Explain what u need help with cuz i think i know this
We have that
<span>A' (2,1)
C(-2,2)-------> </span>Using the transformation-----> C' (-2+2,2+1)----> C' (0,3)
with
A' (2,1) and C' (0,3)
find the distance <span>C'A'
d=</span>√[(y2-y1)²+(x2-x1)²]----> d=√[(3-1)²+(0-2)²]----> d=√8----> 2√2 units
the answer is
the distance C'A' is 2√2 units
The answer is 3x(-12) I 'am 100% show..
(i) To show that a piecewise function is continuous at a point, we need to show that the left hand and right hand limit "agree" with each other. In other words, we want:

Now, since we're given the constraints and the equation of each constraint, we notice that 6061^+ is a number that is slightly bigger than 6061. So we use the second equation. Do you see why?
In much the same way, 6061^- is a number that is slightly smaller than 6061. So we use the first equation. Again, do you see why? (Hint: look at the conditions on x for each equation).
So finally, computing each limit means just "plugging" 6061 into their respective equations. That is:


Since your limits match, we say that, at the point x = 6061, T(x) IS continuous.
(ii) Repeat the process above with x = 32473.
(iii) Find a point of discontinuity just means your right hand and left hand limits do not match -- I'm not an economist, so I may not be of much help with the latter part of the question!