Question Number 203858 by Mastermind last updated on 30/Jan/24 $$\mathrm{Classsify}\:\mathrm{the}\:\mathrm{critical}\:\mathrm{points}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{x}^{\mathrm{2}} \mathrm{y}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{y}^{\mathrm{3}} \:−\:\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{2} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{in}\:\mathrm{advance}! \\ $$ Commented…
Question Number 203859 by Mastermind last updated on 30/Jan/24 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{surface}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{z}\:=\:\mathrm{xy} \\ $$$$\mathrm{has}\:\mathrm{neither}\:\mathrm{a}\:\mathrm{maximum}\:\mathrm{nor}\:\mathrm{a}\:\mathrm{minimum}\:\mathrm{point} \\ $$ Commented by AST last updated on 30/Jan/24 $${For}\:{x},{y}\rightarrow\infty;\:{z}\rightarrow\infty \\…
Question Number 203819 by zahaku last updated on 29/Jan/24 $$\frac{\mid{x}^{\mathrm{2}} −\mathrm{3}{x}\mid}{{x}^{\mathrm{2}} −\mathrm{9}}\:\:{draw}\:{up}\:{the}\:{variaton} \\ $$$${table}\:{of}\:{this}\:{function} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 203750 by patrice last updated on 27/Jan/24 Answered by witcher3 last updated on 27/Jan/24 $$\frac{\mathrm{4}\left(\mathrm{k}+\mathrm{2}\right)−\mathrm{k}}{\mathrm{k}\left(\mathrm{k}+\mathrm{2}\right)\mathrm{2}^{\mathrm{k}} }=\frac{\mathrm{4}}{\mathrm{k}.\mathrm{2}^{\mathrm{k}} }−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{2}\right)\mathrm{2}^{\mathrm{k}} }=\frac{\mathrm{1}}{\mathrm{k}.\mathrm{2}^{\mathrm{k}−\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{2}\right)\mathrm{2}^{\mathrm{k}} }=\mathrm{V}_{\mathrm{k}} −\mathrm{V}_{\mathrm{k}+\mathrm{2}} \\ $$$$\mathrm{s}_{\mathrm{n}}…
Question Number 203736 by Tawa11 last updated on 26/Jan/24 A conductor of length 500cm carried a current of 0.95A when kept in a magnetic field…
Question Number 203701 by Noorzai last updated on 26/Jan/24 Answered by mr W last updated on 26/Jan/24 $$\mathrm{0}<\frac{\mathrm{2}^{{x}} }{{x}!}=\frac{\mathrm{2}×\mathrm{2}×\mathrm{2}×\mathrm{2}×….×\mathrm{2}}{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×…×{x}}<\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} \\ $$$$\mathrm{0}<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{{x}} }{{x}!}<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{0}…
Question Number 203693 by Numsey last updated on 26/Jan/24 Answered by MM42 last updated on 26/Jan/24 $${I}=\int\sqrt{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}{dx}=\mathrm{2}\sqrt{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}\:+{c} \\ $$ Commented by Frix last updated on…
Question Number 203695 by Numsey last updated on 26/Jan/24 Commented by BaliramKumar last updated on 26/Jan/24 $$\mathrm{Step}\:\mathrm{III}\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{27}\right)}\left(\mathrm{1}.\mathrm{5}\right) \\ $$ Terms of Service Privacy Policy Contact:…