How to get started with the GCSE exams platform.

# Getting Started

Ek​\=21​mv2

This is a very simple editor that helps you craft your document.

mathrm{E}\_{\\mathrm{k}}=\\frac{1}{2} \\mathrm{mv}^2 this is it

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\$\\mathrm{E}\_{\\mathrm{k}}=\\frac{1}{2} \\mathrm{mv}^2\$

**\* This is just a pre-formatted sample documentFeel free to delete this text and start all over again**

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$$

A\_r \\text { of Nitrogen }=\\frac{(99.63 \\times 14)+(0.37 \\times 15)}{99.63+0.37}=\\frac{1394.82+5.55}{100.0}=\\frac{1400.37}{100.0}=14.0

$$

You can also embed and HTML

`<iframe width="560" height="315" src="https://www.youtube.com/embed/NxKTy\_ymTjc" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>`

`\\mathrm{E}\_{\\mathrm{k}}=\\frac{1}{2} \\mathrm{mv}^2`

\\mathrm{E}\_{\\mathrm{k}}=\\frac{1}{2} \\mathrm{mv}^2

`A\_r \\text { of Nitrogen }=\\frac{(99.63 \\times 14)+(0.37 \\times 15)}{99.63+0.37}=\\frac{1394.82+5.55}{100.0}=\\frac{1400.37}{100.0}=14.0`

`\$A\_r\$ of Nitrogen \$=\\frac{(99.63 \\times 14)+(0.37 \\times 15)}{99.63+0.37}=\\frac{1394.82+5.55}{100.0}=\\frac{1400.37}{100.0}=14.0\$`

`$$`

`A\_r \\text { of Nitrogen }=\\frac{(99.63 \\times 14)+(0.37 \\times 15)}{99.63+0.37}=\\frac{1394.82+5.55}{100.0}=\\frac{1400.37}{100.0}=14.0`

`$$`

`\\begin{equation}`

`A\_r \\text { of Nitrogen }=\\frac{(99.63 \\times 14)+(0.37 \\times 15)}{99.63+0.37}=\\frac{1394.82+5.55}{100.0}=\\frac{1400.37}{100.0}=14.0`

`\\end{equation}`

<h1 class="PlaygroundEditorTheme__h1" dir="ltr">Getting Started</h1> Ek​=21​mv2 This is a very simple editor that helps you craft your document.mathrm{E}_{\mathrm{k}}=\frac{1}{2} \mathrm{mv}^2 this is itTips <ol class="PlaygroundEditorTheme__ol1"><li value="1" class="PlaygroundEditorTheme__listItem">Use Keyboard shortcuts</li><li value="2" class="PlaygroundEditorTheme__listItem">Highlight a text to see formatting options</li><li value="3" class="PlaygroundEditorTheme__listItem">Enter a new line and press Tab to see options</li></ol> \(\mathrm{E}_{\mathrm{k}}=\frac{1}{2} \mathrm{mv}^2$ <iframe data-lexical-youtube="I0ex8sS7EVs" width="560" height="315" src="https://www.youtube-nocookie.com/embed/I0ex8sS7EVs" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="true" title="YouTube video"></iframe> * This is just a pre-formatted sample documentFeel free to delete this text and start all over again You can also 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Introduction to the GCSE learning

<h1 class="PlaygroundEditorTheme__h1" dir="ltr">Getting Started</h1> Ek​=21​mv2 This is a very simple editor that helps you craft your document.mathrm{E}_{\mathrm{k}}=\frac{1}{2} \mathrm{mv}^2 this is itTips <ol class="PlaygroundEditorTheme__ol1"><li value="1" class="PlaygroundEditorTheme__listItem">Use Keyboard shortcuts</li><li value="2" class="PlaygroundEditorTheme__listItem">Highlight a text to see formatting options</li><li value="3" class="PlaygroundEditorTheme__listItem">Enter a new line and press Tab to see options</li></ol> \(\mathrm{E}_{\mathrm{k}}=\frac{1}{2} \mathrm{mv}^2$ <iframe data-lexical-youtube="I0ex8sS7EVs" width="560" height="315" src="https://www.youtube-nocookie.com/embed/I0ex8sS7EVs" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="true" title="YouTube video"></iframe> * This is just a pre-formatted sample documentFeel free to delete this text and start all over again You can also Diagrams $$A_r \text { of Nitrogen }=\frac{(99.63 \times 14)+(0.37 \times 15)}{99.63+0.37}=\frac{1394.82+5.55}{100.0}=\frac{1400.37}{100.0}=14.0$$<span 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<h1 class="PlaygroundEditorTheme__h1" dir="ltr">Getting Started</h1> Ek​=21​mv2 This is a very simple editor that helps you craft your document.mathrm{E}_{\mathrm{k}}=\frac{1}{2} \mathrm{mv}^2 this is itTips <ol class="PlaygroundEditorTheme__ol1"><li value="1" class="PlaygroundEditorTheme__listItem">Use Keyboard shortcuts</li><li value="2" class="PlaygroundEditorTheme__listItem">Highlight a text to see formatting options</li><li value="3" class="PlaygroundEditorTheme__listItem">Enter a new line and press Tab to see options</li></ol> \(\mathrm{E}_{\mathrm{k}}=\frac{1}{2} \mathrm{mv}^2$ <iframe data-lexical-youtube="I0ex8sS7EVs" width="560" height="315" src="https://www.youtube-nocookie.com/embed/I0ex8sS7EVs" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="true" title="YouTube video"></iframe> * This is just a pre-formatted sample documentFeel free to delete this text and start all over again You can also Diagrams $$A_r \text { of Nitrogen }=\frac{(99.63 \times 14)+(0.37 \times 15)}{99.63+0.37}=\frac{1394.82+5.55}{100.0}=\frac{1400.37}{100.0}=14.0$$<span 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autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture" allowfullscreen&gt;&lt;/iframe&gt;</code> <code>\mathrm{E}_{\mathrm{k}}=\frac{1}{2} \mathrm{mv}^2</code> \mathrm{E}_{\mathrm{k}}=\frac{1}{2} \mathrm{mv}^2 <code>A_r \text { of Nitrogen }=\frac{(99.63 \times 14)+(0.37 \times 15)}{99.63+0.37}=\frac{1394.82+5.55}{100.0}=\frac{1400.37}{100.0}=14.0</code> <code>$A_r$ of Nitrogen $=\frac{(99.63 \times 14)+(0.37 \times 15)}{99.63+0.37}=\frac{1394.82+5.55}{100.0}=\frac{1400.37}{100.0}=14.0$</code> <code>$$</code><code>A_r \text { of Nitrogen }=\frac{(99.63 \times 14)+(0.37 \times 15)}{99.63+0.37}=\frac{1394.82+5.55}{100.0}=\frac{1400.37}{100.0}=14.0</code><code>$$</code> <code>\begin{equation}</code><code>A_r \text { of Nitrogen }=\frac{(99.63 \times 14)+(0.37 \times 15)}{99.63+0.37}=\frac{1394.82+5.55}{100.0}=\frac{1400.37}{100.0}=14.0</code><code>\end{equation}</code>

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