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)
To solve
if you have |a|=b, solve a=b and a=-b
|4x+12|=16
solve
4x+12=16 and 4x+12=-16
minus 12 both sides
4x=4 and 4x=-28
divide both sides by 4
x=1 and x=-7
Answer:
The first one: the error was that the 3 shouldn’t have been moved but the 5 should have. The correct answer is k=-8
The second one: the error was that they added the 6 and 4, you would only get -10 if the 4 was - the correct answer is -2
You can check your answer by plugging in your answer to the variable and making sure both side equal each other
Step-by-step explanation:
The first one: the 5 should be - from both side so you get k=-8
The second one: the 4 should be + to both sides. And the -6 and +4 should be added together to get n=-2
Remember you always want to get the variables by themselves.
Sorry that was a lot. Hope it helps!
Answer:
C
Step-by-step explanation:
Side Side Angle does not exist.
