This is where the web master tries different techniques.

$$\spadesuit \heartsuit \diamondsuit \clubsuit$$

When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

$$ \gimel \daleth \beth \aleph$$

$$\forall x \in X, \quad \exists y \leq \epsilon$$

$$\frac{n!}{k!(n-k)!} = {n \choose k}$$

$$\sum_{i=1}^{10} t_i$$

$$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$$

$$P(E) = {n \choose k} p^k (1-p)^{ n-k}$$

If your browser supports this element, it should allow you to expand and collapse these details.

Homer J Simpson

Copyright © 2011 Jourdan’s Bridge Club. All Rights Reserved.