= 3(4a^2 + 4a - 3) - 4a^2 + 28
= 12a^2 + 12a - 9 - 4a^2 + 28
= 8a^2 + 12a + 19
= 4a(2a + 3) + 19
The coordinates of the vertices of the image are L' = (-4, 6), M' = (-5, 1) and N' = (-7, 3)
<h3>What are the coordinates of the vertices of the image?</h3>
The vertices of the preimage of the triangle are given as:
L = (4, -6)
M = (5, -1)
N = (7, -3)
The rotation is given as: 180 degrees counterclockwise
<h3 />
The rule of this rotation is
(x, y) => (-x, -y)
So, we have:
L' = (-4, 6)
M' = (-5, 1)
N' = (-7, 3)
Hence, the coordinates of the vertices of the image are L' = (-4, 6), M' = (-5, 1) and N' = (-7, 3)
Read more about rotation at:
brainly.com/question/4289712
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Answer:
0.992
Step-by-step explanation:
do you not have a calculator
Answer:
650
Step-by-step explanation:
651.606881968=P
round: 650
