Daily Scribe Tues. April 27

Today we learned how to find an answer to an equation using coins and pouches. With these problems, you find how many coins are in a pouch. For example...

C= coins P=pouches

cccccccccc=pppcccc


So first you put
3p (3 pouches)+4 (the 4 coins on the side it was with) =10 (total coins on the other side)

Then you do
3p+4=10
-4 -4
__________
3p=6

So you subtract 4 from 4, or in any problem, the number of coins on the pouches side. Then you'd subtract 4 from 10, or 4 from the 10 coins on the other side.

You see after you do the first steps you see 3p=6

So then you'd do

3p=6
________
3 3
_______
1p= 2

So 3 divided by 3 is 1, so you have 1 pouch. Then you do 6 divided by 3 and get 2, so there's 2 coins per pouch.

This is what we learned in class.

Daily Scribe for 4/26/10

Today in class, we began to focus, among other things, upon mantaining an equivalent value between two things.
In other words, we were learning how to keep a balance.

We did a labsheet involving "gold coins" and "diplomatic pouches". The idea was that each pouch contained a certain number of gold coins. On one side, there was a certain amount of gold coins, and perhaps a pouch or two. On the other side was a diferent amonut of coins, and possibly pouches. By eliminating the same number of coins and pouches on each side, we were eventually able to come up with one pouch, and however many gold coins.


For pouches, I am going to use X's. For coins, I am going to use Y's.

************************************************************************

A) y y y y y y y y y y = x x x y y y y

10 y's = 3 x's plus 4 y's. First, you can eliminate 4 y's from each side. The new equation will be:

y y y y y y = x x x

So, 6 y's = 3 x's. This means 1 x = 2 y's.

Therefore, if x means pouches and y means coins, there are 2 coins per pouch.



B) x x x y y y = y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y

3 x's plus 3 y's = 30 y's, so you can eliminate 3 y's from each side.

x x x = y y y y y y y y y y y y y y y y y y y y y y y y y y y

3 x's = 27 y's, so 1 x = 9 y's. Therefore, there are 9 coins per pouch.



B2) x x y y y y = y y y y y y y y y y y y

2 x's and 4 y's = 12 y's. This means that 4 y's can be taken from each side.

x x = y y y y y y y y

2 x's = 8 y's. This means that 1 x = 4 y's. In other words, there are 4 coins per pouch.



B3) x x x = x x y y y y y y y y y y y y y

3 x's = 2 x's and 12 y's, so this time 2 x's can be eliminated from each side.

x = y y y y y y y y y y y y

x = 12 y's, so there are 12 coins per pouch.



B4) x x x y y y = x x y y y y y y y y y y y y

3 x's and 3 y's equals 2 x's and 12 y's. First take 2 x's away from each side.

x y y y = y y y y y y y y y y y y

1 x and 3 y's = 12 y's, so now 3 y's can be taken away from both sides as well.

x = y y y y y y y y y

x = 9 y's. Therefore, there are 9 coins per pouch.



B5) x x y y y y y y y y y y y y y y y y y y y y y = x x x x x y y y

2 x's and 21 y's = 5 x's and 3 y's. First take away 2 x's from each side.

y y y y y y y y y y y y y y y y y y y y y = x x x y y y

21 y's = 3 x's and 3 y's, so now eliminate 3 y's on each side.

y y y y y y y y y y y y y y y y y y = x x x

18 y's = 3 x's, so 6 y's = 1 x. There are 6 coins per pouch.

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That was a run-through of pretty much everything we covered in class today. Hopefully, you have now gained a clearer understanding of equivalent balances!

Signed,

Grace T :)

Daily Scribe

For the last couple of days, we have been doing Problem 4.1 and 4.2 in our Comparing and Scaling book. They are both about proportions.

Problem 4.1 shows us four ways to write a proportion to solve a problem.

A. This gets us to figure out what they did, how it works, and if it is correct.
B. We get to try some proportions and we have to explain to discover if we understand.
C. They give us proportions to solve and figure out if our theories from B are right.
D. We use proportions to solve sides of similar shapes.

Problem 4.2 is about everyday use of proportions.

A. How many miles burns how many calories?
B. How many miles in how many hours?
D. How many teaspoons for each dog?
E. How long is one side of a similar figure?

By now, we all have a pretty good understanding of proportions and know how to set it up and how to solve it.

Today in class we talked about unit rates. A unit rate is a comparison between two numbers.
For example if you could have 120 minutes of fun for $23 or 60 minutes for $11 which one would you pick? You would pick the 60 minutes for $11 because half of 120 is 60 so half of $23 is $11.50. In that case you would rather spend $11 thanhave to spend $11.50.
Another example is driving 21 miles in 30 minutes or 80 miles in 100 minutes. The answer is 21 miles in 30 minutes. If you look at it in fractions 21/30 simplifies to 7/10 and 80 miles in 100 minutes simplifies to 8/10 so the first one is the shorter drive.

In class, on February 26, we worked on problem 3.4. In the problem, you had to find unit rates. For example, you had to find out the cost per orange and how many oranges you could buy with one dollar. You also had to explain what the unit rates meant. In the problem, you had to decide which grocery store had a better buy, CannedStuff or CornerMarket.On Friday, February 26, we worked on problem 3.4.

Daily Scribe

Today in class we worked on problem 3.1 in groups of three. The problems were based on three different types of calculators; fraction calculators, scientific calculators, and graphing calculators. We made a table showing the cost of the number of calculators. After that, we found the amount of calculators a school can get depending on what type of calculator the school wanted and the budget the school had. Lastly, we made an equation for each kind of calculator to show how you would find the cost of any number of calculators ordered.

For homework it is C&S page 40 #1-3, 33 due tomorrow.
It is just like the classwork we did today.

Daily Scribe for 1-29-10

Today in class we were focusing on how advertisers decide what, where, and when to advertise. When is the right time to advertise? Do you want to target a specific audience? What channels should you advertise this probust on? These are all questions that advertisers ask themselves when they are advertising a product.

We finished problem 1.1 and started 1.2. We decided which advertisements would be the best to persuade people to buy them.

To figure out how to decide which advertisement mwould be the most effective, read the advertisement over and ovver and see if the wording is easy to understand and if the numbers are easy to figure out instantly.

To understand what the advertisement is saying, take the number(s) and make an equation or write a sentence that has easy wording.

DEFINATELY MAKE SURE THAT THE INFORMATION IS TRUE!

Then, find other ways you could advertise the information.

By:Sarah Zolondick

Daily Scribe

Today in class we did a Math Reflection Gallery Walk which is like a rough draft for the Math Reflection final. It consisted of 3 questions:

1. How can you tell if two polygons are similar?

Answer: You can tell if two polygons are similar by whether or not they have a scale factor. For example, if one polygon had dimensions of 4 and 6 and another polygon had dimensions of 8 and 12, the 2nd polygon is 2x bigger (length wise) than the 1st polygon so the scale factor would be 2.

2. If two polygons are similar, how can you find the scale factor from one polygon to the other? Describe how you find the scale factor from the smaller figure to the enlarged figure. Then, describe how you find the scale factor from the larger figure to the smaller figure.

Answer: You find the scale factor from the smaller figure to the enlarged figure by figuring how many side lengths of the smaller figure go into the enlarged figure. Like, 4 and 6 with 8 and 12, 4 goes into 8 twice and 6 goes into 12 twice so the scale factor is 2. You can find the scale factor from the enlarged figure to the small figure because it's the reciprocal of the scale factor from small to large, so the scale factor is 1/2.

3. For parts (a)-(c), what does the scale factor between two similar figures tell you about the given measurements?
a. side lengths
b. perimeters
c. areas

Answer:
a. You can use the scale factor to find the similar side lengths because the side lengths would be the scale factor times the original side lengths to get the side lengths of the bigger polygon. For example, if one side is 4, you can do 4*2 to get 8 and 8 would be the length of the similar line.

b. You can use the scale factor to find the perimeter because the perimeter would be the scale factor times the perimeter to get the bigger polygon's perimeter. If the dimensions of the smaller polygon are 4 and 6, the perimeter is 20. And the dimensions of the larger polygon are 8 and 12, so the perimeter is 40. So that means the perimeter is 2x bigger than the smaller polygon, which is the same as the scale factor.

c. You can use the scale factor to find the area because the area of the larger polygon would be the scale factor squared times the area of the smaller area. So if the area of the smaller polygon (4 by 6) was 24, the area of the larger polygon (8 by 12) would be 96 because the scale factor is 2 and the 2 squared is 4 and 4*24 is 96. The way you can check is because 8*12 also equals 96 and its the same number so it is correct.


☺That is what we did in math today!☻

Daily Scribe

Today in class we worked on scale factors and similar shapes. We made similar rectangles and triangles for problems A and B for problem 3.3. We had to make new rectangles with the information that they gave us. One of the problems was to make rectangle A to a new rectangle with the scale factor of 2.5. To do this, you would have to first multiplied to base by 2.5, and then the height by 2.5. The new rectangle should be 2.5 times larger and still be similar to the original rectangle.
Another problem on 3.3 was to to make a new rectangle have and area 9 times larger than triangle B and still be similar. To figure this out, you would have to multiply the dimensions by 3. You would do this because to find out the area of a new shape, you would have to multiply the scale factor by itself, then multiply the area from the original shape by that.


This is what we worked on in class!!!

Today in class, we learned about the scale factor inside bigger shapes like triangles. We figured out a strategy to make perfect smaller shapes. First look at each side of the shape. Find the halfway point and make a dot. Do this for every side of the shape. Now connect the dots from each point so your lines make a smaller version of the original shape. The scale factor would be a fraction, because you are dividing the shape large to small. This would've been dividing the number too, but since the scale factor doesn't work in division, the factor would be less than 1. The area of the new perimeter is the scale factor times the original area squared. Example: the original area=2 and the scale factor from the original to the new is 3. 3 squared is 9, so the new area of the shape is 18. 2•(3^2)= 18.

Scale Factors

Today in class we took some notes on scale facors. We found the shapes that were similar and then we found the difference between the similar figures. The rectangles that were similar were L, J, and N. The difference from L to J was 2, the difference of the area was 4. The difference from J to L was 1/2, the difference of the area was 1/4. The difference from L to N was 3, the difference of the area was 9. The difference from N to L was 1/3, the difference of the area was 1/9. the difference from J to N was 1 1/2 or 3/2. The difference from N to J was 2/3.
We also found the difference between triangles. The triangles that were similar were O, R, and S. The difference from O to R was 2, the difference of the area was 4. The difference from R to O was 1/2, the difference of the area was 1/4. The difference from O to S was 3, the difference of the area was 9. The difference from S to O was 1/3 the difference of the area was 1/9. Tghe difference from R to S was 1 1/2 or 3/2. The difference from S to R was 2/3.

Enlarging Images by Percents

If you want to enlarge something by a percent, the perimerter and the side length will be increased by that percent, but the area will be different. For example: if you want to increase a square of 16 sq in. by 25%, you would need to find the side length so you can find the perimeter. You can multiply that by 4 since there are 4 sides of a square. The side length would be 4 sq in. for the original figure so the perimeter would be 16 sq in. To find the perimeter of the figure enlarged be 25%, you enlarge the sides by 25%. The enlarged figure's side would be 5 sq in. so the perimeter would be 5x4=20 sq in. To find the area, you would multiply the base and the height. That would be 5x5=25 in. sq. Don't forget about Mug Wump, Zug Wump, Lug Wump, Bug Wump, and Glug Wump, the video game! Remember that on the coordinates of the game characters, "start over" means literally pick up your pencil.

Similarities

Today in class, we practiced drawing things that are similar, and we discussed what similar means. To be similar, is to have things in common. When an image is redrawn in a similar way, the first image that is referred to that is used to draw the second drawing is known as the original drawing. The second image that is similar to the original drawing is known as the image drawing. Today in class, we worked on a figure activity, and some things that were similar between the two shapes, is the general shape, the degrees of each angle, and the vertices. What was different about the two shapes, is the lines of each angle get longer or shorter, the size of each shape, the area of each shape, each length of the image is twice the length of the original, and the perimeter of each shape. This is what we did in class on December 3rd, 2009.