Suppose that the functions \( g \) and \( h \) are defined as follows. \( g(x)=(x+2)(x-5) \) \( h(x)=x-1 \) (a) Find \( \left(\frac{g}{h}\right)(-6) \). (b) Find all values that are NOT in the domain of \( \frac{g}{h} \). If there is more than one value, separate them with commas. (a) \( \left(\frac{g}{h}\right)(-6)=\square \) (b) Value(s) that are NOT in the domain of \( \frac{g}{h}: \square \)

To solve the problem, we first need to understand the functions \( g(x) \) and \( h(x) \). Given: \[ g(x) = (x + 2)(x - 5) \] \[ h(x) = x - 1 \] ### Part (a): Find \( \left(\frac{g}{h}\right)(-6) \) First, we need to calculate \( g(-6) \) and \( h(-6) \). 1. Calculate \( g(-6) \): \[ g(-6) = (-6 + 2)(-6 - 5) = (-4)(-11) = 44 \] 2. Calculate \( h(-6) \): \[ h(-6) = -6 - 1 = -7 \] Now, we can find \( \left(\frac{g}{h}\right)(-6) \): \[ \left(\frac{g}{h}\right)(-6) = \frac{g(-6)}{h(-6)} = \frac{44}{-7} = -\frac{44}{7} \] Thus, the answer for part (a) is: \[ \left(\frac{g}{h}\right)(-6) = -\frac{44}{7} \] ### Part (b): Find all values that are NOT in the domain of \( \frac{g}{h} \) The function \( \frac{g}{h} \) is defined as long as \( h(x) \neq 0 \). Therefore, we need to find the values of \( x \) for which \( h(x) = 0 \). Set \( h(x) = 0 \): \[ x - 1 = 0 \implies x = 1 \] Thus, the only value that is NOT in the domain of \( \frac{g}{h} \) is \( x = 1 \). The answer for part (b) is: \[ \text{Value(s) that are NOT in the domain of } \frac{g}{h}: 1 \] ### Final Answers: (a) \( \left(\frac{g}{h}\right)(-6) = -\frac{44}{7} \) (b) Value(s) that are NOT in the domain of \( \frac{g}{h}: 1 \)