*1 a car is driving away from a crosswalk. the distance d (in feet) of the car from the crosswalk t
seconds since the car started moving is given by the formula d = t^2+3.5.
a. as the number of seconds since the car started moving
increases from 1 second to 3 seconds, what is the change
in the car's distance from the crosswalk?

The change in the car's distance is 8 feet

Step-by-step explanation:

* Lets explain how to solve the problem

- A car is driving away from a crosswalk

- The distance d (in feet) of the car from the crosswalk t  seconds

  since the car started moving is given by the formula d = t² + 3.5

- The time increasing from 1 second to 3 seconds

- We need to now the change of the car's distance from the crosswalk

∵ The equation of the distance is d = t² + 3.5

∵ The time is 1 second

∴ d = (1)² + 3.5

∴ d = 1 + 3.5 = 4.5 feet

∵ The time is 3 seconds

∴ d = (3)² + 3.5

∴ d = 9 + 3.5 = 12.5 feet

∵ The change of the distance = d of 3 sec - d of 1 sec

∵ d of 3 sec = 12.5 feet

∵ d of 1 sec = 4.5 feet

∴ The change of the distance = 12.5 - 4.5 = 8 feet

∴ The change in the car's distance is 8 feet

i will take them

step-by-step explanation: