Question Number 84100 by niroj last updated on 09/Mar/20 $$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations}: \\ $$$$\:\:\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}=\:\mathrm{y}−\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} } \\ $$ Answered by TANMAY PANACEA last updated on 09/Mar/20 $${xdy}−{ydx}=−\sqrt{{x}^{\mathrm{2}}…
Question Number 149638 by Samimsultani last updated on 06/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 84101 by niroj last updated on 09/Mar/20 $$\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{reduction}}\:\boldsymbol{\mathrm{formula}} \\ $$$$\:\:\int\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{ax}}} \:\boldsymbol{\mathrm{dx}} \\ $$ Answered by MJS last updated on 09/Mar/20 $$\int{x}^{{n}} \mathrm{e}^{{ax}}…
Question Number 18561 by tawa tawa last updated on 24/Jul/17 $$\mathrm{The}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{an}\:\mathrm{object}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{a}\left(\mathrm{t}\right)\:=\:\mathrm{cos}\left(\mathrm{n}\pi\right),\:\mathrm{and}\:\mathrm{its}\:\mathrm{velocity} \\ $$$$\mathrm{at}\:\mathrm{time}\:\mathrm{t}\:=\:\mathrm{0}\:\mathrm{is}\:\:\:\frac{\mathrm{1}}{\mathrm{2}\pi}.\:\mathrm{Find}\:\mathrm{both}\:\mathrm{the}\:\mathrm{net}\:\mathrm{and}\:\mathrm{the}\:\mathrm{total}\:\mathrm{distance}\:\mathrm{traveled}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{first}\:\:\mathrm{1}.\mathrm{5}\:\mathrm{seconds}. \\ $$ Commented by tawa tawa last updated on 24/Jul/17…
Question Number 149634 by puissant last updated on 06/Aug/21 $$\mathrm{Trouver}\:\mathrm{toutes}\:\mathrm{les}\:\mathrm{fonctions}\:\mathrm{f}:\mathbb{N}\rightarrow\mathbb{R}^{+} \\ $$$$\mathrm{telque}\:\forall\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{N}, \\ $$$$\mathrm{f}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)=\mathrm{f}\left(\mathrm{a}^{\mathrm{2}} \right)+\mathrm{f}\left(\mathrm{b}^{\mathrm{2}} \right)\:\mathrm{et}\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$ Answered by Ar Brandon last…
Question Number 149628 by DELETED last updated on 06/Aug/21 Answered by DELETED last updated on 06/Aug/21 $$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p}} }\:=\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\:+\:\frac{\mathrm{2}}{\mathrm{6}}\:=\frac{\mathrm{3}}{\mathrm{6}}\:\rightarrow\mathrm{R}_{\mathrm{p}} =\frac{\mathrm{6}}{\mathrm{3}}=\mathrm{2}\Omega \\ $$$$\mathrm{I}_{\mathrm{p}}…
Question Number 18556 by aplus last updated on 24/Jul/17 Answered by Tinkutara last updated on 25/Jul/17 $$\mathrm{V}\:=\:\pi{r}^{\mathrm{2}} {h} \\ $$$$\mathrm{Total}\:\mathrm{surface}\:\mathrm{area},\:{A}\:=\:\mathrm{2}\pi{r}^{\mathrm{2}} \:+\:\mathrm{2}\pi{rh} \\ $$$$=\:\mathrm{2}\pi{r}^{\mathrm{2}} \:+\:\mathrm{2}\frac{\pi{r}^{\mathrm{2}} {h}}{{r}}\:=\:\mathrm{2}\pi{r}^{\mathrm{2}}…
Question Number 149625 by puissant last updated on 06/Aug/21 $$\left.\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{1}}{\mathrm{a}+\mathrm{sin}\left(\mathrm{t}\right)}\mathrm{dt}\:,\:\mathrm{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \frac{\mathrm{1}}{\mathrm{2}+\mathrm{sin}\left(\mathrm{t}\right)}\mathrm{dt}.. \\ $$ Answered by ArielVyny last updated on 06/Aug/21…
Question Number 149621 by mathdanisur last updated on 06/Aug/21 $$\mathrm{if}\:\:\:\mathrm{2}^{\boldsymbol{{x}}} =\mathrm{3}^{\boldsymbol{{y}}} =\mathrm{7}^{\boldsymbol{{z}}} =\sqrt[{\mathrm{3}}]{\mathrm{42}} \\ $$$$\mathrm{find}\:\:\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=? \\ $$ Answered by iloveisrael last updated on 06/Aug/21 $$\:\mathrm{If}\:\mathrm{2}^{\mathrm{x}}…
Question Number 18551 by mondodotto@gmail.com last updated on 24/Jul/17 Answered by Tinkutara last updated on 24/Jul/17 $$\sqrt{{a}\sqrt{{a}\sqrt{{a}}}}\:=\:\sqrt{{a}\sqrt{{a}^{\frac{\mathrm{3}}{\mathrm{2}}} }}\:=\:=\sqrt{{a}^{\frac{\mathrm{7}}{\mathrm{4}}} }\:=\:{a}^{\frac{\mathrm{7}}{\mathrm{8}}} \\ $$$$\mathrm{Remember}:\:\sqrt{{a}\sqrt{{a}\sqrt{{a}\sqrt{{a}\sqrt{{a}…}}}}}\:=\:{a}^{\frac{\mathrm{2}^{{n}} \:−\:\mathrm{1}}{\mathrm{2}^{{n}} }} \\ $$$$\forall\:{n}\:\geqslant\:\mathrm{1},\:\mathrm{where}\:\mathrm{there}\:\mathrm{are}\:{n}\:\mathrm{square}\:\mathrm{roots}.…