No, only their positions will change.
ENE plus 90 degrees (clockwise) is SSE.
Rotating the graph y = x² clockwise 90° about the origin gives the graph of: y² = x → y = ±√x Removing the negative part leaves: y = √x (Note: it is convention that the radical symbol (√) means the positive square root.)
270 degrees is 3/4 of the way around the circle. Ir is the same as rotating it 90 degrees (1/4) of the way clockwise. Turn it so anything that was pointing straight up would be pointing to the right.
3 quarters clockwise is 270 degrees clockwise or 90 degrees anti(counter)-clocwise
rotate it 90 degrees
You dont, its just 90 degrees 3 times..
I dont really know if this is right but i think to do this problem you have to take a point then rotate the paper counter clockwise around the origin then you have a new point which is called a prime. Then reflect it over the y axis on the graph.
No, only their positions will change.
ENE plus 90 degrees (clockwise) is SSE.
stick your arms straight out in front of you. Pretend that's twelve o'clock then move one of your arms to three o'clock. Bring the other arm and turn your body to three o'clock. you have just moved 90 degrees clockwise.
Rotating the graph y = x² clockwise 90° about the origin gives the graph of: y² = x → y = ±√x Removing the negative part leaves: y = √x (Note: it is convention that the radical symbol (√) means the positive square root.)
Imagine a clock: a circle is 360 degrees, so every 5 minutes is 30 degrees. If you started at 1pm and rotated it 90 degrees it would be 1.15pm
Switch the x and y coordinates and multiply the first first coordinate (the new x coordinate) by -1
It rotates 90 degrees.
270 degrees is 3/4 of the way around the circle. Ir is the same as rotating it 90 degrees (1/4) of the way clockwise. Turn it so anything that was pointing straight up would be pointing to the right.
3 quarters clockwise is 270 degrees clockwise or 90 degrees anti(counter)-clocwise