The stopping distance D in feet for a car traveling at x miles per hour is given by d(x)= (1/12)x^2+(11/9)x. Determine the driving speeds that correspond to stopping distances between 300 and 500 feet, inclusive. Round speeds to the nearest mile per hour.

x^2/12 + 11x/9 = 300

Multiply both sides by 36:
3x^2 + 44x = 10800
3x^2 + 44x - 10800 = 0
Use Quadratic Formula:
X = 53.1 mi/h.

the distance d in feel that it take s an automobile to stop if it is traveling S miles per hour is given by:

S=21d
Find the distance it would take an automobile traveling 60mph to stop.

The stopping distance d in feet for a car traveling at speed of s miles per hour depends on car road conditions. here are two possible stopping distance formulas: d=3s and d=0.05s^2+s. Write and solve an equation to answer the question, "for what speed(s) do the two function predict the same stopping distance

To determine the driving speeds that correspond to stopping distances between 300 and 500 feet, we need to solve the inequality:

300 ≤ D(x) ≤ 500,

where D(x) represents the stopping distance in feet at a given speed x.

The stopping distance function is given by d(x) = (1/12)x^2 + (11/9)x.

Let's first solve the inequality 300 ≤ D(x):

300 ≤ (1/12)x^2 + (11/9)x.

To solve this inequality, we first get rid of the fractions by multiplying the entire inequality by the least common denominator, which in this case is 36:

36 * 300 ≤ 36 * [(1/12)x^2 + (11/9)x].

10,800 ≤ 3x^2 + 44x.

Next, we subtract 3x^2 + 44x from both sides to get:

3x^2 + 44x - 10,800 ≥ 0.

Now, let's solve the other inequality: D(x) ≤ 500.

(1/12)x^2 + (11/9)x ≤ 500.

To solve this inequality, we subtract 500 from both sides:

(1/12)x^2 + (11/9)x - 500 ≤ 0.

Now, we have two quadratic inequalities to solve. To solve them, we can use different methods, such as factoring, completing the square, or using the quadratic formula.

In this case, let's use the quadratic formula to solve each inequality separately.

For the inequality 3x^2 + 44x - 10,800 ≥ 0:

The quadratic formula is given by x = [-b ± √(b^2 - 4ac)] / (2a), where a = 3, b = 44, and c = -10,800.

x = [-44 ± √(44^2 - 4(3)(-10,800))] / (2(3)).

Simplifying further, we get:

x = [-44 ± √(1936 + 129,600)] / 6.

x = [-44 ± √131,536] / 6.

x = [-44 ± 362] / 6.

This gives two solutions:

x₁ = (-44 + 362) / 6 = 53. (rounded to the nearest whole number)
x₂ = (-44 - 362) / 6 = -68. (rounded to the nearest whole number)

For the inequality (1/12)x^2 + (11/9)x - 500 ≤ 0:

Using the quadratic formula again with a = 1/12, b = 11/9, and c = -500:

x = [-(11/9) ± √((11/9)^2 - 4(1/12)(-500))] / (2(1/12)).

x = [-(11/9) ± √(121/81 + 10,000/12)] / (1/6).

x = [-(11/9) ± √(121/81 + 833.33)] / (1/6).

x = [-(11/9) ± √(321.42)] / (1/6).

This gives two solutions:

x₃ = [-(11/9) + √(321.42)] / (1/6). (rounded to the nearest whole number)
x₄ = [-(11/9) - √(321.42)] / (1/6). (rounded to the nearest whole number)

To summarize the results:

x₁ = 53 mph
x₂ = -68 mph
x₃ = 46 mph
x₄ = -253 mph

Since we are only interested in realistic driving speeds, we can disregard the negative solutions.

Therefore, the driving speeds that correspond to stopping distances between 300 and 500 feet, inclusive, rounded to the nearest mile per hour, are:

- Nearest driving speed ≤ x ≤ 53 mph.
- The integer values in this range are 0, 1, 2, ..., 53 mph.