1. **State the problem** We have $f(x)=\frac{3}{4}x+10$ and $g(x)=x^2-3$. Find $f\bigl(-2-g(3)\bigr)$. 2. **First find $g(3)$ using the rule for $g$** $$g(3)=3^2-3$$ $$g(3)=9-3$$ $$g(3)=6$$ 3. **Substitute into the input of $f$** The input is $-2-g(3)$. $$-2-g(3)=-2-6$$ $$-2-g(3)=-8$$ 4. **Now evaluate $f(-8)$** Use $f(x)=\frac{3}{4}x+10$. $$f(-8)=\frac{3}{4}(-8)+10$$ 5. **Simplify the fraction term (show cancellation)** $$\frac{3}{4}(-8)=\frac{3}{\cancel{4}}\left(-\cancel{8}\right)$$ $$\frac{3}{4}(-8)=3\cdot(-2)$$ $$\frac{3}{4}(-8)=-6$$ 6. **Finish the addition** $$f(-8)=-6+10$$ $$f(-8)=4$$ 7. **Final answer** $\boxed{4}$