Question Number 204664 by cortano12 last updated on 25/Feb/24 $$\:\:\mathrm{8}+\sqrt{\mathrm{8}^{\mathrm{2}} +\sqrt{\mathrm{8}^{\mathrm{4}} +\sqrt{\mathrm{8}^{\mathrm{8}} +\sqrt{\mathrm{8}^{\mathrm{16}} +\sqrt{…}}}}}\:=\:?\: \\ $$ Answered by Frix last updated on 25/Feb/24 $$=\mathrm{8}+\mathrm{8}\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+…}}}} \\…
Question Number 204666 by Ghisom last updated on 25/Feb/24 $$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{curve}: \\ $$$${f}\left(\theta\right)=\left(\mathrm{e}^{\mathrm{i}\theta} \right)^{\left(\mathrm{e}^{\mathrm{i}\theta} \right)} =\mathrm{e}^{−\theta\mathrm{sin}\:\theta} \mathrm{e}^{\mathrm{i}\theta\mathrm{cos}\:\theta} ;\:−\pi<\theta\leqslant\pi \\ $$$${f}:\:\begin{cases}{{x}\left(\theta\right)=\mathrm{e}^{−\theta\mathrm{sin}\:\theta} \mathrm{cos}\:\left(\theta\mathrm{cos}\:\theta\right)}\\{{y}\left(\theta\right)=\mathrm{e}^{−\theta\mathrm{sin}\:\theta} \mathrm{sin}\:\left(\theta\mathrm{cos}\:\theta\right)}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{area} \\ $$…
Question Number 204699 by AtulKumar last updated on 25/Feb/24 Commented by LuisTony last updated on 25/Feb/24 $${In}\:{spanish}? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 204688 by Davidtim last updated on 25/Feb/24 $${f}\left({x}\right)={sgn}\left({x}\right);\:\:\:\:\:{f}^{'} \left({x}\right)=\frac{{d}}{{dx}}\left[{f}\left({x}\right)\right]=? \\ $$ Answered by Faetmaaa last updated on 27/Feb/24 $$\frac{\mathrm{d}}{\mathrm{d}{x}}\left[{f}\mid_{\left.\right]−\infty,\:\mathrm{0}\left[\right.} \left({x}\right)\right]\:=\:\frac{\mathrm{d}}{\mathrm{d}{x}}\left[{f}\mid_{\left.\right]\mathrm{0},\:+\infty\left[\right.} \left({x}\right)\right]\:=\:\mathrm{0} \\ $$…
Question Number 204689 by Davidtim last updated on 25/Feb/24 $${y}=\mid{f}\left({x}\right)\mid\:\:;\:\:\:\:\frac{{dy}}{{dx}}=? \\ $$ Answered by A5T last updated on 25/Feb/24 $${y}=\sqrt{\left({f}\left({x}\right)\right)^{\mathrm{2}} }=\left[\left({f}\left({x}\right)\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\left({f}\left({x}\right)\right)^{\mathrm{2}} }}×\mathrm{2}{f}\left({x}\right)×{f}'\left({x}\right)=\frac{{f}\left({x}\right){f}^{'}…
Question Number 204690 by Davidtim last updated on 25/Feb/24 $$\sqrt[{{p}}]{{a}^{{x}} \boldsymbol{\div}\sqrt[{{q}}]{{b}^{{y}} \boldsymbol{\div}\sqrt[{{r}}]{{c}^{{z}} }}}=? \\ $$ Answered by A5T last updated on 25/Feb/24 $$\left(\frac{{a}^{{x}} }{\left(\frac{{b}^{{y}} }{{c}^{\frac{{z}}{{r}}}…
Question Number 204691 by Davidtim last updated on 25/Feb/24 $$\sqrt{{a}\boldsymbol{\div}\sqrt{{a}\boldsymbol{\div}\sqrt{{a}\boldsymbol{\div}\centerdot\centerdot\centerdot\boldsymbol{\div}\sqrt{{a}}}}}=? \\ $$ Answered by Faetmaaa last updated on 27/Feb/24 $$\forall{a}\in\boldsymbol{\mathrm{R}}\backslash\left\{\mathrm{0}\right\} \\ $$$${x}\::=\:\sqrt{\frac{{a}}{\:\sqrt{\frac{{a}}{\:\sqrt{\frac{{a}}{\ldots}}}}}} \\ $$$$\Rightarrow{x}=\sqrt{\frac{{a}}{{x}}} \\…
Question Number 204684 by MrGHK last updated on 25/Feb/24 Commented by mr W last updated on 25/Feb/24 $$\Rightarrow{Q}\mathrm{204522} \\ $$ Commented by Davidtim last updated…
Question Number 204686 by Davidtim last updated on 25/Feb/24 $$\sqrt{{a}−\sqrt{{a}−\sqrt{{a}−\centerdot\centerdot\centerdot}}}=? \\ $$ Commented by A5T last updated on 25/Feb/24 $${If}\:\sqrt{{a}−\sqrt{{a}−\sqrt{{a}−…}}}\:{converges},\:{then}\:\sqrt{{a}−{x}}={x} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}−{a}=\mathrm{0}\Rightarrow{x}=\frac{−\mathrm{1}\underset{−} {+}\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}} \\…
Question Number 204687 by Davidtim last updated on 25/Feb/24 $$\sqrt{{a}−\sqrt{{b}−\sqrt{{c}}}}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com