Answer:
I don't matter of how well u do on edmentum diagnostic test for math,because u may have the main of knowledge...........pts
Answer:
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.
The standard equation for a circle is
(x-h)^2 + (y-k)^2 = r^2
where the center is at (h,k)
For this problem, the center is at (3,2). The abscissa and the ordinate are all positive. So this must be located on the first quadrant.
Hence, the answer is <span>It is located in the first quadrant..</span>
Y=-3x-5 would be the slope
12'5" is 149 inches. If 149in is the maximum, then anything above it including 162" cannot pass through safely.