Question Number 187689 by Mingma last updated on 20/Feb/23 Answered by mr W last updated on 21/Feb/23 Commented by mr W last updated on 26/Feb/23…
Question Number 187589 by Mingma last updated on 19/Feb/23 Commented by Mingma last updated on 19/Feb/23 Blue area=? Answered by cortano12 last updated on 19/Feb/23 $$\mathrm{cos}\:\alpha=\frac{\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 187587 by Mingma last updated on 19/Feb/23 Commented by Mingma last updated on 19/Feb/23 Blue area=? Answered by mr W last updated on 19/Feb/23…
Question Number 121793 by ajfour last updated on 11/Nov/20 Commented by ajfour last updated on 13/Nov/20 $${If}\:{O}\overset{\frown} {{DE}}\:\:\:{is}\:{a}\:{sector},\:{and}\:\bigtriangleup{ABC}\: \\ $$$${is}\:{equilateral}\:{with}\:{minimum}\:{area} \\ $$$${has}\:{an}\:{area}\:{equal}\:{to}\:\bigtriangleup{OEF}\:,\:{find}\:\theta. \\ $$$$ \\…
Question Number 187301 by Rupesh123 last updated on 15/Feb/23 Answered by HeferH last updated on 16/Feb/23 $$\:{chord}\:{theorem}: \\ $$$$\mathrm{5}^{\mathrm{2}} \:=\:{x}\left(\mathrm{5}\:+\:\mathrm{3}\right) \\ $$$$\:\frac{\mathrm{25}}{\mathrm{8}}\:=\:{x} \\ $$$$\:\frac{\mathrm{25}}{\mathrm{8}}\:+\:\mathrm{8}\:=\:\mathrm{2}{r} \\…
Question Number 187284 by Rupesh123 last updated on 15/Feb/23 Commented by mr W last updated on 15/Feb/23 $${i}\:{think}\:{the}\:{green}\:{area}\:{is}\:{not}\:{unique}. \\ $$ Commented by mr W last…
Question Number 187253 by Rupesh123 last updated on 15/Feb/23 Answered by MikeH last updated on 15/Feb/23 $$\left(\mathrm{i}\right)\:\mathrm{Area}\:\mathrm{of}\:\bigtriangleup{BCD}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}\right)\left(\mathrm{4}\right)\:\mathrm{sin}\:{x}\:=\:\mathrm{8}\:\mathrm{sin}\:{x} \\ $$$$\:\:\:\:\:\:\:\mathrm{Area}\:\mathrm{of}\:\Delta{ABD}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}\right)\left(\mathrm{4}\right)\mathrm{sin}\:\left(\mathrm{90}−{x}\right)\:=\:\mathrm{6}\:\mathrm{cos}\:{x} \\ $$$$\mathrm{total}\:\mathrm{area}\:=\:\mathrm{area}\:\mathrm{of}\:\Delta{BCD}\:+\:\mathrm{area}\:\mathrm{of}\:\Delta{ABD} \\ $$$$\Rightarrow\:{A}\:=\:\mathrm{6}\:\mathrm{cos}\:{x}\:+\:\mathrm{8}\:\mathrm{sin}\:{x}\:\blacksquare \\ $$…
Question Number 187252 by Rupesh123 last updated on 15/Feb/23 Answered by ajfour last updated on 15/Feb/23 Commented by ajfour last updated on 15/Feb/23 $${e}^{\mathrm{2}} +{f}^{\mathrm{2}}…
Question Number 187176 by Rupesh123 last updated on 14/Feb/23 Answered by cortano12 last updated on 14/Feb/23 $$\frac{\mathrm{2}}{\mathrm{5}}\:=\:\frac{{r}+\mathrm{2}}{{r}+\mathrm{9}}\Rightarrow\mathrm{2}{r}+\mathrm{18}=\mathrm{5}{r}+\mathrm{10} \\ $$$$\Rightarrow\mathrm{3}{r}=\mathrm{8}\:;\:{r}=\frac{\mathrm{8}}{\mathrm{3}}\: \\ $$ Terms of Service Privacy…
Question Number 55914 by peter frank last updated on 06/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19 $$\left({k}−\mathrm{1}\right)\mid\frac{{x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left({k}−\mathrm{1}\right)\left(\frac{\mathrm{4}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\…