Question Number 197034 by Mingma last updated on 06/Sep/23 Answered by TheHoneyCat last updated on 08/Sep/23 $$\mathrm{let} \\ $$$${f}:\begin{cases}{\mathbb{R}}&{\rightarrow}&{\mathbb{R}_{+} }\\{\alpha}&{ }&{\frac{\mathrm{4}\alpha^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}\alpha^{\mathrm{2}} }}\end{cases} \\ $$$$…
Question Number 197002 by hardmath last updated on 05/Sep/23 Commented by Frix last updated on 06/Sep/23 $$\mathrm{See}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{question}\:\mathrm{197011} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 197003 by hardmath last updated on 05/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196964 by Mingma last updated on 05/Sep/23 Answered by Rasheed.Sindhi last updated on 05/Sep/23 $${x}+{y}=−\mathrm{9}\:\wedge\:{x}+\mathrm{2}{y}=−\mathrm{25} \\ $$$$\Rightarrow{x}+\mathrm{2}\left(−\mathrm{9}−{x}\right)=−\mathrm{25} \\ $$$$\:\:\:\:{x}−\mathrm{2}{x}=−\mathrm{25}+\mathrm{18}=−\mathrm{7} \\ $$$$\:\:\:\:\:\:\:{x}=\mathrm{7} \\ $$…
Question Number 196935 by keiner last updated on 03/Sep/23 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196893 by hardmath last updated on 02/Sep/23 Answered by Rasheed.Sindhi last updated on 02/Sep/23 $${x}=\sqrt{\left(−\mathrm{1}+{i}\right)+\sqrt{\left(−\mathrm{1}+{i}\right)+\sqrt{…}}}\: \\ $$$${x}=\sqrt{\left(−\mathrm{1}+{i}\right)+{x}}\: \\ $$$${x}^{\mathrm{2}} =\:\left(−\mathrm{1}+{i}\right)+{x} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{1}−{i}=\mathrm{0}…
Question Number 196885 by hardmath last updated on 02/Sep/23 Answered by MM42 last updated on 02/Sep/23 $$\mathrm{2}{cosx}=\sqrt{\mathrm{2}+\mathrm{2}{cos}\mathrm{2}{x}}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+{cos}\mathrm{4}{x}}} \\ $$$$=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+{cos}\mathrm{8}{x}}}}=….=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}} \\ $$$${let}\:\:{x}=\mathrm{0}\Rightarrow\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}=\mathrm{2}\:\checkmark \\ $$$$ \\ $$…
Question Number 196872 by York12 last updated on 02/Sep/23 $${Let}\:\xi\:{be}\:{a}\:{positive}\:{Root}\:{of}\:{x}^{\mathrm{2}} −\mathrm{2023}{x}−\mathrm{1} \\ $$$${Define}\:{a}\:{sequence}\:\varphi_{{i}} \:{such}\:{That}\:\varphi_{\mathrm{0}} =\mathrm{1} \\ $$$$\varphi_{{n}+\mathrm{1}} =\lfloor\varphi_{{n}} \xi\rfloor,\:{find}\:{The}\:{Remainder}\:{When}\:\varphi_{\mathrm{2023}\:} {is}\:{divided}\:{by}\:\sqrt{\varphi_{\mathrm{2}} } \\ $$ Answered by…
Question Number 196870 by York12 last updated on 02/Sep/23 $${let}\:{b}_{{i}} \wedge\:{a}_{{i}} >\mathrm{0}\:{where}\:{i}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},…,{n}\right\}\&\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({b}_{{i}} \right)=\lambda\:{Prove}\:{that} \\ $$$$\frac{\lambda−\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} \right)}{\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} \right)}\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)+\frac{\lambda−\left({b}_{\mathrm{1}} +{b}_{\mathrm{3}} \right)}{\left({b}_{\mathrm{1}}…
Question Number 196860 by hardmath last updated on 01/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com