What are the boundaries of x and y if ${(x - 2)}^{2} + {(y - 3)}^{2} \geq 16, {(x - 3)}^{2} + (\frac{{(y - 4)}^{2}}{64}) < 1$ $(x-2)^2+(y-3)^2 >=16, (x-3)^2+((y-4)^2/64)<1$ ?

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What are the boundaries of x and y if ${(x - 2)}^{2} + {(y - 3)}^{2} \geq 16, {(x - 3)}^{2} + (\frac{{(y - 4)}^{2}}{64}) < 1$ $(x-2)^2+(y-3)^2 >=16, (x-3)^2+((y-4)^2/64)<1$ ?

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Answer 1 · 2016-06-04T11:05:32

The region defined by the inequations is shown in light blue.

Explanation:

${(x - 2)}^{2} + {(y - 3)}^{2} \geq 16$ $(x - 2)^2 + (y - 3)^2 ge 16$ defines the exterior of a circumference centered at ${2, 3}$ ${2,3}$ with radius $4$ $4$

${(x - 3)}^{2} + \frac{{(y - 4)}^{2}}{64} \leq 1$ $(x - 3)^2 + (y - 4)^2/64 le 1$ defines the interior of an ellipse centered at ${3, 4}$ ${3,4}$ having axes $1, 8$ $1, 8$

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What are the boundaries of x and y if ${(x - 2)}^{2} + {(y - 3)}^{2} \geq 16, {(x - 3)}^{2} + (\frac{{(y - 4)}^{2}}{64}) < 1$ $(x-2)^2+(y-3)^2 >=16, (x-3)^2+((y-4)^2/64)<1$ ?

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What are the boundaries of x and y if (x-2)^2+(y-3)^2 >=16, (x-3)^2+((y-4)^2/64)<1(x−2)2+(y−3)2≥16,(x−3)2+((y−4)264)<1(x-2)^2+(y-3)^2 >=16, (x-3)^2+((y-4)^2/64)<1?

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What are the boundaries of x and y if ${(x - 2)}^{2} + {(y - 3)}^{2} \geq 16, {(x - 3)}^{2} + (\frac{{(y - 4)}^{2}}{64}) < 1$ $(x-2)^2+(y-3)^2 >=16, (x-3)^2+((y-4)^2/64)<1$ ?