Question Number 32745 by vakil vishaWakrma last updated on 01/Apr/18 $$\mathrm{The}\:\mathrm{first},\:\mathrm{second}\:\mathrm{and}\:\mathrm{middle}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{AP}\:\mathrm{are}\:{a},\:{b},\:{c}\:\mathrm{respectively}.\:\mathrm{Their}\:\mathrm{sum}\:\mathrm{is} \\ $$ Answered by Rasheed.Sindhi last updated on 01/Apr/18 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{AP}\:\mathrm{has}\:\mathrm{n}\:\left(\mathrm{odd}\right)\:\mathrm{terms}\:\mathrm{and}\:\mathrm{common} \\ $$$$\mathrm{difference}\:\mathrm{d}.…
Question Number 98058 by PengagumRahasiamu last updated on 11/Jun/20 $$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} −{cx}+{d}=\mathrm{0}\:\mathrm{has}\:\mathrm{roots} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{fourth}\:\mathrm{powers}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots} \\ $$$$\mathrm{of}\:{x}^{\mathrm{2}} +{ax}+{b}=\mathrm{0},\:\mathrm{where}\:{a}^{\mathrm{2}} >\mathrm{4}{b}\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} −\mathrm{4}{bx}+\mathrm{2}{b}^{\mathrm{2}} −{c}=\mathrm{0}\:\mathrm{will}\:\mathrm{be} \\ $$ Answered by…
Question Number 98057 by PengagumRahasiamu last updated on 11/Jun/20 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{of} \\ $$$$\mathrm{3}^{{x}} +\mathrm{4}^{{x}} =\mathrm{5}^{{x}} \:\mathrm{is}\:\_\_\_\_. \\ $$ Answered by Rio Michael last updated on 11/Jun/20…
Question Number 98056 by PengagumRahasiamu last updated on 11/Jun/20 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quadratic} \\ $$$$\mathrm{equation}\:\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({x}−\mathrm{6}\right)^{\mathrm{2}} =\mathrm{0}\:\mathrm{is} \\ $$ Commented by som(math1967) last updated on 11/Jun/20 $$\mathrm{4},\mathrm{5},\mathrm{6}…
Question Number 32418 by hizmzm1 last updated on 24/Mar/18 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{or} \\ $$$${x}^{\mathrm{8}} −\:{x}^{\mathrm{5}} −\:{x}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Commented by hizmzm1 last updated on 27/Mar/18 $$\mathrm{1}^{\mathrm{8}} −\mathrm{1}^{\mathrm{5}}…
Question Number 32382 by hizmzm1 last updated on 24/Mar/18 $$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:{ax}^{\mathrm{2}} +\mathrm{2}{bx}−\mathrm{3}{c}=\mathrm{0}\:\mathrm{has} \\ $$$$\mathrm{no}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{and}\:\left(\frac{\mathrm{3}{c}}{\mathrm{4}}\right)<\:{a}+{b},\:\mathrm{then} \\ $$ Commented by MJS last updated on 24/Mar/18 $$\mathrm{I}\:\mathrm{think}\:\mathrm{that}'\mathrm{s}\:\mathrm{not}\:\mathrm{enough}\:\mathrm{information} \\ $$…
Question Number 32329 by naeems1000 last updated on 23/Mar/18 $$\mathrm{The}\:\mathrm{smallest}\:\mathrm{number}\:\mathrm{which}\:\mathrm{must}\:\mathrm{be} \\ $$$$\mathrm{subtracted}\:\mathrm{from}\:\mathrm{3400}\:\mathrm{to}\:\mathrm{make}\:\mathrm{it}\:\mathrm{a}\: \\ $$$$\mathrm{perfect}\:\mathrm{cube}\:\mathrm{is}\:\_\_\_\_\_. \\ $$ Answered by Joel578 last updated on 23/Mar/18 $$\mathrm{15}^{\mathrm{3}} \:=\:\mathrm{3375}…
Question Number 97737 by bagjamath last updated on 09/Jun/20 $$\mathrm{If}\:\:^{\left({a}^{\mathrm{2}} −{a}\right)} {C}_{\mathrm{2}} =\:^{\left({a}^{\mathrm{2}} −{a}\right)} {C}_{\mathrm{4}} \:,\:\mathrm{then}\:{a}\:= \\ $$ Commented by som(math1967) last updated on 09/Jun/20…
Question Number 97658 by Vishal Sharma last updated on 09/Jun/20 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}+\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{7}}\:\:\mathrm{is} \\ $$ Commented by bemath last updated on 09/Jun/20 $${x}=\frac{\mathrm{2}\pi}{\mathrm{7}}\: \\ $$$$\frac{\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{2sin}\:{x}}×\:\mathrm{2sin}\:{x} \\ $$$$\frac{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{2sin}{x}\:\mathrm{cos}\:\mathrm{2}{x}+\mathrm{2sin}\:{x}\mathrm{cos}\:\mathrm{3}{x}\:}{\mathrm{2sin}\:{x}}\:=…
Question Number 97657 by Vishal Sharma last updated on 09/Jun/20 $$\mathrm{If}\:\:\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta\:=\:\mathrm{0}\:=\:\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta, \\ $$$$\mathrm{then}\:\:\mathrm{cos}\:\mathrm{2}\alpha+\mathrm{cos}\:\mathrm{2}\beta\:= \\ $$ Commented by bemath last updated on 09/Jun/20 $$\left(\mathrm{1}\right)\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta\:=\:\mathrm{0} \\ $$$$\mathrm{2cos}\:\left(\frac{\alpha+\beta}{\mathrm{2}}\right)\:\mathrm{cos}\:\left(\frac{\alpha−\beta}{\mathrm{2}}\right)\:=\:\mathrm{0}…