Country Boy

Why use L'Hopital's rule? Mathematic's answer is much simpler.

Hey, bacon wrapped around anything is a 'thing'!

I suspect "regin" was supposed to be "reign". An "Elite Engineer" certainly should know, as swansont said, that "100% efficiency" is impossible. (Does "swans on tea" refer to swans drinking tea, swans smoking "tea", or swans swimming in tea?)

Your explanation of the Coriolis effect, that the earth moves eastward as something is moving north or south, is wrong- though close. If you were to throw a ball or shoot a rifle northward, the ball or bullet would also move eastward, along with the earth because you are moving eastward with the earth and so impart an eastward velocity component to the ball or bullet. But you are close- it is the change in the rotational speed of the earth that causes the Coriolis effect. But I suppose you will continue to post, say, essentially, "I am right and everyone else is wrong", giving no coherent explanation.

I know people that say that Biology isn't a science, just classifying things! I guess the question is "what do you mean by science'? I would base a definition on the "scientific method"- and by that standard Physics and Chemistry are definitely sciences- Biology, perhaps less so.

It is also due to the "binomial theorem": [math](x+ y)^n= \sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} x^ky^{n-k}[/math] Where [math]\begin{pmatrix}n \\ k\end{pmatrix}[/math] is the "binomial coefficient", [math]\frac{n!}{k!(n-k)!}[/math]. With x= y= 1 that becomes [math]2^n= \sum_{k=0}^\infty \begin{pmatrix}n \\ k\end{pmatrix}[/math] and clearly the binomial coefficient is "symmetric": [math]\begin{pmatrix}n \\ k \end{pmatrix}= \frac{n!}{k!(n-k)!}= \frac{n!}{(n- (n-k))!(n-k)!}= \begin{pmatrix}n \\ n-k\end{pmatrix}[/math]

How could your statement prove anything about a moderator's understanding? You pretty much told us in your first post that you do not know basic physics.

This sentence, "the hand puppet isn't objectively blocking the waveform of the photons moving towards the moon, only affecting the probability that they will arrive at a particular spot. So as we see the shadow disappear in a spot, it is really illuminated by photons that were already there." is pretty much non-sense. No, the photons were NOT already there. And by "affecting the probability that they will arrive at a particular spot" it is effectively making that probability 0. With something as large as a "hand puppet" that is NOT a quantum effect.

What do you mean by "EMR" (I would have thought "electro-magnetic radiation" or "resonance" but those aren't numbers.) Ther was no mention of "EMR" in the original post. The "wave number" is the reciprocal of wavelengh- The number of wave crests per unit length.

Have you ever taken a class in probability? If you did then you should have learned about the "binomial probability distribution". It is called "binomial" because there are two possible outcomes. Often those are called "success" and "failure"- here they are "happy" and "unhappy". If the probability of "success" is p then the probability of "failure" is 1- p and the probability of k "successes" and n- k "failures" in n "trials" is [math]\frac{n!}{k!(n-k)!}p^k (1- p)^{n-k}[/math]. "At least 2 happy" is the same as "not only 0 or 1 happy" so the probability that "at least 2 are happy" is 1 minus (the probability none are happy times the probability that 1 is happy).

I'm not sure what you mean by "area has no relations with perimeter." Certainly for a polygon with n sides there is a simple relationship between area and perimeter. And for a circle there is a simple relationship between area and perimeter. Did you mean that there the relationship between area and perimeter of a polygon changes as the number of sides changes?

Perhaps you do not understand what the words "profound" and "speculation" mean​. To talk about "if an object happens to travel faster than the speed of light" immediately after having said "no object can travel faster than the speed of light" is not "speculation", it is simply non-sense. And non-sense is never "profound".

Working from the bottom up: since k3= k2+ m and k2= k1+ m, k3= k1+ 2m. Then d= c+ k3 so d= c+ k1+ 2m. c= b+ k2 so c= b+ k1+ m and d= b+ 2k1+ 3m. b= a+ k1 so c= a+ 2k1+ m and d= a+ 3k1+ 3m That is: b= a+ k1, c= a+ 2k1+ m, and d= a+ 3k1+ 3m so we can write everything in terms of a, k1, and m. The real problem is those powers of b, c, and d and the fact that the definition of the rest of the terms is not clear- After a+ b^2+ c^3+ d^4, I would expect the next term to be a+ b^2+ c^3+ d^4+ e^5 but you have not defined "e"!

Saying that you violated a law out of ignorance of the law is NOT the same as being "innocent". If you "know you are innocent" even after being apprised of the law, you are being arrogant and claiming to be above the law.

If an object moves with constant acceleration, a, then its speed at time t is [math]v= at+ v_0[/math] where [math]v_0[/math] is its speed at t= 0. The distance it covered in that time is [math]d= (a/2)t^2+ v_0t[/math]. Those are what you are referring to as the "uvast" equations.

If, as you appear to be saying, you are not taking, and have never taken, a course in which this kind of thing is explained, where in the world did you get this question?

I recommend that the first thing you do is read the article!

"5 cards are drawn from a normal deck of cards. a) What are the odds that drawn was 1 five card and 2 three cards?" ​I presume you mean "exactly 1 five card and 2 three cards" and not "at least that many". ​ Can we at least assume you know that there are 52 cards in a "normal deck of cards", 13 in each suit, and 4 of each rank? ​ There are 4 "5" cards so the probability of drawing a "5" on the first draw is 4/52= 1/13. There are then 51 cards left in the deck, 4 of them "3"s. The probability of drawing a "3" on the second draw is 4/51. There are then 50 cards left in the deck, 3 of them "3"s. The probability of a "3" on the next draw is 3/50. There are then 49 cards left in the deck, three of them "5"s and two of them "3"s so there are 52- 3- 2= 47 of them not "5" or "3". The probability of drawing any card other than a "3" or a "5" is 47/49. There are then 48 cards left in the deck, 46 of them neither "5" or "3". The probability the last card drawn is 46/48. Therefore the probability drawing "5", "3", "3", "A", "A", where "A" is "any card other than a 3 or a 5", is (4/52)(4/51)(3/50)(47/49)(6/48). Do the same argument for "5, A, 3, A, 3" or any other order of the same 5 cards to see that the probability of drawing a "5", two "3"s, and two other cards is the same (you have different fractions but just with the numerators rearranged so the product is the same). There are [tex]\frac{5!}{1!2!2!}= \frac{120}{4}= 30[/tex] different arrangements of a "5", two "3"s, and 2 other cards so multiply that previous probability by 30 to get the probability of a "5", two "3"s, and two of any other cards. "b) What are the odds that 3 of the cards were of the same suit?" ​ I presume you mean exactly "3 cards of the same suit" and not "at least 3 cards of the same suit". ​The first card can be of any suit. That leaves 51 cards, 12 of the same suit as the previous card so the probability the second card is of the same suit as the first is 12/51. That leaves 50 cards, 11 of the same suit as the two previous cards so the probability the third card is also of that suit is 11/50. That leaves 49 cards, 39 of which are not of the same suit as the previous three. The probability that the fourth card is of a different suit from the first three is 39/49. That leaves 48 cards, 38 of which are not of the same suit as the first three (but might be of the same suit as the fourth). The probability that the fifth card is not of the same suits as the first three (but might be of the same suit as the fourth) is 38/48. The probability of getting the first three cards of the same suit and the last two of other suits is (12/51)(11/50)(39/49)(38/48). There are [tex]\frac{5!}{3!1!1!}= \frac{120}{6}= 20[/tex] different rearrangements that has three cards of the same suit so we need to multiply by 20 to get the probability of "three cards of the same suit" in any order.

Assuming your equations are (dS / dt) = -S * I* b (dI / dt) = -S * I * b - y * I (dR / dt) = y * I and b and y are constants, then, differentiating the first equation a second time, d^2S/dt^2= -b(dS/dt)I- bS(dI/dt) Replace "dI/dt" with -S*I*b- y*I from the second equation: d^S/dt^2= -b(dS/dt)I- bS(-bS- y)I= -b(dS/dt- bS^2- yS)I we can rewrite the first equation as I= -(dS/dt)/(bS) and putting that into the last equation d^S/dt^2= b(dS/dt- bS^2- yS)(dS/dt)/bS= (1/S)(dS/dt)^2- S(dS/dt)- dS/dt That's reduced to a single equation for S but it is a rather badly non-linear equation. I doubt there will be any method for getting an exact solution for S.

​ That question would be answered by examining their experimental method, equipment, and results, not by just saying "it is very difficult".

No, logic can only be applied to our models of "physical reality" Whether that gives "accurate" results or not depends upon how accurate our models are.

You say "for parents like me"- have you taken such calculations for your wife's heritage into account?

What, exactly, do you mean by "religion"? There are religions whose basic tenet is "do unto others as you would have others do unto you", or versions of that (Kant's "categorical imperative"), with no other dogma. That does not "form boundaries". There are, of course, many artificial boundaries that people set up but phrasing it solely in terms of "religion" is itself polarizing.

You are assuming a number of things. For one thing, who told you that God wants us to "reject all forms of evil"? What makes you think God has commited evil things? Are you assuming the "Judeo-Christian" God? If so you should say that!

Back in my dark past, when I was teaching basic algebra, I always discouraged students from saying "move a number from one side of the equation to another". With something like "ax= b" many students would come up with x= - b/a. Then when I pointed out that the answer should be b/a, they would complain "but I thought that when you moved a number from one side to the other you had to change the sign!". Instead I tried to teach them to think in terms of "reversing" whatever is done to x. In ax= b, x is multiplied by a. To reverse that do the opposite: divide by 3. And, of course, whatever you do to one side you must do to the other: ax/a= x= b/a. If the equation were x+ a= b, then, since a is added to x, we do the opposite: subtract a: x= a- b.