Question Number 213632 by Jamaicalasigan last updated on 11/Nov/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 213628 by Thomaseinstein last updated on 10/Nov/24 Answered by MrGaster last updated on 03/Feb/25 $$\left.=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{4}} \mathrm{sin}\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} \mathrm{cos}\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)\right)\right) \\ $$ Terms of…
Question Number 213624 by hardmath last updated on 10/Nov/24 $$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{5x}\:−\:\mathrm{42}\:=\:\left(\mathrm{x}\:−\:\mathrm{3}\right)\centerdot\mathrm{P}\left(\mathrm{x}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{P}\left(\mathrm{3}\right)\:=\:? \\ $$ Answered by A5T last updated on 10/Nov/24 $$\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{14}\right)−\left({x}−\mathrm{3}\right){P}\left({x}\right)=\mathrm{0} \\…
Question Number 213626 by Tawa11 last updated on 10/Nov/24 Commented by Tawa11 last updated on 10/Nov/24 The area of ABCD is 40.5 square units .…
Question Number 213621 by hardmath last updated on 10/Nov/24 $$\mathrm{a},\mathrm{b}>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:=\:\mathrm{2}\:\:\:\mathrm{and}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{12} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:=\:? \\ $$ Answered by Ghisom last updated on 10/Nov/24…
Question Number 213613 by Spillover last updated on 10/Nov/24 Answered by Frix last updated on 10/Nov/24 $$\mathrm{The}\:\mathrm{universe}\:\mathrm{is}\:\mathrm{all}\:\mathrm{that}\:\mathrm{there}\:\mathrm{is},\:\mathrm{it}'\mathrm{s}\:\mathrm{simply} \\ $$$$\mathrm{expanding}\:\mathrm{into}\:\mathrm{itself}. \\ $$$$\mathrm{We}\:\mathrm{cannot}\:\mathrm{find}\:\mathrm{analogies}\:\mathrm{to}\:\mathrm{this},\:\mathrm{the}\:\mathrm{picture} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{balloon}\:\mathrm{isn}'\mathrm{t}\:\mathrm{accurate}. \\ $$…
Question Number 213614 by Spillover last updated on 10/Nov/24 Commented by Frix last updated on 10/Nov/24 $$\mathrm{A}\:\mathrm{photon}\:\mathrm{is}\:\mathrm{a}\:\mathrm{quantum}\:\mathrm{particle}. \\ $$$$\mathrm{We}\:\mathrm{cannot}\:\mathrm{measure}\:\mathrm{its}\:\mathrm{location}\:\mathrm{because} \\ $$$$\mathrm{we}\:\mathrm{exactly}\:\mathrm{know}\:\left(\mathrm{measured}\right)\:\mathrm{its}\:\mathrm{speed}. \\ $$$$\mathrm{Its}\:\mathrm{speed}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{in}\:\mathrm{every}\:\mathrm{reference} \\ $$$$\mathrm{frame}.\:\mathrm{You}\:\mathrm{cannot}\:\mathrm{stop}\:\mathrm{it}\:\mathrm{and}\:\mathrm{say}\:“\mathrm{here}…
Question Number 213615 by Spillover last updated on 10/Nov/24 Answered by Frix last updated on 10/Nov/24 $$\mathrm{The}\:\mathrm{photon}. \\ $$$$\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{light}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{in}\:\mathrm{every} \\ $$$$\mathrm{reference}\:\mathrm{frame}.\:\mathrm{This}\:\mathrm{means}\:\mathrm{no}\:\mathrm{matter}\:\mathrm{how} \\ $$$$\mathrm{fast}\:\mathrm{the}\:\mathrm{observer}\:\mathrm{moves},\:\mathrm{the}\:\mathrm{photons}\:\mathrm{are} \\ $$$$\mathrm{always}\:\mathrm{moving}\:\mathrm{away}\:\mathrm{from}\:\mathrm{him}\:\mathrm{with}\:\mathrm{about}…
Question Number 213604 by issac last updated on 10/Nov/24 $$\mathrm{show}\:\mathrm{that}\:\:\int_{\boldsymbol{\mathcal{C}}} \:{e}^{{z}^{\mathrm{3}} } \:\mathrm{d}{z}=\mathrm{0} \\ $$$$\mathrm{where}\:\boldsymbol{\mathcal{C}}\:\mathrm{is}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}. \\ $$$$\:\:\:\: \\ $$$$\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\int_{\:{C}_{\mathrm{1}} } {f}\left({z}\right)\mathrm{d}{z}\:,\:\int_{\:{C}_{\mathrm{2}} } {f}\left({z}\right)\mathrm{d}{z}…
Question Number 213606 by CrispyXYZ last updated on 10/Nov/24 $$\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}}…