Regularity or Randomness - page 30
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How could it not be? For starters, divide the Euro\dollar figures by the Pound\dollar figures at a given second, the USD is declining. You get the Euro/Pound figure.
With any other triangles it will be the same.
No, not like that.
AUD/CAD, for example.
and there are plenty of such examples
No, it's not.
AUD/CAD, for example.
and there are plenty of such examples
I suspect if you divide the AUD/xxx figures by the CAD\xxx figures, you get the AUD/CAD figures. The currency xxxx is your choice.
I suspect if you divide the AUD/xxx figures by the CAD\xxx figures, you get the AUD/CAD figures. The currency xxxx is your choice.
No, it's not.
I've already done the math.
No, it's not.
I've already done the math.
Didn't take the trouble to check. Euro/Yen=124.367, Euro/Franc=1.117, Frank/Yen=111.321. Dividing 124.367 by 111.321 equals 1.117, check.
Didn't take the trouble to check. Euro/Yen=124.367, Euro/Franc=1.117, Frank/Yen=111.321. Divide 124.367 by 111.321 equals 1.117, check.
no it isn't
I already did the math.
aud/cad=(aud/usd)*(usd/cad)
No magic, mathematics)
aud/cad=(aud/usd)*(usd/cad)
no magic, maths)
aud/cad=(aud/usd)*(usd/cad)
no magic, maths)
Thank you for your support. I even complicated it by showing it without a dollar.
ok
then how would the conditions of this problem be phrased?
That's where it started...
otherwise you would have solved the dead-end problem.ok
then how will the conditions of this problem be formulated?
that's where it started...
otherwise we'd be solving a dead-end problem.No, further on, on your own. Next, both time and brainpower are spent, and it pays to do so. Moreover, I don't understand the terms of the problem.