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Aleksandr [31]
3 years ago
10

A car’s stopping distance in feet is modeled by the equation d(v)=2.15vsquared over 58.4f, where v is the initial velocity of th

e car in miles per hour and f is a constant related to friction. If the initial velocity of the car is 47 mph and f = 0.34, what is the approximate
Mathematics
2 answers:
Alexus [3.1K]3 years ago
7 0

Answer:(i dont know the answer but this person stopped the question early idk if it matters but the rest of it is)

approximate stopping distance of the car

A: 21 feet

B: 21 miles

C: 239 feet

D: 239 miles

i think that the answer was 239 feet

Step-by-step explanation:


frutty [35]3 years ago
3 0
According to your description, you can simply plug in all the numbers:

d(47) = 2.15 * 45^2 / (58.4*0.34) = 219.27 m
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Consider the equation x = 2+4y-5, it can be expressed as x = 4y-3.

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Since, Equivalent equations are those equations which have exactly same solution. We can say that the two equations are equivalent if solution of one equation is the solution of other equation.

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6 0
3 years ago
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12. Make equation and plug in

13. Proportion

14. Finding a constant rate, and then manipulate




12. Where x is miles driven, and y is answer, and z is days
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