Question Number 160806 by HongKing last updated on 06/Dec/21 Answered by MJS_new last updated on 06/Dec/21 $${A}=\mathrm{0}\vee{A}=\mathrm{5} \\ $$$$ \\ $$$${y}={A}−{x} \\ $$$${x}^{\mathrm{3}} +\left({A}−{x}\right)^{\mathrm{3}} =\mathrm{5}{x}\left({A}−{x}\right)…
Question Number 95269 by Farruxjano last updated on 24/May/20 Commented by bobhans last updated on 24/May/20 Which is the solution to this question? Commented by Farruxjano last updated on 24/May/20 $$\boldsymbol{{Sorry}}\:\boldsymbol{{I}}\:\boldsymbol{{wanted}}\:\boldsymbol{{to}}\:\boldsymbol{{send}}\:\boldsymbol{{another}}\:…
Question Number 95262 by 6532 last updated on 24/May/20 $$\mathrm{3x}^{\mathrm{2}} +\mathrm{5x}^{\mathrm{4}} −\mathrm{7} \\ $$$$\mathrm{plz}\:\mathrm{help}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{equation} \\ $$ Answered by i jagooll last updated on 24/May/20 $$\mathrm{do}\:\mathrm{you}\:\mathrm{meant}\:\mathrm{3x}^{\mathrm{2}}…
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Question Number 160768 by cortano last updated on 06/Dec/21 $$\:\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}−\mathrm{18}\:=\:\mathrm{0}\: \\ $$$$\:\mathrm{x}\in\mathbb{R}\:,\:\mathrm{x}=? \\ $$ Commented by yeti123 last updated on 06/Dec/21 $${x}\:=\:\mathrm{3} \\ $$…
Question Number 160777 by HongKing last updated on 06/Dec/21 $$\mathrm{3x}^{\mathrm{3}} \:+\:\mathrm{x}^{\mathrm{2}} \:+\:\left(\mathrm{m}\:+\:\mathrm{2}\right)\centerdot\mathrm{x}\:+\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\mathrm{equation}\:\mathrm{root}\:\:\mathrm{x}_{\mathrm{1}} \:;\:\mathrm{x}_{\mathrm{2}} \:;\:\mathrm{x}_{\mathrm{3}} \\ $$$$\mathrm{and}\:\:\mathrm{x}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} } \\ $$$$\mathrm{find}\:\:\boldsymbol{\mathrm{m}}=? \\ $$…
Question Number 95232 by bobhans last updated on 24/May/20 Commented by bobhans last updated on 24/May/20 $$\mathrm{i}\:\mathrm{want}\:\mathrm{compare}\:\mathrm{with}\:\mathrm{my}\:\mathrm{answer} \\ $$ Commented by bobhans last updated on…
Question Number 95222 by bobhans last updated on 24/May/20 $$\begin{cases}{{x}^{\mathrm{2}} \:+\:{x}\:\sqrt[{\mathrm{3}\:\:}]{{xy}^{\mathrm{2}} }\:=\:\mathrm{80}\:}\\{{y}^{\mathrm{2}} \:+\:{y}\:\sqrt[{\mathrm{3}\:\:}]{{x}^{\mathrm{2}} {y}}\:=\:\mathrm{5}\:}\end{cases} \\ $$$${find}\:{x}\:{and}\:{y}\: \\ $$ Answered by john santu last updated on…
Question Number 160744 by cortano last updated on 05/Dec/21 $$\:\:\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{12}\right)^{\mathrm{3}} +\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{18}\right)^{\mathrm{2}} =\:\mathrm{9}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{x}=?\: \\ $$ Commented by blackmamba last updated on…
Question Number 95182 by i jagooll last updated on 23/May/20 Answered by mr W last updated on 23/May/20 $$\frac{\mathrm{1}}{{S}}+\frac{\mathrm{1}}{{S}+\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${S}^{\mathrm{2}} −\mathrm{6}{S}−\mathrm{20}=\mathrm{0} \\ $$$${S}=\mathrm{3}+\sqrt{\mathrm{29}}\approx\mathrm{8}.\mathrm{4}\:{h} \\…