If ΔABC ≅ ΔEDF where the coordinates of A(0, 2), B(2, 4), and C(2, −1), what is the measure of DF?
2 answers:
BC = DF
(-1-4)^2 + (2-2)^ 2
(-5)^2 + (0)^2
Square root 25 is 5
C is the answer. Can you mark me as brianliest? Really appreciate it!
Answer:
C. 5
Step-by-step explanation:
Since Triangle ABC is congruent to EDF it mean that the sides are the same so the length of BC is congruent to the length of DF:
distance formula:
d = √(x2 - x1)^2 + (y2 - y1)^2
d = √(2 - 2)^2 + (-1 - 4)^2
d = √(0)^2 + (-5)^2
d = √0 + 25
d = √25
d = 5
You might be interested in
Answer:
The overall change in field position is 7yds, 0 - 5 + 12 = 7, if that isn't a "sum of integers" I apologize.
Step-by-step explanation:
<span>g = 9.8 m/s2 and PE = m × g × h.
hope this helps you(:</span>
C=2πr r radius it would be right
Answer:
the answer is the third one
Step-by-step explanation:
Answer
nvm................
