Answer: m = ½
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For this case we must indicate which of the equations shown can be solved using the quadratic formula.
By definition, the quadratic formula is applied to equations of the second degree, of the form:

Option A:

Rewriting we have:

This equation can be solved using the quadratic formula
Option B:

Rewriting we have:

It can not be solved with the quadratic formula.
Option C:

Rewriting we have:

This equation can be solved using the quadratic formula
Option D:

Rewriting we have:

It can not be solved with the quadratic formula.
Answer:
A and C
Answer:
8jbggvvhj was born on a 5
Answer:
yes
Step-by-step explanation:
The line intersects each parabola in one point, so is tangent to both.
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For the first parabola, the point of intersection is ...
y^2 = 4(-y-1)
y^2 +4y +4 = 0
(y+2)^2 = 0
y = -2 . . . . . . . . one solution only
x = -(-2)-1 = 1
The point of intersection is (1, -2).
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For the second parabola, the equation is the same, but with x and y interchanged:
x^2 = 4(-x-1)
(x +2)^2 = 0
x = -2, y = 1 . . . . . one point of intersection only
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If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.
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Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.