Question Number 194637 by manxsol last updated on 12/Jul/23 $$ \\ $$$${x}+{y}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} =? \\ $$$$ \\ $$$$ \\…
Question Number 194634 by York12 last updated on 12/Jul/23 $${a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} ,….,{a}_{{n}} >\mathrm{0}\:{such}\:{that}\:{a}_{{i}} \in\left[\mathrm{0},{i}\right]\: \\ $$$$\forall\:{i}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},…,{n}\right\}\:{prove}\:{that} \\ $$$$\mathrm{2}^{{n}} .{a}_{\mathrm{1}} \left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)…\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…+{a}_{{n}}…
Question Number 194619 by Shrinava last updated on 11/Jul/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}: \\ $$$$−\mathrm{3x}^{\mathrm{3}} \:+\:\mathrm{8x}^{\mathrm{2}} \:−\:\mathrm{6x}\:−\:\mathrm{7}\:=\:\mathrm{0} \\ $$ Answered by Frix last updated on 11/Jul/23…
Question Number 194612 by Abdullahrussell last updated on 11/Jul/23 Commented by TheHoneyCat last updated on 15/Jul/23 $$\left.\mathrm{1}\right)\:\mathrm{Let}\:\alpha=\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}=\mathrm{45} \\ $$$$\mathrm{be}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{basis}\:\mathrm{10}. \\ $$$$\mathrm{Let}\:{S}_{\mathrm{1}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{first}\:\mathrm{sum},\:{S}_{\mathrm{1},\mathrm{0}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all} \\ $$$$\mathrm{unit}\:\mathrm{digits},\:{S}_{\mathrm{1},\mathrm{1}}…
Question Number 194610 by justenspi last updated on 11/Jul/23 $${where}\:{can}\:{I}\:{learn}\:{about}\:{multiple}\:{sigma}\:{notaions} \\ $$$${of}\:{dependent}\:{and}\:{independent}\:{variables} \\ $$$$ \\ $$$${something}\:{like}\:{this} \\ $$$$\underset{\mathrm{1}\leqslant{i}} {\sum}\underset{<{j}} {\sum}\underset{<{k}\leqslant\mathrm{1}} {\sum}\left({i}+{j}+{k}\right)=\lambda \\ $$$${find}\:\lambda \\ $$$${I}\:{want}\:{to}\:{know}\:{what}\:{to}\:{study}…
Question Number 194586 by Shrinava last updated on 10/Jul/23 $$\mathrm{abc}\:=\:\mathrm{e}^{\mathrm{3}} \:+\:\mathrm{d}^{\mathrm{3}} \:+\:\mathrm{f}^{\mathrm{3}} \\ $$$$\mathrm{edf}\:=\:\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} \\ $$$$\mathrm{find}:\:\mathrm{abc}\:\:\mathrm{and}\:\:\mathrm{edf}\: \\ $$ Commented by Shrinava last updated…
Question Number 194579 by MM42 last updated on 10/Jul/23 $${if}\:\:\:\:{u}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right] \\ $$$$\:{then}\:\:\:{u}_{{n}+\mathrm{1}} ={u}_{{n}} +{u}_{{n}−\mathrm{1}} \:\:\:?\:\:\:\:\:;\:\:\:{n}=\mathrm{0},\mathrm{1},\mathrm{2},.. \\ $$ Answered by Frix last updated…
Question Number 194573 by Aduragbemi last updated on 10/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194559 by MM42 last updated on 09/Jul/23 $${repeat}\:{question} \\ $$$${Shiw}\:{that}\:: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(\frac{\mathrm{1}}{\mathrm{2}{i}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{i}}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{n}+{i}}\:\:? \\ $$ Answered by witcher3 last updated…
Question Number 194526 by mathlove last updated on 09/Jul/23 $$\frac{{f}\left({x}+\mathrm{1}\right)}{{f}\left({x}\right)}={x}^{\mathrm{2}\:\:\:\:\:\:\:\:} \:\:\:\:{f}\left({x}\right)=? \\ $$$$\frac{{f}\left(\mathrm{6}\right)}{{f}\left(\mathrm{3}\right)}=? \\ $$ Answered by JDamian last updated on 09/Jul/23 $$\frac{{f}\left(\mathrm{6}\right)}{{f}\left(\mathrm{3}\right)}=\frac{{f}\left(\mathrm{6}\right)}{{f}\left(\mathrm{3}\right)}×\frac{{f}\left(\mathrm{5}\right)}{{f}\left(\mathrm{5}\right)}×\frac{{f}\left(\mathrm{4}\right)}{{f}\left(\mathrm{4}\right)}= \\ $$$$=\frac{{f}\left(\mathrm{6}\right)}{{f}\left(\mathrm{5}\right)}×\frac{{f}\left(\mathrm{5}\right)}{{f}\left(\mathrm{4}\right)}×\frac{{f}\left(\mathrm{4}\right)}{{f}\left(\mathrm{3}\right)}=\mathrm{5}^{\mathrm{2}}…