If you were given a pie what is the maximum number of pieces you can produce from 4, 5, and 10 cuts? Keep in mind, that the slices do not have to be the same size and the cuts do not necessarily have to go through the center of the pie, but the cuts do have to be straight and go all the way across the pie. Include any diagrams you used to find the solution such as an In-Out table, or any patterns you found.

Process-

The first thing I did to try to find my solution was to finish the In-Out table given, which already told us the maximum number of pieces that could be made with 1, 2 and 3 cuts. So I drew two circles, and drew in four cuts in one and five cuts in another to find the maximum number of pieces that could be produced. After several circles for each I found that the maximum number of pieces you can produce from four cuts is 11, and the maximum number of pieces for five cuts is 15. In-Out Table

In-X (Number of cuts)Out-Y (Maximum # of pieces)
12
24
37
411
516

So instead of trying to do the same thing t find out the maximum number of pieces for 10 cuts, I started looking for a pattern. I found that the difference between four and two is 2, the difference between seven and four is 3, the difference between eleven and seven is 4, and finally the difference between sixteen and eleven is 5. Based, on these results from my In-Out table I found out that one more value is added to the previous addend to come up with the next value. However, this wouldn't work that well because if I was to find the maximum number of pieces that can be produced from 50 cuts I would have to do a lot of tedious work to finally reach 50 cuts. This is because the independent variable is the number of cuts, not the maximum number of pieces. Finding the pattern was easy, but the challenging part was finding the formula. I tried finding something that had to do with the way the actual pie was cut, like how...

...in all possible pairs of two, like so: bales 1 &2, bales 1 &3, bales 1 & When weighed, they were weighed in all possible pairs of two, like so: bales 1&2, bales 1&3, bales 1&4, bales 1&5, bales 2&3, bales 2&4, bales 2&5, bales 3&4, bales 3&5, and bales 4&5. There are also 10 different weights and they are: 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91. My task is to find out how much each bale weighs. When I am done looking for solutions, I might have to look back over the problem to see if I can find some easier way to find the weights.
Process
Well I worked with 3 other people to get the answer. So we knew that the numbers had to be in in the high thirty’s to high forty’s. So we just started to try number until we got a few of the pairs to match up to the weights. Well we went through all the number s between 35 and 50 and what we did from there was just eliminate numbers that didn’t work, so that how we found the five individual weights for each bale. So we did this on the white board, but what we did first after we got a good feeling about the numbers we had left was plug them into the bales. We lined up the bales in sort of a circle shape and drew arrows with the numbers on them to see what to numbers equal what weight. As you can see we found all the number with the arrows that have to weights on them.
Solution
If you don’t understand it from the picture of our work I will explain it...

...was going to make a chart 1-100 of all the patterns and numbers.
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100
Touchdown rest Field Goals = 3,8,13,18,23,28,33
I noticed that one the digits got into the teens there was a pattern of hitting all the numbers ending with 3 and 8.
Field Goal rest Touchdowns = 5,8,11,14,17,20,23,26,29
This pattern is a little bit more tricky, but i noticed that its just like the threes it goes up three every number but instead of zero being the starting number it is five.
Two Touchdown rest Field Goals = 3,6,11,16,21,26,31
After three the pattern hits every number ending in 1 and 6
Two Field Goals rest Touchdowns = 5,10,13,16,19,22,25,28
Like the others that have rest touchdowns they go up by three this patter has a start of 10.
Three Touchdown rest Field Goals= 3,6,9,14,19,24, 29,
after the pattern reaches nine the pattern hits every number ending with 4 and 9.
Three Field Goals rest Touchdowns= 5,10,15,18,21,24,27,30,
This ends up giving you numbers that are already taken.
Four Touchdown rest Field Goals= 3,6,9,12,17,22,
This pattern after hitting the number 9 the ending numbers will hit every 2 and 7
Solution:...

...POW 8
Problem Statement-
For this POW, our task was to find the best formula for finding the area of any polygon that is formed on a geoboard. In order to do this, there are two formulas given to help you. One tells how to get the area of a polygon based on the number of pegs on the boundary. This works as an In-Out table, where In is the amount of pegs on the boundary, and Out is the area. The other formula tells how to get the area by having a polygon with exactly four pegs on the boundary. All you do is tell her how many pegs are on the inside. Using her formula, she can give you the area immediately. Using these two formulas for reference, we have to find a formula that anyone can use to find the area of any polygon by knowing the number of pegs on the boundary and interior.
Process-
The first two equations by Freddie and Sally were a preparation for the final, building up the information in order to find the "superformula".
For Freddie's equation, I drew an In-Out table, with the number of pegs on the boundary for the In/x value, and the area of the polygon for the Out/y value. With this table set up, I made some different polygons on geo-paper (attached), and plugged them into the table. I found the area for all of them by looking at the figure. However, just looking at the figure was not very accurate, so I looked for a pattern between the in and the out, and quickly found a formula that made sense, and I worked...

...T-Table, with a drawing of the figure, the number of Pegs (in), and the Area (out). I looked for a pattern between the in and the out, and quickly found one that made sense, and I worked it into a formula. I got X/2-1=Y. Where X is IN (number of pegs) and Y is OUT (Area). This works in all shapes with no interior pegs, like Freddie described. I attached this T-Table.
For Sally I followed my luck of the 3 column T-Table, and drew another with the same guidelines. The figure, the interior pegs (in), and the area (out). After I filled in a few figures, and their properties, I noticed a pattern, and not long after, a formula, which worked for them. It was X+1=Y. This T-Table is also attached.
Now...the next was not so easy. Frashy's required a long thought process, and several hours thinking it over, logically. I thought that this next equation would be a combination of the two, it would have to incorporate what I had found out from both of the above. Especially the first. So I thought to myself what this equation, or formula, would have to include. And realized there wasn't 1 variable, but 2. Because it has the variable from the first, and the second problem. 1: The number of pegs on the border, and 2: The number of pegs on the interior. So this means that there are 2 IN's. And operations on the two variables will give me my out. So then I went to the T-Table, the perfect tool for seeing patterns. This time I had 4...

...IMP2POW 3: Divisor Counting.
I. Problem statement:
This POW is all about finding information and patterns about the way divisors of certain numbers are found and expressed. In this POW when we talk about divisors we usually are counting the number of divisors that a number has. The divisor is a number that a number can be divided by, of course every number is divisible by every other number but in these problems we are only talking about whole, positive numbers. Every number is divisible evenly by one or itself so every number has at least 2 divisors. Numbers that have only 2 divisors are called prime numbers, only uneven numbers can be prime numbers, all even numbers except two have at least 3 divisors, the number itself, 1 and 2.
the task was to find information about different questions about divisors, to find patterns and to make our own question.
II. Questions:
What kinds of numbers have exactly 3 divisors?, 4 divisors? and so on.
Do bigger numbers necessarily have more divisors?
Is there a way to figure out how many divisors one million has, how many one billion has and so on without actually counting all the different ones? (in other words is there a system for counting divisors)
What is the smallest number with 20 divisors?
Choose your own interesting question: Do all the rules for divisors work the same with negative numbers?
III....

...Pow2
Problem Statement:
There’s a standard 8 x 8 checkerboard made up by 64 small squares. Each square is able to combine with others squares to make other squares of different sizes. Our job is to find out how many squares there’s in total. Once you get all the number of squares get all the number of squares and feel confident with your answer you next explain how to find the number of squares on any size checkerboard. You will know you have the answer when no matter how what size board you have you can give a clear description of to easily compute the total number of squares. So basically what you’re doing is finding the total number of squares in a 8 x 8 checkerboard and pretty much finding an equation on how to find the total number of squares on any size board.
Process:
Alright so how do you find this answer? It’s not as hard as you think it is. Although it is kind of tricky on how to get the answer. What my first concern was, “This is going to take forever. I’m never going to get done.” I honestly thought I was going to get the wrong answer. So my first method was counting all the squares. I knew this was going to take forever. Then I literally was so close to giving up until I looked at...

...1. To find my conclusions I had to think about each part of the problem. When you know that one thing means you go on to the next part. When you figure out what that means you have to see how the two statements are related. If they are related then you can deduce a conclusion that makes sense.
2. Here are my conclusions for the 6 problems on page 7.
1. a. No medicine is nice
b. Senna is a medicine
Here I deduced that Senna is not a nice medicine. I think this because the first statement says that “no medicine is nice.” That tells me that all medicines are not nice. The second statement says “Senna is a medicine”. That statement is straight forward. When you put them together you can decide that Senna is a medicine and medicines are not nice. So Senna is not nice.
2. a. All shillings are round
b. These coins are round
Here I decided that no now conclusions can be drawn. The first statement says “All shillings are round.” That statement is clear. The second statement says “These coins are round.” This tells you the coin they have are round. When you put these statements together you can see some flaws. They say these coins but you don’t know if any of these coins are shillings. They can be other coins that are round. So you cannot deduce anything.
These coins are
3. a. Some pigs are wild
b. All pigs are fat
Here I decided that there are no conclusions that can be made. The first...

...“A Sticky Gum Problem” POW 4
Problem statement:
The next scenario is very similar. In this one, Ms. Hernandez passed a different gumball machine the next day with three different colors Once again her twins each want a gumball of the same color, and each gumball is still one cent. What is the most amount of money that Ms. Hernandez would have to spend in order to get each of her daughters the same color gumball?
In the last scenario, Mr. Hodges and his triplets pass the same gumball machine that Ms. Hernandez and her twins passed in scenario two. This time, each of Mr. Hodges children wants the same color gumball out of the three-color gumball machine. What is the most amount of money that Mr. Hodges would have to spend on his triplets in order to get them each the same color gumballs?
Process:
In the process of solving the first question I drew up different color gumballs.
These are both the different colored gumballs. If this was the outcome after spending two pennies on two gumballs, then the next gumball would have to be one of the previous color gumballs that already came out of the machine.
For the next scenario it was a little trickier that the first problem. Since there were still only two children involved and there were three colored gumballs it wasn’t too hard. Once again, I drew up the three different colored gumballs in the gumball machine. The gumballs were red, white, and...