For the nonzero numbers $a$, $b$, and $c$, define $$
\text{{J}}(a,b,c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
$$Find $\text{{J}}(2,12, 9)$.
We have \[\text{{J}}(2,12, 9)=\frac{2}{12} + \frac{12}{9} + \frac{9}{2} =
\frac{1}{6} + \frac{4}{3} + \frac{9}{2} = \frac{1 + 8 + 27}{6} = \frac{36}{6} = \boxed{6}.\]
6
Level 2
Algebra
Problem: For the nonzero numbers $a$, $b$, and $c$, define $$
\text{{J}}(a,b,c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
$$Find $\text{{J}}(2,12, 9)$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find J(2, 12, 9), we need to substitute the values of a, b, and c into the defined function J. 

Start with the definition: J(a, b, c) = a/b + b/c + c/a. 

Substituting the values:
J(2, 12, 9) = 2/12 + 12/9 + 9/2.

Now, calculate each term individually:
1. 2/12 simplifies to 1/6.
2. 12/9 simplifies to 4/3.
3. 9/2 remains as is.

Next, we need a common denominator to add these fractions. The least common multiple of 6, 3, and 2 is 6. 

Convert each term:
1. 1/6 is already in terms of the common denominator.
2. 4/3 can be converted to (4 * 2)/(3 * 2) = 8/6.
3. 9/2 can be converted to (9 * 3)/(2 * 3) = 27/6.

Now, add the fractions:
1/6 + 8/6 + 27/6 = (1 + 8 + 27)/6 = 36/6.

Finally, simplify 36/6 to get the final answer, which is 6.<|END_OF_LATENT|> The final answer is $6$.