Acar is speeding at 60 miles per hour along a winding coastal highway. (the posted speed limit is 45 miles perhour). the speeding car passes another car going the speed limit at the same time as they pass a police officer. the police officer begins to chase the speeding car 5 minutes after the speeding car passes. the police officer's vehicle pursues the speeder at the rate of 75 miles per hour. the police officer eventually catches up to the speeder and pulls them over.

how long will the speeder be pulled over for it they are pulled over until the car going the speed limit passes them?

8 minutes 20 seconds. First, lets determine who many miles per minute each vehicle moves by dividing each speed by 60. Speeder = 60 / 60 = 1 mile per minute. Police = 75 / 60 = 1.25 miles per minute. Other car = 45 / 60 = 0.75 miles per minute. Since the speeding car moves for 5 minutes before the police start to chase, that means that the speeding car will now be 5 + T miles down the road with T being the time the police has been chasing. The police will be 1.25 T. We're looking for when those two equations equal each other. So 5 + T = 1.25 T Subtract T from both sides 5 = 0.25T Divide both sides by 0.25 20 = T So it will take the police officer 20 minutes to catch up to the speeder. And they will both have traveled a total distance of 25 miles from the point where the speeder passed the police car. Now we need to figure out how far the law obeying car has moved during those 25 minutes. So 25 * 0.75 = 18.75 miles. The distance the law obeying car needs to travel to catch up to the police officer then becomes 25 - 18.75 = 6.25 miles. The number of minutes that the law obeying car needs to travel that distance is 6.25 / 0.75 = 8.333 Which is 8 minutes 20 seconds.

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

f(-2) = -3f(1) = 6 . . . estimated from f(x) = (x+2)^2-3, which matches points on the graph (-1,-2) and (0, 1)

so the change in y-value for function f is (6 )) = 9.

the table tells us

g(-2) = 1g(1) = 31

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

The answers 5 so there’s the you need trust me it’s right