The answer should be 28/32
We use the distance formula for this problem.
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
The distance between point (-2,-2) and point (-2,4).
d = √[(⁻2 - ⁻2)² + (4 - ⁻2)²] = 6 units
Then, compute for 20% of 6 units:
Distance traveled = 6(0.2) = 1.2 units
Use 1.2 units as distance and the starting point (-2,-2). The x-coordinate should still be at -2 because the distance is a straight line as shown in the picture.
1.2 = √[(-2 - ⁻2)² + (y - ⁻2)²]
Solving for y,
y = -0.8
The point is found at (-2,-0.8). This is located at quadrant 3. As to the distance traveled, that would be: 1.2*6 = 6 miles. Thus, the answer is C.
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.
8 can go into 17 twice with one left over so
the answer is 2 1/8
Hope this helps!!!
Answer:
The probability that the pair consists of a female junior and a male sophomore is 0.096
Step-by-step explanation:
No of male juniors=4
No of female juniors=10-4=6
No of female sophomores=5
No of male sophomores=13-5=8
As per question,
Probability that the pair consists of a female junior and
a male sophomore =
=0.096
Therefore, the probability that the pair consists of a female junior and a male sophomore is 0.096