Question Number 19433 by Tinkutara last updated on 11/Aug/17 $$\frac{\sqrt{\mathrm{5}\:+\:\mathrm{12}{i}}\:+\:\sqrt{\mathrm{5}\:−\:\mathrm{12}{i}}}{\:\sqrt{\mathrm{5}\:+\:\mathrm{12}{i}}\:−\:\sqrt{\mathrm{5}\:−\:\mathrm{12}{i}}}\:= \\ $$ Answered by ajfour last updated on 11/Aug/17 $$=\frac{\left(\sqrt{\mathrm{5}+\mathrm{12i}}+\sqrt{\mathrm{5}−\mathrm{12i}}\right)^{\mathrm{2}} }{\mathrm{24i}} \\ $$$$=\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{25}+\mathrm{144}}}{\mathrm{24i}}=\frac{\mathrm{3}}{\mathrm{2i}}=−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{i}\:. \\ $$…
Question Number 150489 by mathdanisur last updated on 12/Aug/21 $$\underset{\:\mathrm{0}} {\overset{\:\mathrm{2}} {\int}}\:\mid\mathrm{x}\mid\:\mathrm{x}^{\left[\boldsymbol{\mathrm{x}}+\mathrm{1}\right]} \:\mathrm{sgn}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 12/Aug/21 $$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}} \mid{x}\mid.{x}^{\left[{x}+\mathrm{1}\right]}…
Question Number 19413 by Tinkutara last updated on 10/Aug/17 $$\mathrm{How}\:\mathrm{many}\:\mathrm{integer}\:\mathrm{pairs}\:\left({x},\:{y}\right)\:\mathrm{satisfy} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{4}{y}^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:−\:\mathrm{2}{x}\:−\:\mathrm{4}{y}\:−\:\mathrm{8}\:=\:\mathrm{0}? \\ $$ Commented by mrW1 last updated on 11/Aug/17 $$\left(\mathrm{0},−\mathrm{1}\right) \\…
Question Number 150486 by mathdanisur last updated on 12/Aug/21 $$\boldsymbol{\mathrm{x}}\:\in\:\mathbb{R} \\ $$$$\mid\mathrm{x}\:-\:\mathrm{1}\mid\:+\:\mid\mathrm{x}\:+\:\mathrm{3}\mid\:+\:\mid\mathrm{x}\:-\:\mathrm{5}\mid \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{given} \\ $$$$\mathrm{expression} \\ $$ Commented by MJS_new last updated on 12/Aug/21…
Question Number 150482 by mathdanisur last updated on 12/Aug/21 $$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{4}} \:\frac{\boldsymbol{\pi}}{\mathrm{7}}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{4}} \:\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{7}}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{4}} \:\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{7}}}\:\:=\:\:? \\ $$ Commented by liberty last updated on 13/Aug/21 $$=\mathrm{374} \\ $$…
Question Number 19403 by Tinkutara last updated on 10/Aug/17 $$\mathrm{Let}\:{P}\left({n}\right)\:=\:\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{3}\right)\left({n}\:+\:\mathrm{5}\right)\left({n}\:+\:\mathrm{7}\right)\left({n}\:+\:\mathrm{9}\right). \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{integer}\:\mathrm{that}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{divisor}\:\mathrm{of}\:{P}\left({n}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{even} \\ $$$$\mathrm{integers}\:{n}? \\ $$ Commented by RasheedSindhi last updated on 12/Aug/17…
Question Number 84941 by Power last updated on 17/Mar/20 Commented by mr W last updated on 17/Mar/20 $${use}\:{method}\:{in}\:{Q}\mathrm{84782}\:{to}\:{find}\:{the} \\ $$$${last}\:{two}\:{digits} \\ $$ Answered by mr…
Question Number 19405 by $@ty@m last updated on 10/Aug/17 $${Convert}\:{i}\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}}\:{into}\:{polarform}. \\ $$ Answered by Tinkutara last updated on 10/Aug/17 $$\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{2}}\:−\:\mathrm{1}}{\mathrm{2}}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{2}}\:+\:{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\right) \\ $$ Commented by $@ty@m…
Question Number 150469 by mathdanisur last updated on 12/Aug/21 $$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{4}^{\boldsymbol{\mathrm{x}}} }{\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}}\:\:\:\mathrm{find}\:\:\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{17}}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{16}}{\mathrm{17}}\right)\:\overset{?} {=} \\ $$ Answered by gsk2684 last updated on 12/Aug/21 $${f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)\:=\frac{\mathrm{4}^{{x}} }{\mathrm{4}^{{x}} +\mathrm{2}}+\frac{\mathrm{4}^{\mathrm{1}−{z}}…
Question Number 19391 by chernoaguero@gmail.com last updated on 10/Aug/17 Commented by dioph last updated on 12/Aug/17 $$\mathrm{Have}\:\mathrm{you}\:\mathrm{tried}\:\mathrm{Newton}'\mathrm{s}\:\mathrm{method} \\ $$ Terms of Service Privacy Policy Contact:…