Question Number 7809 by Tawakalitu. last updated on 16/Sep/16 $${x}\:+\:{y}\:+\:{z}\:=\:\mathrm{1}\:\:\:\:\:\:………\:\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{37}\:\:\:\:\:……..\:\left({ii}\right) \\ $$$${x}^{\mathrm{3}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{3}} \:=\:\mathrm{91}\:\:\:\:\:……..\:\left({iii}\right) \\ $$$$ \\ $$$${Solve}\:{simultaneously}.\: \\…
Question Number 138879 by peter frank last updated on 19/Apr/21 Answered by Dwaipayan Shikari last updated on 19/Apr/21 $${e}^{{i}\theta\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+..{n}\right)} =\mathrm{1}\Rightarrow{e}^{{i}\theta\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} ={e}^{\mathrm{2}\pi{im}} \:\:\left({m}\in\mathbb{Z}\right) \\ $$$$\Rightarrow\theta=\frac{\mathrm{4}\pi{m}}{{n}\left({n}+\mathrm{1}\right)} \\…
Question Number 138872 by peter frank last updated on 19/Apr/21 Answered by bramlexs22 last updated on 19/Apr/21 $$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\left(\mathrm{A}−\mathrm{7}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{Ax}−\mathrm{8} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{x}^{\mathrm{3}} +\left(\mathrm{A}−\mathrm{7}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{Ax}−\mathrm{8}=\mathrm{0} \\…
Question Number 7803 by Tawakalitu. last updated on 16/Sep/16 Answered by Rasheed Soomro last updated on 16/Sep/16 $$\left(\frac{\mathrm{1}×\mathrm{3}×\mathrm{5}×…×\mathrm{2015}}{\mathrm{2}×\mathrm{4}×\mathrm{6}×…×\mathrm{2016}}\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$\left(\frac{\mathrm{2015}!/\left(\mathrm{2}.\mathrm{4}.\mathrm{6}…..\mathrm{2014}\right.}{\mathrm{2}.\mathrm{4}.\mathrm{6}…..\mathrm{2016}}\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$\left(\frac{\mathrm{2015}!}{\left(\mathrm{2}.\mathrm{4}.\mathrm{6}….\mathrm{2014}\right)\left(\mathrm{2}.\mathrm{4}.\mathrm{6}…..\mathrm{2016}\right)}\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$\left(\frac{\mathrm{2015}!}{\left(\mathrm{2}.\mathrm{4}.\mathrm{6}….\mathrm{2014}\right)^{\mathrm{2}} \left(\mathrm{2015}.\mathrm{2016}\right)}\right)^{\mathrm{1}/\mathrm{4}}…
Question Number 7791 by Tawakalitu. last updated on 15/Sep/16 Commented by prakash jain last updated on 15/Sep/16 $$\mathrm{What}\:\mathrm{is}\:{d}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}\:\mathrm{above}? \\ $$ Commented by Chantria last updated…
Question Number 138863 by bramlexs22 last updated on 19/Apr/21 $$\mid\mathrm{2x}−\mathrm{1}\mid\:\leqslant\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{x}+\mathrm{2}}}\:+\:\mathrm{1} \\ $$ Answered by lyubita last updated on 19/Apr/21 $$-\:\mathrm{2}\:<\:{x}\:\leqslant\:\mathrm{2} \\ $$ Commented by bramlexs22…
Question Number 138850 by bramlexs22 last updated on 19/Apr/21 $${x}^{−\mathrm{log}\:\left({x}\right)+\mathrm{4}} \:<\:\frac{\mathrm{1}}{\mathrm{16}}{x}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 73308 by wo1lxjwjdb last updated on 10/Nov/19 $${what}\:{are}\:{the}\:{solutions} \\ $$$${of}\:\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}}={n}\:{where}\:{n}\in\mathbb{N} \\ $$ Commented by MJS last updated on 10/Nov/19 $${n}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{k}} +\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{k}}…
Question Number 138835 by mathdanisur last updated on 18/Apr/21 $${x}^{\mathrm{3}} −\mathrm{9}{x}+\mathrm{1}=\mathrm{0} \\ $$ Commented by mr W last updated on 18/Apr/21 $${x}_{{k}} =\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\:\mathrm{18}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\:\:\:{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\…
Question Number 138827 by mathdanisur last updated on 18/Apr/21 $${Solve}\:{for}\:{real}\:{numbers}: \\ $$$$\begin{cases}{−{a}^{\mathrm{1}} −{b}^{\mathrm{2}} −{c}^{\mathrm{3}} =\mathrm{2024}^{\mathrm{12}} }\\{−{a}^{−\mathrm{1}} −{b}^{−\mathrm{2}} −{c}^{−\mathrm{3}} =\mathrm{2024}^{−\mathrm{12}} }\\{{a}^{\mathrm{1}} {b}^{\mathrm{2}} {c}^{\mathrm{3}} =\mathrm{2024}^{\mathrm{24}} }\end{cases} \\…