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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 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 Number 42934 by naka3546 last updated on 05/Sep/18 Commented by naka3546 last updated on 05/Sep/18 $${solve}\:{it}\:\:{without}\:\:{complex}\:\: \\ $$ Commented by MJS last updated on…
Question Number 108466 by mathdave last updated on 17/Aug/20 Commented by Her_Majesty last updated on 17/Aug/20 $${why}\:{not}\:{simply}\:{insert}\:{in}\:{the}\:{result}\:{of} \\ $$$${question}\:\mathrm{108416}\:??? \\ $$ Terms of Service Privacy…
Question Number 173993 by AgniMath last updated on 22/Jul/22 Answered by som(math1967) last updated on 22/Jul/22 $$\:{xy}+{yz}+{zx}+\mathrm{2}{xyz} \\ $$$$=\frac{{ab}}{\left({b}+{c}\right)\left({c}+{a}\right)}\:+\frac{{bc}}{\left({c}+{a}\right)\left({a}+{b}\right)} \\ $$$$\:\:+\frac{{ca}}{\left({a}+{b}\right)\left({b}+{c}\right)}+\frac{\mathrm{2}{abc}}{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)} \\ $$$$=\frac{{ab}\left({a}+{b}\right)+{bc}\left({b}+{c}\right)+{ca}\left({c}+{a}\right)+\mathrm{2}{abc}}{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)} \\ $$$$=\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}\:\:\:\:\:\bigstar…
Question Number 108456 by Khalmohmmad last updated on 17/Aug/20 Commented by Rizwankhan last updated on 17/Aug/20 $${how}\:{did}\:{you}\:{added}\:{this}\:{page}\:{sir}? \\ $$ Answered by bemath last updated on…
Question Number 42908 by naka3546 last updated on 04/Sep/18 $${Solve}\: \\ $$$$\:\:\:\:\:\mathrm{21}^{{a}} \:+\:\mathrm{28}^{{b}} \:\:=\:\:\mathrm{35}^{{c}} \\ $$$${if} \\ $$$${a},\:{b},\:\:{and}\:\:{c}\:\:{are}\:\:{positive}\:\:{integers}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…
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}! \\…