First of all, 0.01 Bitcoin is 0.01 Bitcoin, or one hundredth of Bitcoin.

So if people were willing to pay $ 100,000 for a naïve Bitcoin economy, the math would suggest that they would be willing to pay one-hundredth of that amount for a hundredth of bitcoins, which is $ 1,000.

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FireHeaven
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. [Vn5socks.net] Automatic update 24h / 24, 7/7 – good socks at 11h15 LIVE ~ 47.93.251.207:3001 | 0,062 | Ottawa | ON | k1y4h7 | Canada | Checked at vn5socks.net
LIVE ~ 140.82.59.139:31125 | 0,191 | San Jose | CA | 95131 | United States | Checked at vn5socks.net
LIVE ~ 125.65.92.104:7302 | 0.24 | Chengdu | 32 | Unknown | China | Checked at vn5socks.net
LIVING ~ 169.239.221.90:50802 | 0,417 | Unknown | Unknown | Unknown | Unknown | Checked at vn5socks.net
LIVE ~ 173.208.251.50:24089 | 0.22 | Kansas City | MO | 64116 | United States
Evaluation: 5

I came across many digital features and have reviewed many VBA codes available on the net.

The main function that is called repeatedly is the function that converts numbers from 001 to 999 because this is the main function for conversion under the English enumeration system.
I've seen that this main function is sometimes unnecessarily divided into sub-functions to convert one, tens and 20's into 90's.

I have developed the following simple VBA function that takes an entry as a number in a string format from "001" at "999" and return the output as a string.
The function uses the dash "-" for numbers, for example Forty-two.

The function is easily convertible into other programming languages.

With your help, I'm looking for to further improve or simplify the function, if possible.

Function Do999 (ThreeDigits As String)
& # 39; ---------------------------------------------
& # 39; Converts the numeric string from 001 to 999 into words
• Use dash for numbers between 21 and 99 for British and American English
& # 39; Mohsen Alyafei October 17, 2018
& # 39; On Entry: NumIn MUST be a string of 3-digit digits "001" to "999"
& # 39; On Exit: String of numbers in English words
& # 39; ---------------------------------------------
Dim Ones (), Ten (), dash As String, h As String, t As String, N1 integer, N2 integer
Ones = Array ("", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven" , "Twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen")
Ten = table ("", "", "twenty", "thirty", "forty", "fifty", "sixty", "sixty", "sixty", "eighty", "eighty", "hundred ")
& # 39; The following line is optional for English words (UK, US)
If the right (ThreeDigits, 1) <> "0" Then dash = "-" dash according to the English spelling
& # 39; ------------ Code starts here ------------
& # 39; Get the hundreds (N1) and the tens (N2)
N1 = Left (ThreeDigits, 1): N2 = Right (ThreeDigits, 2)
If N2> 19 Then t = Tens (Val (Mid (ThreeDigits, 2, 1))) & dash & Ones (Val (Right (ThreeDigits, 1))) Otherwise t = Ones (N2)
Do999 = Trim (IIf (N1> 0, Ones (N1) and tens (10), "") & "" & t)
End function

n = 101 is prime, with only 1 zero in between.
How many zeros do we have to add to n to become prime again? 1001 is not prime 10001 is not prime 100..001 ?
What is the minimum number of zeros> 1 for this number to be prime?

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FireHeaven
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. [Vn5socks.net] Automatic update 24 hours a day, 7 days a week – good socks LIVE ~ 139.59.117.118:10080 | 0,044 | Unknown | Unknown | Unknown | Australia | Checked at vn5socks.net
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LIVE ~ 138.68.59.157:1210 | 0,188 | Wilmington | DE | 19880 |
Evaluation: 5

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I am asked to calculate the line integral of the vector field $ K = (x-y, y-z, x + z) $ along one way on the unit sphere centered on $ (0,0,0) $. The path starts with $ (1,0,0) $ at $ (0,1,0) $ (section (1)) and then $ (0,1,0) $ at $ (0,0,1) $ (section (2)), the $ (0,0,1) $ return to $ (1,0,0) $ (section 3)).

I do not have (yet) any solution to this exam question, so I would like you to check my calculations. Thank you.

I have therefore divided the integral into 3 parts.

(1): here, $ z = $ 0. We can set the path with $ ( cos ( theta), sin ( theta), 0) $ or $ theta in [0, frac{pi}{2}]$.

The derivative is $ (- sin ( theta), cos ( theta), 0) $

(2): here, $ x = $ 0. We can set the path with $ (0, cos ( theta), sin ( theta)) $ or $ theta in [0, frac{pi}{2}]$.

The derivative is $ (0, – sin ( theta), cos ( theta)) $ (3): here, $ y = 0 $. We can set the path with $ ( sin ( theta), 0, cos ( theta)) $ or $ theta in [0, frac{pi}{2}]$.

The derivative is $ ( cos ( theta), 0, – sin ( theta)) $

Each time, we have the integral of $ 0 at $ frac { pi} {2} $ compared to $ theta $

(1): $ int_ {0} ^ { frac { pi} {2}} ( cos ( theta) – sin ( theta), sin ( theta) -0, cos ( theta) +0 ) cdotp (- sin ( theta), cos ( theta), 0) d theta $

(2): $ int_ {0} ^ { frac { pi} {2}} (0- cos ( theta), cos ( theta) – sin ( theta), 0 + sin ( theta) ) cdotp (0, – sin ( theta), cos ( theta)) d theta $

(3): $ int_ {0} ^ { frac { pi} {2}} ( sin ( theta) -0,0- cos ( theta), sin ( theta) + cos ( theta) ) cdotp ( cos ( theta), 0, – sin ( theta)) d theta $

So we get: $$ int_ {0} ^ { frac { pi} {2}} – sin ( theta) cos ( theta) + sin ^ 2 ( theta) + cos ( theta) sin ( theta) theta $$$$ + int_ {0} ^ { frac { pi} {2}} – sin ( theta) cos ( theta) + sin ^ 2 ( theta) + cos ( theta) sin ( theta) theta $$$$ + int_ {0} ^ { frac { pi} {2}} sin ( theta) cos ( theta) – sin ^ 2 ( theta) – cos ( theta) sin ( theta) theta $$ $$ = int_ {0} ^ { frac { pi} {2}} sin ^ 2 ( theta) theta = frac { pi} {4} $$

So, the result seems to make sense to me. I mean that's not an ugly result, but I still do not know if it's okay. (What I find really boring, is that teachers who bring no solution to their exams are also the ones who ask the most difficult questions, relatively speaking -_-)

In addition, we could of course use Stoke's theorem in this question, finding $ rot (K) $ and the cross product of the two partial derivatives of some parametrization, and then an integration over the entire region, but are you in agreement that it is much more complicated than if we were simply evaluating the three different paths?