Paul lives 6 miles west and 3 miles north of school. What is the direct distance from Paul's house to school?
2 answers:
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Answer:
The points (x,y,z) that respond to Ir-r0I =1, are all that describes the form with:
-1+x₀<x<1+x₀
-1+y₀<y<1+y₀
-1+z₀<z<1+z₀
Step-by-step explanation:
All points required in this problem came from applying the definition of modulus of a vector:
Ir-r0I =1.
And
(16x4-9)(x2-5)2 (16x4-25)(x4-9)
(16x4-9)(x2-5)2 (16x4-25)(x4-9)
<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.
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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.