Question Number 152697 by mathdanisur last updated on 31/Aug/21 $$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\Sigma\:\frac{\left(\mathrm{r}_{\boldsymbol{\mathrm{a}}} +\mathrm{r}_{\boldsymbol{\mathrm{b}}} \right)\left(\mathrm{r}_{\boldsymbol{\mathrm{a}}} +\mathrm{r}_{\boldsymbol{\mathrm{c}}} \right)}{\mathrm{h}_{\boldsymbol{\mathrm{b}}} +\mathrm{h}_{\boldsymbol{\mathrm{c}}} }\:\geqslant\:\frac{\mathrm{9r}}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy…
Question Number 21622 by Joel577 last updated on 29/Sep/17 $$\mathrm{If}\:\mathrm{sec}\:{x}\:+\:\mathrm{tan}\:{x}\:=\:\mathrm{2012} \\ $$$$\mathrm{then}\:\mathrm{2011}\left(\mathrm{cosec}\:{x}\:+\:\mathrm{cot}\:{x}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left({A}\right)\:\mathrm{2011} \\ $$$$\left({B}\right)\:\mathrm{2012} \\ $$$$\left({C}\right)\:\mathrm{2013} \\ $$$$\left({D}\right)\:\frac{\mathrm{2011}}{\mathrm{2013}} \\ $$$$\left({E}\right)\:\frac{\mathrm{2013}}{\mathrm{2012}} \\ $$ Answered…
Question Number 152678 by mathdanisur last updated on 31/Aug/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equations}: \\ $$$$\mathrm{a}^{\mathrm{3}} -\mathrm{6a}^{\mathrm{2}} +\mathrm{15a}-\mathrm{17}=\mathrm{0}\:\:\:\mathrm{and} \\ $$$$\mathrm{a}^{\mathrm{3}} -\mathrm{6a}^{\mathrm{2}} +\mathrm{15a}-\mathrm{11}=\mathrm{0} \\ $$ Answered by mr W last…
Question Number 87130 by jagoll last updated on 03/Apr/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{for}\:\mathrm{the}\:\mathrm{curve}\: \\ $$$$\mathrm{r}\:=\:\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{at}\:\theta\:=\frac{\pi}{\mathrm{4}}\:? \\ $$ Commented by mr W last updated on 03/Apr/20 $${y}={r}\:\mathrm{sin}\:\theta=\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{sin}\:\theta \\ $$$${x}={r}\:\mathrm{cos}\:\theta=\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\theta…
Question Number 152663 by mr W last updated on 31/Aug/21 $$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} -\mathrm{x}+\mathrm{3}=\mathrm{0}\:\mathrm{has}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{a},\:\mathrm{b}\:\mathrm{and}\:\mathrm{c}. \\ $$$$\mathrm{determine}\:\mathrm{the}\:\mathrm{monic}\:\mathrm{polynomial}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\:\mathrm{a}^{\mathrm{5}} ,\:\mathrm{b}^{\mathrm{5}} \:\mathrm{and}\:\:\mathrm{c}^{\mathrm{5}} . \\ $$$$\left[{Q}\mathrm{152396}\right] \\ $$ Answered by…
Question Number 21578 by ANTARES_VY last updated on 28/Sep/17 $$\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{whole}}\:\:\boldsymbol{\mathrm{part}}\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{A}}? \\ $$$$\boldsymbol{\mathrm{A}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}}+……+\frac{\mathrm{1}}{\:\sqrt{\mathrm{9999}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{10000}}}. \\ $$ Commented by Tinkutara last updated on 28/Sep/17 $$\mathrm{Almost}\:\mathrm{similar}\:\mathrm{to}\:\mathrm{KVS}\:\mathrm{JMO}\:\mathrm{2016}. \\ $$$$\mathrm{Solutions}\:\mathrm{are}\:\mathrm{available}\:\mathrm{on}\:\mathrm{net}. \\…
Question Number 152625 by Dandelion last updated on 30/Aug/21 $${x}^{\mathrm{4}} +{c}_{\mathrm{3}} {x}^{\mathrm{3}} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{0}} =\mathrm{0} \\ $$$$\mathrm{for}\:{c}_{{n}} \in\mathbb{R}\:\mathrm{this}\:\mathrm{can}\:\mathrm{have} \\ $$$$\mathrm{4}\:\mathrm{unique}\:\mathrm{zeros}\:\in\mathbb{R} \\ $$$$\mathrm{2}\:\mathrm{unique}\:\mathrm{zeros}\:+\:\mathrm{1}\:\mathrm{double}\:\mathrm{zero}\:\in\mathbb{R} \\…
Question Number 87093 by M±th+et£s last updated on 02/Apr/20 $$\lfloor\frac{{x}−\mathrm{1}}{\mathrm{4}}\rfloor+\lfloor\frac{{x}−\mathrm{2}}{\mathrm{3}}\rfloor=\lfloor\frac{{x}−\mathrm{3}}{\mathrm{2}}\rfloor \\ $$ Commented by M±th+et£s last updated on 02/Apr/20 $${slove}\:{the}\:{equation} \\ $$ Commented by MJS…
Question Number 152626 by mathdanisur last updated on 30/Aug/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{1}}{\mathrm{x}-\mathrm{1}}\:+\:\frac{\mathrm{2}}{\mathrm{x}-\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{x}-\mathrm{3}}\:+\:\frac{\mathrm{4}}{\mathrm{x}-\mathrm{4}}\:=\:\mathrm{2x}^{\mathrm{2}} -\mathrm{5x}-\mathrm{4} \\ $$ Answered by MJS_new last updated on 30/Aug/21 $${x}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{2}}…
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