Question

Write down an (in)equality which describes the solid ball of radius 3 centered at (7,1,−1). It should have a form like x2+y2+(z−2)2−4>=0, where you use one of the following symbols ≤,<,=,≥,>.

Answer 1

Answer:

$(x-7)^2+(y-1)^2+(z+1)^2-9\le 0$

Step-by-step explanation:

A ball is the solid inside a surface which is a sphere.

The equation of a sphere has the form:

$(x-h)^2+(y-k)^2+(z-s)^2= r^2$

Where (h,k,s) is the center and r is the radius.

When we plug the given center (7,1,-1) and radius 3 we get:

$(x-7)^2+(y-1)^2+(z+1)^2= 3^2$

Then we turn this into an inequality: Since the ball is inside the surface we must use the $\le$ symbol.

$(x-7)^2+(y-1)^2+(z+1)^2\le 9$

Then we just subtract 9 from both sides to get:

$(x-7)^2+(y-1)^2+(z+1)^2-9\le 0$