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:
(3x^2 + 2)(x + 4).
Step-by-step explanation:
3x3 + 12x2 + 2x + 8
3x^2(x + 4) + 2(x + 4) The x + 4 is common to the 2 groups so we have:
(3x^2 + 2)(x + 4).
The least is 0,015
hope this will help you
Answer:
m(arc FE) = 166°
Step-by-step explanation:
By theorem of intersecting chords and tangent of a circle,
Measure of the angle formed by a chord and tangent intersecting on the circle measure the half of the intercepted arc.
m(∠GFE) = 
83° = 
m(arc FE) = 166°