Question Number 194928 by cortano12 last updated on 19/Jul/23 $$\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{9}}−\sqrt{\mathrm{16}{x}^{\mathrm{2}} −\mathrm{8}}\:=?\right. \\ $$ Answered by horsebrand11 last updated on 20/Jul/23…
Question Number 194896 by cortano12 last updated on 19/Jul/23 $$\:\:\:\:\:\:\:\cancel{ } \\ $$ Answered by AST last updated on 19/Jul/23 $$\left({x}^{\mathrm{2}} −{f}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} {f}\left({x}\right)=\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} {f}\left({x}\right)…
Question Number 194915 by York12 last updated on 19/Jul/23 $$ \\ $$$$\frac{\mathrm{3}}{{x}−\mathrm{3}}+\frac{\mathrm{5}}{{x}−\mathrm{5}}+\frac{\mathrm{7}}{{x}−\mathrm{17}}+\frac{\mathrm{19}}{{x}−\mathrm{19}}={x}^{\mathrm{2}} −\mathrm{11}{x}−\mathrm{4} \\ $$ Commented by BaliramKumar last updated on 19/Jul/23 $$\mathrm{typo}\:\:\:\mathrm{7}\rightarrow\mathrm{17} \\ $$…
Question Number 194899 by kapoorshah last updated on 19/Jul/23 Answered by Frix last updated on 19/Jul/23 $${y}={px}\wedge{z}={qx} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{1}−{pq}}=\frac{\mathrm{8}}{{p}^{\mathrm{2}} −{q}}=\frac{\mathrm{10}}{{q}^{\mathrm{2}} −{p}} \\…
Question Number 194914 by Mingma last updated on 19/Jul/23 Answered by a.lgnaoui last updated on 20/Jul/23 $$\mathrm{distance}\:\mathrm{betwen}\:\mathrm{Dand}\:\mathrm{E}\:=\mathrm{distance}\:\mathrm{betwen}\: \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{D};\:\:\:\mathrm{so}\:\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{of}\:\mathrm{point} \\ $$$$\:\mathrm{4}^{\mathrm{9}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\boldsymbol{\mathrm{A}} \\ $$$$ \\…
Question Number 194877 by SajaRashki last updated on 18/Jul/23 $${without}\:{calculator}\:{Prove}:\:\sqrt[{\mathrm{6}}]{\pi^{\mathrm{5}} +\pi^{\mathrm{4}} }<{e} \\ $$ Commented by SajaRashki last updated on 24/Jul/23 $${without}\:{calculator}\:{Prove}:\:\sqrt[{\mathrm{6}}]{\pi^{\mathrm{5}} +\pi^{\mathrm{4}} }<{e} \\…
Question Number 194879 by tri26112004 last updated on 18/Jul/23 Answered by cortano12 last updated on 18/Jul/23 $$\:\:=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{2}{u}\right)^{{u}} −\mathrm{1}}{\left(\mathrm{1}+\mathrm{3}{u}\right)^{{u}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\left[\:{u}\:=\frac{\mathrm{1}}{{x}}\:\right] \\ $$$$\:{then}\:{apply}\:{L}'{Hopital}\: \\…
Question Number 194878 by oMwarimu last updated on 18/Jul/23 Answered by Rasheed.Sindhi last updated on 18/Jul/23 $${a}+{d}={ar}^{\mathrm{3}} \:\:\wedge\:\:{a}+\mathrm{9}{d}={ar}^{\mathrm{6}} \\ $$$${a}\left({r}^{\mathrm{3}} −\mathrm{1}\right)={d}\:\wedge\:{a}\left({r}^{\mathrm{6}} −\mathrm{1}\right)=\mathrm{9}{d} \\ $$$${a}=\frac{{d}}{{r}^{\mathrm{3}} −\mathrm{1}}\:\wedge\:{a}=\frac{\mathrm{9}{d}}{{r}^{\mathrm{6}}…
Question Number 194888 by Abdullahrussell last updated on 18/Jul/23 Answered by aleks041103 last updated on 18/Jul/23 $${The}\:{combined}\:{ages}\:{of}\:{Mary}\:{and}\:{Ann}\:{is}\:\mathrm{44}\:{years}. \\ $$$$\left(\:{M}\left({t}_{\mathrm{1}} \right)+{A}\left({t}_{\mathrm{1}} \right)=\mathrm{44}\:\right) \\ $$$${Mary}\:{is}\:{twice}\:{as}\:{old}\:{as}\:{Ann}\:{was} \\ $$$$\left(\:{M}\left({t}_{\mathrm{1}}…
Question Number 194891 by cortano12 last updated on 18/Jul/23 $$\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{3}^{+} } {\mathrm{lim}}\:\left(\frac{\sqrt{{x}}−\sqrt{{x}−\mathrm{3}}−\sqrt{\mathrm{3}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}\:\right)=? \\ $$ Answered by MM42 last updated on 18/Jul/23 $${lim}_{{x}\rightarrow\mathrm{3}} \:\left(\frac{\sqrt{{x}}−\sqrt{\mathrm{3}}}{\:\sqrt{{x}−\mathrm{3}}×\sqrt{{x}+\mathrm{3}}}\:−\:\frac{\sqrt{{x}−\mathrm{3}}}{\:\sqrt{{x}−\mathrm{3}}×\sqrt{{x}+\mathrm{3}}}\:\right) \\…