Question Number 17279 by tawa tawa last updated on 03/Jul/17 $$\mathrm{Is}\:\:\mathrm{cosh}^{\mathrm{2}} \left(\mathrm{3x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}\:+\:\mathrm{cos}\left(\mathrm{6x}\right)\right]\:\:?????? \\ $$ Commented by mrW1 last updated on 03/Jul/17 $$\mathrm{cosh}^{\mathrm{2}} \:\mathrm{3x}=\left(\frac{\mathrm{e}^{\mathrm{3x}} +\mathrm{e}^{−\mathrm{3x}} }{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 148341 by mathdanisur last updated on 27/Jul/21 $${Solve}\:{the}\:{inequality}: \\ $$$$\left(\mathrm{5}\:-\:\mid{x}\mid\:\right)^{−\:\frac{\mathrm{1}}{\mathrm{3}}} \:\centerdot\:\left({x}^{\mathrm{2}} \:-\:\mathrm{4}\right)\:<\:\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148339 by mathdanisur last updated on 27/Jul/21 $${arccos}\:\left({cos}\:\mathrm{9}\right)\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 27/Jul/21 $$\left.{x}\:=\:\mathrm{arccos}\left(\mathrm{cos9}\right)\right) \\ $$$$\left.{x}\:=\:\mathrm{arccos}\left(\mathrm{cos}\left(\mathrm{9}−\mathrm{2}\pi\right)\right)\right),\:\mathrm{9}−\mathrm{2}\pi\:\in\left[\mathrm{0},\pi\right] \\ $$$${x}\:=\:\mathrm{9}−\mathrm{2}\pi \\…
Question Number 17260 by Umar math last updated on 03/Jul/17 $${for}\:{a},{b},{c}>\mathrm{0}\:{prove}\:{that} \\ $$$$\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} \geqslant\mathrm{3}\left({a}+{b}+{c}\right){abc} \\ $$ Commented by prakash jain last updated on 03/Jul/17 $$\mathrm{As}\:\mathrm{expained}\:\mathrm{in}\:\mathrm{answer}\:\mathrm{by}\:\mathrm{18}?\mathrm{1}.…
Question Number 148333 by mathdanisur last updated on 27/Jul/21 $$\left(\sqrt[{\mathrm{3}}]{{x}}\:+\:\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{15}} \\ $$$${Find}\:{the}\:{limit}\:{that}\:{does}\:{not}\:{inclued} \\ $$$${the}\:{variable}\:\boldsymbol{{x}}\:{in}\:{the}\:{opening}\:{of}\:{the} \\ $$$${binomial}. \\ $$ Answered by qaz last updated on 27/Jul/21…
Question Number 148321 by mathdanisur last updated on 27/Jul/21 Answered by dumitrel last updated on 27/Jul/21 $${p}\left({x}\right)={x}^{\mathrm{2021}} −{x}−\mathrm{1},{x}>\mathrm{0} \\ $$$${p}\left({a}\right)=\mathrm{0};{p}\left({x}\right)\:\nearrow\:{for}\:{x}\in\left({a},\infty\right);{a}>\mathrm{1} \\ $$$${if}\:{b}\geqslant{a}\Rightarrow{f}\left({b}\right)\geqslant{f}\left({a}\right)=\mathrm{0}\Rightarrow{b}^{\mathrm{2021}} \geqslant{b}+\mathrm{1}\Rightarrow \\ $$$${b}^{\mathrm{4042}}…
Question Number 148314 by Tawa11 last updated on 27/Jul/21 $$\mathrm{solve}:\:\:\:\:\:\:\mathrm{x}^{\mathrm{x}} \:\:\:=\:\:\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{How}\:\mathrm{can}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{solution} \\ $$ Commented by Tawa11 last updated on 27/Jul/21 $$\mathrm{what}\:\mathrm{is}\:\:\:\:\:\:\mathrm{W}_{\mathrm{n}} \left(−\:\mathrm{ln}\:\mathrm{4}\right) \\…
Question Number 148300 by mathdanisur last updated on 26/Jul/21 $${lg}^{\mathrm{2}} \left(\mathrm{10}{x}\right)\:+\:{lg}\left(\mathrm{10}{x}\right)\:=\:\mathrm{6}\:-\:{lg}\left({x}\right) \\ $$$${x}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 26/Jul/21 $$\mathrm{lg}^{\mathrm{2}} \left(\mathrm{10}{x}\right)+\mathrm{lg}\left(\mathrm{10}{x}\right)\:=\:\mathrm{6}−\mathrm{lg}\left({x}\right) \\…
Question Number 148301 by mathdanisur last updated on 26/Jul/21 $${Solve}\:{for}\:{equation}: \\ $$$$\mathrm{4}{sin}^{\mathrm{2}} \left({x}\right)\:+\:{sin}\left(\mathrm{2}{x}\right)\:=\:\mathrm{2} \\ $$ Answered by mr W last updated on 27/Jul/21 $$\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right)+\mathrm{sin}\:\mathrm{2}{x}=\mathrm{2} \\…
Question Number 148289 by saly last updated on 26/Jul/21 Commented by saly last updated on 26/Jul/21 $$\:{hepl}\:{me} \\ $$ Answered by mr W last updated…