Question Number 212506 by hardmath last updated on 15/Oct/24 $$\mathrm{Find}:\:\:\:\:\:\:\sqrt[{\mathrm{4}}]{−\:\frac{\mathrm{18}}{\mathrm{1}\:+\:\boldsymbol{\mathrm{i}}\:\sqrt{\mathrm{3}}}} \\ $$ Answered by Frix last updated on 15/Oct/24 $$−\frac{\mathrm{18}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}}=−\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}=\mathrm{9e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\left(\mathrm{9e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} =\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{6}}} =\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}…
Question Number 212500 by ajfour last updated on 15/Oct/24 $${Can}\:{we}\:{exactly}\:{find}\:{r}_{{max}} \left({a}\right). \\ $$$${r}^{\mathrm{2}} +\frac{\mathrm{2}{rt}}{{a}}\left({t}+\mathrm{2}\sqrt{{ar}}\right)={t}^{\mathrm{2}} \:\:\:\:\forall\:{t}\:{is}\:{parameter} \\ $$ Answered by mr W last updated on 15/Oct/24…
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Question Number 212496 by efronzo1 last updated on 15/Oct/24 $$\:\mathrm{If}\:\mathrm{1}.\mathrm{1}!+\mathrm{3}.\mathrm{2}!+…+\left(\mathrm{2n}−\mathrm{1}\right).\mathrm{n}!\: \\ $$$$\:=\:\mathrm{a}.\left(\mathrm{n}+\mathrm{1}\right)!+\mathrm{b}\left(\mathrm{1}!+\mathrm{2}!+…+\left(\mathrm{n}+\mathrm{1}\right)!\right)+\mathrm{c}\: \\ $$$$\:\mathrm{for}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{integers}\:\mathrm{number} \\ $$$$\:\mathrm{find}\:\mathrm{2a}−\mathrm{b}+\mathrm{3c}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 212499 by efronzo1 last updated on 15/Oct/24 $$\:\:\mathrm{Given}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{and}\:\mathrm{d}\:\mathrm{are}\:\mathrm{reals}\: \\ $$$$\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{10}}\\{\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} =\mathrm{10}\:}\\{\mathrm{ab}+\mathrm{cd}=\mathrm{0}}\end{cases} \\ $$$$\:\:\mathrm{Find}\:\mathrm{ac}\:+\:\mathrm{bd}. \\ $$ Answered by ajfour…
Question Number 212492 by MrGaster last updated on 15/Oct/24 $$\mathrm{Given}\:\mathrm{that}\:{x}\:\in\:\mathbb{R}\left(\right. \\ $$$$\:\left(\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{real}\:\mathrm{numbers}\right)\mathrm{d} \\ $$$$\mathrm{an}\:{n}\:\in\:\mathbb{N}^{\ast} \\ $$$$\left(\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{naturalu}\right. \\ $$$$\left.\mathrm{nmbers}\right)\:\mathrm{prove}\:\mathrm{that}\:: \\ $$$$\mid\mathrm{cos}\:{x}\mid+\mid\mathrm{cos}\:\mathrm{2}{x}\mid+\mid\mathrm{cos}\:\mathrm{3}{x}\mid+\ldots+\mid\mathrm{cos}\left({n}+\mathrm{1}\right){x}\mid\geq\frac{{n}}{\mathrm{2}} \\ $$ Terms of Service…
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Question Number 212477 by Durganand last updated on 14/Oct/24 Answered by mehdee7396 last updated on 14/Oct/24 $$\frac{{sin}\mathrm{2}{acosa}−{sinacos}\mathrm{2}{a}}{{sin}\mathrm{2}{asina}} \\ $$$$=\frac{{sina}}{{sin}\mathrm{2}{asina}}=\frac{\mathrm{1}}{{sin}\mathrm{2}{a}}\:\:\checkmark \\ $$ Answered by A5T last…