Question Number 209707 by alcohol last updated on 19/Jul/24 $${a},{b}\:\in\mathbb{C}\::\:{a}\overset{−} {{b}}\:+\:{b}\:=\:\mathrm{0}\:{f}\::\:{z}'\:=\:{a}\overset{−} {{z}}\:+\:{b}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{such}\:{that}\:{f}\left({M}\right)\:=\:{M}' \\ $$$$\mathrm{1}.\:{let}\:{z}_{{A}} \:=\:{z}\:{and}\:{z}_{{A}'} \:=\:{z}'\:{and}\:{f}\left({A}\right)\:=\:{A} \\ $$$${show}\:{that}\:\mathrm{2}{Re}\left(\overset{−} {{b}z}\right)\:=\:{b}\overset{−} {{b}} \\ $$$$\left({A}\:{is}\:{the}\:{set}\:{of}\:{invariant}\:{points}\:{and}\right. \\…
Question Number 209732 by MM42 last updated on 19/Jul/24 $$\left.{Q}\right)\:{The}\:{collection}\:{A}=\left\{\mathrm{12},\mathrm{13},\mathrm{15},\mathrm{18},\mathrm{23},\mathrm{24},\mathrm{25},\mathrm{26}\right\}\&\:{B}\subseteq{A} \\ $$$${if}\:\:{m},{M}\:\in{B}\:\:;\:{m}={min}\:\&\:{M}\:={max}\:\&\:\:{nm}=\mathrm{10}{k} \\ $$$${which}\:{number}\:{of}\:\:{B}\:: \\ $$$$\left.\mathrm{1}\left.\right)\left.\mathrm{5}\left.\mathrm{9}\:\:\:\:\:\:\:\mathrm{2}\right)\mathrm{60}\:\:\:\:\:\:\:\mathrm{3}\right)\mathrm{61}\:\:\:\:\:\:\mathrm{4}\right)\mathrm{62} \\ $$$$ \\ $$ Answered by mr W last…
Question Number 209718 by Ismoiljon_008 last updated on 19/Jul/24 Answered by mr W last updated on 19/Jul/24 $${a}_{{n}} =\frac{\mathrm{1}}{{n}!×\left({n}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)!} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}{a}_{{n}} =\left(\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{2024}!}−\frac{\mathrm{1}}{\mathrm{2025}!}\right. \\…
Question Number 209735 by hardmath last updated on 19/Jul/24 $$\mathrm{tan}\left(\mathrm{3x}\right)\:\:+\:\:\mathrm{tan}\left(\mathrm{5x}\right)\:\:=\:\:\mathrm{2} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by a.lgnaoui last updated on 20/Jul/24 $$\mathrm{voir}\:\mathrm{reponse}\:\mathrm{en}\:\mathrm{bas} \\ $$$$\:\left(\mathrm{a}\:\mathrm{l}\:\mathrm{exeption}\:\:\mathrm{d}\:'\mathrm{ereur}\:\mathrm{de}\:\mathrm{calcul}\right) \\…
Question Number 209719 by hardmath last updated on 19/Jul/24 $$\mathrm{If}: \\ $$$$\mathrm{7}^{\mathrm{243}} \:\:=\:\:\overline {\mathrm{a}…\mathrm{bc}} \\ $$$$\mathrm{Find}: \\ $$$$\mathrm{b}\centerdot\mathrm{c}\:=\:? \\ $$ Answered by MM42 last updated…
Question Number 209712 by mr W last updated on 19/Jul/24 Commented by mr W last updated on 19/Jul/24 $${general}\:{case} \\ $$$${OA}={a} \\ $$$${OB}={b}\:>\:{a} \\ $$$$\angle{AOB}=\theta\:<\frac{\pi}{\mathrm{2}}…
Question Number 209694 by Frix last updated on 18/Jul/24 $${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\alpha^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} +{yz}+{z}^{\mathrm{2}} =\beta^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} +{zx}+{x}^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \\ $$$$\mathrm{Find}\:{x}+{y}+{z}\:\mathrm{for}\:{x},\:{y},\:{z}\:\in\mathbb{R}^{+} \\ $$…
Question Number 209662 by depressiveshrek last updated on 18/Jul/24 $$\int\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} {x}\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}\right)}{dx} \\ $$ Answered by ARUNG_Brandon_MBU last updated on 18/Jul/24 $${I}=\int\frac{{dx}}{\mathrm{sin}^{\mathrm{2}} {x}\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}\right)} \\…
Question Number 209695 by hardmath last updated on 18/Jul/24 $$\mathrm{If}: \\ $$$$\left(\mathrm{a}\:+\:\mathrm{b}\right)\centerdot\sqrt{\mathrm{2}}\:=\:\mathrm{7}\centerdot\left(\mathrm{a}−\mathrm{b}−\mathrm{4}\right) \\ $$$$\mathrm{Find}: \\ $$$$\left(\mathrm{2a}\:+\:\mathrm{b}\right)\:=\:? \\ $$ Answered by mr W last updated on…
Question Number 209691 by efronzo1 last updated on 18/Jul/24 $$\:\:\:\int\left(\mathrm{2x}^{\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{7}} \right)\left(\mathrm{log}\:_{\mathrm{2}} \:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{7}\right)\right)\mathrm{e}^{\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{5}} \:\mathrm{dx}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com