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      <title>Module 9: Quadratics by Anna Bradley</title>
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      <description>By:Anna Bradley P:4</description>
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      <pubDate>2017-04-27 16:59:11 UTC</pubDate>
      <lastBuildDate>2017-04-27 17:06:49 UTC</lastBuildDate>
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         <title>Linear</title>
         <author>bradleya7547</author>
         <link>https://padlet.com/bradleya7547/z8q30aahc0pz/wish/168697926</link>
         <description><![CDATA[<div><br></div><div><br></div><div>n | f(n)<br>0 | 2<br>1 | 6<br>2 | 10<br>3 | 14</div><div><br></div><div>Pattern: linear because it has a constant pattern</div><div><br></div><div>Recursive:</div><div>f(0)=2</div><div>f(n)=f(n-1)+ 4</div><div><br></div><div>Explicit:</div><div>f(n)=4x+2</div>]]></description>
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         <pubDate>2017-04-27 17:00:49 UTC</pubDate>
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         <title>Exponential</title>
         <author>bradleya7547</author>
         <link>https://padlet.com/bradleya7547/z8q30aahc0pz/wish/168698077</link>
         <description><![CDATA[<div><br></div><div><br></div><div>n | f(n)<br>0 | 2<br>1 | 4<br>2 | 8<br>3 | 16</div><div><br></div><div>This is exponential because it has a constant ratio.</div><div><br></div><div>Recursive: </div><div>f(0)=2</div><div>f(n)=f(n-1)*2</div><div><br>Explicit: f(n)=(2)(2)^n</div>]]></description>
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         <pubDate>2017-04-27 17:01:18 UTC</pubDate>
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         <title>Quadratic</title>
         <author>bradleya7547</author>
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         <description><![CDATA[<div><br></div><div><br></div><div>n | f(n)<br>0 | 1<br>1 | 2<br>2 | 5<br>3 | 10</div><div><br></div><div>This is quadratic because the 2nd difference is constant.</div><div><br></div><div>Recursive: </div><div>f(0)=1</div><div>f(n)=f(n-1)+2n-1</div><div><br>Explicit: f(n)=n^2+1</div>]]></description>
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         <pubDate>2017-04-27 17:01:39 UTC</pubDate>
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         <title>How to solve quadratic problems</title>
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         <pubDate>2017-04-27 17:02:16 UTC</pubDate>
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