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lim-x-x-2-1-x-1-

Question Number 197407 by mathlove last updated on 16/Sep/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\mathrm{1}}=? \\ $$ Answered by Rasheed.Sindhi last updated on 16/Sep/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\mathrm{1}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}}…

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-197393

Question Number 197393 by mathlove last updated on 16/Sep/23 Answered by Frix last updated on 16/Sep/23 $$\frac{\mathrm{3}×\mathrm{5}\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{6}} \mathrm{3}}}+\mathrm{3}×\mathrm{7}\sqrt[{\mathrm{3}}]{\mathrm{2}×\mathrm{3}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{3}^{\mathrm{4}} }}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{2}} \mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{3}} \mathrm{3}}+\mathrm{2}×\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}×\mathrm{5}^{\mathrm{3}} }}}= \\…

please-check-my-answer-x-2y-5-dx-2x-y-4-dy-0-X-x-a-amp-Y-y-b-X-a-2Y-2b-5-dX-2X-2a-Y-b-4-dY-0-a-2b-5-2a-b-4-a-1-b-2-X-2Y-dX-2X-Y-dY-0-Y-XV-dY-dX-V-X-dV-dx-X-2XV-2X-XV-V-X

Question Number 197389 by uchihayahia last updated on 16/Sep/23 $$ \\ $$$${please}\:{check}\:{my}\:{answer} \\ $$$$\:\left({x}−\mathrm{2}{y}+\mathrm{5}\right){dx}+\left(\mathrm{2}{x}−{y}+\mathrm{4}\right){dy}=\mathrm{0} \\ $$$$\:{X}={x}+{a}\:\&\:{Y}={y}+{b} \\ $$$$ \\ $$$$\:\left({X}−{a}−\mathrm{2}{Y}+\mathrm{2}{b}+\mathrm{5}\right){dX}+\left(\mathrm{2}{X}−\mathrm{2}{a}−{Y}+{b}+\mathrm{4}\right){dY}=\mathrm{0} \\ $$$$\:-{a}+\mathrm{2}{b}=-\mathrm{5} \\ $$$$\:-\mathrm{2}{a}+{b}=-\mathrm{4} \\…

Question-197419

Question Number 197419 by universe last updated on 16/Sep/23 Answered by cortano12 last updated on 17/Sep/23 $$\:\:\:\:\begin{cases}{\left(\mathrm{2a}\right)^{\mathrm{ln}\:\mathrm{a}} \:=\:\left(\mathrm{bc}\right)^{\mathrm{ln}\:\mathrm{b}} }\\{\mathrm{b}^{\mathrm{ln}\:\mathrm{2}} \:=\:\mathrm{a}^{\mathrm{ln}\:\mathrm{c}} \:}\end{cases} \\ $$$$\:\:\:\:\:\begin{cases}{\mathrm{ln}\:\left(\mathrm{2a}\right).\:\mathrm{ln}\:\left(\mathrm{a}\right)\:=\:\mathrm{ln}\:\left(\mathrm{b}\right).\:\mathrm{ln}\:\left(\mathrm{bc}\right)}\\{\mathrm{ln}\:\left(\mathrm{b}\right).\:\mathrm{ln}\:\left(\mathrm{2}\right)=\:\mathrm{ln}\:\left(\mathrm{c}\right).\:\mathrm{ln}\:\left(\mathrm{a}\right)}\end{cases} \\ $$$$\:\:\Rightarrow\mathrm{ln}\:\mathrm{a}\:\left\{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{a}\right\}=\:\mathrm{ln}\:\mathrm{b}\:\left\{\mathrm{ln}\:\mathrm{b}+\mathrm{ln}\:\mathrm{c}\:\right\}…