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HallsofIvy

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About HallsofIvy

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  1. I don't know where you got this idea! The last ice age, "when most of the planet was covered in ice" ended about 3,000,000 years ago (perhaps you just dropped ",000"). There was a period, sometimes called "the little ice age" of lower than normal tempartures but no ice sheets that only ended about 200 years ago (that's why the painting "Washington crossing the Delaware" shows ice in the river).
  2. HallsofIvy

    Question?

    It's hard to believe you are serious! From your other posts I had thought that you were past third or fourth grade arithmetic- which is basically what this is. I would hope that you know that 1x1= 1, 2x1= 2, and 1- 1= 0. And if 1- ?= ? the, adding "?" to both sides 1= 2?. What number, multiplied by 2, gives1?
  3. "Vast organisms"? Blue whales, today, are much larger than dinosaurs were!
  4. That's not what "consistency" means. Saying the Peano axiom are consistent means that the Peano axioms cannot be used to prove both statement "p" and "not p". Saying that the Peano axioms cannot be used to prove its own consistency means that the axioms cannot be used to prove that statement.
  5. This isn't really relevant to the main point here, but most people don't die because of "events". They die because their body wears out. And that would happen in any universe.
  6. If I remember correctly, a region in $R^2$ is "regular" if its boundary is a simple closed curve. (And a curve is "simple" if it does not cross itself.) Yes, the rectangular region $a\le x\le b$, $c\le y \le d$ is a "regular region". As to the integral, $\int_{u(x)}^{v(x)} f(x,y)dy$, Assuming that f is an "integrable" function of y, then $\int_{u(x)}^{v(x)} f(x,y)dy$ is a function of x, F(x). I might think of the "x" as a "parameter" in the integral but as a variable in f(x,y) and in F(x).
  7. The only thing "The Architect" has definitely shown is that he is NOT a mathematician! (And I am inclined to hope he is also NOT an architect- I would hate to have to live on an upper floor or a building he designed.)
  8. Do you understand that this primarily an exercise in READING COMPEHENSION? (Which is an extremely important skill in programming or software engineering). There is little mathematics or programming involved.
  9. How does that contradict what I said? A person who is on the dole still needs money- staying alive is not living in a satisfying way!
  10. If they had offered "Effing Science" when I was in school, I certainly would have taken it! That is, if "Effing" means what I think it does.
  11. I remember when I was very young thinking about this. Suppose we just let people go into a store and TAKE whatever they want! Since everyone can have whatever they want, no one would complain. Then as I got older I discovered a very important, very disturbing fact- work is hard! Most people would not want to work if they did not have to have money! What if we could just go into a store and take whatever we want, but discovered that there was not anything in the store? Who would plant, grow, and harvest the vegetables? That's hard work! If farmers did not need the money they wouldn't do it! If tailor's did not need the money they wouldn't make shirts, slacks, suits!
  12. "I am a layman trying to understand above theorems. This could be a stupid question." There is no such thing as a stupid question! "Does these theorems imply that we actually cannot prove that 2+2 = 4???" Gosh, I may have to reconsider! No, Godel's theorem say that, given any set of axioms large enough to encompass the properties of the non-negative integers there must exist some theorem that can neither be prove nor disproved. It does not say that a specific theorem cannot be proved. In fact, if we were able to identify a specific theorem that can not be proved nor disproved, we can always extend the axioms, perhaps by adding that theorem itself as an axiom, so that theorem can be proved. Of course, there would then be still another theorem that cannot be proved nor disproved.
  13. So you are arguing that NO ONE can criticize your paper because YOU did not include a "working roadmap" or "outline"?
  14. For a "free projectile" (one that has an initial force but no continuing thrust), and ignoring air resistance or other friction, the acceleration is the constant -g, the acceleration due to gravity. Since the gravity, set into an "xy- coordinate system" with positive y upward and positive x to the right, is <0, -g>, the velocity, at any time t, is <vx, vy- gt> where "vx" is the initial velocity in the x-direction and "vy" is the initial velocity in the y-direction. Taking the initial speed to be "v" at angle $\theta$ to the horizontal, the initial velocity is $\left< v cos(\theta), v sin(\theta)\right>$ so we have the velocity at time t to be $\left<v cos(\theta), v sin(\theta)- gt\right>$ and, integrating that with respect to time, and taking the initial position to be (0, 0), the position at time t is $\left<v cos(\theta)t, v sin(\theta)t- \frac{1}{2}gt^2\right>. Now, suppose, after time t, the projectile is at point (w, h). That is, that the projectile is distance w from the initial point, horizontally, and at height h (both of which may be positive or negative). Then we must have $v cos(\theta)t= w$ and $v sin(\theta)t- \frac{1}{2}gt^2= h$. How we proceed depends upon exactly what the problem is. If we are given initial speed and angle of the projectile we can immediately calculate w and h. If we are given w and h and are asked to find the necessary angle and initial speed to achieve that, we can solve the first equation for t, $t= \frac{w}{v cos(\theta)}$ and put that into the second equation to get an equation, $w tan(\theta)- \frac{1}{2}\frac{gw^2}{v^2cos^2(\theta)}$ relating v and $\theta$ (there will be more than one correct answer).
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