Question Number 18349 by Yozzzzy last updated on 19/Jul/17 $${Consider}\:{the}\:{iteration} \\ $$$${x}_{{k}+\mathrm{1}} ={x}_{{k}} −\frac{\left[{f}\left({x}\right)\right]^{\mathrm{2}} }{{f}\left({x}_{{k}} +{f}\left({x}_{{k}} \right)\right)−{f}\left({x}_{{k}} \right)},\:\:\:\:\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},… \\ $$$${for}\:{the}\:{solution}\:{of}\:{f}\left({x}\right)=\mathrm{0}.\:{Explain}\:{the} \\ $$$${connection}\:{with}\:{Newton}'{s}\:{method},\:{and}\:{show} \\ $$$${that}\:\left({x}_{{k}} \right)\:{converges}\:{quadratically}\:{if}\:{x}_{\mathrm{0}}…
Question Number 18342 by mondodotto@gmail.com last updated on 19/Jul/17 Answered by alex041103 last updated on 19/Jul/17 $$\frac{{x}+\mathrm{4}}{{x}+\mathrm{1}}=\frac{{x}+\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{3}}{{x}+\mathrm{1}}=\mathrm{1}+\frac{\mathrm{3}}{{x}+\mathrm{1}} \\ $$$$\frac{{x}−\mathrm{2}}{{x}−\mathrm{4}}=\frac{{x}−\mathrm{4}}{{x}−\mathrm{4}}+\frac{\mathrm{2}}{{x}−\mathrm{4}}=\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{4}} \\ $$$$\Rightarrow\frac{\mathrm{3}}{{x}+\mathrm{1}}<\frac{\mathrm{2}}{{x}−\mathrm{4}} \\ $$$${The}\:{inequality}\:{makes}\:{sense}\:{when} \\ $$$${x}\neq−\mathrm{1};\mathrm{4}…
Question Number 149415 by abdurehime last updated on 05/Aug/21 $$\mathrm{show}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{sint}−\mathrm{tcost}}{\mathrm{t}^{\mathrm{3}} }\right)^{\mathrm{2}} \mathrm{dt}=\frac{\Pi}{\mathrm{15}} \\ $$ Commented by abdurehime last updated on 05/Aug/21 $$\mathrm{please}\:\mathrm{help}\:\mathrm{me}???? \\…
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Question Number 83874 by Power last updated on 07/Mar/20 Commented by niroj last updated on 07/Mar/20 $$\:\:\:\mathrm{xy}\left(\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)=\:\mathrm{24}….\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{10}……\left(\mathrm{ii}\right) \\ $$$$\:\:\mathrm{multipy}\:\mathrm{by}\:\mathrm{xy}\:\mathrm{in}\:\left(\mathrm{ii}\right)\mathrm{then}\:\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\:…
Question Number 149408 by mathdanisur last updated on 05/Aug/21 $${cos}\left(\frac{\pi}{\mathrm{2}}\:-\:\frac{\mathrm{1}}{\mathrm{2}}\:{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}\right)\:=\:? \\ $$ Commented by liberty last updated on 05/Aug/21 $$\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}\right)=\mathrm{sin}\:\mathrm{u} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{u}\:\Rightarrow\mathrm{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{2u} \\ $$$$\mathrm{cos}\:\mathrm{2u}=\frac{\mathrm{4}}{\mathrm{5}}\Rightarrow\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{u}=\frac{\mathrm{4}}{\mathrm{5}}…
Question Number 149410 by mathdanisur last updated on 05/Aug/21 $$\frac{{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{20}} \:+\:{log}_{\mathrm{2}} \:\mathrm{20}\:\centerdot\:{log}_{\mathrm{2}} \:\mathrm{5}\:-\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{5}} }{{log}_{\mathrm{2}} \:\mathrm{20}\:+\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{5}}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 149405 by mathdanisur last updated on 05/Aug/21 $${if}\:\:\:\int{f}\left({x}\right){dx}\:=\:{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −{c} \\ $$$${find}\:\:\:{f}\left(−\mathrm{2}\right)\:=\:? \\ $$ Commented by liberty last updated on 05/Aug/21 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3x}^{\mathrm{2}} −\mathrm{6x}…
Question Number 83871 by jagoll last updated on 07/Mar/20 $$\mathrm{If}\:\mathrm{equation}\: \\ $$$$\begin{cases}{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{10}}\\{\mathrm{x}+\mathrm{2y}=\:\mathrm{5z}}\end{cases} \\ $$$$\mathrm{has}\:\mathrm{solution}\:\mathrm{is}\:\left(\mathrm{a},\mathrm{b},\mathrm{c}\right).\: \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{2b}+\mathrm{3c}\: \\…
Question Number 18333 by ajfour last updated on 19/Jul/17 Commented by ajfour last updated on 19/Jul/17 $$\mathrm{In}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{Q}.\mathrm{18327} \\ $$ Terms of Service Privacy Policy Contact:…