Using the equation representing the height of the firework (h = -16t^2 + v↓0 + h↓0), algebraically determine the extreme value of f(t) by completing the square and finding the vertex. interpret what the value represents in this situation.

y = (-1/15)(x - 3)2 + 15

Step-by-step explanation:

The general equation of a parabola is

y = a(x - h)2 + k

If the halfway line if the origin, i.e. (x, y) = (0, 0), then the ball is kicked at (-12, 0) and it lands at (18, 0). By symmetry, the ball reaches its greatest height halfway between the starting and ending x values, or at x = 3. So, we know three points on this parabola: (-12, 0), (18, 0) and (3, 15).

Let us plug the first point into the generalized equation:

y = (x - h)2 + k

0 = (-12 - h)2 + k

So, k = -(-h - 12)2

or

k = -(h + 12)2, since we can factor out -1 and it becomes squared.

We can also plug (18, 0) into the generalized equation, and get:

0 = (18 - h)2 + k

So, k = -(18 - h)2

Since we have two expressions both equal to k, we can set them equal to each other and eliminate k:

-(h + 12)2 = -(18 - h)2

We can cancel out the negative signs and square out the terms on each side of the last equation.

h2 + 24h + 144 = h2 - 36h + 324

We can cancel out h2 and then solve for h.

24h + 144 = -36h + 324

60h = 180

h = 3

Therefore, the generalized equation becomes

y = a(x - 3)2 + k

We know that if x = 3, y = 15 from out third point.

15 = a(3 - 3)2 + k

So, k = 15

Thus, the equation becomes

y = a(x - 3)2 + 15

We can then pick one of the other points, plug in the corresponding x and y and then solve for a. I choose (18, 0).

0 = a(18 - 3)2 + 15

0 = 152a + 15

Divide this equation through by 15.

0 = 15a + 1

or

a = -1/15

Therefore, the generalized equation is

y = (-1/15)(x - 3)2 + 15

sup

step-by-step explanation:

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)