Here it is given that AB || CD
< EIA = <GJB
Now
∠EIA ≅ ∠IKC and ∠GJB is ≅ ∠ JLD (Corresponding angles)
∠EIA ≅ ∠GJB then ∠IKC ≅ ∠ JLD (Substitution Property of Congruency)
∠IKL + ∠IKC 180° and ∠DLH + ∠JLD =180° (Linear Pair Theorem)
So
m∠IKL + m∠IKC = 180° ....(1)
But ∠IKC ≅ ∠JLD
m∠IKC = m∠JLD (SUBTRACTION PROPERTY OF CONGRUENCY)
So we have
m∠IKL + m∠JLD = 180°
∠IKL and ∠JLD are supplementary angles.
But ∠DLH and ∠JLD are supplementary angles.
∠IKL ≅ ∠DLH (CONGRUENT SUPPLEMENTS THEOREM)
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.
Answer:
Step-by-step explanation:
Answer:
What grade are you?
Step-by-step explanation:
Cause it's hard even gòogle can't answer it.
Parallelogram has two pairs of parallel sides. Obviously the left and lights are well and grandpa parallel because they’ll give me other thing that marks them is that they both have the same one that means that the sides that they intersect, the top and the bottom, or also parallel because they are the same distance