Question Number 18366 by ketto last updated on 19/Jul/17 Answered by Tinkutara last updated on 19/Jul/17 $$−\mathrm{4}\:<\:\mathrm{5}\:−\:\mathrm{3}{x}\:\leqslant\:\mathrm{17} \\ $$$$−\mathrm{9}\:<\:−\mathrm{3}{x}\:\leqslant\:\mathrm{12} \\ $$$$−\mathrm{12}\:\leqslant\:\mathrm{3}{x}\:<\:\mathrm{9} \\ $$$$−\mathrm{4}\:\leqslant\:{x}\:<\:\mathrm{3} \\ $$$${x}\:\in\:\left[−\mathrm{4},\:\mathrm{3}\right)…
Question Number 18357 by tawa tawa last updated on 19/Jul/17 $$\mathrm{From}\:\mathrm{the}\:\mathrm{topic}\:\mathrm{transformer} \\ $$$$ \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\mathrm{e}\:=\:\sqrt{\mathrm{2}}\:\varepsilon\:\mathrm{cos}\left(\omega\mathrm{t}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18349 by Yozzzzy last updated on 19/Jul/17 $${Consider}\:{the}\:{iteration} \\ $$$${x}_{{k}+\mathrm{1}} ={x}_{{k}} −\frac{\left[{f}\left({x}\right)\right]^{\mathrm{2}} }{{f}\left({x}_{{k}} +{f}\left({x}_{{k}} \right)\right)−{f}\left({x}_{{k}} \right)},\:\:\:\:\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},… \\ $$$${for}\:{the}\:{solution}\:{of}\:{f}\left({x}\right)=\mathrm{0}.\:{Explain}\:{the} \\ $$$${connection}\:{with}\:{Newton}'{s}\:{method},\:{and}\:{show} \\ $$$${that}\:\left({x}_{{k}} \right)\:{converges}\:{quadratically}\:{if}\:{x}_{\mathrm{0}}…
Question Number 83861 by Umar last updated on 07/Mar/20 $$\mathrm{An}\:\mathrm{object}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{7kg}\:\mathrm{is}\:\mathrm{sliding}\:\mathrm{down} \\ $$$$\mathrm{a}\:\mathrm{frictionless}\:\mathrm{20m}\:\mathrm{inclined}\:\mathrm{plane}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{object}\:\mathrm{when}\: \\ $$$$\mathrm{it}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{ground}. \\ $$ Commented by mr W last updated on…
Question Number 18322 by Tinkutara last updated on 18/Jul/17 $$\mathrm{The}\:\mathrm{pulley}\:\mathrm{arrangements}\:\mathrm{are}\:\mathrm{identical}. \\ $$$$\mathrm{The}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rope}\:\mathrm{is}\:\mathrm{negligible}.\:\mathrm{In} \\ $$$$\left(\mathrm{a}\right),\:\mathrm{the}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{lifted}\:\mathrm{up}\:\mathrm{by}\:\mathrm{attaching} \\ $$$$\mathrm{a}\:\mathrm{mass}\:\left(\mathrm{2}{m}\right)\:\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rope}. \\ $$$$\mathrm{In}\:\left(\mathrm{b}\right),\:{m}\:\mathrm{is}\:\mathrm{lifted}\:\mathrm{up}\:\mathrm{by}\:\mathrm{pulling}\:\mathrm{the} \\ $$$$\mathrm{other}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rope}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\mathrm{downward}\:\mathrm{force}\:{F}\:=\:\mathrm{2}{mg}.\:\mathrm{In}\:\mathrm{which} \\ $$$$\mathrm{case},\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:{m}\:\mathrm{is}\:\mathrm{more}? \\…
Question Number 18320 by Tinkutara last updated on 18/Jul/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:{ABC}\:\mathrm{with}\:\mathrm{fixed}\:\mathrm{base}\:{BC}, \\ $$$$\mathrm{the}\:\mathrm{vertex}\:{A}\:\mathrm{moves}\:\mathrm{such}\:\mathrm{that}\:\mathrm{cos}\:{B}\:+ \\ $$$$\mathrm{cos}\:{C}\:=\:\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}\:.\:\mathrm{If}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{denote} \\ $$$$\mathrm{the}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathrm{opposite}\:\mathrm{to}\:\mathrm{the}\:\mathrm{angles}\:{A},\:{B}\:\mathrm{and}\:{C} \\ $$$$\mathrm{respectively},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{b}\:+\:{c}\:=\:\mathrm{4}{a} \\ $$$$\left(\mathrm{2}\right)\:{b}\:+\:{c}\:=\:\mathrm{2}{a}…
Question Number 83849 by Rio Michael last updated on 06/Mar/20 $$\mathrm{Gven}\:\mathrm{that}\:{y}\:=\:{e}^{−{x}} \mathrm{sin}{bx}\:,\mathrm{where}\:{b}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant},\mathrm{show}\:\mathrm{that} \\ $$$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}\frac{{dy}}{{dx}}\:+\:\left(\mathrm{1}\:+\:{b}^{\mathrm{2}} \right){y}\:=\:\mathrm{0}. \\ $$ Commented by niroj last updated on…
Question Number 83850 by Rio Michael last updated on 06/Mar/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:{f},\:\mathrm{defined}\:\mathrm{by} \\ $$$$\:{f}\left({x}\right)\:=\:\frac{{x}}{\mathrm{1}+\:{x}^{\mathrm{2}} }\:,\:{x}\in\mathbb{R} \\ $$ Commented by mathmax by abdo last updated on 06/Mar/20…
Question Number 149383 by fotosy2k last updated on 05/Aug/21 Commented by fotosy2k last updated on 05/Aug/21 $${please}\:{help}\:{me} \\ $$ Answered by Ar Brandon last updated…
Question Number 149378 by fotosy2k last updated on 05/Aug/21 Answered by Ar Brandon last updated on 05/Aug/21 $${S}=\psi\left(\mathrm{4}\right)−\psi\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}+\psi\left(\mathrm{1}\right)−\psi\left(\mathrm{1}\right)=\frac{\mathrm{11}}{\mathrm{6}} \\ $$$${S}=\left(\mathrm{1}−\cancel{\frac{\mathrm{1}}{\mathrm{4}}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\cancel{\frac{\mathrm{1}}{\mathrm{5}}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\cancel{\frac{\mathrm{1}}{\mathrm{6}}}\right)+\left(\cancel{\frac{\mathrm{1}}{\mathrm{4}}}−\cancel{\frac{\mathrm{1}}{\mathrm{7}}}\right)+\centerdot\centerdot\centerdot+ \\ $$$$\:\:\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{11}}{\mathrm{6}} \\ $$ Answered…