Question Number 105181 by mohammad17 last updated on 26/Jul/20 | ||
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$${graph}\:{the}\:{function}\:{x}^{\mathrm{2}} +\left({y}−\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{1} \\ $$ | ||
Answered by bemath last updated on 26/Jul/20 | ||
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Commented by mr W last updated on 27/Jul/20 | ||
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Commented by Dwaipayan Shikari last updated on 26/Jul/20 | ||
Commented by Dwaipayan Shikari last updated on 26/Jul/20 | ||
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Commented by mr W last updated on 26/Jul/20 | ||
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$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} \leqslant{x}^{\mathrm{2}} {y}^{\mathrm{3}} \\ $$ | ||
Commented by malwaan last updated on 27/Jul/20 | ||
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