If r(x) = 2 – x2 and w(x) = x – 2, what is the range of (wºr) (x)
(-∞,0]
(-∞,2]
[0,∞)
[2,∞)

(-inf,2]

Step-by-step explanation:

I replaced r(x) with 2-x^2

I replace the x in w(x)=x-2 with 2-x^2

You can graph this to find the range.

But since this is a quadratic (the graph is a parabola), I'm going to find the vertex to help me to determine the range.

The vertex is at x=-b/(2a).  Once I find x, I can find the y that corresponds to it by using y=-x^2-x+2.

Comparing ax^2+bx+c to -x^2-x+2 tells me a=-1, b=-1, and c=2.

So the vertex is at x=1/(2*-1)=-1/2.

To find the y-coordinate that corresponds to that I will not plug in -1/2 in place of x into -x^2-x+2.

This gives me

-(-1/2)^2-(-1/2)+2

-1/4 +  1/2  +2

Find a common denominator which is 4.

-1/4 +  2/4  +8/4

8/4

2.

So the highest y value is 2 ( I know tha parabola is upside down because a=negative number)

That mean then range is 2 or less than 2.

So the answer an interval notation is (-inf,2]

uhhh idk u yolo?

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v