Question Number 129147 by Adel last updated on 13/Jan/21 $$\mathrm{cot}^{\mathrm{2}} \boldsymbol{\pi}/\mathrm{7}+\mathrm{cot}^{\mathrm{2}} \mathrm{2}\boldsymbol{\pi}/\mathrm{7}+\mathrm{cot}^{\mathrm{2}} \mathrm{3}\boldsymbol{\pi}/\mathrm{7}=? \\ $$ Commented by benjo_mathlover last updated on 13/Jan/21 $$\:\mathrm{ans}\::\:\mathrm{5} \\ $$…
Question Number 129145 by Dwaipayan Shikari last updated on 13/Jan/21 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{1}+…}}}}}} \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 63602 by pavithra last updated on 06/Jul/19 $$\mathrm{32}{x}^{\mathrm{3}} −\mathrm{48}{x}^{\mathrm{2}} −\mathrm{22}{x}−\mathrm{3}=\mathrm{0} \\ $$ Answered by MJS last updated on 06/Jul/19 $${x}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{11}}{\mathrm{16}}{x}−\frac{\mathrm{3}}{\mathrm{32}}=\mathrm{0} \\…
Question Number 129132 by BHOOPENDRA last updated on 13/Jan/21 Answered by Dwaipayan Shikari last updated on 13/Jan/21 $$\mathscr{L}^{−\mathrm{1}} \left(\frac{\mathrm{6}{as}}{\left(\mathrm{9}{s}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}\right)=\frac{\mathrm{1}}{\mathrm{9}}\mathscr{L}^{−\mathrm{1}} \left(\frac{\mathrm{2}\left(\frac{{a}}{\mathrm{3}}\right){s}}{\left({s}^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{3}}\right)^{\mathrm{2}} \right)}\right)=\frac{{t}}{\mathrm{9}}{sin}\left(\frac{{at}}{\mathrm{3}}\right) \\…
Question Number 63597 by aliesam last updated on 05/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 63596 by Tawa1 last updated on 05/Jul/19 $$\mathrm{The}\:\mathrm{surnames}\:\mathrm{of}\:\mathrm{40}\:\mathrm{students}\:\mathrm{in}\:\mathrm{a}\:\mathrm{class}\:\mathrm{were}\:\mathrm{arranged}\:\mathrm{in} \\ $$$$\mathrm{alphabetical}\:\mathrm{order}.\:\mathrm{16}\:\mathrm{of}\:\mathrm{the}\:\mathrm{surnames}\:\mathrm{begin}\:\mathrm{with}\:\mathrm{O}\:\mathrm{while} \\ $$$$\mathrm{9}\:\mathrm{of}\:\mathrm{the}\:\mathrm{surnames}\:\mathrm{begin}\:\mathrm{with}\:\mathrm{A}.\:\:\mathrm{14}\:\mathrm{of}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{alphabet}\:\mathrm{do}\:\mathrm{not}\:\mathrm{appear}\:\mathrm{as}\:\mathrm{the}\:\mathrm{first}\:\mathrm{letter}\:\mathrm{of}\:\mathrm{any}\:\mathrm{surname}. \\ $$$$ \\ $$$$\left(\mathrm{i}\right)\:\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{surname}\:\mathrm{of}\:\mathrm{a}\:\mathrm{child}\:\mathrm{picked} \\ $$$$\mathrm{at}\:\mathrm{random}\:\mathrm{from}\:\mathrm{the}\:\mathrm{class}\:\mathrm{begins}\:\mathrm{with}\:\mathrm{either}\:\mathrm{A}\:\mathrm{or}\:\mathrm{O} \\ $$$$\left(\mathrm{ii}\right)\:\:\mathrm{If}\:\mathrm{more}\:\mathrm{than}\:\mathrm{one}\:\mathrm{surname}\:\mathrm{begins}\:\mathrm{with}\:\mathrm{a}\:\mathrm{letter}\:\mathrm{besides}\:\mathrm{A} \\…
Question Number 129130 by bounhome last updated on 13/Jan/21 $${solve}\:: \\ $$$$\:\:{y}''+\mathrm{4}{y}={cos}\mathrm{2}{x} \\ $$$$\: \\ $$ Answered by liberty last updated on 13/Jan/21 $$\:\mathrm{Homogenous}\:\mathrm{solution}\:\mathrm{y}_{\mathrm{h}} =\mathrm{C}_{\mathrm{1}}…
Question Number 129131 by Boucatchou last updated on 13/Jan/21 $${Solve}\:\:\mathrm{134}^{{x}+\mathrm{1}} =\mathrm{16}^{{x}} −\mathrm{768} \\ $$ Commented by Dwaipayan Shikari last updated on 13/Jan/21 $${No}\:{solution}\:{in}\:\:{x}\in\mathbb{R} \\ $$…
Question Number 129129 by Adel last updated on 13/Jan/21 Commented by MJS_new last updated on 13/Jan/21 $$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{e}}\:\mathrm{but}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{yet}\:\mathrm{show}\:\mathrm{it} \\ $$ Commented by Adel last updated on…
Question Number 129126 by benjo_mathlover last updated on 13/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{y}\:=\:\mathrm{xy}^{\mathrm{5}} \: \\ $$ Answered by liberty last updated on 13/Jan/21 $$\:\mathrm{let}\:\mathrm{v}\:=\:\mathrm{y}^{−\mathrm{4}} \:;\:\frac{\mathrm{dv}}{\mathrm{dx}}\:=−\mathrm{4y}^{−\mathrm{5}\:} \frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{y}^{\mathrm{5}}…