Question Number 114943 by ZiYangLee last updated on 22/Sep/20 $$\mathrm{If}\:\:^{{m}} {C}_{\mathrm{1}} =\:^{{n}} {C}_{\mathrm{2}} \:, \\ $$$$\mathrm{express}\:{m}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n}. \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 49061 by Pk1167156@gmail.com last updated on 02/Dec/18 $$\mathrm{The}\:\mathrm{numerical}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\mathrm{tan}\:\left(\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\:−\:\frac{\pi}{\mathrm{4}}\right)\:\:\mathrm{is} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18 $${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)={tan}^{−\mathrm{1}}…
Question Number 114422 by Aina Samuel Temidayo last updated on 19/Sep/20 $$\mathrm{The}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\: \\ $$$$\:\mid\frac{{x}+\mathrm{1}}{{x}}\mid+\mid{x}+\mathrm{1}\mid=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mid{x}\mid}\:\:\mathrm{is} \\ $$ Commented by bemath last updated on 19/Sep/20 $$\frac{\mid{x}+\mathrm{1}\mid}{\mid{x}\mid}+\mid{x}+\mathrm{1}\mid\:=\:\frac{\mid{x}+\mathrm{1}\mid^{\mathrm{2}}…
Question Number 114420 by Aina Samuel Temidayo last updated on 19/Sep/20 $$\mathrm{Solution}\:\mathrm{of}\:\:\mid\:{x}−\mathrm{1}\:\mid\geqslant\mid\:{x}−\mathrm{3}\:\mid\:\mathrm{is} \\ $$ Commented by bemath last updated on 19/Sep/20 $${short}\:{cut} \\ $$$$\Leftrightarrow\:\left({x}−\mathrm{1}+{x}−\mathrm{3}\right)\left({x}−\mathrm{1}−{x}+\mathrm{3}\right)\geqslant\mathrm{0} \\…
Question Number 48736 by Pk1167156@gmail.com last updated on 28/Nov/18 $$\mathrm{If}\:\:{P}\:\left({A}\cup{B}\right)=\:\frac{\mathrm{3}}{\mathrm{4}},\:{P}\left(\bar {{A}}\right)=\frac{\mathrm{2}}{\mathrm{3}},\:\mathrm{then}\:{P}\:\left(\bar {{A}}\cap{B}\right)= \\ $$ Answered by Abdulhafeez Abu qatada last updated on 28/Nov/18 $${p}\left({A}\cup{B}\right)\:=\:{p}\left({A}\right)\:+\:{p}\left({B}\right)\:−\:{p}\left({A}\cap{B}\right) \\…
Question Number 48735 by Pk1167156@gmail.com last updated on 28/Nov/18 $$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{2}^{{x}+\mathrm{2}} \centerdot\:\mathrm{3}^{\frac{\mathrm{3}{x}}{{x}−\mathrm{1}}} =\:\mathrm{9}\:\mathrm{are}\:\mathrm{given}\:\mathrm{by} \\ $$ Answered by Smail last updated on 28/Nov/18 $$\mathrm{2}^{{x}+\mathrm{2}} .\mathrm{3}^{\frac{\mathrm{3}{x}}{{x}−\mathrm{1}}}…
Question Number 48543 by Pk1167156@gmail.com last updated on 25/Nov/18 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{times}\:\mathrm{the}\:\mathrm{digit}\:\mathrm{3}\:\mathrm{will} \\ $$$$\mathrm{be}\:\mathrm{written}\:\mathrm{when}\:\mathrm{listing}\:\mathrm{the}\:\mathrm{integers} \\ $$$$\mathrm{from}\:\mathrm{1}\:\mathrm{to}\:\mathrm{1000}\:\mathrm{is} \\ $$ Answered by mr W last updated on 25/Nov/18 $${one}−{digit}−{numbers}:…
Question Number 48529 by Pk1167156@gmail.com last updated on 25/Nov/18 $$\mathrm{If}\:\:\mathrm{cos}\:{A}+\mathrm{cos}\:{B}={m}\:\mathrm{and}\:\mathrm{sin}\:{A}+\mathrm{sin}\:{B}={n} \\ $$$$\mathrm{where}\:{m},\:{n}\:\neq\mathrm{0},\:\mathrm{then}\:\mathrm{sin}\:\left({A}+{B}\right)\:\mathrm{is}\:\mathrm{equal} \\ $$$$\mathrm{to} \\ $$ Answered by math1967 last updated on 25/Nov/18 $$\frac{\mathrm{2}{mn}}{{m}^{\mathrm{2}} +{n}^{\mathrm{2}}…
Question Number 48528 by Pk1167156@gmail.com last updated on 25/Nov/18 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\sqrt{\mathrm{3}}\:\mathrm{cot}\:\mathrm{20}°−\:\mathrm{4}\:\mathrm{cos}\:\mathrm{20}°\:\:\mathrm{is} \\ $$ Answered by math1967 last updated on 25/Nov/18 $$\:\frac{\sqrt{\mathrm{3}}{cos}\mathrm{20}−\mathrm{4}{cos}\mathrm{20}{sin}\mathrm{20}}{{sin}\mathrm{20}}\:\:\: \\ $$$$=\frac{{tan}\mathrm{60}{cos}\mathrm{20}−\mathrm{2}{sin}\mathrm{40}}{{sin}\mathrm{20}} \\ $$$$=\frac{\frac{{sin}\mathrm{60}{cos}\mathrm{20}}{{cos}\mathrm{60}}−\mathrm{2}{sin}\mathrm{40}}{{sin}\mathrm{20}} \\…
Question Number 48527 by Pk1167156@gmail.com last updated on 25/Nov/18 $$\mathrm{If}\:\:{xy}\:+\:{yz}\:+\:{zx}\:=\:\mathrm{1},\:\mathrm{then} \\ $$$$\mathrm{tan}^{−\mathrm{1}} {x}\:+\:\mathrm{tan}^{−\mathrm{1}} {y}\:+\:\mathrm{tan}^{−\mathrm{1}} {z}\:=\: \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18 $${tan}^{−\mathrm{1}}…