Water is shot straight up out of a water soaker toy. The quadratic function y = –16x2 + 32x models the
height in feet of a water droplet after x seconds. How
long is the water droplet in the air?

Help!

The water droplet is in the air for two seconds.

Step-by-step explanation:

You can solve this by finding the times at which its height is zero.  The lesser of the two values will be upon launch, and the greater upon landing:

y = -16x² + 32x

Let y = 0

-16x² + 32x = 0

factor -16 out:

x² - 2x = 0

complete the square:

x² - 2x + 1 = 1

and solve for x

(x - 1)² = 1

x - 1 = ±1

x = 1 ± 1

x = 0, 2

So the water is at a height of zero at zero seconds and two seconds.  It's thus in the air for two seconds.

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation: