Question Number 208789 by Tawa11 last updated on 23/Jun/24 $$\mathrm{Why}\:\mathrm{is}\:\mathrm{surface}\:\mathrm{tension}\:\mathrm{formula}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{2L} \\ $$$$\mathrm{that}\:\mathrm{is},\:\:\:\:\mathrm{surface}\:\mathrm{tension}\:\:=\:\:\frac{\mathrm{F}}{\mathrm{2L}} \\ $$$$\mathrm{why}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{2L}. \\ $$$$\mathrm{Where}\:\mathrm{did}\:\mathrm{the}\:\mathrm{2}\:\mathrm{come}\:\mathrm{from}? \\ $$$$ \\ $$$$\mathrm{Example}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{force}\:\mathrm{required}\:\mathrm{to}\:\mathrm{lift}\:\mathrm{a}\:\mathrm{needle} \\ $$$$\mathrm{4cm}\:\mathrm{long}\:\mathrm{off}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{water},\:\mathrm{if}\:\mathrm{surface} \\…
Question Number 208759 by Tawa11 last updated on 22/Jun/24 Commented by Tawa11 last updated on 22/Jun/24 $$\mathrm{Will}\:\mathrm{this}\:\mathrm{question}\:\mathrm{have}\:\mathrm{workings}? \\ $$$$\mathrm{Based}\:\mathrm{on}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}. \\ $$$$ \\ $$$$\mathrm{Or}\:\mathrm{answer}\:\mathrm{is}\:\:\mathrm{E} \\ $$…
Question Number 208741 by Tawa11 last updated on 22/Jun/24 Answered by mr W last updated on 22/Jun/24 Commented by mr W last updated on 22/Jun/24…
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Question Number 208645 by Mastermind last updated on 20/Jun/24 $$\mathrm{Solve}\:: \\ $$$$\mathrm{2x}_{\mathrm{1}} \:−\:\lambda_{\mathrm{1}} \:−\:\mathrm{5}\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{2}} \:−\:\lambda_{\mathrm{1}} \:−\:\mathrm{2}\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{3}} \:−\:\mathrm{3}\lambda_{\mathrm{1}} \:−\:\lambda_{\mathrm{2}} \:=\:\mathrm{0}…
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Question Number 208520 by Tawa11 last updated on 17/Jun/24 Answered by mr W last updated on 17/Jun/24 Commented by mr W last updated on 17/Jun/24…
Question Number 208367 by Mastermind last updated on 13/Jun/24 $$\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}}…