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iren [92.7K]
3 years ago
13

Would you round 14.47 to 14 or 15??

Mathematics
2 answers:
zhuklara [117]3 years ago
6 0

Answer:

15

Step-by-step explanation:

If the first digit is 4 or under then you either leave it or round it down, if the digit is 5 or over then round up no matter what. If there is a digit in the hundreds place then the same rule applies as well, in this example you can round the .47 to a .5 which is 14.5 and since the digit is 5 we round it to 15.

torisob [31]3 years ago
5 0

Answer:

14

Step-by-step explanation:

When rounding to a whole number, you round down if the decimal in the 10th's place is below 5, you round up if the decimal is 5 or above.  

Ex:  12.4  rounds down to 12 because there is a 4 in the 10th's

       12.5 rounds up to 13 because there is a 5 in the 10th's place

        14.47 rounds down because there is a 4 in the 10th's place, the 7 in the hundredth's place would only affect the 4 in the 10th's place when rounding.  It wouldn't affect the 14.  

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Bas_tet [7]

Answer:

a² + 2ab + b²

Explanation:

⇒ (a + b)²

⇒ (a + b)(a + b)    

[apply distributive method: (a + b) (c + d) = ac + ad + bc + bd]

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1 year ago
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Bond [772]

Answer:

  1. 3log(10) -2log(5) ≈ 1.60206
  2. no; rules of logs apply to any base. ln(x) ≈ 2.302585×log(x)
  3. no; the given "property" is nonsense

Step-by-step explanation:

<h3>1.</h3>

The given expression expression can be simplified to ...

  3log(10) -2log(5) = log(10^3) -log(5^2) = log(1000) -log(25)

  = log(1000/25) = log(40) . . . . ≠ log(5)

  ≈ 1.60206

Or, it can be evaluated directly:

  = 3(1) -2(0.69897) = 3 -1.39794

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__

<h3>2.</h3>

The properties of logarithms apply to logarithms of any base. Natural logs and common logs are related by the change of base formula ...

  ln(x) = log(x)/log(e) ≈ 2.302585·log(x)

__

<h3>3.</h3>

The given "property" is nonsense. There is no simplification for the product of logs of the same base. There is no expansion for the log of a sum. The formula for the log of a power does apply:

  \log(a)\log(b)=\log(a^{\log(b)})=\log(b^{\log(a)})

Numerical evaluation of Mr. Kim's expression would prove him wrong.

  log(3)log(4) = (0.47712)(0.60206) = 0.28726

  log(7) = 0.84510

  0.28726 ≠ 0.84510

  log(3)log(4) ≠ log(7)

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