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:
85
Step-by-step explanation:
I hope my answer help you
Answer:
Step-by-step explanation:
y=-3x²-4x+7...(*)
3x+2y=18...(**)
use (*) put the value of y in (**) :
3x +2(-3x²-4x+7)=18
3x-6x²-8x +14 -18 = 0
-6x²-5x- 4 = 0
6x²+5x+4 = 0
delta = b² - 4ac a = 6 b=5 c =4
delta = 5² - 4(6)(4)= 25 -96= -71 negatif
so no reals solution
Multiply the whole thing by (3a/3a) to get (6a)/(3+1)=6a/4=3a/2, so ya, 2nd is correct