Question Number 49462 by Pk1167156@gmail.com last updated on 07/Dec/18 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{2}\mid{x}\mid^{\mathrm{2}} −\:\mathrm{7}\mid{x}\mid\:+\:\mathrm{6}=\mathrm{0}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18 $$\mid{x}\mid={x}\:\:{when}\:{x}>\mathrm{0} \\ $$$$\:\:\:\:\:=−{x}\:\:{when}\:{x}<\mathrm{0}…
Question Number 49460 by Pk1167156@gmail.com last updated on 07/Dec/18 $$\mathrm{If}\:\:{x}=\mathrm{1}+{i}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$${x}^{\mathrm{3}} −{ix}+\mathrm{1}−{i}=\mathrm{0}\:,\:\mathrm{then}\:\mathrm{the}\:\mathrm{other}\:\mathrm{real} \\ $$$$\mathrm{root}\:\mathrm{is} \\ $$ Commented by Kunal12588 last updated on 07/Dec/18 $${i}\:{do}\:{it}\:{with}\:{normal}\:{real}\:{way}.\:{so}\:{it}…
Question Number 49461 by Pk1167156@gmail.com last updated on 07/Dec/18 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{for}\:{x}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} −\mid{x}\mid−\mathrm{2}=\mathrm{0}\:\:\mathrm{is} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18 $$\mid{x}\mid={x}\:\:{x}>\mathrm{0} \\ $$$$\:\:\:\:\:=−{x}\:{x}<\mathrm{0}…
Question Number 49459 by Pk1167156@gmail.com last updated on 07/Dec/18 $$\mathrm{If}\:\mathrm{tan}\:\theta=\frac{\mathrm{cos}\:\mathrm{9}°+\mathrm{sin}\:\mathrm{9}°}{\mathrm{cos}\:\mathrm{9}°−\mathrm{sin}\:\mathrm{9}°}\:\mathrm{then}\:\theta\:=\:\_\_\_ \\ $$ Answered by math1967 last updated on 07/Dec/18 $${tan}\theta=\frac{{cos}\mathrm{9}+{cos}\left(\mathrm{90}−\mathrm{9}\right)}{{cos}\mathrm{9}−{cos}\left(\mathrm{90}−\mathrm{9}\right)} \\ $$$${tan}\theta=\frac{{cos}\mathrm{9}+{cos}\mathrm{81}}{{cos}\mathrm{9}−{cos}\mathrm{81}} \\ $$$${tan}\theta=\frac{\mathrm{2}{cos}\mathrm{45}.{cos}\mathrm{36}}{\mathrm{2}{sin}\mathrm{45}.{sin}\mathrm{36}} \\…
Question Number 114943 by ZiYangLee last updated on 22/Sep/20 $$\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} \\ $$$$\Leftrightarrow\:\left({x}−\mathrm{1}+{x}−\mathrm{3}\right)\left({x}−\mathrm{1}−{x}+\mathrm{3}\right)\geqslant\mathrm{0} \\…
Question Number 48735 by Pk1167156@gmail.com last updated on 28/Nov/18 $$\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 $$\mathrm{2}^{{x}+\mathrm{2}} .\mathrm{3}^{\frac{\mathrm{3}{x}}{{x}−\mathrm{1}}}…
Question Number 48736 by Pk1167156@gmail.com last updated on 28/Nov/18 $$\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) \\…