A ball is thrown into the air at an initial velocity of 18 meters per second from an
initial height of 10 meters. The equation that models the path of the ball is given
by h = -4.9t2 + 18t + 10. What is the maximum height of the ball?

The maximum height is 26.5 meters.

Step-by-step explanation:

I know this graph is facing downwards because of the negative leading coefficient, this means that the vertex is a maximum not a minimum. In order to find the maximum height, we would need to take the derivative and equate it to 0, this will give us the vertex. I got that t = .

This is the time in which the ball was at the vertex or maximum. Now, in order to find the height we would need to plug it in back to the formula and this would give us the height. In this case it is 26.5 meters.

Hope this helps

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

answer: 0+0=0

step-by-step explanation:

that is the only answer that can make 0