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.
Step-by-step explanation:
Refer to the attachment.
I hope it helps:)
(x,y)
x=input
y=output
example
we see
g=(1,2)
theefor
g(1)=2
a.
f(4)=1
g(1)=2
g(f(4))=2
b.
g(-2)=4
f(4)=1
f(g(-2))=1
c.
f(3)=5
g(5)=5
f(5)=0
f(g(f(3)))=0
The factors of 117 are: 1, 3, 9, 13, 39, and 117
The factors of 99 are: 1, 3, 9, 11, 33, and 99
The factors of 126 are: 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, and 126
The greatest common factors of 117, 99, and 126 are: 1, 3, and 9. And according to your answer choices your answer is C. 9