Headings
# H1
## H2
### H3
#### H4
##### H5
###### H6
H1
H2
H3
H4
H5
H6
Paragraphs
This is a paragraph.
I am still part of the paragraph.
New paragraph.
This is a paragraph. I am still part of the paragraph.
New paragraph.
Image
Web Image
Local Image
Block Quotes
> This is a block quote
This is a block quote
Code Blocks
```javascript
// Fenced **with** highlighting
function doIt() {
    for (var i = 1; i <= slen ; i^^) {
        setTimeout("document.z.textdisplay.value = newMake()", i*300);
        setTimeout("window.status = newMake()", i*300);
    }
}
```
function doIt() {
    for (var i = 1; i <= slen ; i^^) {
        setTimeout("document.z.textdisplay.value = newMake()", i*300);
        setTimeout("window.status = newMake()", i*300);
    }
}
Tables
| Colors        | Fruits          | Vegetable         |
| ------------- |:---------------:| -----------------:|
| Red           | *Apple*         | [Pepper](#Tables) |
| ~~Orange~~    | Oranges         | **Carrot**        |
| Green         | ~~***Pears***~~ | Spinach           |
| Colors | Fruits | Vegetable | 
|---|---|---|
| Red | Apple | Pepper | 
| Oranges | Carrot | |
| Green | Spinach | 
List Types
Ordered List
1. First item
2. Second item
3. Third item
- First item
- Second item
- Third item
Unordered List
- First item
- Second item
- Third item
- First item
- Second item
- Third item
Math
$$
evidence\_{i}=\sum\_{j}W\_{ij}x\_{j}+b\_{i}
$$
$$
AveP = \int_0^1 p(r) dr
$$
When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
evidencei=∑jWijxj+bi
AveP=∫10p(r)dr
When a≠0, there are two solutions to (ax^2 + bx + c = 0) and they are x=−b±√b2−4ac2a.
Emoji
This is a test for emoji. 😄 🙈 😸 🍉

