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Question Number 100767 by Rio Michael last updated on 28/Jun/20

$$\mathrm{sketch}{x}^2=y^3\mathrm{and}{x}^2+y^2=16$$$$\mathrm{and}\mathrm{hence}\mathrm{solve}(x^2-y^3)(x^2+y^2-16)⩾0$$

Answered by john santu last updated on 28/Jun/20

Answered by john santu last updated on 28/Jun/20

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