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:
x=y+c/a
Step-by-step explanation:
Answer:
(x, y) = ( 1/2, -3 )
Step-by-step explanation:
y = -3
3/2 - 1 - x + y = - 3
3/2 - 1 - x - 3 = - 3
x = 1/2
solution : (x , y ) = ( 1/2, -3)
here we are gonna see why is that answer :
-3 = 3/2 -1 - 1/2 - 3 = -3
-3 = -3 = -3
i hope this helps
Answer:
A. x= 3, y= -2
Step-by-step explanation:
[2] 2x= 4y+14
[1] 3y+5x=9
Solve equation [2] for the variable x
[2] 2x = 4y + 14
[2] x = 2y + 7
// Plug this in for variable x in equation [1]
[1] 5•(2y+7) + 3y = 9
[1] 13y = -26
// Solve equation [1] for the variable y
[1] 13y = - 26
[1] y = - 2
// By now we know this much :
x = 2y+7
y = -2
// Use the y value to solve for x
x = 2(-2)+7 = 3
good luck, I hope this helps!!!