A point is reflected over the y-axis, but the x-coordinate does not change. What is the location of the point? (0, 3) (-1, 0) (1
, 1) (-1, -1)
2 answers:
Answer:
Step-by-step explanation:
When we apply a reflection in a point over an axis this is what happens, especially if we consider, a reflexion of those points over y-axis
A(0,-3) B(-1,0) C(1,1) D(-1,-1)
Since the reflection over y-axis equals to (-x,y) we are going to have a change. But the question says we're not going to have any change at all. Calculating to prove it:
(-x,y) Reflection across y-axis
A(0,-3) then A'(0,-3)
B(-1,0) then =B'(1,0)
C(1,1) then C'(-1,1)
D(-1,-1) then D'(1,-1)
Hence the point is A, whose A' did not change its x-coordinate across the y-axis after reflected.
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Answer:
d = -1/3, 0
Step-by-step explanation:
Subtract the constant on the left, take the square root, and solve from there.
(6d +1)^2 + 12 = 13 . . . . given
(6d +1)^2 = 1 . . . . . . . . . .subtract 12
6d +1 = ±√1 . . . . . . . . . . take the square root
6d = -1 ±1 . . . . . . . . . . . .subtract 1
d = (-1 ±1)/6 . . . . . . . . . . divide by 6
d = -1/3, 0
_____
Using a graphing calculator, it is often convenient to write the function so the solutions are at x-intercepts. Here, we can do that by subtracting 13 from both sides:
f(x) = (6x+1)^ +12 -13
We want to solve this for f(x)=0. The solutions are -1/3 and 0, as above.
