Question Number 217015 by hardmath last updated on 26/Feb/25 $$\mathrm{If} \\ $$$$\mathrm{f}\left(\mathrm{2x}\:+\:\mathrm{1}\right)\:=\:\mathrm{3x}\:+\:\mathrm{5} \\ $$$$\mathrm{Find} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 26/Feb/25…
Question Number 216983 by hardmath last updated on 26/Feb/25 Answered by mr W last updated on 26/Feb/25 $$\overset{\rightarrow} {\boldsymbol{{A}}}=\left({a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,..,{a}_{{n}} \right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{B}}}=\left({b}_{\mathrm{1}}…
Question Number 216998 by hardmath last updated on 26/Feb/25 Commented by mr W last updated on 26/Feb/25 $${ceva}'{s}\:{theorem} \\ $$ Commented by hardmath last updated…
Question Number 216919 by Abdullahrussell last updated on 24/Feb/25 Answered by Wuji last updated on 24/Feb/25 $$\alpha^{\mathrm{3}} −\mathrm{3}\alpha^{\mathrm{2}} +\mathrm{5}\alpha−\mathrm{17}=\mathrm{0}−−−−−\mathrm{1} \\ $$$$\beta^{\mathrm{3}} −\mathrm{3}\beta^{\mathrm{2}} +\mathrm{5}\beta+\mathrm{11}=\mathrm{0}−−−−−\mathrm{2} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)…
Question Number 216914 by hardmath last updated on 24/Feb/25 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{1}}{\mathrm{1001}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1002}}\:\:+\:\:…\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2000}}\:\:>\:\:\frac{\mathrm{5}}{\mathrm{8}} \\ $$ Answered by MrGaster last updated on 24/Feb/25 $$\underset{{k}=\mathrm{1001}} {\overset{\mathrm{2000}} {\sum}}\frac{\mathrm{1}}{{k}}>\frac{\mathrm{5}}{\mathrm{8}} \\…
Question Number 216900 by MathematicalUser2357 last updated on 24/Feb/25 Commented by MathematicalUser2357 last updated on 24/Feb/25 Is this right? (Part 2) - Complex number to the power of complex number Sorry for solving too complicated. But I was trying to solve clearly. Added more equivalent solution step from the last solution step and graph. The red one is the fixed one. Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216842 by ArshadS last updated on 22/Feb/25 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers}\:\:\mathrm{x},\:\mathrm{y}\:\:\mathrm{that}\:\mathrm{satisfy} \\ $$$$\mathrm{the}\:\:\mathrm{system}\:\: \\ $$$$\mathrm{xy}\:+\:\mathrm{x}\:+\:\mathrm{y}=\mathrm{71}\: \\ $$$$\mathrm{x}^{\mathrm{2}} \mathrm{y}\:+\:\mathrm{xy}^{\mathrm{2}} =\mathrm{880} \\ $$ Answered by Frix last updated…
Question Number 216841 by hardmath last updated on 22/Feb/25 $$\mathrm{Prove}\:\mathrm{that}:\:\:\:\:\:\delta\left(\mathrm{n}\right)\:=\:\underset{\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{n}}}} {\sum}\:\boldsymbol{\varphi}\left(\mathrm{d}\right)\:\boldsymbol{\tau}\left(\frac{\mathrm{n}}{\mathrm{d}}\right) \\ $$$$\boldsymbol{\delta}\left(\mathrm{n}\right)\:=\:\underset{\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{n}}}} {\sum}\:\mathrm{d}\:\:\:,\:\:\:\boldsymbol{\tau}\left(\mathrm{n}\right)\:=\:\underset{\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{n}}}} {\sum}\:{l}\:\:\:\mathrm{and}\:\:\:\varphi-\mathrm{Eyler}.\mathrm{f} \\ $$ Answered by MrGaster last updated on 23/Feb/25 $$\mathrm{Let}\:{f}\left({n}\right)=\underset{{d}\mid{n}}…
Question Number 216836 by hardmath last updated on 22/Feb/25 $$\mathrm{Find}: \\ $$$$\frac{\left(\mathrm{1}\:+\:\mathrm{tan1}°\right)\left(\mathrm{1}\:+\:\mathrm{tan2}°\right)…\left(\mathrm{1}\:+\:\mathrm{tan44}°\right)}{\left(\mathrm{1}−\mathrm{tan46}°\right)\left(\mathrm{1}−\mathrm{tan47}°\right)…\left(\mathrm{1}−\mathrm{tan89}°\right)}\:=\:? \\ $$ Answered by BaliramKumar last updated on 22/Feb/25 $$\mathrm{1} \\ $$ Answered…
Question Number 216785 by universe last updated on 20/Feb/25 $$\:\:\:\mathrm{given}\:\mathrm{the}\:\mathrm{recursive}\:\left\{\mathrm{a}_{\mathrm{n}} \right\}\:\mathrm{define}\:\mathrm{by}\:\mathrm{setting} \\ $$$$\:\:\mathrm{a}_{\mathrm{1}\:} \:\in\:\left(\mathrm{0},\mathrm{1}\right)\:\:\:,\:\:\:\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{a}_{\mathrm{n}} \left(\mathrm{1}−\mathrm{a}_{\mathrm{n}} \right)\:\:\:,\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\:\:\mathrm{prove}\:\mathrm{that}\:\:\left(\mathrm{1}\right)\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{na}_{\mathrm{n}} =\:\mathrm{1} \\ $$$$\:\:\left(\mathrm{2}\right)\:\:\mathrm{b}_{\mathrm{n}} \:=\:\mathrm{n}\left(\mathrm{1}−\mathrm{na}_{\mathrm{n}} \right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{incresing}\:\mathrm{sequence}…