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dangina [55]
3 years ago
7

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?
Mathematics
1 answer:
Margaret [11]3 years ago
8 0
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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Answer:

20. x = 5, y = -2

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Consecutive angles are supplementary, so

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\left\{\begin{array}{l}11x-10y=1\\ \\17x-y=175\end{array}\right.

From the second equation

y=17x-175

Substitute it into the first equation:

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22. Opposite angles in the parallelogram are congruent, so

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Consecutive angles are supplementary, so

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Solve this system of two equations:

\left\{\begin{array}{l}2x-3y=-7\\ \\3x+7y=185\end{array}\right.

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x=-3.5+1.5y

Substitute it into the second equation:

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\left\{\begin{array}{l}2x-4y=-18\\ \\3x-2y=13\end{array}\right.

Multiply the second equation by 2 and subtract it from the first equation:

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Substitute it into the first equation:

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3 0
3 years ago
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