Question Number 83473 by M±th+et£s last updated on 02/Mar/20 $${solve}\:{in}\:{R} \\ $$$${sin}\left(\pi{ln}\left({x}\right)\right)+{cos}\left(\pi{ln}\left({x}\right)\right)=\mathrm{1} \\ $$ Answered by mind is power last updated on 02/Mar/20 $$\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{sin}\left(\pi{ln}\left({x}\right)\right)+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{cos}\left(\pi{ln}\left({x}\right)\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\…
Question Number 149008 by Samimsultani last updated on 02/Aug/21 Commented by bramlexs22 last updated on 02/Aug/21 $$\sqrt{\mathrm{5}+\sqrt{\mathrm{24}}\:}\:=\:\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}\:=\:\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\: \\ $$$$\sqrt{\mathrm{5}−\sqrt{\mathrm{24}}}\:=\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}} \\ $$$$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{x}} −\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)^{\mathrm{x}} \:=\:\mathrm{40}\sqrt{\mathrm{6}} \\ $$$$\mathrm{let}\:\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{x}}…
Question Number 148997 by gsk2684 last updated on 02/Aug/21 $${find}\:{number}\:{of}\:{rational}\:{terms}\: \\ $$$${in}\:{the}\:{expansion}\:{of}\:\left(\sqrt[{\mathrm{5}}]{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)^{\mathrm{15}} . \\ $$$${prove}\:{sum}\:{of}\:{all}\:{irrarional}\:{terms}\: \\ $$$${is}\:{greater}\:{than}\:{sum}\:{of}\:{all}\:{rational} \\ $$$${terms}. \\ $$ Terms of Service Privacy…
Question Number 148994 by gsk2684 last updated on 02/Aug/21 $$\left(\mathrm{102}\right)^{\mathrm{4}} \\ $$$${easy}\:{way}\:{to}\:{caculate} \\ $$ Answered by iloveisrael last updated on 02/Aug/21 $$\left(\mathrm{100}+\mathrm{2}\right)^{\mathrm{4}} =\underset{\mathrm{n}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\mathrm{C}_{\mathrm{k}}…
Question Number 148987 by mathdanisur last updated on 02/Aug/21 $${if}\:\:\:{tg}\left(\mathrm{0},\mathrm{5}{x}\right)\:=\:−\mathrm{2} \\ $$$${find}\:\:\:\frac{{sin}\left({x}\right)\:+\:\mathrm{2}}{{cos}\left({x}\right)\:-\:\mathrm{3}}\:=\:? \\ $$ Answered by EDWIN88 last updated on 02/Aug/21 $$\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{2}}{x}\right)=−\mathrm{2}\:\rightarrow\begin{cases}{\frac{\mathrm{1}}{\mathrm{2}}{x}\:{in}\:\mathrm{2}^{{nd}} \:{quadrant}}\\{\frac{\mathrm{1}}{\mathrm{2}}{x}\:{in}\:\mathrm{4}^{{th}} \:{quadrant}}\end{cases} \\…
Question Number 148986 by mathdanisur last updated on 02/Aug/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{tg}^{\mathrm{2}} {x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{6}} }{\mathrm{5}{sin}^{\mathrm{2}} {x}^{\mathrm{3}} }\:=\:? \\ $$ Answered by EDWIN88 last updated on 02/Aug/21…
Question Number 148975 by mathdanisur last updated on 01/Aug/21 Commented by Kamel last updated on 02/Aug/21 $${yes}\:{sorry}\:\angle{ABD}=\mathrm{72}° \\ $$ Answered by Kamel last updated on…
Question Number 148947 by mathdanisur last updated on 01/Aug/21 $$\mathrm{615}\:+\:{x}^{\mathrm{2}} \:=\:\mathrm{2}^{\boldsymbol{{y}}} \:\:\:;\:\:\:{x};{y}\in\mathbb{N} \\ $$$$\Rightarrow\:{x};{y}\:=\:? \\ $$ Answered by dumitrel last updated on 02/Aug/21 $${if}\:{y}=\mathrm{2}{k}+\mathrm{1}\Rightarrow{u}\left(\mathrm{2}^{{y}} \right)\in\left\{\mathrm{2},\mathrm{8}\right\}\Rightarrow…
Question Number 148946 by Tawa11 last updated on 01/Aug/21 $$\mathrm{Sum}\:\mathrm{to}\:\:\mathrm{n}\:\:\mathrm{term}:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:\:\:+\:\:\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:\:+\:\:…\:\:\:\mathrm{to}\:\:\mathrm{n}. \\ $$ Answered by Olaf_Thorendsen last updated on 02/Aug/21 $$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)\left(\mathrm{3}{n}+\mathrm{3}\right)} \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty}…
Question Number 83378 by M±th+et£s last updated on 01/Mar/20 $${x}+{y}+{z}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{3} \\ $$$${find} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}}…