Question Number 196723 by mr W last updated on 30/Aug/23 $${prove}\:{that}\:{the}\:{curve}\: \\ $$$$\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }+\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{4}\: \\ $$$${is}\:{an}\:{ellipse}\:{and}\:{find}\:{its}\:{semi} \\ $$$${major}\:{axis}\:{and}\:{semi}\:{minor}\:{axis}. \\ $$ Answered by…
Question Number 196275 by AROUNAMoussa last updated on 21/Aug/23 Answered by MM42 last updated on 21/Aug/23 $$\langle{ABD}=\mathrm{35}\Rightarrow{AD}={AB} \\ $$$$\langle{BCD}=\mathrm{55} \\ $$$$\frac{{AB}}{{AC}}=\frac{{sinx}}{{sin}\mathrm{130}}\:\:\:\&\:\:\frac{{AD}}{{AC}}=\frac{{sin}\left(\mathrm{55}−{x}\right)}{{sin}\mathrm{65}} \\ $$$$\Rightarrow{sinx}×{sin}\mathrm{65}=\mathrm{2}{sin}\mathrm{65}×{cos}\mathrm{65}×{sin}\left(\mathrm{55}−{x}\right) \\ $$$$\Rightarrow{sinx}=\mathrm{2}{cos}\mathrm{65}{sin}\left(\mathrm{55}−{x}\right)={sin}\left(\mathrm{55}−{x}+\mathrm{65}\right)+{sin}\left(\mathrm{55}−{x}−\mathrm{65}\right)…
Question Number 196225 by mr W last updated on 20/Aug/23 Commented by mr W last updated on 20/Aug/23 $${find}\:{the}\:{area}\:{of}\:{red}\:{part}\:{in}\:{the} \\ $$$${parallelogram}. \\ $$ Answered by…
Question Number 196021 by universe last updated on 16/Aug/23 Answered by mr W last updated on 17/Aug/23 Commented by mr W last updated on 17/Aug/23…
Question Number 195896 by AROUNAMoussa last updated on 12/Aug/23 Answered by HeferH last updated on 13/Aug/23 Commented by HeferH last updated on 13/Aug/23 $$\mathrm{180}°−\mathrm{4}{x}\:+\:\mathrm{40}°+\mathrm{2}{x}\:=\:\mathrm{180}° \\…
Question Number 195892 by AROUNAMoussa last updated on 12/Aug/23 Answered by mr W last updated on 13/Aug/23 $$\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}…