In the
plane, we have
everywhere. So in the equation of the sphere, we have

which is a circle centered at (2, -10, 0) of radius 4.
In the
plane, we have
, which gives

But any squared real quantity is positive, so there is no intersection between the sphere and this plane.
In the
plane,
, so

which is a circle centered at (0, -10, 3) of radius
.
Answer:
Please include the table below and the question
Step-by-step explanation:
Answer:
12^2 + 4x
Step-by-step explanation:
We need to solve the area of these two rectangles separately.
Equation: 4x(3x+1) --> 12x + 4
Area: 12x^2 + 4x
Hope this helps.
A: -5-(-13) = 8
b: -4-8 = -12
D: a^8 b^12
Answer:
45.2
Step-by-step explanation: