Answer:
The points (x,y,z) that respond to Ir-r0I =1, are all that describes the form 
with:
-1+x₀<x<1+x₀
-1+y₀<y<1+y₀
-1+z₀<z<1+z₀
Step-by-step explanation:
All points required in this problem came from applying the definition of modulus of a vector:
Ir-r0I =1.
Answer:
i cant see it can you get a batter pic
Step-by-step explanation:
9 and a half hours. That is kind of a long time
<span>The correct answer is D) g(x) = -(x + 2)^2. The given formula F(x) = x^2 creates a parabola that is open at the top. To reflect this figure across the x-axis and have it open at the bottom, the y-position of the figure on the coordinate system for every x value, which is F(x) = y = x^2 has to be inverted. This is done by negating y and respectively x^2, so to reflect the figure on the x-axis the formula would now look like this: F(x) = -y = -x^2. To move any parabola two units to the left and thereby have its root be at -2, you would simply subtract -2 from every x-position of the figure in the coordinate system. For an inverted parabola like this one the value to move it on the x-axis has to be added instead and this results in the formula from answer D: g(x) = -(x+2)^2</span>
Enlargement. the image on the right is larger, the angles are the same, and the side lengths are proportional
