State the x-values where the limit does not exist.

9514 1404 393

Step-by-step explanation:

There are vertical discontinuities in the function at x=-2, x=2, and x=4. When looking at limits, we find the function values to be different either side of those discontinuities at x = -2 and at x = 2. Thus, limits do not exist at those points (x = {-2, 2}).

At x=4, the discontinuity is in the function definition, not in the values left or right of that point. Hence the limit does exist at x=4, but the function is not continuous there.

y = x + 95

y = 42 + 95

y=137 mm hg

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3