Question Number 193848 by Mingma last updated on 21/Jun/23 Answered by MM42 last updated on 21/Jun/23 $$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{6}}×…×\:?\:×\frac{\mathrm{2024}}{\mathrm{2025}}\: \\ $$ Answered by witcher3 last updated on…
Question Number 193796 by Rajpurohith last updated on 20/Jun/23 $${Let}\:{G}\:{be}\:{a}\:{finite}\:{group},{f}\:{be}\:{an}\:{automorphism}\:{of}\:{G} \\ $$$${such}\:{that}\:{f}\left({x}\right)={x}\:\Rightarrow{x}={e}\:. \\ $$$${Then}\:{prove}\:{that}, \\ $$$$\left(\boldsymbol{{i}}\right)\forall{g}\in{G},\:\exists{x}\in{G}\:{such}\:{that}\:{g}={x}^{−\mathrm{1}} {f}\left({x}\right). \\ $$$$\left(\boldsymbol{{ii}}\right){If}\:\forall{x}\in{G}\:,\:{f}\left({f}\left({x}\right)\right)={x}\:\Rightarrow\:{G}\:{is}\:{Abelian}. \\ $$$$ \\ $$ Terms of…
Question Number 193793 by Mingma last updated on 20/Jun/23 Commented by MM42 last updated on 20/Jun/23 $${yes}.{you}\:{are}\:{right} \\ $$$${it}\:{was}\:{my}\:{mistake}.{thank}\:{you} \\ $$ Commented by Frix last…
Question Number 193794 by Rajpurohith last updated on 20/Jun/23 $${Let}\:{H}\:{be}\:{a}\:{subgroup}\:{of}\:\left(\mathbb{R},+\right)\:{such}\:{that}\:{H}\cap\left[−\mathrm{1},\mathrm{1}\right]\: \\ $$$${contains}\:{a}\:{non}\:{zero}\:{element}. \\ $$$${Prove}\:{that}\:{H}\:{is}\:{cyclic}. \\ $$ Answered by TheHoneyCat last updated on 28/Jun/23 $$ \\…
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Question Number 193714 by Rupesh123 last updated on 18/Jun/23 Commented by Frix last updated on 18/Jun/23 $$\sqrt{\mathrm{17}\pm\sqrt{\mathrm{288}}}=\sqrt{\mathrm{17}\pm\mathrm{12}\sqrt{\mathrm{2}}}=\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{7}\pm\sqrt{\mathrm{48}}}=\sqrt{\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}}}=\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{2} \\ $$…
Question Number 193707 by Mingma last updated on 18/Jun/23 Answered by Subhi last updated on 18/Jun/23 $$\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}}…
Question Number 193688 by Mingma last updated on 18/Jun/23 Answered by cortano12 last updated on 18/Jun/23 $$\:\mathrm{4}\left(\mathrm{2p}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}.\left(−\mathrm{1}\right).\left(−\mathrm{2p}−\mathrm{31}\right)<\mathrm{0} \\ $$$$\:\mathrm{4p}^{\mathrm{2}} +\mathrm{4p}+\mathrm{1}−\mathrm{2p}−\mathrm{31}<\mathrm{0} \\ $$$$\:\mathrm{4p}^{\mathrm{2}} +\mathrm{2p}−\mathrm{30}<\mathrm{0} \\…
Question Number 193687 by Mingma last updated on 18/Jun/23 Commented by Frix last updated on 18/Jun/23 $$\mathrm{Squaring}\:\&\:\mathrm{transforming}\:\mathrm{3}\:\mathrm{times}\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}^{\mathrm{8}} −\frac{\mathrm{10}{x}^{\mathrm{7}} }{\mathrm{3}}+\frac{\mathrm{143}{x}^{\mathrm{6}} }{\mathrm{27}}−\frac{\mathrm{854}{x}^{\mathrm{5}} }{\mathrm{81}}+\frac{\mathrm{4352}{x}^{\mathrm{4}} }{\mathrm{243}}−\frac{\mathrm{4678}{x}^{\mathrm{3}} }{\mathrm{243}}+\frac{\mathrm{611}{x}^{\mathrm{2}}…