POW #12

Problem Statement:

Here is a lattice polygon: a polygon whose vertices lie on the points of a square lattice.

Problem Statement:

Here is a lattice polygon: a polygon whose vertices lie on the points of a square lattice.

Assuming that the points are evenly spaced apart, at intervals of 1 unit, what is the area of the polygon? There is a simple way to find the areas, however complicated the polygon is. For any lattice polygon, the area can be calculated from the number of boundary points and interior points using a formula.

Process:

The first things that we did as a class was to ask any clarifying questions that we had about the problem. We then all took to drawing twenty lattice polygons each person, along with drawing the polygons we wrote down the boundary point each drawing had along with the interior point. We also had to find the area of each one, without trying to find the formula just yet. As a class we then put together the data that we collected in groups based on how many interior points each polygon had. With those in the groups we then took the group of the polygons with the interior points of one and grafted them. By grafting them we then found the line of best fit, which then we used to find a formula that would help us find the area of a polygon with zero interior point. For which we got A=1/2B-1

After that we were given a sheet that had all of the interior points groups we were left to keep working as a class and figure out how to keep going. We all kinda started doing our own thing, I started working with Gisselle and we went through the data to see if any looked like they were wrong. After a while of kinda being stuck on what to do,because we weren't sure how to graph the data sets because of the 3 variables. We saw that Elizabeth and Karalie were doing the same thing that we had done the first time with the data set of 1 interior point. We went then to start graphing, as a class we then all had to make a graph for one of the data set of interior points. We then all came the next day and shared the formulas that we got for each of the graphed that we did. Wanting to make sure that they were right because we had gotten different formulas for the same data sets we tested out all of the formulas to find the wrong ones. We all found a polygon that could prove the formulas incorrect except for one. The formulas that was right for its data set was for two interior points; A=1/2 B +2. From there we noticed that there was a pattern and that the number of interior points you have then that was how much it made the 'jump'. For example; A=1/2 B -1 for zero and then when you go up by 2 interior points you get A=1/2 B +1 , the pattern is the add of the 2 to the -1 form there we said that the pattern meant I - 1. That mean the the number of interior points you have you subtract one. We then made sure that it worked with every polygon. We even got stuck with one of the polygons that Lilly drew because we were calculating the area wrong. We ended up working everything out right and we made sure that it worked.

Solution:

A = 1/2 B + I -1

Process:

The first things that we did as a class was to ask any clarifying questions that we had about the problem. We then all took to drawing twenty lattice polygons each person, along with drawing the polygons we wrote down the boundary point each drawing had along with the interior point. We also had to find the area of each one, without trying to find the formula just yet. As a class we then put together the data that we collected in groups based on how many interior points each polygon had. With those in the groups we then took the group of the polygons with the interior points of one and grafted them. By grafting them we then found the line of best fit, which then we used to find a formula that would help us find the area of a polygon with zero interior point. For which we got A=1/2B-1

After that we were given a sheet that had all of the interior points groups we were left to keep working as a class and figure out how to keep going. We all kinda started doing our own thing, I started working with Gisselle and we went through the data to see if any looked like they were wrong. After a while of kinda being stuck on what to do,because we weren't sure how to graph the data sets because of the 3 variables. We saw that Elizabeth and Karalie were doing the same thing that we had done the first time with the data set of 1 interior point. We went then to start graphing, as a class we then all had to make a graph for one of the data set of interior points. We then all came the next day and shared the formulas that we got for each of the graphed that we did. Wanting to make sure that they were right because we had gotten different formulas for the same data sets we tested out all of the formulas to find the wrong ones. We all found a polygon that could prove the formulas incorrect except for one. The formulas that was right for its data set was for two interior points; A=1/2 B +2. From there we noticed that there was a pattern and that the number of interior points you have then that was how much it made the 'jump'. For example; A=1/2 B -1 for zero and then when you go up by 2 interior points you get A=1/2 B +1 , the pattern is the add of the 2 to the -1 form there we said that the pattern meant I - 1. That mean the the number of interior points you have you subtract one. We then made sure that it worked with every polygon. We even got stuck with one of the polygons that Lilly drew because we were calculating the area wrong. We ended up working everything out right and we made sure that it worked.

Solution:

A = 1/2 B + I -1

Evaluation:

The problem to me was really educational because it was something that was different from what we normally do. It was challenging enough to where I knew that I could solve it but it wasn't going to be easy. The work that we had to do was something that I enjoy doing because it wasn't really riddled, more straightforward with the equations and math.

The problem to me was really educational because it was something that was different from what we normally do. It was challenging enough to where I knew that I could solve it but it wasn't going to be easy. The work that we had to do was something that I enjoy doing because it wasn't really riddled, more straightforward with the equations and math.

POW #13

The Absent- Minded Teller

Problem Statement :

An absent-minded teller switched the dollars and cents when she cashed a check for Mr. Brown, giving him dollars instead of cents and cents instead of dollars. After buying a five-cent newspaper, Brown discovered that he had exactly twice as much left as his original check. What was the amount of the check?

Process:

To start this problem I was confused I was confused with the wording of the problem not knowing what exactly it was asking me to do. After asking some clarifying questions I got started by trying to find an equation that would work in finding the right amount of money. I first thought that if you had the equation ; o - 0.05 = t O being the original amount of the check and then TO = twice the amount of the check. We then as a class got that different equations and compared all of them. We all had pretty different equations so we went back to trying to use all of the equations and make a the right one that actually worked. The next equation I got was, c=2b+(2 x cents + 5) . That equation ended up not working but then got more equations from that last few that I had gotten. I then had; D+C = D +(C - 0.05) from there we looked at it again as a class. I kept just moving number around to see if any look right or if i plugged in a number it would work. As a class we then came up with the equation of ; 2 ( 100D + C ) = (100 C + D )-5

Having the equation I started to think that maybe the answer could be $35.60 because the cents was double the amount of dollars after you subtract 5. Plugging it into the equation we then saw that it wasn’t right but it was close. Some other combinations I tried were: $35.87

$40.80

$45.90

$43.87

$31.60

$42.86

Solution :

I got that the amount of the check was $31.63. I used the equation 2(100D+C)=(100C+D)-5 . It works because after he cashed the and spent 5 cents he had twice as much as the original check. So the cents would be .63 and that is twice the amount of dollars the original check had and the dollars would be $31, so that is half of the original cents.

The Absent- Minded Teller

Problem Statement :

An absent-minded teller switched the dollars and cents when she cashed a check for Mr. Brown, giving him dollars instead of cents and cents instead of dollars. After buying a five-cent newspaper, Brown discovered that he had exactly twice as much left as his original check. What was the amount of the check?

Process:

To start this problem I was confused I was confused with the wording of the problem not knowing what exactly it was asking me to do. After asking some clarifying questions I got started by trying to find an equation that would work in finding the right amount of money. I first thought that if you had the equation ; o - 0.05 = t O being the original amount of the check and then TO = twice the amount of the check. We then as a class got that different equations and compared all of them. We all had pretty different equations so we went back to trying to use all of the equations and make a the right one that actually worked. The next equation I got was, c=2b+(2 x cents + 5) . That equation ended up not working but then got more equations from that last few that I had gotten. I then had; D+C = D +(C - 0.05) from there we looked at it again as a class. I kept just moving number around to see if any look right or if i plugged in a number it would work. As a class we then came up with the equation of ; 2 ( 100D + C ) = (100 C + D )-5

Having the equation I started to think that maybe the answer could be $35.60 because the cents was double the amount of dollars after you subtract 5. Plugging it into the equation we then saw that it wasn’t right but it was close. Some other combinations I tried were: $35.87

$40.80

$45.90

$43.87

$31.60

$42.86

Solution :

I got that the amount of the check was $31.63. I used the equation 2(100D+C)=(100C+D)-5 . It works because after he cashed the and spent 5 cents he had twice as much as the original check. So the cents would be .63 and that is twice the amount of dollars the original check had and the dollars would be $31, so that is half of the original cents.