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1 answer:
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Answer:
The region represented by the equation is a full sphere of radius √3 centered in the origin of coordinates.
Step-by-step explanation:
<em>In a plane xy, the equation that represents a circle with center in the origin, of radius r is</em>
<em>in R³, or a space xyz, we can represent a sphere with its center in the origin, and of radius r, with the equation</em>
So, in this problem we have that
which means that the sphere has a radius of √3.
<u>Finally, our equation is an inequality</u>, and the sphere is equal to, and less than, the calculated radius.
Therefore, the sphere is "full" from the surface to its center.
The first thing you should do is graph the following lines
2x + 3y = 8
x-2y = -3
x = 0
y = 0
After you have graphed them, you should proceed to evaluate points in the xy plane that meet the following restrictions:
2x + 3y≤8, x-2y≥-3, x≥0, y≥0
The resulting region is the region "R" shown in the attached graph.
2x + 3y = 8
x-2y = -3
x = 0
y = 0
After you have graphed them, you should proceed to evaluate points in the xy plane that meet the following restrictions:
2x + 3y≤8, x-2y≥-3, x≥0, y≥0
The resulting region is the region "R" shown in the attached graph.