d(v)=2.15v^2/64.4f. Determine the value of *f*, rounded to the nearest hundredth.
 f= 0.39 Choose the quadratic model for the situation.

d(v) = 2.15v^2/.039 

d(v) = 2.15v^2/64.79

d(v) = 2.15v^2/25.116

To determine which quadratic model for \( d(v) \) is correct based on the given value of \( f = 0.39 \), we can start by using the original equation provided:

\[
d(v) = \frac{2.15v^2}{64.4f}
\]

Substitute \( f = 0.39 \) into the equation:

\[
d(v) = \frac{2.15v^2}{64.4 \times 0.39}
\]

Now, calculate \( 64.4 \times 0.39 \):

\[
64.4 \times 0.39 = 25.116
\]

This gives the quadratic model:

\[
d(v) = \frac{2.15v^2}{25.116}
\]

Therefore, the correct quadratic model is:

\[
d(v) = \frac{2.15v^2}{25.116}
\]

### Do you want further details or have any questions?

### Related Questions:
1. What is the significance of the constant 2.15 in the equation?
2. How do different values of \( f \) affect the quadratic model?
3. Can you graph the function \( d(v) = \frac{2.15v^2}{25.116} \)?
4. How would you solve for \( v \) if given a specific value for \( d(v) \)?
5. What are the real-world applications of such quadratic models?

### Tip: Always ensure your constants and units are clearly defined in modeling equations to avoid confusion.

d(v)=2.15v^2/64.4f. Determine the value of *f*, rounded to the nearest hundredth. f= 0.39 Choose the quadratic model for the situation: d(v) = 2.15v^2/.039, d(v) = 2.15v^2/64.79, d(v) = 2.15v^2/25.116

Learn how to determine the correct quadratic model for the equation d(v) = 2.15v^2 / 64.4f when f = 0.39. This problem covers key algebraic concepts such as substitution and rational expressions. Suitable for students in Grades 9-11, it includes a detailed step-by-step explanation of the solution.

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