Question Number 80733 by key of knowledge last updated on 05/Feb/20 $$\mathrm{x}^{\mathrm{2}} =\mathrm{2}^{\mathrm{x}} \Rightarrow\mathrm{x}=? \\ $$ Commented by mr W last updated on 05/Feb/20 $${x}^{\mathrm{2}}…
Question Number 146261 by mathdanisur last updated on 12/Jul/21 $$\mathrm{4}\:{cos}\:\left(\mathrm{50}°\right)\:-\:\frac{\mathrm{1}}{{sin}\:\left(\mathrm{20}°\right)}\:=\:? \\ $$ Answered by bemath last updated on 12/Jul/21 $$\:\frac{\mathrm{4cos}\:\mathrm{50}°.\mathrm{cos}\:\mathrm{70}°−\mathrm{1}}{\mathrm{sin}\:\mathrm{20}°}\:= \\ $$$$\:\frac{\mathrm{2}\left(\mathrm{cos}\:\mathrm{120}°+\mathrm{cos}\:\mathrm{20}°\right)−\mathrm{1}}{\mathrm{sin}\:\mathrm{20}°}\:= \\ $$$$\:\frac{−\mathrm{1}+\mathrm{2cos}\:\mathrm{20}°−\mathrm{1}}{\mathrm{sin}\:\mathrm{20}°}\:=\: \\…
Question Number 146253 by mathdanisur last updated on 12/Jul/21 $$\left(\mathrm{4}{x}^{\mathrm{4}} \:-\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{17}\right)_{\boldsymbol{{max}}} \:=\:? \\ $$ Answered by iloveisrael last updated on 12/Jul/21 $$\mathrm{f}\left(\mathrm{x}\right)_{\mathrm{min}} =\:\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{17}…
Question Number 146244 by mathdanisur last updated on 12/Jul/21 $$\left(\sqrt{\mathrm{5}}\:-\:\mathrm{2}\right)^{\frac{\mathrm{2}{x}^{\mathrm{2}} -\mathrm{1}}{\mathrm{2}}} \:\:=\:\left(\sqrt{\mathrm{5}}\:+\:\mathrm{2}\right)^{\frac{\mathrm{3}{x}-\mathrm{7}}{\mathrm{4}}} \:\:\Rightarrow\:\:{x}=? \\ $$ Answered by gsk2684 last updated on 12/Jul/21 $${use}\:\sqrt{\mathrm{5}}−\mathrm{2}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}+\mathrm{2}} \\ $$$${compare}\:{exponents}…
Question Number 146242 by mathdanisur last updated on 12/Jul/21 $$\mathrm{4}\:{sin}\left(\mathrm{50}°\right)\:-\:\frac{\mathrm{1}}{{cos}\left(\mathrm{20}°\right)}\:=\:? \\ $$ Answered by gsk2684 last updated on 12/Jul/21 $$=\frac{\mathrm{4cos}\:\mathrm{40}^{\mathrm{0}} \mathrm{cos}\:\mathrm{20}^{\mathrm{0}} −\mathrm{1}}{\mathrm{cos}\:\mathrm{20}^{\mathrm{0}} } \\ $$$$=\frac{\mathrm{2}\left(\mathrm{cos}\:\mathrm{60}^{\mathrm{0}}…
Question Number 80706 by TawaTawa last updated on 05/Feb/20 Answered by mind is power last updated on 05/Feb/20 $${we}\:{use}\:{sin}\:{relation}\Rightarrow \\ $$$$\frac{{BD}}{{sin}\left(\frac{{A}}{\mathrm{2}}\right)}=\frac{{AB}}{{sin}\left(\angle{BDA}\right)} \\ $$$$\frac{{CD}}{{sin}\left(\frac{{A}}{\mathrm{2}}\right)}=\frac{{AD}}{{sin}\left(\mathrm{180}−\angle{BDA}\right)}=\frac{{AC}}{{sin}\left(\angle{BDA}\right)} \\ $$$$\Rightarrow\frac{{BD}}{{CD}}=\frac{{AB}}{{AC}}=\frac{{c}}{{b}}…
Question Number 146236 by mathdanisur last updated on 12/Jul/21 $$\int\:\frac{\mathrm{3}{x}\:+\:\mathrm{1}}{{x}}\:{dx}\:=\:? \\ $$ Answered by puissant last updated on 12/Jul/21 $$=\int\mathrm{3}+\frac{\mathrm{1}}{\mathrm{x}}\mathrm{dx} \\ $$$$=\mathrm{3x}+\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{k}.. \\ $$ Terms…
Question Number 80702 by john santu last updated on 05/Feb/20 $$\begin{cases}{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\mathrm{34}}\\{\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{\:\sqrt{{y}}}=\mathrm{23}−\frac{\mathrm{1}}{\:\sqrt{{xy}}}\:}\end{cases} \\ $$$${find}\:{the}\:{solution}. \\ $$ Commented by mind is power last updated on 05/Feb/20 $${let}\:\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{\:\sqrt{{y}}}={u}…
Question Number 146235 by mathdanisur last updated on 12/Jul/21 $${if}\:\:\:{arg}\:{z}_{\mathrm{1}} \:=\:\frac{\mathrm{4}\pi}{\mathrm{3}}\:\:\:{and}\:\:\:{arg}\:{z}_{\mathrm{1}} ^{\mathrm{2}} \:\centerdot\:{z}\:=\:\frac{\mathrm{7}\pi}{\mathrm{6}} \\ $$$${find}\:\:\:{arg}\:{z}\:=\:? \\ $$ Answered by gsk2684 last updated on 12/Jul/21 $${arg}\:{z}_{\mathrm{1}}…
Question Number 146230 by mathdanisur last updated on 12/Jul/21 $$\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}\:-\:\mathrm{1}\:+\:\frac{\mathrm{1}}{{cos}\left({x}\right)}\:=\:\mathrm{1}\:\Rightarrow\:{x}=? \\ $$ Answered by qaz last updated on 12/Jul/21 $$\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}+\mathrm{sec}\:\mathrm{x}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sec}\:\mathrm{x}−\mathrm{2}=\mathrm{0}…