STOPPING DISTANCE The formula d = 0.05s2 + 1.1s estimates the minimum stopping distance d in feet for a car traveling s miles per hour. If a car stops in 200 feet, what is the fastest it could have been traveling when the driver applied the brakes? Use the quadratic formula to find your answer.

Approximately 0.343miles/hour

Step-by-step explanation:

Given the stopping distance of a car expressed as d = 0.05s² + 1.1s

If a car stops in 200 feet, the fastest it could have been traveling when the driver applied the brakes can be gotten by substituting S = 200ft into the equation above and solving the resulting quadratic equation.

Converting 200ft to miles

1 foot = 0.000189 mile

200feet = 200×0.000189

= 0.0378miles

The equation becomes:

0.0378 = 0.05s² + 1.1s

Multiplying through by 10000

378 = 500s²+11000s

500s²+11000s-378 = 0

Using the general formula

s = -b±√b²-4ac/2a

a = 500 b = 11000, c = -378

s = -11000±√11000²-4(500)(-378)/2(500)

s = -11000±√121,000.000+756000/1000

s = -11000±√121,756,000/1000

s = -11000±11,034.3/1000

s = -11000+11,034.3/1000

s = 34.3/1000

s = 0.0343miles/hour

Disregarding the negative value, the fastest it could have been traveling when the driver applied the brakes will be approximately 0.0343miles/hour.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.