Question Number 120780 by mathace last updated on 02/Nov/20 $${solve}\:{x}^{\mathrm{2}^{{x}} } =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented by Anuragkar last updated on 02/Nov/20 $${Use}\:\:{Lambert}\:{W}\:{function}\:{to}\:{work}\:{it}\:{out}…… \\ $$ Commented…
Question Number 186305 by mustafazaheen last updated on 03/Feb/23 $$\mathrm{log}\:\left(\frac{\mathrm{3}.\bar {\mathrm{2}}}{\mathrm{3}.\bar {\mathrm{1}}}\right)\:\:\:\:\:\:\:{find}\:\mathrm{Characteristic}? \\ $$ Commented by MJS_new last updated on 03/Feb/23 $$\frac{\mathrm{3}+\frac{\mathrm{2}}{\mathrm{9}}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{9}}}=\frac{\mathrm{29}}{\mathrm{28}}=\frac{\mathrm{29}}{\mathrm{2}^{\mathrm{2}} \mathrm{7}} \\ $$$$\mathrm{log}\:\frac{\mathrm{29}}{\mathrm{2}^{\mathrm{2}}…
Question Number 186300 by mnjuly1970 last updated on 03/Feb/23 $$\:\:\: \\ $$$$\:\:\:\:\:\mathrm{solve}\:\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$ \\ $$$$\:\:\:\lfloor\:\:\mathrm{2log}_{\:\mathrm{8}} \left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\rfloor\:=\:\mathrm{log}_{\:\mathrm{4}} \left({x}\:\right)+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$ Answered by MJS_new…
Question Number 186292 by ajfour last updated on 03/Feb/23 $$\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −{px}^{\mathrm{3}} −{x}^{\mathrm{2}} +\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right){x}+\frac{{p}}{\mathrm{3}}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +{ax}+{h}\right)\left({x}^{\mathrm{2}} +{bx}+{k}\right)=\mathrm{0} \\ $$$${a}+{b}=−{p} \\ $$$${h}+{k}+{ab}=−\mathrm{1} \\…
Question Number 186285 by Shrinava last updated on 03/Feb/23 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{R}\:\left(\mathrm{m}\:,\:\mathrm{n}\right)\:\leqslant\:\mathrm{C}_{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{m}}} \\ $$$$\mathrm{Here}\:\:\mathrm{R}\:\:\mathrm{states}\:\mathrm{the}\:\:\mathrm{Ramsey}\:\:\mathrm{theory} \\ $$ Commented by mr W last updated on 03/Feb/23…
Question Number 55198 by ajfour last updated on 19/Feb/19 $${x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}−\mathrm{32}=\mathrm{0} \\ $$$${Find}\:{at}\:{least}\:{one}\:{root}. \\ $$ Commented by arvinddayama00@gmail.com last updated on 19/Feb/19 $${all}\:{velue}\:{of}\:{x}=?…
Question Number 186265 by mnjuly1970 last updated on 02/Feb/23 $$ \\ $$$$\:\:\:{f}\left({x}\right)=\:\sqrt{\:{x}\:−{a}}\:\:+\:\sqrt{\mathrm{3}{a}\:−{x}}\:\:\:{with}\:\left(\:{a}>\mathrm{0}\right) \\ $$$$\:\:\:\:{is}\:\:{given}\:.{If}\:\:,\:\:{f}_{\:{max}} \:.\:{f}_{\:{min}} \:=\:\sqrt{\mathrm{32}} \\ $$$$\:\:\:\:\:{find}\:\:,\:\:\:\:\:\:\:''\:\:\:{a}\:\:''\:\:=\:? \\ $$ Answered by mahdipoor last updated…
Question Number 55195 by behi83417@gmail.com last updated on 19/Feb/19 $$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\sqrt{\mathrm{2}}\boldsymbol{\mathrm{b}}=\mathrm{0},\boldsymbol{\mathrm{has}}\:\mathrm{2}\:\boldsymbol{\mathrm{roots}}:\boldsymbol{\mathrm{c}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{d}},\boldsymbol{\mathrm{also}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{cx}}+\sqrt{\mathrm{2}}\boldsymbol{\mathrm{d}}=\mathrm{0},\boldsymbol{\mathrm{has}}\:\mathrm{2}\:\boldsymbol{\mathrm{roots}}:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{b}}.\boldsymbol{\mathrm{such}} \\ $$$$\boldsymbol{\mathrm{that}}:\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{b}},\:\boldsymbol{\mathrm{c}},\:\boldsymbol{\mathrm{d}},\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{defferent}}\:\boldsymbol{\mathrm{non}}\:\boldsymbol{\mathrm{zero}}\: \\ $$$$\boldsymbol{\mathrm{numbers}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{value}}\left(\boldsymbol{\mathrm{s}}\right)\:\boldsymbol{\mathrm{for}}:\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{c}}^{\mathrm{2}} +\boldsymbol{\mathrm{d}}^{\mathrm{2}} . \\…
Question Number 186256 by Shrinava last updated on 02/Feb/23 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathbb{R}\:\left(\mathrm{m}\:,\:\mathrm{n}\right)\:\leqslant\:\mathbb{C}_{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{m}}} \\ $$$$ \\ $$ Commented by mr W last updated on 02/Feb/23…
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