When reflecting across the Y axis, the Y values remain the same.
Now if you were reflecting across Y = 0, the x values would just be inverse ( opposite signs).
So this triangle if reflected across Y = 0 the new vertices would be (4,4) (2,3) and (5,2)
Now since the reflection line is y = -1, which is a one unit shift to the left of y = 0, subtract 1 unit from each X value.
The locations are now: A'(3,4), B'(1,3) and C'(4,2)
Answer:
99
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
using the power of elemination,looking at the first equation,we singled out the top 2, now looking at the 2nd equation, we see the y intercept is 2 meaning it has to be graph c
Area=LW
<span>10x^2-29x-21=area
factor and those are the lengh and width
(2x-7)(5x+3)
perimiter=2(L+W)
P=2(2x-7+5x+3)
P=2(7x-4)
P=14x-8
answer is D
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