Answer:
90deg
Step-by-step explanation:
If the angles are supplementary, that means mA + mB = 180*
we know that
mA = (2x + 8)deg
mB = (2x + 8)deg = mA
mA + mB = 180deg
mB + mB = 2mB = 180deg
mB = 90deg
Note, that we did not need to compute x even, because the angles were the same.
You have not provided the diagram/coordinates for point Q, therefore, I cannot provide an exact answer.
However, I can help you with the concept.
When rotating a point 90° counter clock-wise, the following happens:
coordinates of the original point: (x,y)
coordinates of the image point: (-y,x)
Examples:
point (2,5) when rotated 90° counter clock-wise, the coordinates of the image would be (-5,2)
point (1,9) when rotated 90° counter clock-wise, the coordinates of the image would be (-9,1)
point (7,4) when rotated 90° counter clock-wise, the coordinates of the image would be (-4,7)
Therefore, for the given point Q, all you have to do to get the coordinates of the image is apply the transformation:
(x,y) .............> are changed into.............> (-y,x)
Hope this helps :)
Answer:
-4a + 7a = + 3a
But you don't actually need a pkus sign 3a is fine too. You are basically sum - 4 and 7. Let me know if you still confused.
Answer:
(x, y) = (0, 3)
Step-by-step explanation:
(x, y) = (0, 3) is in both tables. Hence, that point is a solution to the system of equations.
