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smacker124

some math twisters?

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muhauhau

 

SI prevails again.

hehe

 

Is America the only country using Imperial stilll?

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wow! I`m impressed! nice work (now if only I understood most of it) :)

 

it just seemed to be a bit like; imagine 1kilo = 2.2 pounds. you have half a kilo, how many pounds is it?

1.1 pounds sure.

so I didn`t see a problem in converting from one unit to another halving it, and re-applying it back to my original units?

esp since F to C has a known formula and is`nt an arbitrary measurement, there is a direct corellation between them.

maybe I`m missing something?

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What you're missing is that celsius and Farenheit are relative scales. So are pounds and kilos.

 

Kelvin is an absolute scale.

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rellative scales?

a kilo is a kilo and ever so shall be as is a meter or a foot, if I`m given a scale in meters and am asked to halve it, I may If I wish convert to feet halve that and convert back to meters agin and still be correct.

the temp was in C (not Kelvin or Farenheight) and so either is wrong or right dependants upon interpretation :)

personaly I have no arg with Faf`s answer, In fact I thought it was cool. there`s not a issue with that, and for BOTH to be right??? well that doesn`t make sense either unless both answers are the same, SOooo it falls down to interpretaion OR one is wrong?

I`m not botherd about the wrong bit, more the reasoning behind it :)

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Oh jesus f corbett.

 

Actually, using kilos and pounds as examples of a relative scale is not a great way to go, since they share a common zero value which is the same for each in a very real sense. One could argue that they are absolute if negative mass and mass deficit did not exist.

 

However, c and F are very much relative. One cannot have a system with an energy deficit below 0k. Only scales that share 0k as their base zero value are absolute, and you cannot convert between scales that do not share a base zero and still get a meaningful result.

 

ps - "relative scale" does not mean that a meter is sometimes less or more than a meter. It means it is a relational measurement that is not derived from first principles.

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I still don't see how relating cold and temerature works. 0 K is the maximum amount of coldness. Halving the amount between a value and it's maximum doesn't equal doubling the value. I presume you would have to find a point at which cold reaches zero, assign an arbitrary scale, then double the value between it and the given temperature. This depends on there actually being a zero amount of cold.

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The assumption is that "if it is twice as cold tomorrow" is a euphemism meaning "if there is half the heat energy tomorrow".

 

Like I've said from the start, this isn't the only interpretation. But it's the only one that allows a meaningful answer.

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ok... what if the temp was 1c and was gunna be twice as cold next day?

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You'd still use Kelvin as your scalar, because it's the only scale out of the three that uses "no heat energy" as the zero value.

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well my point was that is it was +1c and twice as cold next day would it be .5c? or -130 something?

or even if it was 10c and twice as cold next day? etc....

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If 0c was the least amount of heat energy you could possibly have, then yes - "twice as cold as 1c" would be 0.5c.

 

But it isn't. 0K is the least amount of heat energy you can measure, so whether you work it out in c, F or K you need to use 0K as your zero reading.

 

Simplest possible analogy - it's like using the "tare" button on an electronic balance.

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Pearl Question Revised:

 

I will now address the error of calculation(s), which I noticed:

 

The Question:

 

“33 pearls, the middle pearl is the largest and most expensive of all. Starting from one end, each pearl is worth $100 more than the one before, up to the middle. From the other end, each pearl is worth $150 more than the one before, up to the middle. The string of pearls is worth $65,000. What is the value of the middle pearl?”

 

This is not an easy question, so I will take it slowly:

 

There are exactly 16 pearls to either side of the pearl in the center. The end where each pearl is worth $100 more than the previous one the left side, and the end where each pearl is worth $150 more than the previous one the right side. (That's just so it's easier for me to write about them.) So I've cut the necklace into 3 parts: the left side, the middle pearl, and the right side. Each of the two sides has 16 pearls. And the middle is One Pearl, the most expensive.

 

Say that the left side has value L, the middle pearl has value M, and the right side has value R. Then we know that L + M + R = $65000.

 

L:

We know that the pearl in the left side closest to the middle has value M-100. And the next one has value M-200, on to the last pearl that has value M-1600 (the easy way to imagine this is to draw a picture of the whole necklace). So we have: L = (M-100) + (M-200) + (M-300) + ... + (M-1600).

 

Collect terms here to get the M's separate: L = 16M - (100+200+300+...+1600)

L = 16M - 100(1+2+3+...+16).

 

What’s 1+2+...+16?

 

1+2+3+...+16 is the 16th triangle number. The general form for the nth triangle number is 1/2 * (n) * (n+1). Group the sum instead as 1+16 + 2+15 + 3+14 + 4+13 +…..

 

Each of these number conformations, will have a sum equal to 17, (the general sum is n+1), and there will be exactly 8 of them (in the general case, n/2). So the sum will be (n+1) * n/2, which is another way of stating that formula I just gave you.

 

So:

 

L = 16M - 100(1+2+...+16) L = 16M - 100(17)(16/2) L = 16M - 100*136 L = 16M - 13600.

 

R:

 

The pearls of R are worth M-150, M-300, and so on, all the way down to M - 16*150, which is M-2400. So the total value of the right side is: R = (M-150) + (M-300) + ... + (M-2400) R = 16M - (150+300+...+2400) R = 16M - 150(1+2+...+16) R = 16M - 150*136 R = 16M - 20400 .

 

We now need to apply the values for L and R into our most expensive necklace finding (MENFE) equation for the value of the necklace:

 

L+M+R = 65000

16M - 13600 + M + 16M - 20400 = 65000

33M - 34000 = 65000

33M = 99000

M = 3000 (9000/33).

 

So the most expensive middle pearl is $3000.

 

NOW IN TERMS OF X FORMAT:

 

x = (x-100) (x-150) <-- 2x – 250

| | (x-200) (x-300) <-- 2x – 500

| | (x-300) (x-450) <-- 2x – 750

| | (x-400) (x-600) <-- 2x - 1000

| | | |

| | | |

(x-100n) (x-150n) <-- 2x - 250n

 

Add them all together and for the 2n + 1 pearls you get

 

Total value = x + n(2x) –250[1 + 2 + ... + n]

 

= (2n+1)x - 250[n(n+1)/2]

 

We have 33 pearls, so n=16; and the total value is $65,000.

 

$65,000 = (2*16 + 1)x - 250[16(17)/2]

= 33x - $34,000

 

$65,000 + $34,000 = $99,000

 

$99,000 / 33 = $3000.

 

The above area Vending is where you had the calculation error, you forgot to add the $65,000, to the $34,000.

 

This gives you a value of x of $3000. Same as the above, and my previous equation.

 

Wolfson.

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ah, addition, my old nemisis -- we meet again. Perhaps you have been vicotrious this time, but next time, i shall add you correctly!!!!! muhahahahaha!!!!!!!!!!

 

cool, thanks wolfson

:D

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a=2 and each letter proceeding is 2 times the letter preceding (ie b=2a, c=2b). What is the value of (x-a)(x-b)(x-c) . . . (x-z)?

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(x-a)(x-b)(x-c) (x-y)(x-z) = x^26

 

- (Sn PT >>>1 per interval) x^25

 

The total number of terms in the expansion is 2^26 = 67108864

 

And the nth term for the un terms would be (geometric progression with a ratio of 2).

 

A(x) = Sum_{n >= 1} a(n)*x^n = x / Product_{n >= 1}(1-(-x)^n)^((-1)^n*a(n)

 

 

Proof: x prod_{n>0} (1-x^(4n-2))^a(2n-1)/(1-x^n)^a(n).

 

n%4!=2,0,a(n/2))

 

sum_{k=1.Infinity) B(x^k)/k )

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Do we know the values of any of these letters yes, there are multiples of different sets so then we can look for a pattern by starting small and working our way up:

 

(x-a)(x-b) = x^2

- ax - bx

+ ab

 

(x-a)(x-b)(x-c) = x^3

- cx^2 - ax^2 - bx^2

+ cax + bcx + abx

- abc

 

(x-a)(x-b)(x-c)(x-d) = x^4

- dx^3 - cx^3 - ax^3 + dcx^2 + adx^2 - bx^3

+ dbx^2 + cax^2 + bcx^2 + abx^2

- dcax - dbcx - dabx - abcx

+ dabc

 

See the pattern? For 22 terms the first line would be x^22, and then every combination of 1 letter (26 of them) times x^21 would be subtracted, and then every combination of 2 letters times x^20 would be added, and then every combination of 3 letters times x^19 would be subtracted, and so on and so on until in the last line every combination of 26 letters (1 of them) times x^0 (which is 1) would be added.

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Oh and BTW we get those values from your Geometric progression!!!!!!!

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