Question Number 195322 by Rupesh123 last updated on 30/Jul/23 Answered by HeferH last updated on 31/Jul/23 $${say}\:{k}\:{is}\:{the}\:{incenter}\:{length} \\ $$$$\:{k}\:=\:\frac{{A}}{{s}} \\ $$$$\:{s}\:=\:\frac{{AB}+{AC}+{BC}}{\mathrm{2}} \\ $$$$\:{A}\:=\:\frac{\mathrm{3}{k}\centerdot{BC}}{\mathrm{2}} \\ $$$$\:\mathrm{1}\:=\:\frac{\mathrm{3}{BC}}{{AB}+{AC}+{BC}}…
Question Number 195331 by Shlock last updated on 30/Jul/23 Answered by mr W last updated on 06/Aug/23 $${say}\:{a}={pc},\:{b}={qc} \\ $$$$\left({p}+{q}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\mathrm{1}\right)=\mathrm{10} \\ $$$$\frac{{a}+{b}}{{c}}={p}+{q} \\ $$$$\left({p}+{q}\right)_{{max}} =?…
Question Number 195315 by cortano12 last updated on 30/Jul/23 $$\:\:\:\:\:\:\Subset \\ $$ Answered by horsebrand11 last updated on 30/Jul/23 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} \mathrm{a}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{a}+\mathrm{tan}\:^{\mathrm{4}} \mathrm{a}\:\mathrm{tan}\:^{\mathrm{2}}…
Question Number 195330 by Rupesh123 last updated on 30/Jul/23 Commented by AST last updated on 30/Jul/23 $$\Rightarrow\left({abc}\right)^{\mathrm{3}} =\mathrm{10}^{\mathrm{45}} \Rightarrow{abc}=\mathrm{10}^{\mathrm{15}} \\ $$ Terms of Service Privacy…
Question Number 195292 by Rupesh123 last updated on 29/Jul/23 Answered by MM42 last updated on 29/Jul/23 $$\frac{\left({y}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1}\:\:\Rightarrow{O}'\left(\mathrm{0},\mathrm{1}\right)\:,\:\:{A}\left(\mathrm{0},\mathrm{3}\right)\:\:,\:\:{A}'\left(\mathrm{0},−\mathrm{1}\right) \\ $$$$\Rightarrow{min}\left\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right\}=\mathrm{1}\:\checkmark\: \\ $$…
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 $$\bigtriangleup{CMB}\:\cong\:\bigtriangleup{AMD}\:\:\Rightarrow\:\angle{IJM}\:=\:\mathrm{60}° \\…
Question Number 195288 by CrispyXYZ last updated on 29/Jul/23 $${x},\:{y},\:{z}\in\mathbb{R}_{+} , \\ $$$${P}\:=\:\frac{{x}}{{x}\:+\:{y}}\:+\:\frac{{y}}{{y}\:+\:{z}}\:+\:\frac{{z}}{{z}\:+\:{x}}, \\ $$$${Q}\:=\:\frac{{y}}{{x}\:+\:{y}}\:+\:\frac{{z}}{{y}\:+\:{z}}\:+\:\frac{{x}}{{z}\:+\:{x}}, \\ $$$${Q}\:=\:\frac{{z}}{{x}\:+\:{y}}\:+\:\frac{{x}}{{y}\:+\:{z}}\:+\:\frac{{y}}{{z}\:+\:{x}}. \\ $$$${f}\left({x},\:{y},\:{z}\right)=\mathrm{max}\left\{{P},\:{Q},\:{R}\right\},\:\mathrm{find}\:{f}_{\mathrm{min}} . \\ $$ Commented by Frix…
Question Number 195291 by Matica last updated on 29/Jul/23 $$\:\:{it}\:{is}\:{given}\:{a},{b},{c}\:\in\:\mathbb{N}^{\ast} \:\:{and}\:\:{ab}<{c}\:.\:{Prove}\:{that}\:{a}+{b}\leqslant{c}. \\ $$ Answered by Frix last updated on 29/Jul/23 $${c}={ab}+\mathrm{1} \\ $$$${a}+{b}\leqslant{ab}+\mathrm{1} \\ $$$$\mathrm{Let}\:{a}<{b}…
Question Number 195290 by Matica last updated on 29/Jul/23 $$\:{A}\:{professor}\:{said}\:\:\mathrm{0}\mid\mathrm{0}\:{because}\:\mathrm{0}=\:\mathrm{0}×{a}+\mathrm{0}\:\:\:,\:{a}\in\:\mathbb{N}.\:{Can}\:{you}\:{prove}? \\ $$ Answered by Frix last updated on 29/Jul/23 $${a}\mid{b}\:\Leftrightarrow\:\frac{{b}}{{a}}\in\mathbb{Z}\:\mathrm{but}\:\frac{\mathrm{0}}{\mathrm{0}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\Rightarrow\:\mathrm{0}\mid\mathrm{0}\:\mathrm{is}\:\mathrm{meaningless} \\ $$ Terms of Service…
Question Number 195301 by SajaRashki last updated on 29/Jul/23 $$\begin{cases}{{x}^{\mathrm{2}} +{y}=\mathrm{11}}\\{{x}+{y}^{\mathrm{2}} =\mathrm{7}}\end{cases}\Rightarrow\:{x},{y}=? \\ $$ Answered by AST last updated on 30/Jul/23 $${y}=\mathrm{11}−{x}^{\mathrm{2}} \Rightarrow\left(\mathrm{11}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{7}−{x}…\left({i}\right)…