Question Number 212522 by Nadirhashim last updated on 16/Oct/24 $$\:\:\boldsymbol{{in}}\:\boldsymbol{{how}}\:\boldsymbol{{many}}\:\boldsymbol{{ways}}\:\boldsymbol{{we}} \\ $$$$\boldsymbol{{can}}\:\boldsymbol{{distribute}}\:\mathrm{6}\:\boldsymbol{{distinct}} \\ $$$$\boldsymbol{{balls}}\:\boldsymbol{{in}}\:\mathrm{3}\:\boldsymbol{{identical}}\:\boldsymbol{{boxes}} \\ $$ Answered by mehdee7396 last updated on 16/Oct/24 $$\:\mathrm{6}/\mathrm{0}/\mathrm{0}\:{or}\:\mathrm{5}/\mathrm{1}/\mathrm{0}\:{or}\:\mathrm{4}/\mathrm{2}/\mathrm{0}\:\:{or}\:\mathrm{4}/\mathrm{1}/\mathrm{1}\:{or}\:\mathrm{3}/\mathrm{3}/\mathrm{0}/\:{or}\:\mathrm{3}/\mathrm{2}/\mathrm{1}\:{or}\:\mathrm{2}/\mathrm{2}/\mathrm{2} \\…
Question Number 212519 by MrGaster last updated on 16/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{certificate}: \\ $$$$\:\:\:\:\:\left({x}−\mathrm{1}\right)\mid\left({x}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}\right)\mathrm{and}\left({x}+\mathrm{1}\right)\mid\left({x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}\right) \\ $$$$\mathrm{All}\:\mathrm{established} \\ $$$$\left[\mathrm{2024}.\mathrm{10}.\mathrm{16}\right] \\ $$ Answered by mathmax last updated…
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 $$−\frac{\mathrm{18}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}}=−\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}=\mathrm{9e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\left(\mathrm{9e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} =\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{6}}} =\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}…
Question Number 212508 by hardmath last updated on 15/Oct/24 Commented by hardmath last updated on 15/Oct/24 $$\mathrm{Find}: \\ $$$$ \\ $$Find the modulus of the…
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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…
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 212479 by hardmath last updated on 14/Oct/24 $$\begin{cases}{\mathrm{2x}\:+\:\mathrm{3y}\:−\:\mathrm{z}\:=\:\mathrm{7}}\\{\mathrm{4x}\:−\:\mathrm{y}\:+\:\mathrm{2z}\:=\:\mathrm{1}}\\{−\mathrm{x}\:+\:\mathrm{5y}\:+\:\mathrm{3z}\:=\:\mathrm{14}}\end{cases}\:\:\:\:\:\mathrm{find}:\:\:\mathrm{x},\mathrm{y},\mathrm{z}\:=\:? \\ $$ Answered by A5T last updated on 14/Oct/24 $$\mathrm{3}\left({ii}\right)+\left({i}\right)\Rightarrow\mathrm{14}{x}+\mathrm{5}{z}=\mathrm{10}…\left({iv}\right) \\ $$$$\mathrm{5}\left({ii}\right)+\left({iii}\right)\Rightarrow\mathrm{19}{x}+\mathrm{13}{z}=\mathrm{19}…\left({v}\right) \\ $$$$\mathrm{13}\left({iv}\right)−\mathrm{5}\left({v}\right)\Rightarrow\mathrm{87}{x}=\mathrm{35}\Rightarrow{x}=\frac{\mathrm{35}}{\mathrm{87}}\Rightarrow{z}=\frac{\mathrm{76}}{\mathrm{87}}\Rightarrow{y}=\frac{\mathrm{205}}{\mathrm{87}} \\…
Question Number 212470 by Spillover last updated on 14/Oct/24 Answered by Ar Brandon last updated on 14/Oct/24 $$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{1}−{x}\right)\mathrm{ln}{x}}{\mathrm{1}−{x}^{\mathrm{6}} }{dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}−{x}^{\mathrm{6}}…
Question Number 212432 by hardmath last updated on 13/Oct/24 $$\frac{\mathrm{9}}{\mathrm{2}\centerdot\mathrm{4}}\:+\:\frac{\mathrm{9}}{\mathrm{4}\centerdot\mathrm{6}}\:+…+\:\frac{\mathrm{9}}{\mathrm{4n}\centerdot\left(\mathrm{n}\:+\:\mathrm{1}\right)}\:=\:\frac{\mathrm{15}}{\mathrm{8}} \\ $$$$\mathrm{Find}:\:\:\boldsymbol{\mathrm{n}}\:=\:? \\ $$ Answered by Ar Brandon last updated on 13/Oct/24 $$\mathrm{9}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{15}}{\mathrm{8}}…