Answer
$10.9087$
Work Step by Step
The Taylor approximation for degree $n$ centred at $a=0$ can be written as:
$p(x)=1-\dfrac{x}{2}-\dfrac{x^2}{8}-\dfrac{x^3}{16}$----
Re-write the expression as: $\sqrt {119}=\sqrt {121-2}=11\sqrt {1-\dfrac{2}{121}}$
Now, we have:
$p(\dfrac{2}{121})=1-\dfrac{(\dfrac{2}{121})}{2}-\dfrac{(\dfrac{2}{121})^2}{8}-\dfrac{(\dfrac{2}{121})^3}{16}=0.9917$
Therefore, $11\sqrt {1-\dfrac{2}{121}}=(11)(0.9917)=10.9087$
