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Question Number 11285 by uni last updated on 19/Mar/17 $$\frac{{sin}\mathrm{20}}{{cos}\mathrm{80}−{tan}\mathrm{30}×{sin}\mathrm{80}}=? \\ $$ Answered by mrW1 last updated on 19/Mar/17 $$=\frac{{sin}\mathrm{20}}{{cos}\mathrm{80}−\frac{\mathrm{sin}\:\mathrm{30}}{\mathrm{cos}\:\mathrm{30}}×{sin}\mathrm{80}} \\ $$$$=\frac{{sin}\mathrm{20}×\mathrm{cos}\:\mathrm{30}}{\mathrm{cos}\:\mathrm{30}×{cos}\mathrm{80}−\mathrm{sin}\:\mathrm{30}×{sin}\mathrm{80}} \\ $$$$=\frac{{sin}\mathrm{20}×\mathrm{cos}\:\mathrm{30}}{\mathrm{cos}\:\mathrm{110}} \\…
Question Number 142358 by rs4089 last updated on 30/May/21 $$\int_{\mathrm{0}} ^{\infty} {x}^{{n}−\mathrm{1}} {log}_{{e}} \left(\mathrm{1}−{x}\right){dx} \\ $$ Answered by mathmax by abdo last updated on 30/May/21…
Question Number 11284 by suci last updated on 19/Mar/17 $${a}+\mathrm{2}{b}+\mathrm{3}{c}….\left(\mathrm{1}\right) \\ $$$$−\mathrm{3}{a}−\mathrm{4}{b}+\mathrm{2}{c}….\left(\mathrm{2}\right) \\ $$$$\mathrm{2}{a}−{b}−{c}….\left(\mathrm{3}\right) \\ $$$${a}=…?? \\ $$$${b}=…?? \\ $$$${c}=…?? \\ $$ Terms of Service…
Question Number 11283 by suci last updated on 19/Mar/17 $$\mathrm{2}{a}+{b}+\mathrm{4}{c}….\left(\mathrm{1}\right) \\ $$$${a}+\mathrm{3}{c}….\left(\mathrm{2}\right) \\ $$$$−\mathrm{3}{a}−\mathrm{4}{b}−{c}….\left(\mathrm{4}\right) \\ $$$${a}=..?? \\ $$$${b}=..?? \\ $$$${c}=..?? \\ $$ Terms of Service…
Question Number 76819 by Rio Michael last updated on 30/Dec/19 $$\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{foci}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ellipse} \\ $$$$\:\:\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{9}}\:+\:\frac{{y}^{\mathrm{2}} }{\mathrm{4}}\:=\:\mathrm{1}\:\mathrm{is} \\ $$$$\mathrm{A}.\:\left(\mathrm{4},\mathrm{0}\right) \\ $$$$\mathrm{B}.\:\left(\mathrm{9},\mathrm{0}\right) \\ $$$$\mathrm{C}.\:\left(\mathrm{5},\mathrm{0}\right) \\ $$$$\mathrm{D}.\:\left(\sqrt{\mathrm{5}}\:,\:\mathrm{0}\right) \\ $$…
Question Number 76817 by Rio Michael last updated on 30/Dec/19 $$\mathrm{A}\:\mathrm{compound}\:\mathrm{pendulum}\:\mathrm{ocsillates} \\ $$$$\mathrm{through}\:\mathrm{angles}\:\theta\:\mathrm{about}\:\mathrm{its}\:\mathrm{equilibrium} \\ $$$$\mathrm{position}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{8}{a}\theta^{\mathrm{2}} \:=\:\mathrm{9}{g}\:{cos}\theta,\:{a}>\mathrm{0}.\:\mathrm{its}\:\mathrm{period}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\mathrm{2}\pi\sqrt{\frac{\mathrm{8}{a}}{\mathrm{9}{g}}} \\ $$$$\mathrm{B}.\:\frac{\mathrm{3}\pi}{\mathrm{8}}\sqrt{\frac{{a}}{{g}}} \\ $$$$\mathrm{C}.\:\mathrm{2}\pi\sqrt{\frac{\mathrm{9}{g}}{\mathrm{8}{a}}} \\…
Question Number 142349 by mnjuly1970 last updated on 30/May/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{n}+\mathrm{2}\right)\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}^{{n}} }=? \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 142348 by mathdanisur last updated on 30/May/21 $$\frac{\sqrt[{\mathrm{6}}]{\mathrm{4}−\sqrt{\mathrm{15}}}}{\:\sqrt{\mathrm{4}−\sqrt{\mathrm{15}}}\:\centerdot\:\sqrt[{\mathrm{3}}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:=\:? \\ $$ Answered by bramlexs22 last updated on 30/May/21 $$\:\frac{\sqrt[{\mathrm{6}\:}]{\mathrm{4}−\sqrt{\mathrm{15}}}}{\:\sqrt[{\mathrm{6}\:}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{3}} }.\sqrt[{\mathrm{3}\:}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:= \\ $$$$\:\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{6}}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{2}} }.\sqrt[{\mathrm{3}\:}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:= \\…