The correct classification for the polygon would be a concave hexagon.
Answer:
<h2>

</h2>
Step-by-step explanation:
Given,







Comparing with A • 
A = 
Hope this helps...
Good luck on your assignment...
Answer:
J+2J=12 so 3J=12 so J=4
Step-by-step explanation:
Use J to represent Jake's miles.
We know that Gloria's miles (G) is twice J, so G=2J
We also know that Jake and Gloria together ran 12 miles.
So we know, J+2J=12 or 3J=12 meaning J=4
So Jake runs 4 miles, and since Gloria is two times that she ran 8 miles.
"He starts both trains at the same time. Train A returns to its starting point every 12 seconds and Train B returns to its starting point every 9 seconds". Basically, what you need to do is find the least common multiple. The least common multiple of 12 and 9 is 36, so the least amount of time, in seconds, that both trains will arrive at the starting points at the same time is 36 seconds.