Question Number 150056 by ajfour last updated on 09/Aug/21 Commented by ajfour last updated on 09/Aug/21 $${If}\:{the}\:{blue}\:{region}\:{is}\:{a}\:{square} \\ $$$${of}\:{maximum}\:{area},\:{find}\:{a} \\ $$$${and}\:{side}\:{s}\:{of}\:{the}\:{square}.\:\:\: \\ $$ Answered by…
Question Number 84512 by 698148290 last updated on 13/Mar/20 Answered by jagoll last updated on 14/Mar/20 $$\mathrm{tangent}\:\mathrm{equation}\:\mathrm{at}\:\mathrm{point} \\ $$$$\mathrm{P}\left(\mathrm{2a}+\mathrm{2t}\:,\:\frac{\mathrm{at}^{\mathrm{2}} }{\mathrm{2}}\right)\:\mathrm{to}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\left(\mathrm{x}−\mathrm{2a}\right)^{\mathrm{2}} \:=\:\mathrm{2ay}\: \\ $$$$\Rightarrow\:\mathrm{2t}\left(\mathrm{x}−\mathrm{2a}\right)\:=\:\mathrm{ay}\:+\:\frac{\left(\mathrm{at}\right)^{\mathrm{2}}…
Question Number 84441 by Power last updated on 13/Mar/20 Commented by Power last updated on 13/Mar/20 $$\mathrm{sir}\:\mathrm{please}\:\mathrm{solution} \\ $$ Answered by mr W last updated…
Question Number 18904 by Satyamtt last updated on 01/Aug/17 $${Solve}\:{the}\:{triangle}\:{in}\:{which}\:{a}=\left(\sqrt{\mathrm{3}}+\mathrm{1}\right),\: \\ $$$${b}=\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\:{and}\:\angle{C}=\mathrm{60}°. \\ $$$$ \\ $$ Answered by behi.8.3.4.1.7@gmail.com last updated on 01/Aug/17 $${c}^{\mathrm{2}} ={a}^{\mathrm{2}}…
Question Number 18823 by rish@bh last updated on 30/Jul/17 $$\mathrm{If}\:\mathrm{P}\equiv\left(\mathrm{2},\mathrm{1}\right)\:\mathrm{and}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{x}−\mathrm{axis} \\ $$$$\mathrm{and}\:\mathrm{y}=\mathrm{x}\:\mathrm{respectively}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{PA}+\mathrm{PB}+\mathrm{AB}\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{find} \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{B}. \\ $$ Commented by ajfour last updated on 30/Jul/17…
Question Number 149796 by mr W last updated on 07/Aug/21 Commented by mr W last updated on 07/Aug/21 $${solution}\:{to}\:{Q}\mathrm{148806} \\ $$ Commented by mr W…
Question Number 18642 by tawa tawa last updated on 26/Jul/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{coordinate}\:\mathrm{axis}\:\mathrm{and}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{graph},\:\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\:+\:\mathrm{1},\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\:\left(\mathrm{0},\:\mathrm{1}\right) \\ $$ Answered by Tinkutara last updated on 26/Jul/17 $$\frac{{dy}}{{dx}}\:=\:\frac{{x}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{2}} \\…
Question Number 84157 by redmiiuser last updated on 10/Mar/20 $${if}\:{a}\:{circle}\:{having}\:{an}\: \\ $$$${equation}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{8}{y}=\mathrm{0} \\ $$$${is}\:{intersected}\:{at}\:{A}\:{and}\:{B} \\ $$$${by}\:{x}+{y}=\mathrm{1}.{find}\:{the}\: \\ $$$${equation}\:{of}\:{the}\:{circle} \\ $$$${on}\:{AB}\:{as}\:{diameter} \\ $$ Answered…
Question Number 149634 by puissant last updated on 06/Aug/21 $$\mathrm{Trouver}\:\mathrm{toutes}\:\mathrm{les}\:\mathrm{fonctions}\:\mathrm{f}:\mathbb{N}\rightarrow\mathbb{R}^{+} \\ $$$$\mathrm{telque}\:\forall\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{N}, \\ $$$$\mathrm{f}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)=\mathrm{f}\left(\mathrm{a}^{\mathrm{2}} \right)+\mathrm{f}\left(\mathrm{b}^{\mathrm{2}} \right)\:\mathrm{et}\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$ Answered by Ar Brandon last…
Question Number 149462 by liberty last updated on 05/Aug/21 Answered by Olaf_Thorendsen last updated on 05/Aug/21 $$\mathrm{A}\left(−\mathrm{3},\mathrm{0}\right)\:\mathrm{B}\left(\mathrm{1},−\mathrm{1}\right)\:\mathrm{C}\left(\mathrm{0},\mathrm{3}\right)\:\mathrm{D}\left(−\mathrm{1},\mathrm{3}\right) \\ $$$$\left(\mathrm{AC}\right)\::\:{y}\:=\:{x}+\mathrm{3} \\ $$$$\left(\mathrm{BD}\right)\::\:{y}\:=\:−\mathrm{2}{x}+\mathrm{1} \\ $$$$\mathrm{P}\:=\:\left(\mathrm{AC}\right)\cap\left(\mathrm{BD}\right)\:=\:\left(−\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{7}}{\mathrm{3}}\right) \\ $$$$\mathrm{P}\:=\:\left(\mid\mathrm{PA}\mid+\mid\mathrm{PB}\mid+\mid\mathrm{PC}\mid+\mid\mathrm{PD}\mid\right)_{{min}}…