Question

NO ONE IS HELPING ON THIS SITE, PLEASE HELP MEEEEE

Answer 1

Answer:

I don't know the answer but try using 900gle lens and scanning it on homework setting

Answer 2

Answer:

Range is 0 ≤  h ≤ 32

Step-by-step explanation:

The height is a function of time and is described as h(t) = -8t² + 32t

h is the y-axis and t is the x axis, so we can write this as y = -8x² + 32 x.

This describes a parabola that opens downward, since the coefficient of the square term is negative. The maximum point for height will be at the vertex of the parabola.

y = ax² + bx + c

The x coordinate of the vertex is at -b/2a

a = -8, b = 32, c = 0

-b/2a = -32/-16 = 2

Maximum height is at x = 2 seconds (we can see this on the graph)

At 2 seconds, y = -8(2²) + 32(2) = -32 + 64 = 32

Again, we can see this on the graph.

The x intercepts are the points where y = 0

0 = -8x² + 32x = 8x(-x +4)

y will be 0 when x is 0 or 4 (confirmed by graph)

So, the height (y) = 0 at time (x) = 0 sec, 4 sec

0 ≤ y ≤ 32

y = h, so 0 ≤ h ≤ 32

I hope this helps.