## Thursday, December 16, 2010

### 4.4 Question 14

LauraKathleen 9-05

" Eliza is building a model of the canvas tent her family uses in Behchoko, NWT. The model will have a peak height of 12 cm. The actual tent floor measures 2.4 m by 3 m. The walls are 1.5 m high and the peak height is 2.4 m.

a) What scale factor will Eliza need to use for her model?

b) The front of the tent is a pentagon. Calculate the dimensions of this polygon on the model.

c) Calculate the other dimensions of the tent model. "

It's always smart to highlight or underline the key points in a question :)

a) To find the scale factor, divide the actual by the reduction. Here's a picture to help:

Step 1: Convert

2.4 m X 100 = 240 cm.

240 cm/ 12 cm = 20 cm

The scale factor for this problem is 20 cm.

b) It gives the bottom length of the pentagon, so naturally, it's the easiest to calculate.

Solution:

3 m/ 100 = 300 cm

300 cm/20 = 15 cm (Divide by 20 because that is the scale factor)

The rest of the sides of the pentagon are the same length, so if you can figure out one, you have them all! For this one, use the Pythagorean Theory.
(a squared + b squared = c squared)

a squared + b squared = c squared

1.5 X 1.5 + 0.9 X 0.9 = c squared

2.25 + 0.81 = c squared

3.06 = c

square root of 3.06 = 1.75 (approx.)

1.75/ 20 = 0.0875 (Divide by 20 because 20 is the scale factor)

0.0875 X 100 = 8.75 cm

Sorry if this is wrong! Please correct me! THANKS :D

Brianna

MERRY CHRISTMAS

## Tuesday, December 14, 2010

### Mary Jane's Scribepost for December 14, 2010

In today's math class, we started off with playing math wars in which the face cards were equal to 17 and the operation was addition. After that, we then talked about regular polygons.

Regular Polygon: when a shape has equal interior angle and equal side lengths, it is called a regular polygon.

Mr. Backe, then drew a picture that was identified as a five sided regular polygon. The diagram below shows that the two polygons are similar (~). The ratio for each side of the pentagon will have a ratio of 4:8 or 8:4 if the measurements were 4 and 8.

The next shape we were worked with was similar to the shapes in the diagram above although it is connected and some angles are not the same as the one above. The shape looked something like this:

First we put the down the ratios of the corresponding angles and sides which will eventually help solve our problem.

There were two methods we did to find the missing side lengths. The first method was to divide the only corresponding sides that have measurements for both small and big polygon and use that as a scale factor. In this case the ratio is 2/5 and when you divide 2 and 5 you get 0.4 which will then be your scale factor. You then use that scale factor to find the other missing side lengths by multiplying their corresponding side by 0.4. This goes for all of them.

The second method we did was to use a proportional expression to solve the missing side lengths. We used the variable x to represent the missing side lenghts. All the work is shown in the diagram above.

HOMEWORK:
• use the provided diagram above to find the perimeter of the small polygon.
• 4.4 Practise: 3,5,6
• 4.4 Apply: All
• 4.4 Extend: 13-17

*NOTE: click the picture for a larger view.

## Monday, December 13, 2010

### Connor's Scribepost for December 13, 2010

Today in class we learned more on corresponding angles and how to find the length of sides.

First we corrected the tests.

Then we moved onto corresponding angles.

The corresponding angles are the sides that overlap when the triangle is flipped over the line of reflection.

The two sides that are marked blue are corresponding and the two sides marked yellow are corresponding.
After we reviewed the what corresponding angles are we went onto a quetion involving it.

The question was to find the value of X on the larger triangle.

First you must write the corresponding angles formula.

Next convert the two angle fractions with the measurements given in the question. (Image Below)

Next you must cross multiply.
4.7 (x) = 4.7x
13.7 (7) = 94.5

The answer to this question is x = 20.11

Next we did another question involving overlapping triangles.

The question was asking you to find the height of the ramp if it gives you the height of a support beam.
We need to convert the (cm) to (m)
50cm = 0.5m
Again we must find the side formulas.
(CD/AB) = CE/AE = (DE/BE)
Take the first and third side ^ because that is the measurements that is gives us.
50/x = 175/85
then convert to metres.
0.5/x = 1.75/(1.75+0.85)
Then we cross multiply.
1.75x = 0.5(2.6)
1.75x/1.75= 1.3/1.75
x= o.74m
Then we convert it into cm again which is 74.29cm.
The height of the ramp is 74.29cm.
Sorry if you cant understand some of the math written is hard to understand. I tried the best i could in the time I could. Comment if you have any questions.
The next person for the scribe is Maryjhane

## Sunday, December 12, 2010

### Elijah's Scribepost for December 10 2010

In class today, we classified triangles by its sides, and angles. We also learned more about SAS (side, angle, side).
Classifying Triangles by side:
• Scalene: All Sides are different
• Isosceles: 2 sides are the same.
• Equilateral: 3 sides are equal.

Classifying Triangles by Angle
• Right angle: One angle equals 90°
• Obtuse angle: One angle's greater than 90°
• Acute angle: All angle's are less than 90°
• Equiangular: All angles equal 60°

Other Tringle facts

• All triangles have an interior measure that sums to 180°
• Right triangle:

• Equiangular :

• Obtuse :
• Acute :

SAS (side, angle, side):
If sides and angles are proportional the triangles are similar.

Then we got to solve a problem: Find the b and the h.

Solve:
The next problem is based on a right triangle. Find the a b and c
Solve:

Homework :
Extend 18 and 20
Homework book/ Extra practise

## Thursday, December 9, 2010

### Princess' scribepost for December 9, 2010

HELLO(:

So, today in class we learned a little about Similar Triangles. We learned how to determine if two triangles are similar or not. Also, we learned how to use similar triangles to determine a missing side length.

We learned a few terms:
"" means "is congruent to"
Congruent- same shape, same size

"~" means "similar to"
Similar- proportional in size

Corresponding Angles/Sides- have the same relative position in a figure

Then we learned that two triangles are similar if:
• two pairs of corresponding angles are congruent (therefore the third pair of corresponding angles are also congruent) WHY? Because the angles of each triangle are equal to 180°.
• the three pairs of corresponding sides are proportional.
After that, we looked at a pair of triangles and figured out if they were similar or not.
*notice that the vertices are always going to be capitalized*

To find out if they were similar or not, we found out if the side lengths were proportional.

Since they are all proportional, the pair of triangles are similar.

Then, we looked at another type or triangle called an Isosceles Triangle.

With this triangle we learned how to find missing side lengths.
For example:
a= 3
e= 6
f= 10
h= ?

We did a few more:
d= 7
h= 5
a= 4
f= ?

A to D= 6
e= 9
f= 12
c= ?
a= ?
h= ?

HOMEWORK:

• Practice - odds or even
• Apply - odds or even

## Monday, December 6, 2010

### Allysa's Scribepost for December 6, 2010

In class today, we talked about Scale:Proportional reasoning and we learned how to find a variable's number using "Cross-Multiplication".

A scale is a image:actual.

This is what we started with. I crossed out the A over A because it's not a valid operation and we may not do that.
Also, I put the ticks in the corner of the letter's because they're primes.

Now for the example, it's just like converting a decimal into a fraction. Except your replacing the variable x, into the answer.

Our math teacher also gave us a few questions to practice with. Here are the two:

<-- That is just an easier way to show it.

- Show You Know(Both)
- Practice(odds or evens)
- Apply(13-16 OR 15-19)
- Extend(20 or 22 and 21)

Some of the questions may show this:
3:5
____
^^You are supposed to measure that line.

And most importantly, there's a TEST ON WEDNESDAY on Scale and Scale Factors.

I choose PRINCESS to do the scribe next!

## Thursday, December 2, 2010

### Mary Jane's Scribepost for December 2, 2010

Hello Everyone! For today's math class, we first had to double the so called "carlos" diagram using tangram shapes. We then went on with talking about dilation and reduction.

Dilation: an enlargement of an object, shape or design. (my own definition, feel free to correct me)

For example:

We also talked about scale factors and scales (ratios).

Scale Factor: The constant factor by which all dimensions of an object are enlarged or reduced in a scale drawing.

Diagram above: It shows the enlargement and the reduction of the square. When the square is enlarged, the area will increase from 36 units(squared) to 144 units(squared) and the side length will increase from 6 units to 12 units. In this case you are what is called doubling (multiplying by 2) and the scale factor is 2. The scale for the enlargement will be 1:2. When the square is reduced, you are basically multiplying the bigger square by 1/2 or 0.5. In this case you are halving and the scale factor is 1/2 or 0.5. The scale for the reduction will be 2:1.

Don't forget to do:

Chapter 4.1 (Page 130-135)
Show You Know (page 132 & 134)
Practice: Questions 4,6,7
Apply: Questions 9,11,12

Extend: Questions 14 & 17

## Sunday, November 28, 2010

### 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

-------------------------------------------------------------------------------------------------

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

## Tuesday, November 23, 2010

### Camille's Scribepost

Today in math class we had to find the answer to say for example 8²= 64 the answer is 4 because we look for the number that is in the ones place and also find the pattern. We also had to figure out some math problems.

examples:

Circle radius is 4 cm. How much of the square is not covered by the circle?
Formula: (2r)² -Ï€ r²
(2r)² -Ï€ r²
(2•4)² -Ï€ 4²
8²-Ï€ 16
64-16.Ï€
= 13.73 cm²
First do the bracket. r= 4. So 2 multiplied by 4 equals 8². then 4² = 16. 8² = 64. So 64 subtract by 16 multiplied by Ï€. equals 13.73 cmÏ€

Side is 12 cm. How much of the square is not covered by the circle?
Formula: s² - Ï€ ( s/2) ²
12²-Ï€ (12/2)²
12²-Ï€ 6²
144-Ï€ 36
= 30.90 cm²

S²=12². Then you do the brackets, so 12 divided by 2 all squared equals 6². 12² equals 144 and 6² equals 36. So 144- Ï€ x 36
We also learned what a coefficient is. It is the number you multpiy by. Like 4x, 4 is the coefficient and x is variable.
Oh remember to comment If I made any mistakes! Thank you!
by the way sorry for the layout. blogger isnt cooperating with me.

## Sunday, November 21, 2010

### Jem's scribepost for november 19 2010

Oh Harro guys ! How was your weekend? Good okay. Anyways. . . As of last Wednesday to Friday we've been learning about "EXPONENTS". At the moment we should already know or have a better understanding on:
1. the different exponent laws
2. how to identify different exponents
3. how to solve different exponent type questions
REMINDER: The pictures aren't in their best qualities so just click on the images to get a better view of the work I'm showing you guys! Thanks!
Our lesson on Friday focused on the three other laws that we didn't cover on Thursday so stay tuned because here i am to tell you what has been going on, on Thursday. IF YOU WEREN'T IN CLASS THEN YOU BETTER READ THIS! or else. . . i don't know. I better get this started so here we go.

Power of a Product Law-
When a product is raised to a power, you can rewrite the product by applying the exponent to each factor. Which means you use the same power for both numbers.
(a x b)2 = a2 x b2

Power of a Quotient Law-
When a quotient is raised to an exponent, you can rewrite each number in the quotient with the same exponent. Which means rewrite the the same quotient as many times as the power tells you too.
(a÷b)³ = a³÷b³
(a/b) ³ = a³/b³

Negative Exponent Law-
One is the numerator and the denominator is the positive exponent with the same base. (*Honestly I'm still kind of confused about how this law works so, if i do something wrong just tell me. And Mr.backe I need help!)
a¯¹ = 1/a¹
a-² = 1/a²

Okay, guys. . Thanks for spending your time to look at my blog post. Feel free to comment on how well/poorly my post was. Don't be afraid to make some suggestions on how I can make this blog more better. *DON'T FORGET TO DO YOUR HOMEWORK OR BACKE WILL BLOW IT :)*
• Stash it :All tests signed
Self Evaluation
Problem of The Week 3 and 4
Foldable (divide/add/subtract/ multiply fractions and decimals), Different groups of Numbers
• Textbook and homework book questions
• Study all the time
• Comment at blogs
Enjoy the 4 hours you have left of your week end. Okay okay ? Okay cool. Bye guys :)
CAMILLE CONCEPCION, your up next !

## Thursday, November 18, 2010

### Seppe's scribe post for November 18th

Today in class, we moved to the next chapter. Mr. Backe taught us power and exponents.

A power is a quick way of writing repeated same number multiplication.

Base is the number used as a factor for repeated multiplication.
Exponent is the number of times you multiply the base in a power by itself.

22 = two squared = (lh) = 2x2
23 = two cubed = (lhw) = 2x2x2

You could also say -two to the exponent of 2 or 3
-two to the second/third power

Ordinality describes position. eg. 1st, 2nd, 3rd.....

Here are some examples of how you do it....

24= 2x2x2x2 = 16
23= 2x2x2 = 8
22= 2x2 = 4
21= 2 = 2
20= 1

34= 3x3x3x3 = 81
33= 3x3x3 = 27
32= 3x3 = 9
31= 3 = 3
30= 1

Laws

Zero Exponent Law- any base to the exponent of zero is 1. (X0=1)
Exponent Law of One
- any base the exponent of 1 is itself. (X1=X)
Product Law- any powers with the same base when multiplied, add the exponents to get the new power. (73 x 75 = 78), (122 x 123 x 124 x 12=1210), or (am x an=am+n)

***Be careful when adding the exponents. Always remember that a number without an exponent (12) has a hidden one (1)

Quotient Law- when any powers with the same base are divided, subtract the exponents to get the new power. (44 / 42 = 42), (57 / 53 = 54), (am / an = am-n)

Power of a Power Law- when one (1) is raised to any exponent, it will always equal 1.

Power of a Power Law example...
14= 1x1x1x1 = 1
13= 1x1x1 = 1
12= 1x1 = 1
11= 1 = 1
10= 1

Here's a little video for you to watch :)

## Wednesday, November 17, 2010

### Question 12

12. In the table, a positive number shows how many hours the time location in ahead of the time in London, England. A negative number shows how many hours the time is behind the time in London.

*First of all: remember that in adding and subtracting fractions, you must find a common denominator first (if needed).

a) How many hours is the time in St. John's ahead of the time in Brandon?

To find the answer, I subtracted the amount of time in Brandon from the amount of
time in St. Johns. A diagram below shows you how I solved it, and therefore the time in St. John's is 2 and 1/2 hours ahead of the time in Brandon.

b) How many hours is the time in Victoria behind the time in Mumbai?
The process in the same as the last question, I subtracted the time in Victoria from the one in Mumbai. Below, it shows that time in Victoria is 13 and a half hours behind from that of Mumbai.

c) Determine and interpret the time difference between Tokyo and Kathmandu.
Once again, I subtracted the two times of both places. The diagram below shows that Tokyo is 3 and 1/4 hours ahead of Kathmandu.

d) Determine and interpret the time difference between Chatham Islands and St. John's.
I subtracted the amount of time in Chatham Islands to that of St. John's. The diagram below will show that Chatham Islands is 16 and 1/4 hours ahead.

e) In which location is the time exactly halfway between the times in Istanbul and Alice Springs?
The time in Kathmandu, Nepal is halfway between times in the other two locations. How did I know that? Because I added the times in Istanbul and Alice Springs (refer to chart above) and then I divided their sum by 2 because we are finding the halfway point. (Half= 2)