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.
Walter I believe is the correct answer
Step 1: Simplify both sides of the equation.
Step 2: Subtract 6x from both sides.
- 65x²+390x+585−6x=6x−6x
- 65x²+384x+585=0
For this equation: a=65, b=384, c=585
Step 3: Use quadratic formula with a=65, b=384, c=585
- x=−b±√b2−4ac/2a
- x=−(384)±√(384)2−4(65)(585)/
- 2(65)
- x=−384±√−4644/130
Therefore, There are no real solutions.
In order to match each of the pairs you need to divide the numerator by the denominator to make it into a decimal.
Below are the pairs:
21/25 = 84%
13/20 = 65%
2/5 = 40%
3/4 = 75%
3/5 = 60%
<span>85 x 63 = 5355
hope it helps
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