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 $$\:\mathrm{6}/\mathrm{0}/\mathrm{0}\:{or}\:\mathrm{5}/\mathrm{1}/\mathrm{0}\:{or}\:\mathrm{4}/\mathrm{2}/\mathrm{0}\:\:{or}\:\mathrm{4}/\mathrm{1}/\mathrm{1}\:{or}\:\mathrm{3}/\mathrm{3}/\mathrm{0}/\:{or}\:\mathrm{3}/\mathrm{2}/\mathrm{1}\:{or}\:\mathrm{2}/\mathrm{2}/\mathrm{2} \\…
Question Number 212516 by RojaTaniya last updated on 16/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212518 by efronzo1 last updated on 16/Oct/24 $$\:\:\:\cancel{\underbrace{\downharpoonleft}\underline{}\:} \\ $$ Answered by golsendro last updated on 16/Oct/24 $$\:\:\:\cancel{\underline{\underbrace{\pm}} }=\:\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} +\mathrm{a}\: \\ $$$$\:\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\:\sqrt{\mathrm{x}−\mathrm{a}}\:+\mathrm{a}\:…
Question Number 212519 by MrGaster last updated on 16/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\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 212513 by MrGaster last updated on 16/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{certificate}: \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}{n}} −\mathrm{1} \\ $$$$\mathrm{Can}\:\mathrm{always}\:\mathrm{be}\:{x}+\mathrm{1}\:\mathrm{be}\:\mathrm{divisible}\:\:\:\:\:\:\:\:\: \\ $$ Answered by A5T last updated on 16/Oct/24 $${x}\equiv−\mathrm{1}\left({mod}\:{x}+\mathrm{1}\right)\Rightarrow{x}^{\mathrm{2}{n}}…
Question Number 212514 by ajfour last updated on 16/Oct/24 Answered by mr W last updated on 18/Oct/24 Commented by mr W last updated on 17/Oct/24…
Question Number 212515 by efronzo1 last updated on 16/Oct/24 $$\:\:\mathrm{The}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{natural}\: \\ $$$$\:\:\:\mathrm{numbers}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{with}\:\mathrm{x},\mathrm{y}\:\leqslant\:\mathrm{33}\:\mathrm{that}\: \\ $$$$\:\:\:\mathrm{satisfy}\:\mathrm{5}\:\mid\:\mathrm{3}^{\mathrm{x}^{\mathrm{y}−\mathrm{1}} } \:+\:\mathrm{2}^{\mathrm{y}^{\mathrm{2x}} } \:\mathrm{is}\:…\: \\ $$$$\:\:\left(\mathrm{A}\right)\:\mathrm{295}\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{296}\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{297}\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{298}\:\:\:\left(\mathrm{E}\right)\:\mathrm{299} \\ $$ Commented by A5T…
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…
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…