Question Number 58896 by cesar.marval.larez@gmail.com last updated on 03/May/19 $$\left.\mathrm{11}\right)\:\:\mathrm{lg}_{\mathrm{4}} \mathrm{lg}_{\mathrm{4}} \mathrm{lg}_{\mathrm{2}} \mathrm{16}−\mathrm{lg}_{\mathrm{2}} \mathrm{lg}_{\mathrm{2}} \sqrt{\mathrm{3}} \\ $$$$\mathrm{12}.\:\left(\mathrm{5lg}_{\mathrm{3}} \mathrm{3}−\mathrm{lg}_{\mathrm{4}} \mathrm{1}\right)^{\mathrm{2}} +\frac{\frac{\mathrm{1}}{\mathrm{lg}_{\mathrm{2}} \mathrm{8}}×\mathrm{lg}_{\mathrm{3}} \mathrm{27}}{\mathrm{lg}_{\sqrt{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}}} \\ $$…
Question Number 189952 by SAMIRA last updated on 25/Mar/23 $$\sqrt{\mathrm{2}}\:\mathrm{sin}\left(\mathrm{x}\right)\:+\mathrm{cos}\left(\mathrm{x}\right)=\:−\sqrt{\mathrm{2}}\: \\ $$ Answered by cortano12 last updated on 25/Mar/23 $$\:\sqrt{\mathrm{3}}\:\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\mathrm{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{cos}\:\mathrm{x}\right)=−\sqrt{\mathrm{2}} \\ $$$$\:\sqrt{\mathrm{3}}\:\left(\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\theta+\mathrm{cos}\:\mathrm{x}\:\mathrm{sin}\:\theta\right)=−\sqrt{\mathrm{2}} \\ $$$$\:\:\mathrm{sin}\:\left(\mathrm{x}+\theta\right)=−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\:=\mathrm{sin}\:\left(\mathrm{180}°+\alpha\right) \\…
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Question Number 58853 by cesar.marval.larez@gmail.com last updated on 30/Apr/19 Answered by MJS last updated on 01/May/19 $$\frac{\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}^{\mathrm{3}} \:\frac{\pi}{\mathrm{2}}\:+\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{3}}}{\mathrm{tan}\:\frac{\pi}{\mathrm{3}}\:+\mathrm{cos}\:−\mathrm{2}\pi}=\frac{\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{1}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$ Terms of Service Privacy…
Question Number 124387 by liberty last updated on 03/Dec/20 $$\:{Solve}\:{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{7}={x}−\mathrm{3}\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}} \\ $$$${for}\:{x}\epsilon{R}\:. \\ $$ Answered by benjo_mathlover last updated on 03/Dec/20 Terms of…
Question Number 58851 by hhghg last updated on 30/Apr/19 $$\mathrm{6}+\mathrm{2}×\mathrm{3} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 189916 by mustafazaheen last updated on 24/Mar/23 $$\mathrm{when}\:\:\:\:\mathrm{sinx}×\mathrm{cosx}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{find}\:\:\:\:\:\:\:\mathrm{sinx}+\mathrm{cosx}=? \\ $$ Answered by aminitindas last updated on 24/Mar/23 $$ \\ $$$$\boldsymbol{\mathrm{Ans}}: \\…
Question Number 189919 by horsebrand11 last updated on 24/Mar/23 $$\:{If}\:{x},{y}\:{are}\:{real}\:{numbers}\:{satisfy} \\ $$$$\:\frac{{x}+\mathrm{40}}{{y}}+\frac{\mathrm{569}}{{xy}}=\frac{\mathrm{26}−{y}}{{x}}\:,\:{then}\:{xy}=? \\ $$ Answered by cortano12 last updated on 24/Mar/23 $$\Rightarrow\left(\mathrm{x}+\mathrm{20}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{13}\right)^{\mathrm{2}} =\mathrm{0} \\…
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Question Number 189886 by mnjuly1970 last updated on 23/Mar/23 $$ \\ $$$$\:\:{e}^{\:{x}} \:>\:\mathrm{1}+\:{x}\:\:\:\:\left(\forall\:{x}\:>\mathrm{0}\:\right) \\ $$$$\:\:\:{set}:\:{x}=\sqrt{\frac{\pi}{{e}}}\:−\mathrm{1} \\ $$$$\:\:\:\:{e}^{\:\frac{\sqrt{\pi}}{\:\sqrt{{e}}}\:−\mathrm{1}} >\:\frac{\sqrt{\pi}}{\:\sqrt{{e}}}\:\Rightarrow\:{e}^{\:\frac{\sqrt{\pi}}{\:\sqrt{{e}}}} \:>\:\sqrt{\pi}\: \\ $$$$\:\:\:\:\:\:\left(\:{e}^{\:\frac{\sqrt{\pi}}{\:\sqrt{{e}}}} \right)^{\:\sqrt{{e}}} >\:\sqrt{\pi}\:^{\:\sqrt{{e}}} \:\Rightarrow\:{e}^{\:\sqrt{\pi}} \:>\:\sqrt{\pi\:}\:^{\:\sqrt{{e}}}…