## Meadows and Malls

In this unit, we were tasked with finding the cheapest solution to a set of constraints that were for land use. The people of a town were trying to find out how much of various land would go to development and recreation. The first task was making all of the constraints quantifiable and fins the maximum for each. These constraints ended up resulting in several systems of equations, so we learned about matrices in order to easily solve the constraints and then implement them in a cost equation. Using logic we were able to also see which constraints were possible based on needed constraints and the cost equation. Once we eliminated impossible constraints, we used matrices to find the final cost of certain combinations of constraints in order to see which solution would be the cheapest.

The first two pieces are showing how we solved the unit problem. The first one is where we were eliminating constraints that weren’t possible so that we could then implement constraints that could work into a matrice to see if it could give a cost. While this helped get rid of some constraints, there was still a lot left over that we had to check by solving the problem with them. The second piece shows how we used the possible constraints to figure out the cost of the projects. For instance, we would plug in the constraints into a matrix and the use it’s inverse to multiply the constraints by the costs and then we could find to the total cost.

The third piece is a worksheet on solving systems of equations it was pretty easy because I used elimination to solve the system. Elimination is where you make the coefficient of one of the variables the same and then add or subtract the systems to get rid of the variable. Then, you can use two-step algebra to find the answer to one of the variables and plug it into an equation to find the other.

The fourth piece is an assignment on solving systems of equations with more than two variables. In order to solve these systems, I used matrices so that I didn’t have to solve the equation in a long tedious way.

The fifth piece is showing a worksheet on adding, subtracting, and multiplying matrices. We did this all by hand and was really difficult compared to using a calculator.

The sixth piece is of a problem I found difficult. I was a simplified version of the unit problem using cookies. I wasn’t really able to figure it out and I got frustrated a lot because it didn’t make sense. Looking back I feel like I would be able to better understand it and it’s a good reminder of the growth I had in this unit.

The seventh and final piece is the POW #5. I really liked this POW because we were trying to solve addition and subtraction problems by replacing letters with numbers. This was really challenging and fun because it was a lot of trial and error as well as logic.

The first two pieces are showing how we solved the unit problem. The first one is where we were eliminating constraints that weren’t possible so that we could then implement constraints that could work into a matrice to see if it could give a cost. While this helped get rid of some constraints, there was still a lot left over that we had to check by solving the problem with them. The second piece shows how we used the possible constraints to figure out the cost of the projects. For instance, we would plug in the constraints into a matrix and the use it’s inverse to multiply the constraints by the costs and then we could find to the total cost.

The third piece is a worksheet on solving systems of equations it was pretty easy because I used elimination to solve the system. Elimination is where you make the coefficient of one of the variables the same and then add or subtract the systems to get rid of the variable. Then, you can use two-step algebra to find the answer to one of the variables and plug it into an equation to find the other.

The fourth piece is an assignment on solving systems of equations with more than two variables. In order to solve these systems, I used matrices so that I didn’t have to solve the equation in a long tedious way.

The fifth piece is showing a worksheet on adding, subtracting, and multiplying matrices. We did this all by hand and was really difficult compared to using a calculator.

The sixth piece is of a problem I found difficult. I was a simplified version of the unit problem using cookies. I wasn’t really able to figure it out and I got frustrated a lot because it didn’t make sense. Looking back I feel like I would be able to better understand it and it’s a good reminder of the growth I had in this unit.

The seventh and final piece is the POW #5. I really liked this POW because we were trying to solve addition and subtraction problems by replacing letters with numbers. This was really challenging and fun because it was a lot of trial and error as well as logic.

## POW 5

Process/Solution:

For number one there is a three digit number that is subtracted by a one digit number to get a two digit number (ABB-A=DD). Also, the two letters in the three digits are the same so it’s also the same number. The lowest possible three digit number matches the letter combination is 100. This also correlation with the first letter of the three digit number and the letter that is the answer being the same. If 100 is the three digit number then the two digit number must be 99. That means that the answer to the problem is 100-1=99.

For number two there is two two-digit number added together to equal a three digit number (SS+EE=SST). First, we need to figure out what the S equals. The highest a two digit number added together can go is 198. Also, the three-digit number is in the hundreds. S must be one in order to fit that constraint. This means that the three digit number is most likely to be 110 and the first two digit number is 11. So if 11 is subtracted from 110 the number is 99. This corresponds with the second number being two of the same number. This means the answer is 11+99=110.

For number three it’s again two two-digit numbers added together to get a three digit number. However, this one is different (AB+BC=ADE) there are two letters that are the same in the different sequences. The first number in the three digit number must be 1 (see explanation in the previous solution). This means the first two digit number has a 1 as well. Also, the two two-digit numbers have the same number for the first and second number in the sequence. B cannot equal 9 because it makes the three digit number go into the 110s. 7 also doesn’t work because it ends up being too low. Therefore B must equal 8. So AB is 18 and BC is in the ’80s. There are five answers to this problem. I found that the BC Cannot be more than 89 or less than 84. Also, it couldn’t be 88. So the answers are 18+84=102, 18+85=103, 18+86=104, 18+87=105, and 18=89=107.

For number four there are two three-digit numbers added together to equal a four-digit number (BOO+FUN=FEST). I found two major constraints. First B must be 9 in order to reach a four-digit number. Second F must be 1 in order to get the lowest four digit number. I did a lot of guesses and check for the answer. I found that the only one that worked was 955+132=1087.

For number five I did AEE-DA=D. Like in the first question A must equal 1. So then the Number is in the hundreds. The highest number that can have a single digit number added to it to get a three digit number is 91. So D must equal 9. This means that the answer is 100-91=9.

Evaluation:

This problem was pretty easy. It only took me a few minutes to solve each one. Despite it being easy, I had a lot of fun solving the problem. The hardest question was number five. As it turns out breaking codes is easier than making them.

For number one there is a three digit number that is subtracted by a one digit number to get a two digit number (ABB-A=DD). Also, the two letters in the three digits are the same so it’s also the same number. The lowest possible three digit number matches the letter combination is 100. This also correlation with the first letter of the three digit number and the letter that is the answer being the same. If 100 is the three digit number then the two digit number must be 99. That means that the answer to the problem is 100-1=99.

For number two there is two two-digit number added together to equal a three digit number (SS+EE=SST). First, we need to figure out what the S equals. The highest a two digit number added together can go is 198. Also, the three-digit number is in the hundreds. S must be one in order to fit that constraint. This means that the three digit number is most likely to be 110 and the first two digit number is 11. So if 11 is subtracted from 110 the number is 99. This corresponds with the second number being two of the same number. This means the answer is 11+99=110.

For number three it’s again two two-digit numbers added together to get a three digit number. However, this one is different (AB+BC=ADE) there are two letters that are the same in the different sequences. The first number in the three digit number must be 1 (see explanation in the previous solution). This means the first two digit number has a 1 as well. Also, the two two-digit numbers have the same number for the first and second number in the sequence. B cannot equal 9 because it makes the three digit number go into the 110s. 7 also doesn’t work because it ends up being too low. Therefore B must equal 8. So AB is 18 and BC is in the ’80s. There are five answers to this problem. I found that the BC Cannot be more than 89 or less than 84. Also, it couldn’t be 88. So the answers are 18+84=102, 18+85=103, 18+86=104, 18+87=105, and 18=89=107.

For number four there are two three-digit numbers added together to equal a four-digit number (BOO+FUN=FEST). I found two major constraints. First B must be 9 in order to reach a four-digit number. Second F must be 1 in order to get the lowest four digit number. I did a lot of guesses and check for the answer. I found that the only one that worked was 955+132=1087.

For number five I did AEE-DA=D. Like in the first question A must equal 1. So then the Number is in the hundreds. The highest number that can have a single digit number added to it to get a three digit number is 91. So D must equal 9. This means that the answer is 100-91=9.

Evaluation:

This problem was pretty easy. It only took me a few minutes to solve each one. Despite it being easy, I had a lot of fun solving the problem. The hardest question was number five. As it turns out breaking codes is easier than making them.

## Orchard Problem

In the Orchard Hideout Unit, the central problem was to figure out where the last line of sight was in an orchard with a radius of 50 trees. Also, to find out how much time it will take for the tree to grow and finally block that line of sight. To figure out how to answer these questions the formulas for the area and circumference of a circle. Also, how to reverse the formulas to find the radius is the area of circumference is already given. This is important because knowing these formulas can help find how big the tree needs to be to hit the line of sight. These formulas are used at the end once the distance between the line of sight and tree is known. Another thing that needs to be known is the formulas for the distance between points and the midpoint between two points. These are important because it can give insight into finding lengths of lines to create similar triangles and eventually find the distance from the tree to the line of sight. Once the distance is found then the formulas for area and circumference can be used as well as their reverse versions to find the radius. It’s also important to know how to recognize similar triangles and use proportions to find the missing side length of a triangle. Finding the missing side length will help find the distance between the tree and the last line of sight.

The pieces in this portfolio are p. 119, p.114, and POW 3. Page 119 is in there because it is where the answer to the unit problem is as well as the work to get to the answer. Page 114 is in there because it shows a small scale version of the last line of sight and how to solve it. This helped because the methods for solving p.114 were used to solve the unit problem. POW 3 is in there mainly because it was the only POW I properly solved and take the most pride in. It isn’t connected in methods for solving the unit problem, but it’s more recognizing patterns and making formulas.

The pieces in this portfolio are p. 119, p.114, and POW 3. Page 119 is in there because it is where the answer to the unit problem is as well as the work to get to the answer. Page 114 is in there because it shows a small scale version of the last line of sight and how to solve it. This helped because the methods for solving p.114 were used to solve the unit problem. POW 3 is in there mainly because it was the only POW I properly solved and take the most pride in. It isn’t connected in methods for solving the unit problem, but it’s more recognizing patterns and making formulas.