid\mathrm{0}\leqslant{y}\leqslant{f}\left({x}\right),−\mathrm{2}\leqslant{x}\leqslant\mathrm{0}\right\} \\ $$$${is}\:{v}\:{then}\:{what}\:{the}\:{value}\:{of} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\mathrm{4}{x}\:{f}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}\right)\:{dx}\:. \\ $$$$\left({A}\right)\:\mathrm{5}{u}+\mathrm{4}{v}\:\:\:\:\:\left({D}\right)\mathrm{2}{u}+{v}…
Question Number 7532 by Yozzia last updated on 02/Sep/16 $${Find}\:{x}_{{n}} \:\left({n}\in\mathbb{Z}\right)\:{satisfying}\:{x}_{\mathrm{0}} =\mathrm{0},\:{x}_{\mathrm{1}} =\mathrm{1}\:{and} \\ $$$${x}_{{n}+\mathrm{1}} ={x}_{{n}} \sqrt{{x}_{{n}−\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}}+{x}_{{n}−\mathrm{1}} \sqrt{{x}_{{n}} ^{\mathrm{2}} +\mathrm{1}}\:{for}\:{n}\geqslant\mathrm{1}. \\ $$ Commented…