An object is launched directly in the air at a speed of 32 feet per second from a platform located 8 feet in the air. The motion of the object can be modeled using the function f(t)=−16t2+32t+8, where t is the time in seconds and f(t) is the height of the object. How long, in seconds, will the object be in the air before hitting the ground? Round your answer to the nearest hundredth of a second. Do not include units in your answer.

2.225

Step-by-step explanation:

To find how long the object will be in the air, we just need to find the time when the object reaches the ground, that is, f(t) = 0

So, we have that:

0 = −16t2+32t+8

2t2-4t-1 = 0

Using Bhaskara's formula:

Delta = b2 - 4ac = 16 + 8 = 24

sqrt(Delta) = 4.899

t = (-b + sqrt(Delta)) / 2a

t = (4 + 4.899) / 4

t = 2.2248

Rounding to nearest hundredth, we have t = 2.225

the answer is b.

step-by-step explanation:

identity property is where 0+whatever number=whatever number. basically 0+ a number.

step-by-step explanation:

step 1: solve one of the equations for one of the variables. let's solve the first equation for y:

step 2: substitute that equation into the other equation, and solve for x.

step 3: substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.