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 
15.5in -0.75in= 14.75in
No, because the left over space on the paper with the border is 14.75in wide and the design is 14.5in wide.
Answer:
36 cm squared is the anwer
Answer: Chuck's travel at a rate of 52mph
Step-by-step explanation:
For Chuck's trip:
D=RT
104= (R+4)T
T= 104 / (R+4)
For Dana's trip:
96 = RT
T= 96/R
Set both equation for Chuck's and Dana together
104/(R+4) =96/R
Then we cross multiply
96(R+4) = 104R
96R + 384 = 104R
104R - 96R = 384
8R = 384
To get R, divide both side by 8
8R/8 = 384/8
R= 48mph
This means Dana's speed is 48mph
Chuck's speed will be: 48mph+4mph = 52mph
Answer:
$2000 I hope it's helpful for you
