Question 1207345: The distance. S metres, in which a car can stop is related to its speed, 'km/h by
S=pV+ qt² where p and q are constants. A car travelling at 10 km/h can stop in a distance to the second house is 10: 15 respectively. It is estimated that in 6 years, the value of the first house will increase by 35% and that of the second house will increase by $30,825.00. If the new ratio is 3: 4, find the original value of the first house of 5 metres and at 20 km/h in a distance of 12 metres. Calculate the distance in which a car travelling at 30 km/h can stop
Answer by ikleyn(52795)   (Show Source):
You can put this solution on YOUR website! .
The distance. S metres, in which a car can stop is related to its speed, 'km/h by
S=pV+ qt² where p and q are constants. A car travelling at 10 km/h can stop in a distance to the second house is 10: 15 respectively. It is estimated that in 6 years, the value of the first house will increase by 35% and that of the second house will increase by $30,825.00. If the new ratio is 3: 4, find the original value of the first house of 5 metres and at 20 km/h in a distance of 12 metres. Calculate the distance in which a car travelling at 30 km/h can stop
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Soup of words, or, in other terms, simply GIBBERISH.
The pieces of two different problems are mixed in one post.
Throw it to the closest garbage bin.
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