Question Number 194089 by a.lgnaoui last updated on 27/Jun/23 $$\mathrm{Determiner}\:\mathrm{le}\:\mathrm{rayon}\:\boldsymbol{\mathrm{r}} \\ $$ Commented by a.lgnaoui last updated on 27/Jun/23 Commented by cherokeesay last updated on…
Question Number 194088 by cortano12 last updated on 27/Jun/23 Answered by horsebrand11 last updated on 27/Jun/23 $$\:\mathrm{y}^{\mathrm{2}} =\:\left(\mathrm{2}+\left(\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\right)\left(\mathrm{1}−\left(\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\right) \\ $$$$\:\mathrm{let}\:\mathrm{x}+\mathrm{x}^{\mathrm{2}} =\:\mathrm{u} \\ $$$$\:\mathrm{y}^{\mathrm{2}}…
Question Number 194059 by Rupesh123 last updated on 27/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194058 by Mastermind last updated on 27/Jun/23 $$\mathrm{Ques}.\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Let}\:\mathrm{G}\:=\:\mathrm{C}_{\mathrm{5}} \:×\:\mathrm{C}_{\mathrm{25}} \:×\:\mathrm{C}_{\mathrm{625}} .\:\mathrm{Determine} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{elements}\:\mathrm{of}\:\mathrm{each}\:\mathrm{order}\:\mathrm{in}\:\mathrm{G} \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{List}\:\mathrm{the}\:\mathrm{abelian}\:\mathrm{groups}\:\mathrm{of}\:\mathrm{order}\:\mathrm{16} \\ $$$$\mathrm{and}\:\mathrm{of}\:\mathrm{order}\:\mathrm{27}\:\mathrm{up}\:\mathrm{to}\:\mathrm{Isomorphism}.…
Question Number 194085 by 073 last updated on 27/Jun/23 $$\mathrm{f}\left(\mathrm{x}\right)−\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{f}'\left(\mathrm{6}\right)=? \\ $$$$\mathrm{solution}?? \\ $$ Answered by qaz last updated on 27/Jun/23 $${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}…
Question Number 194086 by mathlove last updated on 27/Jun/23 $${f}\left({x}\right)=\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{3}\:\:{then}\:{a}_{\mathrm{0}\:} {equle}\:{to} \\ $$$$\left.{a}\left.\right)\:\:\:{p}\left(\mathrm{5}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\right)\:\:\:\:\mathrm{7} \\ $$$$\left.{c}\left.\right)\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}\right)\:{not}\:{defined} \\ $$ Commented by Frix last updated on 27/Jun/23…
Question Number 194015 by Rupesh123 last updated on 26/Jun/23 Answered by witcher3 last updated on 26/Jun/23 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\mathrm{f}_{\mathrm{k}} \mathrm{x}^{\mathrm{k}} \\ $$$${f}_{{k}+\mathrm{1}} ={f}_{{k}} +{f}_{{k}−\mathrm{1}} ;\mathrm{f}_{\mathrm{1}} =\mathrm{f}_{\mathrm{2}}…
Question Number 194037 by Subhi last updated on 26/Jun/23 $$ \\ $$$${Let}\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ….{a}_{{n}} \in{R}^{+} ,\:{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…..{a}_{{n}} =\mathrm{1} \\ $$$${prove}\:{that}: \\ $$$$\frac{{a}_{\mathrm{1}} }{\mathrm{2}−{a}_{\mathrm{1}} }+\frac{{a}_{\mathrm{2}}…
Question Number 194034 by MrGHK last updated on 26/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194029 by Rupesh123 last updated on 26/Jun/23 Commented by Rupesh123 last updated on 26/Jun/23 Perfect �� Answered by Subhi last updated on 26/Jun/23 $${AD}\:=\:\sqrt{\mathrm{2}{Q}^{\mathrm{2}}…