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 and D
Step-by-step explanation:
Have a great day!
/(OvO)/ *.':☆ *': .✧
Answer:
see explanation
Step-by-step explanation:
Given
4
- 5a² + 1 = 0
Use the substitution u = a², then equation is
4u² - 5u + 1 = 0
Consider the product of the coefficient of the u² term and the constant term
product = 4 × 1 = 4 and sum = - 5
The factors are - 4 and - 1
Use these factors to split the u- term
4u² - 4u - u + 1 = 0 ( factor the first/second and third/fourth terms )
4u(u - 1) - 1(u - 1) = 0 ← factor out (u - 1) from each term
(u - 1)(4u - 1) = 0
Equate each factor to zero and solve for u
u - 1 = 0 ⇒ u = 1
4u - 1 = 0 ⇒ 4u = 1 ⇒ u = 
Convert u back into terms of a, that is
a² = 1 ⇒ a = ± 1
a² =
⇒ a = ± 
Solutions are a = ± 1 , a = ± 
Answer:
D. 
Step-by-step explanation:
we are given

Firstly, it is reflected over x-axis
we know that for reflection about x-axis , we can replace y as -y
we get


now, it is vertical shift up by 4 units
so, we can add 4 to y-value
we get


Well if I did the math correctly it should be B) <span>9√5</span>