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Question-197501

Question Number 197501 by SANOGO last updated on 19/Sep/23 Answered by witcher3 last updated on 19/Sep/23 $$\left(\mathrm{i}\right)\Rightarrow\left(\mathrm{ii}\right) \\ $$$$\exists\mathrm{x}\in\mathrm{A},\exists\epsilon^{\ast} >\mathrm{0}\:\:\beta\left(\mathrm{x},\epsilon\right)\in\mathrm{A},\mathrm{Boules}\:\mathrm{ouvert} \\ $$$$\mathrm{soit}\:\mathrm{B}\subset\mathrm{X},\overset{−} {\mathrm{B}}=\mathrm{X} \\ $$$$\mathrm{pour}\:\mathrm{x}\in\mathrm{X}\:\:\exists\mathrm{u}_{\mathrm{n}}…

lim-x-3-cos3-cosx-x-3-

Question Number 197496 by sciencestudentW last updated on 19/Sep/23 $$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{{cos}\mathrm{3}−{cosx}}{{x}−\mathrm{3}}=? \\ $$ Answered by cortano12 last updated on 20/Sep/23 $$\:=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:\left(\frac{\mathrm{x}+\mathrm{3}}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{3}−\mathrm{x}}{\mathrm{2}}\right)}{−\mathrm{2}\left(\frac{\mathrm{3}−\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\mathrm{sin}\:\left(\frac{\mathrm{x}+\mathrm{3}}{\mathrm{2}}\right)\:.\:\underset{{x}\rightarrow\mathrm{3}}…

Question-197452

Question Number 197452 by sonukgindia last updated on 18/Sep/23 Answered by Frix last updated on 18/Sep/23 $${t}=\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{7} \\ $$$$\sqrt{{t}+\mathrm{4}}−\mathrm{2}=\sqrt[{\mathrm{3}}]{\mathrm{2}{t}} \\ $$$$\mathrm{Obviously}\:{t}=−\mathrm{4}\vee{t}=\mathrm{0} \\ $$$${t}=−\mathrm{4}\:\Rightarrow\:{x}=−\frac{\mathrm{3}}{\mathrm{5}}\vee{x}=\mathrm{1} \\…

Question-197396

Question Number 197396 by sonukgindia last updated on 16/Sep/23 Commented by Frix last updated on 16/Sep/23 $$\mathrm{We}\:\mathrm{had}\:\mathrm{this}\:\mathrm{several}\:\mathrm{times}\:\mathrm{before}. \\ $$$${t}=\sqrt{\mathrm{tan}\:{x}}\:\Rightarrow\:\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\ $$$$\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{by}\:\mathrm{decomposing} \\ $$…

Question-197373

Question Number 197373 by sciencestudentW last updated on 15/Sep/23 Answered by Tokugami last updated on 17/Sep/23 $$\mathrm{6}\:\frac{\cancel{\mathrm{km}}}{\cancel{\mathrm{h}}}×\frac{\mathrm{1000}\:\mathrm{m}}{\mathrm{1}\:\cancel{\mathrm{km}}}×\frac{\mathrm{1}\:\cancel{\mathrm{h}}}{\mathrm{60}\:\mathrm{m}}=\mathrm{100}\:\frac{\mathrm{m}}{\mathrm{min}} \\ $$$$\mathrm{8}\:\frac{\cancel{\mathrm{km}}}{\cancel{\mathrm{h}}}×\frac{\mathrm{1000}\:\mathrm{m}}{\mathrm{1}\:\cancel{\mathrm{km}}}×\frac{\mathrm{1}\:\cancel{\mathrm{h}}}{\mathrm{60}\:\mathrm{m}}=\frac{\mathrm{400}}{\mathrm{3}}\:\frac{\mathrm{m}}{\mathrm{min}} \\ $$$$\frac{\mathrm{400}}{\mathrm{3}}{t}=\mathrm{250}+\mathrm{100}{t} \\ $$$$\frac{\mathrm{400}}{\mathrm{3}}{t}−\mathrm{100}{t}=\mathrm{250}+\mathrm{100}{t}−\mathrm{100}{t} \\ $$$$\frac{\mathrm{100}}{\mathrm{3}}{t}×\frac{\mathrm{3}}{\mathrm{100}}=\mathrm{250}×\frac{\mathrm{3}}{\mathrm{100}}…