Country Boy

Please tell us exactly what the question was! If it was simply to find [math]\int_{-2}^2 x^3 dx[/math] then the answer is, indeed, 0. But if the question was to find the area of the region bounded by the x-axis and the graph of [math]x^3[/math] then you need to do it in two parts because those graphs cross at x= 0. For x> 0, the graph of [math]y= x^3[/math] is above the graph of y= 0 so the area is [math]\int_0^2 (x^3- 0) dx[/math]. For x< 0 the graph of [math]y= x^3[/math] is below the graph of [math]y= 0[/math] so the area is [math]\int_{-2}^0 (0- x^3) dx= -\int_{-2}^0 x^3 dx[/math].

It is a linear equation, of the form "An= B" with A= 3/19 and B= -8. It is not a linear function so cannot be graphed- though you could think of it a point on the graph y= (3/19)x at y= -8.

Since no basket can be empty, the first thing you want to do is put one apple in each basket. That leaves n-k apples to be distributed among k baskets. You could put all n-k apples in any one of the k baskets- there are k ways to do that. You could put n-k-1 apples in one basket, one apple in another basket. There are k choices for the first basket, k- 1 choices for the second basket so there are k(k-1) ways to do that. Continue.

You are ​interpreting ​it as 36/[6(2+ 2+ 2)]= 36/[(6)(6)= 36/36= 1. Others are interpreting it as (36/6)(2+ 2+ 2)= (6)(6)= 36. In fact, the original "36/6(2+ 2+ 2)" is just badly written- it is ambiguous and the standard "PEMDAS" or "BODMAS" does not clarify it.

There are situations, such as a very high, though finite, potential barrier, where the probability the particle exists is very, very low, though non-zero. You would have to decide how low that probability must be before you say "the particle is not there".

To write the first column of A, (.8, .2), as a linear combination of x1 and x2 we need to find numbers a and b such that (.8, .2)= a(.6, .4)+ b(1, -1)= (.6a+ b. .4a- b). That gives the two equation .6a+ b= .8 and .4a- b= .2. Adding the two equations eliminates b: (.6+ .4)a= a= .8+ .2= 1.0. With a= 1. .6a+ b= .6+ b= .8 so b= 0.2. It is 0.8x+ 0.2x2. Are you clear on what "x1" and "x2" are? Equation (1) writes (0.8, 0.2) as a linear combination of x1= (0.6, 0.4) and x2= (1, -1). That means we need to find numbers, a and b, such that a(0.6, 0.4)+ b(1, -1)= (0.6a+ b. 0.4a- b)= (0.8, 0.2). That gives the two equations 0.6a+ b= 0.8 and 0.4a- b= 0.2. Adding the two equations eliminates b: (0.6+ 0.4)a= a= 0.8+ 0.2= 1.0. Since a= 1, 0.6a+ b= 0.6+ b= 0.8 so b= 0.8- 0.6= 0.2. (0.8, 0.2)= 0.8(0.6, 0.4)+ 0.2(1, -1) is correct.

You learn mathematics by doing mathematics. not by seeing someone else do mathematics!

That's how a diesel engine works! Instead of having a sparkplug to ignite the gas mixture, the piston compresses the gas mixture enough that the increase in termperature ignites it. There is no expenditure of energy to create a spark and, once the engine has started, the energy from the previous ignition cycle is used to compress the next cycle. Of course, diesel engines are notoriously hard to start!

I said​ that any multiple of (-2, 1) is a solution. Clearly (2, -1) is a solution because it is -1 times (-2, 1). In fact, we could say that all solutions are multiples of (2, -1). However, it is wrong to say that (2, -1) is the solution. It is a solution- one of an infinite number of solutions.

The equation is [math]\begin{bmatrix}1 & 2 \\ 2 & 4 \end{bmatrix}\begin{bmatrix}y \\ z\end{bmatrix}= \begin{bmatrix}y+ 2z \\ 2y+ 4z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/math]. The determinant of the matrix is 0 which means it does not have an inverse. That, in turn, means that there is no unique solution to this equation. Either there is no solution or there are an infinite number of solutions. Obviously [math]\begin{bmatrix} y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/math] is a solution so there must be an infinite number of solutions. To find them, I would write this matrix equation as the system of equations y+ 2z= 0 and 2y+ 4z= 0. Dividing the second equation by 2 gives y+ 2z= 0, the same as the first equation. Any y and z that satisfy y+ 2z= 0, which is the same as y= -2z, will satisfy both equations and so [math]\begin{bmatrix}-2z \\ z\end{bmatrix}[/math] will satisfy the matrix equation for z any number. You could also write that as [math]z\begin{bmatrix}-2 \\ 1 \end{bmatrix}[/math] or "any multiple of [math]\begin{bmatrix}-2 \\ 1 \end{bmatrix}[/math]".

Seeing that combination of logarithm, square root, and powers of x, I very much doubt that it can be solved algebraically.

Provided neither x- y nor x+ y is equal to 0. That is, provided neither x= y nor x= -y. If x= y or x= -y, the expression is not defined.

Yes, S=ut+.5at^2 is the distance traveled in time t- from 0 to t seconds if u is in units of "distance per second" and a is in units of "distance per second per second". The distance traveled "in the nth second" is the distance traveled from 0 up to n seconds minus the distance traveled in n-1 seconds. That is, (un+ .5an^2)- (u(n-1)+ .5a(n- 1)^2)= un+ .5an^2- (un- u+ .5an^2- an+ .5a)= u+ an- .5a since the "un" and ".5an^2" terms cancel.

You missed the Incan prophesy that the world will end on Dec. 21, 2012.

I'm getting into this late but I question the basis of the initial post. The Chinese and Japanese had technology that was, in many ways more advanced than the Europeans and they also conquered other peoples. The mongol invasion which put all of eastern Europe under their thumb and reached into France, until they simply left, was due to a technological advance- they had the stirrup which Europeans did not. And the arabs, with their own technology, conquered all the way from Arabia through Africa and southern Europe.

The definition of energy is really a combination of things- largely defined in order to maintain "conservation of energy". The simplest definition of energy is that of "work" : force times distance. Of course, applying force to something may cause it to move so we add "kinetic energy"- 1/2 mass times velocity squared. Of course there is friction which will slow an object, causing heat, so we add heat energy. "Energy" is really a "book keeping" device!

I would normally intepret "A- 1" for A a linear transformation as "A- I". But I suspect that you mean [math]A^{-1}[/math] (if you don't want to use Latex, write A^-1), the inverse matrix. You should know that, to prove "[math]T^{-1}Z[/math] is a subspace of vector space W", you need to prove two things: 1) If u and v are in [math]T^{-1}Z[/math] then u+ v is in[math]T^{-1}Z[/math]Z. 2) if u is in [math]T^{-1}Z[/math] and a is a number, then au is in [math]T^{-1}Z[/math]. If u and v are in [math]T^{-1}Z[/math] then T(u) and T(v) are both in Z. [math]T(u+ v)= Tu+ Tv[/math] by definition of "linear transformation". Since Z is a subspace, Tu+ Tv is in Z. So u+ v is in [math]T^{-1}Z[/math]. You can do (2).

"Suppose there is an observer standing on a body A and there is another body B at a distance. Now this observer sees that B is coming closer towards him. From this observation, can he conclude that whether A is moving towards B or B is moving towards A ?" Relative to A, B is moving and A is not. Relative to B. A is moving and B is not. You cannot say anything is moving without saying relative to what.

"I understand they were cancer cells and had assumed cancer cells were kind of function less, why is it the case that they can still be infected with such viruses?" Why would whether or not cancer cells have a "function" or not have anything to do with being infected by a virus? It is not as if the virus cared!

First, what is this "law of infinite probability"? I have never heard of such a thing. Second, if a basketball player has a probability "p" or making a free throw then he has a probability of 1- p of missing a free throw. Assuming that every shot is independent of every other shot, the probabliity of missing 100 free throws in a row is [math](1- p)^{100}[math]. If p is close to 1, that will be very close to 0 but not 0. I don't know what "functionally 0" means but, mathematically, there is always a probability that a good free thrower will miss 100 shots in a row.

Given A= B= C, we have immediately that A= B and that B= C. It follows from that that A= C. It is a matter of taste whether you think that last equation is "included" in the original rather than just "implied" by it.

Added: I have found two definitions of "anti-compact": 1) a subset A in a topological space X is anti-compact if every covering of A by closed sets has a finite subcover. 2) anti-compact means that the only compact subsets of X are the finite ones.

Yes, you should use the Archimedian property here. To start, look at y- x and show that there exist an integer n such that n(y- x) is larger than 1.

Thank you for giving the definition of "pseudo-compact". Now, it would help if you would give the definition of "anti-compact'!