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Blizzard [7]
3 years ago
13

If the length of the side of a square is 15 inches. What is the area of the square?

Mathematics
2 answers:
kirill115 [55]3 years ago
6 0

Answer:

225 inch^2

Step-by-step explanation:

15×15=225

exis [7]3 years ago
6 0

Answer:

225 In²

Step-by-step explanation:

15*15=225 because the square has all sides the same length.

***********************\\*\\*\\*BRAINLIEST\\*PLEASE\\*\\*\\*************************************

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200% of what number is 350?
ExtremeBDS [4]
We have, 200% × x = 350
or,
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Multiplying both sides by 100 and dividing both sides by 200,
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If you are using a calculator, simply enter 350×100÷200, which will give you the answer.
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Which of the following is an example of a variable expense? And why
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Answer: Zander Runs, 13.125 miles. You take the mixed number and change it into an improper fraction. So you get 63/4 then multiply that by 5/6 and get 13.125 miles there is your answer. I hope this helps you

Step-by-step explanation:

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3 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.

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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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