Question Number 69589 by ajfour last updated on 25/Sep/19 $${x}^{\mathrm{3}} +{px}−{ry}+{qz}+{a}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} +{rx}+{qy}−{pz}+{b}=\mathrm{0} \\ $$$${z}^{\mathrm{3}} −{qx}+{py}+{rz}+{c}=\mathrm{0} \\ $$$${solve}\:{for}\:{x},{y},{z},\:{in}\:{terms}\:{of} \\ $$$${p},{q},{r},\:{a},{b},{c}. \\ $$ Terms of…
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Question Number 135119 by oooooooo last updated on 10/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 135117 by oooooooo last updated on 10/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 135108 by bramlexs22 last updated on 10/Mar/21 $$\left(\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}}\:\right)^{\mathrm{x}^{\mathrm{2}} } \:=\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}+\mathrm{1}} \\ $$ Answered by MJS_new last updated on 10/Mar/21 $${x}^{\mathrm{2}} ={x}+\mathrm{1}\:\Leftrightarrow\:\frac{{x}−\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\…
Question Number 4037 by Filup last updated on 27/Dec/15 $$\exists{x}\in\mathbb{Z}\exists{n}\in\mathbb{R}:{n}^{{x}} =\mathrm{2}{n} \\ $$$$\mathrm{Does}\:{x}\:{exist}? \\ $$ Commented by Filup last updated on 27/Dec/15 $${n}={x}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{2}}…
Question Number 4036 by Rasheed Soomro last updated on 27/Dec/15 $$\mathrm{Twin}\:\mathrm{of}\:\mathrm{Q}#\mathrm{3943} \\ $$$$\mathrm{Three}\:\mathrm{circles}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane} \\ $$$$\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{region} \\ $$$$\mathrm{is}\:\mathrm{produced}\:\mathrm{which}\:\mathrm{is}\:\mathrm{not}\:\mathrm{included} \\ $$$$\mathrm{in}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circles}.\:\mathrm{Determine}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{this}\:\mathrm{region}. \\ $$$$\underset{−} {\mathrm{Circles}\:\mid\:\mathrm{Radii}\:\mid\:\mathrm{Centers}\:} \\…
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Question Number 4013 by Yozzii last updated on 26/Dec/15 $${Find}\:{x}\:{such}\:{that}\:\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}\right)^{{x}^{\mathrm{2}} −\mathrm{21}{x}+\mathrm{10}} =\mathrm{1}. \\ $$ Commented by 123456 last updated on 26/Dec/15 $${x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}=\mathrm{1}\Leftrightarrow{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{4}=\mathrm{0}…