Recent content by Outlined

thanks for your reply, however with most of this kind of quesitons you do not have to make assumptions. In fact, some of this kind of exercises have the option (it is always multiple choice) "cannot say".

ROFL both joke and truth When mathematicians think about abstract structures they think about the properties those structures have. For groups they think about the order, whether it is abelian, etc. For vector-spaces they think about the dimension (in fact, two vectorspaces over the field F...

let me give you a counterexample. p1 is the number of people who work in Q1 pr is the number of people who work in Q2, Q3 and Q4. Than the average costs per employee per year equals 1/4 * 1/p1 * 41412 + 3/4 *1/ pr * (167040 - 41412) = 1392 Two free variables so c1 can be everything!

thanks for your reply, however the employees who work in Q1 are not necessarily the same as the one who work in Q2, Q3 or Q4. So you have to be cautious drawing conclusion from that.

They say the answer is 41412 / (1392 / 4) = 119. I think you cannot say as they gave the average cost per employee per year so you know nothing about the cost per employee per year in the 1st Q so you cannot decide how many employees you got that quarter.

check what you wrote down on your own paper is the same as in the examination. Is matrix A the same as on your examination? Really? Check it again! check results, found an eigenvector and eigenvalue? Check whether the computation Ax = lambda * x holds. Use tests: a polynomial of degree 4 and...

Given 10 men, how many ways can you pick 5 men out of it? Given 12 women, how many ways can you pick 5 women out of it? Given 5 men and 5 women, how many couples can you make?

Look at the ring Z/nZ, this is a field <=> n is prime This is easy to see, is n is not a prime then n = km for n, m > 1 so k nor m are invertible in Z/nZ. So Z/nZ is not a field. On the other hand if n is prime then the equation an + bk = 1 always has a solution for a and b so k is...

the answer is n \choose k should be in your book.

If all your coefficients are real then for any root holds the complex conjugate is also a root. In your case j is a root so \bar{j} is a root too. Next step: Devide x^{5} -9x^{4} +24x^{3} -24x^{2} +23x -15 by (x - 1)(x - j)(x + j) = (x - 1)(x^{2} + 1) = x^{3} - x^{2} + x - 1.

Follows straight from the definition of basis: x_{k+1} + \ker{T}, \dots , x_{n} + \ker{T} are linearly independent in V / \ker{T}; x_{k+1} + \ker{T}, \dots , x_{n} + \ker{T} span whole V / \ker{T}. Tell me which part you have problems with.

use this rule (a + b)c = ac + bc repeatedly. you also have aa'= 0 hope this helps

It is quite easy, let T be the torsion subgroup. It is easy to proove Q/Z \subseteq T. If we have Q/Z \subset T then there is a real number r \notin Q such that r + Z has finite order. This implies r \in Q. Contradiction so Q/Z = T.

When you write log you have to define the base, which makes it a more complicated function than the ln, which has e as a base. btw: why would one want to use a log with 10 as a base? From what I have seen ln is almost always a good solution.

You can not be sure (2A + I)-1 exists! I would try to write it in the form ( . . . )(x - b) = 0.