10 points! i need filling this in for school asap!

Given : EFGH is a parallelogram.

Prove: EG bisects HF and HF bisects EG.

Since, a parallelogram has two pairs of congruent and parallel sides.

Therefore, In parallelogram EFGH,

EF≅GH and EH≅FG

Also, EF║GH and EH║FG

Since,In triangles EKF and GKH,

EF≅HG

Since,  EF║HG

⇒∠FEK ≅ ∠HGK and ∠EFK ≅ ∠ GHK ( When two parallel lines are cut by a transversal alternative interior angles are congruent )

⇒ Δ EKF ≅ Δ GKH  ( ASA congruence postulate )

⇒ EK ≅ GK  and FK ≅ HK ( CPCTC)

⇒ EG bisects HF and HF bisects EG ( Definition of bisector)

Hence proved.

step-by-step explanation:

we have

using a graphing tool

see the attached figure

this is a quadratic equation (vertical parabola) open up

the vertex is a > vertex

the domain of the function is all real > interval (-∞,∞)

the range of the function is the > [-6.125,∞)

the function has two x->   and

the function is increasing over the > (0.25,∞)

the function is decreasing over the > (-∞,0.25)

answer:

1.14285714286

step-by-step explanation:

divide 40 into 35 and you will get your answer