Someone please hurry and help with number 4 and 6
2 answers:
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Answer:
No invariant point
Step-by-step explanation:
Hello!
When we translate a form, in this case a polygon We must observe the direction of the vector. Since our vector is:
1) Let's apply that translation to this polygon, a square. Check it below:
2) The invariant points are the points that didn't change after the transformation, simply put the points that haven't changed.
Examining the graph, we can see that no, there is not an invariant point, after the translation. There is no common point that belongs to OABC and O'A'B'C' simultaneously. All points moved.
Let be the randomly selected lengths. Without loss of generality, suppose
where are independent random variables with the same uniform distribution on [0, 1].
By their mutual independence, we have
so that the joint density function is
where is the cube with vertices at (0, 0, 0) and (1, 1, 1).
Consider the plane
with . This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)
The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region , since
. So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.
Then we have
