When I walked into the classroom on Tuesday there was immediately a problem on the board that we all had to solve independently. The problem was like so... If there is a dartboard with three sections labeled with the numbers 1, 5, and 10, how many different sums can you obtain if you hit the board using three darts.
The first way that I tried to tackle this problem was to make a list of all the possible outcomes starting with the number one.
After I did this, I moved onto the number 5, and then onto the number 10.
In this list I had a section labeled what the sum of the numbers added together was, to make sure that I had no repeats.
The list ended up looking like so....
Then after we all figured out the problem using our own methods, the class discussed the importance of using clear tables when solving problems.
The next problem that we did in class was a word problem. The word problem had five different types of cars, all finishing a race at different times. We were given clues on how to figure out the order in which the cars finished the race.
In this problem I tried to solve it in my head and was thinking to myself that there had to be a better way to solve this problem. After I talked to the girls that were sitting at the table with me, I saw the best way to solve this problem. Using a number line...
Using a number line is a great way to solve problems that have a time factor in them, because they can also act as a time line, using the tick marks to indicate seconds.
The last problem that my class did on Tuesday was one that I was more familiar with, It was an algebra problem.
There were four hexagons attached together side to side, and we needed to figure out a way to calculate the perimeter of the hexagons, without counting the sides.
The way that I solved the problem was by counting the sides of the beginning and ending hexagons, since there were five sides showing on both of them, and two end pieces, I mulpiplied 5x2. Then I counted the middle pieces, and how many sides they had showing, and multplied 2x4.
Then I added together (5x2)+(4x2) and got an answer of 18.
Then we talked as a class, and I herd of other ways that people solved this problem. Some people started out with knowing that each hexagon had to have at least four sides, and then added the sides necessary from there. Like so....
There were four total hexagons so they started out with 4x4. Then they knew that the two end hexagons had one extra side, and added two to their original total.
(4x4) + 2 = 18
And then someone else did the same process, but started out with six sides on each hexagon and subtracted the necessary sides. The way that they figured out how many to subtract was by how many sides the middle hexagons shared, then multiplied that number by two, because there were two sides being shared.
(6x4) - (3x2) = 18
Thursday, January 17th
On Thursday in class we started to learn about sets, and we started off with Venn Diagrams.
We learned some Venn Diagram lingo first to better understand how to group items together.
A u B = (A union B) is the set of everything in set A or B, or Both.
A n B = (A intersect B) is the set of what set A and B both share, what they have in common.
The term Complement means opposite, and is expressed by a line over the set.
From this information we broke into small groups and began to play a game using different sizes and shapes.There were two teams, and a Venn Diagram. We had to pick a card to express the set that was going to be placed in each circle of the diagram.

