Recent content by konthelion

What you are trying to find is the cumulative distribution function (cdf) which is defined by: P(X \leq x) = \sum_{y \leq x} p(y) To use different numbers... say you want to find P(X \leq 2) , it would be P(X \leq 2) = p(0) + p(2) =... but P(X < 2) = p(0) \neq p(0) + p(2) = P(X \leq...

If p is a prime where the order of G is 2p, then G has a normal subgroup, J, with order p (if p is not equal to 2). Since J is normal, then it is contained in the intersection of all Sylow p-subgroups in G (I thought this was by definition... I don't know). Thus, it must be contained in at...

Do you see a pattern between the numbers? Try to write it in summation form i.e. \sum_{i=1}^n \left\{... \right\}; after you see the pattern, you should know what n is. Then you can find the sum.

The "slant height" of a triangle is the height of one of its sloping sides, in this case the side where the leaning tower of pizza is leaning towards. So, yes the length of the slant is the "slant height". The height of the leaning tower would be different (which was what you thought it was).

Let \left\{ A_{i} \right\}_{i \in I} be your indexed set of family. Do you mean this \bigcap_{i=1} A_i = \left\{ x : \forall i \in I: x \in A_i \right\} ?

What did you get for \frac{N_{x}(x,y) - M_{y}(x,y)}{M(x,y)}? If your integrating factor say, \mu is a function of x only, then \frac{d \mu}{dx} = \frac{N_{x}(x,y) - M_{y}(x,y)}{M(x,y)} \mu From here you can solve for the exact equation after multiply the integrating factor by the...

Since tr(AB) = tr(BA) (sorry wrong terminology), so just apply the definition of an orthogonal matrix.

If you think that it is false, you can provide a counterexample. So, if f : [a, b] \to R and let A = [a, b] is bounded, is it necessarily true that f(A) is bounded?

You need to show your work. But for starters, one way that you can approach this problem is to solve for "A+B" in both equations in terms of "C". Then you can solve for "C" using these two equations. Last, solve for A and B.

You made a mistake here, this should be -7.

A\(A\ B) = B For A\B, Let x \in A \wedge x \notin B Then, A \ (A \ B) implies that x \in A \wedge (x \notin A \vee x \in B) by DeMorgan's Law Then, by associative law, (x \in A \wedge x \notin A) \vee ( x \in A \wedge x \in B) \implies ( x \in A \vee x \in B) So, A \cup B = B if...

It is (t-3) not (t-2)

You have a typo. It's a 3 not a 2. \int \frac{2t}{ (t-3)^2}dt =\int \frac{2}{t-3}dt + \int \frac{6}{(t-3)^2}dt Also, you did not integrate the right term correctly, it's a degree of "-2" not "-1"

cos2x = 2cos^2x-1 = 1-2sin^2x=cos^2x-sin^2x I believe you mean sin^2x = 1 - cos^2 x?

You should get A = 2 and B - 3A = 0