The functions \( f, g \), and \( A \) are defined as follows. \[ f(x)=|12 x|-15 \quad g(x)=\sqrt{-3 x-5} \quad A(x)=\frac{x}{9+x^{2}} \] Find \( f\left(-\frac{3}{4}\right), g(-3) \), and \( A(-3) \). Simplify your answers as much as possible. \[ \left(-\frac{3}{4}\right)=\square \] \( g(-3)=\square \) \( x(-3)=\square \)

To find \( f\left(-\frac{3}{4}\right) \), we substitute \( -\frac{3}{4} \) into the function \( f(x) \). \[ f\left(-\frac{3}{4}\right) = |12\left(-\frac{3}{4}\right)| - 15 \] Calculating inside the absolute value: \[ 12\left(-\frac{3}{4}\right) = -9 \quad \Rightarrow \quad | -9 | = 9 \] Therefore, \[ f\left(-\frac{3}{4}\right) = 9 - 15 = -6 \] So, \[ f\left(-\frac{3}{4}\right) = -6 \] Next, we find \( g(-3) \): \[ g(-3) = \sqrt{-3(-3) - 5} = \sqrt{9 - 5} = \sqrt{4} = 2 \] Thus, \[ g(-3) = 2 \] Finally, we evaluate \( A(-3) \): \[ A(-3) = \frac{-3}{9 + (-3)^2} = \frac{-3}{9 + 9} = \frac{-3}{18} = -\frac{1}{6} \] Therefore, \[ A(-3) = -\frac{1}{6} \] In summary: \[ f\left(-\frac{3}{4}\right) = -6 \] \[ g(-3) = 2 \] \[ A(-3) = -\frac{1}{6} \] To complete the answers: \[ f\left(-\frac{3}{4}\right) = \square (-6) \] \[ g(-3) = \square (2) \] \[ A(-3) = \square \left(-\frac{1}{6}\right) \]