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Category: Algebra

Question-172262

Question Number 172262 by Mikenice last updated on 25/Jun/22 Answered by Rasheed.Sindhi last updated on 25/Jun/22 $$\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}…}}}\:=\sqrt{{n}}\:>\mathrm{0} \\ $$$$\left(\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}…}}}\:\right)^{\mathrm{3}} =\left(\sqrt{{n}}\:\right)^{\mathrm{3}} \\ $$$$\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}…}}\:=\left(\sqrt{{n}}\:\right)^{\mathrm{3}} \\ $$$$\mathrm{3}\sqrt{{n}}=\left(\sqrt{{n}}\:\right)^{\mathrm{3}} \\…

The-LCM-and-the-GCF-of-three-intergers-are-180-and-3-respectively-Two-numbers-are-45-and-60-What-is-the-third-number-

Question Number 41167 by Cheyboy last updated on 02/Aug/18 $${The}\:{LCM}\:{and}\:{the}\:{GCF}\:{of}\:{three} \\ $$$${intergers}\:{are}\:\mathrm{180}\:{and}\:\mathrm{3}\:{respectively}. \\ $$$${Two}\:{numbers}\:{are}\:\mathrm{45}\:{and}\:\mathrm{60}. \\ $$$${What}\:{is}\:{the}\:{third}\:{number} \\ $$ Answered by MJS last updated on 02/Aug/18…

If-in-ABC-m-BAC-gt-90-then-cosA-cosB-cosC-F-aR-

Question Number 172214 by Shrinava last updated on 24/Jun/22 $$\mathrm{If}\:\:\mathrm{in}\:\:\bigtriangleup\mathrm{ABC}\:,\:\mathrm{m}\left(\sphericalangle\mathrm{BAC}\right)>\mathrm{90}°\:\:\mathrm{then}: \\ $$$$\mathrm{cosA}\:+\:\mathrm{cosB}\:\mathrm{cosC}\:=\:\frac{\mathrm{F}}{\mathrm{aR}} \\ $$ Commented by mr W last updated on 24/Jun/22 $${please}\:{check}!\:{maybe}\:{you}\:{meant} \\ $$$$\mathrm{cosA}\:+\:\mathrm{cosB}\:\mathrm{cosC}\:=\:\frac{{aR}}{{F}}…

Question-106655

Question Number 106655 by Algoritm last updated on 06/Aug/20 Commented by prakash jain last updated on 06/Aug/20 $${x}_{\mathrm{1}} ={kx}_{\mathrm{2}} \\ $$$${k}^{\mathrm{9}} =−\mathrm{1}\Rightarrow{k}=−\mathrm{1}\left({e}^{{i}\mathrm{2}\pi{m}/\mathrm{9}} \right)\:\:\left({m}=\mathrm{0}..\mathrm{8}\right)\:\left({I}\right) \\ $$$${k}^{\mathrm{15}}…

Question-172186

Question Number 172186 by Mikenice last updated on 23/Jun/22 Commented by kaivan.ahmadi last updated on 24/Jun/22 $${A}={x}^{{x}^{{x}^{….} } } \\ $$$${li}\underset{{x}\rightarrow\mathrm{1}} {{m}}\frac{{lnxA}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:\overset{{Hop}} {=}\:\:\:{li}\underset{{x}\rightarrow\mathrm{1}} {{m}}\:\frac{\frac{\mathrm{1}}{{x}}{A}+{lnx}.{A}'}{\mathrm{2}{x}}=…