Approximate 1935√ , correct to one decimal place, by using a sequence of values that gets closer and closer to 5–√ .(1 point)

1 answer

To approximate \( \sqrt{1935} \) using a sequence of values that get closer to \( \sqrt{1935} \) (which is approximately 44.04), we can use a method like successive averaging or the Newton-Raphson method for roots.

However, since we want to approach this using trial values starting with our estimated \( \sqrt{44} = 6.63 \), let's consider values around it until we refine our estimate by averaging.

  1. Start with a guess \( x_0 = 44 \).
  2. Calculate \( \frac{1935}{x_0} \) and average: \[ x_1 = \frac{x_0 + \frac{1935}{x_0}}{2} = \frac{44 + \frac{1935}{44}}{2} \]

Calculating that:

  • \( \frac{1935}{44} \approx 43.86 \),
  • so \[ x_1 = \frac{44 + 43.86}{2} \approx \frac{87.86}{2} \approx 43.93. \]
  1. Now repeat the process with the new guess \( x_1 = 43.93 \): \[ x_2 = \frac{43.93 + \frac{1935}{43.93}}{2} \approx \frac{43.93 + 44.00}{2} \approx \frac{87.93}{2} \approx 43.965. \]

  2. Continue one more step: \[ x_3 = \frac{43.965 + \frac{1935}{43.965}}{2}. \] First, calculate \( \frac{1935}{43.965} \approx 43.980 \). Then, \[ x_3 \approx \frac{43.965 + 43.980}{2} \approx \frac{87.945}{2} \approx 43.9725. \]

The more iterations you do, the closer your approximation will be to \( \sqrt{1935} \).

To stop and finalize: Taking \( 44.0 \), we see from the steps that as they get closer we would note this rounded value \( 44.0 \).

Thus, the approximation of \( \sqrt{1935} \), correct to one decimal place, is \( 44.0 \).