Question Number 209098 by Spillover last updated on 01/Jul/24 Answered by A5T last updated on 02/Jul/24 $$\mathrm{4}=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{\mathrm{2}{g}}\Rightarrow{sin}\theta=\frac{\sqrt{\mathrm{8}{g}}}{{v}_{\mathrm{0}} } \\ $$$${R}={v}_{\mathrm{0}} {cos}\theta×\mathrm{2}{t}={v}_{\mathrm{0}} \sqrt{\mathrm{1}−\frac{\mathrm{8}{g}}{{v}_{\mathrm{0}}…
Question Number 209066 by Tawa11 last updated on 01/Jul/24 Commented by Tawa11 last updated on 01/Jul/24 Answered by A5T last updated on 01/Jul/24 $${cos}\mathrm{48}°=\frac{{XA}}{\mathrm{19}}\Rightarrow{XA}=\mathrm{19}{cos}\mathrm{48}° \\…
Question Number 209099 by Spillover last updated on 01/Jul/24 Answered by mr W last updated on 02/Jul/24 $${say}\:{particle}\:\mathrm{2}\:{starts}\:{time}\:{T}\:\:{later} \\ $$$${than}\:{particle}\:\mathrm{1}.\: \\ $$$$ \\ $$$${particle}\:\mathrm{2}\:{at}\:{time}\:{t}: \\…
Question Number 209062 by hardmath last updated on 01/Jul/24 Answered by som(math1967) last updated on 01/Jul/24 $$\:{x}^{{y}^{{z}} } =\mathrm{1} \\ $$$$\:{x},{y},{z}\:\in{N}\:\Rightarrow\mathrm{1}^{{y}^{{z}} } =\mathrm{1}\:\Rightarrow{x}=\mathrm{1} \\ $$$$\:{y}^{{z}^{{x}}…
Question Number 209059 by hardmath last updated on 01/Jul/24 $$\mathrm{Compare}: \\ $$$$\mathrm{8}!\:\:\:\mathrm{and}\:\:\:\mathrm{8}!! \\ $$ Commented by mr W last updated on 01/Jul/24 $${if}\:\frac{{a}}{{b}}=\mathrm{1}\:\Rightarrow{a}={b} \\ $$$$\frac{\cancel{\mathrm{8}!}!}{\cancel{\mathrm{8}!}}=\mathrm{1}!=\mathrm{1}\:\Rightarrow\mathrm{8}!!=\mathrm{8}!\:\:\:\:\:\underset{\mid<>\mid}…
Question Number 209021 by hardmath last updated on 30/Jun/24 k part, that is, l > a1+a2+…+ak is given, that is, we have enough seats. ”…
Question Number 209055 by hardmath last updated on 30/Jun/24 Commented by hardmath last updated on 30/Jun/24 $$\mathrm{dear}\:\mathrm{professor}\:\mathrm{mr}\:\mathrm{W} \\ $$ Commented by mr W last updated…
Question Number 209023 by Spillover last updated on 30/Jun/24 Commented by Spillover last updated on 01/Jul/24 $${let}\:{u}=\frac{\mathrm{4}{x}}{\mathrm{1}+\mathrm{5}{x}}\:\:\:\:\:\:\frac{{du}}{{dx}}=\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{5}{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{d}}{{dx}}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}{x}}{\mathrm{1}+\mathrm{5}{x}}\right)=\frac{\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{5}{x}\right)^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{\mathrm{4}{x}}{\mathrm{1}+\mathrm{5}{x}}\right)^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{1}+\mathrm{10}{x}+\mathrm{25}{x}^{\mathrm{2}} } \\…
Question Number 209016 by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 02/Jul/24 Answered by Spillover last updated on 02/Jul/24 Terms of…
Question Number 208980 by lmcp1203 last updated on 30/Jun/24 $${please}\:.\:\:\:\:\:{find}\:\:\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} } {mod}\:\mathrm{23}\:\:\:\:\:\:\:\:{thanks}. \\ $$ Answered by A5T last updated on 30/Jun/24 $$\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} } {mod}\left(\mathrm{23}\right)\equiv\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} \left[{mod}\:\phi\left(\mathrm{23}\right)\right]}…