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        <title><![CDATA[Pre-Calculus : Weblog]]></title>
        <description><![CDATA[The weblog for Pre-Calculus, hosted on College-Cram.]]></description>
        <generator>Elgg</generator>
        <link>http://www.college-cram.com/study/precalculus/weblog/</link>        
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            <title><![CDATA[Pre-Calculus Homework Help]]></title>
            <link>http://www.college-cram.com/study/precalculus/weblog/-1</link>
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            <pubDate>Fri, 09 May 2008 22:37:09 GMT</pubDate>
		<dc:subject><![CDATA[pre-calculus]]></dc:subject>
		<dc:subject><![CDATA[pre-algebra]]></dc:subject>
		<dc:subject><![CDATA[polynomials]]></dc:subject>
		<dc:subject><![CDATA[matrix algebra]]></dc:subject>
		<dc:subject><![CDATA[logarithms]]></dc:subject>
		<dc:subject><![CDATA[homework help]]></dc:subject>
		<dc:subject><![CDATA[graphs and lines]]></dc:subject>
		<dc:subject><![CDATA[functions]]></dc:subject>
		<dc:subject><![CDATA[free worksheets]]></dc:subject>
		<dc:subject><![CDATA[free worksheet]]></dc:subject>
		<dc:subject><![CDATA[equations and inequalities]]></dc:subject>
		<dc:subject><![CDATA[counting and probability]]></dc:subject>
		<dc:subject><![CDATA[conic sections]]></dc:subject>
		<dc:subject><![CDATA[complex numbers]]></dc:subject>
		<dc:subject><![CDATA[bottomless worksheets]]></dc:subject>
		<dc:subject><![CDATA[bottomless worksheet]]></dc:subject>
		<dc:subject><![CDATA[pre-calculus homework help]]></dc:subject>
		<dc:subject><![CDATA[pre-calculus worksheet]]></dc:subject>
		<dc:subject><![CDATA[pre-calculus worksheets]]></dc:subject>
		<dc:subject><![CDATA[systems of equations]]></dc:subject>
		<dc:subject><![CDATA[bottomless]]></dc:subject>
            <description><![CDATA[<p>Pre-calculus students from colleges and high schools can get homework help here from our Pre-calculus resources. Whatever the reason for their difficulties -- bad textbook, missed classes, or whatever -- we have the Cramlets to help pre-calculus students get the homework help they need:</p><ul><li><a href="http://www.college-cram.com/study/search/index.php?tag=pre-algebra&amp;owner=21">pre-algebra</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=graphs+and+lines&amp;owner=21">graphs and lines</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=functions&amp;owner=21">functions</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=equations+and+inequalities&amp;owner=21">equations and inequalities</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=complex+numbers&amp;owner=21">complex numbers</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=polynomial&amp;owner=21">polynomial</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=logarithm&amp;owner=21">logarithm</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=angle+functions&amp;owner=21">angle functions</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=systems+of+equations&amp;owner=21">systems of equations</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=matrix+algebra&amp;owner=21">matrix algebra</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=counting+and+probability&amp;owner=21">counting and probability</a></li><li><a href="http://www.college-cram.com/study/search/index.php?tag=conic+sections&amp;owner=21">conic sections</a></li></ul><p>Try our resources for yourself, and you find that getting <a href="http://www.college-cram.com/study/help/presentations/1114"  title="better grades in less time"><strong>better grades in less time</strong></a> isn&#39;t that hard at all!&nbsp;</p>]]></description>
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        <item>
            <title><![CDATA[Using the Graphing Method]]></title>
            <link>http://www.college-cram.com/study/precalculus/weblog/using-the-graphing-method</link>
            <guid isPermaLink="true">http://www.college-cram.com/study/precalculus/weblog/using-the-graphing-method</guid>
            <pubDate>Thu, 09 Aug 2007 19:53:34 GMT</pubDate>
		<dc:subject><![CDATA[math help]]></dc:subject>
		<dc:subject><![CDATA[math]]></dc:subject>
		<dc:subject><![CDATA[graphing method]]></dc:subject>
		<dc:subject><![CDATA[graphing]]></dc:subject>
		<dc:subject><![CDATA[Using the Graphing Method]]></dc:subject>
		<dc:subject><![CDATA[system of equations]]></dc:subject>
		<dc:subject><![CDATA[Algebra]]></dc:subject>
            <description><![CDATA[<div id="question"><p>Dear Professor Cram:</p><p>What is the solution to the system of equations: x = y and x = -y?</p><p>KG, US Navy</p></div><div class="icon"><img src="http://www.college-cram.com/ask/Using-The-Graphing-Method.gif"  border="0"  alt="See the graphs of both equations."  title="See the graphs of both equations."  width="250"  height="250"  align="left" /></div><p>KG, thank you for this algebra question.</p><p>The easiest way to solve this system of equations problem is by using the graphing method.</p><p>(Use the illustration to the left for a helpful guide.) If we graph the first equation, we get a diagonal line passing through the origin sloping upwards. (See the blue line in the illustration.) Graphing the second equation gives us a diagonal passing through the origin and sloping downwards. (See the pink line in the illustration.)</p><p>From visual inspection, we can see that the lines intersect at one place only -- the origin. Thus, the solution to this system is (0,0).</p><div id="signature"><p>Good studying,</p><p>Professor Cram</p></div>]]></description>
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        <item>
            <title><![CDATA[Systems of Linear Equations Word Problem]]></title>
            <link>http://www.college-cram.com/study/precalculus/weblog/systems-of-linear-equations-word-problem</link>
            <guid isPermaLink="true">http://www.college-cram.com/study/precalculus/weblog/systems-of-linear-equations-word-problem</guid>
            <pubDate>Thu, 09 Aug 2007 19:44:30 GMT</pubDate>
		<dc:subject><![CDATA[Systems of Linear Equations Word Problem]]></dc:subject>
		<dc:subject><![CDATA[Systems of Linear Equations]]></dc:subject>
		<dc:subject><![CDATA[Substitution Method]]></dc:subject>
		<dc:subject><![CDATA[Math]]></dc:subject>
		<dc:subject><![CDATA[Linear Equations]]></dc:subject>
		<dc:subject><![CDATA[Word Problem]]></dc:subject>
		<dc:subject><![CDATA[Algebra]]></dc:subject>
            <description><![CDATA[<div id="question"> <p>Dear Professor Cram:</p> <p>Jerry was 1/3 as young as his grandfather 15 years ago. If the sum of their ages is 110, how old is Jerry&#39;s grandfather? What is the easiest way to calculate this? Thanks</p>  <p>Kaye B, Riverdale</p> </div>  <p>Thanks for your word-problem question, Kaye. This is a system of linear equations problem,  	which we&#39;ll solve using the Substitution Method. (Learn more about the   Systems of Linear Equations Substitution Method.) Let&#39;s review the facts first:</p> <ul><li>Jerry&#39;s age today is J</li><li>Grandpa&#39;s age today is G</li><li>Jerry&#39;s age 15 years ago is (J-15)</li><li>Grandpa&#39;s age 15 years ago is (G-15)</li></ul>  <p>In equation form, then, we can express their ages 15 years ago and today like so:</p> <ul><li>#1: J - 15 = (G-15)/3 (age 15 years ago)</li><li>#2: J + G = 110  (age today)</li></ul>  <p>Solve equation #2 for grandpa&#39;s age by subtracting J from both sides, yielding:</p> <ul>#2: G = 110 - J</ul>  <p>Now substitute this definition of G into equation #1 so we can solve for J:</p> <ul>#1: J-15 = (<strong>(110-J)</strong>-15)/3, or<br /> #1: J-15 = (95-J)/3, or<br /> #1: 3(J-15) = (95-J), or<br /> #1: 3J-45 = 95-J, or<br /> #1: 3J = 95-J+45, or<br /> #1: 3J+J = 95+45, or<br /> #1: 4J = 140, or<br /> #1: J = 35</ul>  <p>So if J=35, then that means Jerry&#39;s age today is 35 years old. Using equation #2, that means  	Grandpa&#39;s age today is 110-35, or 75 years old.</p> 	 <p>The final step in solving this problem is one that many students forget: checking your answer! Let&#39;s put our values for J and G into equations #1 and #2 and make sure they work:</p>  <ul>#1: 35-15 = (75-15)/3, or<br /> #1: 20 = (60)/3, or<br /> #1: 20 = 20 <strong>CORRECT</strong></ul> <ul>#2: 75 = 110-35, or<br /> #2: 75 = 75 <strong>CORRECT</strong></ul>  <p>The correct answer, then, is that Jerry&#39;s grandfather is 75 years old.</p>  <div id="signature"><p>No substitution for success,</p> <p>Professor Cram</p></div>]]></description>
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        <item>
            <title><![CDATA[Factoring Trinomials Cramlet Review]]></title>
            <link>http://www.college-cram.com/study/precalculus/weblog/factoring-trinomials-cramlet-review</link>
            <guid isPermaLink="true">http://www.college-cram.com/study/precalculus/weblog/factoring-trinomials-cramlet-review</guid>
            <pubDate>Tue, 25 Oct 2005 05:00:00 GMT</pubDate>
		<dc:subject><![CDATA[Algebra]]></dc:subject>
		<dc:subject><![CDATA[College Algebra]]></dc:subject>
		<dc:subject><![CDATA[Factoring]]></dc:subject>
		<dc:subject><![CDATA[Factoring Trinomials]]></dc:subject>
		<dc:subject><![CDATA[Math]]></dc:subject>
		<dc:subject><![CDATA[Trinomials]]></dc:subject>
		<dc:subject><![CDATA[Factoring Trinomials Cramlet Review]]></dc:subject>
            <description><![CDATA[<p>Every College Algebra textbook tries to explain how to factor trinomials, but they fail  	miserably. Why? Because there isn&#39;t a specific step-by-step way to do it, that&#39;s why.</p>  <p>Look at a typical algebra textbook. It gives one or two examples, and tells you what the  	answers are, but it doesn&#39;t really explain how you get them. It&#39;s like when I was a kid  	learning how to tie my shoes with the other kids. The adult told us to cross the laces,  	twist one underneath and pull, make a loop and hold it, then take the other lace and.  	do something. (This part I had a really hard time getting. Showing me once or twice  	definitely didn&#39;t help, nor did showing me shoes that were perfectly tied.) Eventually  	I got it, but that was after trying over and over again and asking if I&#39;d gotten it  	right yet.</p>  <p>The same process holds true for factoring trinomials. There are guidelines for doing  	it, but no hard-and-fast steps to follow every time. The only way to learn it properly   	is to practice repeatedly, referring back to the guidelines, until you get it.</p>  <p>The   <span class="titleCram"><strong>Factoring Trinomials</strong></span> 	 Cramlet&trade; has several features that make it effective where textbooks are not.</p> <ul><li>The <strong>See Tutorial</strong> button gives a good overview of the techniques you can use  	to factor trinomials, walking you through an example complete with typical missteps  	until you find the correct answer. This tutorial is available at any time while you&#39;re  	running the Cramlet&trade;.</li><li>The <strong>New Problem</strong> button instantly gives you another problem to solve. Unlike  	some programs that use random number generation, though, all these problems actually  	have a solution so you&#39;re not wasting your time.</li><li>This Cramlet&trade; also sports a <strong>Show Hints</strong> button, available at any time  	in the program, which displays the guidelines covered in the tutorial. These  	reminders are very helpful if you get stuck.</li><li><strong>Check my Answer</strong> takes your guesses, compares them with the correct answer, and  	tells which coefficients and signs you got right. This is really useful where you  	might pick the right numbers, but still get the wrong answer because you put the minus sign  	in the wrong place. You can keep trying and clicking this button until you get the  	problem right.</li></ul> <p>Factoring Trinomials is one of our most popular Cramlets&trade;, and countless students  	have used it to master this concept. Of course, its best feature is that, unlike a  	textbook, it is available <strong>for free</strong> to every student who signs up for  	a <a href="http://www.college-cram.com/study/_invite/register.php"  title="Free Registration">free registration</a>.</p>  <div id="signature"><p>Good luck and learn stuff,</p> <p>Professor Cram</p></div>]]></description>
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