The answer is b. If your problem is y=1/2x-3
U just add the exponents.
b^8 x b^4 = b^12
hope this helps :)
"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.
I don't know you have to use a measuring tool to figure it out ruler helps sorry try a ruler of a angle scale
