A red ball and a green ball are simultaneously tossed into the air. The red ball is given an initial velocity of 96 feet per second, and its height, t seconds after it is tossed is (-16t^2 + 96t) feet. The green ball is given an initial velocity of 80 feet per second, and its height, t seconds after it is tossed is (-16t^2 +80t) feet.

1: Write the algebraic expression in simplest form that represents the difference in heights between the two balls at any time, t.
2: How much higher is the red ball after 2 seconds?

Your friend is taking an algebra class that began one week after ours. Explain to your friend how to graph linear inequalities with two variables

To find the difference in heights between the two balls at any time, t, we need to subtract the height of the green ball from the height of the red ball.

1: Algebraic expression representing the difference in heights between the two balls at any time, t:
Difference in heights = (Red ball height) - (Green ball height)
Difference in heights = (-16t^2 + 96t) - (-16t^2 + 80t)

To simplify this expression, we can remove the parentheses and combine like terms:
Difference in heights = -16t^2 + 96t + 16t^2 - 80t

The terms "-16t^2" and "+16t^2" cancel out each other, leaving us with:
Difference in heights = 96t - 80t

Now, we can further simplify by combining like terms:
Difference in heights = 16t

Therefore, the algebraic expression in simplest form that represents the difference in heights between the two balls at any time, t, is 16t.

2: To find how much higher the red ball is after 2 seconds, we need to substitute t = 2 into the expression for the red ball's height:

Red ball height at t = 2:
(-16 * (2)^2 + 96 * (2))
(-16 * 4 + 192)
(-64 + 192)
128 feet

Therefore, the red ball is 128 feet higher than the green ball after 2 seconds.