<u>We are given:</u>
The function: y = -16t² + 64
where y is the height from ground, t seconds after falling
<u>Part A:</u>
when the droplet would hit the ground, it's height from the ground will be 0
replacing that in the given function:
0 = -16t² + 64
16t² = 64 [adding 16t² on both sides]
t² = 4 [dividing both sides by 16]
t = 2 seconds [taking square root of both sides]
<u>Part B:</u>
for second droplet,
height from ground = 16 feet
time taken = t seconds
acceleration due to gravity = 10 m/s²
initial velocity = 0 m/s
h = ut + (1/2)at² [second equation of motion]
16 = (0)(t) + (1/2)(10)(t²)
16 = 5t²
t² = 16/5
t = 1.8 seconds (approx)
Therefore, the second droplet takes the least amount time to hit the ground
The answer is probably C or D
Answer:
Step-by-step explanation:
<span> x x
----------------- - --------------------
x^2 -4x + 4 </span><span>x^2-3x+2
</span> x x
= ---------------- - --------------------
(x -2)^2 (x - 2)(x - 1)
x (x-1) - x(x +2)
= ----------------------------
(x -2)^2 (x - 1)
x^2 -x -x^2 - 2x
= --------------------------
(x -2)^2 (x - 1)
-3x
= -----------------------
(x -2)^2 (x - 1)
answer
<span>missing term in the equation: -3x</span>
I’m pretty sure the answer would be A, sorry if i’m incorrect, have a great day :)
