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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}}…

Question-197365

Question Number 197365 by mokys last updated on 15/Sep/23 Answered by witcher3 last updated on 15/Sep/23 $$\mathrm{x}=\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}}\right)^{\mathrm{1}−\mathrm{p}} \left(\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}}\right)^{\mathrm{p}} }{\left(\frac{\mathrm{2}}{\mathrm{1}+\mathrm{t}}\right)^{\mathrm{3}} }.\frac{\mathrm{2}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} }\mathrm{dt}…

calcul-n-1-oo-1-n-2n-1-n-n-1-

Question Number 197349 by SANOGO last updated on 14/Sep/23 $${calcul}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+{oo}} {\sum}}\left(−\mathrm{1}\right)^{{n}\:} \frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)} \\ $$ Answered by MM42 last updated on 14/Sep/23 $$\frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}…