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:
B
Step-by-step explanation:
1) lets use elimination to solve this:
multiply 10x+4y=24 with 3 in order to get "4y" into "12y"
3(10x+4y=24)
30x+12y=72
now subtract both equations to eliminate 'y"
30x+12y=72
-
6x+12y=48
---------------------
24x=24
x=1
now substitute 1 in the above equation:
6(1)+12y=48
6+12y=48
12y=42
y=7/2
Answer:
66.3
Step-by-step explanation:
i just did it on deltamath
Answer:
0.5
Step-by-step explanation:
we see that A is 1 of 2 choices, so theoretically, P(A) = 1/2 = 0.5
<h2>
Answer:</h2>
<u>The answer is </u><u>(C) 4</u>
<h2>
Step-by-step explanation:</h2>
According to the rule of BODMAS
We will first do division then Multiplication then addition and in the last we do subtraction
So
12 ÷ 2 + 4 – 2 × 3 will be done as
= (12 ÷ 2) + 4( – 2 × 3)
= 6+4-6
= 4