Answer: A & C
<u>Step-by-step explanation:</u>
HL is Hypotenuse-Leg
A) the hypotenuse from ΔABC ≡ the hypotenuse from ΔFGH
a leg from ΔABC ≡ a leg from ΔFGH
Therefore HL Congruency Theorem can be used to prove ΔABC ≡ ΔFGH
B) a leg from ΔABC ≡ a leg from ΔFGH
the other leg from ΔABC ≡ the other leg from ΔFGH
Therefore LL (not HL) Congruency Theorem can be used.
C) the hypotenuse from ΔABC ≡ the hypotenuse from ΔFGH
at least one leg from ΔABC ≡ at least one leg from ΔFGH
Therefore HL Congruency Theorem can be used to prove ΔABC ≡ ΔFGH
D) an angle from ΔABC ≡ an angle from ΔFGH
the other angle from ΔABC ≡ the other angle from ΔFGH
AA cannot be used for congruence.
Answer:
Step-by-step explanation:
Given
--- domain
Required
Determine the range
To do this, we simply substitute the values of n in the given expression
So, we have:
Hence, the range is:
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