Answer:
I think the answer is 12.5
Answer:
Let A1=a1+a2+a3, A2=a2+a3+a4, and so on, A10=a10+a1+a2. Then A1+A2+⋯+A10=3(a1+a2+⋯+a10)=(3)(55)=165, so some Ai≥165/10=16.5, so some Ai≥17.
Step-by-step explanation:
The answer is B (3, 9), (1,-5), (3,3)
Answer: I’m pretty sure the answer is S
