The Taylor series is defined by:

Let a = 0.
Then its just a matter of finding derivatives and determining how many terms is needed for the series.
Derivatives can be found using product rule:

Do this successively to n = 6.

Plug in x=0 and sub into taylor series:

If more terms are needed simply continue the recursive derivative formula and add to taylor series.
The answer for this question would be true
Sqrt(50) = sqrt(25*2) = sqrt(25) * sqrt(2) = 5*sqrt 2, answer is A
Answer:
It is compared
Step-by-step explanation: