The coordinates of the pre-image of point F' is (-2, 4)
<h3>How to determine the coordinates of the pre-image of point F'?</h3>
On the given graph, the location of point F' is given as:
F' = (4, -2)
The rule of reflection is given as
Reflection across line y = x
Mathematically, this is represented as
(x, y) = (y, x)
So, we have
F = (-2, 4)
Hence, the coordinates of the pre-image of point F' is (-2, 4)
Read more about transformation at:
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Hey mate !!
Here's your answer !!
2w^2 - 11w = -12
2w^2 -11w + 12 = 0
2w^2 - 8w - 3w + 12 = 0
2w ( w - 4) -3 ( w - 4) = 0
(2w - 3) ( w - 4) = 0
Hence
2w - 3 = 0
2w = 3
w = 3/2
w - 4 = 0
w = 4
Hence value of w are 4 and 3/2
Hope this helps!!
Cheers!!
Area of a sector is x over 360 * pi r squares so in this case x would be 210 so it would be 210 over 360 and the pi r squared would be 24 pi and when we times that our answer is 14 pi
Answer:14 pi
Answer:
72
Step-by-step explanation:
RQS + KLM = 180 degrees, because RQS = QLK, and QLK + KLM = 180 degrees (because they are on the same line).
Substitute the values for RQS and KLM:
x-36 + x = 180 -->
2x-36=180 -->
2x = 144 -->
x=72
They are up and down and left and right
