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.
<h3>
Answer: 130</h3>
Explanation:
Let x be the unknown angle we want to find.
Let y be adjacent and supplementary to x. This means x+y = 180
Let z also be adjacent and supplementary to x. So x+z = 180 also
Subtracting the two equations leads to y-z = 0 and y = z. So effectively we've proven the vertical angle theorem.
Since the supplementary angles to x add to 100, we know that y+z = 100. Plug in y = z and solve for z
y+z = 100
z+z = 100
2z = 100
z = 100/2
z = 50
Therefore,
x+z = 180
x+50 = 180
x = 180-50
x = 130
3(2x+2)=3x-15
mutiply the bracket by 3
(3)(2x)(3)(2)= 6x+6
6x+6= 3x-15
move 3x to the other side
sign changes from +3x to -3x
6x-3x+6= 3x-3x-15
6x-3x+6= -15
3x+6= -15
move +6 to the other side
sign changes from +6 to -6
3x+6-6= -15-6
3x= -21
divide by 3 for both sides
3x/3= -21/3
x= -7
Answer: x= -7
Answer:
It Mike who ate the most pizza
Step-by-step explanation: