Question Number 212560 by ajfour last updated on 17/Oct/24 $${Exact}\:{solution}\:{just}\:{to}\:{this}\:{please}: \\ $$$$\mathrm{9}{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{3}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right) \\ $$ Commented by Frix last updated on 18/Oct/24…
Question Number 212552 by mnjuly1970 last updated on 17/Oct/24 $$ \\ $$$$\:\overset{\mathrm{Q}:} {\:}\:\mathrm{In}\:\mathrm{A}\overset{\Delta} {\mathrm{B}C}\::\:\:\:{cos}\left(\mathrm{A}\right)\:+{cos}\left(\mathrm{B}\:\right)+\:\mathrm{2}{cos}\left(\mathrm{C}\:\right)=\:\mathrm{2} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{show}\:\mathrm{that}\::\:\:\:{a}\:+\:{b}\:=\:\mathrm{2}{c}\:\:\:\:\:\:\:\:\:\:\:\blacksquare\: \\ $$$$ \\ $$ Answered by Ghisom…
Question Number 212576 by MATHEMATICSAM last updated on 17/Oct/24 $$\mathrm{If}\:\frac{{x}^{\mathrm{2}} \:−\:{yz}}{{a}^{\mathrm{2}} \:−\:{bc}}\:=\:\frac{{y}^{\mathrm{2}} \:−\:{zx}}{{b}^{\mathrm{2}} \:−\:{ca}}\:=\:\frac{{z}^{\mathrm{2}} \:−\:{xy}}{{c}^{\mathrm{2}} \:−\:{ab}}\:\mathrm{then}\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{{x}}{{a}}\:=\:\frac{{y}}{{b}}\:=\:\frac{{z}}{{c}}\:. \\ $$ Terms of Service Privacy Policy…
Question Number 212550 by mnjuly1970 last updated on 17/Oct/24 $$ \\ $$$$\:\:\:\:{the}\:{following}\:{equation}\:{has} \\ $$$$\:\:\:\:\:{no}\:{root}\:.\:{find}\:{the}\:{relationship} \\ $$$$\:\:\:{between}\:{a}\:,\:{b}\:,\:{c}\:: \\ $$$$\:\:\:\:\mathrm{1}:\:\:{c}\leqslant\mathrm{2} \\ $$$$\:\:\:\:\mathrm{2}:\:{c}\:>\mathrm{2} \\ $$$$\:\:\:\:\mathrm{3}:\:{c}\:>{ab} \\ $$$$\:\:\:\:\mathrm{4}:\:{c}\leqslant\:{ab} \\…
Question Number 212541 by hardmath last updated on 16/Oct/24 $$\mathrm{Find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$\mathrm{sin}\left(\frac{\mathrm{88}\pi^{\mathrm{2}} }{\mathrm{x}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{3x}\right)} \\ $$ Answered by Frix last updated on 17/Oct/24 $$\mathrm{sin}\:\frac{\mathrm{88}\pi^{\mathrm{2}} }{{x}}\:\in\left[−\mathrm{1},\:\mathrm{1}\right] \\…
Question Number 212531 by RojaTaniya last updated on 16/Oct/24 Answered by a.lgnaoui last updated on 17/Oct/24 $$\boldsymbol{\mathrm{x}}=\mathrm{1}+^{\mathrm{7}} \sqrt{\mathrm{2}}\:+\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} +\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)^{\mathrm{3}} +\left(^{\mathrm{7}} \sqrt{\mathrm{2}\:}\:\right)^{\mathrm{4}} +\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)^{\mathrm{5}}…
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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}…