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:
m < 5
Step-by-step explanation:
10m-5 < 45
Add 5 to both sides
10m - 5 + 5 < 45 + 5
Simplify
10m < 50
Divide both sides by 2
10m/10 < 50/10
Simplify
m < 5
<u><em>Kavinsky</em></u>
Since this point is located in the fourth quadrant you have to rotate it 270 degrees counterclockwise which would end up being in third quadrant. which means the new point would be (-6,-9)
Answer:
sin 77°
cos 49°
tan 22°
Step-by-step explanation:
use your calculator with trig functions to find the arcsin, arccos, and arctan
To start with, write both fractions with same denominator, i.e LCM of 6 and 7
Thus,
4/7=24/42
1/6=7/42
This means 4/7 is greater than 1/6
The answer is thus a
