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(...) David (no "!") understood what I was refering to - technically everything Canada did was supposed to be approved by the Queen until that last vestige of empire was officially expunged a couple of decades ago. I was embarrassed for my Canadian (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.debate)
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| | Re: Buses & Beggers [Re: Vote against/for...]
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(...) In what way are they *SPECIFICALLY* linked. -John (22 years ago, 11-Nov-02, to lugnet.off-topic.debate)
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| | Re: Test #2
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(...) (22 years ago, 11-Nov-02, to lugnet.off-topic.test.foo, lugnet.off-topic.test.bar)
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| | Test #2
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I hate Computers (22 years ago, 11-Nov-02, to lugnet.off-topic.test.bar)
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| | Test
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Just a test! (22 years ago, 11-Nov-02, to lugnet.off-topic.test.foo)
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| | Re: Comparative freedom
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(...) If it was Québec, I am not a bit surprised. The Crown had assumed compromises to protect the french heritage of the province. Obviously, they would rather keep the privilege. (...) Was that clause used often, btw? (...) The Crown declared war. (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.debate)
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| | Re: Yet another math problem
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(...) SNIP (...) Per Dave's last message, y is unknown (can't be 0), which Jeff has solved to 855/11. -Rob. (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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| | Re: Yet another math problem
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(...) Hey, that's super! And I finally found an online reference at the U of Georgia site, so I can probably handle these in the future. Thanks for the clear explanation and the solution. Dave! (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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| | Re: Comparative freedom
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(...) Someone once said that America needed a revolution to become their own country--Canada just 'evolved' into our own, and not without a few snags ourself. PET (Pierre Trudeau) 'decides' to get our constitution over here from Britain--got 9 (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.debate)
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| | Re: Yet another math problem
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(...) Oops. Hang on there Adrian, read for content. The five points as given in the initial problem (including 0,0) will not solve to a simple parabolic equation. Eliminating 0,0 does give us the solution specified above. Bleh. Adrian (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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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)
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| | 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)
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| | 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)
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| | 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)
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| | 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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| | Living in America--was Re: Comparative freedom
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In lugnet.off-topic.debate, Bruce Schlickbernd writes: <snip> (...) *If* I were to live in the USofA... (and that's a mighty big *if*)... I'd want to live in sunny SoCal... I love it when Craiggers does the "What's the weather like in Buffalo this (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.debate)
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| | Testing
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tset Fredrik (22 years ago, 11-Nov-02, to lugnet.off-topic.test)
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| | Re: Yet another math problem
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(...) Ugh - I can't believe it takes a math problem to get me to post for the first time in weeks. The formula for a parabola is: y = ax^2 + bx + c You only need three points to define a parabola, but since you've made yours symmetrical about the (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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| | Re: Yet another math problem
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(...) Are you looking for the defining equation of the form y = Ax2+Bx+c ?? or for something else? If the former wouldn't you just solve for "A" and "B" given that "C" is known to be zero (since you said that 0,0 is a point and is the vertex it (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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| | Yet another math problem
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This one should be pretty simple for you math-literate folks out there, but it's giving me a dreadful time... I have five points and am trying to define the parabola that contains them (if such exists). The points are: (0,0) (which is also the (...) (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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