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6.05 Module Six Activity

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Ayanna .

on 25 March 2014

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Transcript of 6.05 Module Six Activity

6.05 Module Six Activity
By: Naomie and Ayanna

Working with Formulas
Question 1:
Find the perimeter of your figure. (In the example image from step 1, you would find the perimeter of the building and patio.)
Finding the perimeter of an object, or figure, is a matter of adding up all the sides together to get the total boundary amount. With that taken into account, after you’ve counted the square units of each side, one by one, all there’s left to do is add all the sides up. The equation is 14 + 10 + 5 + 8 + 2 + 12 + 9 + 20 +2 + 10 = 92. The perimeter is

Question 2:
Find the area of your figure. (In the example image from step 1, you would find the area of the building and patio.)
To find the area of the parking garage we need to separate the overall shape into the shapes it is composed of, which happens to be 3 rectangles. We must figure out their areas, and then add them all up. Here is the work:
10*14= 140
108+56+140= 304
The area of the figure is

Question 3:
Use the key of 1 square = 2 feet for your top view picture. What would be the actual perimeter and area of your building?
If 1 square on the grid is equivalent to 2 feet then the actual perimeter would be 184 feet. I found this number by taking the perimeter I calculated from before and multiplying it by 2. 92 X 2 =
Now the area would be 608. I got this by again multiplying it by 2. 304 X 2 =


Use the image below to answer the question:
Point A, located at (-8, 3), is the location of the storage closet and point C, located at (12, -4), is the director's office. Find point B, the location of the art supplies, if it is 2 over 5 the distance from point A to point C.
First, to solve this problem you want to look for the distance between the x values on the number line, so the distance between -8 and 12 on the number line; this 20. Next, you want to look for the distance between the y values, so the distance from 3 to -4 is 7. Now take those two numbers to set equations that multiply them by 2/5:
x coordinate: 2/5 X 20/1 = 40/5 = 8
y coordinate: 2/5 X 7/1 = 14/5 = 2.8
The problem asks us to solve 2/5 from point A to point C. Therefore, we must start on the coordinate of point A and do the moves mathematically from there. Hence we have A’s coordinate (-8, 12). The slope is negative, going right (add to the x coordinate) and going down (subtract from the y coordinate). (-8 + 8, 3 – 2.8). After doing these calculations we see that point B lies on the coordinate (0, .2). In other words, the location of the art supplies is at (0, .2)

Part 1: Provide a two-paragraph response to the prompt below which you will post on the discussion board. Each paragraph must contain a minimum of 4 sentences. Remember that your paragraphs should be written in a formal style that is clear and coherent.
Paragraph 2 topic: An enlarged view of the daycare patio is shown below that contains two congruent triangles, one of which has been rotated. Using this image, prove that perpendicular lines have opposite and reciprocal slopes. You must show all of your work to receive credit.
Read your partner's response to their paragraph and compose a response to the information they provided. When writing, make sure to respond and reflect on their point of view. Do you notice any incorrect statements or reasoning? Is there any evidence they provide that is overstated or inaccurate? Your response should be a minimum of one paragraph and the paragraph should contain a minimum of 4 sentences.
Naomie's Response to Ayanna's paragraph:
Response to Your Paragraph Reflection
The point of view in which the reflection is written in is very good, because it allows someone to easily follow the steps to this proof if they wanted to prove two lines parallel on their own. Reading the reflection, anyone could easily see that for two lines to be parallel they must have congruent slopes. The calculations are correct and the evidence seems to be very accurate. All statements are detailed and prevalent, having to do with the paragraph topic, and all statements are useful.
Ayanna's Response to Naomie's paragraph.
Your point of view was great. I liked how you explained when the lines would be parallel and when they would be perpendicular. It is a good reminder to the reader if he or she cannot remember how to determine whether two lines are parallel or perpendicular. You also used good transition words. I didn’t notice anything wrong with your paragraph, and I agree with your solution and work.

Paragraph 1 topic: An enlarged view of the daycare patio is shown below that contains two similar triangles. Using this image, prove that parallel lines have the same slope. You must show all of your work to receive credit.
The coordinates of the line above is (6,10) and (2,14). The coordinates for the line below is (6,2) and (0,8). To prove that these lines are parallel, you must calculate their slope. Let’s call the above line “Line A” and the line below “Line B”. First set up Line A’s coordinates in the slope formula which is M= y2-y1/x2-x1.  M=14-10/2-6. This turns into 4/-4  -1. So the slope of Line A is negative one. Now onto the slope of Line B. M=y2-y1/x2=x1  M= 8-2/0-6. The solution is 6/-6  -1. The slope of Line B is also negative one. Since both have the slope of -1, the two lines are parallel.
My Paragraph Reflection
We are given a line 1, whose endpoints are A (0, 8) and B (6, 2). We are also given a line 2, whose endpoints are A’ (4, 6) and B’ (-2, 0). To determine whether lines 1 and 2 are perpendicular to each other, we must determine the slope of each line using the slope formula; if the slopes are the same the lines are parallel, and if they are opposite reciprocals of each other the lines are perpendicular. Henceforth, the slope of line 1: m= 2-8/ 6-0 = -6/6 = -1. The slope of line 2 is: m= 0-6/ -2-4 = -6/-6 = 1. The slopes of the two lines are opposites (one has a positive slope and the other negative) and you could say that they are reciprocals of each other (-1/1 and 1/1). Because of their slope relation, lines 1 and 2 are perpendicular to each other.
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