Question Number 62942 by Tawa1 last updated on 27/Jun/19 $$\mathrm{Make}\:\:\mathrm{r}\:\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formular}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{S}\:\:=\:\:\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}} \:−\:\mathrm{1}\right)}{\mathrm{r}\:−\:\mathrm{1}} \\ $$ Commented by Tawa1 last updated on 27/Jun/19 $$\mathrm{Sir},\:\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{this} \\ $$$$\:\:\:\:\:\:\:…
Question Number 128458 by SLVR last updated on 07/Jan/21 Commented by BHOOPENDRA last updated on 07/Jan/21 $${Givenf}\left({f}\left(\mathrm{1}\right)\right)=\mathrm{0},{f}\left({f}\left(\mathrm{2}\right)\right)=\mathrm{0} \\ $$$${i}.{e}\:{equation}\:{f}\left({x}\right)=\mathrm{0} \\ $$$${has}\:{two}\:{root}\:{f}\left(\mathrm{1}\right){andf}\left(\mathrm{2}\right) \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)=−\alpha\:{and}\:{f}\left(\mathrm{1}\right).{f}\left(\mathrm{2}\right)=\beta \\ $$$${so}\:\mathrm{5}+\mathrm{3}\alpha+\mathrm{2}\beta=−\alpha…
Question Number 128459 by SLVR last updated on 07/Jan/21 $${For}\:{any}\:{complex}\:{number}\:{z},{z}^{{n}} =\bar {{z}}\:{has}\:\left({n}+\mathrm{2}\right)\:{solutions}\:{How}??? \\ $$ Answered by mr W last updated on 10/Jan/21 $${let}\:{z}={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)={re}^{{i}\theta} \\ $$$${z}^{{n}}…
Question Number 128457 by SLVR last updated on 07/Jan/21 Commented by BHOOPENDRA last updated on 07/Jan/21 $${x}^{\mathrm{5}} =\frac{\mathrm{133}{x}−\mathrm{78}}{\mathrm{133}−\mathrm{78}{x}} \\ $$$$\mathrm{78}{x}^{\mathrm{6}} −\mathrm{133}{x}^{\mathrm{5}} +\mathrm{133}−\mathrm{78}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{78}{x}^{\mathrm{4}}…
Question Number 62895 by Tawa1 last updated on 26/Jun/19 Commented by Tony Lin last updated on 26/Jun/19 $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=\pm\sqrt{\frac{\mathrm{1}}{\mathrm{2}}},\:{b}=\pm\sqrt{\frac{\mathrm{1}}{\mathrm{2}}},\:{c}=\pm\sqrt{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\because\:\mid{ab}\mid<\mid{bc}\mid=\mid{ac}\mid…
Question Number 128388 by enter last updated on 06/Jan/21 $${y}=−{sin}\left(\frac{\Pi}{\mathrm{2}}+\mathrm{2}{x}\right)+\mathrm{2}{cos}\left(\mathrm{5}{x}+\Pi\right) \\ $$$${y}_{{min}} =\:?? \\ $$ Commented by TheSupreme last updated on 07/Jan/21 $${y}=−{cos}\left(\mathrm{2}{x}\right)−\mathrm{2}{cos}\left(\mathrm{5}{x}\right)\geqslant−\mathrm{3} \\ $$$${y}=−\mathrm{3}\:{x}=\mathrm{10}{k}\pi…
Question Number 62844 by alphaprime last updated on 25/Jun/19 $${Let}\:{p}\left({x}\right)\:=\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:\:{be}\:{such}\:{that}\:{p}\left({x}\right)\:{takes}\:{real}\:{values} \\ $$$${for}\:{real}\:{values}\:{of}\:{x}\:{and}\:{non}−{real}\:{values}\:{for}\:{non}−{real} \\ $$$${values}\:{of}\:{x}\:.\:{Prove}\:{that}\:{a}\:=\:\mathrm{0}\:{and}\:{find}\:{all} \\ $$$${possible}\:{values}\:{of}\:{c}. \\ $$ Answered by ajfour last updated on…
Question Number 128371 by aliibrahim1 last updated on 06/Jan/21 Commented by MJS_new last updated on 06/Jan/21 $${A}_{{n}} =\frac{\mathrm{752}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:−\mathrm{752}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}}}{\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}+{k}}} \\…
Question Number 128368 by I want to learn more last updated on 06/Jan/21 $$\mathrm{If}\:\:\:\:\mathrm{a}_{\mathrm{1}} \:=\:\:\mathrm{2},\:\:\:\:\mathrm{a}_{\mathrm{2}} \:\:=\:\:\mathrm{3},\:\:\:\:\:\:\mathrm{a}_{\mathrm{n}\:\:+\:\:\mathrm{2}} \:\:=\:\:\mathrm{a}_{\mathrm{n}\:\:+\:\:\mathrm{1}} \:\:+\:\:\frac{\mathrm{a}}{\mathrm{2}},\:\:\:\:\:\:\mathrm{find}\:\:\mathrm{a}_{\mathrm{n}} \\ $$ Answered by mr W last…
Question Number 128369 by I want to learn more last updated on 06/Jan/21 $$\mathrm{If}\:\:\:\mathrm{u}_{\mathrm{1}} \:\:+\:\:\mathrm{u}_{\mathrm{2}} \:\:+\:\:\mathrm{u}_{\mathrm{3}} \:\:+\:\:…\:\:+\:\:\mathrm{u}_{\mathrm{n}} \:\:\:=\:\:\:\mathrm{2n}^{\mathrm{2}} \:\:+\:\:\mathrm{n}\:\:\:\mathrm{is}\:\mathrm{an}\:\mathrm{AP}. \\ $$$$\mathrm{Find}\:\:\:\:\:\:\:\:\mathrm{u}_{\mathrm{1}} \:\:+\:\:\mathrm{u}_{\mathrm{2}} \:\:+\:\:\mathrm{u}_{\mathrm{3}} \:\:+\:\:…\:\:+\:\:\mathrm{u}_{\mathrm{2n}\:\:−\:\:\mathrm{2}} \:\:+\:\:\mathrm{u}_{\mathrm{2n}\:\:−\:\:\mathrm{1}}…