That means our price markup is $67.00-$50.00=$17.00
Answer:49 sec
Step-by-step explanation:
Given
Maximum speed to reach is 183.58 mi/h
Length of course is 5 mi
acceleration rate is defined by 60mi/h in 4 sec
therefore acceleration(a)


To reach a speed of 183.58mi/h with an acceleration of 
Using equation of motion
v=u+at


t=0.00339 hours
t=12.23 s reach maximum speed
To complete course it takes


t=0.01360 hour
or 
Answer:
f(18)=334
Step-by-step explanation:
- f(x)=13x+100
- f(18) = 13(18)+100
- f(18)=234+100
- f(18)=334
108 is the answer for this question