- Wed Jul 01, 2015 12:07 am
- Forum: Discrete Math
- Topic: solve this difficult number sequence
- Replies:
**3** - Views:
**51**

There does appear to be another instance of this question, proposed at nearly the same point in time five years past. The second instance ( here ) may be a repost of the poker-forum original which, as noted in the previous reply, may have contained an error -- and indeed may have have been a jape. T...

- Tue Jun 16, 2015 5:51 pm
- Forum: Discrete Math
- Topic: i need this to be proved really QUICK!
- Replies:
**1** - Views:
**115**

Prove that every tree graph has at least two vertexes with degree of one. There are many discussions of this exercise. You may find some of the following to be helpful: ⋅ StackExchange: Tree has at least two vertices of degree 1 ⋅ Homework Solutions (see Exercise 1) ⋅&nbs...

- Fri Jun 12, 2015 7:39 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: problem how to find solutions to hyperbola equation
- Replies:
**1** - Views:
**63**

A problem I have asks for the equations for the asymptotes of a hyperbola with this equation: [(x-3)^2/81] - [(y+7)^2/49] = 1 The answer given is: (-18, -7); (0,-7) I don't understand this You are quite correct that these points cannot possibly be the required equations of asymptotes. In fact, thes...

- Wed May 20, 2015 12:39 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Finding an nth degree polynomial
- Replies:
**1** - Views:
**157**

Suppose the division of f(x) by x-5 gives a quotient Q(x) and remainder R of 23. (b) Suppose Q(9) = 4; find f(9) One may "plug in" to what has already been determined by the conditions of the exercise. The exercise provides that: . . . f(x)\, =\, q(x)(x\, -\, 5)\, ...

- Wed May 20, 2015 12:29 pm
- Forum: Intermediate Algebra
- Topic: Substituting a Zero into a negative variable
- Replies:
**8** - Views:
**247**

so i'm trying to substitute this (1x^2 + 0x + 1 = 0) into the quadratic formula here is a reference link for the quadratic formula: http://postimg.org/image/8wx31enwf/ the part i'm not sure about is.. substituting 0 where -b is... what exactly happens What happens is that, where the Formula has a &...

- Mon May 04, 2015 2:14 am
- Forum: Uncategorized
- Topic: How did Babylonians divide circle into 360 ?
- Replies:
**2** - Views:
**251**

How did ancient people divide circle into 360 ? equal parts It is quite possible that nobody knows; this information may have been lost. But one method might have started with inscribing a pentagon using only a compass and straightedge . This would split the circumference into five equal arcs, each...

- Wed Mar 25, 2015 11:57 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Having difficulty with exponent equation
- Replies:
**1** - Views:
**439**

6561^{2x^2\, -\, x}\, =\, 27 This is what I started: 81^((x^2)-x)=3^3 Some steps appear to be missing. I believe, between the two lines above, you have converted the 6561 into 81^2. However, since the 2 does not factor out of 2x^2 - x, it cannot be "taken out" and applied to the base sepa...

- Fri Feb 27, 2015 4:04 pm
- Forum: Discrete Math
- Topic: Find two proofs for the identity
- Replies:
**4** - Views:
**778**

Using a combinatorial argument, prove the following identity: \binom{2n}{2}\,=\,2\,\binom{n}{2}\,+\,n^2 To use a "combinatorial argument", one must consider sets of elements. On the left-hand side: Consider a set with 2n elements, and pick two elements at a time. This may be done in C(2n,...

- Sun Feb 15, 2015 3:05 am
- Forum: Calculus
- Topic: Finding the derivative of arctan
- Replies:
**1** - Views:
**518**

y=arctan sqrt[(1+x)/(1-x)] what i did: let u= sqrt[(1+x)/(1-x)] du=1/2 [(1-x)/(1+x)]^1/2 [{(1+x)(-1)-(1-x)}/(1+x)^2]dx I solved this problem many times, but I don't get the right answer.... Where did you take the derivative of the inverse tangent? What did you get for your answer? What is "the...

- Fri Jan 30, 2015 7:37 pm
- Forum: Matrix (Linear) Algebra
- Topic: is A C B^T = B C^T A^T ?
- Replies:
**1** - Views:
**559**

I don't understand how A C B^T + B C^T A^T = 2 A C B^T This may be rearranged as: A C B T + B C T A T = A C B T + A B C T This results in: B C T A T = A B C T Two properties of transpositions are that the transpose of a transpose is the original matrix, and that the transposition of a product resul...