Answer:
12x^4-35x^3+45x^2-13x-15
Step-by-step explanation:
some issues with the signa
You can solve the pair of equations graphically by finding where the two lines intersect/meet. The point where they intersect is the solution to both of the equations.
Slope-intercept form: y = mx + b
(m is the slope, b is the y-intercept, or the point where x = 0 ---> (0 , y))
y = 7x - 9
m = 7
y-intercept = -9 -----> (0, -9)
y = 3x - 1
m = 3
y-intercept = -1 ------> (0, -1)
I'm not really sure what the last sentence of the question is asking, so you could clarify the question if you need help
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
.