Elaine's Scribe Post

Hey everybody! In class, we went over what was due on Monday. They were:

-the exponent foldable 1 (this foldable is the one Mr. Backe handed to us. It had the list of the different laws. Here, we write an explanation of the laws, give examples of it, and on the back you write three BEDMAS questions and answer it.)

-the exponent foldable 2 (this is the foldable that we folded into thirds and then into fifths. Here, we write the definition, examples, and not examples of POWER, EXPONENT, BASE, COEFFICIENT, and EXPONENTIAL FORM.)

- the self -assessment of chapter 3

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Then, Mr. Backe wrote questions on the board for us to answer.They were:

QUESTION 1 & 2

In math class, we were all assigned a question to do on the blog. I got two questions, 1 and 2, which are on the history link on page 71.

First off, this history link is about UNIT FRACTIONS. A unit fraction, is a fraction that always has a numerator of 1. If you are trying to show a fraction that has a numerator that is larger than 1, you have to express it by adding separate unit fractions.

Questions:

1. Express each of the following as the sum of two unit fractions.

2. Describe any strategies that helped you to complete #1.

Elaine's Scribe Post for November 3, 2010

Hey everyone!!! Well in class, we were all assigned questions to do on the blog and I got 14 & 15

14. Identify a decimal number between each of the following pairs of rationa

l numbers.

- to answer these questions, first, I converted each p

air into decimal form and t

hen found a decimal number that is in between the two numbers.

Here is how I answered 14. a):

These are the answers that I got from the other questions :

And remeber! To get the decimals in between the pair of numbers, make sure that the decimal place value isn't greater than

the largest decimal of the pair or less that the smallest decimal of the pair.

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15. What is a decimal number between each of the following pairs of rational numbers?

- to get to the answer, it is basically, the same steps as question 14, but an extra step is added, you have to convert the mixed numbers into improper fractions, then into decimals.

Here are the steps to how I got my answer for a):

All my other answers are:

Thanks for reading my scribe!!!! Sorry if the pictures are too small, and tell me in the comment box if I made any mistakes :)

Elaine's Scribepost for October 5, 2010

Hi you guys! We each were assigned a question to do for the blog and the question I got was the one with the picture of the house.

The question was :

The surface area of walls, roofs, doors, and windows all affect the amount of heat loss from a house. Calculate the surface area for the doors,windows, walls, and roof for the house below. Express each answer to the nearest tenth of a square metre.

wi

ndo

ws = 80cm x 60cm each

doors = 205cm x 82

cm each

-back of the house is identical to the front

-sides of the house are identical to each other

I started off with the windows. Since there are 3 window in the front and 2 window on the side, and the opposite sides are equal there would be 10 windows all together [(3+2)(2)]

Here is how I got the area of the windows:

First, I converted the centimetres into metres by dividing it by 100

80/100= 0.8m

60/100= 0.6m

SA = 10(lw)

= 10(0.8 x 0.6)

= 10 (0.48)

= 4.8 squared m

Now, I will calculate the area of the doors. Since, the back and front of the house are both identical, there would be 2 doors :

I also had to convert the cm in m:

SA = 2 (lw)

= 2 (2.05 x 0.82)

= 2 ( 1.68)

= 3.36 squared m

For me, finding the area of the walls got a little tricky. What I did was find the sides of the walls, then subtracted the area of the doors and windows:

SA = 2 (lw) = 2 (lw)

= 2 ( 4.8 x 2.5) + (2.5 x 7)

= 2 (12) + 2 (17.5)

= 24 + 35

= 59 squared m

- after adding everything, together, I had to subtract the area taken up by the windows and doors.

-the area of the doors was 3.36 squared m

- but for the windows, only 8 are on the walls so I had to find the surface area of 8 windows:

SA = 8 (0.8 x 0.6)

= 8 (0.48)

= 3.84 squared m

SA = 59 - 3.36 - 3.84

= 51.8 squared m

Lastly, I have to find the area of the roof I did that using two different steps, the rectangles, then the triangles:

SA = 2(lw)

= 2 (7 x 4)

= 2 (28)

= 56 squared m

now I find the area of the triangles:

-SA =2 (b x h/ 2)

- but since the formula is multiplied and divided by two, they both cancel out

-SA = b x h

SA = b x h

= 4.8 x 3.2

= 15.36 squared m

now I add the two:

SA = 56 + 15.36

= 71.36 squared m

-but after adding the two, I have to subtract the area of the two windows

-SA= 2 (0.8 x 0.6)

= 2 (0.48)

= 0.96

SA = 71.36 - 0.96

= 70.04 squared m

After all the work that's done, the answers are:

Windows = 4.8 squared metres

Doors = 3.36 squared metres

Walls = 51.8 squared metres

Roof = 70.04 squared metres

That is how I got all my answers. Sorry if it's kind of confusing and THANKS FOR READING MY SCRIBE