Decrease £280 by 15%
1 answer:
You might be interested in
Answer:
( -1 , 26)
Step-by-step explanation:
So since you need to find the other endpoint, you would follow these steps:
1.)
2.) 8 = 9 + x ( you just multiplied the 2 to the 4 to get 8)
3.) -1 = x (just solve it like a regular equation, so just subtract 9 on both sides to get rid of it and that leaves you with -1 = x)
You took the x values of both points and put them in the equation.
And its the same for y
1.)
2.) 16 = -10 + y
3.) 26 = y (you added the 10 on both sides because the 10 was negative and that took the 10 out and so it left you with 26 = y)
<h3>Answer:</h3>
Any 1 of the following transformations will work. There are others that are also possible.
- translation up 4 units, followed by rotation CCW by 90°.
- rotation CCW by 90°, followed by translation left 4 units.
- rotation CCW 90° about the center (-2, -2).
The order of vertices ABC is clockwise, as is the order of vertices A'B'C'. Thus, if reflection is involved, there are two (or some other even number of) reflections.
The orientation of line CA is to the east. The orientation of line C'A' is to the north, so the figure has been rotated 90° CCW. In general, such rotation can be accomplished by a single transformation about a suitably chosen center. Here, we're told there is <em>a sequence of transformations</em> involved, so a single rotation is probably not of interest.
If we rotate the figure 90° CCW, we find it ends up 4 units east of the final position. So, one possible transformation is 90° CCW + translation left 4 units.
If we rotate the final figure 90° CW, we find it ends up 4 units north of the starting position. So, another possible transformation is translation up 4 units + rotation 90° CCW.
Of course, rotation 90° CCW in either case is the same as rotation 270° CW.
_____
We have described transformations that will work. What we don't know is what is in your drop-down menu lists. There are many other transformations that will also work, so guessing the one you have available is difficult.