Question Number 208367 by Mastermind last updated on 13/Jun/24 $$\mathrm{1}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\:\:\:\:\:\left(\mathrm{a}\right)\:\left\{\mathrm{x}\::\:−\mathrm{3}\:<\:\mathrm{x}\:<\:\mathrm{7}\right\} \\ $$$$\:\:\:\:\:\left(\mathrm{b}\right)\:\left\{\mathrm{x}\::\:\mathrm{2}\:\leqslant\:\mathrm{x}\:\leqslant\:\mathrm{6}\right\}\:\cup\:\left\{−\mathrm{3}\:\leqslant\:\mathrm{x}\:\leqslant\:−\mathrm{1}\right\} \\ $$$$\:\:\:\:\:\left(\mathrm{c}\right)\:\left\{\mathrm{x}\::\:−\mathrm{2}\:\leqslant\:\mathrm{x}\:<\:\mathrm{5}\right\}\:\cup\:\left\{\mathrm{1}\:<\:\mathrm{x}\:\leqslant\:\mathrm{7}\right\} \\ $$$$ \\ $$$$\mathrm{2}.\:\mathrm{Let}\:\mathrm{I}=\left(\mathrm{a},\:\mathrm{b}\right).\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{I}\:\mathrm{is}\:\mathrm{measurable} \\ $$$$\mathrm{and}\:\mathrm{m}\left(\mathrm{I}\right)\:=\:\mathrm{L}\left(\mathrm{I}\right). \\ $$ Terms…
Question Number 208312 by messele last updated on 11/Jun/24 $${lim}_{{x}\rightarrow\mathrm{0}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}\:=\:{log}\:{a}} \\ $$ Answered by mathzup last updated on 11/Jun/24 $${let}\:{f}\left({x}\right)={a}^{{x}} \:={e}^{{xlna}} \:\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1}\:{and} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}}…
Question Number 208277 by Mastermind last updated on 10/Jun/24 Answered by Rasheed.Sindhi last updated on 10/Jun/24 $$\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{5}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{5}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}\end{vmatrix}\: \\ $$$${R}_{\mathrm{2}} ={R}_{\mathrm{2}} −\mathrm{2}{R}_{\mathrm{1}} \\ $$$$=\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\mathrm{1}}&{\:\:\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{-\mathrm{5}}\\{\mathrm{1}}&{\mathrm{5}}&{\mathrm{2}}&{\mathrm{0}}&{\:\:\mathrm{0}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{2}}&{\:\:\mathrm{0}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\:\:\mathrm{2}}\end{vmatrix}\: \\ $$$${R}_{\mathrm{4}}…
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Question Number 207951 by MWSuSon last updated on 31/May/24 $$\mathrm{Anybody}\:\mathrm{with}\:\mathrm{knowledge}\:\mathrm{or}\:\mathrm{books}\:\mathrm{on} \\ $$$$\mathrm{mathemtical}\:\mathrm{modeling}? \\ $$ Commented by TheHoneyCat last updated on 01/Jun/24 Modeling of what? shapes? physical phenomenon? dynamical system? statistical modeling? it is hard to find math that doesn't show up in any sort of modeling, so I think you'll get better results by being slightly more precise... Commented by MWSuSon…
Question Number 207805 by udaythool last updated on 27/May/24 $$\mathrm{I}'\mathrm{ve}\:\mathrm{changed}\:\mathrm{my}\:\mathrm{handset},\:\mathrm{now}\:\mathrm{unable}\:\mathrm{to}\:\mathrm{view}\:\mathrm{saved}\:\mathrm{equations}. \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{access}\:\mathrm{saved}\:\mathrm{equatios}\:\mathrm{in}\:\mathrm{new}\:\mathrm{handset}? \\ $$ Commented by mr W last updated on 27/May/24 $${with}\:{your}\:{old}\:{device}: \\ $$$${save}\:{your}\:{data}\:{using}…
Question Number 207692 by qeemah last updated on 23/May/24 Commented by qeemah last updated on 23/May/24 $${pleasehelpmewiththeTobitregression} \\ $$$$ \\ $$ Terms of Service Privacy…
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Question Number 206591 by necx122 last updated on 19/Apr/24 $${I}\:{have}\:{a}\:{background}\:{in}\:{building}\:{design} \\ $$$${but}\:{I}\:{have}\:{great}\:{interest}\:{in}\:{Mathematics}. \\ $$$${That}'{s}\:{one}\:{major}\:{reason}\:{why}\:{I}\:{joined} \\ $$$${this}\:{platform}\:{cause}\:{I}\:{did}\:{little}\:{or}\:{no} \\ $$$${maths}\:{in}\:{my}\:{undergraduate}\:{days}\:{as}\:{a} \\ $$$${building}\:{design}\:{student}. \\ $$$$ \\ $$$${This}\:{forum}\:{has}\:{contributed}\:{greatly}\:{to} \\…
Question Number 206452 by langatfredrick last updated on 15/Apr/24 Answered by A5T last updated on 15/Apr/24 $${If}\:{remainder}\:{means}\:\mid{x}−{y}\mid,\:{then}: \\ $$$$\frac{{t}_{{x}} }{{x}}={x};\frac{{t}_{{y}} }{{y}}={y}\Rightarrow{t}_{\mid{x}−{y}\mid} =\mid{t}_{{x}} −{t}_{{y}} \mid=\mid{x}^{\mathrm{2}} −{y}^{\mathrm{2}}…