Question Number 11334 by uni last updated on 20/Mar/17 $$\mathrm{z}+\mid\mathrm{z}\mid=\mathrm{9}−\mathrm{3i}\Rightarrow\mathrm{Re}\left(\mathrm{z}\right) \\ $$ Answered by mrW1 last updated on 20/Mar/17 $${z}={a}+{bi} \\ $$$$\mid{z}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\…
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Question Number 142379 by Rexzie last updated on 30/May/21 $${Show}\:{that}\:\mathrm{1}+\mathrm{3}{n}<{n}^{\mathrm{2}} \:{for}\:{every}\:{positive}\:{integer}\:{n}\geqslant\mathrm{4} \\ $$ Commented by mr W last updated on 30/May/21 $${n}\geqslant\mathrm{4} \\ $$$${n}^{\mathrm{2}} \geqslant\mathrm{4}{n}=\mathrm{3}{n}+{n}\geqslant\mathrm{3}{n}+\mathrm{4}>\mathrm{3}{n}+\mathrm{1}…
Question Number 11302 by uni last updated on 19/Mar/17 $$\frac{\mathrm{sin10x}−\mathrm{sin6x}−\mathrm{sin2x}}{\mathrm{sin9x}−\mathrm{sin7x}−\mathrm{sinx}}=? \\ $$ Answered by sandy_suhendra last updated on 20/Mar/17 $$=\frac{\left(\mathrm{sin10x}−\mathrm{sin6x}\right)−\mathrm{sin2x}}{\left(\mathrm{sin9x}−\mathrm{sin7x}\right)−\mathrm{sinx}} \\ $$$$=\frac{\mathrm{2cos8xsin2x}−\mathrm{sin2x}}{\mathrm{2cos8xsinx}−\mathrm{sinx}} \\ $$$$=\frac{\mathrm{sin2x}\left(\mathrm{2cos8x}−\mathrm{1}\right)}{\mathrm{sinx}\left(\mathrm{2cos8x}−\mathrm{1}\right)} \\…
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Question Number 11285 by uni last updated on 19/Mar/17 $$\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}} \\ $$$$=\frac{{sin}\mathrm{20}×\mathrm{cos}\:\mathrm{30}}{\mathrm{cos}\:\mathrm{30}×{cos}\mathrm{80}−\mathrm{sin}\:\mathrm{30}×{sin}\mathrm{80}} \\ $$$$=\frac{{sin}\mathrm{20}×\mathrm{cos}\:\mathrm{30}}{\mathrm{cos}\:\mathrm{110}} \\…
Question Number 142348 by mathdanisur last updated on 30/May/21 $$\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}}}}\:= \\ $$$$\:\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{6}}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{2}} }.\sqrt[{\mathrm{3}\:}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:= \\…
Question Number 11274 by uni last updated on 18/Mar/17 $$\frac{{sin}\mathrm{20}+\sqrt{\mathrm{3}}×{cos}\mathrm{20}}{{cos}\mathrm{10}}=? \\ $$ Answered by bahmanfeshki1 last updated on 18/Mar/17 $${sin}\:\mathrm{20}+\sqrt{\mathrm{3}}×{cos}\mathrm{20}=\mathrm{2}×{cos}\left(\mathrm{30}−\mathrm{20}\right)=\mathrm{2}×{cos}\mathrm{10} \\ $$$$\Rightarrow\frac{\mathrm{2}{cos}\mathrm{10}}{{cos}\mathrm{10}}=\mathrm{2} \\ $$ Answered…