Question Number 176891 by Shrinava last updated on 27/Sep/22 Answered by mr W last updated on 27/Sep/22 $$\mathrm{sin}\:{x}\:\mathrm{cos}\:\mathrm{3}{x}\:\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:\left(\frac{\mathrm{cos}\:{x}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:{x}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}}{\mathrm{2}}\right) \\ $$$$\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{cos}\:{x}\:\left(\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:{x}\right) \\ $$$$\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{3}\:=\:\mathrm{cos}^{\mathrm{2}}…
Question Number 111344 by Study last updated on 03/Sep/20 $$\sqrt{\mathrm{3}}!=? \\ $$ Commented by Dwaipayan Shikari last updated on 03/Sep/20 $$\mathrm{1}.\mathrm{585000}… \\ $$ Commented by…
Question Number 111343 by Study last updated on 03/Sep/20 $${i}!=? \\ $$ Commented by Dwaipayan Shikari last updated on 03/Sep/20 $$\Gamma\left(\mathrm{1}+{i}\right)=\mathrm{0}.\mathrm{4980}−\mathrm{0}.\mathrm{1549}{i} \\ $$ Terms of…
Question Number 176846 by Ar Brandon last updated on 27/Sep/22 $${m}\in\mathbb{Z} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{leading}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$$\mathrm{P}\left({x}\right)=\mathrm{4}{x}^{\frac{\mathrm{13}}{{m}−\mathrm{5}}} −\mathrm{6}{x}^{\mathrm{25}−\mathrm{2}{m}} +\mathrm{4}{x}^{\mathrm{13}} +\mathrm{5}{x}^{\mathrm{10}} −\mathrm{4}\:? \\ $$ Answered by mr W…
Question Number 176773 by Ar Brandon last updated on 26/Sep/22 $$\mathrm{Let}\:{a},\:{b},\:{c}\:>\mathrm{0}.\:\mathrm{If}\: \\ $$$$\frac{{a}−{b}}{{b}}>\mathrm{7},\:\frac{{b}+{c}}{{c}}<\mathrm{8},\: \\ $$$$\mathrm{Find}\:\mathrm{min}\left({a}+{b}+{c}\right)\: \\ $$ Commented by Ar Brandon last updated on 26/Sep/22…
Question Number 111208 by Aziztisffola last updated on 02/Sep/20 $$\:\mathrm{let}\:{p}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{s}.\mathrm{t}\:{p}\geqslant\mathrm{7}\:\mathrm{and}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=\underset{{p}−\mathrm{1}\:{times}} {\mathrm{333}……\mathrm{3}} \\ $$$$\:\mathrm{Show}\:\mathrm{that}\:\mathrm{11}\mid{a}. \\ $$ Commented by mr W last updated on 02/Sep/20…
Question Number 176750 by Tawa11 last updated on 26/Sep/22 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{general}\:\mathrm{form}\:\mathrm{of} \\ $$$$\alpha^{\mathrm{n}} \:\:\:+\:\:\:\beta^{\mathrm{n}} \:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\alpha^{\mathrm{n}} \:\:\:−\:\:\:\beta^{\mathrm{n}} \:\:\:\:\:???? \\ $$$$ \\ $$$$\mathrm{e}.\mathrm{g}:\:\:\:\:\alpha^{\mathrm{3}} \:\:\:+\:\:\:\beta^{\mathrm{3}} \:\:\:\:=\:\:\:\:\left(\alpha\:\:\:+\:\:\:\beta\right)\left[\left(\alpha\:\:\:+\:\:\:\beta\right)^{\mathrm{2}} \:\:\:−\:\:\:\mathrm{3}\alpha\beta\right] \\ $$…
Question Number 176746 by mr W last updated on 26/Sep/22 $${a}_{{n}+\mathrm{2}} −\mathrm{5}{a}_{{n}+\mathrm{1}} +\mathrm{6}{a}_{{n}} =\mathrm{3}{n}+\mathrm{5}^{{n}} \\ $$$${a}_{\mathrm{1}} =\mathrm{1},\:{a}_{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:{a}_{{n}} \\ $$ Answered by FongXD…
Question Number 176727 by Shrinava last updated on 25/Sep/22 $$\mathrm{In}\:\:\:\bigtriangleup\mathrm{ABC}\:\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{relationship} \\ $$$$\mathrm{holds}: \\ $$$$\mathrm{6r}\:\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{r}_{\boldsymbol{\mathrm{a}}} }{\mathrm{s}\:+\:\mathrm{n}_{\boldsymbol{\mathrm{a}}} }\:\:+\:\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\mathrm{n}_{\boldsymbol{\mathrm{a}}} \:\geqslant\:\mathrm{3s} \\ $$ Terms of Service Privacy…
Question Number 176718 by Frix last updated on 25/Sep/22 $${x}^{\mathrm{3}} +{y}^{\mathrm{3}} ={z}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +{z}^{\mathrm{3}} ={y}^{\mathrm{2}} \\ $$$${y}^{\mathrm{3}} +{z}^{\mathrm{3}} ={x}^{\mathrm{2}} \\ $$$$\mathrm{obviously}\:{x}={y}={z}=\mathrm{0}\vee\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{trying}\:\mathrm{to}\:\mathrm{totally}\:\mathrm{solve}\:\mathrm{it} \\…