1- ASA for all triangle mentioned. The have one side adjacent to 2 angles
Answer: don't know sorry
Step-by-step explanation: YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET !!!!!!!!!!!!!!!!!!!!!!!!
Have a nice day Bye! :)
F(x) = 2x - 4
f(2 ≤ x) = 2(2 ≤ x) - 4
f(x ≥ 2) = 2(x ≥ 2) - 4
f(x ≥ 2) = 2(x) ≥ 2(2) - 4
f(x ≥ 2) = 2x ≥ 4 - 4
f(x ≥ 2) = 2x ≥ 0
f(x ≥ 2) = x ≥ 0
f(x) = 2x - 4
f(x ≤ 6) = 2(x ≤ 6) - 4
f(x ≤ 6) = 2(x) ≤ 2(6) - 4
f(x ≤ 6) = 2x ≤ 12 - 4
f(x ≤ 6) = 2x ≤ 8
f(x ≤ 6) = x ≤ 4
Nope. Perimeter = sum of all sides
P = 20 + 11 + 20 + 11 = 62 and not 64
So, it is not possible.
