Question

Approximate 1131√
, correct to one decimal place, by using a sequence of values that gets closer and closer to 31−−√
.(1 point)

240,060,336.2

167,595,781

628,390

628,389.7

Answer 1

The problem text looks garbled. Two plausible interpretations and their results:

- If you meant √1131: use Newton's iteration x_{n+1} = (x_n + 1131/x_n)/2 (it converges to √1131). Starting x0 = 33:
x1 ≈ 33.636, x2 ≈ 33.6319 → √1131 ≈ 33.6319, so to one decimal place 33.6.

- If you meant √31: √31 ≈ 5.567764… so to one decimal place 5.6.

Which of these was intended?