In which choice is $$ y $$ a nonlinear function of $$ x $$ ? A $$ y=\frac{x}{5}-3 $$ B $$ y=4-3 x $$ C $$ y=\frac{6}{x-4}+7 $$ D $$ y=\frac{x+1}{8}+9 $$

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To determine which choice has $$ y $$ as a nonlinear function of $$ x $$, we need to analyze each equation.

1. **Choice A:** $$ y = \frac{x}{5} - 3 $$
   - This is a linear function because it can be expressed in the form $$ y = mx + b $$ where $$ m = \frac{1}{5} $$ and $$ b = -3 $$.

2. **Choice B:** $$ y = 4 - 3x $$
   - This is also a linear function, as it can be rewritten as $$ y = -3x + 4 $$.

3. **Choice C:** $$ y = \frac{6}{x-4} + 7 $$
   - This function is nonlinear because it involves a variable in the denominator, which creates a hyperbola.

4. **Choice D:** $$ y = \frac{x+1}{8} + 9 $$
   - This is a linear function, as it can be expressed as $$ y = \frac{1}{8}x + \frac{1}{8} + 9 $$.

From this analysis, the only choice where $$ y $$ is a nonlinear function of $$ x $$ is:

**Choice C: $$ y = \frac{6}{x-4} + 7 $$**.
Choice C: $$ y = \frac{6}{x-4} + 7 $$ is a nonlinear function of $$ x $$.

In which choice is \( y \) a nonlinear function of \( x \) ? A \( y=\frac{x}{5}-3 \) B \( y=4-3 x \) C \( y=\frac{6}{x-4}+7 \) D \( y=\frac{x+1}{8}+9 \)

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