Answer:
On average, cars enter the highway during the first half hour of rush hour at a rate 97 per minute.
Step-by-step explanation:
Given that, the rate R(t) at which cars enter the highway is given the formula

The average rate of car enter the highway during first half hour of rush hour is the average value of R(t) from t=0 to t=30.

![=[100(t-0.0001\frac{t^3}{3})]_0^{30}](https://tex.z-dn.net/?f=%3D%5B100%28t-0.0001%5Cfrac%7Bt%5E3%7D%7B3%7D%29%5D_0%5E%7B30%7D)
![=100[(30-0.0001\frac{30^3}{3})-(0-0.0001\frac{0^3}{3})]](https://tex.z-dn.net/?f=%3D100%5B%2830-0.0001%5Cfrac%7B30%5E3%7D%7B3%7D%29-%280-0.0001%5Cfrac%7B0%5E3%7D%7B3%7D%29%5D)
=2901
The average rate of car is 

=97
On average, cars enter the highway during the first half hour of rush hour at a rate 97 per minute.
There is no solution
Hope this helps :)
The answer k is n divided by 6
Using the points (0,14) and (10,0), the gradient of the line is (14-0)/(0-5) which simplifies to -7/5.
From these points, we can also see that when the x value is zero, the y value is 14. Therefore the y-intercept is 14.
We can then put this into the equation of a straight line, y=mx+c:
y = (-7/5)x + 14
Answer:
CSA =220sq.cm
Step-by-step explanation: