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Q: What clockwise rotation produces the same image as a counterclockwise rotation of 220 degrees?
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Related questions

Why doesn't the direction of rotation clockwise and counterclockwise matter when the angle of rotation is 180?

Because 180 degrees clockwise is the same as 180 degrees counterclockwise.


What is the image of the point (43) if the rotation is 90 degrees?

The answer will depend on whether the rotation is clockwise or counterclockwise.


180 degrees counterclockwise rotation?

Fomula(work with both clockwise/counterclockwise):(-x,-y)


What is the image of point (-1-2) if the rotation is 90 degrees?

The answer will depend on whether the rotation is clockwise or counterclockwise.


What is the image of (1 -6) for a 180 counterclockwise rotation about the origin?

It is (-1, 6).Also, if the rotation is 180 degrees, then clockwise or anticlockwise are irrelevant.It is (-1, 6).


What is 90 degrees counterclockwise?

Clockwise means turning to your right, counterclockwise is to the left.


What is the answer to rotate 180 degrees counterclockwise?

The same as 180 degrees clockwise. What do you mean "the answer to"?


What do you if both the coordinates are negative when doing a rotation of 90 degrees?

The answer will depend on whether the rotation is clockwise or anti-clockwise.


How do you get an ordered pair of 90 degrees?

Assume we want to find the ordered pair after 90° counterclockwise rotation. From (x,y), we have (-y,x). If we want to find the ordered pair after 90° clockwise rotation, then from (x,y) we have (y, -x)


90 degrees clockwise rotation?

It is 1/4 of a turn


Rule for 90 degrees counterclockwise rotation?

(x,y)-> (-y,x)


What is the image of a point 4 3 if the rotation is 270 degrees?

A measure of rotation MUST state whether it is clockwise or anti-clockwise. Unless the rotation is 0 degrees (ie no rotation) or 180 degrees (the two are the same). It must also specify the centre of rotation. Since you have not bothered to share these crucial bits of information, I cannot provide a more useful answer.