Answer:
Time t = 2 seconds
It will reach the maximum height after 2 seconds
Completed question;
Amir stands on a balcony and throws a ball to his dog who is at ground level. The ball's height, in meters above the ground, after t seconds that Amir has thrown the ball is given by:
H (t) = -(t-2)^2+9
many seconds after being thrown will the ball reach its maximum height?
Step-by-step explanation:
The equation of the height!
h(t) = -(t-2)^2 + 9 = -(t^2 -4t +4) + 9
h(t) = -t^2 +4t -4+9
h(t) = -t^2 + 4t +5
The maximum height is at dh/dt = 0
dh/dt = -2t +4 = 0
2t = 4
t = 4/2 = 2
Time t = 2 seconds
It will reach the maximum height after 2 seconds
Answer:
the third one
Step-by-step explanation:
<h3>
Answer: -16n-41</h3>
Work Shown:
2(-n-3) -7(5+2n)
2(-n)+2(-3) - 7(5)-7(2n) ... distribute
-2n - 6 - 35 - 14n
(-2n-14n) + (-6-35)
-16n - 41
Answer:
A ≈ 35.3 units²
Step-by-step explanation:
Calculate the radius CB using Pythagoras' identity in the right triangle.
CB² + AB² = AC²
CB² + 6² = 9²
CB² + 36 = 81 ( subtract 36 from both sides )
CB² = 45 = r²
Then area of quarter circle is
A = 
× πr² = × π × 45 ≈ 35.3
Answer:
5.6 days
Step-by-step explanation:
We are given;
Initial Mass; N_o = 25 g
Mass at time(t); N_t = 25/2 = 12.5 (I divide by 2 because we are dealing with half life)
k = 0.1229
Formula is given as;
N_t = N_o•e^(-kt)
Plugging in the relevant values;
12.5 = 25 × e^(-0.1229t)
e^(-0.1229t) = 12.5/25
e^(-0.1229t) = 0.5
(-0.1229t) = In 0.5
-0.1229t = -0.6931
t = -0.6931/-0.1229
t = 5.6 days
