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Question Number 44291 by peter frank last updated on 25/Sep/18 $${withiout}\:{using}\:{calculator}\:{find}\:{approximate}\:{for} \\ $$$$\sqrt{\mathrm{9}.\mathrm{01}} \\ $$ Commented by maxmathsup by imad last updated on 25/Sep/18 $$\sqrt{\mathrm{9},\mathrm{01}}=\sqrt{\mathrm{9}+\mathrm{0},\mathrm{01}}=\sqrt{\mathrm{9}+\mathrm{10}^{−\mathrm{2}}…
Question Number 44267 by rahul 19 last updated on 25/Sep/18 $${If}\:{x}\epsilon\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:,\:\mathrm{1}\right)\:,{differentiate}\:\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right). \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18 $${x}={cos}\alpha\:\:\:{dx}=−{sin}\alpha\:{d}\alpha \\ $$$${y}={cos}^{−\mathrm{1}}…
Question Number 44208 by Tinkutara last updated on 23/Sep/18 Commented by tanmay.chaudhury50@gmail.com last updated on 24/Sep/18 $${some}\:{one}\:{throw}\:{light}…{the}\:{meaning}\:{of}\:{twice} \\ $$$${differentiable}\:{function}… \\ $$$${f}\left({x}\right)={A}\left({x}−{a}\right)\left({x}−{e}\right)+{Cx}^{\mathrm{2}} +{D}\:\:{is}\:{it}\:{correcet}\:{to}\:{assume} \\ $$ Commented…
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Question Number 109736 by mnjuly1970 last updated on 25/Aug/20 $$\:\:\:\:{please}\:{solve}\:: \\ $$$$\:\: \\ $$$${lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}\:+\sqrt{\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\sqrt[{\mathrm{4}}]{\mathrm{4}}\:+……\sqrt[{\mathrm{2}{n}}]{\mathrm{2}{n}}}{\mathrm{3}{n}−\mathrm{4}} \\ $$$$=??? \\ $$ Answered by Her_Majesty last updated on…
Question Number 175275 by mnjuly1970 last updated on 25/Aug/22 Answered by Rasheed.Sindhi last updated on 25/Aug/22 $$\left(−\mathrm{5},\:{x}+\mathrm{2}{y}\right)\cup\left(\mathrm{2}{x}−{y}−\mathrm{4},\:\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(−\mathrm{5},{a}\right)\cup\left({a},\mathrm{3}\right) \\ $$$${a}={x}+\mathrm{2}{y}=\mathrm{2}{x}−{y}−\mathrm{4}=\frac{−\mathrm{5}+\mathrm{3}}{\mathrm{2}}=−\mathrm{1} \\ $$$${a}=−\mathrm{1} \\ $$$${x}+\mathrm{2}{y}=−\mathrm{1}\:\wedge\:\mathrm{2}{x}−{y}−\mathrm{4}=−\mathrm{1}…