Question Number 195892 by AROUNAMoussa last updated on 12/Aug/23 Answered by mr W last updated on 13/Aug/23 $$\mathrm{1}+\sqrt{\left(\mathrm{4}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{1}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{4}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} }+\mathrm{2} \\ $$$$=\mathrm{7}+\mathrm{4}\sqrt{\mathrm{2}} \\…
Question Number 195880 by aawb_247 last updated on 12/Aug/23 Commented by aawb_247 last updated on 12/Aug/23 $${anyone}\:{can}\:{help}..\:{what}\:{a}\:{value}\:{of}\:{x}? \\ $$ Answered by mr W last updated…
Question Number 195806 by Mingma last updated on 11/Aug/23 Answered by som(math1967) last updated on 11/Aug/23 Commented by som(math1967) last updated on 11/Aug/23 $$\mathrm{13}^{\mathrm{2}} −{a}^{\mathrm{2}}…
Question Number 195740 by universe last updated on 09/Aug/23 Answered by Frix last updated on 09/Aug/23 $$\mathrm{Triangle}\:\Rightarrow\:{a}+{b}>{c}\wedge{a}+{c}>{b}\wedge{b}+{c}>{a} \\ $$$$\mathrm{Let}\:{b}=\left({u}−{v}\right){a}\wedge{c}=\left({u}+{v}\right){a} \\ $$$$\Rightarrow\:{u}>\frac{\mathrm{1}}{\mathrm{2}}\wedge−\frac{\mathrm{1}}{\mathrm{2}}<{v}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left({u},\:{v}\right)=\frac{\mathrm{1}}{\mathrm{2}{u}}+\frac{{u}−{v}}{{u}+{v}+\mathrm{1}}+\frac{{u}+{v}}{{u}−{v}+\mathrm{1}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\leqslant{f}\left({u},\:{v}\right)<\mathrm{2}…
Question Number 195688 by universe last updated on 07/Aug/23 Answered by mr W last updated on 07/Aug/23 Commented by York12 last updated on 08/Aug/23 $${please}\:{bro}\:{try}\:{to}\:{remember}…
Question Number 195344 by Mingma last updated on 31/Jul/23 Answered by kapoorshah last updated on 31/Jul/23 $${cos}\:\alpha\:=\:{cos}\:\alpha \\ $$$$\frac{{a}^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}.{a}.\mathrm{6}}\:=\:\frac{{b}^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}.{b}.\mathrm{6}}…
Question Number 195289 by Rupesh123 last updated on 29/Jul/23 Answered by HeferH last updated on 31/Jul/23 Commented by HeferH last updated on 31/Jul/23 $$\bigtriangleup{CMB}\:\cong\:\bigtriangleup{AMD}\:\:\Rightarrow\:\angle{IJM}\:=\:\mathrm{60}° \\…
Question Number 195287 by Mingma last updated on 29/Jul/23 Answered by MM42 last updated on 29/Jul/23 $${CD}={DE}\Rightarrow\angle{C}\mathrm{2}=\angle{E}\:\:\:\&\:\angle{D}\mathrm{1}+\angle{E}=\mathrm{60} \\ $$$$\angle{C}\mathrm{1}=\mathrm{60}−\angle{C}\mathrm{2}=\mathrm{60}−\angle{E}=\angle{D}\mathrm{1} \\ $$$$\frac{{CD}}{{Sin}\mathrm{60}}=\frac{\mathrm{2}}{{SinC}\mathrm{1}}\:\:\&\:\:\frac{{DE}}{{Sin}\mathrm{120}}=\frac{{BE}}{{SinD}\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{{SinC}\mathrm{1}}=\frac{{BE}}{{SinD}\mathrm{1}}\Rightarrow{BE}=\mathrm{2}\:\checkmark \\ $$…
Question Number 195273 by Shlock last updated on 28/Jul/23 Commented by necx122 last updated on 30/Jul/23 $${are}\:{the}\:{three}\:{red}\:{lines}\:{equal}? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 194998 by a.lgnaoui last updated on 22/Jul/23 $$\mathrm{Resolution}\:\mathrm{du}\:\mathrm{probldme}\:\mathrm{pose}\:\mathrm{par}\: \\ $$$$\mathrm{sonukgindia}\:\left(\mathrm{16}.\mathrm{7}.\mathrm{2023}\right) \\ $$$$\mathrm{voir}\:\:\mathrm{Q194819} \\ $$$$\bigtriangleup\boldsymbol{\mathrm{ABC}}\:\:\boldsymbol{\mathrm{AM}}=\boldsymbol{\mathrm{AN}}=\boldsymbol{\mathrm{AD}}\mathrm{cos}\:\frac{\boldsymbol{\alpha}}{\mathrm{2}} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{AC}}=\boldsymbol{\mathrm{AM}}+\boldsymbol{\mathrm{MC}}=\mathrm{17}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AN}}+\boldsymbol{\mathrm{NB}}\:\:=\mathrm{18}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\:\:\boldsymbol{\mathrm{AB}}−\boldsymbol{\mathrm{AC}}=\mathrm{1}=\boldsymbol{\mathrm{NB}}−\boldsymbol{\mathrm{MC}}\:\:\:\left(\mathrm{3}\right) \\ $$$$\bigtriangleup\boldsymbol{\mathrm{CDE}}\:\:\:\mathrm{cos}\:\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{CM}}}{\boldsymbol{\mathrm{CD}}}=\frac{\boldsymbol{\mathrm{CE}}}{\boldsymbol{\mathrm{CD}}}\:\Rightarrow\boldsymbol{\mathrm{CM}}=\boldsymbol{\mathrm{CE}} \\ $$$$\bigtriangleup\boldsymbol{\mathrm{BDE}}\:\:\:\mathrm{cos}\:\frac{\boldsymbol{\mathrm{B}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{BE}}}{\boldsymbol{\mathrm{BD}}}=\frac{\boldsymbol{\mathrm{BN}}}{\boldsymbol{\mathrm{BD}}}\Rightarrow\boldsymbol{\mathrm{BN}}=\boldsymbol{\mathrm{BE}} \\…