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Question-213615

Question Number 213615 by Spillover last updated on 10/Nov/24 Answered by Frix last updated on 10/Nov/24 $$\mathrm{The}\:\mathrm{photon}. \\ $$$$\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{light}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{in}\:\mathrm{every} \\ $$$$\mathrm{reference}\:\mathrm{frame}.\:\mathrm{This}\:\mathrm{means}\:\mathrm{no}\:\mathrm{matter}\:\mathrm{how} \\ $$$$\mathrm{fast}\:\mathrm{the}\:\mathrm{observer}\:\mathrm{moves},\:\mathrm{the}\:\mathrm{photons}\:\mathrm{are} \\ $$$$\mathrm{always}\:\mathrm{moving}\:\mathrm{away}\:\mathrm{from}\:\mathrm{him}\:\mathrm{with}\:\mathrm{about}…

show-that-C-e-z-3-dz-0-where-C-is-any-simple-closed-contour-Evaluate-the-integral-C-1-f-z-dz-C-2-f-z-dz-where-f-z-y-x-3x-2-i-C-3-OA-z-y-x-iy-iy-0-y-1-C-1-AB-z-x

Question Number 213604 by issac last updated on 10/Nov/24 $$\mathrm{show}\:\mathrm{that}\:\:\int_{\boldsymbol{\mathcal{C}}} \:{e}^{{z}^{\mathrm{3}} } \:\mathrm{d}{z}=\mathrm{0} \\ $$$$\mathrm{where}\:\boldsymbol{\mathcal{C}}\:\mathrm{is}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}. \\ $$$$\:\:\:\: \\ $$$$\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\int_{\:{C}_{\mathrm{1}} } {f}\left({z}\right)\mathrm{d}{z}\:,\:\int_{\:{C}_{\mathrm{2}} } {f}\left({z}\right)\mathrm{d}{z}…

ABC-2a-b-2c-Find-the-minimum-of-3-sin-C-1-tan-A-

Question Number 213606 by CrispyXYZ last updated on 10/Nov/24 $$\bigtriangleup{ABC}.\:\mathrm{2}{a}+{b}=\mathrm{2}{c}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of} \\ $$$$\frac{\mathrm{3}}{\mathrm{sin}\:{C}}\:+\:\frac{\mathrm{1}}{\mathrm{tan}\:{A}}. \\ $$ Answered by Ghisom last updated on 10/Nov/24 $$\mathrm{wlog}\:{b}=\mathrm{1}\:\Rightarrow\:{c}={a}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}}…

Find-3-3-4-3-3-5-3-1-2-3-3-7-3-1-6-27-5-

Question Number 213589 by hardmath last updated on 09/Nov/24 $$\mathrm{Find}: \\ $$$$\frac{\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{4}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{5}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{1}}{\mathrm{2}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{7}}\right)\centerdot…\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{27}^{\mathrm{5}} }\:=\:? \\ $$ Answered by issac last updated on 09/Nov/24 $$\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{15}} \underset{{h}=\mathrm{1}} {\overset{\mathrm{15}}…