Question Number 128048 by bramlexs22 last updated on 04/Jan/21 $$\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{8n}+\mathrm{3}}\:=? \\ $$ Answered by Olaf last updated on 04/Jan/21 $$\mathrm{L}\left(\lambda,\alpha,{s}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{\mathrm{2}{i}\pi\lambda{n}}…
Question Number 62494 by Tawa1 last updated on 21/Jun/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62489 by Tawa1 last updated on 21/Jun/19 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}\:−\:\mathrm{x}}\:+\:\sqrt{\mathrm{2}\:+\:\mathrm{x}}}{\:\sqrt{\mathrm{2}\:−\:\mathrm{x}}\:−\:\sqrt{\mathrm{2}\:+\:\mathrm{x}}}\:\:=\:\:\mathrm{3} \\ $$ Answered by som(math1967) last updated on 22/Jun/19 $$\frac{\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}+\sqrt{\mathrm{2}−{x}}−\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}−\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}}=\frac{\mathrm{3}+\mathrm{1}}{\mathrm{3}−\mathrm{1}}\:\:\bigstar \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{2}−{x}}}{\mathrm{2}\sqrt{\mathrm{2}+{x}}}=\frac{\mathrm{4}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}=\mathrm{4}\:\:\bigstar\bigstar \\…
Question Number 128007 by ajfour last updated on 03/Jan/21 $${Can}\:{anyone}\:{find}\:{any}\:{error}\:{in} \\ $$$${my}\:{attempt}\:{to}\:{improve}\:{upon} \\ $$$${the}\:{Cardano}'{s}\:{cubic}\:{formula}.. \\ $$$${or}\:{as}\:{an}\:{alternate}\:{to}\:{the} \\ $$$${trigonometric}\:{solution}… \\ $$$${here}\:{it}\:{goes}: \\ $$$$\:\:\:\:{X}^{\mathrm{3}} −{pX}−{q}=\mathrm{0} \\ $$$${let}\:\:\frac{{X}}{\:\sqrt{{p}}}\:=\:{x}\:;\:\:{and}\:{with}\:\frac{{q}}{{p}\sqrt{{p}}}\:=\:{c}\:,…
Question Number 62468 by bshahid010@gmail.com last updated on 21/Jun/19 Commented by mathmax by abdo last updated on 21/Jun/19 $${let}\:\mathrm{3}^{{x}} \:={t}\:\:\:\Rightarrow\left({k}−\mathrm{2}\right){t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{3}>\mathrm{0}\:\:\:\:\forall{t}>\mathrm{0}\:\Rightarrow\:\Delta^{'} <\mathrm{0}\:\Rightarrow \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}\left({k}−\mathrm{2}\right)<\mathrm{0}\:\Rightarrow\mathrm{1}−\mathrm{3}{k}\:+\mathrm{6}\:<\mathrm{0}\:\Rightarrow\mathrm{7}−\mathrm{3}{k}\:<\mathrm{0}\:\Rightarrow{k}>\frac{\mathrm{7}}{\mathrm{3}}…
Question Number 62449 by Tawa1 last updated on 21/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{digit}\:\mathrm{in}\:\:\:\:\mathrm{2}^{\mathrm{50}} \\ $$ Commented by Tawa1 last updated on 21/Jun/19 $$\mathrm{And}\:\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{a}\:\mathrm{general}\:\mathrm{nth}\:\mathrm{number}\:\mathrm{of}\:\mathrm{term}\:\mathrm{in}\:\mathrm{any}\:\mathrm{number}\:\mathrm{and}\:\:\mathrm{powers} \\ $$ Answered by Rasheed.Sindhi…
Question Number 62452 by Tawa1 last updated on 21/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\:\:\mathrm{2014}!\:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\:\mathrm{2017} \\ $$ Answered by Rasheed.Sindhi last updated on 21/Jun/19 $${Wilson}'{s}\:{Theorm}: \\ $$$$\:\:\:\:\:\:\:\left({p}−\mathrm{1}\right)!\equiv−\mathrm{1}\left({mod}\:{p}\right)\::\:{p}\in\mathbb{P} \\ $$$$\:\:\:\because\:\:\:\:\mathrm{2017}\in\mathbb{P} \\…
Question Number 62439 by mathsolverby Abdo last updated on 21/Jun/19 $${sove}\:{inside}\:{Z}/\mathrm{3}{Z}\:{the}\:{systeme} \\ $$$$\begin{cases}{\mathrm{5}{x}+\mathrm{7}{y}\:=\mathrm{10}}\\{\mathrm{2}{x}+\mathrm{5}{y}\:=\mathrm{8}}\end{cases} \\ $$$$ \\ $$ Commented by arcana last updated on 22/Jun/19 $$\mathbb{Z}/\mathrm{3}\mathbb{Z}=\mathbb{Z}_{\mathrm{3}}…
Question Number 127968 by liberty last updated on 03/Jan/21 $$\:\mathrm{How}\:\mathrm{many}\:\mathrm{x}\epsilon\mathbb{R}\:\mathrm{satisfy}\:\sqrt[{\mathrm{7}}]{\mathrm{x}}\:−\sqrt[{\mathrm{5}}]{\mathrm{x}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{x}}\:−\sqrt{\mathrm{x}}\: \\ $$ Answered by MJS_new last updated on 03/Jan/21 $${x}=\mathrm{0}\vee{x}\approx.\mathrm{0117154773}\vee{x}=\mathrm{1} \\ $$ Terms of Service…
Question Number 62428 by Tawa1 last updated on 21/Jun/19 Commented by mathsolverby Abdo last updated on 21/Jun/19 $${this}\:{question}\:{is}\:{done}\:{take}\:{a}\:{look}\:{at}\:{the} \\ $$$${platform} \\ $$ Commented by Tawa1…