Simplified: x^3+5x^2-2x+1
in standard form: x^3+5x^2-2x+1
Involving the second and no higher power of an unknown quantity or variable.
The distance between Car A and car B,When Car A crosses the start line is:
distance =speed car B* time
distance=(15 m/s)(3 s)=45 m
Distance traveled by car A =x, (when the car B is at the same distance from the start line)
time of car A=t
x=10 m/st ⇒ x=10t (1)
Distance traveled by car B=x
time of car B=t-3
x=15(t-3) (2)
With the equations (1) and (2) we make a system of equations:
x=10t
x=15(t-3)
We solve this system of equations:
10t=15(t-3)
10t=15t-45
-5t=-45
t=-45 / -5
t=9
t-3=9-3=6
x=10 t=10 (9)=90
Answer: The time would be 9 seconds for Car A and 6 seconds for car B and the distance would be 90 meters.
Answer: 629.44
Step-by-step explanation:
562 times .2 = 112.4 that’s the 20% discount. 562-112.4 = 449.6 times 1.4 = 629.44
Answer:
3951>3519>3191
Step-by-step explanation:
hope you understood how
