Answer:
62.5%
Step-by-step explanation:
im right
So, the best way to do this is translate it to clockwise. 90 degrees counterclockwise is equal to 270 degrees clockwise. So, basically, to rotate, you would follow the following format for each point-
(X,Y) -> (-Y,X)
Now, you do it for each of the points.
A= (-5,5), so A' would be (-5,-5)
B= (-1,5), so B' would be (-5,-1)
C= (-5,4), so C' would be (-4,-5)
D= (-1,4) so D' would be (-4,-1)
Notice, how all the points end up in the square below it. Each quadrant has a specific number. The top right is quadrant 1, the top left is quadrant 2, the bottom left is quadrant 3, and the bottom right is quadrant 4. If you are rotating 270 degrees clockwise, you move to the right, like a clock. That puts the new rectangle in quadrant 3. That is a way to check your work.
Now, just so you know for future reference, the following are also different formats for different problems--
A 90 degree Clockwise rotation about the origin will be (X,Y) -> (Y, -X) *Note, -x just stands for the opposite. Say your original x is a negative number. Then the prime (new) x will be positive.
A 180 degree Clockwise rotation about the origin would be (X,Y) -> (-X,-Y) *Note, -y also stands for the opposite.
A 270 degree clockwise rotation about the origin would be (X,Y) -> (-Y,X).
For translating---
90 degrees Clockwise = 270 degrees Counter
270 degrees Clockwise = 90 degrees Counter
Hope this helped!
9514 1404 393
Answer:
- Translate P to E; rotate ∆PQR about E until Q is coincident with F; reflect ∆PQR across EF
- Reflect ∆PQR across line PR; translate R to G; rotate ∆PQR about G until P is coincident with E
Step-by-step explanation:
The orientations of the triangles are opposite, so a reflection is involved. The various segments are not at right angles to each other, so a rotation other than some multiple of 90° is involved. A translation is needed in order to align the vertices on top of one another.
The rotation is more easily defined if one of the ∆PQR vertices is already on top of its corresponding ∆EFG vertex, so that translation should precede the rotation. The reflection can come anywhere in the sequence.
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<em>Additional comment</em>
The mapping can be done in two transformations: translate a ∆PQR vertex to its corresponding ∆EFG point; reflect across the line that bisects the angle made at that vertex by corresponding sides.
Answer:
no because 3/20 is eguivilent to 0.15 whitch is half of 0.3
I think it's 21/100 because I found the common denominator by multiplying by ten so I multiplied 1 times ten to get 10. Does that help?
