Question Number 147810 by mathdanisur last updated on 23/Jul/21 $$\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\:\centerdot\:\boldsymbol{{cos}}\left(\mathrm{3}\boldsymbol{{x}}\right)=\boldsymbol{{cos}}\left(\mathrm{5}\boldsymbol{{x}}\right)\:\centerdot\:\boldsymbol{{cos}}\left(\mathrm{7}\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{x}}\:=\:? \\ $$ Answered by gsk2684 last updated on 23/Jul/21 $$\mathrm{2}\:\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{7}{x} \\ $$$$\mathrm{cos}\:\left(\mathrm{3}{x}+{x}\right)+\mathrm{cos}\:\left(\mathrm{3}{x}−{x}\right)=\mathrm{cos}\:\left(\mathrm{7}{x}+\mathrm{5}{x}\right)+\mathrm{cos}\:\left(\mathrm{7}{x}−\mathrm{5}{x}\right) \\…
Question Number 82273 by TawaTawa last updated on 19/Feb/20 Commented by mr W last updated on 19/Feb/20 $$\frac{{h}}{\mathrm{12}}=\frac{\mathrm{6}}{\mathrm{8}} \\ $$$$\Rightarrow{h}=\mathrm{9} \\ $$ Commented by TawaTawa…
Question Number 16736 by Tinkutara last updated on 26/Jun/17 $$\mathrm{The}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equiangular} \\ $$$$\mathrm{octagon}\:\mathrm{are}\:\mathrm{rational}\:\mathrm{numbers}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{octagon}\:\mathrm{has}\:\mathrm{a}\:\mathrm{symmetry} \\ $$$$\mathrm{center}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 16735 by Tinkutara last updated on 26/Jun/17 $$\mathrm{Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:…,\:{a}_{{n}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{of}\: \\ $$$$\mathrm{an}\:\mathrm{equiangular}\:\mathrm{polygon}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{if} \\ $$$${a}_{\mathrm{1}} \:\geqslant\:{a}_{\mathrm{2}} \:\geqslant\:…\:\geqslant\:{a}_{{n}} ,\:\mathrm{then}\:\mathrm{the}\:\mathrm{polygon}\:\mathrm{is} \\ $$$$\mathrm{regular}. \\ $$ Terms…
Question Number 16734 by Tinkutara last updated on 26/Jun/17 $$\mathrm{An}\:\mathrm{equiangular}\:\mathrm{polygon}\:\mathrm{with}\:\mathrm{an}\:\mathrm{odd} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{sides}\:\mathrm{is}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{polygon}\:\mathrm{is}\:\mathrm{regular}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147800 by 0731619 last updated on 23/Jul/21 Answered by Olaf_Thorendsen last updated on 23/Jul/21 $$\mathbb{R}\:=\:\underset{{n}\in\mathbb{Z}} {\cup}\mathrm{E}_{{n}} ,\:\mathrm{E}_{{n}} \:=\:\left[{n},{n}+\mathrm{1}\left[\right.\right. \\ $$$$\forall{x}\in\mathrm{E}_{{n}} ,\:\left\{{x}\right\}\:=\:{x}−\lfloor{x}\rfloor\:=\:{x}−{n} \\ $$$$\forall{x}\in\mathrm{E}_{{n}}…
Question Number 147803 by cesarL last updated on 23/Jul/21 $$\int\frac{\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{4}} }}{dx} \\ $$ Commented by lyubita last updated on 23/Jul/21 $${Mr}\:{Rasheed},\:{you}'{re}\:{right}… \\ $$…
Question Number 82265 by Ali Yousafzai last updated on 19/Feb/20 $${f}\boldsymbol{{actorize}}: \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{6}\right)−\mathrm{3}{x}^{\mathrm{2}} \\ $$$$ \\ $$ Answered by MJS last updated on 19/Feb/20 $$={x}^{\mathrm{4}}…
Question Number 147799 by Eric002 last updated on 23/Jul/21 $$\int\frac{\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{dx} \\ $$ Answered by Khalmohmmad last updated on 24/Jul/21 Terms of Service…
Question Number 16723 by tawa tawa last updated on 25/Jun/17 $$\mathrm{Solve}:\:\mathrm{2}^{\mathrm{x}} \:+\:\mathrm{48}\:=\:\mathrm{16x} \\ $$ Answered by mrW1 last updated on 26/Jun/17 $$\mathrm{2}^{\mathrm{x}} =\mathrm{16}\left(\mathrm{x}−\mathrm{3}\right) \\ $$$$\frac{\mathrm{2}^{\mathrm{x}}…