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.
Let's go though the choices.
Major axis will get shorter. TRUE, CHECK ME
As the plane moves closer to the vertex the ellipse stays the same shape but shrinks.
Ellipse will become a circle. FALSE
As long as the plane is tilted, the ellipse can't be a circle.
Ellipse will degenerate to a point. TRUE, CHECK ME
Eventually the plane gets to the vertex, at which point (ha!) we have a single point.
Nappes? Is that a word? FALSE
The plane needs a steeper angle, steeper than the generators, before it can pass through both halves.
The generators are the lines through the vertex that form the cone. The plane isn't steep enough to hold any of them. FALSE
Where an, an-1,a2, a1, a0 are constants. We call the term containing the highest power of x the leading term, and we call an the leading coefficient. The degree of the polynomial is the power of x in the leading term. We have already seen degree 0, 1, and 2 polynomials which were the constant, linear, and quadratic functions, respectively. Degree 3, 4, and 5
HOPE THIS HELPS YOU..... :)
