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<channel>
	<title>S3 Maths Scribe-post Blog</title>
	<atom:link href="http://edubuzz.org/blogs/s3scribeposts/feed/" rel="self" type="application/rss+xml" />
	<link>http://edubuzz.org/blogs/s3scribeposts</link>
	<description>A learning blog for the best ever S3 maths class</description>
	<pubDate>Wed, 03 Dec 2008 16:50:34 +0000</pubDate>
	<generator>http://wordpress.org/?v=2.5.1</generator>
	<language>en</language>
			<item>
		<title>Revision Plan</title>
		<link>http://edubuzz.org/blogs/s3scribeposts/2008/02/24/revision-plan/</link>
		<comments>http://edubuzz.org/blogs/s3scribeposts/2008/02/24/revision-plan/#comments</comments>
		<pubDate>Sun, 24 Feb 2008 12:48:24 +0000</pubDate>
		<dc:creator>Mr Jones</dc:creator>
		
		<category><![CDATA[scribeposts]]></category>

		<guid isPermaLink="false">http://edubuzz.org/blogs/s3scribeposts/2008/02/24/revision-plan/</guid>
		<description><![CDATA[Here it is:
S4 Revision Calendar
and here&#8217;s a pdf to print off:
s4 revision plan
]]></description>
			<content:encoded><![CDATA[<p>Here it is:</p>
<p><a href="http://www.google.com/calendar/embed?src=85rg095rcngkvtmdkrt009k93k%40group.calendar.google.com&amp;ctz=Europe/London">S4 Revision Calendar</a></p>
<p>and here&#8217;s a pdf to print off:</p>
<p><a href="http://edubuzz.org/blogs/s3scribeposts/files/2008/02/s4-revision.pdf" title="s4 revision plan">s4 revision plan</a></p>
]]></content:encoded>
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		<item>
		<title>Find the opposite side using tangents</title>
		<link>http://edubuzz.org/blogs/s3scribeposts/2007/05/02/find-the-opposite-side-using-tangents/</link>
		<comments>http://edubuzz.org/blogs/s3scribeposts/2007/05/02/find-the-opposite-side-using-tangents/#comments</comments>
		<pubDate>Wed, 02 May 2007 17:38:41 +0000</pubDate>
		<dc:creator>nb04thoms</dc:creator>
		
		<category><![CDATA[scribeposts]]></category>

		<guid isPermaLink="false">http://edubuzz.org/blogs/s3scribeposts/2007/05/02/find-the-opposite-side-using-tangents/</guid>
		<description><![CDATA[all the pics i put on here didn&#8217;t meet security measures, so il have to explain it.
if you have a right angle triangle and the adjent-hypotinuse angle is 30 degrees
adjent line is 40 cm
the opposite is X
the working you have to show is as folows
tan 30= opp/adj
tan 30= X/40
X= 40xtan 30
= 23
for the next scribe i pick [...]]]></description>
			<content:encoded><![CDATA[<p>all the pics i put on here didn&#8217;t meet security measures, so il have to explain it.</p>
<p>if you have a right angle triangle and the adjent-hypotinuse angle is 30 degrees</p>
<p>adjent line is 40 cm</p>
<p>the opposite is X</p>
<p>the working you have to show is as folows</p>
<p>tan 30= opp/adj</p>
<p>tan 30= X/40</p>
<p>X= 40xtan 30</p>
<p>= 23</p>
<p>for the next scribe i pick conor</p>
]]></content:encoded>
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		<item>
		<title>angle language</title>
		<link>http://edubuzz.org/blogs/s3scribeposts/2007/02/21/angle-language/</link>
		<comments>http://edubuzz.org/blogs/s3scribeposts/2007/02/21/angle-language/#comments</comments>
		<pubDate>Wed, 21 Feb 2007 17:07:57 +0000</pubDate>
		<dc:creator>nb04thoms</dc:creator>
		
		<category><![CDATA[scribeposts]]></category>

		<guid isPermaLink="false">http://edubuzz.org/blogs/s3scribeposts/2007/02/21/angle-language/</guid>
		<description><![CDATA[There are different types of angle but ones I have to talk about are:
complementary angles: complementary angle are angles whose mesument adds up to be 90 degrees or a right angle, shown below:)

supplementary angles: supplementary angles are 2 angles whose sum adds up to be 180 degrees , as shown below;)
     [...]]]></description>
			<content:encoded><![CDATA[<p><font color="#ff3300">There are different types of angle but ones I have to talk about are:</font></p>
<p><u><b><font color="#ff3300">complementary angles:</font></b></u> <font color="#0000ff">complementary angle are angles whose mesument adds up to be 90 degrees or a right angle, shown below:)</font></p>
<p><img src="http://edubuzz.org/blogs/s3scribeposts/files/2007/02/angles-2.thumbnail.gif" alt="angles-2.gif" height="164" width="279" /></p>
<p><font color="#ff3300"><u><b>supplementary angles:</b></u> </font><font color="#0000ff">supplementary angles are 2 angles whose sum adds up to be 180 degrees , as shown below;)</font></p>
<p><font color="#0000ff">   <img src="http://edubuzz.org/blogs/s3scribeposts/files/2007/02/angles.thumbnail.gif" alt="angles.gif" height="182" width="299" />  </font></p>
<p><u><b><font color="#0000ff"><font color="#ff0000">vertically opposite angles</font></font><font color="#ff0000">:</font></b></u><font color="#0000ff"> vertically opposite angles are when you get an X shape and all the opposite angles are the same, i had a pictue but it wouldent go on where i wanter so ill show you another way:) </font></p>
<p><font color="#0000ff">           \  <sub><sup>*</sup></sub>  /</font></p>
<p><font color="#0000ff">             \  /</font></p>
<p><font color="#0000ff">(angle a) X (angle b)                angle a is the same ammount              /  \                             of degrees as angle b</font></p>
<p><font color="#0000ff">           / <sup><sub>**</sub></sup> \          </font></p>
<p><font color="#0000ff">         </font></p>
<p><font color="#0000ff">    angle <sup><sub>*</sub></sup> is the same as angle <sub><sup>** </sup></sub>      </font></p>
<p><font color="#0000ff">there you go hope you enjoyed my blog</font></p>
<p><font color="#0000ff">i pick josh ha ha ha&#8230;&#8230;&#8230;&#8230;</font></p>
<p><font color="#0000ff">&#8230;&#8230;&#8230;only joking, i pick blair &gt;:) suffer!</font></p>
]]></content:encoded>
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		</item>
		<item>
		<title>Distance-time graphs</title>
		<link>http://edubuzz.org/blogs/s3scribeposts/2007/02/07/distance-time-graphs/</link>
		<comments>http://edubuzz.org/blogs/s3scribeposts/2007/02/07/distance-time-graphs/#comments</comments>
		<pubDate>Wed, 07 Feb 2007 18:54:27 +0000</pubDate>
		<dc:creator>nb04turnr</dc:creator>
		
		<category><![CDATA[scribeposts]]></category>

		<guid isPermaLink="false">http://edubuzz.org/blogs/s3scribeposts/2007/02/07/distance-time-graphs/</guid>
		<description><![CDATA[Today we studied distance-time graphs. We had to find answers too questions that involved finding an average speed using the equation S= D/T (speed = distance over time)
Aswell as finding out answers from a graph.
Too find average speed you would first find the distance on a graph at a certain then you would find the [...]]]></description>
			<content:encoded><![CDATA[<p>Today we studied distance-time graphs. We had to find answers too questions that involved finding an average speed using the equation S= D/T (speed = distance over time)<br />
Aswell as finding out answers from a graph.</p>
<p>Too find average speed you would first find the distance on a graph at a certain then you would find the time at the bottom. e.g</p>
<p>Distance=15km Time=30mins or 0.5</p>
<p>S=D/T<br />
S=15(divided by)0.5<br />
S=30mph</p>
]]></content:encoded>
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		</item>
		<item>
		<title>Finding Right Angled Triangles</title>
		<link>http://edubuzz.org/blogs/s3scribeposts/2007/01/18/34/</link>
		<comments>http://edubuzz.org/blogs/s3scribeposts/2007/01/18/34/#comments</comments>
		<pubDate>Thu, 18 Jan 2007 18:04:34 +0000</pubDate>
		<dc:creator>Josh</dc:creator>
		
		<category><![CDATA[scribeposts]]></category>

		<guid isPermaLink="false">http://edubuzz.org/blogs/s3scribeposts/2007/01/18/34/</guid>
		<description><![CDATA[Many geometric shapes contain right angles and right-angled triangles.
You may find it helpful to sketch the right angled triangle being used.
Find a right angled trianngle out of a shape and use this example and theorem
xsq =5q +8sq
&#62;xsq =25+64
&#62;xsq =89
&#62;x=89sqroute =9.4 (to 1 d.p)
*the sq is squared and sqroute is canna obvious but its square route
I [...]]]></description>
			<content:encoded><![CDATA[<p>Many geometric shapes contain right angles and right-angled triangles.</p>
<p>You may find it helpful to sketch the right angled triangle being used.</p>
<p>Find a right angled trianngle out of a shape and use this example and theorem</p>
<p>xsq =5q +8sq<br />
&gt;xsq =25+64<br />
&gt;xsq =89<br />
&gt;x=89sqroute =9.4 (to 1 d.p)</p>
<p>*the sq is squared and sqroute is canna obvious but its square route</p>
<p>I PICK STUART!!!!HAHAHAHAHAHA</p>
]]></content:encoded>
			<wfw:commentRss>http://edubuzz.org/blogs/s3scribeposts/2007/01/18/34/feed/</wfw:commentRss>
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		<item>
		<title>Finding the Short Side</title>
		<link>http://edubuzz.org/blogs/s3scribeposts/2007/01/11/finding-the-short-side/</link>
		<comments>http://edubuzz.org/blogs/s3scribeposts/2007/01/11/finding-the-short-side/#comments</comments>
		<pubDate>Thu, 11 Jan 2007 19:29:19 +0000</pubDate>
		<dc:creator>Zoe</dc:creator>
		
		<category><![CDATA[scribeposts]]></category>

		<guid isPermaLink="false">http://edubuzz.org/blogs/s3scribeposts/2007/01/11/finding-the-short-side/</guid>
		<description><![CDATA[At the start of the lesson our aiims were:
 1. To become confident about using Pythagoras&#8217;s technique.
 2. To Learn how to find the missing shorter side.
ok so u all saw sorrel&#8217;s amazing post on pythagoras&#8217; theorem. i will just add a tiny bit more. Yesterday in maths mr jones taught us how to find [...]]]></description>
			<content:encoded><![CDATA[<p>At the start of the lesson our aiims were:<br />
 1. To become confident about using Pythagoras&#8217;s technique.<br />
 2. To Learn how to find the missing shorter side.<br />
ok so u all saw sorrel&#8217;s amazing post on pythagoras&#8217; theorem. i will just add a tiny bit more. Yesterday in maths mr jones taught us how to find a SHORT side of a right angled triangle.<br />
It&#8217;s pretty much the same as before except, you TAKE AWAY the squere of the non-hypotenuse side from the squere of the hypotenuse. </p>
<p>EXAMPLE</p>
<p> A ( hypotenuse) = 40cm </p>
<p>B = 20cm</p>
<p>C = x </p>
<p>so we would do this :</p>
<p>40 sq = 20 sq + x sq<br />
so x sq = 40 sq - 20 sq<br />
           = 1600  (40 times 40) - 400 (20 times 20)<br />
           =1200<br />
so x sq = 1200<br />
so x     = √1200<br />
          = 34.64 (approx)</p>
<p>Conclusion</p>
<p>So x, one of the two sides that isn&#8217;t the hipotonuse meaasures 34.64 centimetres. </p>
<p>ps I nomonate Stuart !! </p>
]]></content:encoded>
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		<title>The Theory of Pythagoras</title>
		<link>http://edubuzz.org/blogs/s3scribeposts/2007/01/09/the-theory-of-pythagoras/</link>
		<comments>http://edubuzz.org/blogs/s3scribeposts/2007/01/09/the-theory-of-pythagoras/#comments</comments>
		<pubDate>Tue, 09 Jan 2007 20:59:56 +0000</pubDate>
		<dc:creator>Sorrel</dc:creator>
		
		<category><![CDATA[scribeposts]]></category>

		<guid isPermaLink="false">http://edubuzz.org/blogs/s3scribeposts/2007/01/09/the-theory-of-pythagoras/</guid>
		<description><![CDATA[
We call C the Hypotenuse of this right angled triangle.
The Hypotenuse is allways opposite of the right angle!  This is the process to work out the length of C!
Using the length of A draw a square,(all the sides of the square are the same length). repeat this on the b and c sides of the triangle   :

Square A and B [...]]]></description>
			<content:encoded><![CDATA[<p><img src="http://www.albanyconsort.com/theorems/Theorems_files/image001.png" alt="" width="410" height="241" /></p>
<p>We call <strong>C </strong>the <strong>Hypotenuse </strong>of this right angled triangle.</p>
<p>The Hypotenuse is allways opposite of the right angle!  This is the process to work out the length of <strong>C!</strong></p>
<p>Using the length of <strong>A</strong> draw a square,(all the sides of the square are the same length). repeat this on the b and c sides of the triangle   :<br />
<img src="http://www.euclideanspace.com/maths/geometry/trig/pythagoras.gif" alt="" /></p>
<p>Square A and B if taken apart fit into C like this:</p>
<p><img src="http://home.c2i.net/greaker/comenius/9899/pythagoras/Image32.jpg" alt="" width="437" height="455" /></p>
<p>So if A was 9 and B was 7 we would know that both B squared and a squared together fit into C.</p>
<p>Therefore Pythagoras&#8217;s thearem say&#8217;s :</p>
<p>A squared + B squared = C squared</p>
<p>So to work out the length of C your sum would look like this&#8230;&#8230;&#8230;(s means squared)</p>
<p>Cs=  AS + BS</p>
<p>Cs= 9S + 7S</p>
<p>Cs= 81 + 49</p>
<p>Cs= 130</p>
<p>C= Square root of 130</p>
<p><strong>c= 11</strong></p>
<p>p,s i have forgotten how to put in hyper links so if some one good tell me i will add one in !</p>
<p>pps i choose zoe !</p>
]]></content:encoded>
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		<title>The Difference Between Direct Proportion And Inverse Proportion</title>
		<link>http://edubuzz.org/blogs/s3scribeposts/2006/12/10/the-difference-betwen-direct-proportion-and-inverse-proportion/</link>
		<comments>http://edubuzz.org/blogs/s3scribeposts/2006/12/10/the-difference-betwen-direct-proportion-and-inverse-proportion/#comments</comments>
		<pubDate>Sun, 10 Dec 2006 20:19:46 +0000</pubDate>
		<dc:creator>nb04wallb</dc:creator>
		
		<category><![CDATA[scribeposts]]></category>

		<guid isPermaLink="false">http://edubuzz.org/blogs/s3scribeposts/2006/12/10/the-difference-betwen-direct-proportion-and-inverse-proportion/</guid>
		<description><![CDATA[To tell the difference between Inverse an Direct proportion you need to make sure that you have understood the concept of them both first of all.
Direct proportion = if  you double one you have to double the other
Inverse proportion = if you double one then you halve the other
So what would this be either [...]]]></description>
			<content:encoded><![CDATA[<p>To tell the difference between Inverse an Direct proportion you need to make sure that you have understood the concept of them both first of all.</p>
<p>Direct proportion = if  you double one you have to double the other</p>
<p>Inverse proportion = if you double one then you halve the other</p>
<p>So what would this be either Inverse or Direct:</p>
<p>5 boxes of bulbs contain 40 bulbs. How many bulbs would 6 boxes contain?</p>
<p>You will see the difference if you do it right because if it is a inverse question then the answer will be lower an if it&#8217;s Direct then it will be higher than the original  price .</p>
<p>The next Person i choose to do the scribe is : Sorrel</p>
]]></content:encoded>
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		</item>
		<item>
		<title>Inverse Proportion</title>
		<link>http://edubuzz.org/blogs/s3scribeposts/2006/12/05/inverse-proportion/</link>
		<comments>http://edubuzz.org/blogs/s3scribeposts/2006/12/05/inverse-proportion/#comments</comments>
		<pubDate>Tue, 05 Dec 2006 09:11:32 +0000</pubDate>
		<dc:creator>nb04dunsa</dc:creator>
		
		<category><![CDATA[scribeposts]]></category>

		<guid isPermaLink="false">http://edubuzz.org/blogs/s3scribeposts/2006/12/05/inverse-proportion/</guid>
		<description><![CDATA[Mr Jones asked us to answer this:
It takes 6 men 10 days to build a wall
How long would it take 5 men?
What we did was made a table that looked like this
                                            Men                 Days
                                              6                      10
  He told us to find out              1                      60
what 1 man was and how         5                      12
many days it would take them
so we did.        
We worked [...]]]></description>
			<content:encoded><![CDATA[<p>Mr Jones asked us to answer this:</p>
<p>It takes 6 men 10 days to build a wall</p>
<p>How long would it take 5 men?</p>
<p>What we did was made a table that looked like this</p>
<p>                                            Men                 Days</p>
<p>                                              6                      10</p>
<p>  He told us to find out              1                      60</p>
<p>what 1 man was and how         5                      12</p>
<p>many days it would take them</p>
<p>so we did.        </p>
<p>We worked this out by doing</p>
<p>6 X 10 = 60.</p>
<p>then because we knew what one man was we could work out what 5 men would be and we did this by doing</p>
<p>60/5 = 12   </p>
<p>He told us a quote</p>
<p>&#8220;Double one, half the other&#8221;  </p>
<p>Here is one that you can try</p>
<p>TWO PEOPLE take 6 HOURS each to deliver a batch of leaflets.</p>
<p>Working at the same rate, how long would it take THREE PEOPLE?</p>
<p>Member find one person and then its easier!. </p>
<p>The next person i chose to do the scribepost is Becky</p>
]]></content:encoded>
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		<title>Loans &#38; Hire Purchase</title>
		<link>http://edubuzz.org/blogs/s3scribeposts/2006/11/30/loans-hire-purchase/</link>
		<comments>http://edubuzz.org/blogs/s3scribeposts/2006/11/30/loans-hire-purchase/#comments</comments>
		<pubDate>Thu, 30 Nov 2006 09:56:42 +0000</pubDate>
		<dc:creator>Katie</dc:creator>
		
		<category><![CDATA[scribeposts]]></category>

		<guid isPermaLink="false">http://edubuzz.org/blogs/s3scribeposts/2006/11/30/loans-hire-purchase/</guid>
		<description><![CDATA[Today in maths we did Hire purchase and Loans.
Example 1
Loans
£1200 Loan
6 Repayments = £215
Total Repayments =
How much more does the loan cost than the paying cash?
We worked this out by finding the total repayments, and you do this by multiplying 6 by £215.
So the total amount of repayments needed is £ 1290
The loan costs £90 [...]]]></description>
			<content:encoded><![CDATA[<p>Today in maths we did Hire purchase and Loans.</p>
<p>Example 1</p>
<p>Loans</p>
<p>£1200 Loan</p>
<p>6 Repayments = £215</p>
<p>Total Repayments =</p>
<p>How much more does the loan cost than the paying cash?</p>
<p>We worked this out by finding the total repayments, and you do this by multiplying 6 by £215.</p>
<p>So the total amount of repayments needed is £ 1290</p>
<p>The loan costs £90 more than paying in cash.</p>
<p>Hire Purchase</p>
<p>With hire purchase sometimes a deposit needs to be made.</p>
<p>This payment must be paid before the goods can be taken home,</p>
<p>Often this deposit is a percentage of the cash price .</p>
<p>Example 2</p>
<p>Hire purchase</p>
<p>Three piece suite</p>
<p>Cash price £1200</p>
<p>Or 10% deposit and 12 monthly instalments of £99.99</p>
<p>= 10% of £1200 = £120</p>
<p>12 x £99.99 = £1199.88</p>
<p>£1199.88 + £120 = £1319.88</p>
<p>For the next post i choose Linzi</p>
]]></content:encoded>
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