The ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
<u>Explanation:</u>
We need to find the time at which the ball will be at height 19 feet.
Equation:
h = 3 + 34t - 16t²
19 = 3 + 34t - 16t²
16 = -16t² + 34t
-16t² + 34t - 16 = 0
On solving the equation, we get
t1 = 0.7 s and t2 = 1.42s
Therefore, the ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
After 1 hour he has driven 40 miles
Greetings from Brasil...
Let's apply the given formula:
A = (1/2)·B·H
The base of this polygon (in this case, the triangle) is B
B = X² - 2X + 6
The height of this polygon is H and is H
H = X + 4
Applying these values (B and H) in the given formula.....
A = (1/2)·B·H
A = (1/2)·(X² - 2X + 6)·(X + 4)
A = (1/2)·(X³ + 2X² - 2X + 24)
A = (X³/2) + X² - X + 12
OR
A = (X³ + 2X² - 2X + 24)/2