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 $$\:=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:\left(\frac{\mathrm{x}+\mathrm{3}}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{3}−\mathrm{x}}{\mathrm{2}}\right)}{−\mathrm{2}\left(\frac{\mathrm{3}−\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\mathrm{sin}\:\left(\frac{\mathrm{x}+\mathrm{3}}{\mathrm{2}}\right)\:.\:\underset{{x}\rightarrow\mathrm{3}}…
Question Number 197499 by cherokeesay last updated on 19/Sep/23 Answered by HeferH last updated on 19/Sep/23 $$\: \\ $$$$\:\frac{{x}}{{r}}\:=\:\frac{\mathrm{4}{r}}{\mathrm{5}{r}}\:\:\Rightarrow\:{x}\:=\:\frac{\mathrm{4}{r}}{\mathrm{5}}\: \\ $$$$\:{Green}\:=\:{Sqr}/\mathrm{2}\:−\:\frac{\mathrm{4}{r}}{\mathrm{5}}\:\centerdot\:\mathrm{4}{r}\:\centerdot\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\:{Green}\:=\:\frac{\mathrm{16}{r}^{\mathrm{2}} }{\mathrm{2}}\:−\frac{\mathrm{16}{r}^{\mathrm{2}} }{\mathrm{10}}\:=\:\frac{\mathrm{4}\centerdot\mathrm{16}{r}^{\mathrm{2}}…
Question Number 197482 by cortano12 last updated on 19/Sep/23 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{3}}\:+\mathrm{2}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{1}}\:\right)=?\: \\ $$ Answered by Frix last updated on 19/Sep/23 $${f}\left({x}\right)=\frac{\sqrt{\mathrm{3}}{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}}…
Question Number 197483 by cortano12 last updated on 19/Sep/23 $$\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\:\sqrt{\mathrm{1}−\mathrm{x}}}\:=? \\ $$ Answered by Frix last updated on 19/Sep/23 $$=\frac{−\frac{\pi}{\mathrm{2}}}{\:\sqrt{\mathrm{1}+\infty}}=\mathrm{0} \\ $$ Terms…
Question Number 197476 by universe last updated on 19/Sep/23 $$ \\ $$$$\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\:\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}}{{x}^{\mathrm{4}} +\mathrm{1}}+………+\frac{\mathrm{2}^{{n}} }{{x}^{\mathrm{2}^{{n}} } +\mathrm{1}}\:\:=\:?? \\ $$ Commented by witcher3 last…
Question Number 197479 by pticantor last updated on 19/Sep/23 $$\boldsymbol{{find}}: \\ $$$$ \\ $$$$\:\:\:\:\:\:\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}}}\:\boldsymbol{{U}}_{\boldsymbol{{n}}} \:=\sqrt[{\mathrm{3}}]{\boldsymbol{{n}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{{n}}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{\boldsymbol{{n}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{n}}^{\mathrm{2}} }\: \\ $$ Commented by Frix…
Question Number 197475 by mnjuly1970 last updated on 19/Sep/23 $$ \\ $$$$\:\:\:\:\:\:{x}\:,\:{y}\:\in\:\mathbb{R}\:\:\:, \\ $$$$\:\:\:\:{x}^{\:\mathrm{2}} \:+\:{xy}\:=\:\mathrm{12} \\ $$$$\:\:\:\:\:{y}^{\:\mathrm{2}} \:+\:\mathrm{2}{xy}\:=\:\mathrm{7}\: \\ $$$$\:\:\:\:−−−−−\:\:\:{x}\:,\:{y}\:=? \\ $$ Answered by Sutrisno…
Question Number 197452 by sonukgindia last updated on 18/Sep/23 Answered by Frix last updated on 18/Sep/23 $${t}=\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{7} \\ $$$$\sqrt{{t}+\mathrm{4}}−\mathrm{2}=\sqrt[{\mathrm{3}}]{\mathrm{2}{t}} \\ $$$$\mathrm{Obviously}\:{t}=−\mathrm{4}\vee{t}=\mathrm{0} \\ $$$${t}=−\mathrm{4}\:\Rightarrow\:{x}=−\frac{\mathrm{3}}{\mathrm{5}}\vee{x}=\mathrm{1} \\…
Question Number 197469 by dragan91 last updated on 18/Sep/23 $$ \\ $$$${solve}\:{limits}\:{for}\:{functions} \\ $$$${f}\left({x}\right)={cos}\left({sgn}\left(\mathrm{1}/{x}\right)\right) \\ $$$${f}\left({x}\right)={sgn}\left({cos}\left(\mathrm{1}/{x}\right)\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Terms…
Question Number 197470 by Mastermind last updated on 18/Sep/23 $${Solve}\:{the}\:{following}\:{equation} \\ $$$${x}\:+\:\mathrm{2}{y}\:+\:\mathrm{2}{z}\:=\:\mathrm{0} \\ $$$$\mathrm{2}{x}\:+\:{y}\:−\:\mathrm{2}{z}\:=\mathrm{0} \\ $$$$\mathrm{3}{x}\:+\:\mathrm{4}{y}\:−\:\mathrm{6}{z}\:=\mathrm{0} \\ $$$$\mathrm{3}{x}\:−\:\mathrm{11}{y}\:+\:\mathrm{12}{z}\:=\:\mathrm{0} \\ $$ Answered by MathedUp last updated…