Country Boy

No "planet" is going to have that kind of density. You appear to be talking about a "neutron star".

If an air parcel rises through a pipe, NOTHING happens to the temperature. Unless there is some other external effect that you haven't mentioned, the temperature stays the same.

Expansion is NOT work! It might do work if the expansion is against some force. Further, the expansion as it rises is not automatic. You apparently are assuming less air pressure at higher altitudes- so there is no force and no work done.

Then the obvious presumption is for both.

Here, you are told that (2, 3) is a point on the graph of y= f(x). That, of course, refers to x= 2, y= 3. A "corresponding point" for, say, f(x+ 3) would be (a, b) such that x+ 3= a and f(x+ 3)= b.

Suppose you are given value A with "uncertainty" x%. That means the true value can be any where from (1- 0.ox)A to (1+ 0.0x)A. Similarly given value B with "uncertainty" y% means that the true value can be anywhere from (1- 0.0y)B to (1+ 0.0y)B. The sum of the two can then be any where from (adding the two lowest possible values) (1- 0.0x)A+ (1- 0.0y)B= (A+ B)- 0.0xA- 0.0yB to (adding the largest possible values) (1+ 0.0x)A+ (1+ 0.0y)B= (A+ B)+ 0.0xA+ 0.0yB- The two "uncertainties" add. There is an engineer's "rule of thumb" that "when quantities add their absolute uncertainties add. When quantities are multiplied, their "relative uncertainties" (percentage uncertainties) add". The latter is not exact- it ignores the interaction of the of the uncertainties. For example, suppose A has a measured value of 200 with a 10% relative uncertainty so absolute uncertainty of 20. The actual value can be anywhere from 200- 20= 180 to 200+ 20= 220. Suppose B has a measured value of 100 with a 5% uncertainty so absolute uncertainty of 5. The actual value can be anywhere from 100- 5= 95 to 100+ 5= 105. The A+ B has a calculated value of 200+ 100= 300 but can be any where from 180+ 95= 275 to 220+ 105= 325. That is, [tex]A+ B= 300\pm 25[/tex]. The absolute uncertainty is 25= 20+ 5, the sum of the two absolute uncertainties.

"Unproctored"? So you want to be able to just tell your college that you did well in it? What, exactly, do you mean by "unproctored"?

This is incorrect. If "0.25= (2.10 e5J)/(kcal)" then multiply both sides by (kcal) to get 0.25(kcal)= (2.10 e5J) and divide both sides by 0.25: kcal= (2.10e5 J)/(0.25)= 8.50e^5= 850000 kcal.

Strictly speaking there is NO use of Calculus in computer science. It is possible to use computer programming to approximately solve problems in Calculus but I would not call that "using Calculus in computer science.

I don't understand this. The "axis of shaft" is a line segment while the "cg of object" is a point. They can't "coincide".

Yes, it is possible to find the inverse of a 4 by 4 or 5 by 5 matrix (or a 100000 by 100000 matrix). Surely you have learned that every "non-singular" matrix (having non-zero determinant) has an inverse. Whether it is 'hard' or not depends upon what you call 'hard'. It certainly is tedious! The simplest way to find the inverse of such a matrix, I think, is to use "row reduction" to reduce the given matrix to the identity matrix while applying the same operations to the identity matrix. The reason this works is that such row operation (add a multiple of one row to another, multiply a row by a number, swap two rows) corresponds to multiplication by an "elementary matrix"- one that is created by applying that same row operation to an identity matrix. If multiplying matrix "A" by a sequence of elementary matrices gives the identity matrix, then their product is the inverse matrix to A. And multiplying the identity matrix by all of them is there product. Here is a simple four by four example: Suppose $A= \begin{bmatrix}1 & 3 0 & 0 \\ -1 & 4 & 1 & 0 \\ 2 & 3 & 2 & 1\\ 0 & 2 & 1 & 0\end{bmatrix}$. I write it "side-by-side" the identity matrix: [math]\begin{bmatrix}1 & 3 0 & 0 \\ -1 & 4 & 1 & 0 \\ 2 & 3 & 2 & 1\\ 0 & 2 & 1 & 0 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}[/math] There is already a "1" leading the first row and a "0" leading the fourth row so the "reduce" the first column I can "add the first row to the second row" and "subtract twice the first row from the third row": [math]\begin{bmatrix}1 & 3 0 & 0 \\ 0 & 7 & 1 & 0 \\ 0 & -3 & 2 & 1\\ 0 & 2 & 1 & 0 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ -3 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}[/math] There is a "7" in "second row second column" so I need to divide that row by 7 to get a "1" there. Once I have done that I can add 3 times the new second row to the third row and subtract 2 times the new second row from the fourth row to get "0" below the "1": [math]\begin{bmatrix}1 & 3 0 & 0 \\ 0 & 1 & \frac{1}{7} & 0 \\ 0 & 0 & \frac{17}{7} & 1\\ 0 & 0 & \frac{5}{7} & 0 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0 \\ \frac{1}{7} & \frac{1}{7} & 0 & 0 \\ -\frac{18}{7} & \frac{3}{7} & 1 & 0 \\ \frac{2}{7} & \frac{2}{7} & 0 & 1\end{bmatrix}[/math]. As I said, it is tedious (and error prone) so I am going to stop here. You can continue with the last two rows and columns yourself if you wish to see that you actually do get the inverse matrix.

"What I would do is first determine the translation necessary to move one vertex of the first triangle to the corresponding vertex of the second triangle. Then determine the rotation necessary to map the other two vertices to their corresponding vertices. The translation and rotation can be written as a single matrix multiplication using "quaternions".

What semi-circle? You have already been told that the equation of a circle with center at (a, b) and radius r is (x- a)^2+ (y- b)^2= r^2. The equation of the semi-circle consisting of the right half of that circle is (x- a)^2+ (y- b)^2= r^2 with the restriction that x> a. Similarly, the left half is the same but with x< a, the top is with y> b and the bottom y< b.

Could I just point out that "the Hebrew for 'I am'" is NOT anything like "pi"? The whole basis for this thread is untrue.

Of course, the specific value depends upon the units you are using. In the "mks", "meter-kilogram-second", system, the acceleration due to gravity, g, is, on the surface of the earth, approximately 9.8 meters per second squared. In "cgs", "centimeter, gram, second, it is 980 centimeters per second squared, and in the "English" system, it is 32.2 feet per second squared. That number is only an average- even on the surface of the earth it varies from place to place, largely changing with altitude but there are places inside the earth that are denser than others so even at the same altitude g may vary slightly. And, of course, on different planets, with different masses and radii, g is different. You may be confusing "g" with "Newton's universal gravitational constant, "G", used in his formula, [math]F= \frac{GMm}{r^2}[/math], for the gravitational force between two objects of masses M and m with distance r between their centers. If a mass m is on a planet with mass M and distance r from the center of the planet, then it feels gravitational force $F= \frac{GMm}{r^2}$ and, since "F= ma", would fall with acceleration [tex]g= a= \frac{GM}{r^2}[/tex]. The mass of the earth and the average radius of the earth is such that that works out to g= 9.8 meters per second squared.

You understand, I hope, that posting this survey on this particular web-site will give a biased, pro-science, result.

One error is that taking n= k+ 1 in cos(2nx) gives cos(2(k+ 1)x)= cos((2k+ 2)x), not cos((2k+1)x) as you have.

I assume x, y, and z must be integers. First x+ z- y= 0. If x= 0 then z- y= 0 so z= y. (0, 0, 0), (0, 1, 1), (0, 2, 2), ..., (0, 15, 15) If x= 1 then z- y= -1 so z= y- 1 (1, 1, 0), (1, 2, 1), (1, 3, 2), ..., (1, 15, 14) If x= 2 then z- y= 2 so z= y- 2 (2, 2, 0), (2, 3, 1), (2, 4, 2), ...., (2, 14, 12) continue. ,

Yep, there's the problem! You are misspelling the "common names". If you look up "benzene", not "bezene", you will find that Its "IUPAC" name is "benzene"! If you look up "toluene", not "toulene", you will find the "IUPAC" name is "methylbenzene". And, finally if you look up "naphthalene" rather than "naphthalene", you will find that the "IUPAC" name is, again, "naphthalene". (The "systematic IUPAC name" (I confess I don't know that that means) can be "Bicyclo[4.4.0]deca-1,3,5,7,9-pentaene" or "Bicyclo[4.4.0]deca-2,4,6,8,10-pentaene").

I would think that people huddling together for warmth is an example of "transference of body energy"!

You say that as if they don't normally go together!

You've written "an annotated bibliography with tons of sources" on what? How can you research enough to have "tons of sources" if you do not yet know what you are researching?

How old are you?

The general solution to the quadratic equation, $ax^2+ bx+ c= 0$ is NOT $\frac{-b}{2a}\pm\sqrt{b^2- 4ac}$. It is $\frac{-b\pm \sqrt{b^2- 4ac}}{2a}$.

Well, that sort of thing happens with every measurement. If you put a thermometer in a bowl of water to measure its temperature, you change the temperature. When you put an air-gauge on a tire to measure its air pressure, you let some air out so change the air pressure by measuring it.