Car A, traveling at a constant 10 m/s, crosses the start line 3 seconds before Car B, who is traveling at a constant 15 m/s. At
what time and distance, respectively, will the two cars be at the same distance from the start line?
1 answer:
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.
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My guess would be A ! Hope this helps
She will get 3 because if you multiply 1/8 and 3 you will get an answer of 3/8
Answer:
-5/2
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(-29-(-9))/(15-7)
m=(-29+9)/8
m=-20/8
simplify
m=-5/2
And your answer will be: m=3
Answer:
X = 1 , -2
Step-by-step explanation:
Given in question as,
(log4(x² + x))² = 0.25
log4(x² + x) = 0.5
Applying log base property if log(a)X = b, then X =a^b
Or, x² + x = 4^0.5
x² + x = 2
x² + x - 2 = 0
x²+2x-x-2 = 0
(x + 2) (x - 1) = 0
So, x = 1 and -2 Answer