Question Number 214002 by RoseAli last updated on 24/Nov/24 $${find}\:{the}\:{integers}\:{x}\:{that}\:{satisfies}\:{a}\:{congruence}\:\mathrm{3}{x}=\mathrm{4}\:\left({mod}\:\mathrm{11}\right)\:. \\ $$ Answered by Rasheed.Sindhi last updated on 24/Nov/24 $$\mathrm{3}{x}\equiv\mathrm{4}\left({mod}\:\mathrm{11}\right) \\ $$$$\mathrm{3}{x}\equiv\mathrm{4}+\mathrm{11}\left({mod}\:\mathrm{11}\right) \\ $$$$\mathrm{3}{x}\equiv\mathrm{15}\left({mod}\:\mathrm{11}\right) \\…
Question Number 213953 by efronzo1 last updated on 22/Nov/24 Answered by mehdee7396 last updated on 22/Nov/24 $${let}\:\:\:{f}\left({x}\right)=\frac{{ax}+{b}}{{cx}+{d}}\Rightarrow{f}\left({f}\left({x}\right)\right)=\frac{{a}\frac{{ax}+{b}}{{cx}+{d}}+{b}}{{c}\frac{{ax}+{b}}{{cx}+{d}}+{d}} \\ $$$$=\frac{\frac{{a}^{\mathrm{2}} {x}+{ab}}{{cx}+{d}}+{b}}{\frac{{acx}+{bc}}{{cx}+{d}}+{d}}=\frac{\left({a}^{\mathrm{2}} +{bc}\right){x}+{ab}+{bd}}{\left({ac}+{cd}\right)+{bc}+{d}^{\mathrm{2}} } \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{bc}=\mathrm{1}\:\:\&\:\:{ac}+{cd}=\mathrm{1}\:\&\:\:{ab}+{bd}=\mathrm{1}\:\:\&\:\:{bc}+{d}^{\mathrm{2}}…
Question Number 213956 by Ari last updated on 22/Nov/24 Answered by A5T last updated on 22/Nov/24 $${Let}\:{the}\:{numbers}\:{be}:\:\left({x}−\mathrm{2},{x}−\mathrm{1},{x},{x}+\mathrm{1},{x}+\mathrm{2}\right) \\ $$$${x}=\mathrm{5}{a}+\mathrm{2}=\mathrm{7}{b}+\mathrm{1}=\mathrm{9}{c}=\mathrm{11}{d}−\mathrm{1}=\mathrm{13}{e}−\mathrm{2} \\ $$$$\mathrm{7}{b}+\mathrm{1}\equiv\mathrm{2}\left({mod}\:\mathrm{5}\right)\Rightarrow{b}\equiv\mathrm{3}\left({mod}\:\mathrm{5}\right)\Rightarrow{b}=\mathrm{5}{f}+\mathrm{3} \\ $$$$\Rightarrow\mathrm{7}\left(\mathrm{5}{f}+\mathrm{3}\right)+\mathrm{1}\equiv\mathrm{0}\left({mod}\:\mathrm{9}\right)\Rightarrow{f}\equiv\mathrm{4}\left({mod}\:\mathrm{9}\right)\Rightarrow{f}=\mathrm{9}{g}+\mathrm{4} \\ $$$$\Rightarrow\mathrm{7}{g}\equiv\mathrm{2}\left({mod}\:\mathrm{11}\right)\Rightarrow{g}\equiv\mathrm{5}\left({mod}\:\mathrm{11}\right)\Rightarrow{g}=\mathrm{11}{h}+\mathrm{5}…
Question Number 213939 by polymathAntunes last updated on 22/Nov/24 Commented by Frix last updated on 22/Nov/24 $$\mathrm{1}.\:\mathrm{All}\:{x}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}, \\ $$$$\:\:\:\:\:\mathrm{constants}\:\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:\mathrm{side} \\ $$$$\:\:\:\:\:\frac{{x}}{\mathrm{4}}+\mathrm{20}=\frac{{x}}{\mathrm{3}}\:\:\:\:\:\mid−\frac{{x}}{\mathrm{3}}−\mathrm{20} \\ $$$$\:\:\:\:\:\frac{{x}}{\mathrm{4}}−\frac{{x}}{\mathrm{3}}=−\mathrm{20} \\ $$$$\mathrm{2}.\:\mathrm{Common}\:\mathrm{denominator}\:\mathrm{and}\:\mathrm{add}…
Question Number 213841 by mnjuly1970 last updated on 18/Nov/24 $$ \\ $$$$\:\:{Find}\:{the}\:{vertical}\:{asymptots} \\ $$$$\: \\ $$$$\:\:{of}\:\:,\:\:\:{f}\left({x}\right)=\:\mathrm{tan}\left(\frac{\:\pi}{\mathrm{2}{x}\:+\:\mathrm{2}}\:\right)\:\:{in}\: \\ $$$$\: \\ $$$$\:\:\:\:\:\left[\:\mathrm{0}\:\:,\:\:\:\mathrm{4}\:\right] \\ $$$$\:−−−−−−−−−−−−− \\ $$$$ \\…
Question Number 213821 by hardmath last updated on 17/Nov/24 $$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)^{\frac{\mathrm{sinx}}{\mathrm{x}\:−\:\mathrm{sinx}}} \:\:=\:\:? \\ $$ Answered by mehdee7396 last updated on 17/Nov/24 $${lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{{sinx}}{{x}}−\mathrm{1}\right)\frac{{sinx}}{{x}−{sinx}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}}…
Question Number 213803 by muallimRiyoziyot last updated on 17/Nov/24 Commented by Frix last updated on 17/Nov/24 $$\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{solutions}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{exact}\:\mathrm{form}\:\mathrm{is}\:\mathrm{not}\:\mathrm{useable}. \\ $$$${x}\approx\mathrm{1}.\mathrm{32848492}\pm.\mathrm{570204126i} \\ $$ Commented by…
Question Number 213791 by hardmath last updated on 16/Nov/24 $$\mathrm{If}\:\:\:\mathrm{x}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{x}}\:−\:\frac{\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}}}\:\:=\:\:\mathrm{10} \\ $$$$\mathrm{Find}\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}}\:−\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}}}\:\:+\:\:\mathrm{3}\:\:=\:\:? \\ $$ Commented by muallimRiyoziyot last updated on 19/Nov/24 $${x}−\mathrm{8}=\sqrt[{\mathrm{3}}]{{x}}+\mathrm{2}+\frac{\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{{x}}} \\ $$$$\left(\sqrt[{\mathrm{3}}]{{x}}−\mathrm{2}\right)\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }+\mathrm{2}\sqrt[{\mathrm{3}}]{{x}}+\mathrm{4}\right)=\frac{\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}}…
Question Number 213751 by hardmath last updated on 15/Nov/24 $$\mathrm{m}\:;\:\mathrm{n}\:\in\:\mathbb{Z}_{+} \\ $$$$\mathrm{2m}^{\mathrm{2}} \:+\:\mathrm{n}^{\mathrm{2}} \:−\:\mathrm{mn}\:=\:\mathrm{54} \\ $$$$ \\ $$$$\mathrm{1}.\:\left(\mathrm{m};\mathrm{n}\right)=? \\ $$$$\mathrm{2}.\:\left(\mathrm{m};\mathrm{n}\right)=? \\ $$$$……………. \\ $$ Commented…
Question Number 213726 by hardmath last updated on 14/Nov/24 $$\mathrm{ax}\:=\:\mathrm{by}\:=\:\mathrm{cz}\:=\:\mathrm{36} \\ $$$$\frac{\mathrm{1}}{\mathrm{x}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{y}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{z}}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 14/Nov/24 $${a}=\frac{\mathrm{36}}{{x}},\:{b}=\frac{\mathrm{36}}{{y}},\:{c}=\frac{\mathrm{36}}{{z}} \\…