Given: klmn is a trapezoid, kf =10 mf ║ lk aklmf = afmn find: kn

The trapezoid is shown in the picture attached.

We know:
LK // MF
KF = 10
A(KLMF) = A(FMN)

Since LM // KF (because they are bases of the trapezoid) and LK // MF because is given in the hypothesis, KLMF is a parallelogram.
The area of a parallelogram is given by the base times the height.
A(KLMF) = b × h
               = 10 × h

The area of a triangle is given by the formula:
A(FMN) = (b × h) / 2
             = (FN × h) / 2

We know that the two areas are congruent, therefore:
A(KLMF) = A(FMN)
10 × h = FN × h / 2

The two "h" cancel out because they are the same and we can solve for FN:
10 = FN / 2
FN = 20

Now we can calculate:
KN = KF + FN = 10 + 20 = 30

Hence, KN is 30 units long.

Idon't see any graph, however the graph of the function : f(x) = x²-2x+3 is a parabola: 1) open upward (since a> 0)2) it intersects y-axis in 3 (y=3)3) its axis of symmetry is x = +1 ( -b/2a)4) it's minimum is y = -2, so its vertex is (1,-2)5) it doesn't intersect the x-axis since discriminant < 0