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Svetllana [295]
4 years ago
9

Rectangle A has length 12 and width 8. Rectangle B has length 15 and width 10. Rectangle C has length 30 and width 15.

Mathematics
1 answer:
Anna007 [38]4 years ago
7 0

Answer:

  yes; 1.25

Step-by-step explanation:

The length to width ratios of the rectangles are ...

  A: 12/8 = 1.5

  B: 15/10 = 1.5

  C: 30/15 = 2.0

__

Rectangles A and B have the same aspect ratio, so are similar. Rectangle B is a scaled copy of A with a scale factor of 10/8 = 1.25.

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Step-by-step explanation:

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3 years ago
Find the distance :
faltersainse [42]
1. You would do 45 × 3 = 135 miles
2. 630 ÷ 2 = 315. 630 + 315 = 945 miles
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I hope this helps!
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3 years ago
Find the two numbers . State the smallest number?
alexgriva [62]

Answer:

7

Step-by-step explanation:

x+y=30

x-y=16

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3 years ago
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What is the simplified form of the expression
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4 years ago
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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.

__

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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3 years ago
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