Question Number 226771 by Spillover last updated on 14/Dec/25 Answered by Frix last updated on 14/Dec/25 $${a}\neq\pm\mathrm{1} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{say}\:{a}>\mathrm{0}\:\mathrm{because} \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{d}\theta}{\mathrm{1}−\mathrm{2}{a}\mathrm{cos}\:\theta\:+{a}^{\mathrm{2}} }=\underset{\mathrm{0}} {\overset{\pi}…
Question Number 226766 by mr W last updated on 13/Dec/25 Answered by mahdipoor last updated on 13/Dec/25 $$\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b}\:\Rightarrow \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}+\mathrm{1}\:\mathrm{or}\:−\mathrm{2x}−\mathrm{3} \\ $$$$\mathrm{but}\:\mathrm{its}\:\mathrm{only}\:\mathrm{answer}? \\ $$$$\mathrm{all}\:\mathrm{function}\:\mathrm{can}\:\mathrm{show}\:\mathrm{as}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{m}}…
Question Number 226755 by Hanuda354 last updated on 13/Dec/25 Answered by TonyCWX last updated on 13/Dec/25 $${A}_{{Green}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}^{\mathrm{2}} \right)\left(\frac{\pi}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}^{\mathrm{2}} \right)\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\mathrm{4}^{\mathrm{2}} \right)=\frac{\mathrm{40}}{\mathrm{3}}\pi−\mathrm{16}\sqrt{\mathrm{3}} \\ $$$${A}_{{Blue}\:} =\:\left(\frac{\mathrm{12}−\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{2}\pi}{\mathrm{12}}\right)\left(\mathrm{8}^{\mathrm{2}} \right)=\mathrm{64}−\mathrm{16}\sqrt{\mathrm{3}}−\frac{\mathrm{32}}{\mathrm{3}}\pi…
Question Number 226732 by Spillover last updated on 12/Dec/25 Answered by TonyCWX last updated on 12/Dec/25 $$\mathrm{Characteristic}\:\mathrm{Equation}: \\ $$$$\lambda^{\mathrm{2}} +\mathrm{2}\lambda−\mathrm{8}=\mathrm{0}\:\Rightarrow\:\lambda_{\mathrm{1}} =−\mathrm{4}\:\mathrm{and}\:\lambda_{\mathrm{2}} =\mathrm{2} \\ $$$$ \\…
Question Number 226745 by fantastic2 last updated on 12/Dec/25 $${if}\:\frac{{x}}{{lm}−{n}^{\mathrm{2}} }=\frac{{y}}{{mn}−{l}^{\mathrm{2}} }=\frac{{z}}{{nl}−{m}^{\mathrm{2}} } \\ $$$${then}\:{show}\:{lx}+{my}+{nz}=\mathrm{0} \\ $$$$ \\ $$ Commented by fantastic2 last updated on…
Question Number 226743 by MrAjder last updated on 12/Dec/25 $$ \\ $$$${Prove}:\frac{\mathrm{1}}{\mathrm{2}{ne}}<\frac{\mathrm{1}}{{e}}−\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)^{{n}} <\frac{\mathrm{1}}{{ne}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 226736 by ajfour last updated on 12/Dec/25 Commented by ajfour last updated on 12/Dec/25 $${Lets}\:{attempt}\:{the}\:\mathrm{25}\:{million}\:{dollar}\: \\ $$$${question}\:{that}\:{i}\:{solved}\:{then}\: \\ $$$${satisfactorily}\:{with} \\ $$$${sir}\:{MJS},\:{still}\:{in}\:{another}\:{better} \\ $$$${way}!…
Question Number 226738 by MrAjder last updated on 12/Dec/25 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Prove}}: \\ $$$$\underset{{v}\rightarrow\infty} {\mathrm{lim}}\sqrt{{n}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{e}^{−{n}} \left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\prod}}\begin{pmatrix}{\mathrm{2}{n}}\\{{k}}\end{pmatrix}^{\frac{\mathrm{1}}{\mathrm{2}{n}}} −\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}^{{k}} }{{k}!}\right)\right)=\frac{\mathrm{1}}{\:\sqrt{\pi}}\left(\frac{{e}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\right) \\ $$ Answered by TonyCWX…
Question Number 226719 by AgniMath last updated on 11/Dec/25 Commented by Ghisom_ last updated on 11/Dec/25 see there: https://math.stackexchange.com/questions/69519/closed-form-for-a-pair-of-continued-fractions Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 226728 by hardmath last updated on 11/Dec/25 $$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\frac{\mathrm{dx}}{\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:=\:? \\ $$ Answered by Ghisom_ last updated on 12/Dec/25 $$\mathrm{there}'\mathrm{s}\:\mathrm{an}\:\mathrm{easy}\:\mathrm{but}\:\mathrm{boring}\:\mathrm{solution}\:\mathrm{using} \\ $$$${t}=\mathrm{tan}\:{x}…