Question Number 205849 by Davidtim last updated on 31/Mar/24 $${if}\:{a}^{{a}} ={b}^{{b}} \:\:;\:{a}={b}\:\:{is}\:{it}\:{true}? \\ $$$${if}\:{it}\:{is}\:{true}\:{then}\:{prove}\:{it}. \\ $$ Commented by mr W last updated on 31/Mar/24 $${not}\:{true}\:{for}\:\mathrm{0}<{a},\:{b}<\mathrm{1}.…
Question Number 205833 by MathedUp last updated on 31/Mar/24 $$\mathrm{can}'\mathrm{t}\:\mathrm{Solve}\:\mathrm{Differantial}\:\mathrm{Equation} \\ $$$$\mathrm{Diff}\:\mathrm{Equa}\::\:\left(\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}{t}}\right)^{\mathrm{2}} +\mathrm{4}{y}\left({t}\right)=\mathrm{8}{t}^{\mathrm{2}} −\mathrm{32}{t}+\mathrm{28}…. \\ $$$$\mathrm{Sadly}\:\mathrm{it}'\mathrm{s}\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{obtain}\:\mathrm{an}\:\mathrm{exact} \\ $$$$\mathrm{closed}−\mathrm{form}\:\mathrm{expression}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Solution}\:\mathrm{of}\:\mathrm{Diff}\:\mathrm{Equa} \\ $$$$\mathrm{But}\:\mathrm{if}\:\mathrm{the}\:\mathrm{Runge}−\mathrm{Kutta}\:\mathrm{method}\:\mathrm{is}\:\mathrm{used}. \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{at}\:\mathrm{any}\:\mathrm{one}\:\mathrm{point}\:\mathrm{can}\:\mathrm{be}\:\mathrm{estimated} \\ $$ Commented…
Question Number 205825 by tri26112004 last updated on 31/Mar/24 $$\left[{f}'\left({x}\right)\right]^{\mathrm{2}} +\mathrm{4}{f}\left({x}\right)=\mathrm{8}{x}^{\mathrm{2}} −\mathrm{32}{x}+\mathrm{28} \\ $$$$\Rightarrow{f}\left({x}\right)=¿ \\ $$ Commented by mr W last updated on 31/Mar/24 $${f}\left({x}\right)={x}^{\mathrm{2}}…
Question Number 205784 by MathedUp last updated on 30/Mar/24 Commented by MathedUp last updated on 30/Mar/24 $$\:\mathrm{meijier}\:\mathrm{G}\:\mathrm{function}….. \\ $$$$\mathrm{OMG}…. \\ $$ Commented by MathedUp last…
Question Number 205775 by SANOGO last updated on 30/Mar/24 $${calcu}/\:\:\:\:{limit}/{n}\rightarrow+{oo} \\ $$$$\:\:\int_{\mathrm{0}} ^{+{oo}} {arctan}\left(\frac{{x}}{{n}}\right){e}^{−{x}} {dx} \\ $$ Answered by MathedUp last updated on 30/Mar/24 $$\mathrm{let}'\mathrm{s}\:\mathrm{consider}\:{F}\left({s}\right)=\int_{\mathrm{0}}…
Question Number 205749 by mokys last updated on 29/Mar/24 $${whats}\:{the}\:{suficient}\:{condition}\:{to}\:{became}\:{the}\:{question}\: \\ $$$$ \\ $$$$\sigma^{\mathrm{2}} \left(\mathrm{1}−{a}_{{i}} \right)\left[\lambda_{{i}} \left(\mathrm{1}+{a}_{{i}} \right)−\left(\mathrm{1}−{a}_{{i}} \right)\left(\mathrm{1}−{d}_{{i}} \right)+\left(\lambda_{{i}} +{k}\right)\right]\:<\:\mathrm{0}\: \\ $$ Terms of…
Question Number 205685 by MathedUp last updated on 27/Mar/24 Answered by TonyCWX08 last updated on 27/Mar/24 $$\frac{\mathrm{1}}{\mathrm{10}}{g} \\ $$ Commented by MathedUp last updated on…
Question Number 205684 by MathedUp last updated on 27/Mar/24 $$\mathrm{Figure}\:\mathrm{Shows}\:\mathrm{that}\:\mathrm{Object}\:\boldsymbol{\mathrm{A}}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{to} \\ $$$$\mathrm{Object}\:\boldsymbol{\mathrm{B}}\:\mathrm{by}\:\mathrm{thread}\:\mathrm{along}\:\mathrm{the}\:\mathrm{Slope} \\ $$$$\mathrm{it}\:\mathrm{shows}\:\mathrm{a}\:\mathrm{costant}\:\mathrm{acceleration}\:\mathrm{motion} \\ $$$$\mathrm{the}\:\mathrm{mass}\:\mathrm{of}\:\boldsymbol{\mathrm{A}}\:\mathrm{of}\:\boldsymbol{\mathrm{B}}\:\mathrm{are}\:\mathrm{3m}\:,\:\mathrm{2m} \\ $$$$\mathrm{respectively}\:\mathrm{and}\:\mathrm{when}\:\boldsymbol{\mathrm{A}}\:\mathrm{communicates}\:\mathrm{from}\:\mathrm{point}\:\boldsymbol{\mathrm{P}}\:\mathrm{to}\:\boldsymbol{\mathrm{Q}}\: \\ $$$$\boldsymbol{\mathrm{B}}'{s}\:\mathrm{the}\:\mathrm{decrease}\:\mathrm{in}\:\mathrm{potential}\:\mathrm{energy}\:\mathrm{is}\:\mathrm{10}\:\:\mathrm{times} \\ $$$$\mathrm{the}\:\mathrm{decrease}\:\mathrm{in}\:\mathrm{Kinetic}\:\mathrm{energy}\:\mathrm{of}\:\boldsymbol{\mathrm{B}}\: \\ $$$$\mathrm{find}\:\mathrm{accerate}\:\mathrm{of}\:\boldsymbol{\mathrm{A}}\:\left(\mathrm{3}\:\mathrm{point}\right) \\…
Question Number 205682 by naka3546 last updated on 27/Mar/24 Commented by naka3546 last updated on 28/Mar/24 $$\mathrm{Thank}\:\mathrm{you} \\ $$ Commented by Frix last updated on…
Question Number 205681 by naka3546 last updated on 27/Mar/24 Answered by mr W last updated on 27/Mar/24 $${a}=\mathrm{1}\:{cm} \\ $$$${R}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{6}} \\ $$$${r}=\frac{\sqrt{\mathrm{3}}{b}}{\mathrm{6}} \\ $$$$\frac{{b}}{{a}}=\frac{\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}−\mathrm{2}{R}}{\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{3}} \\…