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 193792 by Rupesh123 last updated on 20/Jun/23 Answered by a.lgnaoui last updated on 21/Jun/23 Commented by a.lgnaoui last updated on 21/Jun/23 Terms of…
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 $$ \\…
Question Number 193757 by Rupesh123 last updated on 19/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 193756 by Shlock last updated on 19/Jun/23 Answered by talminator2856792 last updated on 19/Jun/23 $$\:\:\mathrm{7}\left(\mathrm{7}{a}\:+\:{b}\right)\:+\:{c}\:=\:\mathrm{7}\left(\mathrm{40}\right)\:+\:\mathrm{6} \\ $$$$\:\: \\ $$$$\:\:\rightarrow\:{c}\:=\:\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\:\:\rightarrow\:{c}\:=\:\mathrm{6} \\ $$$$\:\:…
Question Number 193759 by Mastermind last updated on 19/Jun/23 $$\mathrm{Ques}.\:\mathrm{5}\: \\ $$$$\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:\mathrm{a},\mathrm{b}\:\mathrm{are}\:\mathrm{any}\:\mathrm{elements}\:\mathrm{of}\:\:\mathrm{a}\: \\ $$$$\mathrm{group}\:\left(\mathrm{G},\:\ast\right),\:\mathrm{then}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{y}\ast\mathrm{a}=\mathrm{b} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{in}\:\left(\mathrm{G},\:\ast\right). \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{6}\: \\ $$$$\left.\:\:\:\:\:\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{set}\:\mathrm{G}\:\mathrm{of}\:\mathrm{all}\:\mathrm{non}-\mathrm{zero}\: \\ $$$$\mathrm{complex}\:\mathrm{numbers},\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{under} \\…
Question Number 193785 by aba last updated on 19/Jun/23 $${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{6}}×\frac{\mathrm{7}}{\mathrm{8}}×\frac{\mathrm{9}}{\mathrm{10}}×…×\frac{\mathrm{99}}{\mathrm{100}}<\frac{\mathrm{1}}{\mathrm{10}}\:\:\: \\ $$ Answered by MM42 last updated on 20/Jun/23 $${a}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{6}}…×\frac{\mathrm{99}}{\mathrm{100}}< \\ $$$$<\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{6}}{\mathrm{7}}×…×\frac{\mathrm{98}}{\mathrm{99}}×\frac{\mathrm{100}}{\mathrm{101}}= \\…
Question Number 193754 by Mingma last updated on 19/Jun/23 Answered by talminator2856792 last updated on 19/Jun/23 $$\:\:\mathrm{let}\:{R},\:{r}\:\:\mathrm{represent}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{big}\:\mathrm{semicircle}\:\:\: \\ $$$$\:\:\mathrm{and}\:\mathrm{small}\:\mathrm{semicricle},\:\mathrm{respectively}. \\ $$$$\:\:\mathrm{by}\:\mathrm{pythagorean}\:\mathrm{theorem}: \\ $$$$\:\:{R}\:=\:\sqrt{\left(\frac{{r}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\:\left(\mathrm{3}{r}\right)^{\mathrm{2}} \:}…
Question Number 193749 by Mingma last updated on 19/Jun/23 Answered by som(math1967) last updated on 19/Jun/23 Commented by som(math1967) last updated on 19/Jun/23 $${AD}={DC}={AC}=\mathrm{12}{cm} \\…
Question Number 193750 by thean last updated on 19/Jun/23 Answered by Subhi last updated on 19/Jun/23 $${only}\:{one}\:{number}\:{can}\:{be}\:{zero}\:{to}\:{give}\:{the}\:{mini}\:{value} \\ $$$${suppose}\:{b}=\mathrm{0}\:\Rrightarrow\:{a}+{c}=\mathrm{2} \\ $$$$\frac{{a}}{\mathrm{3}{a}^{\mathrm{2}} }+\frac{{c}}{\mathrm{3}{c}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right) \\ $$$$\left({a}+{b}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\geqslant\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}}…