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Last edited by tony123 (2007-11-22 20:22:52)
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I'm interesting how to solve this can anyone solve this
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let
http://gyan.talkacademy.com.np
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The simplest answer is x, y, z = 1.
If you want x, y, z to have different values and you want those values to work for all three equations, then here's the messy way...
We know:
So if you pick a value for X and Y, you might be able to work out what the corresponding Z value is.
For example, I've picked x=1, y=2. If we substitute these values and simplify then, we end up with:
We know √z = 3 - √y - √x from the first equation, so we can substitute in that:
Because of the plus-minuses and the square roots, I think the right side might have 256 possible combinations (some combos will have the same value, though). Since z²+z is quadratic, I think that each of those combinations could have two answers.
So... rather than bother with any of the above, x=1, y=1, z=1.
Trillian: Five to one against and falling. Four to one against and falling Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still cant cope with is therefore your own problem.
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thnx bro i will try to solve this in my way thnx
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Mitrovica My City
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