Question Number 210297 by Spillover last updated on 05/Aug/24 $${Prove}\:{the}\:{theorem}. \\ $$$${A}\:{non}\:{empty}\:{subset}\:{W}\:\:{of}\:{a}\:{vector}\:{space}\:{V}\left({F}\right) \\ $$$${is}\:{the}\:{subset}\:{of}\:{V}\:\:{if}\:\:{and}\:{only}\:{if} \\ $$$$\alpha{W}_{\mathrm{1}} +\beta{W}_{\mathrm{2}} \:\in{W}\:\:\forall\alpha,\beta\:\in\:{F}\:\:{and}\:{W}_{\mathrm{1}} ,{W}_{\mathrm{2}} \:\in{W} \\ $$ Terms of Service…
Question Number 210295 by ibrahimmatematic last updated on 05/Aug/24 Commented by Frix last updated on 05/Aug/24 $$\mathrm{Question}\:\mathrm{210289} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 210289 by Abdullahrussell last updated on 05/Aug/24 Answered by Frix last updated on 05/Aug/24 $$\mathrm{There}'\mathrm{s}\:\mathrm{one}\:\mathrm{number}\:{x}\:\mathrm{with}\:\mathrm{nice}\:\mathrm{properties}: \\ $$$${x}^{\mathrm{2}} ={x}+\mathrm{1},\:{x}^{\mathrm{3}} =\mathrm{2}{x}+\mathrm{1}, \\ $$$$\frac{\mathrm{1}}{{x}}={x}−\mathrm{1},\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{2}−{x},\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{2}{x}−\mathrm{1}…
Question Number 210309 by Spillover last updated on 06/Aug/24 $${For}\:{what}\:{value}\:{of}\:{p}\:{does}\:{the}\:{series} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{n}} }{\left(\mathrm{2}+{e}^{\mathrm{2}{n}} \right)^{{p}} }\:\:\:\:\:{converge} \\ $$ Commented by klipto last updated on…
Question Number 210308 by Spillover last updated on 05/Aug/24 $${Evaluate}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{2}{y}^{\mathrm{4}} }{{y}^{\mathrm{3}} −{y}^{\mathrm{2}} +{y}−\mathrm{1}}{dy} \\ $$ Commented by Frix last updated on 06/Aug/24 $$\mathrm{Simply}\:\mathrm{expand}\:\mathrm{it}:…
Question Number 210311 by Spillover last updated on 05/Aug/24 $${Let}\:{a}_{\mathrm{1}} =\mathrm{1}\:\:\:{a}_{\mathrm{2}} =\mathrm{2}^{\mathrm{1}} \:\:\:\:{a}_{\mathrm{3}} =\mathrm{3}^{\left(\mathrm{2}^{\mathrm{1}} \right)} \:\:{a}_{\mathrm{4}} =\mathrm{4}^{\left(\mathrm{3}^{\left(\mathrm{2}^{\mathrm{1}} \right)} \right)} \\ $$$${find}\:{the}\:{last}\:{two}\:{digits}\:{of}\:{a}_{\mathrm{23}} \:{and}\:{so}\:{on} \\ $$ Answered…
Question Number 210310 by Spillover last updated on 05/Aug/24 $${Let}\:{a}\:{be}\:{the}\:{unique}\:{real}\:{zero}\:{of}\:{x}^{\mathrm{3}} +{x}+\mathrm{1}. \\ $$$${find}\:{the}\:{simplest}\:{possible}\:{way}\:{to}\:{write}\: \\ $$$$\frac{\mathrm{18}}{\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{as}\:{polynomial}\:{expression}\:{in}\:\:{a} \\ $$$${with}\:{ratio}\:{coefficients} \\ $$ Commented by Frix last…
Question Number 210307 by Spillover last updated on 05/Aug/24 $$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }\:\right)} \\ $$$$ \\ $$ Answered by Spillover last updated on…
Question Number 210261 by klipto last updated on 04/Aug/24 $$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\frac{\boldsymbol{\mathrm{sinAcosA}}−\boldsymbol{\mathrm{sinBcosB}}}{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{B}}}=\boldsymbol{\mathrm{tan}}\left(\boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{B}}\right) \\ $$ Answered by efronzo1 last updated on 04/Aug/24 $$\:\:\:\frac{\mathrm{sin}\:\mathrm{A}\:\mathrm{cos}\:\mathrm{A}−\mathrm{sin}\:\mathrm{B}\:\mathrm{cos}\:\mathrm{B}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{A}−\mathrm{sin}\:^{\mathrm{2}}…
Question Number 210263 by Spillover last updated on 04/Aug/24 Answered by Frix last updated on 04/Aug/24 $$\mathrm{With}\:{t}=\mathrm{tan}\:\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:\mathrm{and}\:\alpha=\mathrm{ln}\:\mathrm{12}\:\mathrm{we}\:\mathrm{get} \\ $$$$−\frac{\mathrm{4}}{\mathrm{3}}\int\frac{{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} \mathrm{sin}\:\alpha\:−\mathrm{2}{t}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)−\mathrm{sin}\:\alpha\right)}{dt}= \\ $$$$\mathrm{Let}\:{u}=\mathrm{cos}\:\alpha\:\wedge\:{v}=\mathrm{sin}\:\alpha \\…