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Question-195292

Question Number 195292 by Rupesh123 last updated on 29/Jul/23 Answered by MM42 last updated on 29/Jul/23 $$\frac{\left({y}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1}\:\:\Rightarrow{O}'\left(\mathrm{0},\mathrm{1}\right)\:,\:\:{A}\left(\mathrm{0},\mathrm{3}\right)\:\:,\:\:{A}'\left(\mathrm{0},−\mathrm{1}\right) \\ $$$$\Rightarrow{min}\left\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right\}=\mathrm{1}\:\checkmark\: \\ $$…

x-y-z-R-P-x-x-y-y-y-z-z-z-x-Q-y-x-y-z-y-z-x-z-x-Q-z-x-y-x-y-z-y-z-x-f-x-y-z-max-P-Q-R-find-f-min-

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…

A-professor-said-0-0-because-0-0-a-0-a-N-Can-you-prove-

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…

x-2-y-11-x-y-2-7-x-y-

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)…

Question-195287

Question Number 195287 by Mingma last updated on 29/Jul/23 Answered by MM42 last updated on 29/Jul/23 $${CD}={DE}\Rightarrow\angle{C}\mathrm{2}=\angle{E}\:\:\:\&\:\angle{D}\mathrm{1}+\angle{E}=\mathrm{60} \\ $$$$\angle{C}\mathrm{1}=\mathrm{60}−\angle{C}\mathrm{2}=\mathrm{60}−\angle{E}=\angle{D}\mathrm{1} \\ $$$$\frac{{CD}}{{Sin}\mathrm{60}}=\frac{\mathrm{2}}{{SinC}\mathrm{1}}\:\:\&\:\:\frac{{DE}}{{Sin}\mathrm{120}}=\frac{{BE}}{{SinD}\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{{SinC}\mathrm{1}}=\frac{{BE}}{{SinD}\mathrm{1}}\Rightarrow{BE}=\mathrm{2}\:\checkmark \\ $$…

y-sin-x-cos-x-2-4sin-x-2y-cos-x-y-tan-x-

Question Number 195263 by dimentri last updated on 28/Jul/23 $$\:\:\:\:\:\:\begin{cases}{{y}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\:=\:\mathrm{2}}\\{\mathrm{4sin}\:{x}−\mathrm{2}{y}\:\mathrm{cos}\:{x}\:=\:{y}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\mathrm{tan}\:{x}\:=?\: \\ $$ Answered by cortano12 last updated on 28/Jul/23 $$\:\:\:\:\:\:\begin{cases}{\:\cancel{\underline{\underbrace{ }}}}\end{cases} \\ $$…