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4 years ago

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

Mathematics

1 answer:

Sedbober [7]4 years ago

4 0

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.

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<h3>What are equivalent expressions?</h3>

Equivalent expressions are simply known as expressions with the same solution but different arrangement.

Given the index expressions;

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This can be written as;

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2 years ago

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4 years ago

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4 years ago

Which statement best describes f(x)= -2 √x-7 +1 –6 is in the domain of f(x) but not in the range of f(x). –6 is not in the domai

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Answer:

B. -6 does not belong to the domain but belongs to the range of f(x).

Step-by-step explanation:

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That is, the domain is {x | x≥ 7}.

Now as we have,

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4 years ago