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:
503 $1 tickets sold.
Step-by-step explanation:
Use two equations
Let x = number of $1 tickets sold
Let y = number of $1.50 tickets sold
x + y = 739
1x + (1.5)y = 857
First equation ==> y = 739 - x
Plug this into the second equation
x + (1.5)(739 - x) = 857
x + 1108.5 - 1.5x = 857
- 0.5x = -251.5
x = 503
There were 503 $1 tickets sold.
To find the number of $1.50 tickets, just plug this value of x into either one of the equations.
(503) + y = 739 (739 - 503 = 236)
y = 236
There were 236 $1.50 tickets sold.
Step-by-step explanation:
First, replace f(x) with y . ...
Replace every x with a y and replace every y with an x .
Solve the equation from Step 2 for y . ...
Replace y with f−1(x) f − 1 ( x ) . ...
Verify your work by checking that (f∘f−1)(x)=x ( f ∘ f − 1 ) ( x ) = x and (f−1∘f)(x)=x ( f − 1 ∘ f ) ( x ) = x are both true.
Answer: No
Step-by-step explanation:
The distance hiked of Mark is represented by 2*t + 100
and the distance hiked of Zoe is represented by 2*t
here, you can see that the slope of both equations is equal, this means that in the same lapse of time, Zoe and Mark displace the same amount, but because Mark started earlier, he has a y-intercept bigger than zero, so for every value of t, the distance that Mark hiked will be higher than the one of Zoe.
We can represent this by:
2*t + 100 > 2*t
100 > 0
So the distance hiked by Mark is always bigger than the one of Zoe
