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The-base-of-triangle-passes-through-a-fixed-point-p-a-b-and-its-sides-are-respectively-bisected-at-right-angles-by-the-line-x-y-0-and-x-9y-if-the-locus-of-the-third-vartex-is-a-circle-then-find-its

Question Number 43005 by mondodotto@gmail.com last updated on 06/Sep/18 $$\mathrm{The}\:\mathrm{base}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{passes}\:\mathrm{through} \\ $$$$\mathrm{a}\:\mathrm{fixed}\:\mathrm{point}\:\mathrm{p}\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{and}\:\mathrm{its}\:\mathrm{sides} \\ $$$$\mathrm{are}\:\mathrm{respectively}\:\mathrm{bisected}\:\mathrm{at}\:\mathrm{right}\:\mathrm{angles}\: \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{line}\:\mathrm{x}+\mathrm{y}=\mathrm{0}\:\mathrm{and}\:\mathrm{x}=\mathrm{9y} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{third}\:\mathrm{vartex}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}. \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{its}\:\mathrm{equation}. \\ $$ Commented by mondodotto@gmail.com…

Question-108498

Question Number 108498 by mohammad17 last updated on 17/Aug/20 Answered by Aziztisffola last updated on 17/Aug/20 $$\left.\mathrm{1}\right)\:\mathrm{y}=\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{x}} =\:\mathrm{e}^{\mathrm{xln}\left(\mathrm{x}+\mathrm{2}\right)} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\mathrm{ln}\left(\mathrm{x}+\mathrm{2}\right)+\frac{\mathrm{x}}{\mathrm{x}+\mathrm{2}}\right)\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{x}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\left(−\mathrm{1}\right)=−\mathrm{1} \\ $$ Commented…

Question-173993

Question Number 173993 by AgniMath last updated on 22/Jul/22 Answered by som(math1967) last updated on 22/Jul/22 $$\:{xy}+{yz}+{zx}+\mathrm{2}{xyz} \\ $$$$=\frac{{ab}}{\left({b}+{c}\right)\left({c}+{a}\right)}\:+\frac{{bc}}{\left({c}+{a}\right)\left({a}+{b}\right)} \\ $$$$\:\:+\frac{{ca}}{\left({a}+{b}\right)\left({b}+{c}\right)}+\frac{\mathrm{2}{abc}}{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)} \\ $$$$=\frac{{ab}\left({a}+{b}\right)+{bc}\left({b}+{c}\right)+{ca}\left({c}+{a}\right)+\mathrm{2}{abc}}{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)} \\ $$$$=\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}\:\:\:\:\:\bigstar…

How-many-pairs-of-a-b-c-d-so-that-a-b-c-d-which-a-b-c-d-positive-integers-

Question Number 42909 by naka3546 last updated on 04/Sep/18 $${How}\:\:{many}\:\:{pairs}\:{of}\:\:\left({a},\:{b},\:{c},\:{d}\right)\:\:{so}\:\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:{a}!\:+\:\:{b}!\:\:+\:\:{c}!\:\:=\:\:{d}! \\ $$$${which}\:\:\:{a},\:{b},\:{c},\:{d}\:\:\in\:\:{positive}\:\:{integers}\:. \\ $$ Answered by MJS last updated on 04/Sep/18 $$\mathrm{only}\:\mathrm{1}:\:\mathrm{2}!+\mathrm{2}!+\mathrm{2}!=\mathrm{3}! \\…