Tuesday, January 26, 2010

Jennifer's ScribePost on Percents :3

Hello Everyone!

Today I'll be showing you the answers for Questions 5 and 6!

Question 5

Part A

Part B

Part C

Now Moving On To Question 6

Part A

Part B

Part C

Thanks For Reading Folks! Now Please Enjoy Some Videos We've done in class.

textbook questions 3 on page 142


Use mental math to determine each of the following.

300% of 2000

you have to fid 100% 100%=2000

to get 300% you multiply by 3 so you have to multiply 2000 by 3

100% multiplied by 3 = 300%

2000 multiplied by 3 =6000

so 300% of 2000 is 6000

Percents foldable 4.3, jan.26/10, tamika816

This goes in the percents foldable under 4.3 percent of a number.

Part A

Q1) Explain how you could use mental math to find each of the following:
a. 300% of 40. b. 0.5% of 120. c. 10.5% of 80.


I use a ratio table to get my answer usually. In this one I went straight to multiplying 100% to 300% because it is an easy number to find. Then i multiply 40 by 3.


You can make the number a decimal by dividing it by 100. Then i multiplied the percent by 10, so its an easier number to work with. I took 1.2 then multiplied it by 5 which makes 6. And last divided it back down by 10.


I took a long way, but if you use numbers you work good with then it still works. I made 10% by dividing 100 by 10. Then I made 0.5% by dividing 1% by 2 and added 10% and 0.5% together to make 10.5%.

Q2) Describe 2 ways to find 6% of 129.


one way is that you can go from 100 to 1(on a ratio table). Then multiply 1 by 5 to get 5 and add the 5% and 1% together to get 6%.
--Or you can divide 120 by 100% to get 1.2. Then multiply 1.2 with 6% which gives you 7.2.

Part B

Mr. Harbeck gave us this extra question.

Q3) How do you find 10%, 1%, 25%, 5%, 150%, and 0.5%?

- 10%= 100/10.
- 1%= 100/100.
- 25%= 100/4 OR 100/2=50, then 50/2=25.
- 5%= 100/10=10, then 10/2=5.
- 150%= 100/2=50, then 50+100=150.
- 0.5%= 100/100=1, then 1/2=0.5.

Thanx for reading!!!!
Pleaze leave comments!!!! =) =D =]

Tuesday, January 19, 2010

percents and problem solving

to help you solve this problem.. you have to represent the part and the total..

a.) 75km/50km=1.5X100=150 km


thanks for reading!
please comment if i got it wrong..
to find the questions go to www.mytextbook.ca

Monday, January 18, 2010

Jomari's Scribepost January 18,2010

Today in class, we learned about percents

We went over the 3 show you knows in the book.

We learned about how we can get fractions from percents(percents are fractions out of 100 that can be simplified), percents from decimals (multiply the decimal by 100) and decimals from fractions (divide the numerator by the denominator ) .
After we went over these questions, we had to copy the key ideas in the 4.2 Fractions, Decimals, Percents section of our foldable
This is what we had to write:

After we finished copying the key ideas section in our foldable, we had to do the first 6 questions in our textbook. You can find these questions at mytextbook.ca

Here is a video to help you understand more about percents:

I hope that you found my scribepost somewhat helpful to you.
Please Comment

Scribepost, January 18, 2010

4. Convert each fraction to a decimal and
a percent.

( click image )
5. Rewrite each fraction as a decimal and
a percent.

( click image )
6. Convert each decimal to a percent and
a fraction.
( click image )


Thursday, January 14, 2010

Scribepost Gordon January 14

Sorry this is late but lets get right down to the work.

Question 2
a)Explain how to show more then 100% using 100 squares.
b)Explain how to fill in less then 1% using 100 squares.

a)To fill in a percent more then 100 you simply us
e more then one square when filling it in.
eg: for 130% you would fill in 100 squares on one of the 100 squa
re sheets and then 30 of the 100 squares on a second square

b) To fill in a percent less the 100 you simply only fill a portion of 1 of the 100 possible squares.

6. Represent each percent on a grid
a) 125%
b) 10 1/2%
c) 0.4%
d) 262%
e) 7/8%
f) 45.6%

I scanned my answers for this one sorry there a little messy.

11. The land of Alberta is 113% the land of Saskatchewan. Show how the land area compares using hundred grids.

A simple grid can show the difference in land:
Saskatchewan Alberta

14. a) If 200 squares were used instead of 100 squares to represent 100%, how would you show 0.25%?

b) If 400 squares were used instead of 100 squares to represent 100%, how would you show 0.75%?

(My scanner stoped working after that last scan so i just took this picture from mytextbook.ca)

a) If we used 200 squares instead of 100 we would need to double the numerator because the denominator (100) was doubled so when representing what would normally appear as 1/4 would now appear as 1/2 but still be equal to 1/4.

b) If we used 400 squares instead of 100 we would need to quadruple the numerator because the denominator (100) was quadrupled so when representing what would normally appear as 3/4 would now appear as 3 but still be equal to 3/4.

And that was my (slightly late) scribepost I hope it was informative.

Deniel's Scribepost January 14

Okay I wanted to do a scribe post so the questions I got to do were 15,10,7,3 but anyway here are my answers:

Question 3
Shindi commented to a friend that “some percents would be easier to show
if we shaded the parts that were not included in the percent.” Explain what
she means. Which percents are easier to show using Shindi’s method? Why?

I think she meant was an example like this: (sorry no pic)

Shindi meant was that on a graph pretend the answer was 54% but the reason why she said that was that the person shaded in 46 meaning that what Shindi meant about the answer or the other reason was that the first answer (46%) was the real answer but it maybe was acutally 54% but anyway yeah.

Question 7
Represent the percent in each statement on a grid.

a) Attendance at the fall fair increased by 3.2% this year.

The Fall fair increased by 3.2%.

b) The average mass of a Singapura cat is about 0.13% of the mass of a Siberian tiger.
The Siberian Tiger would be 0.13%

c) The length of the Yukon River is about 230% of the length of the Fraser River.

The length of the Yukon river is 230%.
Question 10
Why might a scientist studying water pollution work with percents less than one?

Because the scientist might have found out that the pollution is less than 1 %

Question 15
Show how hundred grid(s) could be used to represent a very small percent, such as 0.000 0125%.

In my opinion for this answer I just guessed but her is my solution:

On a graph it would show (sorry no pic :/) 100 so it would show as 10=1 as we keep going we get to 10 dividing it by 0.001 we would get 0.0001 so as we keep going 0.00001 we gotta reach 0.0000125 so here is my answer to that question:

0.1/ 10 = 0.01 / 10 = 0.001
then replace the one making it 125 so 0.125/ 10 = 0.0125
then 0.0125/ 10 = 0.00125 / 10 = 0.000125 / 10 = 0.0000125 and theres your answer

(sorry no pic I did this today and some questions i had no time to think)

Wednesday, January 13, 2010

Althea's Scribepost, January 13, 2010

Today I'm assigned to do question 4, 8, 9, and 16.
First, I'll do question 4

Question no. 8
Question no. 9

* I'm not sure with my answer, that's why I looked at the back of the book, and the book mentioned that the answer is:

Question no. 16

Thanks for reading..
Please leave a comment..

Tuesday, January 12, 2010

Jecelyn's scribepost

Okay so today we were introduced to percents.

In class we had to use 10x10 grids for solving percent problems.

For example if we had to shade in 65% of the grid we would shade 65 squares. Why? because percents are out of 100. We also had to show if we messed with a square so then we just make some sort of pop out to show what we did.

Well Harbeck tricked a lot of people when he said to fill in 3/4 of the grid a lot of people thought that 3/4 was 75% of the square so the shaded in 75% of the square but it actually meant that it only 3/4 or 75% of 1 square.

In class we also made a foldable.

In the foldable at the first flap we had to find the definition of percent and fractional percent. Under that we had to draw out what 180% , 0.6% and 12 3/4% looks like on a grid. We just have to draw it out and show a pop up where we messed with the square.

Percent- is a way of expressing a number as a fraction of 100.
Fractional Percent- a percent that includes a portion of a percent.

Thursday, January 7, 2010

Pythagoras (late)

this question is from page 110. Question 3

3. Walter walks across a rectangular fi eld in a
diagonal line. Maria walks around two sides
of the fi eld. They meet at the opposite corner.

a) How far did Maria walk?
300+120= 420m

b) How far did Walter walk?
c2=104400 or 323m

c) Who walked farther? By how much?
Maria walked farther by 97 m

Reuben's scribe

pg.110 Q.4
Find the height of the pole where the guy wire is attached to the nearest tenth of a metre.
Sorry there is no pic.

If the guy wire(c,10m) is the hypotenuse and the ground(a,2m) is one of the legs , then the pole(b) is the second leg.To do this we need the work.

c2 -a2=c2
square root of 96 =9.78 something,something,something,something
Round to the nearest meter to get 9.8m2
Again sorry about no pics I couldn't get onto my textbook (server lost) or something like that

Wednesday, January 6, 2010

Scribe Jan 6 2010!

7. - What is the height of the wheelchair ramp? Give your answer to the nearest tenth of a centimeter.
a² = c²- b²
a² = 80cm² - 79cm²
a² = (80x80) - (79x79)
a² = 6400cm - 6241cm
a² = 159cm
√a² = √159
a = 12.6cm
The height of the ramp is 12.6 cm tall.

Jasmine's Scribepost January 6, 2010


7) What is the height of the wheelchair ramp? Give your answer to the nearest tenth of a centimeter.
(Sorry I didn't put the 'a,b,&c)
c² - b² = a²

80² - 79² = a²
(80x80) - (79x79) = a²
6400 - 6241 = a²
159 = a²
√159 = √a
12.6cm= a

Telisa's Scribepost for January 6, 2010

Question #3

Walter walks across a rectangular field in a diagonal line. Maria walks around two sides of the field. They meet at the opposite corner.

a) How far did Maria walk ?
I didn't really know how to find the length of what Maria walked so I added together the side lengths...

120 + 300 = 420
Maria walked 420m

b) How far did Walter walk ?

I used the Pythagorean relationship to find the answer...

a² + b² = c²

120² + 300² = c²

( 12o x 120) + ( 300 x 300) = c²

14400 + 90000 = c²

10 4400 = c²

√ 104400 = √ c²

323.1 = c

Walter walked 323.1m.


c) who walked farther ? by what distance ?

Maria walked farther by ... 96.9m

420 - 323.1 = 96.9m


Question : Purple

Work : Blue

Answer : Bold

Me : Grey

Sorry if the picture is hard to read.

Thank you for reading my scribe...

please criticize and comment

Warren's Scribepost for January 5th 2010

Question 10

What is the minimum distance the player
at third base has to throw the ball to get
the runner out at first base? Express your
answer to the nearest tenth of a metre

Okay, soo we need to find the hypotenuse







The shortest distance the 3rd basemen would have to thorw would be 38.18 meters.

Alex's pythogorean scribe

OK. so for my scribe, I decided to do question 14 on page 113 so here it is.
14).the question for this is find the missing side length of each triangle.

a)c+b=d-each of these letters are supposed to be squared
12m+5m=d-each of these letters are supposed to be squared
(12mx12m)+(5mx5m)=d-the d is supposed to be squared
144+25=d-the d is supposed to be squared
169=d-the d is supposed to be squared
169=d-the d is supposed to be squared and their should be a square root over the 169 and d

b)w-t=v-each of these letters are supposed to be squared
15cm-9cm=v-each of these letters are supposed to be squared
(15cmx15cm)-(9cmx9cm)=v-the v is supposed to be squared
225-81=v-the v is supposed to be squared
144=v-the v is supposed to be squared
144=v-the v is supposed to be squared and their should be a square root over the 144 and v

So A's side length is 13m and B's side length is 12cm.

Jecelyn's Scribepost Jan 6 , 2010

I picked number 4 on page 110

The question was: Find the height of the pole where the guy wire is attached to the nearest tenth or a metre.

So the height of the pole is 9.79.

Well hope my scribe post made sense. Please comment :D

Brianna's Scribe Post January 6, 2010

Question 12

Sarah has a vegetable garden in the shape of a right triangle. She wants to put fencing around it to keep the rabbits away.

a) What total length of fencing does she need? Give your answer to the nearest hundredth of a metre.

c2- b2=a2

42-32 =a2



the square root of 7 is 2.6 which equals a2.

Sarah would need 9.6m of fencing.

b) If fencing costs $2/m, what would be the total cost of the fencing?

The fencing would cost $19.20, I got this answer by multiplying 9.6m by 2.

Gordons Scribepost Jan 6 2010

For my scribe post question i volenteered to do number 9.
(Sorry its a little late)

This question asked what the diagnol length of a square (or two triangles right next to each other) was if it had two sides (legs) with the measurments:
With c as the diagnol length.
So to discover the length we use the formula a²+b²+c²

And now we begin the long math equation:
So now we simply write a simple sentince answer
The diagnol length of the rectangle is 13.4m.
Again I apologize for this post being a little late, and remember that tomorrow is the last day of pythagoras and we need 8 of the math books questions done.

Deniel's scribpost 8-16 January 6th 2010 Question 16#


Hey guys okay I really don't know if I was to do this question or not today but ill do it anyway...
The question I chose was Question 16 and it was the hard question but i figured it out anyway so here is my solution:
What is the length of the red diagonal in the box? Express your answer to the nearest tenth of a millimetre.

Here is my answer to this problem:
(Sorry if i didn't add any the letters a,b,c because it was hard putting them on the picture.)

When I was estimating I tried to figure out the problem by seeing what the question meant so here's what I did:
I multiplied the numbers by 2 making it 218 so I squared that number making it 14.8 because I rounded it by a tenth which makes this my solution but my answer is gonna be next.

All I did was:
A²+B²+C² = ???

5mm² + 7mm² + 12mm² =???
So I multiplied them making an answer: (5x5)+(7x7)+(12x12)=218

So I squared the number making an answer which was: √218 = 14.7 rounding it making it 14.8
so that is my answer for Question 16#

Tuesday, January 5, 2010

Question 15


Here is question 15:


To find the answer, we have to use the Pythagorean relationship. From looking at the graph we can tell that the legs, (green) have lengths of 2cm and 4cm. We have to find the length of the hypotenuse, (red). Our formula is: a squared + b squared = c squared.

a squared + b squared = c squared

2cm squared + 4cm squared = c squared

(2cm x 2cm) + (4cm x4cm) = c squared

4cm + 16cm = c squared

20cm= c squared

By doing this we have figured out the area of the hypothetical square attached to the hypotenuse. Now we have to find the square root of that number to find the length of the hypotenuse.

the square root of 20= 4.4 (approx.)

The length of the line is approximately 4.4cm.

Thanks for reading! Please comment :)


Pg.105-Question 12 1/5/10

For those that can't see whats written the question was.... 12. The hypotenuse of the triangle cuts the circle in half. What is the diameter of the circle? Express your answer to the nearest tenth of a centimetre.

And the right answer was 8.60 cm

Jomari's Scribepost, January 5, 2010

For this question, I am supposed to find the length of the red line(C). I can do this by making a right triangle with the red line as the hypotenuse. The legs would meet each other at 5,2. The vertical leg (A) would be 4 units long and the horizontal leg (B) would be 2 units long. I can use this information to first, find out the length of C² and then find the length of C.

A² + B² =
4 cm² + 2 cm² = C²
(4 cm x 4 cm)+( 2cm x 2cm) = C²

16 cm + 4cm = C²
20 cm = C²
√ 20 cm = √ C²
4.5 cm= C

The red line has a length of 4.5 centimeters.

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