Question Number 214916 by Spillover last updated on 23/Dec/24 Answered by A5T last updated on 23/Dec/24 $$\mathrm{2017}^{\mathrm{2017}^{\mathrm{2017}} } \equiv\mathrm{1}^{\mathrm{2017}^{\mathrm{2017}} } =\mathrm{1}\left({mod}\:\mathrm{16}\right) \\ $$$$\mathrm{2017}^{\mathrm{2017}^{\mathrm{2017}} } \equiv\mathrm{142}^{\mathrm{2017}^{\mathrm{2017}}…
Question Number 214888 by Emmanuel07 last updated on 22/Dec/24 Commented by mr W last updated on 23/Dec/24 $${i}\:{got} \\ $$$${a}_{{n}} =\mathrm{cot}\:\left\{\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}−\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\frac{\pi}{\mathrm{8}}\right]\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}−\frac{\mathrm{3}\pi}{\mathrm{8}}\right)−\frac{\pi}{\mathrm{8}}\right]\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right\}…
Question Number 214860 by hardmath last updated on 21/Dec/24 $$\mathrm{Find}: \\ $$$$\mathrm{1}+\:\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}!}\:\:+\:\:\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}\:\:+\:\:…\:\:+\:\:\frac{\mathrm{n}^{\mathrm{3}} }{\mathrm{n}!}\:\:=\:\:? \\ $$ Commented by mr W last updated on 22/Dec/24…
Question Number 214805 by universe last updated on 20/Dec/24 $$\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} −\mathrm{y}^{\mathrm{3}} +\mathrm{x}^{\mathrm{5}} \:+\mathrm{show}\:\mathrm{that} \\ $$$$\:\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{has}\:\mathrm{neither}\:\mathrm{a}\:\mathrm{maximum}\:\mathrm{nor}\:\mathrm{a}\: \\ $$$$\:\:\mathrm{minimum}\:\mathrm{at}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$ Answered by TonyCWX08 last updated…
Question Number 214779 by MrGaster last updated on 19/Dec/24 $$ \\ $$$$\mathrm{seek}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{3}}+\mathrm{2}\:\mathrm{sin}\:\mathrm{2}{x}}{\:\sqrt{\mathrm{3}}+\mathrm{2}\:\mathrm{sin}\:{x}}=\sqrt{\mathrm{3}}\mathrm{sin}\:{x}+\frac{\mathrm{cos2}{x}}{\mathrm{2}\:\mathrm{cos}\:{x}} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\:\mathrm{between}\:\mathrm{0}\:\mathrm{and}\frac{\pi}{\mathrm{2}}. \\ $$$$ \\ $$ Terms of Service Privacy Policy…
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Question Number 214759 by MATHEMATICSAM last updated on 19/Dec/24 $$\mathrm{If}\:{x}\:=\:{cy}\:+\:{bz},\:{y}\:=\:{cx}\:+\:{az}\:\mathrm{and} \\ $$$${z}\:=\:{bx}\:+\:{ay}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{x}}{\:\sqrt{\mathrm{1}\:−\:{a}^{\mathrm{2}} }}\:=\:\frac{{y}}{\:\sqrt{\mathrm{1}\:−\:{b}^{\mathrm{2}} }}\:=\:\frac{{z}}{\:\sqrt{\mathrm{1}\:−\:{c}^{\mathrm{2}} }}\:. \\ $$ Answered by MathematicalUser2357 last updated on…
Question Number 214785 by mathlove last updated on 19/Dec/24 $$\left({x}−\mathrm{3}\right)^{\mathrm{4}} =\mathrm{7}^{\mathrm{4}} \\ $$$${x}=? \\ $$ Answered by Ismoiljon_008 last updated on 19/Dec/24 $$\:\:\:\mid\:{x}\:−\:\mathrm{3}\:\mid\:=\:\mathrm{7} \\ $$$$\:\:\:{x}\:−\:\mathrm{3}\:=\:\mathrm{7}…
Question Number 214753 by BaliramKumar last updated on 18/Dec/24 Commented by BaliramKumar last updated on 19/Dec/24 $${solve}\:{by}\:{computer}\:{programming}\: \\ $$$${for}\:{any}\:{primitive}\:{triplets} \\ $$$${ex}.\:\:\:\:\left({l},\:{b},\:{h}\right)\:\equiv\:\left(\mathrm{2},\:\mathrm{2},\:\mathrm{1}\right)\:\:\&\:\left({a},\:{b},\:{c}\right)\:\equiv\:\left(\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right) \\ $$ Terms of…
Question Number 214658 by ChantalYah last updated on 15/Dec/24 $$\left.\mathrm{1}\right)\:\mathrm{The}\:\mathrm{function}\:\mathrm{H}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{H}\left(\mathrm{x}\right)\:=\mathrm{3cosh}\frac{\mathrm{x}}{\mathrm{3}}+\mathrm{sinh}\frac{\mathrm{x}}{\mathrm{3}}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\lambda\:\mathrm{for}\:\mathrm{which}\:\mathrm{H}\left(\mathrm{ln}\lambda^{\mathrm{3}} \right)=\mathrm{4} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Prove}\:\mathrm{that}\:\int_{\mathrm{2}} ^{\mathrm{4}} \:\frac{\mathrm{6x}+\mathrm{1}}{\left(\mathrm{2x}−\mathrm{3}\right)\left(\mathrm{3x}−\mathrm{2}\right)}\mathrm{dx}=\:\mathrm{ln}\:\mathrm{10}. \\ $$$$\left.\mathrm{3}\right)\mathrm{Show}\:\mathrm{that}\:\frac{\mathrm{sin}\theta\:+\:\mathrm{sin2}\theta}{\mathrm{1}+\:\mathrm{cos}\theta+\:\mathrm{cos2}\theta}\equiv\:\mathrm{tan}\theta \\ $$$$\left.\mathrm{4}\right)\:\mathrm{If}\:\mathrm{z}=\mathrm{cos}\theta+\:\mathrm{i}\:\mathrm{sin}\theta,\:\mathrm{Show}\:\mathrm{that}\:\mathrm{z}+\frac{\mathrm{1}}{\mathrm{z}}=\mathrm{2cos}\theta\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{z}^{\mathrm{n}} +\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{n}} }=\mathrm{2cos}\:\mathrm{n}\theta\:\mathrm{hence}\:\mathrm{or}\:\mathrm{otherwise}\:\mathrm{show}\:\mathrm{that}\:…