Final Paper

The final paper should include all they have discovered in the previous week. I want all of their information here. They then should go on to answer the bus problem.

Rate (mph)	Time (hrs)
25	200
30	166.66
35	142.85
40	125
50	100
55	90.909
60	83.333

We can see this problem is not linear, exponential, or quadratic. When we graph this situation we get a graph that looks like it might be exponential decay but there is another copy of the graph in quadrant 3. This makes it not exponential. We can see that if I multiply the rate and the time the distance is always 5000 so the equation is 5000/r=t. Anytime we divide by a variable (r) we get an inverse relationship. Inverse relationships are the opposite of linear equations that go through (0,0).

Task 3 Quadratic Relationships

Length	Area of Rectangle
0	0
1	9
2	16
3	21
4	24
5	25
6	24
7	21
8	16
9	9
10	0

We can see this is not a linear relationship because the changes in areas are not constant. They increase to a certain area and then decrease by the same areas on the way back down. We can see the relationship is not exponential because the areas are not multiplied by the same number every time and they don't keep increasing or decreasing. The table goes up and then comes back down. We can also see if we take two differences those differences are constant. This is a characteristic of quadratic relationships.

The equation is Length x width. We have to write the equation for the area of the rectangle in terms of L. A= L(10-L). The graph is not a straight line that increases or decreases so it is not linear. The graph doesn't increase slowly and then increase fast, so it is not exponential. The graph is quadratic because of the u-shape or the arch. It has a maximum point and has a line of symmetry.

Task #2a-bExponential Relationships

Number of Cuts	Number of Ballots
1	2
2	4
3	8
4	16
5	32
6	64
7	128
8	256
9	512
10	1024

The relationship between number of cuts and number of ballots is not linear because the number of ballots do not increase at a constant rate. You can see the difference between 2 and 4 is 2, the difference between 4 and 8 is 4. In order for it to be linear the differences have to be the same. This is an example of an exponential relationship because as the cut number increase the amount of ballots double the amount from before. Notice, if Holly makes one cut she has 2, if she makes two cuts the ballots increase to 4 (which is double), she makes another cut the ballots double to 8, double that is 16 double that is 32. Each time the number of ballots multiplies by 2. We know the relationship is exponential so we use the equation Y=a(b)^x . Where a is the starting value and b is the growth factor. The equation is y=1(2)^x .

The graph is not linear because it does not increase at a constant rate and is curved. This is an example of an exponential growth relationship. The graph starts increasing slowly, but then increases at a faster and faster rate.

Task 2b

Number of Cuts	Area of Ballots
0	88
1	44
2	22
3	11
4	5.5
5	2.75
6	1.375
7	.6875
8	.34375
9	.171875
10	.085938

The relationship between number of cuts and number of ballots is not linear because the number of ballots do not decrease at a constant rate.You can see the difference between 88 and 44 is -44, the difference between 44 and 22 is -22. In order for it to be linear the differences have to be the same. This is an example of an exponential relationship because as the cut number increase the area of ballots decrease by 1/2 the amount from before. Notice, if Holly makes one cut she has and area of 44, if she makes two cuts the ballots decrease to 22 (which is half), she makes another cut the ballots decrease to 11, half that is 5.5 half that is 2.75. Each time the number of ballots multiplies by 1/2. We know the relationship is exponential so we use the equation Y=a(b)^x . Where a is the starting value and b is the decay factor. The equation is y=88(1/2)^x .

The graph is not linear because it does not decrease at a constant rate and is curved. This is an example of an exponential growth relationship. The graph starts decreasing fast, but then decreases slower and slower.

Task #1 Linear Relationships

Figure Number	Perimeter
1	3
2	4
3	5
4	6
5	7
6	8
7	9
8	10
9	11
10	12

This is a linear relationship because as the figure number increases the perimeter of the triangle train increases by one more. This is a characteristic of linear relationships. In order for it to be linear the change in the y-values must be a constant number. In this case the change is always one.

In order to graph this relationship I had to write the equation. I found the relationship to be linear so I used the slope-intercept form y=mx + b where m was the constant rate of change (1) and b was the y-intercept, where the whole graph started at (0,2). I came up with y=1x+2. The graph is linear because it is a straight line. You can see from one point to another the change between the y's and x's are constant.