The formula given in the image is:

$v = \sqrt{2.6r}$

where:

• $v$ is the maximum safe speed (in miles per hour),
• $r$ is the radius of curvature (in feet).

From the problem, we know that: $r = 740 \text{ feet}$

Now, substituting this value into the formula:

$v = \sqrt{2.6 \times 740}$

Let's calculate this:

$v = \sqrt{1924} \approx 43.87$

Since the problem asks you to round the answer down to the nearest whole number, the maximum safe speed is:

$v = 43 \text{ miles per hour}.$

Would you like further details or have any questions about this solution?

Here are five related questions you can explore:

1. What would happen if the radius of curvature increased?
2. How is the factor $2.6$ derived in this formula?
3. What safety factors might engineers consider when designing highway ramps?
4. How does the speed change if the ramp’s radius decreases?
5. Can the formula be applied to any type of curve or are there limitations?

Tip: Always consider safety margins in real-world applications; rounding down ensures that the speed is conservative for safety.