Thank you for your comment. This video shows/proves that 0.999... is equal to 9/9, which reduces to 1. The pattern with repeating decimals that consistently have one digit repeating (e.g., 0.22222...) is that they are equal to the number being repeated over 9. This extra example of 0.22222... is equivalent to 2/9. More information on how/why 0.999999... = 1 can be found through the following links: - en.wikipedia.org/wiki/0.999... - www.purplemath.com/modules/howcan1.htm

@@audracrist8451 multiplying 0.9... by 10, by adding a 9 in front of the decimal point is an error, not a proof.

I agree. I get 47.9, which should be rounded to 48, not 47. Thanks for pointing that out.

What updated info are you looking for?

Mikhael, This is a tricky one. But, let's start with projectile motion - we break the motion into x and y components that are independent of each other. "g" will not affect motion in the x direction, only the y. For this problem, to make the math easier, we created a coordinate system in 3 directions - the vt, 1/2 gt^2, and r directions. So, g is only affecting the motion in the 1/2 gt^2 direction. All three are independent of each other.

We're finding average velocity, which is displacement over time. Displacement is a vector, so directions count. Typically we assign left, down and west as a negative displacement and right, up and east as positive. So, the first displacement to the left is a negative number: -60 m. Then we algebraically add that to the right displacement which is + 60. We get zero, so the avg velocity is zero. Another way of looking at it, is the person went 60 m left, and then 60 m right - winding up in the same place the person started. So final position minus initial position is zero - displacement is zero, so zero average velocity when you divide 0 by time.

Poyraz, sorry for the late answer, but you're absolutely correct, and the presentation was actually fixed last week. Thank you. John