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igor_vitrenko [27]
3 years ago
11

1. -4(3)2 + 12(3) + 11

Mathematics
2 answers:
Elenna [48]3 years ago
8 0

Answer:

23

Step-by-step explanation:

kherson [118]3 years ago
4 0
I believe the answer is 23
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mrs_skeptik [129]

9514 1404 393

Answer:

  10.49

Step-by-step explanation:

Since we know 110 = 10² +10, we can make a first approximation to the root as ...

  √10 ≈ 10 +10/21 . . . . . where 21 = 1 + 2×integer portion of root

This is a little outside the desired approximation accuracy, so we need to refine the estimate. There are a couple of simple ways to do this.

One of the best is to use the Babylonian method: average this value with the value obtained by dividing 110 by it.

  ((220/21) + (110/(220/21)))/2 = 110/21 +21/4 = 881/84 ≈ 10.49

An approximation of √110 accurate to hundredths is 10.49.

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The other simple way to refine the root estimate is to carry the continued fraction approximation to one more level.

For n = s² +r, the first approximation is ...

  √n = s +r/(2s+1)

An iterated approximation is ...

  s + r/(s +(s +r/(2s+1)))

The adds 's' to the approximate root to make the new fraction denominator.

For this root, the refined approximation is ...

  √110 ≈ 10 + 10/(10 +(10 +10/21)) = 10 +10/(430/21) = 10 +21/43 ≈ 10.49

_____

<em>Additional comment</em>

Any square root can be represented as a repeating continued fraction.

  \displaystyle\sqrt{n}=\sqrt{s^2+r}\approx s+\cfrac{r}{2s+\cfrac{r}{2s+\dots}}

If "f" represents the fractional part of the root, it can be refined by the iteration ...

  f'=\dfrac{r}{2s+f}

__

The above continued fraction iteration <em>adds</em> 1+ good decimal places to the root with each iteration. The Babylonian method described above <em>doubles</em> the number of good decimal places with each iteration. It very quickly converges to a root limited only by the precision available in your calculator.

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Step-by-step explanation:

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