Question Number 208951 by hardmath last updated on 28/Jun/24 $$\mathrm{2}^{\mathrm{2024}} \::\:\mathrm{2024}\:=\:…\:\left(\mathrm{Remainder}\:=\:?\right) \\ $$ Answered by A5T last updated on 28/Jun/24 $$\left(\mathrm{2}^{\mathrm{11}} \right)^{\mathrm{184}} \:\:\overset{\mathrm{2024}} {\equiv}\:\mathrm{24}^{\mathrm{184}} \\…
Question Number 208945 by hardmath last updated on 28/Jun/24 $$\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{x}\:−\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{3x}\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{3x}\:=\:\mathrm{0} \\ $$$$\left[\:\mathrm{0}\:;\:\frac{\pi}{\mathrm{2}}\:\right] \\ $$$$\mathrm{Find}:\:\:\mathrm{x}\:=\:? \\ $$ Answered by Berbere last updated on…
Question Number 208912 by dimentri last updated on 27/Jun/24 $$\:\:\:\underbrace{\Subset} \underbrace{ \cancel{} }\pi \\ $$ Commented by Frix last updated on 27/Jun/24 $$\mathrm{33} \\ $$…
Question Number 208913 by efronzo1 last updated on 27/Jun/24 Answered by A5T last updated on 27/Jun/24 $$\frac{{sin}\beta}{\mathrm{6}}=\frac{\mathrm{1}}{{AC}};\frac{{sin}\left(\mathrm{60}−\beta\right)}{\mathrm{3}}=\frac{\mathrm{1}}{{AC}} \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{60}−\beta\right)}{{sin}\beta}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\beta\approx\mathrm{40}.\mathrm{8934} \\ $$$$\Rightarrow{sin}\beta\approx\mathrm{0}.\mathrm{654654};\:{we}\:{can}\:{then}\:{find}\:{AC}=\frac{\mathrm{6}}{{sin}\beta} \\ $$$${AC}\approx\mathrm{9}.\mathrm{165151} \\ $$…
Question Number 208931 by byaw last updated on 27/Jun/24 Answered by mr W last updated on 27/Jun/24 $$\left({a}\right) \\ $$$${R}={m}\left({g}+{a}\right)=\mathrm{0}.\mathrm{5}×\left(\mathrm{10}+\mathrm{2}\right)=\mathrm{6}\:{N} \\ $$$$ \\ $$$$\left({b}\right)\left({i}\right) \\…
Question Number 208915 by Tawa11 last updated on 27/Jun/24 Answered by mr W last updated on 27/Jun/24 Commented by mr W last updated on 27/Jun/24…
Question Number 208908 by alcohol last updated on 26/Jun/24 $${I}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{a}} \left(\mathrm{1}−{t}\right)^{{b}} {dt} \\ $$$${Note}\::\:{I}\left({a},{b}\right)\:=\:\frac{{a}}{{b}+\mathrm{1}}{I}\left({a}−\mathrm{1},{b}+\mathrm{1}\right) \\ $$$${show}\:{that} \\ $$$$\bullet\:{I}\left({a}+\mathrm{1},\:{b}\right)\:+\:{I}\left({a},{b}+\mathrm{1}\right)\:=\:{I}\left({a},\:{b}\right) \\ $$$$\bullet\:{find}\:{B}\left({a}+\mathrm{1},\:{b}+\mathrm{1}\right)\:{interms}\:{of}\:{B}\left({a},{b}\right) \\ $$$$\bullet\:{use}\:{I}\left({a},{b}\right)\:=\:{I}\left({b},{a}\right)\:{and}\:{deduce}\:{I}\left(\frac{\mathrm{5}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right) \\…
Question Number 208892 by hardmath last updated on 26/Jun/24 $$\mathrm{Find}: \\ $$$$\sqrt{−\mathrm{16}}\:\:\centerdot\:\:\sqrt{−\mathrm{9}}\:\:=\:\:? \\ $$ Commented by Adeyemi889 last updated on 26/Jun/24 $$ \\ $$$$\sqrt{−\mathrm{16}}\:=\:\sqrt{\:\left(\mathrm{16}\right)\left(−\mathrm{1}\right)}\:=\sqrt{−\mathrm{1}}\:×\sqrt{\mathrm{16}\:} \\…
Question Number 208876 by Tawa11 last updated on 26/Jun/24 The 𝚌a𝚕𝚎𝚗𝚍𝚊𝚛 𝚘𝚏 𝚝𝚑𝚎 𝚢𝚎𝚊𝚛 2024 𝚒𝚜 𝚝𝚑𝚎 𝚜𝚊𝚖𝚎 𝚏𝚘𝚛 𝙰.2044 𝙱.2032 𝙲.2040 𝙳.2036 Commented by…
Question Number 208909 by lepuissantcedricjunior last updated on 26/Jun/24 $$\:\:\:\:\:\boldsymbol{{soit}}\:\boldsymbol{{la}}\:\boldsymbol{{fonction}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{x}}\:\:\boldsymbol{{definie}} \\ $$$$\boldsymbol{{sur}}\:\mathbb{R}\:\boldsymbol{{on}}\:\boldsymbol{{note}}\:\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{f}}^{−\mathrm{1}} \left(\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{alors}}\:\boldsymbol{{que}}\:\:\boldsymbol{{la}}\:\boldsymbol{{primitive}}\:\boldsymbol{{G}}\left(\boldsymbol{{x}}\right)=\int_{\mathrm{0}} ^{\boldsymbol{{x}}} \boldsymbol{{g}}\left(\boldsymbol{{t}}\right)\boldsymbol{{dt}} \\ $$ Commented by Ghisom last updated…