Question Number 52412 by Tawa1 last updated on 07/Jan/19 Commented by tanmay.chaudhury50@gmail.com last updated on 08/Jan/19 Commented by tanmay.chaudhury50@gmail.com last updated on 08/Jan/19 Commented by…
Question Number 52406 by Tawa1 last updated on 07/Jan/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\:\:\:\:\mathrm{26}\:+\:\mathrm{5}\sqrt{\mathrm{3}} \\ $$ Answered by MJS last updated on 07/Jan/19 $$\mathrm{no}\:\mathrm{exact}\:\mathrm{solution} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}} \\ $$$$\left({a}+{b}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} =\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}…
Question Number 117934 by Ar Brandon last updated on 14/Oct/20 $$\mathrm{Let}\:{f}\::\:\left[\mathrm{1},\infty\right)\rightarrow\left[\mathrm{2},\infty\right)\:\mathrm{be}\:\mathrm{the}\:\mathrm{function}\:\mathrm{defined}\:\mathrm{by} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)={x}+\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{If}\:\mathrm{g}\::\:\left[\mathrm{2},\infty\right)\rightarrow\left[\mathrm{1},\infty\right),\:\mathrm{is}\:\mathrm{a}\:\mathrm{function}\:\mathrm{such}\:\mathrm{that}\:\left(\mathrm{g}\circ{f}\right)\left({x}\right)={x} \\ $$$$\mathrm{for}\:\mathrm{all}\:{x}\geqslant\mathrm{1}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{g}\left({t}\right)=\frac{{t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$ Answered by Lordose last updated…
Question Number 52397 by Joel578 last updated on 07/Jan/19 Commented by Joel578 last updated on 07/Jan/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 183424 by Shrinava last updated on 25/Dec/22 $$\mathrm{Among}\:\mathrm{the}\:\mathrm{natural}\:\mathrm{numbers}\:\mathrm{not} \\ $$$$\mathrm{greater}\:\mathrm{than}\:\mathrm{22}\:,\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{modulus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{any}\: \\ $$$$\mathrm{two}\:\mathrm{of}\:\mathrm{the}\:\mathrm{3}\:\mathrm{randomly}\:\mathrm{selected}\:\mathrm{numbers} \\ $$$$\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{5}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{which}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{following}? \\ $$$$\left.\mathrm{a}\left.\right)\left.\mathrm{1}\left.\mathrm{5}/\mathrm{22}\:\:\:\mathrm{b}\right)\mathrm{1}/\mathrm{7}\:\:\:\mathrm{c}\right)\mathrm{1}/\mathrm{22}\:\:\:\mathrm{d}\right)\mathrm{1} \\ $$ Terms…
Question Number 183381 by Shrinava last updated on 25/Dec/22 $$\frac{\mathrm{3n}^{\mathrm{5}} \:+\:\mathrm{4n}^{\mathrm{4}} \:−\:\mathrm{7n}^{\mathrm{3}} \:+\:\mathrm{5n}^{\mathrm{2}} \:−\:\mathrm{5}}{\mathrm{n}\:+\:\mathrm{1}} \\ $$$$\mathrm{There}\:\mathrm{can}\:\mathrm{be}\:\mathrm{no}\:\mathrm{residue}: \\ $$$$\left.\mathrm{a}\left.\right)\left.\mathrm{0}\left.\:\left.\:\:\mathrm{b}\right)\mathrm{2}\:\:\:\mathrm{c}\right)\mathrm{4}\:\:\:\mathrm{d}\right)\mathrm{5}\:\:\:\mathrm{e}\right)\mathrm{9} \\ $$ Commented by Shrinava last updated…
Question Number 183380 by Shrinava last updated on 25/Dec/22 $$\mathrm{a}>\mathrm{0}\:,\:\mathrm{b}>\mathrm{0} \\ $$$$\begin{cases}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} \:+\:\left(\mathrm{y}−\mathrm{7}\right)^{\mathrm{2}} \:=\:\mathrm{a}^{\mathrm{2}} }\\{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} \:+\:\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{b}^{\mathrm{2}} }\end{cases} \\ $$$$\mathrm{Find}:\:\:\:\left(\mathrm{a}+\mathrm{b}\right)_{\boldsymbol{\mathrm{min}}} \:=\:? \\ $$ Commented by…
Question Number 117828 by redouaneee last updated on 13/Oct/20 $$\left.\mathrm{1}\right)\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)=\sqrt{\mathrm{3}}×\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$$=\mathrm{3}−\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$$=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{2}{x}+\sqrt{\mathrm{3}}\right)\left(\mathrm{2}{x}−\sqrt{\mathrm{3}}\right)=\left(\mathrm{2}{x}\right)^{\mathrm{2}} −\mathrm{2}{x}\sqrt{\mathrm{3}}+\mathrm{2}{x}\sqrt{\mathrm{3}}−\mathrm{3} \\ $$$$=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{3} \\ $$ Answered by MJS_new…
Question Number 183366 by Shrinava last updated on 25/Dec/22 $$\mathrm{6}\:\mathrm{of}\:\mathrm{the}\:\mathrm{23}\:\mathrm{given}\:\mathrm{points}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{lie}\:\mathrm{on}\:\mathrm{a}\:\mathrm{circle}.\:\mathrm{Let}\:\boldsymbol{\mathrm{n}}\:\mathrm{be}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{circles}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{at}\:\mathrm{least}\:\mathrm{3}\:\mathrm{of} \\ $$$$\mathrm{these}\:\mathrm{points}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{number}\:\mathrm{of}\:\boldsymbol{\mathrm{n}}? \\ $$ Commented by mr W last…
Question Number 183363 by Shrinava last updated on 25/Dec/22 $$\mathrm{Find}: \\ $$$$\mathrm{2003}\centerdot\mathrm{2005}^{\mathrm{3}} −\mathrm{2004}\centerdot\mathrm{2002}^{\mathrm{3}} \\ $$ Answered by Frix last updated on 25/Dec/22 $$\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} −\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{8}{x}^{\mathrm{3}}…