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To solve for \( z \) when \( y = 14 \) and \( x = 7 \), given the variation equation:

\[
z = 4.7 \cdot x^3 \cdot y^2
\]

we can substitute the values of \( x \) and \( y \) into the equation as follows:

1. Substitute \( x = 7 \) and \( y = 14 \) into the equation:

   \[
   z = 4.7 \cdot (7)^3 \cdot (14)^2
   \]

2. Calculate each component step-by-step:
   - \( 7^3 = 343 \)
   - \( 14^2 = 196 \)

3. Substitute these values back:

   \[
   z = 4.7 \cdot 343 \cdot 196
   \]

4. Multiply the values:

   \[
   z = 4.7 \cdot 67228
   \]

5. Finally, multiply by \( 4.7 \):

   \[
   z = 316971.6
   \]

Therefore, the answer is:

\[
z = 316971.6
\]

Would you like further details on any part of this calculation?

### Related Questions:
1. How does joint variation differ from direct variation?
2. How do you derive the constant of proportionality?
3. What are some examples of real-world applications of joint variation?
4. How would the equation change if \( z \) varied with the fourth power of \( x \)?
5. What steps would you take if asked to find \( x \) or \( y \) instead of \( z \)?

**Tip:** Always double-check your substitutions and calculations, especially when dealing with exponents, as small errors can lead to large discrepancies in answers.

Suppose that z varies jointly with the cube of x and the square of y. Find the constant of proportionality k if z = 601.6 when y = 4 and x = 2. Using the k from above, write the variation equation in terms of x and y. Using the k from above, find z given that y = 14 and x = 7.

Learn how to solve a joint variation problem where z varies with the cube of x and the square of y. This example uses values y = 4 and x = 2 to find the constant of proportionality k, then applies it to solve for z when y = 14 and x = 7. Suitable for high school students in Grades 9-12.

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