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      <title>Nicholas Cunningham P1 by Nicholas Cunningham</title>
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      <language>en-us</language>
      <pubDate>2020-03-20 23:06:27 UTC</pubDate>
      <lastBuildDate>2020-03-21 02:46:15 UTC</lastBuildDate>
      <webMaster>hello@padlet.com</webMaster>
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      <item>
         <title>Dance Floor For My Sister&#39;s Wedding</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468576059</link>
         <description><![CDATA[<div>By: Nicholas Cunningham</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-03-20 23:09:28 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468576059</guid>
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      <item>
         <title>Introduction</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468576591</link>
         <description><![CDATA[<div>My sister is planning her wedding, and I was shocked to find out that for which she wanted me to create a rectangular dance floor. Yay! I will need to experiment with different shapes and measurements. However, I was only given 24 square tiles, each with an area of one square meter. So I am going to have to factor that into my computations, in order to accommodate her needs.</div>]]></description>
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         <pubDate>2020-03-20 23:10:31 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468576591</guid>
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      <item>
         <title>Factor 24 for Combinations</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468578407</link>
         <description><![CDATA[<div>Our first step is to find the factors of 24, more specifically whole numbers only, since we only have whole tiles to lay, in order to make layouts for the floor. 24 is a composite number, so since it can be divided by two (in half), the numerical combination of factors that contains the biggest dividend other than 24 would be, of course, two, and 12. Down from 12 and up from two, the only other whole factors are eight, six, four, and three. Exclusively together, it is eight by three, and six by four, vice versa (three by eight, and four by six). Not to mention, the most simple would be one by 24 on its own.</div>]]></description>
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         <pubDate>2020-03-20 23:14:24 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468578407</guid>
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      <item>
         <title>Table and Graph of above Data</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468583163</link>
         <description><![CDATA[<div>Now, we have to graph our gathered information, as well as put it in a table to observe any patterns we notice with the factors of the number of tiles we were given.</div>]]></description>
         <enclosure url="https://www.desmos.com/calculator/lqqdocqhg4" />
         <pubDate>2020-03-20 23:23:57 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468583163</guid>
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      <item>
         <title>Patterns from Graph and Table</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468587871</link>
         <description><![CDATA[<div>One of the most obvious patterns we see is that once we get to the to the point that divides the graph evenly in half (the line y=x), as soon as we cross it, the values of the x and y ordered pairs from the first side switch spots, as in the y- and x- values flip, because of the fact that 24 divided by either of the numbers would give you the other number. We also see this with just the bare numbers we first used (i.e. {3,8} becomes {8,3}, etcetera), which are also displayed on the table. In addition, it becomes apparent that this function is exponential, due to the curve, and the x- and y- values contain the same numbers when we pass the midpoint of the curve (approximately 4.899, since that is the square root of 24 where the two values would be the same). Finally, the graph is the same in the fourth quadrant, just with opposite values since every value is negative, and it is the same exact shape and has the same "bounds" as the initial graph, bounds being rationally but loosely defined.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-03-20 23:34:11 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468587871</guid>
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      <item>
         <title>Equation of Data</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468593167</link>
         <description><![CDATA[<div>We know that we only have 24 tiles, so we start there - f(x)=24. However, the number of tiles must be divided by a length and a width, so whatever the x-value is as a factor of 24, the y-value must be the factor you need to multiply it by in order to actually get 24 for the product. But we don't have that number yet, so we must reverse that process of multiplication needed, and the opposite of that is division, so we would use 24/ that number to get y. This leaves us with the equation of f(x)=24/x</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-03-20 23:46:20 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468593167</guid>
      </item>
      <item>
         <title>Asymptotes of Graph</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468595111</link>
         <description><![CDATA[<div>24 is a (positive) real number. This means it can be divided by any other real number, and technically the only number that is a real number, yet cannot be a divisor, is zero. We could divide 24 by a wide, in fact, an infinite array of positive numbers, and any would work and give a positive number as the answer, which would be the other factor of 24. However, when you use zero, you do not get an answer, because zero means none, as in you aren't dividing by anything, which is undefined. This means that zero is never a factor for any number except zero. Therefore, the horizontal asymptote is y=0, and the vertical asymptote is also x=0. These are the same for the negative graph as well. We also even see this data, since on the graph they only get so close to 0-values on either axis, but they never touch, only get infinitely smaller, or larger going the other way.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-03-20 23:50:58 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468595111</guid>
      </item>
      <item>
         <title>My sister has just told me something I did not know before -  she wants me to use edge pieces now as well. These will go around the final dance floor, as in we need to now involve the actual perimeter of the rectangle we create</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468610313</link>
         <description><![CDATA[]]></description>
         <pubDate>2020-03-21 00:30:18 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468610313</guid>
      </item>
      <item>
         <title>Finding the Smallest Perimeter</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468639570</link>
         <description><![CDATA[<div>The first thing we are going to do, in order to save money, is find which rectangle will produce the smallest perimeter, so we don’t have to pay as much for the edge pieces. Since the inverse rectangles have the same perimeter, we only have to count the original shapes. 24x1 has a perimeter of 50, 12x2 has a perimeter of 28, 8x3 has a perimeter of 22, and finally, 6x4 has a perimeter of 20. The rectangle with the smallest perimeter now is going to be the 6x4, while the largest will be our 24x1. The reason for this is because the smaller to medium side lengths work better, because more of the square tiles are on the inside, and the ones that are just long have more actual edge pieces, so the perimeter is bigger.</div>]]></description>
         <pubDate>2020-03-21 02:00:13 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468639570</guid>
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      <item>
         <title>Graph of Data from Perimeter</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468644961</link>
         <description><![CDATA[<div>Now, we have to graph our gathered information, as well as put it in a table to observe any patterns we notice with the factors of the number of tiles we were given.</div>]]></description>
         <enclosure url="https://www.desmos.com/calculator/1myqxeneji" />
         <pubDate>2020-03-21 02:18:41 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468644961</guid>
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      <item>
         <title>Patterns Noticed from Graph and Table</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468645765</link>
         <description><![CDATA[<div>This was a very weird graph, I have never seen anything like it. It appeared to be a quadratic function at first, but the evidence suggests it was indeed an exponential. In terms of the graphs actual appearance, it looks exactly like a parabola, just stretched to the bottom left. Another weird thing about this, is that is you zoom all the way out, it looks like a straight, linear line. This already lets us know, it vertically flattens at zero, but it never flattens going out, and this is the same for the negative function, and it is cool because we see that when you have varying lengths, it doesn’t matter which, because the other number in that ordered pair has the same output, like how the x- values eight and three both have the output of 22, because even though either of those is the length, the other one must be the width, but they also both have the same perimeter switched. Also, it appears as though the first half (left side of the y=x line) of the graph is the same as the original tile graph we made, but the other side does a completely new thing.</div>]]></description>
         <pubDate>2020-03-21 02:21:36 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468645765</guid>
      </item>
      <item>
         <title>Equation of Graph and Data</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468649964</link>
         <description><![CDATA[<div>In order to obtain our equation for this perimetric graph, we have to be able to get the same output with two different numbers because they are both factors of 24, just switched, which is, again, the same as a parabola. So if we start with our f(x)=24/x, that is going to give us our other factor in the ordered pair with whatever x was. Since this is a rectangle, there are two sides with this length, across from each other. So let’s double that value to get f(x)=2(24/x). Now, we still have to get the actual length, because the equation we used got the width, since we divided by the length, but now, the x- value itself IS the length, so we double that on its own because there are two sides of that size, which we then add to the widths, giving us the final equation of f(x)=2(24/x)+2x.</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-03-21 02:36:10 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468649964</guid>
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      <item>
         <title>Aymptotes of Our Perimeter List</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468650241</link>
         <description><![CDATA[<div>When we look at the actual graph, the vertical asymptotes remain the same, since the first halves of the graph are the same as the original. This means it vertically flattens at 0, making the vertical asymptotes x=0. Like we discussed previously, the line that goes out from the graph goes on forever, in fact, it goes back upwards, creating an upward ‘v,’ which doesn’t ever flatten, but goes in an almost straight line, traveling diagonally forever. This means, surprisingly, we have no horizontal asymptotes!</div>]]></description>
         <pubDate>2020-03-21 02:37:03 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468650241</guid>
      </item>
      <item>
         <title>Conclusions</title>
         <author>2089748</author>
         <link>https://padlet.com/2089748/q6pis7onjerd/wish/468652285</link>
         <description><![CDATA[<div>Well, I am very happy to announce I already have all of the work planned out for my sister’s wedding to make her dance floor the best it can be! I have officially decided to make the floor six meters long, by four meters wide, with a perimeter of 20 to save money, and not only that, but it also disperses the space across the floor most evenly. She is going to be very pleased with her floor!</div>]]></description>
         <enclosure url="" />
         <pubDate>2020-03-21 02:43:53 UTC</pubDate>
         <guid>https://padlet.com/2089748/q6pis7onjerd/wish/468652285</guid>
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