Triangle:
39 degrees, 33 degrees, 108 degrees
Side opposite the 108 degrees is 360 feet.
length asked for is d
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27. twice as many as 16 would be 32 and 32-5 is 27
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.
Step-by-step explanation:
5²=4²+x²
25=16+x²
25-16=x²
9=x²
√9=√x²
3=x
A
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