X = -3.
The distance from p(-9, 0, 0) is
d = sqrt((x+9)^2 + y^2 + z^2)
The distance from q(3,0,0) is
d = sqrt((x-3)^2 + y^2 + z^2)
Let's set them equal to each other.
sqrt((x+9)^2 + y^2 + z^2) = sqrt((x-3)^2 + y^2 + z^2)
Square both sides, then simplify
(x+9)^2 + y^2 + z^2 = (x-3)^2 + y^2 + z^2
x^2 + 18x + 81 + y^2 + z^2 = x^2 - 6x + 9 + y^2 + z^2
18x + 81 = - 6x + 9
24x + 81 = 9
24x = -72
x = -3
So the desired equation is x = -3 which defines a plane.
Answer:
1. d/a+c=d
2. (m+21)/5=n
3. (1/2+2q)*4=p or 2+8q=p
4. (p-2a)/2pi=r
5. {[(5c+1)/2]+c}/3=a
Answer:
1 is 8x + 16 = 6x
Step-by-step explanation:
Answer:
Step-by-step explanation:
We can combine the -7x and the -2x and the left side to get -9x and the 23 and 13 on the right side to get 36.
This gives us
-9x + 12 = 36
Subtacting 12 from both sides gives us
-9x = 24
dividing by -9 gives us
x = - 8/3
Answer:
i want to say the answer is C, i'm sorry if its incorrect, but i'm 50/50 that its C