Answer:
Left column:
1 & 2
right column:
1 & 2
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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:
Adult=58
Step-by-step explanation:
c=child, a=adult
6.4c+9.7a=1145 equation 1
c+a=149 equation 2
a=149-c modified equation 2 to isolate a
6.4c+9.7(149-c)=1145 substitute value of a from equation 1 into equation 2
6.4c+1445.3-9.7c=1145
-3.3c=-300.3
c=91
solve for a
c+a=149
91+a=149
a=58
Check answer:
6.4c+9.7a=1145
6.4(91)+9.7(58)=1145
582.40+562.60=1145
1145=1145
System b should be (-3,2)!
