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.
A.3n+4+3n+4+4n
=3n+3n+4n+4+4
=10n+8
B.11n+4+n-12
=11n+n+4-12
=12n-8
C.6(6n-2)
=36n-12
D.4(3n-2)
=12n-8
E.4n+22-12+8n
=4n+8n+22-12
=12n+10
so,B and D are the expressions that are equivalent to 12n-8.
Answer:
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Answer:
x=10 and x=-9
Step-by-step explanation:
Factor:
(x-10)(x+9)=0
x-10=0
x=10
and
x+9=0
x=-9
