Here are the solutions to the problems: 9. Use the first 3 terms in the expansion of $\sqrt{4-3x}$ to approximate $\sqrt{4.03}$ to 3 decimal places. Step 1: Relate $\sqrt{4-3x}$ to $\sqrt{4.03}$. We need $4-3x = 4.03$. $$-3x = 4.03 - 4$$ $$-3x = 0.03$$ $$x = \frac{0.03}{-3}$$ $$x = -0.01$$ Step 2: Write $\sqrt{4-3x}$ in the form $(a+b)^n$. $$\sqrt{4-3x} = (4-3x)^{1/2}$$ Here, $a=4$, $b=-3x$, and $n=\frac{1}{2}$. Step 3: Expand using the binomial theorem $(a+b)^n = a^n + n a^{n-1}b + \frac{n(n-1)}{2!} a^{n-2}b^2 + \dots$ The first three terms are: Term 1: $a^n = 4^{1/2} = 2$ Term 2: $n a^{n-1}b = \frac{1}{2} (4)^{1/2 - 1} (-3x) = \frac{1}{2} (4)^{-1/2} (-3x) = \frac{1}{2} \cdot \frac{1}{\sqrt{4}} (-3x) = \frac{1}{2} \cdot \frac{1}{2