<span>
Exercise #1:
Point H = (–2, 2)
Point J = (–2, –3)
Point K = (3, –3)
It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it. Then
you'd immediately see what's going on.
Notice that points H and J have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.
</span><span>Notice that points J and K have different x-coordinates but
the same y-coordinate, so they're on the same horizontal line.
Notice that point-J is on both the horizontal line and the vertical
line, so the lines meet there, and they're perpendicular.
Point-J is one corner of the square.
H is another corner of the square. It's 5 units above J.
K is another corner of the square. It's 5 units to the right of J.
The fourth corner is (2, 3) ... 5 to the right of H,
and 5 above K.
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Exercise #2:
</span><span>Point H = (6, 2)
Point J = (–2, –4)
Point K = (-2, y) .
</span><span>It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it. Then
you'd immediately see what's going on.
</span><span>Notice that points J and K have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.
We need K to connect to point-H in such a way that it's on
the same horizontal line as H. Then the vertical and horizontal
lines that meet at K will be perpendicular, and we'll have the
right angle that we need there to make the right triangle.
So K and H need to have the same y-coordinate.
H is the point (6, 2). So K has to be up at (2, 2) .
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Exercise #3:
</span>
<span>Point H = (-6, 2)
Point J = (–6, –1)
Point K = (4, 2) .
</span>
<span>It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it. Then
you'd immediately see what's going on.
This exercise is exactly the same as #1, except that it's a
rectangle instead of a square. It's still make of horizontal
and vertical lines, and that's all we need to know in order
to solve it.</span><span>
Notice that points H and J have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.
</span><span>Notice that points H and K have different x-coordinates but
the same y-coordinate, so they're on the same horizontal line.
Notice that point-H is on both the horizontal line and the vertical
line, so the lines meet there, and they're perpendicular.
Point-H is one corner of the rectangle.
J is another corner of the rectangle. It's 3 units below H.
K is another corner of the square. It's 4 units to the right of H.
The fourth corner is (2, -1) ... 4 to the right of J,
and 3 below K.
</span>
Answer:confounding variable
Step-by-step explanation:
I believe you're asking about the one-sided limit,
Recall the definition of absolute value:
• 
if 
• 
if 
Since we're approaching -4 from the left, we're effectively focusing on a domain of 
or 
. So, by the definition above, we have 
. Then in the limit, we have
=(100-40)/2-30
=60/2-30
=30-30
=0
