Question Number 140446 by EnterUsername last updated on 07/May/21 $$\mathrm{If}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{vertices}\:{Z}_{\mathrm{1}} ,\:{Z}_{\mathrm{2}} \:\mathrm{and}\:{Z}_{\mathrm{3}} \:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{absolute}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{number} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\lambda{i}\:\:\begin{vmatrix}{{Z}_{\mathrm{1}} }&{\bar {{Z}}_{\mathrm{1}} }&{\mathrm{1}}\\{{Z}_{\mathrm{2}} }&{\bar {{Z}}_{\mathrm{2}} }&{\mathrm{1}}\\{{Z}_{\mathrm{3}} }&{\bar {{Z}}_{\mathrm{3}}…
Question Number 9372 by geovane10math last updated on 03/Dec/16 $$\mathrm{I}\:\mathrm{was}\:\mathrm{able}\:\mathrm{to}\:\mathrm{discover}\:\mathrm{the}\:\mathrm{conditions}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{irrational}\:\mathrm{numbers}\:\mathrm{be}\:\mathrm{an} \\ $$$$\mathrm{integer}\:\mathrm{and}\:\mathrm{the}\:\mathrm{conditions}\:\mathrm{for}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{be}\:\mathrm{a}\:\mathrm{finite}\:\mathrm{decimal}. \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{can}\:\mathrm{not}\:\mathrm{do}\:\mathrm{the}\:\mathrm{same}\:\mathrm{for}\:\mathrm{periodic} \\ $$$$\mathrm{tithe}. \\ $$$$\mathrm{Someone}\:\mathrm{can}\:\mathrm{help}\:\mathrm{me},\:\mathrm{please}? \\ $$$$ \\…
Question Number 9370 by tawakalitu last updated on 03/Dec/16 Answered by geovane10math last updated on 03/Dec/16 $$\mathrm{There}\:\mathrm{are}\:\mathrm{infinite}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sizes}\:\mathrm{10} \\ $$$$\mathrm{and}\:\mathrm{15}.\:\mathrm{It}\:\mathrm{depends}\:\mathrm{of}\:\mathrm{inclination}\:\mathrm{of}\:{X}\hat {{Z}M}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{w}\:\mathrm{depends}\:\mathrm{of}\: \\ $$$${X}\hat {{Z}M}.…
Question Number 140443 by EnterUsername last updated on 07/May/21 $$\mathrm{If}\:{z}_{\mathrm{2}} /{z}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{and}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{non}-\mathrm{zero}\:\mathrm{real} \\ $$$$\mathrm{numbers},\:\mathrm{then}\:\mid\left({az}_{\mathrm{1}} +{bz}_{\mathrm{2}} \right)/\left({az}_{\mathrm{1}} −{bz}_{\mathrm{2}} \right)\mid\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\_\_\_\_\_. \\ $$ Answered by mr W last…
Question Number 74904 by necxxx last updated on 03/Dec/19 Commented by mr W last updated on 03/Dec/19 $${no}\:{unique}\:{solution} \\ $$ Terms of Service Privacy Policy…
Question Number 140442 by EnterUsername last updated on 07/May/21 $$\mathrm{If}\:{z}_{\mathrm{1}} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{2}} \mid\neq\mathrm{1}\:\mathrm{and} \\ $$$$\mid\left({z}_{\mathrm{1}} −\mathrm{2}{z}_{\mathrm{2}} \right)/\left(\mathrm{2}−{z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\mid=\mathrm{1},\:\mathrm{then}\:\mid{z}_{\mathrm{1}} \mid\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\_\_\_\_\_. \\ $$ Answered by…
Question Number 9367 by tawakalitu last updated on 02/Dec/16 $$\mathrm{Solve}\:\mathrm{simultaneously}. \\ $$$$\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{4xy}\:+\:\mathrm{3y}^{\mathrm{2}} \:=\:\mathrm{3}\:\:\:………\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{3x}\:+\:\mathrm{3y}\:=\:\mathrm{4}\:\:\:\:…….\:\left(\mathrm{ii}\right) \\ $$ Answered by mrW last updated…
Question Number 74900 by chess1 last updated on 03/Dec/19 Commented by chess1 last updated on 03/Dec/19 $$\mathrm{solve}\:\mathrm{equation} \\ $$ Commented by MJS last updated on…
Question Number 140428 by byaw last updated on 07/May/21 $$\mathrm{If}\:\mathrm{4}{x}=\mathrm{3}\left(\mathrm{Mod}\:\mathrm{6}\right),\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{four}\:\mathrm{values}\:\mathrm{of}\:{x}. \\ $$ Commented by mr W last updated on 07/May/21 $$\mathrm{4}{x}\:{is}\:{even}. \\ $$$${but}\:{when}\:{it}\:{is}\:{divided}\:{by}\:\mathrm{6}\:{and}\:{the}…
Question Number 140430 by ajfour last updated on 07/May/21 $$\left(\mathrm{1}+\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\left\{\left({x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:−{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}{x}\right)\left({x}+\mathrm{3}\right)\right\}^{\mathrm{2}} \\ $$$${Any}\:{good}\:{non}-{zero}\:{real}\:{solution} \\…