Question Number 195435 by universe last updated on 02/Aug/23 $$\sqrt{\mathrm{5}+\mathrm{1}\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{8}+\mathrm{4}\sqrt{…}}}}}\:\:=\:\:\:? \\ $$ Answered by mr W last updated on 02/Aug/23 $$\mathrm{3}=\sqrt{\mathrm{9}}=\sqrt{\mathrm{5}+\mathrm{4}} \\ $$$$\mathrm{4}=\mathrm{1}×\mathrm{4}=\mathrm{1}×\sqrt{\mathrm{16}}=\mathrm{1}×\sqrt{\mathrm{6}+\mathrm{10}} \\ $$$$\mathrm{10}=\mathrm{2}×\mathrm{5}=\mathrm{2}×\sqrt{\mathrm{25}}=\mathrm{2}×\sqrt{\mathrm{7}+\mathrm{18}}…
Question Number 195428 by mathlove last updated on 02/Aug/23 $$\mathrm{3}^{{a}} =\mathrm{7}^{{b}} =\mathrm{441} \\ $$$$\frac{{ab}}{{a}+{b}}=? \\ $$ Answered by Frix last updated on 02/Aug/23 $$\mathrm{441}=\mathrm{3}^{\mathrm{2}} \mathrm{7}^{\mathrm{2}}…
Question Number 195404 by universe last updated on 01/Aug/23 Answered by AST last updated on 01/Aug/23 $$\left(\frac{\mathrm{127}^{\mathrm{2}} }{{xy}}+\frac{{x}}{{y}}+\frac{{y}}{{x}}\right)\left(\mathrm{127}+{x}+{y}\right) \\ $$$$=\frac{\mathrm{127}^{\mathrm{3}} }{{xy}}+\frac{\mathrm{127}^{\mathrm{2}} }{{y}}+\frac{\mathrm{127}^{\mathrm{2}} }{{x}}+\frac{\mathrm{127}{x}}{{y}}+\frac{{x}^{\mathrm{2}} }{{y}}+{x}+\frac{\mathrm{127}{y}}{{x}}+{y}+\frac{{y}^{\mathrm{2}} }{{x}}…
Question Number 195407 by mustafazaheen last updated on 01/Aug/23 $$\mathrm{f}'\left(\mathrm{2}\right)=\mathrm{g}'\left(\mathrm{2}\right)=\mathrm{g}\left(\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\left(\mathrm{fog}\right)'\left(\mathrm{2}\right)=? \\ $$ Answered by gatocomcirrose last updated on 01/Aug/23 $$\left(\mathrm{fog}\right)'\left(\mathrm{2}\right)=\mathrm{f}'\left(\mathrm{g}\left(\mathrm{2}\right)\right)\mathrm{g}'\left(\mathrm{2}\right)=\mathrm{2f}'\left(\mathrm{2}\right)=\mathrm{4} \\ $$ Answered…
Question Number 195391 by mathlove last updated on 01/Aug/23 $${f}^{\mathrm{2}} \left({x}\right)+\mathrm{2}{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{15} \\ $$$${f}\left({x}\right)=? \\ $$ Answered by MM42 last updated on 01/Aug/23 $$\left({f}+\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{2}}…
Question Number 195390 by mathlove last updated on 01/Aug/23 $${which}\:{prime}\:{number}\:{between} \\ $$$${the}\:\:\mathrm{20}\:\:{and}\:\:\:\:\mathrm{1000}\:\:\: \\ $$ Answered by BaliramKumar last updated on 01/Aug/23 $$\mathrm{1}\:\mathrm{to}\:\mathrm{1000}\:=\:\mathrm{168}\:\mathrm{prime}\:\mathrm{numbers} \\ $$$$\mathrm{1}\:\mathrm{to}\:\mathrm{20}\:=\:\mathrm{8}\:\mathrm{prime}\:\mathrm{numbers} \\…
Question Number 195401 by mnjuly1970 last updated on 01/Aug/23 $$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{calc}\underset{\underset{\Downarrow} {\Downarrow}} {\mathrm{u}late} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{lim}_{\:\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\left(\:\:\mathrm{sin}\left(\mathrm{x}\:\right)\right)^{\:\mathrm{tan}^{\:\mathrm{2}} \left(\mathrm{x}\:\right)} \:\:\:\:=\:?\:\:\:\:\:\:\:\:\begin{array}{|c|c|}\\\\\hline\end{array} \\ $$$$…
Question Number 195385 by Rodier97 last updated on 01/Aug/23 $$ \\ $$$$ \\ $$$$\mathrm{lim}_{{x}\Rightarrow\mathrm{a}^{+} } \:\:\:\:\frac{\sqrt{{x}}\:−\sqrt{\mathrm{a}}\:−\sqrt{{x}−\mathrm{a}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:\:;\:\:\mathrm{a}\:>\:\mathrm{0} \\ $$ Answered by cortano12 last updated…
Question Number 195393 by Erico last updated on 01/Aug/23 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{\:\:\boldsymbol{{x}}\:\:}]{\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)^{{x}} }{{n}}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\sqrt[{\boldsymbol{{n}}}]{\mathrm{C}_{\mathrm{2}\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{n}}} } \\ $$ Answered by witcher3 last updated…
Question Number 195395 by cortano12 last updated on 01/Aug/23 Answered by kapoorshah last updated on 01/Aug/23 $${Any}\:{number}\:{power}\:\mathrm{0}\:{is}\:\mathrm{1} \\ $$$$ \\ $$$$\left(\mathrm{1}\:+\:\infty\:+\:\infty\:+\:…\:+\:\infty\right)^{\mathrm{0}} \:=\:\mathrm{1} \\ $$ Terms…