Initial velocity of the plane is Vo = 0.
acceleration a = 1.3 m/s2
total distance = 2.5 km = 2500m
time taken to reach 2.5 km with 1.3m/s^2 acceleration = t
S = Vo t + 0.5 a t^2
2500 = 0 + (0.5*1.3* t^2)
t^2 = 3846.15
t = 62 s
the maximum velocity plan can reach within 62 s is Vt
Vt = Vo + a t
Vt = 0 + (1.3*62)
Vt = 80.6 m/s
Since 80.6 m/s is greater than 75 m/s, plane can use this runway to takeoff with required speed.
Answer:
The values of 
and 
are 
and 
, respectively.
Step-by-step explanation:
The statement is equivalent to the following mathematic expression:
(1)
By definition of the perfect square trinomial:
And by direct comparison we have the following system:
(2)
(3)
By (3), we solve for :
The values of and are and , respectively.
Answer:
Step-by-step explanation:
f(x)=4x²+19
f(-5) = 4(-5)² + 19 = 4*25 + 19
= 100 + 19 = 119
