Given: klmn is a trapezoid, m∠n=m∠kml, fd=8, lm/kn = 3/5 f∈ kl , d∈ mn , me ⊥ kn kf=fl, md=dn, me=3sqrt5 find: km

KM = √70 ≈ 8.367

Step-by-step explanation:

FD is a midsegment whose length (8) corresponds to the average of the numbers of ratio units of LM : KN = 3 : 5. That is, each of those ratio units must be 2 length units, so KN = 10.

∠MKN = ∠KML = ∠N because the first two angles are alternate interior angles where transversal MK crosses parallel lines ML and KN. The latter two of these angles are equal because the problem statement says so.

These angle relationships make ΔKMN an isosceles triangle, with E the midpoint of the base. That is, KE = 5. We are told ME = 3√5, so the Pythagorean theorem tells us ...

... KM² = KE² +ME² = 5² +(3√5)² = 70

... KM = √70 . . . . . take the square root

never fear papaguy is here

answer

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.