Question Number 211122 by hardmath last updated on 28/Aug/24 $$ \\ $$Is there a special rule for divisibility of four-digit numbers by 13? Answered by…
Question Number 211099 by peter frank last updated on 27/Aug/24 Answered by A5T last updated on 27/Aug/24 $$\mathrm{2}^{{logx}} =\mathrm{2}^{\frac{{log}_{\mathrm{2}} {x}}{{log}_{\mathrm{2}} \mathrm{10}}} ={x}^{\frac{\mathrm{1}}{{log}_{\mathrm{2}} \mathrm{10}}} ={x}^{{log}_{\mathrm{10}} \mathrm{2}}…
Question Number 211004 by RojaTaniya last updated on 26/Aug/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 211029 by BaliramKumar last updated on 26/Aug/24 Answered by Ar Brandon last updated on 26/Aug/24 $${f}\left(\mathrm{2}\right)=\mathrm{5}\:\mathrm{therefore}\:{f}\left({x}\right)\:\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}-\mathrm{zero}\:\mathrm{polynomial}. \\ $$$$\mathrm{Let}\:{f}\left({x}\right)={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +\centerdot\centerdot\centerdot+{a}_{{n}} {x}^{{n}}…
Question Number 211043 by MATHEMATICSAM last updated on 26/Aug/24 Answered by Ghisom last updated on 26/Aug/24 $$\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{9}= \\ $$$$=\left({x}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{5}=\mathrm{2024} \\ $$ Answered…
Question Number 210971 by black_mamba234 last updated on 25/Aug/24 Answered by A5T last updated on 25/Aug/24 $$\left.\mathrm{2}\right)\:\mathrm{2}{x}_{\mathrm{1}} +\mathrm{4}{y}_{\mathrm{1}} −\mathrm{2}{z}_{\mathrm{1}} =\mathrm{2}\left({x}_{\mathrm{1}} +{k}\right)+\mathrm{4}{y}_{\mathrm{1}} −\mathrm{2}\left({z}_{\mathrm{1}} +{k}\right) \\ $$$${f}\left({x}_{\mathrm{1}}…
Question Number 210987 by RojaTaniya last updated on 25/Aug/24 Answered by Frix last updated on 25/Aug/24 $${x}^{\mathrm{6}} −\frac{\mathrm{133}}{\mathrm{78}}{x}^{\mathrm{5}} +\frac{\mathrm{133}}{\mathrm{78}}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}−\frac{\mathrm{2}}{\mathrm{3}}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{6}}{\mathrm{13}}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$ Terms…
Question Number 210996 by RojaTaniya last updated on 25/Aug/24 Answered by Frix last updated on 26/Aug/24 $$\mathrm{Only}\:\mathrm{true}\:\mathrm{for}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solution}\:>\mathrm{0}: \\ $$$${x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{9}{x}=\mathrm{4}{x}^{\mathrm{2}}…
Question Number 210958 by RojaTaniya last updated on 24/Aug/24 Commented by Ghisom last updated on 26/Aug/24 $$\mathrm{I}\:\mathrm{get} \\ $$$${x}=\frac{\mathrm{36}}{\mathrm{25}}\wedge{y}=\frac{\mathrm{64}}{\mathrm{25}} \\ $$$${x}=−\frac{\mathrm{7}}{\mathrm{16}}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{i}\wedge{y}=\frac{\mathrm{63}}{\mathrm{400}}−\frac{\mathrm{27}}{\mathrm{50}}\mathrm{i} \\ $$$${x}=−\frac{\mathrm{7}}{\mathrm{16}}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{i}\wedge{y}=\frac{\mathrm{63}}{\mathrm{400}}+\frac{\mathrm{27}}{\mathrm{50}}\mathrm{i} \\ $$…
Question Number 210961 by RojaTaniya last updated on 24/Aug/24 Answered by A5T last updated on 24/Aug/24 $$\frac{{a}−\mathrm{1}+\mathrm{2}}{{a}−\mathrm{1}}+\frac{{b}−\mathrm{1}+\mathrm{2}}{{b}−\mathrm{1}}+\frac{{c}−\mathrm{1}+\mathrm{2}}{{c}−\mathrm{1}}=\mathrm{10} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}−\mathrm{1}}+\frac{\mathrm{1}}{{b}−\mathrm{1}}+\frac{\mathrm{1}}{{c}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${a}+{b}+{c}=\mathrm{0};{ab}+{bc}+{ca}=\frac{{m}−\mathrm{1}}{{m}};{abc}=\frac{−{m}−\mathrm{1}}{{m}} \\ $$$$\Rightarrow\frac{{ab}+{bc}+{ca}−\mathrm{2}\left({a}+{b}+{c}\right)+\mathrm{3}}{{abc}−{ab}−{bc}−{ac}+{a}+{b}+{c}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\frac{{m}−\mathrm{1}+\mathrm{3}{m}}{{m}}}{\frac{−{m}−\mathrm{1}+\mathrm{1}−{m}−{m}}{{m}}}=\frac{\mathrm{7}}{\mathrm{2}}\Rightarrow\frac{\mathrm{4}{m}−\mathrm{1}}{−\mathrm{3}{m}}=\frac{\mathrm{7}}{\mathrm{2}}…