Answer:
90 degree rotation in the clockwise direction.
Step-by-step explanation:
Point A transforms to A'
- that is x coordinate: 2 ---> 3
and y coordinate 3 ---> -2
So the rotation is clockwise from Quadrant1 to Quadrant 4.
The slope of OA = 3/2 and the slope of OA' = -2/3.
The product of these slopes = 3/2 * -2/3 = -1 so the lines are perpendicular - that is the line has passed through an angle of 90 degrees.
A similar result occurs if we consider points B, C and D.
Answer:
6-b
Step-by-step explanation:
I hope this helps, and have a great day! :D
Answer:
Yes
Step-by-step explanation:
By rounding to the 10ths place, we can easily see that 6.3 is greater than 6.04. 6.3 is already rounded to the 10ths place, but 6.04 rounded to the 10ths place is 6.0.
6.3 is clearly more than 6.0, therefore 6.3 is greater than 6.04.
<span><u><em>The correct answer is:</em></u>
180</span>°<span> rotation.
<u><em>Explanation: </em></u>
<span>Comparing the points D, E and F to D', E' and F', we see that the x- and y-coordinates of each <u>have been negated</u>, but they are still <u>in the same position in the ordered pair. </u>
<u>A 90</u></span></span><u>°</u><span><span><u> rotation counterclockwise</u> will take coordinates (x, y) and map them to (-y, x), negating the y-coordinate and swapping the x- and y-coordinates.
<u> A 90</u></span></span><u>°</u><span><span><u> rotation clockwise</u> will map coordinates (x, y) to (y, -x), negating the x-coordinate and swapping the x- and y-coordinates.
Performing either of these would leave our image with a coordinate that needs negated, as well as needing to swap the coordinates back around.
This means we would have to perform <u>the same rotation again</u>; if we began with 90</span></span>°<span><span> clockwise, we would rotate 90 degrees clockwise again; if we began with 90</span></span>°<span><span> counter-clockwise, we would rotate 90 degrees counterclockwise again. Either way this rotates the figure a total of 180</span></span>°<span><span> and gives us the desired coordinates.</span></span>