9th grade algebra

Part 1. KEY=

y-possible

X-impossible

X(1)

X(2)

y(3)

X(4)

y(5)

y(6)

X(7)

y(8)

y(9)

y(10)

y(11)

y(12)

y(13)

y(14)

y(15)

y(16)

y(17)

y(18)

y(19)

y(20)

y(21)

y(22)

y(23)

y(24)

y(25)

y(26)

y(27)

y(28)

y(29)

y(30)

All higher scores after 7 are possible because there are limited impossible scores. There are very few scores that don’t work because these numbers can add up to just about anything there are infinite possibilities mostly because the numbers can add up to 10 and 9 which can make almost any number after 7.

Part 2.

new scoring system-

field goal-2

touchdown-1

y(1)

y(2)

y(3)

y(4)

field goal-3

touchdown-1

y(1)

X(2)

y(3)

y(4)

y(5)

There is always a highest impossible score unless the numbers are 1 and 2.

Part 3. It’s all just simple multiples of each number in any situation regarding this problem. For example 10 (5+5)+9 (3+3+3)=19

Write up:

In this POW we had to find the highest impossible score with this set of data(field goals=5pts, touchdowns=3 pts). The highest impossible score is 7 and all numbers after 7 are possible. I solved this problem by making a list of numbers and finding out which ones are possible or impossible based on if the sums of 3 or 5 can add up to that number. I tried other combinations of numbers and I found out that every combination of numbers has a highest impossible number besides the data set 1 and 2. As I stated before the highest impossible number is 7. This is because no matter what 5 and 3 will never add up to 7. They can add up too any number above that for example: 8=5+3, 9=3+3+3, 10=5+5. What I found when I tried 2 and 1 however is that there was no rational impossible number. This pow I found to be educationally worthwhile because it was a good refresher on how addition and sums work even in more advanced math. I did infact enjoy this pow (nerd allert) mostly because it was simple and I was assisted by Brendan with methods on how to figure out the problem, other than feeding me the answers.

y-possible

X-impossible

X(1)

X(2)

y(3)

X(4)

y(5)

y(6)

X(7)

y(8)

y(9)

y(10)

y(11)

y(12)

y(13)

y(14)

y(15)

y(16)

y(17)

y(18)

y(19)

y(20)

y(21)

y(22)

y(23)

y(24)

y(25)

y(26)

y(27)

y(28)

y(29)

y(30)

All higher scores after 7 are possible because there are limited impossible scores. There are very few scores that don’t work because these numbers can add up to just about anything there are infinite possibilities mostly because the numbers can add up to 10 and 9 which can make almost any number after 7.

Part 2.

new scoring system-

field goal-2

touchdown-1

y(1)

y(2)

y(3)

y(4)

field goal-3

touchdown-1

y(1)

X(2)

y(3)

y(4)

y(5)

There is always a highest impossible score unless the numbers are 1 and 2.

Part 3. It’s all just simple multiples of each number in any situation regarding this problem. For example 10 (5+5)+9 (3+3+3)=19

Write up:

In this POW we had to find the highest impossible score with this set of data(field goals=5pts, touchdowns=3 pts). The highest impossible score is 7 and all numbers after 7 are possible. I solved this problem by making a list of numbers and finding out which ones are possible or impossible based on if the sums of 3 or 5 can add up to that number. I tried other combinations of numbers and I found out that every combination of numbers has a highest impossible number besides the data set 1 and 2. As I stated before the highest impossible number is 7. This is because no matter what 5 and 3 will never add up to 7. They can add up too any number above that for example: 8=5+3, 9=3+3+3, 10=5+5. What I found when I tried 2 and 1 however is that there was no rational impossible number. This pow I found to be educationally worthwhile because it was a good refresher on how addition and sums work even in more advanced math. I did infact enjoy this pow (nerd allert) mostly because it was simple and I was assisted by Brendan with methods on how to figure out the problem, other than feeding me the answers.

1 2 3 4 - 5 6 7 8

1 2 - 3 4

1 - 2

answer= 3 weighings is the least amount of weighings that is possible to find out which bag is lighter.

write up: In this pow the solution I found is that the economical king would have to measure the gold three times. The first time he would put four bags on each side of the pan scale and then he can eliminate the heavier side. With the remaining bags he can reweigh them with two bags on each side. Again the heavier side can be eliminated leaving two bags, which will be weighed again leaving the lighter bag of them all. Therefore, the least amount of weighings is 3. In this pow I feel like it was VERY easy and the future ones should be more challenging.

1 2 - 3 4

1 - 2

answer= 3 weighings is the least amount of weighings that is possible to find out which bag is lighter.

write up: In this pow the solution I found is that the economical king would have to measure the gold three times. The first time he would put four bags on each side of the pan scale and then he can eliminate the heavier side. With the remaining bags he can reweigh them with two bags on each side. Again the heavier side can be eliminated leaving two bags, which will be weighed again leaving the lighter bag of them all. Therefore, the least amount of weighings is 3. In this pow I feel like it was VERY easy and the future ones should be more challenging.

Sareth Grunert

POW 5 Cutting the Pie

In this pow we are asked to find the maximum pieces of pie for each number of cuts (1-10) and put those numbers into a table. The other half of the problem is proving that the maximum pieces of pie for 3 cuts is 7. To solve this problem I made 5 pages of “pies” and used the appropriate number of cuts to get the most number of slices. (refer to diagrams pg. 1-5) I found that the maximum number of slices with a pie cut 3 times is 7 as originally stated in the question. I however could not find any patterns in the table. The largest number of slices with 10 cuts is 44 pieces. I found this pow challenging and fun, it took me FOREVER though.

1

2

2

4

3

7

4

11

5

15

6

19

7

25

8

38

9

36

10

44

POW 5 Cutting the Pie

In this pow we are asked to find the maximum pieces of pie for each number of cuts (1-10) and put those numbers into a table. The other half of the problem is proving that the maximum pieces of pie for 3 cuts is 7. To solve this problem I made 5 pages of “pies” and used the appropriate number of cuts to get the most number of slices. (refer to diagrams pg. 1-5) I found that the maximum number of slices with a pie cut 3 times is 7 as originally stated in the question. I however could not find any patterns in the table. The largest number of slices with 10 cuts is 44 pieces. I found this pow challenging and fun, it took me FOREVER though.

1

2

2

4

3

7

4

11

5

15

6

19

7

25

8

38

9

36

10

44