Question Number 87989 by john santu last updated on 07/Apr/20 $$\mathrm{find}\:\mathrm{maximum}\:\mathrm{value} \\ $$$$\mathrm{2x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:\mathrm{with}\:\mathrm{constraint} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}+\mathrm{2y}+\mathrm{1}=\mathrm{0}\: \\ $$$$ \\ $$ Answered by…
Question Number 22332 by A1B1C1D1 last updated on 15/Oct/17 Answered by ajfour last updated on 15/Oct/17 $$\frac{{d}\left[{f}\left({x}\right)\right]}{{dx}}=\mathrm{1}+\mathrm{2cos}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{cos}\:{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:{x}=\mathrm{2}{n}\pi\pm\frac{\mathrm{2}\pi}{\mathrm{3}}\:\:{where}\:{n}\in\mathbb{Z}\:. \\ $$$$\:\:\:\:{which}\:{is}\:{equivalent}\:{to} \\ $$$$\:\:\:\:{x}=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\pm\frac{\pi}{\mathrm{3}}\:.…
Question Number 153340 by mnjuly1970 last updated on 06/Sep/21 $$ \\ $$$$\:\:\:\:\:\mathrm{Solve}\:.. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\:,\:\mathrm{y}\:,\:\mathrm{z}\:\in\:\mathbb{R}^{\:+} \:\&\:\:\mathrm{x}+\:\mathrm{y}=\:\mathrm{z} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{K}\::=\:\mathrm{Min}_{\:} \:\left(\frac{\:\mathrm{x}^{\:\mathrm{4}} \:+\:\mathrm{y}^{\:\mathrm{4}} +\:\mathrm{z}^{\:\mathrm{4}} }{\mathrm{x}^{\:\mathrm{2}} \mathrm{y}^{\:\mathrm{2}} }\:\right)\:=\:?\:\:\:\:\:\:\:\:\:\:\blacksquare\:\:\:\:\:\:\:\:\:…
Question Number 153252 by liberty last updated on 06/Sep/21 Answered by MJS_new last updated on 06/Sep/21 $${y}''=\mathrm{0} \\ $$$$\frac{\mathrm{2}{ax}\left({x}^{\mathrm{2}} −\mathrm{3}{b}\right)}{\left({x}^{\mathrm{2}} +{b}\right)^{\mathrm{3}} }=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{0}\vee{x}=\pm\sqrt{\mathrm{3}{b}}\vee{a}=\mathrm{0}\:\left(\mathrm{rejected}\right) \\…
Question Number 87687 by ~blr237~ last updated on 05/Apr/20 $${Let}\:\:{w}=\left[\mathrm{1};\frac{\pi}{{n}}\right]\:,{n}\in\mathbb{N}^{\ast} \: \\ $$$$\:{a}_{{n}} =\underset{{p}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\:\frac{\mathrm{2}{p}+\mathrm{1}}{\mathrm{1}−{w}^{\mathrm{2}{p}+\mathrm{1}} }\:\:\:\:{and}\:\:\:{b}_{{n}} =\underset{{p}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\:\frac{{n}}{\mathrm{1}+{w}^{{p}} }\: \\ $$$${Find}\:\:{all}\:{integer}\:{n}\:{such}\:{as}\:\:{a}_{{n}} ={b}_{{n}} \:…
Question Number 153212 by mnjuly1970 last updated on 05/Sep/21 $$ \\ $$$$\:\:\int_{−\infty} ^{\:\infty} \frac{\:{tan}\left({x}\right).{Arctanh}\left({cos}\left({x}\right)\right)}{{x}}{dx}=? \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 153189 by liberty last updated on 05/Sep/21 Answered by EDWIN88 last updated on 05/Sep/21 $$\:{eq}\:{of}\:{circle}\:\mathrm{16}{x}^{\mathrm{2}} +\mathrm{16}{y}^{\mathrm{2}} +\mathrm{48}{x}−\mathrm{8}{y}−\mathrm{43}= \\ $$$${with}\:{center}\:{point}\:\begin{cases}{{x}=−\frac{\mathrm{48}}{\mathrm{32}}=−\frac{\mathrm{3}}{\mathrm{2}}}\\{{y}=\frac{\mathrm{8}}{\mathrm{32}}=\frac{\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$$${with}\:{radius}\:=\sqrt{\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}−\left(\frac{−\mathrm{43}}{\mathrm{16}}\right)} \\ $$$$\Rightarrow{r}=\sqrt{\frac{\mathrm{36}+\mathrm{1}+\mathrm{43}}{\mathrm{16}}}\:=\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{4}}=\sqrt{\mathrm{5}}\:…
Question Number 22105 by j.masanja06@gmail.com last updated on 11/Oct/17 $${use}\:{the}\:{first}\:{principle}\:{to}\:{find} \\ $$$${value}\:{of} \\ $$$${f}\left({x}\right)=\left({x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$ Answered by $@ty@m last updated on 11/Oct/17 $$\frac{{dy}}{{dx}}=\underset{\delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+\delta{x}\right)−{f}\left({x}\right)}{\delta{x}}…
Question Number 87532 by niroj last updated on 04/Apr/20 $$\:\left(\mathrm{1}\right).\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{general}}\:\boldsymbol{\mathrm{solution}}: \\ $$$$\:\:\:\boldsymbol{\mathrm{y}}=\:\boldsymbol{\mathrm{px}}\:+\boldsymbol{\mathrm{p}}^{\boldsymbol{\mathrm{n}}} \\ $$$$\:\left(\mathrm{2}\right).\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\:\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{2}} \:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:+\:\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\:\left(\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{3}\right)\left(\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{4}\right). \\ $$$$\:\: \\ $$ Commented by…
Question Number 21964 by chernoaguero@gmail.com last updated on 07/Oct/17 Answered by ibraheem160 last updated on 07/Oct/17 $$\mathrm{y}=\mathrm{x}^{\mathrm{2}} \left(\mathrm{2x}−\mathrm{5}\right)^{\mathrm{4}} \\ $$$$\frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{2x},\:\:\frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{8}\left(\mathrm{2x}−\mathrm{5}\right)^{\mathrm{3}} \\ $$$$\: \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{u}\frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v}\frac{\mathrm{du}}{\mathrm{dx}} \\…