Question Number 210208 by Spillover last updated on 02/Aug/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{{e}^{{e}^{{x}} } } \:{e}^{{e}^{{x}} } \:{e}^{{x}} {dx} \\ $$$$ \\ $$ Answered…
Question Number 210194 by Jubr last updated on 02/Aug/24 Commented by Jubr last updated on 02/Aug/24 $${The}\:{numbers}\:{are}\:\mathrm{1}.\mathrm{3}{m}\:{and}\:\mathrm{1}.\mathrm{2}{m} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 210172 by efronzo1 last updated on 01/Aug/24 Answered by a.lgnaoui last updated on 02/Aug/24 $$\frac{\mathrm{sin}\:\mathrm{2}\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{BE}}}=\frac{\mathrm{cos}\:\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{AB}}}\:\:\:\:\left(\measuredangle\mathrm{AED}=\mathrm{90}−\mathrm{2a}\right) \\ $$$$\mathrm{AB}=\frac{\mathrm{DEcos}\:\boldsymbol{\mathrm{a}}}{\mathrm{sin}\:\mathrm{2}\boldsymbol{\mathrm{a}}}.=\frac{\boldsymbol{\mathrm{BE}}}{\mathrm{2sin}\:\boldsymbol{\mathrm{a}}} \\ $$$$\boldsymbol{\mathrm{Aire}}\left(\boldsymbol{\mathrm{ABC}}\right)=\boldsymbol{\mathrm{Air}}\mathrm{e}\left(\boldsymbol{\mathrm{ADFC}}\right)+\boldsymbol{\mathrm{Aire}}\left(\boldsymbol{\mathrm{DBF}}\right) \\ $$$$\boldsymbol{\mathrm{A}}\mathrm{ire}\left(\mathrm{BDE}\right)=\mathrm{Aire}\left(\boldsymbol{\mathrm{BEF}}\right)+\boldsymbol{\mathrm{Aire}}\left(\boldsymbol{\mathrm{BDF}}\right) \\ $$$$\Rightarrow\:…
Question Number 210156 by Spillover last updated on 01/Aug/24 Answered by A5T last updated on 01/Aug/24 $$\left({i}\right)+\mathrm{2}×\left({ii}\right)\Rightarrow{x}=\frac{\mathrm{8}−\mathrm{7}{z}}{\mathrm{7}}=\frac{\mathrm{8}}{\mathrm{7}}−{z} \\ $$$$\left({iii}\right)+\left({i}\right)\Rightarrow\mathrm{7}{x}+\left({a}^{\mathrm{2}} −\mathrm{9}\right){z}={a}+\mathrm{4} \\ $$$$\Rightarrow{x}=\frac{−\left({a}^{\mathrm{2}} −\mathrm{9}\right){z}+{a}+\mathrm{4}}{\mathrm{7}} \\ $$$$\Rightarrow\frac{\mathrm{4}−{a}}{\mathrm{7}}=\frac{\left(\mathrm{16}−{a}^{\mathrm{2}}…
Question Number 210157 by Spillover last updated on 01/Aug/24 Answered by A5T last updated on 01/Aug/24 $${Let}\:{distance}\:{of}\:{vertex},{V},\:{to}\:{centroid},{G},\:{be}\:{GV} \\ $$$$\Rightarrow\frac{{sin}\mathrm{30}°}{{GV}}=\frac{{sin}\mathrm{120}^{°} }{{x}}\Rightarrow{GV}=\frac{\frac{{x}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${H}=\sqrt{{x}^{\mathrm{2}} −{GV}^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{2}}…
Question Number 210142 by Abdullahrussell last updated on 01/Aug/24 Commented by Frix last updated on 02/Aug/24 $$\mathrm{I}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:“\mathrm{nice}''\:\mathrm{solution}: \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i}\:\:\:\:\:{y}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i}\:\:\:\:\:{z}=\mathrm{3} \\ $$$$\Rightarrow\:{x}+{y}+{z}=\mathrm{6} \\ $$$$\mathrm{But}\:\mathrm{there}\:\mathrm{should}\:\mathrm{be}\:\mathrm{more}\:\mathrm{solutions}\:\mathrm{with} \\ $$$${x}+{y}+{z}\in\mathbb{R}…
Question Number 210171 by hardmath last updated on 01/Aug/24 $$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow+\infty} {\mathrm{lim}}\:\:\frac{\mathrm{n}}{\left(\mathrm{n}!\right)^{\mathrm{2}} \:\mathrm{4}^{\boldsymbol{\mathrm{n}}} }\:\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\:\left(\left(\mathrm{2k}−\mathrm{1}\right)^{\mathrm{2}} \:+\:\mathrm{4}\right)\:=\:? \\ $$ Terms of Service Privacy Policy…
Question Number 210155 by Spillover last updated on 01/Aug/24 Answered by A5T last updated on 01/Aug/24 Commented by A5T last updated on 01/Aug/24 $${C}\:{is}\:{the}\:{origin}\:{and}\:{redundant} \\…
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Question Number 210126 by mnjuly1970 last updated on 31/Jul/24 Answered by Frix last updated on 31/Jul/24 $$\mathrm{These}\:\mathrm{substitutions}\:\mathrm{make}\:\mathrm{it}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see} \\ $$$$\mathrm{what}'\mathrm{s}\:\mathrm{going}\:\mathrm{on}: \\ $$$$ \\ $$$$\mathrm{Let}\:{x}=\mathrm{sin}\:\alpha\:\overset{\left[\mathrm{differentiate}\right]} {\Rightarrow} \\…