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 | | Re: Yet another math problem
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| (...) Sorry, but this doesn't go through 0,0. Solving x=0, y=0 gives 0=855/11, which is clearly not true, so this isn't the equation he's looking for either. Adrian -- www.brickfrenzy.com (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
| | | | | Re: Yet another math problem
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| (...) Yes, it does. If you just have the last four points to work with, then you definitely have an upward-turned parabola, symmetrical about the y-axis. Here's what you have to do to find the formula for a parabola, given three points: y = ax^2 + (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
| | | | | Re: Comparative freedom
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| (...) Actually, they did not told the Queen to bugger off... because they did not have to; they got their own PM instead. Given that the Crown has little (if any) effective power, it is even better: getting independent whilst assuring a powerful (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.debate)
| | | | | Re: Yet another math problem
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| (...) Hmm. Well, now that I think of it, the vertex could be x=0 with y as an unknown. Does that change anything? Dave! (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
| | | | | Re: Yet another math problem
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| (...) Oops - I'd better correct myself before else does. Your parabola does NOT exist. It is NOT symmetrical. I solved it for the three points with x > 0 My bad. :( Jeff J (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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