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	<title>Comments on: Addition&#8230;?</title>
	<atom:link href="http://www.mathstudenthelp.info/addition/addition/feed" rel="self" type="application/rss+xml" />
	<link>http://www.mathstudenthelp.info/addition/addition</link>
	<description>Let us help you add it up!</description>
	<pubDate>Mon, 21 May 2012 09:33:13 +0000</pubDate>
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		<title>By: Sean</title>
		<link>http://www.mathstudenthelp.info/addition/addition/comment-page-1#comment-6791</link>
		<dc:creator>Sean</dc:creator>
		<pubDate>Wed, 07 Oct 2009 04:22:59 +0000</pubDate>
		<guid isPermaLink="false">http://www.mathstudenthelp.info/addition/addition#comment-6791</guid>
		<description>3/(4z^3) + 1/(6z^2)

= [3*3 + 2z]/[12z^3]

= [9+2z]/[12z^3] 

It is brought to a common denominator, that is to say 12z^3,  but I am not sure about being simplified.&lt;br&gt;&lt;b&gt;References : &lt;/b&gt;&lt;br&gt;</description>
		<content:encoded><![CDATA[<p>3/(4z^3) + 1/(6z^2)</p>
<p>= [3*3 + 2z]/[12z^3]</p>
<p>= [9+2z]/[12z^3] </p>
<p>It is brought to a common denominator, that is to say 12z^3,  but I am not sure about being simplified.<br /><b>References : </b></p>
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		<title>By: The Game</title>
		<link>http://www.mathstudenthelp.info/addition/addition/comment-page-1#comment-6790</link>
		<dc:creator>The Game</dc:creator>
		<pubDate>Wed, 07 Oct 2009 03:59:59 +0000</pubDate>
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		<description>what??&lt;br&gt;&lt;b&gt;References : &lt;/b&gt;&lt;br&gt;</description>
		<content:encoded><![CDATA[<p>what??<br /><b>References : </b></p>
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	<item>
		<title>By: Cameron</title>
		<link>http://www.mathstudenthelp.info/addition/addition/comment-page-1#comment-6789</link>
		<dc:creator>Cameron</dc:creator>
		<pubDate>Wed, 07 Oct 2009 03:10:59 +0000</pubDate>
		<guid isPermaLink="false">http://www.mathstudenthelp.info/addition/addition#comment-6789</guid>
		<description>The hardest part is getting the least common denominator. In your case, 12 as a number is the LCM of 4 and 6. Then, you also need the most, or highest power of, number of z's. In the first fraction, we have 3. So, the LCD is 12z^3. Now, on top you most multiply by whatever you seemingly &#34;added&#34; to the bottom. For the one on the left you need a 3 and the one on the right you need a 2z.

Therefore, you end up with (9+2z)/(12z^3). In this case, you can not simplify at all!&lt;br&gt;&lt;b&gt;References : &lt;/b&gt;&lt;br&gt;</description>
		<content:encoded><![CDATA[<p>The hardest part is getting the least common denominator. In your case, 12 as a number is the LCM of 4 and 6. Then, you also need the most, or highest power of, number of z&#8217;s. In the first fraction, we have 3. So, the LCD is 12z^3. Now, on top you most multiply by whatever you seemingly &quot;added&quot; to the bottom. For the one on the left you need a 3 and the one on the right you need a 2z.</p>
<p>Therefore, you end up with (9+2z)/(12z^3). In this case, you can not simplify at all!<br /><b>References : </b></p>
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		<title>By: Qwerty Q</title>
		<link>http://www.mathstudenthelp.info/addition/addition/comment-page-1#comment-6788</link>
		<dc:creator>Qwerty Q</dc:creator>
		<pubDate>Wed, 07 Oct 2009 02:31:59 +0000</pubDate>
		<guid isPermaLink="false">http://www.mathstudenthelp.info/addition/addition#comment-6788</guid>
		<description>remember, addition and subtraction need you to make the denominator the same. but whatever you do to the denominator, you have to do to the numerator too.

so:

    3 x 3........................1 x 2
________.........+....________
(4z^3) x 3.................(6x^2) x 2



....9...............................2
_____.........+..........._______
12z^3.........................12z^2


the denominators need to be balanced, then you can simply add the numerators.
i don't think it can get any more simplified than this.&lt;br&gt;&lt;b&gt;References : &lt;/b&gt;&lt;br&gt;</description>
		<content:encoded><![CDATA[<p>remember, addition and subtraction need you to make the denominator the same. but whatever you do to the denominator, you have to do to the numerator too.</p>
<p>so:</p>
<p>    3 x 3&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;1 x 2<br />
________&#8230;&#8230;&#8230;+&#8230;.________<br />
(4z^3) x 3&#8230;&#8230;&#8230;&#8230;&#8230;..(6x^2) x 2</p>
<p>&#8230;.9&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.2<br />
_____&#8230;&#8230;&#8230;+&#8230;&#8230;&#8230;.._______<br />
12z^3&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.12z^2</p>
<p>the denominators need to be balanced, then you can simply add the numerators.<br />
i don&#8217;t think it can get any more simplified than this.<br /><b>References : </b></p>
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