Question Number 141111 by jlewis last updated on 15/May/21 $$\mathrm{A}\:\mathrm{gap}\:\mathrm{must}\:\mathrm{be}\:\mathrm{left}\:\mathrm{between}\:\mathrm{steel}\: \\ $$$$\mathrm{rails}\:\mathrm{to}\:\mathrm{allow}\:\mathrm{for}\:\mathrm{thermal}\:\mathrm{expansion}. \\ $$$$\:\mathrm{How}\:\mathrm{large}\:\mathrm{a}\:\mathrm{gap}\:\mathrm{is}\:\mathrm{needed}\:\mathrm{if}\:\mathrm{the}\: \\ $$$$\mathrm{maximum}\:\mathrm{temperature}\:\mathrm{reached} \\ $$$$\:\mathrm{is}\:\mathrm{50}°\:\mathrm{above}\:\mathrm{the}\:\mathrm{temperature}\:\mathrm{at}\: \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{rails}\:\mathrm{were}\:\mathrm{laid}.\:\mathrm{The}\: \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rail}\:\mathrm{is}\:\mathrm{10m}\:\mathrm{and} \\ $$$$\:\mathrm{the}\:\alpha_{\mathrm{steal}} =\mathrm{12}×\mathrm{10}^{−\mathrm{6}}…
Question Number 141110 by Fikret last updated on 15/May/21 $${sin}^{\mathrm{2}} {x}+{sinx}=\frac{\mathrm{4}}{\mathrm{3}}\:\:\:\:\:\:\:,\:\:\:{x}\in\left[\mathrm{0},\pi\right] \\ $$$${equation}\:\:{sum}\:\:{of}\:{roots}\:{in}\:{range}? \\ $$ Answered by MJS_new last updated on 15/May/21 $${s}^{\mathrm{2}} +{s}−\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{0}\wedge−\mathrm{1}\leqslant{s}\leqslant\mathrm{1}\:\Rightarrow\:{s}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{57}}}{\mathrm{6}} \\…
Question Number 75570 by naka3546 last updated on 12/Dec/19 Commented by naka3546 last updated on 12/Dec/19 $$\frac{\left[\:{blue}\:\:{area}\:\right]}{\left[\:{square}\:\:{area}\:\right]}\:\:=\:\:? \\ $$ Answered by mr W last updated…
Question Number 10035 by Sri rama jeyam last updated on 21/Jan/17 $$\mathrm{is}\:\mathrm{cos}\left(\mathrm{7x}+\mathrm{i5y}\right)\:\mathrm{analytic}\:\mathrm{function}\:\mathrm{or}\:\mathrm{not} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141104 by Opredador last updated on 15/May/21 Answered by hknkrc46 last updated on 15/May/21 $$\left.\begin{matrix}{\sqrt{\mathrm{7}^{\sqrt{\mathrm{63}}} }\:=\:\sqrt{\mathrm{7}^{\mathrm{3}\sqrt{\mathrm{7}}} }\:=\:\mathrm{7}^{\sqrt{\mathrm{7}}} \sqrt{\mathrm{7}^{\sqrt{\mathrm{7}}} }}\\{\mathrm{7}^{\sqrt{\mathrm{7}}} \:=\:\boldsymbol{{u}}}\end{matrix}\right\}\:\frac{\boldsymbol{{u}}\sqrt{\boldsymbol{{u}}}\:−\:\sqrt{\boldsymbol{{u}}}}{\boldsymbol{{u}}\:−\:\mathrm{1}} \\ $$$$=\:\frac{\sqrt{\boldsymbol{{u}}}\left(\boldsymbol{{u}}\:−\:\mathrm{1}\right)}{\boldsymbol{{u}}\:−\:\mathrm{1}}\:=\:\sqrt{\boldsymbol{{u}}}\:=\:\sqrt{\mathrm{7}^{\sqrt{\mathrm{7}}} }…
Question Number 10034 by 25227 last updated on 21/Jan/17 $${cos}\left({z}\right)\:{is}\:{analytic}\:{or}\:{not}\:{explain}\:{me} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141107 by pticantor last updated on 15/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 10033 by ridwan balatif last updated on 21/Jan/17 Answered by mrW1 last updated on 21/Jan/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{cos}\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{cos}\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{cos}^{\mathrm{2}}…
Question Number 75569 by naka3546 last updated on 12/Dec/19 Commented by MJS last updated on 13/Dec/19 $$\mathrm{was}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:\mathrm{one}\:\mathrm{ok}? \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{understand}… \\ $$ Commented by naka3546 last…
Question Number 10031 by ridwan balatif last updated on 21/Jan/17 Commented by ridwan balatif last updated on 21/Jan/17 Terms of Service Privacy Policy Contact: info@tinkutara.com