Answer:

Step-by-step explanation:
The deceleration experimented by the car is the product of kinetic coefficient of friction and gravitational constant:



The initial speed is computed from the following kinematic expression:





Answer:
4/2, 2/7, 1/2, 5/6 is your answer
24x+8x= 32x
32x-11x=21x
21x=-7-14
-7-14=-21
21x=-21
-1