Answer:
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
The first one is 15 mornings
Answer: The graph decreases, then increases. c.The graph remains constant. d.The graph increases, then decreases.
Step-by-step explanation:
First, simplify the equation into y=mx+b form. This would be y=3x/2-15/2. And for the slope of a line perpendicular to the original line, there is a trick. Their slopes should multiply to -1. For 3/2 to multiply to -1, you need -2/3.
