Well, a distance-preserving transformation is called a rigid motion, and the name suggests that it <em>moves the points of the plane around in a rigid fashion.</em>
A transformation is distance-preserving if the distance between the images of any two points and the distance between the two original points are equal.
If that's confusing, I get it; basically if you transform something, the points from the transformation are image points. Take the distance from a pair of image points, and take the distance from a pair of original points, and they should be the same for a <em>rigid </em>motion.
I keep emphasizing this b/c not all transformations preserve distance; a dilation can grow or shrink things. But if you didn't go over dilations, don't say nothin XD
Answer:
yes I try but not any solution come
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First we will convert those radian angles to degrees, since my mind works better with degrees. Let's work one at a time. First,

. If we start at the positive x-axis and measure out 315 we end up in the 4th quadrant with a reference angle of 45 with the positive x-axis. The side across from the reference angle is -1, the side adjacent to the angle is 1, and the hypotenuse is sqrt2. The cotangent of this angle, then is 1/-1 which is -1. As for the second one, converting radians to degrees gives us that

. Sweeping out that angle has us going around the origin more than once and ending up in the first quadrant with a reference angle of 30° with the positive x-axis. The side across from the angle is 1, the side adjacent to the angle is √3, and the hypotenuse is 2. Therefore, the secant of that angle is 2/√3.
Answer:
tbh i really dont know
Step-by-step explanation:
Now, we know that 90°< θ <180°, that simply means the angle θ is in the II quadrant, where sine is positive and cosine is negative.
